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• Consider a central node ’i’ connected to neighboring nodes with springs. • Equations can be written to relate the position of node i to the surrounding nodes and the applied force. The ’x’ and ’y’ values are deflections from the unloaded state. The ’Ks’ value is based on the material stiffness and the geometry of the elements. • This can be combined into a more • If the input forces are known, then the resulting displacements of the nodes can be calculated by inverting the matrix. Consider the matrix for a single node. • If we assume that node ’c’ is fixed in the ’x’ and ’y’ directions, the matrix can reflect this by setting the appropriate matrix rows to zero. • The displacements can then be found by selecting values for the coefficents and solving the matrix. We can select a value of 1000 for the stiffness, and a force of 10 will be applied at point ’a’ in the positive ’x’ direction. • This method generates very large matrices, easily into the millions. • To reduce the matrix size techniques such as symmetry are commonly used. • Strains can be found by calculating the relative displacements of neighboring points. These can then be used to calculate the stresses. • More complex elements are commonly used depending upon the stress conditions, part geometry and other factors. • There are a variety of finite element element methods and applications	{"url":"https://engineeronadisk.com/V2/book_modelling/engineeronadisk-256.html","timestamp":"2024-11-11T02:12:50Z","content_type":"text/html","content_length":"5028","record_id":"<urn:uuid:dc0f7c0a-8f46-46f2-9351-24f1928479e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00194.warc.gz"}
1. Using a diagram drawn on a sort of graph paper with axes (you need 1. Using a diagram drawn on a sort of graph paper with axes (you need to create this), start with Faraday's law and show that Faraday's law leads to an equation relating the spatial derivative of E with the time derivative of B. Rules for diagrams: You must create every part of your diagrams by hand. Do not copy and paste anything. Your diagram should be detailed enough that the reader can (a) identify axes, (b) determine what is an E field and what is a B field, (c) identify where the fields are being evaluated (z or z + Az?), and (d) determine how the diagram leads to positive and negative signs in equations. You may need more than one diagram per section to accomplish this! Fig: 1 Fig: 2	{"url":"https://tutorbin.com/questions-and-answers/from-maxwell-s-equations-to-light-the-assignment-is-to-start-with-maxwell-s-equations-in-integral-form-derive-the-wave","timestamp":"2024-11-14T04:05:00Z","content_type":"text/html","content_length":"65124","record_id":"<urn:uuid:da0e1c16-7c0b-4b4b-9041-5b1266661b40>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00457.warc.gz"}
Machine Learning – Programming Exercise 3: Solved - Ideal coders Multi-class Classification and Neural Networks Machine Learning In this exercise, you will implement one-vs-all logistic regression and neural networks to recognize hand-written digits. Before starting the programming exercise, we strongly recommend watching the video lectures and completing the review questions for the associated topics. To get started with the exercise, you will need to download the starter code and unzip its contents to the directory where you wish to complete the exercise. If needed, use the cd command in Octave/ MATLAB to change to this directory before starting this exercise. You can also find instructions for installing Octave/MATLAB in the “Environment Setup Instructions” of the course website. Files included in this exercise ex3.m – Octave/MATLAB script that steps you through part 1 ex3 nn.m – Octave/MATLAB script that steps you through part 2 ex3data1.mat – Training set of hand-written digits ex3weights.mat – Initial weights for the neural network exercise submit.m – Submission script that sends your solutions to our servers displayData.m – Function to help visualize the dataset fmincg.m – Function minimization routine (similar to fminunc) sigmoid.m – Sigmoid function [?] lrCostFunction.m – Logistic regression cost function [?] oneVsAll.m – Train a one-vs-all multi-class classifier [?] predictOneVsAll.m – Predict using a one-vs-all multi-class classifier [?] predict.m – Neural network prediction function ? indicates files you will need to complete Throughout the exercise, you will be using the scripts ex3.m and ex3 nn.m. These scripts set up the dataset for the problems and make calls to functions that you will write. You do not need to modify these scripts. You are only required to modify functions in other files, by following the instructions in this assignment. Where to get help The exercises in this course use Octave or MATLAB, a high-level programming language well-suited for numerical computations. If you do not have Octave or MATLAB installed, please refer to the installation instructions in the “Environment Setup Instructions” of the course website. At the Octave/MATLAB command line, typing help followed by a function name displays documentation for a built-in function. For example, help plot will bring up help information for plotting. Further documentation for Octave functions can be found at the Octave documentation pages. MATLAB documentation can be found at the MATLAB documentation pages. We also strongly encourage using the online Discussions to discuss exercises with other students. However, do not look at any source code written by others or share your source code with others. 1 Multi-class Classification For this exercise, you will use logistic regression and neural networks to recognize handwritten digits (from 0 to 9). Automated handwritten digit recognition is widely used today – from recognizing zip codes (postal codes) on mail envelopes to recognizing amounts written on bank checks. This exercise will show you how the methods you’ve learned can be used for this classification task. In the first part of the exercise, you will extend your previous implemention of logistic regression and apply it to one-vs-all classification. 1.1 Dataset You are given a data set in ex3data1.mat that contains 5000 training examples of handwritten digits. The .mat format means that that the data has been saved in a native Octave/MATLAB matrix format, instead of a text (ASCII) format like a csv-file. These matrices can be read directly into your program by using the load command. After loading, matrices of the correct dimensions and values will appear in your program’s memory. The matrix will already be named, so you do not need to assign names to them. % Load saved matrices from file load(‘ex3data1.mat’); % The matrices X and y will now be in your Octave environment There are 5000 training examples in ex3data1.mat, where each training example is a 20 pixel by 20 pixel grayscale image of the digit. Each pixel is represented by a floating point number indicating the grayscale intensity at that location. The 20 by 20 grid of pixels is “unrolled” into a 400-dimensional vector. Each of these training examples becomes a single row in our data matrix X. This gives us a 5000 by 400 matrix X where every row is a training example for a handwritten digit image. — (x(1))T — — (x(2))T — — ( The second part of the training set is a 5000-dimensional vector y that contains labels for the training set. To make things more compatible with Octave/MATLAB indexing, where there is no zero index, we have mapped the digit zero to the value ten. Therefore, a “0” digit is labeled as “10”, while the digits “1” to “9” are labeled as “1” to “9” in their natural order. 1.2 Visualizing the data You will begin by visualizing a subset of the training set. In Part 1 of ex3.m, the code randomly selects selects 100 rows from X and passes those rows to the displayData function. This function maps each row to a 20 pixel by 20 pixel grayscale image and displays the images together. We have provided the displayData function, and you are encouraged to examine the code to see how it works. After you run this step, you should see an image like Figure Figure 1: Examples from the dataset 1.3 Vectorizing Logistic Regression You will be using multiple one-vs-all logistic regression models to build a multi-class classifier. Since there are 10 classes, you will need to train 10 separate logistic regression classifiers. To make this training efficient, it is important to ensure that your code is well vectorized. In this section, you will implement a vectorized version of logistic regression that does not employ any for loops. You can use your code in the last exercise as a starting point for this exercise. 1.3.1 Vectorizing the cost function We will begin by writing a vectorized version of the cost function. Recall that in (unregularized) logistic regression, the cost function is To compute each element in the summation, we have to compute hθ(x(i)) for every example i, where hθ(x(i)) = g(θTx(i)) and is the sigmoid function. It turns out that we can compute this quickly for all our examples by using matrix multiplication. Let us define X and θ as — (x(1))T — θ0 — (x(2))T — θ1 … and θ = … . — (x(m))T — θn Then, by computing the matrix product Xθ, we have — (x(1))Tθ — — θT(x(1)) — — (x(2))Tθ — — θT(x(2)) — .. = … . — ( — ( ) — In the last equality, we used the fact that aTb = bTa if a and b are vectors. This allows us to compute the products θTx(i) for all our examples i in one line of code. Your job is to write the unregularized cost function in the file lrCostFunction.m Your implementation should use the strategy we presented above to calculate θTx(i). You should also use a vectorized approach for the rest of the cost function. A fully vectorized version of lrCostFunction.m should not contain any loops. (Hint: You might want to use the element-wise multiplication operation (.*) and the sum operation sum when writing this function) 1.3.2 Vectorizing the gradient Recall that the gradient of the (unregularized) logistic regression cost is a vector where the jth element is defined as To vectorize this operation over the dataset, we start by writing out all the partial derivatives explicitly for all θj, Note that x(i) is a vector, while (hθ(x(i))−y(i)) is a scalar (single number). To understand the last step of the derivation, let βi = (hθ(x(i)) − y(i)) and observe that: where the values βi = (hθ(x(i)) − y(i)). The expression above allows us to compute all the partial derivatives without any loops. If you are comfortable with linear algebra, we encourage you to work through the matrix multiplications above to convince yourself that the vectorized version does the same computations. You should now implement Equation 1 to compute the correct vectorized gradient. Once you are done, complete the function lrCostFunction.m by implementing the gradient. Debugging Tip: Vectorizing code can sometimes be tricky. One common strategy for debugging is to print out the sizes of the matrices you are working with using the size function. For example, given a data matrix X of size 100 × 20 (100 examples, 20 features) and θ, a vector with dimensions 20×1, you can observe that Xθ is a valid multiplication operation, while θX is not. Furthermore, if you have a non-vectorized version of your code, you can compare the output of your vectorized code and non-vectorized code to make sure that they produce the same outputs. 1.3.3 Vectorizing regularized logistic regression After you have implemented vectorization for logistic regression, you will now add regularization to the cost function. Recall that for regularized logistic regression, the cost function is defined Note that you should not be regularizing θ0 which is used for the bias term. Correspondingly, the partial derivative of regularized logistic regression cost for θj is defined as for j = 0 for j ≥ 1 Now modify your code in lrCostFunction to account for regularization. Once again, you should not put any loops into your code. Octave/MATLAB Tip: When implementing the vectorization for regularized logistic regression, you might often want to only sum and update certain elements of θ. In Octave/MATLAB, you can index into the matrices to access and update only certain elements. For example, A(:, 3:5) = B(:, 1:3) will replaces the columns 3 to 5 of A with the columns 1 to 3 from B. One special keyword you can use in indexing is the end keyword in indexing. This allows us to select columns (or rows) until the end of the matrix. For example, A(:, 2:end) will only return elements from the 2nd to last column of A. Thus, you could use this together with the sum and .^ operations to compute the sum of only the elements you are interested in (e.g., sum(z(2:end).^2)). In the starter code, lrCostFunction.m, we have also provided hints on yet another possible method computing the regularized gradient. You should now submit your solutions. 1.4 One-vs-all Classification In this part of the exercise, you will implement one-vs-all classification by training multiple regularized logistic regression classifiers, one for each of the K classes in our dataset (Figure 1). In the handwritten digits dataset, K = 10, but your code should work for any value of K. You should now complete the code in oneVsAll.m to train one classifier for each class. In particular, your code should return all the classifier parameters in a matrix Θ ∈ RK×(N+1) , where each row of Θ corresponds to the learned logistic regression parameters for one class. You can do this with a “for”-loop from 1 to K, training each classifier independently. Note that the y argument to this function is a vector of labels from 1 to 10, where we have mapped the digit “0” to the label 10 (to avoid confusions with indexing). Octave/MATLAB Tip: Logical arrays in Octave/MATLAB are arrays which contain binary (0 or 1) elements. In Octave/MATLAB, evaluating the expression a == b for a vector a (of size m×1) and scalar b will return a vector of the same size as a with ones at positions where the elements of a are equal to b and zeroes where they are different. To see how this works for yourself, try the following code in a = 1:10; % Create a and b b = 3; a == b % You should try different values of b here Furthermore, you will be using fmincg for this exercise (instead of fminunc). fmincg works similarly to fminunc, but is more more efficient for dealing with a large number of parameters. After you have correctly completed the code for oneVsAll.m, the script ex3.m will continue to use your oneVsAll function to train a multi-class classifier. You should now submit your solutions. 1.4.1 One-vs-all Prediction After training your one-vs-all classifier, you can now use it to predict the digit contained in a given image. For each input, you should compute the “probability” that it belongs to each class using the trained logistic regression classifiers. Your one-vs-all prediction function will pick the class for which the corresponding logistic regression classifier outputs the highest probability and return the class label (1, 2,…, or K) as the prediction for the input example. You should now complete the code in predictOneVsAll.m to use the one-vs-all classifier to make predictions. Once you are done, ex3.m will call your predictOneVsAll function using the learned value of Θ. You should see that the training set accuracy is about 94.9% (i.e., it classifies 94.9% of the examples in the training set correctly). You should now submit your solutions. 2 Neural Networks In the previous part of this exercise, you implemented multi-class logistic regression to recognize handwritten digits. However, logistic regression cannot form more complex hypotheses as it is only a linear classifier. In this part of the exercise, you will implement a neural network to recognize handwritten digits using the same training set as before. The neural network will be able to represent complex models that form non-linear hypotheses. For this week, you will be using parameters from a neural network that we have already trained. Your goal is to implement the feedforward propagation algorithm to use our weights for prediction. In next week’s exercise, you will write the backpropagation algorithm for learning the neural network parameters. The provided script, ex3 nn.m, will help you step through this exercise. 2.1 Model representation Our neural network is shown in Figure 2. It has 3 layers – an input layer, a hidden layer and an output layer. Recall that our inputs are pixel values of digit images. Since the images are of size 20×20, this gives us 400 input layer units (excluding the extra bias unit which always outputs +1). As before, the training data will be loaded into the variables X and y. You have been provided with a set of network parameters (Θ(1),Θ(2)) already trained by us. These are stored in ex3weights.mat and will be loaded by ex3 nn.m into Theta1 and Theta2 The parameters have dimensions that are sized for a neural network with 25 units in the second layer and 10 output units (corresponding to the 10 digit classes). % Load saved matrices from file load(‘ex3weights.mat’); % The matrices Theta1 and Theta2 will now be in your Octave % environment % Theta1 has size 25 x 401 % Theta2 has size 10 x 26 Figure 2: Neural network model. 2.2 Feedforward Propagation and Prediction Now you will implement feedforward propagation for the neural network. You will need to complete the code in predict.m to return the neural network’s prediction. You should implement the feedforward computation that computes hθ(x(i)) for every example i and returns the associated predictions. Similar to the one-vs-all classification strategy, the prediction from the neural network will be the label that has the largest output (hθ(x))k. Implementation Note: The matrix X contains the examples in rows. When you complete the code in predict.m, you will need to add the column of 1’s to the matrix. The matrices Theta1 and Theta2 contain the parameters for each unit in rows. Specifically, the first row of Theta1 corresponds to the first hidden unit in the second layer. In Octave/MATLAB, when you compute z(2) = Θ(1)a(1), be sure that you index (and if necessary, transpose) X correctly so that you get a(l) as a column vector. Once you are done, ex3 nn.m will call your predict function using the loaded set of parameters for Theta1 and Theta2. You should see that the accuracy is about 97.5%. After that, an interactive sequence will launch displaying images from the training set one at a time, while the console prints out the predicted label for the displayed image. To stop the image sequence, press Ctrl-C. You should now submit your solutions. Submission and Grading After completing this assignment, be sure to use the submit function to submit your solutions to our servers. The following is a breakdown of how each part of this exercise is scored. Part Submitted File Points Regularized Logisic Regression lrCostFunction.m 30 points One-vs-all classifier training oneVsAll.m 20 points One-vs-all classifier prediction predictOneVsAll.m 20 points Neural Network Prediction Function predict.m 30 points Total Points 100 points You are allowed to submit your solutions multiple times, and we will take only the highest score into consideration. There are no reviews yet. Be the first to review “Machine Learning – Programming Exercise 3: Solved”	{"url":"https://idealcoders.com/product/machine-learning-programming-exercise-3-solved-3/","timestamp":"2024-11-14T23:49:12Z","content_type":"text/html","content_length":"190788","record_id":"<urn:uuid:aba6ed4e-e754-4c15-85d9-26ea7bfdca51>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00145.warc.gz"}
Kiwis and Limes This post assumes that the reader has knowledge of basic vector calculus and physics. In the study of fluid dynamics the term ‘momentum transport’ is thrown around quite often and often in conjunction with the idea of ‘momentum accumulation’; it is a rather difficult concept to understand because it is quite impossible to see momentum. Hopefully this post will help to clarify the concept. We will start with an easy to visualize concept: ‘mass transport’. Let us assume that there is a laminar and steady fluid flow of density $\rho$ and velocity $\mathbf{v}$. We would like to know the rate $K_T$ at which mass is transported in this flow. The answer— \[ K_T = \rho \|\mathbf{v}\| \] Makes sense until you consider the units of $K_T$ and realize that they are $\frac{\text{g}}{\text{cm}^2 \cdot \text{s}}$ which definitely doesn't make much sense. In fact it is difficult to determine at this point what the correct units of the rate of mass transport should even be. To solve this problem we need to consider mass transport from the other side of the equation: ‘mass This post assumes a working knowledge of elementary circuit theory as well as Fourier Analysis. It seems to me that often in an introductory circuits course complex impedance is a major concept but its mathematical basis is never taught. Thus I'd like to take a moment to discuss some of the mathematics surrounding this concept. The resistor is a purely resistive elementary circuit element possessing a time independent current voltage characteristic described by Ohm's Law. \[ V(t) = I(t) R \] The capacitor is a purely reactive elementary circuit element that stores energy in its electric field. \[ I(t) = C \frac{dV(t)}{dt} \] The inductor is a purely reactive elementary circuit element that stores energy in its magnetic field. \[ V(t) = L \frac{dI(t)}{dt} \] What we aim to do is to derive a linear current voltage characteristic for the two reactive circuit elements by reducing the differential equations to algebraic equations. Using the Fourier Transform we can compute a complex quantity in the frequency domain that is analogous to resistance in the time domain.	{"url":"https://kaitocracy.blogspot.com/","timestamp":"2024-11-13T20:37:21Z","content_type":"text/html","content_length":"96137","record_id":"<urn:uuid:bb07de69-d8d1-43dd-92cf-2b19cdddf578>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00740.warc.gz"}
MMRCA 2.0 - Updates and Discussions All junk compared to Rafale and Typhoon. Nothing much has changed for any of the contenders since MMRCA. LCA Mk2 is better than both the Teens. And for India, the Su-35 and Mig-35 offers are superior to them as well. To put things in perspective, the Mig-21 is from the 60s, the Teens are from the 70s, and the 35s are from the 80s. Rafale, Typhoon, Mk2 and Gripen E are from the 2000s. So all the pre-2000s jets are pointless to the IAF, never mind stuff from the 70s. The f-15 literally has the best radar out of the bunch followed by eurofighters captor mk2. The designs maybe of the 70's. The tech is 2010+. Although typhoon is the best plane out of all along with the rafale. One big reason why the IAF chose the Rafale in the first place was that it was designed with the N-strike role in mind. It was meant to replace both the M2000N and the carrier borne Super Etendard in that role. Buying any other ac means spending big bucks on hardening electronics against EMP, etc. The Brits gave up their air launched nuke in favour of sub deterrent so India would have do it on its own dime. Dealing with 4 nations of the EF consortium with their convoluted approval processes (for upgrades, weapons intregration and maintenance issues) is going to be a pain. The Brits have denied weapons licenses to Israel when it needed them most. We only need to deal with the Brits and maybe the italians/spanish. The amount of job creation the typhoon can bring for the British economy will be too big for them to miss. The only issue with the typhoon is the high life cycle cost. The typhoon may not be as good as a strike fighter like the rafale but it is pretty much the best aircraft when it comes to air defence and air superiority among all nato fighters especially with captor mk2. Considering we will be facing j-10,j-16 and j-20 which have huge radar ranges, typhoon is the only one that genuinely fits the bill. Plus all kinds of nato munitions can be integrated like the Gripen. The only thing it might not have is the mica ng but we get the amraams in our arsenal. Maybe a shield. The point I'm making doesn't change though. Since we have our own deterence we don't need a shield nor any weapon. We really need a third light big enough to be its own market (France can't be this one), strong enough to not depend from anyone else (France can be this one but India is not ready for this point), stable enough and not depending from one of the two other superpowers to keep an open world. Who knows, may be Koreans bring their KF-21 to the competition. We only need to deal with the Brits and maybe the italians/spanish Still too much back and forth for day to day issues. France plans to operate the Rafale through 2050 now that SCAF is delayed. The Brits are now focused on Tempest/GCAP with service entry planned for The amount of job creation the typhoon can bring for the British economy will be too big for them to miss The Brits are infamous for being complete US poodles when it comes to foreign policy, defence and trade. Geopolitics always trumps trade and commerce as we're seeing in Ukraine. The only issue with the typhoon is the high life cycle cost License production from raw material stage will cost 2X at minimum. Add the cost of weapons, training and support (+inflation) and the figure is astronomical. The typhoon may not be as good as a strike fighter like the rafale but it is pretty much the best aircraft when it comes to air defence and air superiority among all nato fighters especially with captor mk2 It also fares poorly in terms of strike radius/endurance which is why EF partner nations toyed with the idea of conformal over wing fuel tanks in the past. But seems it went nowhere. Rafale is competitive with the MKI in terms of staying in the fight. That said, Typhoon desperately needs orders beyond small-fry ME nations (Pakistani pilots may even be flying them in KSA, Qatar!) If we get a good deal in terms of unit cost, spares, source codes, weapons + sovereign guarantees of lifetime support (regardless of the 'K' games being played - Kashmir, Khalistan, et all) it might be worth a look. But at the moment, that's a big if.. Break this down under 2025-35 and then 2035- beyond. The capabilities against Pakistan and China have already been established. Now it's all about modernization and some additional numbers here and there, like an extra division or two and new satellite capabilities; fast-paced inductions. Technological inductions are capability driven. Let's hope the IAF gets those between 2035-45 via MRFA, AMCA and drones, whereas the IA can get them before 2035, ie, artillery guns, FRCV, FICV, infantry modernization etc. Net security provider in the IOR under the navy's hands will naturally take time with the induction of SSKs, SSNs and a third carrier with air complement. We are pretty good elsewhere. This is also capability driven. So you already know about our white paper requirements before 2050. All I am saying is the timelines. If a Rafale from an Indian line joining IAF in 2035 is the goal, then just keep TEDBF/AMCA option open. Even if ORCA meets MRFA requirements, it's not going to meet proven credentials. Basically, a Rafale squadron inducted in 2035 is ready to fight in 2036. An ORCA squadron inducted in 2035 is not going to be ready until sometime in the mid-2040s. It takes that much time to learn and train personnel to use tech. It takes 5-8 years to bring a pilot up to expert level. It takes 7-10 years to train an expert level maintenance crew. With the Rafale, all of that's been ready since Furthermore, we will need to push the TEDBF's team's best scientific personnel towards the IN's requirements, 'cause it's being made for them. If we have to diversify labor towards the IAF's needs too, the quality will decrease on the IN's side as the industry prioritises the more profitable and easier IAF requirements. Once again, why buy ORCA when there's AMCA? The f-15 literally has the best radar out of the bunch followed by eurofighters captor mk2. The designs maybe of the 70's. The tech is 2010+. Although typhoon is the best plane out of all along with the rafale. The F-15's radar is of the older ilk. It's based on the F-22 and F-35's radar developments. Whereas the radar we are looking for via MRFA barely even exist today. Since we have our own deterence we don't need a shield nor any weapon. When I say shield, umbrella or footmat, I'm referring to politics, not military. We really need a third light big enough to be its own market (France can't be this one), strong enough to not depend from anyone else (France can be this one but India is not ready for this point), stable enough and not depending from one of the two other superpowers to keep an open world. I don't know why you think how any of that will help France. Let's assume US and France have sanctioned India in 2030. France and India have now become enemies. And let's assume India doesn't need external assistance for its military. Then how do you think things will play out in the larger scheme? I'm actually quite curious about this. Who knows, may be Koreans bring their KF-21 to the competition. I would definitely like to see that. It won't beat the Rafale, but will create new standards with its new gen airframe. It's unlikely to meet our expected schedule either, 'cause the Koreans are still in the process of developing an export model. All I am saying is the timelines. If a Rafale from an Indian line joining IAF in 2035 is the goal, then just keep TEDBF/AMCA option open. If Tedbf is equal to Rafale.. And tedbf can after Rafale production has ended ( assume it) Simply having 300 Rafale equivalent aircrafts are better than having 100 Rafale alone right.. We could upgrade, modify as when needed.. Letting Tedbf fail will become big blow Navy s fighter plans. 3 carrier + shore based plans. Unfortunately except Rafale and F-35, no other jets will fulfill our requirements. F-35 could be offered if Trump comes to power. So let's see....... Too much of strings will be attached Technical complexity will prevent any safeguard measures we can do.. I don't know why you think how any of that will help France. Once again France can't play the regulator role. We're not big enough. But India can. Let's assume US and France have sanctioned India in 2030. I don't think you have understood how strategic reliationship with India is for France. Any sanction against India will be block by France in UN. At the end of the XVIII century we have help US to counter UK much to big for France in the world, In the XVI century we had good reliationships with the otoman empire to counter the holy roman german empire. At the end of the XIX we have tried to have our own empire considering the ottoman and the UK empire. It was not very succesfull considering current US state. The current situation is that any country in the world is on the way to choose between US and China and to accept to be a footmat for one of these both country. If India has the capacity of his sovereignty then a non align choice exist and China and US could not win more power. Once again there is no sign from India to have imperialist behaviour. Perhaps in the furur but not currently. Considering this behaviour then you can maintain viable an open world . That's exactly what numerous countries including France want. It's difficult to say. Both need only 3-4 years of flight testing and both are going to fly for the first time around the same time. Perhaps induction will happen simultaneously. Both are next decade programmes. However, in my opinion, MK2 will be inducted before IUSAV. The capabilities against Pakistan and China have already been established. Now it's all about modernization and some additional numbers here and there, like an extra division or two and new satellite capabilities; fast-paced inductions. Technological inductions are capability driven. Let's hope the IAF gets those between 2035-45 via MRFA, AMCA and drones, whereas the IA can get them before 2035, ie, artillery guns, FRCV, FICV, infantry modernization etc. Net security provider in the IOR under the navy's hands will naturally take time with the induction of SSKs, SSNs and a third carrier with air complement. We are pretty good elsewhere. This is also capability driven. So you already know about our white paper requirements before 2050. I mean we will be losing approximately 200 more aircrafts by 2035-37 max. Is our aim to just replenish those 200 + maybe 100 Mig21s (last 4 bison squadrons).... Or we aim to increase the number beyond 32 squadrons by 2040? That's what I meant. I mean we will be losing approximately 200 more aircrafts by 2035-37 max. Is our aim to just replenish those 200 + maybe 100 Mig21s (last 4 bison squadrons).... Or we aim to increase the number beyond 32 squadrons by 2040? That's what I meant. To fight a one-front war, we need 30 squadrons. To fight a two-front war simultaneously, we need 42.5 squadrons. Due to current geopolitical realities, there is no immediate fear of a two-front war. If we exclude all Mig-21s and LCA Mk1As, we will have 29 squadrons. 3 Mig-29 squadrons were supposed to be phased out, but will be extended by a decade more. 3 Jaguar DARIN II squadrons are expected to be phased out by 2035, so we drop down to 26 squadrons. Another half squadron MKI is expected to be raised after the 13th one; 20x13.5 = 270. So 26.5 in total. With MKI MLU at 84 jets, ie, 21 jets per squadron, it's possible that the 13.5th squadron will not be raised. So with all the Mig-21s and half the Jaguars gone, there is no new phase out until after 2035, when all the Jags (2040), Mig-29 (2042) and Mirages (2042+), a total of 9 squadrons, are phased out. The delay in upgrading the Mirages would mean at least 1 squadron can survive until 2045. So, with 9 new LCA squadrons coming in between 2024 and 2032, we will have 26+9 = 35 squadrons in 2032. If we assume 2 Mk2 and 1 MRFA squadrons are inducted by 2035, then we will have 38 squadrons, by my calculations. The IAF is actually expecting to have all 42.5 squadrons by 2035, by assuming LCA Mk2 inductions will begin in 2031 instead of 2033, giving them at least 4 squadrons, and MRFA could happen equally faster for an additional squadron or two. With 9 squadrons pending, we can assume the remaining 5 MRFA, 4 Mk2 and 7 AMCA squadrons will fill the gap between 2035 and 2045, pushing us to 45 squadrons. Although I say 3 Jaguar squadrons will be phased out over the next few years, it could be 2.5 because half a squadron is a relatively new sea strike version with the EL/M 2032 under the Dragons. It could take a few more years for phase out. I guess the previous plan was the 13.5th MKI squadron will replace it, and this is not necessary with the LCA Mk2, but you never know. Anyway, based on my calculations, we will have 38 squadrons by 2035, whereas IAF believes they will have all 42.5 squadrons by then. I believe they think LCA Mk2 will enter production in 2029 with deliveries beginning in 2031, with all 6 squadrons inducted by 2036, with MRFA halfway done by then, both at 24 per year each. I gave an RFP to contract negotiations timeframe of 3 years for MRFA, I won't be surprised if it's less than that. Here's the full thread from Indranil Roy on why MRFA churning out FAs in the 2030s is a hare brained idea. ...Contd. below	{"url":"https://www.strategicfront.org/forums/threads/mmrca-2-0-updates-and-discussions.1156/page-271","timestamp":"2024-11-01T19:24:29Z","content_type":"text/html","content_length":"274651","record_id":"<urn:uuid:7c9b02b0-b2b5-4c94-ba26-a4bb9678521c>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00445.warc.gz"}
Today in Science History - Quickie Quiz Who said: “A people without children would face a hopeless future; a country without trees is almost as helpless.” Category Index for Science Quotations Category Index Q > Category: Quantum Theory Quantum Theory Quotes (67 quotes) [T]he laws of quantum mechanics itself cannot be formulated … without recourse to the concept of consciousness. From essay by Eugene Wigner, 'The Probability of the Existence of a Self-Reproducing Unit', contributed in M. Polanyi, The Logic of Personal Knowledge: Essays Presented to Michael Polanyi on his Seventieth Birthday, 11th March 1961 (1961), 232. [When thinking about the new relativity and quantum theories] I have felt a homesickness for the paths of physical science where there are more or less discernible handrails to keep us from the worst morasses of foolishness. The Nature Of The Physical World (1928), 343. After the discovery of spectral analysis no one trained in physics could doubt the problem of the atom would be solved when physicists had learned to understand the language of spectra. So manifold was the enormous amount of material that has been accumulated in sixty years of spectroscopic research that it seemed at first beyond the possibility of disentanglement. An almost greater enlightenment has resulted from the seven years of Röntgen spectroscopy, inasmuch as it has attacked the problem of the atom at its very root, and illuminates the interior. What we are nowadays hearing of the language of spectra is a true 'music of the spheres' in order and harmony that becomes ever more perfect in spite of the manifold variety. The theory of spectral lines will bear the name of Bohr for all time. But yet another name will be permanently associated with it, that of Planck. All integral laws of spectral lines and of atomic theory spring originally from the quantum theory. It is the mysterious organon on which Nature plays her music of the spectra, and according to the rhythm of which she regulates the structure of the atoms and nuclei. Atombau und Spektrallinien (1919), viii, Atomic Structure and Spectral Lines, trans. Henry L. Brose (1923), viii. After the German occupation of Holland in May 1940, the [last] two dark years of the war I spent hiding indoors from the Nazis, eating tulip bulbs to fill the stomach and reading Kramers' book “Quantum Theorie des Elektrons und der Strahlung” by the light of a storm lamp. Anyone who is not shocked by the quantum theory has not understood it. [Attributed.] Webmaster is not alone in failing to find a primary source. Regardless of how widely quoted, the few citations to be found merely reference other books in which it is stated without a valid citation. For example, this quote is an epigraph in Eric Middleton, The New Flatlanders (2007), 19, with a note (p.151) citing Niels Bohr, Atomic Physics and Human Knowledge (1958), but Webmaster’s search of that text does not find it. If you know the primary source, or an early citation, please contact Webmaster. Bertrand Russell had given a talk on the then new quantum mechanics, of whose wonders he was most appreciative. He spoke hard and earnestly in the New Lecture Hall. And when he was done, Professor Whitehead, who presided, thanked him for his efforts, and not least for “leaving the vast darkness of the subject unobscured.” Quoted in Robert Oppenheimer, The Open Mind (1955), 102. But how is one to make a scientist understand that there is something unalterably deranged about differential calculus, quantum theory, or the obscene and so inanely liturgical ordeals of the precession of the equinoxes. In Van Gogh, the Man Suicided by Society (1947). As translated in Jack Hirschman (ed.) Artaud Anthology (1965), 149. But, but, but … if anybody says he can think about quantum theory without getting giddy it merely shows that he hasn’t understood the first thing about it! Quoted in Otto R. Frisch, What Little I Remember (1979), 95. By far the most important consequence of the conceptual revolution brought about in physics by relativity and quantum theory lies not in such details as that meter sticks shorten when they move or that simultaneous position and momentum have no meaning, but in the insight that we had not been using our minds properly and that it is important to find out how to do so. 'Quo Vadis'. In Gerald Holton (ed.), Science and the Modern Mind (1971), 84. Formula describing the energy of a photon, in quantum theory, E=Energy, h=Planck’s constant, f=frequency of light. Every new theory as it arises believes in the flush of youth that it has the long sought goal; it sees no limits to its applicability, and believes that at last it is the fortunate theory to achieve the 'right' answer. This was true of electron theory—perhaps some readers will remember a book called The Electrical Theory of the Universe by de Tunzelman. It is true of general relativity theory with its belief that we can formulate a mathematical scheme that will extrapolate to all past and future time and the unfathomed depths of space. It has been true of wave mechanics, with its first enthusiastic claim a brief ten years ago that no problem had successfully resisted its attack provided the attack was properly made, and now the disillusionment of age when confronted by the problems of the proton and the neutron. When will we learn that logic, mathematics, physical theory, are all only inventions for formulating in compact and manageable form what we already know, like all inventions do not achieve complete success in accomplishing what they were designed to do, much less complete success in fields beyond the scope of the original design, and that our only justification for hoping to penetrate at all into the unknown with these inventions is our past experience that sometimes we have been fortunate enough to be able to push on a short distance by acquired momentum. The Nature of Physical Theory (1936), 136. God runs electromagnetics on Monday, Wednesday, and Friday by the wave theory, and the devil runs it by quantum theory on Tuesday, Thursday, and Saturday. In Daniel J. Kevles, The Physicists (1978), 159, citing interview by David Webster, AIP, p.35. I cannot seriously believe in it [quantum theory] because the theory cannot be reconciled with the idea that physics should represent a reality in time and space, free from spooky actions at a distance [spukhafte Fernwirkungen]. Letter to Max Born (3 Mar 1947). In Born-Einstein Letters (1971), 158. I have tried to read philosophers of all ages and have found many illuminating ideas but no steady progress toward deeper knowledge and understanding. Science, however, gives me the feeling of steady progress: I am convinced that theoretical physics is actual philosophy. It has revolutionized fundamental concepts, e.g., about space and time (relativity), about causality (quantum theory), and about substance and matter (atomistics), and it has taught us new methods of thinking (complementarity) which are applicable far beyond physics. My Life & My Views (1968), 48. I like relativity and quantum theories because I don't understand them and they make me feel as if space shifted about like a swan that can't settle, refusing to sit still and be measured; and as if the atom were an impulsive thing always changing its mind. 'Relativity', David Herbert Lawrence, The Works of D.H. Lawrence (1994), 437. If all this damned quantum jumping were really here to stay, I should be sorry, I should be sorry I ever got involved with quantum theory. As reported by Heisenberg describing Schrödinger’s time spent debating with Bohr in Copenhagen (Sep 1926). In Werner Heisenberg, Physics and Beyond: Encounters and Conversations (1971), 75. As cited in John Gribbin, Erwin Schrodinger and the Quantum Revolution. If I get the impression that Nature itself makes the decisive choice [about] what possibility to realize, where quantum theory says that more than one outcome is possible, then I am ascribing personality to Nature, that is to something that is always everywhere. [An] omnipresent eternal personality which is omnipotent in taking the decisions that are left undetermined by physical law is exactly what in the language of religion is called God. As quoted by John D. Barrow in The Universe that Discovered Itself (2000), 171. If the world has begun with a single quantum, the notions of space and would altogether fail to have any meaning at the beginning; they would only begin to have a sensible meaning when the original quantum had been divided into a sufficient number of quanta. If this suggestion is correct, the beginning of the world happened a little before the beginning of space and time. I think that such a beginning of the world is far enough from the present order of Nature to be not at all repugnant. It may be difficult to follow up the idea in detail as we are not yet able to count the quantum packets in every case. For example, it may be that an atomic nucleus must be counted as a unique quantum, the atomic number acting as a kind of quantum number. If the future development of quantum theory happens to turn in that direction, we could conceive the beginning of the universe in the form of a unique atom, the atomic weight of which is the total mass of the universe. This highly unstable atom would divide in smaller and smaller atoms by a kind of super-radioactive process. In a seminal short letter (457 words), 'The Beginning of the World from the Point of View of Quantum Theory', Nature (9 May 1931), 127, 706. If we take quantum theory seriously as a picture of what’s really going on, each measurement does more than disturb: it profoundly reshapes the very fabric of reality. In 1900 however, he [Planck] worked out the revolutionary quantum theory, a towering achievement which extended and improved the basic concepts of physics. It was so revolutionary, in fact, that almost no physicist, including Planck himself could bring himself to accept it. (Planck later said that the only way a revolutionary theory could be accepted was to wait until all the old scientists had died.) (1976). In Isaac Asimov’s Book of Science and Nature Quotations (1988), 324. In Darwin’s theory, you just have to substitute ‘mutations’ for his ‘slight accidental variations’ (just as quantum theory substitutes ‘quantum jump’ for ‘continuous transfer of energy’). In all other respects little change was necessary in Darwin’s theory. In fact, it is often stated that of all the theories proposed in this century, the silliest is quantum theory. Some say that the only thing that quantum theory has going for it, in fact, is that it is unquestionably correct. In Hyperspace: A Scientific Odyssey Through Parallel Universes, Time Warps, and The Tenth Dimension (1994), 262. In the discussion of the. energies involved in the deformation of nuclei, the concept of surface tension of nuclear matter has been used and its value had been estimated from simple considerations regarding nuclear forces. It must be remembered, however, that the surface tension of a charged droplet is diminished by its charge, and a rough estimate shows that the surface tension of nuclei, decreasing with increasing nuclear charge, may become zero for atomic numbers of the order of 100. It seems therefore possible that the uranium nucleus has only small stability of form, and may, after neutron capture, divide itself into two nuclei of roughly equal size (the precise ratio of sizes depending on liner structural features and perhaps partly on chance). These two nuclei will repel each other and should gain a total kinetic energy of c. 200 Mev., as calculated from nuclear radius and charge. This amount of energy may actually be expected to be available from the difference in packing fraction between uranium and the elements in the middle of the periodic system. The whole 'fission' process can thus be described in an essentially classical way, without having to consider quantum-mechanical 'tunnel effects', which would actually be extremely small, on account of the large masses involved. [Co-author with Otto Robert Frisch] Lise Meitner and O. R. Frisch, 'Disintegration of Uranium by Neutrons: a New Type of Nuclear Reaction', Nature (1939), 143, 239. In the history of physics, there have been three great revolutions in thought that first seemed absurd yet proved to be true. The first proposed that the earth, instead of being stationary, was moving around at a great and variable speed in a universe that is much bigger than it appears to our immediate perception. That proposal, I believe, was first made by Aristarchos two millenia ago ... Remarkably enough, the name Aristarchos in Greek means best beginning. [The next two revolutions occurred ... in the early part of the twentieth century: the theory of relativity and the science of quantum mechanics...] Edward Teller with Judith L. Shoolery, Memoirs: A Twentieth-Century Journey in Science and Politics (2001), 562. It did not cause anxiety that Maxwell’s equations did not apply to gravitation, since nobody expected to find any link between electricity and gravitation at that particular level. But now physics was faced with an entirely new situation. The same entity, light, was at once a wave and a particle. How could one possibly imagine its proper size and shape? To produce interference it must be spread out, but to bounce off electrons it must be minutely localized. This was a fundamental dilemma, and the stalemate in the wave-photon battle meant that it must remain an enigma to trouble the soul of every true physicist. It was intolerable that light should be two such contradictory things. It was against all the ideals and traditions of science to harbor such an unresolved dualism gnawing at its vital parts. Yet the evidence on either side could not be denied, and much water was to flow beneath the bridges before a way out of the quandary was to be found. The way out came as a result of a brilliant counterattack initiated by the wave theory, but to tell of this now would spoil the whole story. It is well that the reader should appreciate through personal experience the agony of the physicists of the period. They could but make the best of it, and went around with woebegone faces sadly complaining that on Mondays, Wednesdays, and Fridays they must look on light as a wave; on Tuesdays, Thursdays, and Saturdays, as a particle. On Sundays they simply prayed. The Strange Story of the Quantum (1947), 42. It is a remarkable fact that the second law of thermodynamics has played in the history of science a fundamental role far beyond its original scope. Suffice it to mention Boltzmann’s work on kinetic theory, Planck’s discovery of quantum theory or Einstein’s theory of spontaneous emission, which were all based on the second law of thermodynamics. From Nobel lecture, 'Time, Structure and Fluctuations', in Tore Frängsmyr and Sture Forsén (eds.), Nobel Lectures, Chemistry 1971-1980, (1993), 263. It is not surprising that our language should be incapable of describing the processes occurring within the atoms, for, as has been remarked, it was invented to describe the experiences of daily life, and these consists only of processes involving exceedingly large numbers of atoms. Furthermore, it is very difficult to modify our language so that it will be able to describe these atomic processes, for words can only describe things of which we can form mental pictures, and this ability, too, is a result of daily experience. Fortunately, mathematics is not subject to this limitation, and it has been possible to invent a mathematical scheme—the quantum theory—which seems entirely adequate for the treatment of atomic processes; for visualization, however, we must content ourselves with two incomplete analogies—the wave picture and the corpuscular picture. The Physical Principles of the Quantum Theory, trans. Carl Eckart and Frank C. Hoyt (1949), 11. It would take a civilization far more advanced than ours, unbelievably advanced, to begin to manipulate negative energy to create gateways to the past. But if you could obtain large quantities of negative energy—and that's a big “IF”—then you could create a time machine that apparently obeys Einstein's equation and perhaps the laws of quantum theory. Quoted by J.R. Minkel in 'Borrowed Time: Interview with Michio Kaku', Scientific American (23 Nov 2003). Looking back … over the long and labyrinthine path which finally led to the discovery [of the quantum theory], I am vividly reminded of Goethe’s saying that men will always be making mistakes as long as they are striving after something. From Nobel Prize acceptance speech (2 Jun 1920), as quoted and translated by James Murphy in 'Introduction: Max Planck: a Biographical Sketch', in Max Planck and James Murphy (trans.), Where is Science Going (1932), 23. This passage of Planck’s speech is translated very differently for the Nobel Committee. See elsewhere on this web page, beginning, “When I look back…”. Memories are not recycled like atoms and particles in quantum physics. They can be lost forever. From music video, Marry The Night: The Prelude Pathétique (2011). Quoted in Richard J. Gray II, The Performance Identities of Lady Gaga: Critical Essays (2011), 137. No known theory can be distorted so as to provide even an approximate explanation [of wave-particle duality]. There must be some fact of which we are entirely ignorant and whose discovery may revolutionize our views of the relations between waves and ether and matter. For the present we have to work on both theories. On Mondays, Wednesdays, and Fridays we use the wave theory; on Tuesdays, Thursdays, and Saturdays we think in streams of flying energy quanta or corpuscles. 'Electrons and Ether Waves', The Robert Boyle Lecture 1921, Scientific Monthly, 1922, 14, 158. No other theory known to science [other than superstring theory] uses such powerful mathematics at such a fundamental level. …because any unified field theory first must absorb the Riemannian geometry of Einstein’s theory and the Lie groups coming from quantum field theory… The new mathematics, which is responsible for the merger of these two theories, is topology, and it is responsible for accomplishing the seemingly impossible task of abolishing the infinities of a quantum theory of gravity. In 'Conclusion', Hyperspace: A Scientific Odyssey Through Parallel Universes, Time Warps, and the Tenth Dimension (1995), 326. On quantum theory I use up more brain grease than on relativity. Our scientific work in physics consists in asking questions about nature in the language that we possess and trying to get an answer from experiment by the means at our disposal. In this way quantum theory reminds us, as Bohr has put it, of the old wisdom that when searching for harmony in life one must never forget that in the drama of existence we are ourselves both players and spectators. It is understandable that in our scientific relation to nature our own activity becomes very important when we have to deal with parts of nature into which we can penetrate only by using the most elaborate tools. The Copenhagen Interpretation of Quantum Theory (1958). In Steve Adams, Frontiers (2000), 13. People were pretty well spellbound by what Bohr said… While I was very much impressed by [him], his arguments were mainly of a qualitative nature, and I was not able to really pinpoint the facts behind them. What I wanted was statements which could be expressed in terms of equations, and Bohr's work very seldom provided such statements. I am really not sure how much later my work was influenced by these lectures of Bohr's... He certainly did not have a direct influence because he did not stimulate one to think of new equations. Recalling the occasion in May 1925 (a year before receiving his Ph.D.) when he met Niels Bohr who was in Cambridge to give a talk on the fundamental difficulties of the quantum theory. In History of Twentieth Century Physics (1977), 109. In A. Pais, 'Playing With Equations, the Dirac Way'. Behram N. Kursunoglu (Ed.) and Eugene Paul Wigner (Ed.), Paul Adrien Maurice Dirac: Reminiscences about a Great Physicist (1990), 94. Peter Atkins, in his wonderful book Creation Revisited, uses a … personification when considering the refraction of a light beam, passing into a medium of higher refractive index which slows it down. The beam behaves as if trying to minimize the time taken to travel to an end point. Atkins imagines it as a lifeguard on a beach racing to rescue a drowning swimmer. Should he head straight for the swimmer? No, because he can run faster than he can swim and would be wise to increase the dry-land proportion of his travel time. Should he run to a point on the beach directly opposite his target, thereby minimizing his swimming time? Better, but still not the best. Calculation (if he had time to do it) would disclose to the lifeguard an optimum intermediate angle, yielding the ideal combination of fast running followed by inevitably slower swimming. Atkins concludes: That is exactly the behaviour of light passing into a denser medium. But how does light know, apparently in advance, which is the briefest path? And, anyway, why should it care? He develops these questions in a fascinating exposition, inspired by quantum theory. In 'Introduction to the 30th Anniversary Edition', The Selfish Gene: 30th Anniversary Edition (1976, 2006), xi-xii. Physicists often quote from T. H. White’s epic novel The Once and Future King, where a society of ants declares, “Everything not forbidden is compulsory.” In other words, if there isn't a basic principle of physics forbidding time travel, then time travel is necessarily a physical possibility. (The reason for this is the uncertainty principle. Unless something is forbidden, quantum effects and fluctuations will eventually make it possible if we wait long enough. Thus, unless there is a law forbidding it, it will eventually occur.) In Parallel Worlds: a Journey Through Creation, Higher Dimensions, and the Future of the Cosmos (2006), 136. Professor [Max] Planck, of Berlin, the famous originator of the Quantum Theory, once remarked to me that in early life he had thought of studying economics, but had found it too difficult! Professor Planck could easily master the whole corpus of mathematical economics in a few days. He did not mean that! But the amalgam of logic and intuition and the wide knowledge of facts, most of which are not precise, which is required for economic interpretation in its highest form is, quite truly, overwhelmingly difficult for those whose gift mainly consists in the power to imagine and pursue to their furthest points the implications and prior conditions of comparatively simple facts which are known with a high degree of precision. 'Alfred Marshall: 1842-1924' (1924). In Geoffrey Keynes (ed.), Essays in Biography (1933), 191-2 Quantum theory thus reveals a basic oneness of the universe. It shows that we cannot decompose the world into independently existing smallest units. As we penetrate into matter, nature does not show us any isolated “building blocks,” but rather appears as a complicated web of relations between the various parts of the whole. These relations always include the observer in an essential way. The human observer constitute the final link in the chain of observational processes, and the properties of any atomic object can be understood only in terms of the object’s interaction with the In The Tao of Physics (1975), 68. Quantum theory—at least in the Heisenberg interpretation—describes the way the world works as a literal moment-to-moment emergence of actual facts out of a background of less factual 'potentia.' Quoted in article 'Nick Herbert', in Gale Cengage Learning, Contemporary Authors Online (2002). Scientists, therefore, are responsible for their research, not only intellectually but also morally. This responsibility has become an important issue in many of today's sciences, but especially so in physics, in which the results of quantum mechanics and relativity theory have opened up two very different paths for physicists to pursue. They may lead us—to put it in extreme terms—to the Buddha or to the Bomb, and it is up to each of us to decide which path to take. In The Turning Point: Science, Society, and the Rising Culture (1983), 87. So-called “common sense” is definitely detrimental to an understanding of the quantum realm! As given in an epigraph, without citation, in David M. Harland (ed.), The Big Bang: A View from the 21st Century (2003), ix. Something unknown is doing we don’t know what—that is what our theory amounts to. Expressing the quantum theory description of an electron has no familiar conception of a real form. In The Nature Of The Physical World (1928), 291. The dividing line between the wave or particle nature of matter and radiation is the moment “Now.” As this moment steadily advances through time it coagulates a wavy future into a particle past. The great revelation of the quantum theory was that features of discreteness were discovered in the Book of Nature, in a context in which anything other than continuity seemed to be absurd according to the views held until then. What is Life? (1944), 48. The incomplete knowledge of a system must be an essential part of every formulation in quantum theory. Quantum theoretical laws must be of a statistical kind. To give an example: we know that the radium atom emits alpha-radiation. Quantum theory can give us an indication of the probability that the alpha-particle will leave the nucleus in unit time, but it cannot predict at what precise point in time the emission will occur, for this is uncertain in principle. The Physicist's Conception of Nature (1958), 41. The mathematical framework of quantum theory has passed countless successful tests and is now universally accepted as a consistent and accurate description of all atomic phenomena. The verbal interpretation, on the other hand – i.e., the metaphysics of quantum theory – is on far less solid ground. In fact, in more than forty years physicists have not been able to provide a clear metaphysical model. In The Tao of Physics (1975), 132. The mathematical framework of quantum theory has passed countless successful tests and is now universally accepted as a consistent and accurate description of all atomic phenomena. The verbal interpretation, on the other hand, i.e. the metaphysics of quantum physics, is on far less solid ground. In fact, in more than forty years physicists have not been able to provide a clear metaphysical model. In The Tao of Physics: An Exploration of the Parallels Between Modern Physics (1975), 132. The mathematically formulated laws of quantum theory show clearly that our ordinary intuitive concepts cannot be unambiguously applied to the smallest particles. All the words or concepts we use to describe ordinary physical objects, such as position, velocity, color, size, and so on, become indefinite and problematic if we try to use them of elementary particles. In Across the Frontiers (1974), 114. The modern physicist is a quantum theorist on Monday, Wednesday, and Friday and a student of gravitational relativity theory on Tuesday, Thursday, and Saturday. On Sunday he is neither, but is praying to his God that someone, preferably himself, will find the reconciliation between the two views. In I Am a Mathematician, the Later Life of a Prodigy (1956), 109. The more success the quantum theory has, the sillier it looks. The more success the quantum theory has, the sillier it looks. The plain fact is that there are no conclusions. If we must state a conclusion, it would be that many of the former conclusions of the nineteenth-century science on philosophical questions are once again in the melting-pot. In 'On Free-Will', Physics and Philosophy (1942), 216. Also collected in Franklin Le Van Baumer (ed.), Main Currents of Western Thought (1978), 703. The prevailing trend in modern physics is thus much against any sort of view giving primacy to ... undivided wholeness of flowing movement. Indeed, those aspects of relativity theory and quantum theory which do suggest the need for such a view tend to be de-emphasized and in fact hardly noticed by most physicists, because they are regarded largely as features of the mathematical calculus and not as indications of the real nature of things. Wholeness and the Implicate Order? (1981), 14. The quantum theory of gravity has opened up a new possibility, in which there would be no boundary to space-time and so there would be no need to specify the behaviour at the boundary. There would be no singularities at which the laws of science broke down and no edge of space-time at which one would have to appeal to God or some new law to set the boundary conditions for space-time. One could say: 'The boundary condition of the universe is that it has no boundary.' The universe would be completely self-contained and not affected by anything outside itself. It would neither be created nor destroyed. It would just BE. A Brief History of Time: From the Big Bang to Black Holes (1988), 136. The Theory of Relativity confers an absolute meaning on a magnitude which in classical theory has only a relative significance: the velocity of light. The velocity of light is to the Theory of Relativity as the elementary quantum of action is to the Quantum Theory: it is its absolute core. 'A Scientific Autobiography' (1948), in Scientific Autobiography and Other Papers, trans. Frank Gaynor (1950), 47. The trend of mathematics and physics towards unification provides the physicist with a powerful new method of research into the foundations of his subject. … The method is to begin by choosing that branch of mathematics which one thinks will form the basis of the new theory. One should be influenced very much in this choice by considerations of mathematical beauty. It would probably be a good thing also to give a preference to those branches of mathematics that have an interesting group of transformations underlying them, since transformations play an important role in modern physical theory, both relativity and quantum theory seeming to show that transformations are of more fundamental importance than equations. From Lecture delivered on presentation of the James Scott prize, (6 Feb 1939), 'The Relation Between Mathematics And Physics', printed in Proceedings of the Royal Society of Edinburgh (1938-1939), 59 , Part 2, 122. The words are strung together, with their own special grammar—the laws of quantum theory—to form sentences, which are molecules. Soon we have books, entire libraries, made out of molecular “sentences.” The universe is like a library in which the words are atoms. Just look at what has been written with these hundred words! Our own bodies are books in that library, specified by the organization of molecules—but the universe and literature are organizations of identical, interchangeable objects; they are information systems. In The Cosmic Code: Quantum Physics as the Language of Nature (1983), 255. There are something like ten million million million million million million million million million million million million million million (1 with eighty zeroes after it) particles in the region of the universe that we can observe. Where did they all come from? The answer is that, in quantum theory, particles can be created out of energy in the form of particle/antiparticle pairs. But that just raises the question of where the energy came from. The answer is that the total energy of the universe is exactly zero. The matter in the universe is made out of positive energy. However, the matter is all attracting itself by gravity. Two pieces of matter that are close to each other have less energy than the same two pieces a long way apart, because you have to expend energy to separate them against the gravitational force that is pulling them together. Thus, in a sense, the gravitational field has negative energy. In the case of a universe that is approximately uniform in space, one can show that this negative gravitational energy exactly cancels the positive energy represented by the matter. So the total energy of the universe is zero. A Brief History of Time: From the Big Bang to Black Holes (1988), 129. Background image credit: Lu Viatour, www.lucnix.be (source) There was a time when we wanted to be told what an electron is. The question was never answered. No familiar conceptions can be woven around the electron; it belongs to the waiting list. The Nature Of The Physical World (1928), 290. Undeterred by poverty, failure, domestic tragedy, and persecution, but sustained by his mystical belief in an attainable mathematical harmony and perfection of nature, Kepler persisted for fifteen years before finding the simple regularity [of planetary orbits] he sought… . What stimulated Kepler to keep slaving all those fifteen years? An utter absurdity. In addition to his faith in the mathematical perfectibility of astronomy, Kepler also believed wholeheartedly in astrology. This was nothing against him. For a scientist of Kepler’s generation astrology was as respectable scientifically and mathematically as the quantum theory or relativity is to theoretical physicists today. Nonsense now, astrology was not nonsense in the sixteenth century. In The Handmaiden of the Sciences (1937), 30. We don't know what we are talking about. Many of us believed that string theory was a very dramatic break with our previous notions of quantum theory. But now we learn that string theory, well, is not that much of a break. The state of physics today is like it was when we were mystified by radioactivity. They were missing something absolutely fundamental. We are missing perhaps something as profound as they were back then. Closing address to the 23rd Solvay Conference in Physics, Brussels, Belgium (Dec 2005). Quoted in Ashok Sengupta, Chaos, Nonlinearity, Complexity: The Dynamical Paradigm of Nature (2006), vii. Cite in Alfred B. Bortz, Physics: Decade by Decade (2007), 206. We may be well be justified in saying that quantum theory is of greater importance to chemistry than physics. For where there are large fields of physics that can be discussed in a completely penetrating way without reference to Planck's constant and to quantum theory at all, there is no part of chemistry that does not depend, in its fundamental theory, upon quantum principles. As quoted in Leonard W. Fine, Chemistry (1972), 537. Please contact Webmaster if you know the primary source. What has been learned in physics stays learned. People talk about scientific revolutions. The social and political connotations of revolution evoke a picture of a body of doctrine being rejected, to be replaced by another equally vulnerable to refutation. It is not like that at all. The history of physics has seen profound changes indeed in the way that physicists have thought about fundamental questions. But each change was a widening of vision, an accession of insight and understanding. The introduction, one might say the recognition, by man (led by Einstein) of relativity in the first decade of this century and the formulation of quantum mechanics in the third decade are such landmarks. The only intellectual casualty attending the discovery of quantum mechanics was the unmourned demise of the patchwork quantum theory with which certain experimental facts had been stubbornly refusing to agree. As a scientist, or as any thinking person with curiosity about the basic workings of nature, the reaction to quantum mechanics would have to be: “Ah! So that’s the way it really is!” There is no good analogy to the advent of quantum mechanics, but if a political-social analogy is to be made, it is not a revolution but the discovery of the New World. From Physics Survey Committee, U.S. National Academy of Sciences, National Research Council, 'The Nature of Physics', in report Physics in Perspective (1973), 61-62. As cited in I. Bernard Cohen, Revolution in Science (1985), 554-555. What nature demands of us is not a quantum theory or a wave theory, instead nature demands of us a synthesis both conceptions, which, to be sure, until now still exceeds the powers of thought of the Concluding remark in Lecture (23 Feb 1927) to Mathematisch-physikalische Arbeitsgemeinschaft, University of Berlin, reported in 'Theoretisches und Experimentelles zur Frage der Lichtentstehung', Zeitschr. f. ang. Chem., 40, 546. As translated and cited in Arthur I. Miller, Sixty-Two Years of Uncertainty: Historical, Philosophical, and Physical Inquiries into the Foundations of Quantum Mechanics (2012), 89 & 108. When Einstein has criticized quantum theory he has done so from the basis of dogmatic realism. You believe in the God who plays dice, and I in complete law and order in a world that objectively exists, and which I, in a wildly speculative way, am trying to capture. … Even the great initial success of the quantum theory does not make me believe in the fundamental dice-game, although I am well aware that our younger colleagues interpret this as a consequence of senility. No doubt the day will come when we will see whose instinctive attitude was the correct one. Letter to Max Born (7 Sep 1944). In Born-Einstein Letters , 146. Einstein Archives 8-207. In Albert Einstein, Alice Calaprice, Freeman Dyson, The Ultimate Quotable Einstein (2011), 393-394. Often seen paraphrased as “I cannot believe that God plays dice with the cosmos.” Also see a related quote about God playing dice on the Stephen W. Hawking Quotes page of this website. In science it often happens that scientists say, 'You know that's a really good argument; my position is mistaken,' and then they would actually change their minds and you never hear that old view from them again. They really do it. It doesn't happen as often as it should, because scientists are human and change is sometimes painful. But it happens every day. I cannot recall the last time something like that happened in politics or religion. (1987) -- Carl Sagan Sitewide search within all Today In Science History pages: Visit our Science and Scientist Quotations index for more Science Quotes from archaeologists, biologists, chemists, geologists, inventors and inventions, mathematicians, physicists, pioneers in medicine, science events and technology. 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Mastering the Mann-Whitney U Test: Theory and Practice with Python Let's start with non-parametric tests. These are also called distribution-free tests; they work with data that doesn't fit a normal distribution. We use them when our data is skewed, ordinal, or has outliers. Here, ordinal means a type of data where the order of the data points matters, but not the difference between the data points. For example, the sequence in which participants finish matters in a race, but the exact time difference between each runner does not. The Mann-Whitney U test compares two independent groups when the dependent variable is ordinal or continuous but doesn't follow a normal distribution. It ranks the values from the two groups and adds the ranks. Similar sums of ranks indicate the two groups are not significantly different. The Mann-Whitney U test provides two values: the U-statistic and the p-value. The U-statistic signifies the rank sum difference between two groups regarding their observed data values. In everyday language, the larger the U-statistic, the more separation or difference between the two groups' data. The p_value here means the same as in the T-test: If the p-value is less than 0.05, we conclude that the difference is statistically significant and not due to randomness.	{"url":"https://learn.codesignal.com/preview/lessons/598","timestamp":"2024-11-01T19:55:46Z","content_type":"text/html","content_length":"129914","record_id":"<urn:uuid:cebf1251-3612-4b3c-abbe-6233ba37c278>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00008.warc.gz"}
Gavrinis 1: Its dimensions and geometrical framework – Sacred Number Sciences Gavrinis 1: Its dimensions and geometrical framework This article first appeared in my Matrix of Creation website in 2012 which was attacked, though an image had been made. Some of this material appeared in my Lords of Time book. photo For Wikipedia by Mirabella. Gavrinis and Tables des Marchands are very similar monuments, both in the orientation of their passageways and their identical latitude. Gavrinis is about 3900 metres east of Tables des Marchands but, unlike the latter, has a Breton name based upon the root GVR (gower). Both passageways directly express the difference between the winter solsticeThe extreme points of sunrise and sunset in the year. In midwinter the sun is to the south of the celestial equator (the reverse in the southern hemisphere) and in midsummer the sun is north of that equator, which is above the geographical Equator). sunrise and the lunar maximum moonrise to the South, by designing the passages to allow these luminaries to enter at the exact day of the winter solstice or the most southerly moonrise over many lunar orbits, during the moon’s maximum standstill. Thus both the monuments allow the maximum moon along their passageway whilst the winter solstice sunrise can only glance into their end From Howard Crowhurst’s work on multiple squares, we know that this difference in angle is that between a 3-4-5The side lengths of the “first” Pythagorean triangle, special because the side lengths are successive small primes and, at Carnac, defined the solsticial extremes of the sun. triangle and the diagonal of a square which is achieved directly by the diagonal of a seven square rectangle. Figure 1 The essence of difference between the winter solstice sunrise (as diagonal of 4 by 3 rectangle) and southerly maximum moonrise (as diagonal of a single square), on the horizon, is captured in the diagonal of a seven squares rectangle. The best available plan for Gavrinis is that made by the AAK before the monument was taken apart and rebuilt by archaeologists. Some time back I managed to introduce a scale by comparison with another plan and estimate the length of the passage and chamber is close to 14 metres. Figure 2 below shows how I had placed 14 squares along and 2 across the AAK plan (vol.2) and to this has now been added the path of the sun and the moon along with a seven squares rectangle where the sun is its diagonal. Figure 2 The AAK plan of Gavrinis with my own metre squares, arrows for sun and moon and seven squares diagonal. Note that the 14 metre squares correspond to the number and often the width of the 14 orthostats either side of the passageway. In the same manner, there are two end stones to mark the duality of Sun and Moon, thence 2 times 7 = 14 stones either side. One notices that the left hand passage wall hugs the Moon for the first third and then the moon travels to the centre of the chamber’s end wall – around the junction between the two stones 45 and 46. Then it is the Sun’s arrow that hugs the left wall thereafter, all the way into the end chamber, having only just entered past the right doorjamb of the entrance. It seems true therefore that the distinctive serpentine form of the passageway is defined by the passage of light from either sun or moon at their extremes to the south, and the width of the entrance. The right hand wall of the passage seems to then be defined by the requirement to maintain a reasonably constant width of at least a metre, for human access and the display of engraved stones. The point where the sun and moon would cross is marked by the singular quartz stone numbered as seventh from the entrance doorjamb and apparently signifying the full moon. The designers appear to have used metres since 14 metres speaks of the sevenfold square within the design. However, 3/4 of a metre is the day-inch count for a lunar month and 14 metres times 4/3 gives the number of lunar months as 56/3 or 18 and 2/3rds which is close to the number 18.618 which defines the lunar nodal cycle. The hypothesis therefore emerges that this passageway and chamber embody a significant day-inch count and the location of the first and second thresholds (seuils) appear to mark the day-inch count for the eclipse yearthe time taken (346.62 days) for the sun to again sit on the same lunar node, which is when an eclipse can happen. of 364.6 days and half this, an eclipse season of 173.3 days. The quartz stone (number 7) lies six metres from the entrance and this would be eight lunar months from the entrance, the quartz signifying a full moon at the crossing of solar and lunar light. The total intended length can then be corrected to being 29.53 times 18.618 day inches long, or about 13.96 metres. This length would represent: the number of days for the moon’s nodes to move retrograde by the same distance (on the eclipticThe path of the Sun through the sky along which eclipses of sun and moon can occur, traditionally divided into the 365¼ parts of the solar year, each part then a DAY in angle rather than time.) as is travelled by the sun in one lunar month of 29.53 days. This is because the lunar nodes move 1/18.618 of the apparent movement of the sun. The nodes will always take 18.618 times longer than the sun, over ANY period of time. The length of time represented at Gavrinis turns the moon’s nodal periodUsually referring to the backwards motion of the lunar orbit's nodes over 6800 days (18.618 years), leading to eclipse cycles like the Saros. into a recapitulation of the solar yearFrom Earth: the time in which the sun moves once around the Zodiac, now known to be caused by the orbital period of the Earth around the Sun. in that 12.368 Gavrinis lengths will add up to one complete nodal period of 18.618 years (6800 days). By measuring (somehow) the angular motion of the moon’s nodes and equating it to the lunar month, a long length of time has been created in day-inches that is 18.618 units long, each unit being 3/4 metre (29.53 day-inches). This measurement was possible to make and large enough to be accurate with regards to the difference between 18.666, 18.6 and 18.62. I would suggest that by embedding such a specific “symbolism” within Gavrinis, something of true significance was preserved: that at some stage an actual measurement of the lunar nodes was made over the same angular distance travelled by the sun in one lunar month. This gave the speed of the lunar nodes relative to the sun and identified the correct length for one complete nodal cycle of 18.618 years – since it is that exact ratio of nodal to solar motion which manifests as the nodal period’s length in years. To conclude, I offer an edited version of a 2004 DuVersity.com tour visit to Gavrinis, filmed on DV by Anthony Blake and hosted by Howard Crowhurst. This film preceeded the film of Locmariaquer. We arrived in the nick of time to catch a feery to Gavrinis Island in a time when official tours were having lunch. We had a massive hamper made but had little time to eventually have a picnic near the pier. This is significantly reduced in length and introduced some of the questions about Gavrinis that can take years to digest. Metrological note: the length of Gavrinis, at 14 metres, can be seen in terms of the AAK Pior π: The constant ratio of a circle's circumference to its diameter, approximately equal to 3.14159, in ancient times approximated by rational approximations such as 22/7. cubit3/2 feet of any sort, such as 12/7 {1.714285}, 1.5 Royal feet of 8/7 feet, but sometimes a double foot, such as the Assyrian {9/10} of 1.8 feet. of 84 cm. This gives the length as 14/0.84 = 50/3 = 16 and two thirds Pi cubits. The Pi cubit is virtually identical to the vara of 33This is the number of years for an exact number of 12053 days. This period can be measured using the equinoctal sun and it has come to be known as the lifetime of semi-divine Solar Heroes such as Jesus and Mithras. This period relates geometrically to the 18.618 years of the moon's nodal period. inches so that there is an equation (29.53 day-inches =) 3/4 metre, times 18.618 equals 33 inches, times 16.3. The former gives more scope for interpretation than the latter. One can recognise that many such metrological equations, between units, could have been familiar within the specialists of that time but also, that some we might find were perhaps not known to them. It seems mean to deny the builders a high level of astronomical meaning when they have been organising Gavrinis around the solar and lunar extremes with great accuracy. In conclusion: The megalithic astronomers needed a metrologyThe application of units of length to problems of measurement, design, comparison or calculation. with which to build at all and, since they had developed it through day-inch countingThe practice of counting the days, using inches or other small units, between synodic phenomena such as years or planetary loops. (using a standard inch) then, when monuments were constructed, they were not only a means to an astronomical observation, but also a lasting monument forming an astronomical record for a pre-literate but technically accomplished	{"url":"https://sacred.numbersciences.org/2020/02/02/gavrinis-1-its-dimensions-and-geometrical-framework/","timestamp":"2024-11-09T06:53:34Z","content_type":"text/html","content_length":"136965","record_id":"<urn:uuid:0f5ca7a2-f867-44a4-a91f-47b250715cd9>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00491.warc.gz"}
Magnetic Dipole Moment of A Revolving Electron According to the atomic model of Niel Bohar, the electrons which are having negative charge are revolving around the nucleus of an atom that is positively charged. This movement is in the circular orbit having the radius r. The revolving of electrons is in the closed path that constitutes the electric current. If the motion of electrons is in an anticlockwise direction, then there is a production of current in a clockwise direction. The current I= e/T, and here T is representing a period of revolution of electrons. Here the orbital velocity is represented by V and the following situation can be stated. Due to the motion of electrons in orbits, an orbital magnetic moment will result. If the mass of electron is represented by m, then Here mvr is angular momentum of an electron around the center of nucleus. The value of this angular momentum is 8.8 × 1010 C kg^-1. According to the hypothesis of Bohar, the angular momentum can only have a discrete set of values that are given by the below equation. Here n is the natural number, and h is used to represent Plank’s constant and its value is 6.626 × 10^-34 Js. Here by the substitution of equation 2 from equation 1, the following condition can be stated. The minimum value associated with the magnetic moment can be represented by below equation. This value of eh/4πm, is known as the Bohr magneton. By the substitution of values of m, h, and e, the calculated value of Bohar magneton is 9.27 × 10–24 Am^2. In addition to magnetic momentum that is resulted due to orbital motion, the magnetic moment possessed by electrons is the vector sum of orbital magnetic moment and the spin magnetic moment. The ratio of magnetic dipole moment to the angular momentum of the revolving electron is known as the gyromagnetic ratio. The value of electron orbital factor is always equal to one, so by using quantum mechanical arguments the classical gyrometric ratio can be derived. The example of hydrogen is most significant for this phenomenon. The atom of hydrogen interacts with external magnetic fields such as pull and push between the two bar magnets. If the magnetic moment is in antiparallel direction to the external magnetic field then the associated potential energy is large but if the magnetic moment is in a parallel direction, then this energy is usually small. An intrinsic property of electron is spin magnetic moment. Whereas, the orbital magnetic dipole moment is also associated with the with the revolving electron in the axis.	{"url":"https://www.w3schools.blog/magnetic-dipole-moment-of-a-revolving-electron","timestamp":"2024-11-13T15:12:18Z","content_type":"text/html","content_length":"140468","record_id":"<urn:uuid:28689175-508e-4f31-9334-d1b0dd0084c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00594.warc.gz"}
Direction and Extrapolation in Active Learning Learning as an Iterative Process One key challenge that resonates in numerous learning tasks, is the process of deciding what to learn next. For example, imagine you are trying to learn a new hobby. One way to learn your hobby quickly, would be to keep track of a reading list of books that will allow you to learn from experts. As you keep reading these books from your reading list, you’ll need to decide which of them is worth investing time to cover all the learning material. For example, if your goal is learning chess, there’s no need to repeat a beginner’s book twice, when you can already progress to learning new opening tactics. How can you intelligently choose the next book? Active learning algorithms can methodologically guide you in your learning endeavor. Let’s take our chess learning problem and model it as an Active Learning problem. Suppose that each book can be represented as a set of features that uniquely identify it. Let’s also assume that you can freely order any book in Amazon Books, denote the set of all of these books as \mathcal{X} and assume that there exists some unknown function f(x) that maps features of books to your ability to win chess games. Let’s further denote our reading list, from which we “sample” new books, as \mathcal{S}. Classical Active Learning algorithms try to address the following question: How can we globally learn f by sampling (noisy) observations of it within \mathcal{S}? Uncertainty Sampling So how can we use Active Learning to solve our chess learning problem? First, assume that at each round n the learner has to choose a book, represented by a point x_n \in \mathcal{S}. Given this choice, it observes a noisy measurement y = f(x) + \varepsilon(x) where \varepsilon is independent, but possibly heteroscedastic, noise. Deciding how to sample x_n (typically) depends on all previously collected data \mathcal{D}_{n -1} = {{\{(x_i, y_i)\}}_{i < n}}. In particular, \mathcal{D}_{n -1} = {{\{(x_i, y_i)\}}_{i < n}} is typically used to learn a stochastic process that models our probabilistic hypothesis over the function f. One classical method for Active Learning is “Uncertainty Sampling” ^[1]Lewis, D. D. and Catlett, J. Heterogeneous uncertainty sampling for supervised learning. In Machine learning proceedings 1994. Elsevier, 1994.. Uncertainty Sampling chooses the next query point by greedily maximizing the variance of a stochastic process that models f. More formally, we write x_n = \arg\max_{x \in \mathcal {X}}\mathrm{Var}[y \mid \mathcal{D}_{n - 1}]. Note that if we model f with a Gaussian Process, the computation of Uncertainty Sampling is exact, i.e., the variance after having observed \mathcal{D_{n - 1}} can be computed in closed-form. A more modern view generalizes Uncertainty Sampling by taking an information-theoretic approach. Specifically, one can show that for homoscedastic noise, x_n = \arg\max_{x \in \mathcal{S}} \mathrm{I} (f_{\mathcal{S}}; y \mid \mathcal{D}_{n - 1}) reduces to the classical decision rule of Uncertainty Sampling. When is Directedness Crucial? We can spot two weaknesses of active learning and Uncertainty Sampling for our chess learning problem: • Learning from every single book in Amazon Books is very impractical. • Learning with Uncertainty Sampling is not directed towards learning chess, making it less suitable for choosing chess books. There is no need to read all the books in Amazon’s library. As mentioned before, some books are more relevant than others. In this case, we would want to choose an “area of interest” \mathcal{A} \ subseteq \mathcal{X}, that can guide our learning. Note that we still haven’t specified whether \mathcal{A} \subseteq \mathcal{S} or not. While this difference may seem “innocent” at first, it raises some deep questions regarding the learner’s generalization. We’ll start with the easier case, where \mathcal{A} \subseteq \mathcal{S} (compare with case B below), and address the more challenging case \mathcal{A} \not\subset \mathcal{S} (C and D) later. In the meantime, can you think what makes it more challenging? Active learning with an “area of interest”. Case B corresponds to \mathcal{A} \subseteq \mathcal{S}. Cases A, C and D correspond to \mathcal{A} \not\subset \mathcal{S}. As we saw, Uncertainty Sampling is designed to learn about f globally and therefore does not address the directed case, in which we only care about a specific part of \mathcal{X}, thus learning f globally may be “wasteful”. What should we do to focus our learning in \mathcal{A} more effectively? Enter Information-based Transductive Learning (ITL)^[2] Jonas Hübotter, Bhavya Sukhija, Lenart Treven, Yarden As, & Andreas Krause. (2024). Transductive Active Learning: Theory and Applications. . Compare \arg\max_{x \in \mathcal{S}} \mathrm{I}(f_{\mathcal{S}}; y \mid \mathcal{D}_{n - 1}) and \arg\max_{x \in \mathcal{S}} \mathrm{I} (f_\mathcal{A}; y \mid \mathcal{D}_{n - 1}). Although the two querying rules may appear similar at first glance, a closer examination reveals that they result in significantly different decision rules. We can see that ITL takes one step further than Uncertainty Sampling, and computes the posterior distribution, given that our next query is in \mathcal{A}. This is in contrast to Uncertainty Sampling, which queries x_n based only on a prior distribution, that does account for our current specific knowledge of \mathcal{A}. One important fact about ITL, is that in the undirected case, where our “area of interest” is “everything” (\mathcal{S} \subseteq \mathcal{A}, case A above), under homoscedastic noise, ITL reduces to Uncertainty Sampling (can you think why?). Notably, in the undirected case, prior and posterior sampling are the same(!) but in the directed case they can be drastically different — allowing the learner to focus only on those books that are related to chess. The plot below illustrates how prior and posterior sampling differ in the directed case. Prior Sampling (Uncertainty Sampling) Uncertainty during learning As we see, ITL’s decision rule samples points that reduce uncertainty in \mathcal{A}, leading to improved sample efficiency and better performance during learning and at convergence. In the previous section we saw how learners can focus on specific areas of interest within \mathcal{S} to learn more “personalized” tasks. An orthogonal, but equally important challenge, is concerned with problems that require learners to generalize and extrapolate outside of their direct training data. To illustrate this, consider again our chess learning example. Since you only order books from Amazon Books, and you don’t speak Russian, you cannot be exposed to novel chess openings that only exist in Russian chess literature. How learning only from English books can still help you with playing against Russian-speaking players? How much should we expect to learn about Russian openings by only reading from English chess literature? As we saw before, to some extent, classical active learning theory cannot cover cases in which the learner has to extrapolate outside of its accessible sample space. Such cases often arise not only due lack of access to data or distributional shifts, but also due to safety restrictions we want to impose on the learner (like “censoring books”). Therefore, we need more general tools to handle the aforementioned problems. As discussed before, the decision rule given by \arg\max_{x \in \mathcal{S}} \mathrm{I} (f_\mathcal{A}; y \mid \mathcal{D}_{n - 1}) does not make any specific assumptions on the relationship between \mathcal{A} and \mathcal{S}, and thus, it is general enough to tackle cases where generalization outside of \mathcal{S} is required. But how can we quantify generalization outside of \mathcal{S}? In our recent paper, we derive a novel bound on the uncertainty of points x \in \mathcal{A}. In particular, it can be shown that this uncertainty is composed of irreducible and reducible parts, wherein the irreducible uncertainty can be written as \text{Var}[f(x) \mid f_{\mathcal{S}}]. Observe that even if the learner is given full knowledge of f at every point x \in \mathcal{S}, the irreducible uncertainty of points outside of \mathcal{S} will typically be non-zero. Importantly, irreducible uncertainty is a lower bound to the total uncertainty one can hope to reduce. The good news are that learners can still gain information (and reduce uncertainty) by querying the “right” points in \mathcal{S}. How do you choose these points? One way is to use ITL, which, under some mild assumptions, guarantees that the reducible uncertainty will converge to zero. Let’s modify our previous example to a case that requires extrapolation outside of \mathcal{S}. Prior Sampling (Uncertainty Sampling) Uncertainty during learning Like the previous example, ITL’s directedness towards \mathcal{A} allows it to sample more informative query points, leading to better performance, even when the learner has no direct access to \ Closing Thoughts Our chess learning example demonstrates two crucial aspects of intelligent learning: the ability to focus only on what’s important for the task or goal at hand (“no need to read cooking books if you care about chess”) and the ability to generalize outside of the accessible data (“learning Russian chess openings from English books”). ITL brings a fresh perspective to deal with these challenges. By following a principled information-theoretic approach, we are able to derive a decision rule that is theoretically sound (i.e., does not rely too much on heuristics), has convergence guarantees, and finally, exhibits strong empirical performance. For more details about this work, further experiments on harder problems (quadcopter safe exploration, neural-network fine-tuning) check out our recent arxiv and workshop papers. ↑1 Lewis, D. D. and Catlett, J. Heterogeneous uncertainty sampling for supervised learning. In Machine learning proceedings 1994. Elsevier, 1994. ↑2 Jonas Hübotter, Bhavya Sukhija, Lenart Treven, Yarden As, & Andreas Krause. (2024). Transductive Active Learning: Theory and Applications.	{"url":"https://las.inf.ethz.ch/direction-and-extrapolation-in-active-learning","timestamp":"2024-11-04T07:01:53Z","content_type":"text/html","content_length":"57207","record_id":"<urn:uuid:9eeda350-df1c-4d84-85b6-94f663a85473>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00404.warc.gz"}
Operator Overloading in Rust You can make your own types support arithmetic and other operators, too, just by implementing a few built-in traits. This is called operator overloading. #[derive(Clone, Copy, Debug)] struct Complex<T> { /// Real portion of the complex number re: T, /// Imaginary portion of the complex number im: T, Arithmetic and Bitwise Operators In Rust, the expression a + b is actually shorthand for a.add(b), a call to the add method of the standard library’s std::ops::Add trait. Rust’s standard numeric types all implement std::ops::Add. use std::ops::Add; assert_eq!(4.125f32.add(5.75), 9.875); assert_eq!(10.add(20), 10 + 20); • Bring the Add trait into scope so that its method is visible. The definition of std::ops::Add: trait Add<Rhs = Self> { type Output; fn add(self, rhs: Rhs) -> Self::Output; • The trait Add<T> is the ability to add a T value to yourself. □ If you want to be able to add i32 and u32 values to your type, your type must implement both Add<i32> and Add<u32>. • The trait’s type parameter Rhs defaults to Self, so if you’re implementing addition between two values of the same type, you can simply write Add for that case. use std::ops::Add; // impl Add<Complex<i32>> for Complex<i32> { impl Add for Complex<i32> { type Output = Complex<i32>; fn add(self, rhs: Self) -> Self { Complex { re: self.re + rhs.re, im: self.im + rhs.im, □ We shouldn’t have to implement Add separately for Complex<i32>, Complex<f32>, Complex<f64>, and so on. All the definitions would look exactly the same except for the types involved, so we should be able to write a single generic implementation that covers them all, as long as the type of the complex components themselves supports addition: impl<T> Add for Complex<T> where T: Add<Output = T>, type Output = Self; fn add(self, rhs: Self) -> Self { Complex { re: self.re + rhs.re, im: self.im + rhs.im, ☆ By writing where T: Add<Output=T>, we restrict T to types that can be added to themselves, yielding another T value. □ A maximally generic implementation would let the left- and right-hand operands vary independently and produce a Complex value of whatever component type that addition produces: impl<L, R> Add<Complex<R>> for Complex<L> where L: Add<R>, type Output = Complex<L::Output>; fn add(self, rhs: Complex<R>) -> Self::Output { Complex { re: self.re + rhs.re, im: self.im + rhs.im, ☆ In practice, however, Rust tends to avoid supporting mixed-type operations. You can use the + operator to concatenate a String with a &str slice or another String. However, Rust does not permit the left operand of + to be a &str, to discourage building up long strings by repeatedly concatenating small pieces on the left. • This performs poorly, requiring time quadratic in the final length of the string. • Generally, the write! macro is better for building up strings piece by piece. Rust’s built-in traits for arithmetic and bitwise operators come in 3 groups: unary operators, binary operators, and compound assignment operators. Unary Operators Aside from the dereferencing operator , Rust has two unary operators that can be customized: - and !. Trait name Expression Equivalent expression std::ops::Neg -x x.neg() std::ops::Not !x x.not() All of Rust’s signed numeric types implement std::ops::Neg, for the unary negation operator -; the integer types and bool implement std::ops::Not, for the unary complement operator !. There are also implementations for references to those types. • ! complements bool values and performs a bitwise complement (that is, flips the bits) when applied to integers; it plays the role of both the ! and ~ operators from C and C++. trait Neg { type Output; fn neg(self) -> Self::Output; trait Not { type Output; fn not(self) -> Self::Output; Negating a complex number simply negates each of its components. use std::ops::Neg; impl<T> Neg for Complex<T> where T: Neg<Output = T>, type Output = Complex<T>; fn neg(self) -> Complex<T> { Complex { re: -self.re, im: -self.im, Binary Operators Category Trait name Expression Equivalent expression Arithmetic operators std::ops::Add x + y x.add(y) std::ops::Sub x - y x.sub(y) std::ops::Mul x y x.mul(y) std::ops::Div x / y x.div(y) std::ops::Rem x % y x.rem(y) Bitwise operators std::ops::BitAnd x & y x.bitand(y) std::ops::BitOr x \| y x.bitor(y) std::ops::BitXor x ^ y x.bitxor(y) std::ops::Shl x << y x.shl(y) std::ops::Shr x >> y x.shr(y) All of Rust’s numeric types implement the arithmetic operators. Rust’s integer types and bool implement the bitwise operators. There are also implementations that accept references to those types as either or both operands. Compound Assignment Operators A compound assignment expression is one like x += y or x &= y. In Rust, the value of a compound assignment expression is always (), never the value stored. Many languages have operators like these and usually define them as shorthand for expressions like x = x + y or x = x & y. In Rust, however, x += y is shorthand for the method call x.add_assign(y), where add_assign is the sole method of the std::ops::AddAssign trait: trait AddAssign<Rhs = Self> { fn add_assign(&mut self, rhs: Rhs); All of Rust’s numeric types implement the arithmetic compound assignment operators. Rust’s integer types and bool implement the bitwise compound assignment operators. The built-in trait for a compound assignment operator is completely independent of the built-in trait for the corresponding binary operator. Implementing std::ops::Add does not automatically implement std::ops::AddAssign. Equivalence Comparisons Rust’s equality operators, == and !=, are shorthand for calls to the std::cmp::PartialEq trait’s eq and ne methods: assert_eq!(x == y, x.eq(&y)); assert_eq!(x != y, x.ne(&y)); trait PartialEq<Rhs = Self> Rhs: ?Sized, fn eq(&self, other: &Rhs) -> bool; fn ne(&self, other: &Rhs) -> bool { Since the ne method has a default definition, you only need to define eq to implement the PartialEq trait: impl<T: PartialEq> PartialEq for Complex<T> { fn eq(&self, other: &Complex<T>) -> bool { self.re == other.re && self.im == other.im Implementations of PartialEq are almost always of the form shown: they compare each field of the left operand to the corresponding field of the right. These get tedious to write, and equality is a common operation to support, so if you ask, Rust will generate an implementation of PartialEq for you automatically. Simply add PartialEq to the type definition’s derive attribute: #[derive(Clone, Copy, Debug, PartialEq)] struct Complex<T> { // ... • Rust’s automatically generated implementation is essentially identical to the handwritten code, comparing each field or element of the type in turn. Rust can derive PartialEq implementations for enum types as well. Naturally, each of the values the type holds (or might hold, in the case of an enum) must itself implement PartialEq. Unlike the arithmetic and bitwise traits, which take their operands by value, PartialEq takes its operands by reference. This means that comparing non-Copy values like Strings, Vecs, or HashMaps doesn’t cause them to be moved, which would be troublesome: let s = "d\x6fv\x65t\x61i\x6c".to_string(); let t = "\x64o\x76e\x74a\x69l".to_string(); assert!(s == t); // s and t are only borrowed... // ... so they still have their values here. assert_eq!(format!("{} {}", s, t), "dovetail dovetail"); where Rhs: ?Sized relaxes Rust’s usual requirement that type parameters must be sized types, letting us write traits like PartialEq<str> or PartialEq<[T]>. The traditional mathematical definition of an equivalence relation, of which equality is one instance, imposes three requirements. For any values x and y: 1. If x == y is true, then y == x must be true as well. 2. If x == y and y == z, then it must be the case that x == z. 3. It must always be true that x == x. Rust’s f32 and f64 are IEEE standard floating-point values. According to that standard, expressions like 0.0/0.0 and others with no appropriate value must produce special not-a-number values, usually referred to as NaN values. The standard further requires that a NaN value be treated as unequal to every other value—including itself. Any ordered comparison with a NaN value must return false. So while Rust’s == operator meets the first two requirements for equivalence relations, it clearly doesn’t meet the third when used on IEEE floating-point values. This is called a partial equivalence relation, so Rust uses the name PartialEq for the == operator’s built-in trait. If you write generic code with type parameters known only to be PartialEq, you may assume the first two requirements hold, but you should not assume that values always equal themselves. If you’d prefer your generic code to require a full equivalence relation, you can instead use the std::cmp::Eq trait as a bound, which represents a full equivalence relation: if a type implements Eq, then x == x must be true for every value x of that type. • In practice, almost every type that implements PartialEq should implement Eq as well; f32 and f64 are the only types in the standard library that are PartialEq but not Eq. The standard library defines Eq as an extension of PartialEq, adding no new methods: trait Eq: PartialEq<Self> {} If your type is PartialEq and you would like it to be Eq as well, you must explicitly implement Eq, even though you need not actually define any new functions or types to do so. We could implement it even more succinctly by just including Eq in the derive attribute on the Complex type definition. impl<T: Eq> Eq for Complex<T> {} #[derive(Clone, Copy, Debug, Eq, PartialEq)] struct Complex<T> { // ... Derived implementations on a generic type may depend on the type parameters. With the derive attribute, Complex<i32> would implement Eq, because i32 does, but Complex<f32> would only implement PartialEq, since f32 doesn’t implement Eq. When you implement std::cmp::PartialEq yourself, Rust can’t check that your definitions for the eq and ne methods actually behave as required for partial or full equivalence. They could do anything you like. Rust simply takes your word that you’ve implemented equality in a way that meets the expectations of the trait’s users. Although the definition of PartialEq provides a default definition for ne, you can provide your own implementation if you like. However, you must ensure that ne and eq are exact complements of each other. Users of the PartialEq trait will assume this is so. Ordered Comparisons Rust specifies the behavior of the ordered comparison operators <, >, <=, and >= all in terms of a single trait, std::cmp::PartialOrd: trait PartialOrd<Rhs = Self>: PartialEq<Rhs> Rhs: ?Sized, fn partial_cmp(&self, other: &Rhs) -> Option<Ordering>; fn lt(&self, other: &Rhs) -> bool {} fn le(&self, other: &Rhs) -> bool {} fn gt(&self, other: &Rhs) -> bool {} fn ge(&self, other: &Rhs) -> bool {} • PartialOrd<Rhs> extends PartialEq<Rhs>: you can do ordered comparisons only on types that you can also compare for equality. • The only method of PartialOrd you must implement yourself is partial_cmp. □ When partial_cmp returns Some(o), then o indicates self’s relationship to other: enum Ordering { Less, // self < other Equal, // self == other Greater, // self > other □ If partial_cmp returns None, that means self and other are unordered with respect to each other: neither is greater than the other, nor are they equal. Among all of Rust’s primitive types, only comparisons between floating-point values ever return None: specifically, comparing a NaN (not-a-number) value with anything else returns None. Expression Equivalent method call Default definition x < y x.lt(y) x.partial_cmp(&y) == Some(Less) x > y x.gt(y) x.partial_cmp(&y) == Some(Greater) x <= y x.le(y) matches!(x.partial_cmp(&y), Some(Less) \| Some(Equal) x >= y x.ge(y) matches!(x.partial_cmp(&y), Some(Greater) \| Some(Equal) If you know that values of two types are always ordered with respect to each other, then you can implement the stricter std::cmp::Ord trait: trait Ord: Eq + PartialOrd<Self> { fn cmp(&self, other: &Self) -> Ordering; • The cmp method here simply returns an Ordering, instead of an Option<Ordering> like partial_cmp: cmp always declares its arguments equal or indicates their relative order. • Almost all types that implement PartialOrd should also implement Ord. In the standard library, f32 and f64 are the only exceptions to this rule. Index and IndexMut You can specify how an indexing expression like a[i] works on your type by implementing the std::ops::Index and std::ops::IndexMut traits. Arrays support the [] operator directly, but on any other type, the expression a[i] is normally shorthand for a.index(i), where index is a method of the std::ops::Index trait. However, if the expression is being assigned to or borrowed mutably, it’s instead shorthand for a.index_mut(i), a call to the method of the std::ops::IndexMut trait. Other Operators Not all operators can be overloaded in Rust. Author fungho License CC BY-NC-ND 4.0 LastMod 2023-07-03 18:01:26 (f036e23)	{"url":"https://blog.hofungkoeng.com/post/buildingblock/rust/11_operator_overloading/","timestamp":"2024-11-07T03:56:44Z","content_type":"text/html","content_length":"62644","record_id":"<urn:uuid:aa25d733-9272-4b97-b8d6-9c2a59d9188d>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00346.warc.gz"}
Arithmetic - Definition, Facts & Examples - Cuemath A day full of math games & activities. Find one near you. A day full of math games & activities. Find one near you. The teacher asked the class to find the value of $3+7\times2$ Mia and John were the first two to answer. Who do you think is correct? It's Mia who got it right. Let's proceed to understand why her calculation is correct and what mistake John did. As we do so, we will explore the various concepts of arithmetic math and the operations involved. Lesson Plan 1. What Is Arithmetic? 2. Important Notes on Arithmetic 3. Solved Examples on Arithmetic 4. Challenging Questions on Arithmetic 5. Interactive Questions on Arithmetic What Is Arithmetic? Arithmetic is the branch of mathematics in which we study numbers and relations among numbers using various properties and use them to solve examples. The word ‘Arithmetic’ comes from "arithmos", a Greek word meaning numbers. One of the oldest and fundamental principles of mathematics, Arithmetic is all about numbers and the elementary operations (addition, subtraction, multiplication, and division) that can be performed with those numbers. Arithmetic is all around you. If you take out two ice cubes from the ice tray, how many are left? To find this, you will have to subtract 2 from the total number of slots. If each room in your house has 3 windows, and there are 4 rooms, to find the total windows in the house you will have to multiply 3 with 4. Basic Rules of Arithmetic Let's just briefly go over the basic arithmetics operations. Addition and Subtraction Addition and subtraction are the most basic arithmetics operations. These concepts are building blocks of understanding and operating on numbers. We add and subtract numbers, amounts, and values in our everyday lives. Addition can be visualized as 'putting together' of two or more quantities. In arithmetics mathematics, subtraction means to take away some things from a group. In other words, subtraction is the process of removing things from a group. Multiplication and Division Multiplication is one of the four basic arithmetic operations that can be applied to different math concepts like multiplying and dividing fractions, decimals, rationals, integers, etc. These operations form the building blocks for the other math concepts. And the last of basic arithmetic operations is division. In simple words, the division can be defined as the splitting of a large group into equal smaller groups. Equal to Sign : "=" Equal to sign is used to denote the outcome of the operations on numbers. Inverse Operations • The operations addition and subtractions are inverse operators of each other. • Similarly, the operations multiplication and division are inverse operators of each other. Example 1 When we add 3 to 8 we get 11. i.e. 8+3 = 11 This would also mean: • If we take away 3 from 11, we get 8 • If we take away 8 from 11, we get 3 Example 2 Multiplying 3 with 8 gives 24 i.e 8 x 3 = 24 This would also mean: • If we divide 24 by 3 we get 8 • If we divide 24 by 8 we get 3 When there is more than 1 operator present, there is a rule called DMAS which is to be followed to operate them. As per this rule: When we operate numbers with multiple operators, going from left to right we need to first operate numbers involving division or multiplication followed by operators addition and Let's take the example that we took in the beginning, $3+7\times2$ Here, we have 2 operators, multiplication and addition. So, the first multiplication of 7 and 2 will be done and then the answer will be added to 3 Arithmetic Examples Let's quickly see some arithmetic examples from our day-to-day life. Example 1 Sia plucked 45 flowers from a garden and distributed them equally among 9 of her friends. Can you find, how many flowers did each of her friends receive? As we need to distribute 45 flowers equally among 9 children, we need to divide 45 by 9 \[45\div 9 = 5\] So, each of her friends will receive 5 flowers. Example 2 Mother bought 4 packets of candies each for John and Mia. There were 5 candies inside each packet. Can you find total how many candies were there? Number of candies in 1 packet = 5 So, number of candies in 4 packets = $4\times5=20$ Thus, each child will get 20 candies. Number of children = 2 So, total number of candies = $20\times2=40$ 1. Arithmetic is a branch of mathematics that deals with operations on numbers. 2. There are four basic operations in arithmetic: Addition, Subtraction, Multiplication, and Division. 3. The order of these operations is given by the DMAS rule. Solved Examples A school library has 500 books out of which 120 reference books, 150 non-fiction books and the rest are fiction books. How many fiction books are there in the library? Number of reference books = 120 Number of non-fiction books = 150 Total number of books = 500 \text {Number of fiction books} &= 500 - (120 +150) \\&=500-270\\&=230 $\therefore$ Number of fiction books = 230 Peter has $25. He bought 3 cupcakes and a glass of milkshake. If a cupcake costs $2 and a glass of milkshake costs $7, how much money is he left with? Cost of 1 cupcake = $ 2 Cost of 3 cupcakes $3\times2=$ 6$ Cost of a glass of milkshake = $7 $\therefore$ Total amount spent = 6 + 7 = $13 So, the money left with him = 20 - 13 = $7 $\therefore$ He has $7 left with him A box has some bananas, oranges, and apples. There are 30 bananas and the number of oranges is half of the bananas whereas the number of apples is 5 more than oranges. How many fruits are there in the box? Number of bananas = 30 Number of oranges =$\dfrac{1}{2}$ of 30 = 15 Number of apples = Number of oranges + 5 = 20 Total = 30 + 15 + 20 = 65 $\therefore$There are 65 fruits in the box Here is an image of a set of bowling pins. 1. If we continue with the given arrangement, can you find how many bowling pins will be there in the 10th row? 2. What will be the total number of bowling pins, if we consider the arrangement till the 10th row? Interactive Questions Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result. Let's Summarize Starting from arithmetic definition to arithmetic sequence and arithmetic formula, this section covered various aspects of arithmetic math. You should now be in a good position to identify sequences and patterns on paper, as well as around you, and come up with the required answers. About Cuemath At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in. FAQs on Arithmetic 1. What is the difference between arithmetic and algebra? Arithmetics mathematics revolves around specific numbers and their computations using various basic arithmetic operations. On the other hand, algebra is about the rules and boundaries which stand true for whole numbers, integers, fractions, functions, and all other numbers in general. Algebra is built upon arithmetic math, following the arithmetic definition in all cases. 2. What topics come under arithmetic? With arithmetic definition being extremely vast, there is a wide range of topics that come under its umbrella. They start from the basics like numbers, addition, subtraction, division, and move on to more complex subjects like exponents, variations, sequence, progression, and more. Some of the arithmetic formulas and arithmetic sequence did get covered in this section. 3. What is basic arithmetic math? Basic arithmetic math covers four fundamental operations, which include addition, subtraction, multiplication, and division. In arithmetic math, the properties of numbers are associative, commutative, and distributive. 4. What are the 4 basic mathematical operations? The four basic operations in maths are addition, subtraction, multiplication, and division. 5. Who is the father of arithmetic? Brahmagupta is known as the father of arithmetic. He was a 7th Century Indian Mathematician, and also an astronomer. 6. Why is arithmetic important? Arithmetic involves basic operations related to numbers that are used by every individual on a daily basis. It is an essential building block for advanced mathematics. Learn from the best math teachers and top your exams ● Live one on one classroom and doubt clearing ● Practice worksheets in and after class for conceptual clarity ● Personalized curriculum to keep up with school	{"url":"https://www.cuemath.com/numbers/arithmetic/","timestamp":"2024-11-04T21:01:30Z","content_type":"text/html","content_length":"230485","record_id":"<urn:uuid:e07d8d5f-72d6-4d8f-8805-d4ad03085107>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00008.warc.gz"}
nilpotent matrix Definitions for nilpotent matrix nilpo·tent ma·trix This dictionary definitions page includes all the possible meanings, example usage and translations of the word nilpotent matrix. 1. Nilpotent matrix In linear algebra, a nilpotent matrix is a square matrix N such that N k = 0 {\displaystyle N^{k}=0\,} for some positive integer k {\displaystyle k} . The smallest such k {\displaystyle k} is called the index of N {\displaystyle N} , sometimes the degree of N {\displaystyle N} . More generally, a nilpotent transformation is a linear transformation L {\displaystyle L} of a vector space such that L k = 0 {\displaystyle L^{k}=0} for some positive integer k {\displaystyle k} (and thus, L j = 0 {\displaystyle L^{j}=0} for all j ≥ k {\displaystyle j\geq k} ). Both of these concepts are special cases of a more general concept of nilpotence that applies to elements of rings. 1. Chaldean Numerology The numerical value of nilpotent matrix in Chaldean Numerology is: 5 2. Pythagorean Numerology The numerical value of nilpotent matrix in Pythagorean Numerology is: 3 Find a translation for the nilpotent matrix definition in other languages: • - Select - • 简体中文 (Chinese - Simplified) • 繁體中文 (Chinese - Traditional) • Español (Spanish) • Esperanto (Esperanto) • 日本語 (Japanese) • Português (Portuguese) • Deutsch (German) • العربية (Arabic) • Français (French) • Русский (Russian) • ಕನ್ನಡ (Kannada) • 한국어 (Korean) • עברית (Hebrew) • Gaeilge (Irish) • Українська (Ukrainian) • اردو (Urdu) • Magyar (Hungarian) • मानक हिन्दी (Hindi) • Indonesia (Indonesian) • Italiano (Italian) • தமிழ் (Tamil) • Türkçe (Turkish) • తెలుగు (Telugu) • ภาษาไทย (Thai) • Tiếng Việt (Vietnamese) • Čeština (Czech) • Polski (Polish) • Bahasa Indonesia (Indonesian) • Românește (Romanian) • Nederlands (Dutch) • Ελληνικά (Greek) • Latinum (Latin) • Svenska (Swedish) • Dansk (Danish) • Suomi (Finnish) • فارسی (Persian) • ייִדיש (Yiddish) • հայերեն (Armenian) • Norsk (Norwegian) • English (English) Word of the Day Would you like us to send you a FREE new word definition delivered to your inbox daily? Use the citation below to add this definition to your bibliography: Are we missing a good definition for nilpotent matrix? Don't keep it to yourself... A unsealed B butch C eminent D ravening	{"url":"https://www.definitions.net/definition/nilpotent%2Bmatrix","timestamp":"2024-11-10T17:32:34Z","content_type":"text/html","content_length":"70233","record_id":"<urn:uuid:23b1492e-f002-4761-89fb-d1f4630d01c0>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00376.warc.gz"}
UN1494 Intro to Exp Introduction and Logistics • Lecturer: Emily Tiberi ect2158@columbia.edu. Office hours: TBA or by appointment • Expectations □ your data analysis should be on Python or Mathematica, but not Excel □ Background physics: have taken any physics class at college level □ Time commitment: expected to spend on average less than 10 hours on your lab report each week • Absence Policy □ Attendance includes lab participation and lab report submission □ You are allowed two excused absences and one unexcused absence • Lab sessions: □ Each lab session will begin with a brief recap with the TA. □ Your lab group will collect raw data □ There will be in-lab discussions. Work in groups, analyze data and answer the discussion questions. □ You will have a week to finish the lab report □ bring your laptop, recommend recording raw data using excel and export • Lab report: your lab report should be your own, but all the others will be/can be collaborative □ your first week lab report will only be personal feedback, no grade □ there will be a rubric for lab reports • Grading: no quizzes, 90% lab report, and 10% in-lab discussion and participation (discussion with the TA) Scientific Writing Readers interpret prose more easily when it flows smoothly… From background rationale conclusion • Don’t force the reader to figure out your logic – clearly state the rationale. • Clear writing is also concise writing. The report should be fairly brief (up to 4 pages). Your report should look like: • “References” most of the time empty • Abstract can be there, but is optional. For example • the introduction should usually be no more than half a page in single column □ describe the research question, why it is important, and what approaches (briefly) you used (optional: briefly mention the results if you want) • method: you really want to talk about your setup, including your apparatus and instruments used, any techniques, etc. □ this should be brief in general • results and analysis: the most important section in your report. You should succinctly give □ data obtained, graphs and tables □ how you calculated those quantities (if not obvious) □ comments on uncertainties. Why are certain error bars so big? □ is your result good? • conclusions: should be short. Contain your interpretation of the result, what you expected Other notes for writing these reports • example, annotated lab reports are also provided in Courseworks Error Analysis When we measure any quantity, we cannot expect to measure it exactly = we will have errors/uncertainties in our experiment! Therefore, instead of giving each of your friend $2.3684$ slices of pizza, you might say: \[2.37 \pm 0.01 \quad \mathrm{slices}\] note that • you may want to match the number of significant figures in your number and your uncertainty • usually keep up to 2 sig. figs. Type of Errors There are mainly two types of errors you get in your experiment □ Statistical/Random Errors = no going around this, but are quantifiable ☆ Due to random fluctuations from measurement to measurement ☆ Can be reduced by taking more measurements, computing their mean and std □ Systematic Errors = e.g. how you are measuring ☆ Always bias the data in one direction ☆ “do you best” to identify and correct it ☆ Hard to quantify How do we quantify statistical/random errors? • calculate mean $\bar{x}$, std $\sigma$ • but also Standard Error of the Mean $\sigma_\bar{x}$ = uncertainty we have on our measured average = uncertainty in your lab report \[\sigma_\bar{x} = \frac{s}{\sqrt{N}}\] because each time one more measurement is taken, your mean will change. So here is that measure of “precision in our measurement of the mean” What happens when you have different measurements all with different precisions and you want to combine them into a unique result? For example, consider you are measuring a length with: • 4 different rulers, each time measure once = $x_1\pm \sigma_1, x_2\pm \sigma_2,x_3 \pm \sigma_3,x_4 \pm \sigma_4$ • 4 different rulers, with each ruler you measured a lot of times, hence four averages $\bar{x}1 \pm \sigma{1}, \bar{x}2 \pm \sigma{2}, \bar{x}3 \pm \sigma{3}, \bar{x}4 \pm \sigma{4}$ And you want to report a single length measurement by a weigthed average of them. Then you can do this: \[\bar{x} = \frac{\sum_i w_i x_i}{\sum_i w_i}, \quad w_i = \frac{1}{\sigma_i^2}\] and then your error for this weighted mean is: \[\frac{1}{\sigma^2} = \sum_{i=1}^N \frac{1}{\sigma_i^2}\] (the more general way to derive those uncertainty is in the section: Error Propagation) Confidence Intervals You want to see if a theoretically calculated = true average, $\mu$, falls at a certain distance from the statistical mean=your measured mean, $\bar{x}$ in your experiments. The idea is, say, your measurement follows a Gaussian distribution then we can “estimate” whether a certain result is reasonable: • if within $1\sigma_x$, then in general you can say it is good agreement • within $1\sigma \sim 2\sigma$, it is consistent (but probably need more measures) • beyond that it is not very good note that if the error bar is very large, it does not mean your measurement is automatically good = caveat to mention in your report Error Propagation Given some function $f(x,y,z)$, where each variable has uncertainty $x \pm \sigma_x$ etc, then: \[\sigma_f^2 = \left( \frac{\partial f}{\partial x} \right)^2 \sigma_x^2 + \left( \frac{\partial f}{\partial y} \right)^2 \sigma_y^2 + \left( \frac{\partial f}{\partial z} \right)^2 \sigma_z^2\] which is valid if the errors on x, y and z are uncorrelated. Note: you may want to automate this calculation in your python code. where notice that • relative uncertainty = $1\pm (\sigma_x / \bar{x})$, absolute uncertainty $\sigma_x$ • relative uncertainty is useful when you are doing multiplication Graphical Analysis of Data When plotting data, you want to show at minimum a) uncertainty = error bars; b) axis labels But often you would also want to fit a linear model given this data, i.e. \[y_i = ax_i + b\] where your measurements are $(x_i, y_i)$. Obviously you will not be able to get a perfect fit, so we consider: \[\hat{y}_i = ax_i + b\] and hope to pass as close as possible to the highest number of points = linear regression = minimize least square errors given the error = residuals $\Delta y_i = y_i - \hat{y}_i$, then you consider a \[\min_\beta J = \min_\beta \frac{1}{2N}\sum_{i=1}^N \left( y_i - \hat{y}_i \right)^2 = \min_\beta \frac{1}{2n}\sum_{i=1}^N \Delta y_i^2\] where $\beta = (a, b)$ in this example. • the more general case if to of course consider linear algebra formulation \[\mathbf{y} = \mathbf{X\alpha + \beta} = \mathbf{[1,X] \beta} \equiv \mathbf{X\beta}\] where $\beta = [\alpha_1, \alpha_2…,\alpha_n, \beta]$. This is so that you can more easily find your least square solution. • one interpretation of why your model is not perfect is to consider how $y$ are constructed. Consider a process where you have a random Gaussian noise to generate those data \[\mathbf{y} = \mathbf{X\beta} + \epsilon\] meaning the residuals you have $\Delta y_i = \epsilon_i$, so that if the your model is correct you should see which would be a good way to perform sanity check. But there are cases when you see that plotting $\Delta y_i$ gives (not limited to only those two cases): • in case 1 it could be you are fitting a wrong model, or you have some systematic error • in case 2, you might need a some form of a weighted fit Lab 1 Basically we are re-imagining Galileo’s ramp experiment, where basically we are considering: • motion under constant acceleration • forces will affect the acceleration of a mass (newton’s second law) The apparatus we will be using is the Fritionless Air-Track where Timing is performed by Sonic Ranger. • Sonic Ranger: Measuring Velocity: It simply computes the average velocity over very small time intervals Then, one can then have a measure of the elasticity of a collision by computing the coefficient of restitution \[e = \left\| \frac{v_f}{v_i} \right\| = \begin{cases} 1 & \text{elastic}\\ <1 & \text{non elastic} \end{cases}\] Three main parts 1. Leveling the ramp 2. Measuring the elasticity of the bumper (under constant velocity) □ By comparing the velocity before and after, you can compute elasticity (note to propagate uncertainty to compute $e$) □ note that since $e = \vert v_f / v_i\vert$, then the uncertainty is: \[\sigma_e^2 = \frac{1}{v_i^2}\sigma_{v_f}^2 + \frac{v_f^2}{v_i^4}\sigma^2_{v_i}\] □ will measurements/error of $v_i$ be dependent on $v_f$? In this experiment, theses two are treated as independent variables so intuitively no. But practically you need to be careful about 3. Measuring the acceleration due to gravity (under constant acceleration □ When you incline the ramp, the size of the gravitational force along the ramp is proportional to the angle of the ramp incline for which you can estimate acceleration with both $v(t)=a_x t$ and $x(t)=…$. □ then, once you have measured $a_x$, you can compute your $g$ by \[a_x = g \sin(\theta) = g\frac{h}{L}\] □ Do you expect steeper slopes to be more accurate than shallower ones? Why or why not? ☆ measurement readings error as there is less time when $h$ is high $\implies$ $a_x$ is high $\implies$ less time to collect data $\implies$ more error introduced ☆ if there is systematic uncertainty, then relative uncertainty of height becomes less ☆ the higher velocity means other contributions such as air resistance becomes more significant 4. Challenge: estimate friction by looking at the loss of energy Check: when you doing linear regression for this lab, you are probably going to just do the unweighted version $\implies$ do not need to capture uncertainty Lab 2: Projectile Motion and Conservation of Energy Overview: basically launching a ball of a ramp, so that you can • estimate friction • predicting landing position Primary objectives □ understanding distributions of data □ Gaussian statistics □ write (at least some part) of your derivation in your lab report So how does this work? We are essentially exchanging potential and kinetic energy in a closed system Then, since kinetic energy is \[E_{k} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\] for rolling without slipping, $\omega R = v$. and potential energy is \[E_{p}(r) = -G \frac{M_e m }{r} \approx mgh\] But of course, often we get energy “leaked” to friction, e.g. $W_f$ work done by friction \[E_{k}^{ini} = E_k^{fin} + W_f\] or the canonical form from work-energy theorem \[W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\] Then, the position of the ejected ball can be described by \[x(t) = x_0 + v_{x,0}t\\ y(t) = y_0 + v_{y,0}t - \frac{1}{2}gt^2\] This should summarize all you need. You basically get • $E$ • $\Delta E_p = mg \Delta h$ • $\Delta E_k = (7/10) m v_0^2$ • $W_f = mg \Delta h’$ for $\Delta h’$ being the height at which the ball stops • ..blablabla with which you can find out the final landing position $x(t)$ Note that □ the friction force we are estimating is technically velocity dependent. So measuring friction at one configuration might not mean you have the same friction in later setups □ The experiment is extremely sensitive to the value of $W_f$, so measure it as carefully as possible! Taking Data Basically you will record the position of the landing by having the ball hitting a carbon paper: • draw a marker of where you expect it to land • compare against the actual data Then, we can look at the distribution of the data □ you should expect your data to look like a normal distribution (even if your model is wrong) if you are testing stuff in the same configuration □ error propagation in this lab is onerous. It is then highly recommended that you define \[x_{app} = \frac{D\sqrt{2h_2h_E}}{L};\quad h_E = \frac{10}{7}(\Delta h - \Delta h')\] $h_2$ and $h_E$ are not independent. Therefore you should really calculate $\sigma_u$ for $u = h_2h_e$ Lab 4 Primary Learning Goals: • Weighted linear regressions • Weighted means of population of data And we will be probing atomic structures. Before, we understood the classical macroscopic force between matter, but not really what the constituents of matter were • JJ Thomson shows that matter has constituents that are negatively charged and whose charge/mass ratio is constant = quantization! • proposed the plum pudding model (not quite right, improved by latter models) • today, quantum mechanics! Ann the standard model of elementary particles Experiment: measure $e/m$ by using lorentz force giving a circular motion 1. given a loop current, we can compute the magnetic field created specifically, we will use a Helmholtz coil, which is basically two coils: and $C$ will be given. magnetic field at the center of experimental apparatus. Finally we also want to not have contribution from external field, hence you want to align your apparatus with any external field (very 2. Then, in this field (around the axis of the loop) if we send in a moving charge: \[F = q\vec{v}\times \vec{B} = m \frac{v^2}{R}\] which performs a circular motion with radius $R$. Hence \[\frac{q}{m} = \frac{2v}{R^2 B^2}\] 3. but we don’t have an easy way to measure the velocity $v$. One solution to this problem is to speed the electrons up thanks to a known potential difference \[K_{\mathrm{gain}} = e V = \frac{1}{2}mv^2\] assuming $v_0=0$ is at rest and accelerated across $V$. 4. measure the radius of curvature Finally, plot with a line of best fit to a equation = shows that the equation is linear hence has more physical meaning (e.g. than computing $e/m$ for each of your trial and averaging across) • note that you will need to pick different $I$, one good way is to pick it such that it hits one of the markers • what about the error of the $I$? You will notice that the marker is quite thick. So you can take $I_\min$ and $I_\max$ that hits that marker, and then take $\bar{I}\pm \sigma_I$ as uncertainty Lab 5: Polarization and Interference A few background: • light as EM wave = oscillating E and B fields • Electric and magnetic fields are always perpendicular to each other • Everyday light is usually unpolarized. All directions of the electric field are equally probable □ a polarized light, e.g. electric fields points in one direction only □ so a better definition/visualization of unpolarized light = cannot define a plane where it oscillates in Then how much field will be transmitted is given by a linearly polarized light: \[\|\vec{E}_{trans}\| = \| \vec{E}_0 \| \cos\theta\] Experiment 1 Background Mauls’ law: but all we care is its intensity because that is what our eyes can observe \[I \propto \| \vec{E} \|^2\] Therefore we get Malus’ Law (at the second polarizer) \[I=I_0 \cos^2\theta\] so you should see something like this So how do you measure this? You will get a setup with a rotatable polarizer: So that you can where in the plot you will need to extract at least 20 points = 20 different values of $\cos^2\theta$ Experiment 2 Background: Young’s Double Slit • constructive interference = same propagation direction, same frequency, in phase • destructive interference = same propagation direction, same frequency, but out of phase • (recall that standing wave = opposite propagation direction, same frequency) Here we focus on the constructive and destructive interference the idea is to measure where the peaks/dark spots are. The key insight is that: assumption to make thing easier: two waves are parallel, as $D » d$ • for bright spot to appear, then the $\Delta l$ must be an integer multiple of wavelength = $m\lambda$ • for dark spot to appear, then $\Delta l$ will be $(m+1/2)\lambda$ Then for the positions for the bright spot is How do you measure this? There will be a sensor you can move along $x$, and record intensity $I(x)$ so that you can find $x_m$ Experiment 3 Background: Diffraction: can be thought of as self-interference: where the diffraction single-slit minima occurs at so that you can overlay your single-slit minima envelop on top of your double slit experiment. How did this happen? • in an ideal double slit, all the amplitudes will be constant. In an “ideal” single split, you get your envelope • therefore in the practical double split, the observed intensity is actually an ideal double split $\times$ single slit How would yuo measure this? Once again take measurements by moving the sensor in the transverse direction What are some questions to think about: • What limits the precision of these measurements with light? theory, physical limitations, aberration, diffraction, ambient lights Some tips: 1. For all the three parts: Move the RMS slowly when recording data. 2. For the polarizer part: Try your best to minimize the amount of environmental light coming in. E.g. using the dim light in the room, move components closer to the sensor, etc. 3. For the last part: move the laser closer to the light sensor to see more fringes but remember to record the value of D! Lab 6: Interferometer The idea that we can use light as a precision measurement tool. Recall that some key properties include • interference basically when two waves interfere under different conditions • refraction, governed by snell’s law so that basically, as Maxwell’s equation is “different” in medium, we get \[v = c/n,\quad n \text{ being index of refraction}\] • optical path length: how many “cycles” the light spend during the a physical distance in the case of different medium (but same physical distance), then you can simply calculate the number of wavelength we can fit in each “box” (i.e. physical distance) why would this quantity be useful? Recall that in the previous lab this difference in wavelength relates to constructive/destructive interf. Finally, interferometry, the idea being 1. Split single beam into two, then recombine. 2. Observer sees interference patterns projected onto a small screen. 3. By moving one of the mirrors, the observer can change the path length difference . The consequence is a shift in the interference pattern. so that as you move the $M_1$ length back and forth, you will see a different interference pattern where basically you get two rays because one is being transmitted and other being reflected measure the gravitational wave=stretching/compressing space=changes the physical distance=interference pattern Inteferometers Today Recorded Interference due to Gravitational Wave Alternatively, the beams that are transmitted and reflected+transmitted will have interference = can measure the gap If we change the path length by a certain distance \[2d_m = m \lambda,\] where $d$ is the distance you moved the mirror w.r.t the original position, then you will restore the original intereference pattern. □ be careful that the drawings we did are “assuming” a single light ray, but of course in reality is a “spherical wave front” □ therefore, if you moved $d_{10}$, then you would have seen the bright spot reappeared $10$ times during the time you moved it Experiment: you will use both the Michelson and the Fabry-Perot interferometers How to read the micrometers You will turn the knob and count how many fringes have passed. Then you will do this again: To think about: □ in the end both setup measures the same thing. Which configuration is better? □ sometimes count the first fringe might be hard. It could be better to count the second ring, etc = the largest error is most likely mis-counting the number of fringes. So that should be included in your uncertainty measurement The last part is we can change the optical distance by changing the medium (before we are changing the physical distance) so that we can change the pressure=change index of refraction and observe different interference patterns. Therefore, since the interference patterns re-occur every integer number of wavelengths apart: Then finally from this we can compute $n_{air}$ by □ start turning the knob from the 500 $\mu m$ mark Feedback Received Feedback for Lab 1: Total score: 13.5/20 Overall Communication/Organization rubric items: Full points Data Vis: -1 Plots must have titles as well as captions. Note: A short table or two with some of your measurements would not have been out of place here, but it’s not strictly necessary and in this case, you wouldn’t lose points Data Manipulation & Error Analysis rubric items: -1 - Uses Python, Mathematica, or similar: Yes - Only most relevant equations or analyses are explicitly shown: Yes - Analysis and modeling are connected to physical concepts: Yes - Uncertainties are justified: There is no such thing as “human error.” This will fall under “random error” but never use the phrase “human error” or anything like Discussion rubric items: -3.5 - Comparison of results to expectation with justification: Agreement of measured e with expected value (1) is not reported. When discussing whether results are in agreement with predictions, don’t say that your results are “satisfactory” or “close” or anything like that (e.g. when you talked about your result for b, the y-intercept). When it comes to analyzing results, there’s only one thing that matters. Are your measurements the same as the predicted values within 3 sigma? If yes, they’re in agreement with predictions. If not, there’s a statistically significant error and they’re not in agreement. Once you have established this, you go into discussing the quality of your results (why they were or weren’t in agreement), and then you can potentially use more qualitative descriptions, but only once the quantitative actual analysis has been established. - Discussion of quality (quantitative and qualitative) of results: Whether or not your results are in agreement with expectation, need to comment on why. This hasn’t been done fully. - Sources of errors reference model, assumptions, and technical limitations: Sources of error for e not commented on. - Responds to discussion questions: Not all Narrative flow & context rubric items: -1 - Sufficiently contextualize results and discussion: Yes - Include only necessary and relevant background: Yes. If you want it to be perfect, you need to include just one or two sentences that capture the really big picture (why do we even care about any of the things we investigated in this experiment). You’re really almost there, but think of the intro as the part that will convince the reader that this is worth reading. Think of a scientific journal article where the intro always has something to make the project sound immediately relevant to everyone, no matter their background. - Separate information appropriately: Yes - Summarize concisely: Yes	{"url":"http://jasonyux.com/lectures/2022@columbia/UN1494_Intro_to_Exp.html/","timestamp":"2024-11-05T05:31:25Z","content_type":"text/html","content_length":"45869","record_id":"<urn:uuid:176cbe5c-4d06-4c58-ab9a-72d054081f9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00555.warc.gz"}
Causal Bandits without Graph Learning - RBC Borealis This post is about our paper Causal Bandits without Graph Learning, refer to the full paper for more information. What are Causal Bandits? Stochastic bandits are commonly studied problems aimed at modeling online decision making processes. In stochastic multi-armed bandits an agent selects which of $K$ arms to pull in each of $T$ rounds. After each pull the agent receives stochastic reward $R_t$ coming from distribution $\mathbb{P}_{A_t}(\cdot)$ assuming that $A_t$ denotes the arm pulled at the beginning of the round $t$. In this setting a common goal is to maximize the total expected reward received during the $T$ rounds of interaction, or equivalently to minimize the regret. Denote with $\mu_a$ the expected reward of pulling arm $a\in[K]$, where $[K]={\{1,\dots,K}\}$, then the cumulative regret is defined to be: Figure 1. Standard stochastic $K$-armed bandit as a causal bandit. Causal bandits extend the setting above by allowing the agent to receive additional information at the end of each round. This information comes in the form of samples from the Probabilistic Causal Model (PCM) to which the reward and action variables belong. More specifically, let $\mathcal{G}$ be a directed acyclic graph each node of which is associated with a random variable and let $\mathbb {P}$ be a distribution over the variables in the graph such that the following factorization holds: where $\mathcal{V}$ is the set of all the variables in the graph and $\mathcal{P}(X)$ is the set of parents of $X$. The pair $(\mathcal{G}\\,\mathbb{P})$ is the PCM and in this setting an agent interacts with it by performing interventions over sets of nodes $\mathbf{X}_t\subseteq\mathcal{V}$. An intervention cuts the incoming edges into the nodes in $\mathbf{X}_t$ and sets their values to the values $\mathbf{x}_t$ specified by the intervention. This results in a new probability distribution over the remaining nodes in the graph, denoted by ${\mathbb{P}\{\mathcal{V}\setminus\mathbf{X} _t\mid do (\mathbf{X}_t=\mathbf{x}_t)}\}$. As before, the goal of the agent is to maximize the expected total reward and the associated cumulative regret could be written as Reg_T(\mathcal{G},\mathcal{P})=\max_{\mathbf{X}\subseteq\mathcal{V’}}\max_{\mathbf{x}\in[K]^{\left\| \mathbf{x} \right\|}} \mu_{{\mathbf{X}}={\mathbf{x}}}-\sum_{t=1}^T\mathbb{E}[R_t],\tag{1} where we denoted the set of parent nodes of the reward node by $\mathcal{P}$, assumed that each node in the graph takes on values from the set $[K]$, denoted with $\mathcal{V’}$ the set $\mathcal{V}\ setminus{\{R}\}$, and introduced the notation $\mu_{\mathbf{X}=\mathbf{x}}$ for the expected value of the reward under the intervention $\ do (\mathbf{X}=\mathbf{x})$. In the regret definition we explicitly show the dependence on the graph and the sets of parents of the reward node. In general, the regret depends on the full definition of the PCM, including the probability distribution $\ mathbb{P}$. This dependence is implicit in our notation. At the same time, the proposed algorithm will have dependence of the regret on the set of parent nodes which is why we include it in the notation here. Results from Bandit Literature Standard stochastic $K$-armed bandit could be thought of as a causal bandit with the corresponding graph shown in Figure 1. At the same time, one could treat a causal bandit as a multi-armed bandit with $K^n$ arms. Define $\Delta_{\mathbf{X}=\mathbf{x}}$ as the reward gap of the action $\ do (\mathbf{X}=\mathbf{x})$ from the best action: \Delta_{\mathbf{X}=\mathbf{x}} =\max_{\mathbf{Z}\subseteq\mathcal{V’}}\max_{\mathbf{z}\in[K]^{\left\| \mathbf{Z} \right\|}}\mu_{\mathbf{Z}=\mathbf{z}} -\mu_{\mathbf{X}=\mathbf{x}}. With this we can arrive at the following regret bound for causal bandits[1]: Reg_T(\mathcal{G},\mathcal{P})\preceq\sum_{\mathbf{X}\subseteq\mathcal{V’}}\sum_{\mathbf{x}\in[K]^{\left\|{\mathbf{X}}\right\|}} \Delta_{\mathbf{X}=\mathbf{x}}\left( {1+\frac{\log{T}}{\Delta_{\mathbf Note however, that when $\Delta_{\mathbf{X}=\mathbf{x}}$ is small, the regret bound can become vacuous, since assuming that all the rewards are bounded, we get that the regret cannot grow faster than $\mathcal{O}(T)$. When $\Delta_{\mathbf{X}=\mathbf{x}}$ is small we can simply bound the regret by $\Delta_{\mathbf{X}=\mathbf{x}}T$ and using this [1] arrive at the following bound: Reg_T(\mathcal{G},\mathcal{P})\preceq\sum_{\textcolor{WildStrawberry}{\mathbf{X}\subseteq\mathcal{V’}}}\sum_{\textcolor{WildStrawberry}{\mathbf{x}\in[K]^{\left\|{\mathbf{X}}\right\|}}} \Delta_{\mathbf The dependence on $\textcolor{WildStrawberry}{K^n}$ is very bad and can itself lead to vacuous bound even for moderate values of $K$ and $n$. In our paper we aim to improve this dependence. In particular, [2] show that in the setting described in the previous section, the best intervention in the $PCM$ is over the set of the parent nodes of the reward node. Thus, we propose an algorithm that first discovers the set of parents of the reward node and then uses a standard multi armed bandit algorithm to discover the best interventions over the parents. When the number of parent nodes is smaller than the total number of nodes in the graph, this leads to an improvement. Randomized Parent Search Algorithm To find the parents of the reward node we first consider the case when there is at most one such parent. Consider the example in figure 2 that depicts a run of our algorithm on a graph with four nodes not including the reward node and a single parent of the reward node $P$. At each iteration the algorithm selects a node (shown in $\textcolor{gray}{gray}$) and intervenes on it to estimate the set of descendants of that node and whether the reward node is among these descendants. This is possible to do by comparing the observational distributions of all the nodes and mean reward value with the corresponding distributions and mean value under the intervention. If the current node is an ancestor of the reward node, the algorithm shall remove all the non-descendants of the current node from the consideration during the future iterations and otherwise it should focus only on the non descendants removing all the other nodes. In figure 2 $X_3$ is not an ancestor of $R$ and thus $X_3$ and its descendant $X_2$ should be removed from consideration. Subsequently, $X_1$ is an ancestor of $R$. It is possible that $X_1$ is a parent of $R$ and all its remaining descendants are not ancestors of $R$. Thus, the algorithm shall remember $X_1$ which is indicated by the question mark. At the next iteration there is only $P$ that is left and since it is an ancestor of $R$ that has no descendants, it should be returned as the estimated parent. Figure 2. An example of a run of our algorithm to discover the parent node. When there are multiple parents of the reward node our proposed algorithm discovers them one by one in a reverse topological order. Note that with the procedure described above, the algorithm shall discover a parent of the reward node that has no other parent of the reward node as its descendant. Thus, one could run a similar procedure to discover the next parent that may now have the first discovered parent as its descendant. There is a caveat, however, in that a node might be an ancestor of a previously discovered parent but not a parent of the reward node, see figure 3. In this case, intervening on this node will change the parent node, which in turn will change the reward node. To overcome this issue, our algorithm intervenes on all previously discovered parent nodes in addition to the node under consideration at each iteration. In the end, the procedure halts when there are no more parent nodes that could be discovered. Figure 3. An example of a graph illustrating the need to intervene on previously discovered parents. In practice, to determine the descendants of a node and whether the reward node is among them our algorithm requires to know the distribution and mean reward gaps which we will denote with $\epsilon$ and $\Delta$. These are the smallest absolute values of the difference between the distribution (mean) for intervention when discovering a new parent from the distribution (mean) when discovering the previous parent or observational distribution during the time when trying to discover the first parent. We refer the reader to the paper for the full formal definition. The algorithm compares the empirical values of the gaps with their true values and determines that a node is a descendant of the current node under consideration when the empirical gap is at most $\epsilon/2$. Similarly, it determines that the reward node is a descendant of the current node, when the empirical mean threshold is below $\Delta/2$. The full algorithm is presented in algorithm 1 and maintains • the set of previously discovered parents $\mathcal{\hat{P}}$, • the last discovered ancestor of $R$ denoted by $\hat{P}$ (corresponds to the question mark in figure 2), • the candidate set of nodes $\mathcal{C}$, and the descendants of $\hat{P}$ denoted $\mathcal{D’}$ (corresponds to the set of not crossed out nodes in figure 2), and • a set of nodes $\mathcal{S}$ that should be excluded from consideration when trying to find the next parent. After intervening on the current node and all the nodes in the set of parents found up until the current iteration of the while loop on lines 6-9, the algorithm estimates the descendants of the selected node $X$ on line 10. The test on line 11 determines whether the current node is an ancestor of $R$ and after that the sets $\mathcal{S},\mathcal{C},\mathcal{D’}$ and the value of $\hat{P}$ are updated accordingly. After that if there are no nodes in the candidate set $\mathcal{C}$ then $\hat{P}$ must denote the new parent. If all parents have already been discovered then $\hat{P}$ will take on the $\texttt{null}$ value denoted by $\varnothing$. Regret Bound of our Approach Denote with $\textcolor{MidnightBlue}{\mathbb{E}[N(\mathcal{G},\mathcal{P})]}$ the expected number of distinct node interventions required to discover the parent nodes with our algorithm and $E$ be the event that the set of discovered parent nodes is equal to the true set of parent nodes. We provide a high probability guarantee for the event $E$ to hold and as a result a high probability conditional regret bound of our approach where the expected total reward is conditioned on $E$. Algorithm 1 Full version of $RAPS$ for unknown number of parents Require: Set of nodes $\mathcal{V}$ of $\mathcal{G}$, $\epsilon,\Delta$, probability of incorrect parent set estimate $\delta$ Output: Estimated set of parent nodes $\mathcal{\hat{P}}$ 1: $\mathcal{\hat{P}} \gets\emptyset$, $\mathcal{S} \gets\emptyset$, $\hat{P}\gets\varnothing$, $\mathcal{C} \gets\mathcal{C}$, $\mathcal{D’} \gets\emptyset$ # $\mathcal{D’}$ is the set of descendants of last ancestor of $R$ 2: $B\gets\max\left\{ {\frac{32}{\Delta^2}\log\left( {\frac{8nK(K+1)^n}{\delta}} \right), \frac8{\epsilon^2}\log\left( {\frac{8n^2K^2(K+1)^n}{\delta}} \right)} \right\}$ 3: Observe $B$ samples from $PCM$ and compute $\bar{R}$ and $\hat{P}(X)$ for all $X\in\mathcal{V}$ 4: while $\mathcal{C}\neq\emptyset$ do 5: $\qquad X\sim Unif(\mathcal{C})$ 6: $\qquad$ for $x\in[K]$ and $\mathbb{z}\in[K]^{\left\| {\mathcal{\hat{P}}} \right\|}$ do 7: $\qquad\qquad$ Perform $B$ interventions $do (X=x,\mathcal{\hat{P}}=\mathbf{z})$ 8: $\qquad\qquad$ Compute $\bar{R}^{do (X=x,\mathcal{\hat{P}}=\mathbf{z})}$, $\hat{P}(Y\| do (X=x,\mathcal{\hat{P}}=\mathbf{z}))$ for all $Y\in\mathcal{V}$ 9: $\qquad$ end for 10: $\qquad$Estimate descendants of $X$: \mathcal{D}\gets\biggl\{Y\in\mathcal{V} &\mid \exists x\in[K],\mathbf{z}\in[K]^{\left\| {\mathcal{\hat{P}}} \right\|}, \\ &\text{ such that }\left\| {\hat{P}(Y\|do(\mathcal{\hat{P}}=\mathbf{z})) -\hat{P} (Y\|do(X=x,\mathcal{\hat{P}}=\mathbf{z}))} \right\|>\epsilon/2\biggr\} 11: $\qquad$ if $\exists x\in[K],\mathbf{z}\in[K]^{\left\| {\mathcal{\hat{P}}} \right\|}$ such that $\left\| {\bar{R}^{do(\mathcal{\hat{P}}=\mathbf{z})}-\bar{R}^{do(X=x,\mathcal{\hat{P}}=\mathbf{z})}} \ right\| > \Delta/2$ then 12: $\qquad\qquad \mathcal{C}\gets\mathcal{D}\setminus\left\{ {X} \right\}$ 13: $\qquad\qquad \mathcal{\hat{P}}\gets X, \mathcal{D’}\gets\mathcal{D}$ 14: $\qquad$ else 15: $\qquad\qquad \mathcal{C}\gets\mathcal{C}\setminus \mathcal{D}$ 16: $\qquad\qquad \mathcal{S}\gets\mathcal{S}\cup\mathcal{D}$ 17: $\qquad$ end if 18: $\qquad$ if $\mathcal{C}=\emptyset$ and $\mathcal{\hat{P}}\neq\varnothing$ then 19: $\qquad\qquad \mathcal{\hat{P}}\gets\mathcal{\hat{P}}\cup\left\{ {\hat{P}} \right\}$ 20: $\qquad\qquad \mathcal{S}\gets\mathcal{S}\cup\mathcal{D’}$ 21: $\qquad\qquad \mathcal{C}\gets\mathcal{V}\setminus\mathcal{S}$ 22: $\qquad\qquad \mathcal{\hat{P}}\gets\varnothing$ 23: $\qquad$ end if 24: end while 25: return $\mathcal{\hat{P}}$ Theorem 1. For the learner that uses algorithm 1 and then runs $UCB$ the following bound for the conditional regret holds with probability at least $1-\delta$: Reg_T(\mathcal{G},\mathcal{P}\mid E)\preceq \max\left\{ {\textcolor{Orange}{ \frac1{\Delta^2},\frac1{\epsilon^2}}} \right\} \textcolor{WildStrawberry}{ K^{\left\| \mathcal{P} \right\|+1}}\textcolor {MidnightBlue}{\mathbb{E}[N(\mathcal{G},\mathcal{P})]}\log \left( { \frac{\textcolor{WildStrawberry}{nK^n}}{\delta}} \right) +\sum_{\textcolor{WildStrawberry}{\mathbf{x}\in[K]^{\left\| {\mathcal{P}} \ right\|}}} \Delta_{\mathcal{P}=\mathbf{x}} \left( {1+\frac{\log{T}}{\Delta_{\mathcal{P}=\mathbf{x}}^2} }\right). In the bound above we see improved dependence from $K^n$ to $K^{\left\|{\mathcal{P}}\right\|+1}$ which leads to improved performance of our algorithm when the number of parents is small. We hypothesize that $\max{\{\textcolor{Orange}{1/\Delta^2}, \textcolor{Orange}{1/\epsilon^2}}\}$ dependence is suboptimal and could be improved. For the expected number of distinct node interventions we prove the following result. Theorem 2. Let $\mathcal{P}=\left\{ P \right\}$, then the number of distinct node interventions for any graph $\mathcal{G}$ is equal to: =\sum_{X\in\mathcal{V’}}\frac1{\left\| \mathcal{A}(X)\triangle\mathcal{A}(P)\cup\left\{ {X} \right\} \right\|}, where $\mathcal{A}(X)$ is the set of ancestors of $X$ in the graph induced over the nodes in $\mathcal{V’}$ and $A\triangle B$ stands for the symmetric difference of sets $A$ and $B$, i.e. $A\ triangle B=(A\cup B)\setminus(A\cap B)$. Figure 4. Regret of $RAPS$+$UCB$ vs $UCB$ on a tree graphover $\mathcal{V’}$ with the number of parents ranging from 1 to 3 indicated by the last character in the label. In the paper we provide conditions under which the expected number of distinct node interventions to discover the parent node is sublinear and also extend these conditions to the case when the reward node has multiple parents. Experimental Verification In figure 4 we empirically verify that our approach achieves an improvement. For this figure we let the graph have a binary tree structure for which our theory predicts $\Theta\left({\frac{n}{\log_2 {n}}}\right)$ number of distinct node interventions. While this graph structure does not lead to the best number of distinct node interventions, we still expect to see an improvement since discovering the set of parent nodes drastically reduces the set of arms that $UCB$ has to explore. In this blog post we introduced a causal bandit problem and proposed an algorithm that combines randomized parent search and standard $UCB$ algorithm to achieve improved regret. Check out the full paper for mode details. We would also like to note that possible directions for future work included improving the dependence on $\epsilon$ and $\Delta$, lifting the assumption of no unobserved confounders and proving lower bounds for causal bandits with unknown graph structure. [1] Tor Lattimore and Csaba Szepesv ́ari. Bandit algorithms. Cambridge University Press, 2020. [2] Sanghack Lee and Elias Bareinboim. Structural causal bandits: where to intervene? Advances in Neural Information Processing Systems, 31, 2018.	{"url":"https://rbcborealis.com/research-blogs/causal-bandits-without-graph-learning/","timestamp":"2024-11-07T01:03:07Z","content_type":"text/html","content_length":"197821","record_id":"<urn:uuid:fbd2e999-7fb2-4bc5-bc4c-157af4b499ef>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00593.warc.gz"}
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This is the sixth post in a series on Katie Mann and Kasra Rafi’s paper Large-scale geometry of big mapping class groups. The purpose of this post is to discuss when a Polish group is generated by a coarsely bounded set, and give examples of mapping class groups which are locally coarsely bounded but fail this criterion. The general case We have the following theorem of Rosendal. Theorem 1.2 (Rosendal). Let $G$ be a Polish group. Then $G$ is generated by a coarsely bounded set if and only if $G$ is locally coarsely bounded and not the countably infinite union of a chain of proper, open subgroups. Let’s make sense of this. Note that countable, discrete groups are Polish. Being generated by a coarsely bounded set is the analogue of finite generation for discrete groups, so we should expect that for a countable discrete group (which is automatically locally coarsely bounded, having finite discrete neighborhoods of the identity) being finitely generated is equivalent to not being the countably infinite union of a chain of proper subgroups, which are automatically open. Indeed, suppose $G$ is finitely generated by a set $S$ , and that $G_1 < G_2 < G_3 < \cdots$ is a strictly increasing chain of proper subgroups. Since the union of the $G_i$ is $G$ , each $s \in S$ belongs to some $G_i$ , but then at some finite stage $G_n$ , we have each $s \in S$ contained in $G_n$ , so actually $G_n = G$ . Conversely, supposing every strictly increasing chain of proper subgroups terminates, we show that $G$ is finitely generated. Indeed, we can take a sequence of finitely generated subgroups! Begin with $G_1 = \langle s_1 \rangle$ for some $s_1 \in G$ . At each stage, add $s_{n+1} \notin G_n$ and take $G_{n+1} = \langle s_1,\ldots,s_{n+1}\rangle$ . Since this sequence terminates, we’ve proven that $G$ is finitely generated. (This proof contains the useful fact that every generating set for a finitely generated group contains a \emph{finite} generating set.) A non-example: limit type Remember the “Great Wave off Kanagawa” surface from the previous post? It had genus zero and end space homeomorphic to $\omega^\omega + 1$ in the order topology. Take the connect sum of two copies of that surface; so the genus-zero surface with end space homeomorphic to $\omega^\omega \cdot 2 + 1$ . It looks a little like this: The original “Great Wave” surface has self-similar end space and genus zero, so has coarsely bounded mapping class group. This surface $\Sigma$ has locally coarsely bounded mapping class group, but we will show that $\operatorname{Map}(\Sigma)$ is not generated by a coarsely bounded set. Consider the index-two subgroup $G$ of $\operatorname{Map}(\Sigma)$ comprising those mapping classes that fix pointwise the two maximal ends. We’ll show that $G$ is a countable union of proper open subgroups $G_0 < G_1 < \cdots$ . Since $G$ has index two in $\operatorname{Map}(\Sigma)$ , this will show that $\operatorname{Map}(\Sigma)$ is also a countable union of proper open subgroups $G'_0 < G'_1 < \cdots$ , where each $G'_i$ is obtained from $G_i$ by adding a fixed mapping class $\phi$ that swaps the two maximal ends of $\Sigma$ . (Recall that a subgroup is open if and only if it contains an open neighborhood of the identity, so the $G'_i$ are open since $G$ is open in $\operatorname {Map}(\Sigma)$ and the $G'_i$ are open in $G$ .) To start, consider a simple closed curve $\alpha$ separating $\Sigma$ into two pieces, each containing exactly one maximal end. The identity neighborhood we consider is $U_A$ , where $A$ is an annular subsurface with core curve $\alpha$ . Since $U_A$ is a subgroup of $G$ , we’ll let $G_0 = U_A$ . Since $\operatorname{Map}(\Sigma)$ and hence $G$ is Polish, there is a countable dense subset $\{\phi_i : i \in \mathbb{N}\}$ of $G$ . Any open subgroup containing the $\phi_i$ is in fact all of $G$ , so consider the sequence of open subgroups $G_1 \le G_2 \le \cdots$ , where $G_i = \langle G_0, \phi_1,\ldots, \phi_i \rangle.$ If we can show that each $G_i$ is a proper subgroup of $G$ , we will be done, even though a priori this chain may not be strictly increasing. Consider a maximal end $\xi$ and a neighborhood basis of $\xi$ comprising nested clopen neighborhoods $U_j$ with $U_{j+1} \subset U_j$ , beginning with $U_0$ being the end set of the component of $\Sigma - A$ containing $\xi$ . In the figure we can think of the clopen neighborhoods as coming from the “fluting” process described in the previous post. Thus, $U_0 - U_j$ contains points homeomorphic to $\omega^{j-1} + 1$ but not points homeomorphic to $\ omega^j + 1$ . In plainer words, $U_0 - U_1$ contains isolated planar ends, $U_0 - U_2$ contains ends accumulated by isolated planar ends, $U_0 - U_3$ contains ends accumulated by ends accumulated by isolated planar ends, so on and so forth. Anyway, consider $\phi_1,\ldots,\phi_n$ . Since each $\phi_i$ leaves $\xi$ invariant, we claim that there exists $M$ large such that for all $m \ge M$ , ends homeomorphic to $\omega^m + 1$ contained in $U_m$ actually remain inside $U_m$ under each $\phi_i$ . To see this, note that if there was a sequence of ends $\{\xi_m\}$ with each $\xi_m$ homeomorphic to $\omega^m + 1$ such that each $\xi_m$ was moved outside of $U_m$ by some $\phi_i$ , then since the sequence $\{\xi_m\}$ necessarily converges to $\xi$ , by the pigeonhole principle some $\phi_i$ would have to move $\xi$ . This already shows us that $G_i$ is a proper subgroup of $G$ , since by the classification of surfaces we can move some $\xi_m$ with $m > M$ outside of $U_m$ (and into a neighborhood of the other maximal end). This surface $\Sigma$ is an example of the general phenomenon Mannâ€“Rafi term having end space of “limit type”. The argument we just gave generalizes to show that if $\Sigma$ has limit type (see Definition 6.2 of their paper for a precise definition) then $\operatorname{Map}(\Sigma)$ cannot be generated by a coarsely bounded set. A non-example: infinite rank If $G$ is a finitely generated group, notice that all (a fortiori continuous) quotients of $G$ are finitely generated, and that conversely if $G$ has a quotient that is not finitely generated, then $G$ cannot be finitely generated. The same is true of Polish groups and coarsely bounded generation: if $G$ is a Polish group that has a continuous quotient which is not coarsely boundedly generated, then $G$ is not either. A prime example of such a group as a quotient is the countably infinite group $\bigoplus_{n = 1}^\infty \mathbb{Z}$ . It is possible to build continuous maps to $\bigoplus_{n=1}^\infty \mathbb{Z}$ by using the topology of the end space. Here is one example. Consider the ends $\xi_n$ constructed in the previous post which are pairwise noncomparable. The $\xi_n$ are maximal ends of self-similar surfaces, so have stable neighborhoods. We form a surface by “fluting” together the union of countably infinitely many copies of each $\xi_n$ . Necessarily each collection of ends locally homeomorphic to $\xi_n$ converges to the maximal end of the flute. Now, this surface is self-similar, with genus zero or infinity, hence has coarsely bounded mapping class group. So take the connect sum of two copies of this surface, and call the connect sum $\Sigma$ . As before, take a simple closed curve $\alpha$ that separates $\Sigma$ into two pieces, each one containing a single one of the two maximal ends. Pick one of the maximal ends, call it $\xi$ , and let $U$ be the neighborhood of $\xi$ determined by $\alpha$ . We claim that for each end $\xi_n$ and any mapping class $\phi$ belonging to the index-two subgroup of $\operatorname{Map}(\Sigma)$ fixing $\ xi$ , the number of ends locally homeomorphic to $\xi_n$ mapped into and out of $U$ is finite. Indeed, were either quantity infinite, the same argument as in the previous example shows that $\phi$ would have to move $\xi$ . Anyway, count up the number of ends of type $\xi_n$ moved into $U$ by $\phi$ and subtract the number of ends of type $\xi_n$ moved out of $U$ by $\phi$ . This defines a homomorphism $\ell_n \colon G \to \mathbb{Z}$ . By “shifting a strip of ends locally homeomorphic to $\xi_n$ ”, we can show that $\bigoplus_{n=1}^\infty \ell_n\colon G \to \bigoplus_{n=1}^\infty \mathbb{Z}$ is surjective, and continuous, since if $A$ is an annular subsurface with core curve $\alpha$ , the open set $U_A$ is contained in the kernel of $\bigoplus_{n=1}^\infty \ell_n$ . This surface $\Sigma$ is an example of the general phenomenon Mannâ€“Rafi term having end space of “infinite rank”. The argument we just gave generalizes to show that if $\Sigma$ has infinite rank (see Definition 6.5 for a precise definition) then $\operatorname{Map}(\Sigma)$ cannot be generated by a coarsely bounded set.	{"url":"https://ryleealanza.org/posts/nonexamples-of-coarsely-bounded-generation/","timestamp":"2024-11-13T07:56:56Z","content_type":"text/html","content_length":"12027","record_id":"<urn:uuid:653c8e94-071c-4ccd-9fb9-5a270d266a6d>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00572.warc.gz"}
Short Ton to Metric Tons Converter Enter Short Ton Metric Tons โ Switch toMetric Tons to Short Ton Converter How to use this Short Ton to Metric Tons Converter ๐ ค Follow these steps to convert given weight from the units of Short Ton to the units of Metric Tons. 1. Enter the input Short Ton value in the text field. 2. The calculator converts the given Short Ton into Metric Tons in realtime โ using the conversion formula, and displays under the Metric Tons label. You do not need to click any button. If the input changes, Metric Tons value is re-calculated, just like that. 3. You may copy the resulting Metric Tons value using the Copy button. 4. To view a detailed step by step calculation of the conversion, click on the View Calculation button. 5. You can also reset the input by clicking on button present below the input field. What is the Formula to convert Short Ton to Metric Tons? The formula to convert given weight from Short Ton to Metric Tons is: Weight[(Metric Tons)] = Weight[(Short Ton)] / 1.1023113109 Substitute the given value of weight in short ton, i.e., Weight[(Short Ton)] in the above formula and simplify the right-hand side value. The resulting value is the weight in metric tons, i.e., Weight[(Metric Tons)]. Calculation will be done after you enter a valid input. Consider that a high-performance sports car manufacturing plant processes 10 short tons of aluminum daily. Convert this weight from short tons to Metric Tons. The weight of sports car plant in short ton is: Weight[(Short Ton)] = 10 The formula to convert weight from short ton to metric tons is: Weight[(Metric Tons)] = Weight[(Short Ton)] / 1.1023113109 Substitute given weight of sports car plant, Weight[(Short Ton)] = 10 in the above formula. Weight[(Metric Tons)] = 10 / 1.1023113109 Weight[(Metric Tons)] = 9.0718 Final Answer: Therefore, 10 tn is equal to 9.0718 t. The weight of sports car plant is 9.0718 t, in metric tons. Consider that a premium marble quarry produces 30 short tons of marble per week. Convert this weight from short tons to Metric Tons. The weight of marble quarry in short ton is: Weight[(Short Ton)] = 30 The formula to convert weight from short ton to metric tons is: Weight[(Metric Tons)] = Weight[(Short Ton)] / 1.1023113109 Substitute given weight of marble quarry, Weight[(Short Ton)] = 30 in the above formula. Weight[(Metric Tons)] = 30 / 1.1023113109 Weight[(Metric Tons)] = 27.2155 Final Answer: Therefore, 30 tn is equal to 27.2155 t. The weight of marble quarry is 27.2155 t, in metric tons. Short Ton to Metric Tons Conversion Table The following table gives some of the most used conversions from Short Ton to Metric Tons. Short Ton (tn) Metric Tons (t) 0.01 tn 0.0090718474 t 0.1 tn 0.090718474 t 1 tn 0.9072 t 2 tn 1.8144 t 3 tn 2.7216 t 4 tn 3.6287 t 5 tn 4.5359 t 6 tn 5.4431 t 7 tn 6.3503 t 8 tn 7.2575 t 9 tn 8.1647 t 10 tn 9.0718 t 20 tn 18.1437 t 50 tn 45.3592 t 100 tn 90.7185 t 1000 tn 907.1847 t Short Ton The short ton is a unit of mass commonly used in the United States, equivalent to 2,000 pounds or approximately 907 kilograms. It is often used in industries such as construction, mining, and freight Metric Tons The ton is a unit of mass used in the imperial and U.S. customary systems. There are two main types of tons: the short ton (equal to 2,000 pounds) and the long ton (equal to 2,240 pounds). The ton is commonly used in the context of larger weights, such as the weight of goods, vehicles, or cargo. Frequently Asked Questions (FAQs) 1. What is the formula for converting Short Ton to Metric Tons in Weight? The formula to convert Short Ton to Metric Tons in Weight is: Short Ton / 1.1023113109 2. Is this tool free or paid? This Weight conversion tool, which converts Short Ton to Metric Tons, is completely free to use. 3. How do I convert Weight from Short Ton to Metric Tons? To convert Weight from Short Ton to Metric Tons, you can use the following formula: Short Ton / 1.1023113109 For example, if you have a value in Short Ton, you substitute that value in place of Short Ton in the above formula, and solve the mathematical expression to get the equivalent value in Metric Tons. Weight Converter Android Application We have developed an Android application that converts weight between kilograms, grams, pounds, ounces, metric tons, and stones. Click on the following button to see the application listing in Google Play Store, please install it, and it may be helpful in your Android mobile for conversions offline.	{"url":"https://convertonline.org/unit/?convert=short_ton-ton","timestamp":"2024-11-10T15:23:15Z","content_type":"text/html","content_length":"86798","record_id":"<urn:uuid:55114246-e91d-4bcd-9860-75c31d4b5128>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00669.warc.gz"}
When quoting this document, please refer to the following DOI: 10.4230/LIPIcs.ITCS.2023.73 URN: urn:nbn:de:0030-drops-175767 URL: http://dagstuhl.sunsite.rwth-aachen.de/volltexte/2023/17576/ Hu, Zhuangfei ; Li, Xinda ; Woodruff, David P. ; Zhang, Hongyang ; Zhang, Shufan Recovery from Non-Decomposable Distance Oracles A line of work has looked at the problem of recovering an input from distance queries. In this setting, there is an unknown sequence s ∈ {0,1}^{≤ n}, and one chooses a set of queries y ∈ {0,1}^?(n) and receives d(s,y) for a distance function d. The goal is to make as few queries as possible to recover s. Although this problem is well-studied for decomposable distances, i.e., distances of the form d(s,y) = ∑_{i=1}^n f(s_i, y_i) for some function f, which includes the important cases of Hamming distance, ?_p-norms, and M-estimators, to the best of our knowledge this problem has not been studied for non-decomposable distances, for which there are important special cases such as edit distance, dynamic time warping (DTW), Fréchet distance, earth mover’s distance, and so on. We initiate the study and develop a general framework for such distances. Interestingly, for some distances such as DTW or Fréchet, exact recovery of the sequence s is provably impossible, and so we show by allowing the characters in y to be drawn from a slightly larger alphabet this then becomes possible. In a number of cases we obtain optimal or near-optimal query complexity. We also study the role of adaptivity for a number of different distance functions. One motivation for understanding non-adaptivity is that the query sequence can be fixed and the distances of the input to the queries provide a non-linear embedding of the input, which can be used in downstream applications involving, e.g., neural networks for natural language processing. BibTeX - Entry author = {Hu, Zhuangfei and Li, Xinda and Woodruff, David P. and Zhang, Hongyang and Zhang, Shufan}, title = {{Recovery from Non-Decomposable Distance Oracles}}, booktitle = {14th Innovations in Theoretical Computer Science Conference (ITCS 2023)}, pages = {73:1--73:22}, series = {Leibniz International Proceedings in Informatics (LIPIcs)}, ISBN = {978-3-95977-263-1}, ISSN = {1868-8969}, year = {2023}, volume = {251}, editor = {Tauman Kalai, Yael}, publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik}, address = {Dagstuhl, Germany}, URL = {https://drops.dagstuhl.de/opus/volltexte/2023/17576}, URN = {urn:nbn:de:0030-drops-175767}, doi = {10.4230/LIPIcs.ITCS.2023.73}, annote = {Keywords: Sequence Recovery, Edit Distance, DTW Distance, Fr\'{e}chet Distance} Keywords: Sequence Recovery, Edit Distance, DTW Distance, Fréchet Distance Collection: 14th Innovations in Theoretical Computer Science Conference (ITCS 2023) Issue Date: 2023 Date of publication: 01.02.2023 DROPS-Home \| Fulltext Search \| Imprint \| Privacy	{"url":"http://dagstuhl.sunsite.rwth-aachen.de/opus/frontdoor.php?source_opus=17576","timestamp":"2024-11-08T21:50:22Z","content_type":"text/html","content_length":"7698","record_id":"<urn:uuid:50788e93-f554-4470-b520-0a18d008a5f1>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00830.warc.gz"}
Four Pies Hardware components Adafruit Mini Thermal Receipt Printer × 1 × 1 × 1 Software apps and online services Hand tools and fabrication machines Below a quick overview of the content. • Introduction and showcase video • Pie • Pi • 3.14 • Pies • Result Introduction and showcase video To welcome and celebrate the new Raspberry Pi 4, we’re building a project consisting of four Pi(e)s! Our first pie has two parts, a container and a lid. Both are based on this great 3d model. All we have to do is make it bigger, and add an opening for our pies to come out. Now we have a basic pie, but it does not look the part. Using some glue, felt and coloured balls we add some extra pie-ness. There’s a great tutorial on how to exactly do this right here. Our pie will hold everything we need, and one of them is our Raspberry Pi 4. Configuring and powering is done the usual way. They have a very neat guide on exactly how right here. If you want to know more details about this new Pi, look no more. We want our Pi in our pie to calculate π. To achieve this there’s a bit of python code. def calculatePi(): os.system('echo "scale=2000;4*a(1)"\|bc -l') You can play with the precision by changing the scale. The higher the scale, the longer it will take. We have a Pi within a pie that calculates π. The last step is to add some pies. This requires two things, some ASCII art and a little thermal printer. First of all, our ASCII pie, thanks to this great website! We have our ASCII pie, next is connecting the thermal printer. All we need to do is follow the steps outlined here. You can find the complete code in the ‘Code’ section. Now we have a Raspberry Pi 4, in a pie, calculating π, and printing pies! What better way to welcome the new Raspberry Pi 4 then a Pi(e)fest! Read more	{"url":"https://www.hackster.io/8_Bits_and_a_Byte/four-pies-cf9fc7","timestamp":"2024-11-14T10:58:56Z","content_type":"text/html","content_length":"88540","record_id":"<urn:uuid:313c5de0-b155-45e6-91a6-ac0d8dd2e1c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00379.warc.gz"}
In malignancy research profiling studies have been extensively conducted searching for In malignancy research profiling studies have been extensively conducted searching for genes/SNPs associated with prognosis. method is definitely developed to identify genes that contain important SNPs associated with prognosis. The proposed method has an intuitive Schisantherin A formulation and is recognized using an iterative algorithm. Asymptotic properties are rigorously founded. Simulation demonstrates the proposed method has acceptable overall performance and outperforms a penalization-based meta-analysis method and a regularized thresholding method. An NHL (non-Hodgkin lymphoma) prognosis study with SNP Schisantherin A measurements is definitely analyzed. Genes associated with the three major subtypes namely DLBCL FL and CLL/SLL are recognized. The proposed method identifies genes that are different from alternatives and have important implications and acceptable prediction performance. much smaller than the quantity of SNPs subtypes of the same malignancy and you will find iid observations for subtype as the logarithm of failure time. Denote mainly because the length-covariate vector. The subscript “of subtype Schisantherin A is the intercept ? ?is the vector of regression coefficients and is the error term. Under right censoring we observe ( is the logarithm of censoring time and is Schisantherin A the event indication. Let become the Kaplan-Meier estimator of the distribution function of and for = 2 … are the order statistics of are the connected event indicators. Similarly let become the connected covariate vectors of the ordered subtypes. The overall objective function is definitely where = (SNPs belong to genes. To accommodate partially matched gene units without loss of generality presume that gene is definitely measured only for the 1st subtypes. Denote mainly because the number of SNPs in the is definitely kept to accommodate partially matched SNP units for the same gene. Then contains the regression coefficients for those SNPs in gene across all subtypes. Here notations are more complicated than those in Section 2 to accommodate the “SNP-within-gene??structure and partially matched SNP/gene units. Denote indicates an association between the related gene (SNP) and subtype’s prognosis. Consider the penalty function > 0 is definitely a data-dependent tuning parameter is definitely a constant and accommodates partially matched gene units \|\| · \|\| is the < 1 is the fixed bridge parameter. Statistical properties of the estimate are founded in Supplementary Materials. The above penalty has been designed to tailor our unique data and model characteristics. In our analysis genes are the fundamental functional models. The penalty is the sum of individual terms with one for each gene. For a specific gene two levels of selection need to be carried out. The first is to determine whether it is associated with any subtype whatsoever. This is accomplished using a bridge-type penalty. For any gene associated with at least one subtype the second level of selection is definitely to determine which subtype(s) it is related to. This is accomplished using a Lasso type penalty. The composition of the two penalties can achieve the desired two-level selection. Multiple SNPs may correspond to Schisantherin A the same gene. The effect of gene for subtype is definitely represented from the length-vector as the vector composed of as the matrix composed of = (= diag(as the submatrix of related to = (is definitely a penalty parameter. Proposition 3.1 If minimizes the objective function in (3) if and only if (≥ 0 Mouse monoclonal to IgG2b Isotype Control.This can be used as a mouse IgG2b isotype control in flow cytometry and other applications. for those and may be conducted iteratively. With a fixed has a simple analytic answer. With a fixed minimizes a weighted-group-Lasso type objective function which can be solved using existing algorithms. Motivated by such an observation we propose the following algorithm. Denote = 0. = + 1. Schisantherin A Compute will all become zero and this gene cannot be added back. Therefore over iterations the selected gene units are non-increasing. Properties of the proposed algorithm will also be investigated numerically in Section 4. Research code written in R is definitely available at works.bepress.com/shuangge/45/. 3.4 Tuning parameter selection With bridge-type penalties the value of is usually fixed. Theoretically speaking different ideals of → 1 the bridge estimate behaves similarly to the Lasso estimate. On the other hand as → 0 it behaves similarly to the AIC/BIC penalized estimate. In simulation we experiment with different ideals including 0.5 (which is the most commonly used in the books) 0.7 and 0.9. The result of is comparable to that with various other fines. As → ∞ fewer genes/SNPs are determined. We make use of BIC for tuning parameter selection. With a set minimizes BIC( particularly? on is certainly obtained by installing an AFT model (with least squares estimation) using.	{"url":"https://biospraysehatalami.com/in-malignancy-research-profiling-studies-have-been-extensively-conducted-searching-for/","timestamp":"2024-11-02T14:08:46Z","content_type":"text/html","content_length":"35766","record_id":"<urn:uuid:f202704f-df5e-4979-8568-8c2bada75ba9>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00363.warc.gz"}
Intersubband and intrasubband electronic scattering rates in semiconductor quantum wells Results from calculations of temperature-dependent intrasubband and intersubband electron-electron scattering rates in two subbands in a quasi-two-dimensional quantum well are presented. The screening of the interaction between the electrons plays an important role in determining the magnitude of these scattering rates. In this work we compare the use of different screening models in calculations of electron-electron scattering rates. The screening of the interaction between the electrons, due to the surrounding electron gas, is modeled using the full dynamic, finite-temperature, multisubband dielectric function derived in the random-phase approximation (RPA). A comparison of the scattering rates calculated using the dynamic RPA dielectric function, and the static, long-wavelength approximation to the dielectric function, shows, when plasmon emission is present the necessity of using the dynamic dielectric function to describe the enhancement of the intrasubband scattering rates due to this process. This point is further emphasized by the good agreement between our calculated scattering rates with experimental estimates of electron-electron scattering rates published in the literature [Kim et al., Phys. Rev. Lett. 68, 2838 (1992)]. An approximation that is sometimes used in evaluating the static, long-wavelength approximation to the dielectric function is to assume that all the electrons are in the lowest subband. We point out that the dielectric function derived with this assumption is invalid for interactions involving intersubband transitions. We derive and present the correct expression for the static, long-wavelength screening in this case, which is also analytic and equally simple to evaluate. We also present an analytic result for the dynamic, temperature-dependent, RPA dielectric function derived in the Boltzmann limit. Finally, we compare electron-electron scattering rates calculated with both the dynamic RPA dielectric function, and the dynamic lattice permittivity, allowing for the treatment of coupled plasmon-phonon modes. ©1999 The American Physical Society. Dive into the research topics of 'Intersubband and intrasubband electronic scattering rates in semiconductor quantum wells'. Together they form a unique fingerprint.	{"url":"https://researchportal.hw.ac.uk/en/publications/intersubband-and-intrasubband-electronic-scattering-rates-in-semi","timestamp":"2024-11-06T21:16:28Z","content_type":"text/html","content_length":"57596","record_id":"<urn:uuid:eaddcd18-666b-4c64-84f3-f90b28d6d316>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00873.warc.gz"}
SPPU Information and Cyber Security - December 2015 Exam Question Paper \| Stupidsid SPPU Information Technology (Semester 7) Information and Cyber Security December 2015 Total marks: -- Total time: -- (1) Assume appropriate data and state your reasons (2) Marks are given to the right of every question (3) Draw neat diagrams wherever necessary Solve any one question from Q1 and Q2 1 (a) State the Chinese Remainder theorem with example. 6 M 1 (b) In a public key cryptosystem using RSA, given N=187 and the encryption key (E) as 17, find out the corresponding private key (D). 4 M 2 (a) Draw AES block diagram and explain the steps in detail. 6 M 2 (b) Define following. i) Discrete logarithm ii) Fermat theorem 4 M Solve any one question from Q3 and Q4 3 (a) Explain X.509 standard for Digital Certificate. 6 M 3 (b) Explain permutation and substitution steps in DES algorithm. 4 M 4 (a) Using Euclidean algorithm calculate i) GCD (48, 30) ii) GCD (105, 80) 4 M 4 (b) What problem was Kerberos designed to address. Describe Kerberos realm. 6 M Solve any one question from Q5 and Q6 5 (a) Define IKE protocol and illustrate IKE format in detail. 8 M 5 (b) Discuss SSL with respect to four phases i) Establish security capabilities ii) Server authentication and key exchange iii) Client authentication and key exchange iv) Finish 8 M 6 (a) Explain various categories of Intrusion Detection system (IDS). 8 M 6 (b) How AH and ESP are differs while working under transport and tunnel mode. 8 M Solve any one question from Q7 and Q8 7 (a) Describe the classification of Cyber Crime. 10 M 7 (b) Define cyber security and information security with example. 6 M 8 (a) Explain with example how social engineering is playing wide role in cyber crime. 10 M 8 (b) Write a short note on Indian legal perspective. 6 M Solve any one question from Q9 and Q10 9 (a) What is SQL injection? Explain in detail. 8 M Write short note on: 9 (b) (i) Indian IT act 5 M 9 (b) (ii) Different ways of password cracking 5 M Define the differentiate: 10 (a) Proxy server and an anonymizer. 6 M 10 (b) DOS and DDOS 6 M 10 (c) Virus and worm. 6 M More question papers from Information and Cyber Security	{"url":"https://stupidsid.com/previous-question-papers/download/information-and-cyber-security-13068","timestamp":"2024-11-08T01:54:57Z","content_type":"text/html","content_length":"67359","record_id":"<urn:uuid:3ea537ca-9f72-4c58-afb8-0f8eafee2090>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00887.warc.gz"}
9.2 - Confidence Intervals for a Population Mean \| STAT 100 Over the three-day period from April 1 to April 3, 2015, a national poll surveyed 1500 American households to gauge their levels of discretionary spending. The question asked was how much the respondent spent the day before; not counting the purchase of a home, motor vehicle, or normal household bills. For these sampled households, the average amount spent was $\bar x$ = \$95 with a standard deviation of s = \$185. How close will the sample average come to the population mean? Let's follow the same reasoning as developed in section 9.2 for proportions. We have: \[\text{Sample average} = \text{population mean} + \text{random error}\] The Normal Approximation tells us that the distribution of these random errors over all possible samples follows the normal curve with a standard deviation of $\frac{\sigma}{\sqrt{n}}$. Notice how the formula for the standard deviation of the average depends on the true population standard deviation $\sigma$. When the population standard deviation is unknown, like in this example, we can still get a good approximation by plugging in the sample standard deviation (s). We call the resulting estimate the Standard Error of the Mean (SEM). Standard Error of the Mean (SEM) = estimated standard deviation of the sample average = \[\frac{\text{standard deviation of the sample}}{\sqrt{n}} = \frac{s}{\sqrt{n}}\] In the example, we have s = \$185 so the Standard Error of the Mean = \[\frac{\text{\$185}}{\sqrt{1500}} = \$4.78\] Recap: the estimated daily amount of discretionary spending amongst American households at the beginning of April 2015 was \$95 with a standard error of \$4.78 The Normal Approximation tells us, for example, that for 95% of all large samples, the sample average will be within two SEM of the true population average. Thus, a 95% confidence interval for the true daily discretionary spending would be \$95 ± 2(\$4.78) or\$95 ± \$9.56. Of course, other levels of confidence are possible. When the sample size is large, s will be a good estimate of $\sigma$ and you can use multiplier numbers from the normal curve. When the sample size is smaller (say n < 30), then s will be fairly different from $\sigma$ for some samples - and that means that we need a bigger multiplier number to account for that. (see the optional material on "t-multipliers" in chapter 21). Confidence Intervals for a population mean (n $\ge$) 30 Section For large random samples, a confidence interval for a population mean is given by \[\text{sample mean} \pm z^* \frac{s}{\sqrt{n}}\] where z* is a multiplier number that comes from the normal curve and determines the level of confidence (see Table 9.1 in section 9.2). The equatorial radius of the planet Jupiter is measured 40 times independently by a process that is practically free of bias. These measurements average $\bar x$ = 71492 kilometers with a standard deviation of s = 28 kilometers. Find a 90% confidence interval for the equatorial radius of Jupiter. Note! Note that the equatorial radius of the planet is a fixed number (Jupiter is not changing in size). But measurements are random quantities that might come out different when repeated independently. If the measurement process is unbiased, then repeating the process many times and taking the average gives a better estimate of the true value. Solution: since s = 28 km, the SEM = $\frac{28}{\sqrt{40}}=4.4 km$. With n = 40, using the multiplier number from the normal curve for 90% confidence (z=1.645) will work pretty well so our confidence interval would be: 71492 km ± 1.645(4.4 km) or 71492 km ± 7.3 km How much credit card debt do students typically have when they graduate from Penn State University? A sample of 15 recent Penn State graduates is obtained. Each of these recent graduates is asked to indicate the amount of credit card debt they had at the time of graduation. It turns out that the sample mean was $\bar x$ = \$2430 with a sample standard deviation of s = \$2300. Would it be appropriate to use the method above to find a 99% confidence interval for the average credit card debt for all recent Penn State graduates? Solution: No, with n = 15, using s as an estimate of $\sigma$ would add quite a bit of extra variability; so it would not be appropriate to use the normal curve multiplier associated with 99% confidence (z = 2.576). Also, we can tell from the large value of s relative to the sample average that the data here are quite skewed and so the normal curve would not be a good approximation to the sampling distribution regardless.	{"url":"https://online.stat.psu.edu/stat100/lesson/9/9.2","timestamp":"2024-11-12T23:44:27Z","content_type":"text/html","content_length":"83505","record_id":"<urn:uuid:18fa6372-7370-40b5-aa3f-c27fb93b0174>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00318.warc.gz"}
New theories reveal the nature of numbers A key creative breakthrough occurred when Emory mathematicians Ken Ono, left, and Zach Kent were hiking. As they walked, they noticed patterns in clumps of trees and began thinking about what it would be like to "walk" amid partition numbers. By Carol Clark For centuries, some of the greatest names in math have tried to make sense of partition numbers, the basis for adding and counting. Many mathematicians added major pieces to the puzzle, but all of them fell short of a full theory to explain partitions. Instead, their work raised more questions about this fundamental area of math. Now, Emory mathematician Ken Ono is unveiling new theories that answer these famous old questions. ( Click here to watch a video of Ono's lecture on the topic.) Ono and his research team have discovered that partition numbers behave like fractals. They have unlocked the divisibility properties of partitions, and developed a mathematical theory for “seeing” their infinitely repeating superstructure. And they have devised the first finite formula to calculate the partitions of any number. “Our work brings completely new ideas to the problems,” Ono says. “We prove that partition numbers are ‘fractal’ for every prime. These numbers, in a way we make precise, are self-similar in a shocking way. Our ‘zooming’ procedure resolves several open conjectures, and it will change how mathematicians study partitions.” The problems of partition numbers "have long fascinated mathematicians," Ono says. The work was funded by the American Institute of Mathematics (AIM) and the National Science Foundation. Last year, AIM assembled the world’s leading experts on partitions, including Ono, to attack some of the remaining big questions in the field. Ono, who is a chaired professor at both Emory and the University of Wisconsin at Madison, led a team consisting of Jan Bruinier, from the Technical University of Darmstadt in Germany; Amanda Folsom, from Yale; and Zach Kent, a post-doctoral fellow at Emory. “Ken Ono has achieved absolutely breathtaking breakthroughs in the theory of partitions,” says George Andrews, professor at Pennsylvania State University and president of the American Mathematical Society. “He proved divisibility properties of the basic partition function that are astounding. He went on to provide a superstructure that no one anticipated just a few years ago. He is a Child’s play On the surface, partition numbers seem like mathematical child’s play. A partition of a number is a sequence of positive integers that add up to that number. For example, 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1. So we say there are 5 partitions of the number 4. It sounds simple, and yet the partition numbers grow at an incredible rate. The amount of partitions for the number 10 is 42. For the number 100, the partitions explode to more than 190,000,000. “Partition numbers are a crazy sequence of integers which race rapidly off to infinity,” Ono says. “This provocative sequence evokes wonder, and has long fascinated mathematicians.” By definition, partition numbers are tantalizingly simple. But until the breakthroughs by Ono’s team, no one was able to unlock the secret of the complex pattern underlying this rapid growth. The work of 18th-century mathematician Leonhard Euler (below) led to the first recursive technique for computing the partition values of numbers. The method was slow, however, and impractical for large numbers. For the next 150 years, the method was only successfully implemented to compute the first 200 partition numbers. “In the mathematical universe, that’s like not being able to see further than Mars,” Ono says. A mathematical telescope In the early 20th century, Srinivasa Ramanujan and G. H. Hardy invented the circle method, which yielded the first good approximation of the partitions for numbers beyond 200. They essentially gave up on trying to find an exact answer, and settled for an approximation. “This is like Galileo inventing the telescope, allowing you to see beyond what the naked eye can see, even though the view may be dim,” Ono says. Ramanujan also noted some strange patterns in partition numbers. In 1919 he wrote: “There appear to be corresponding properties in which the moduli are powers of 5, 7 or 11 … and no simple properties for any moduli involving primes other than these three.” The legendary Indian mathematician died at the age of 32 before he could explain what he meant by this mysterious quote, now known as Ramanujan’s congruences. In 1937, Hans Rademacher found an exact formula for calculating partition values. While the method was a big improvement over Euler’s exact formula, it required adding together infinitely many numbers that have infinitely many decimal places. “These numbers are gruesome,” Ono says. In the ensuing decades, mathematicians have kept building on these breakthroughs, adding more pieces to the puzzle. Despite the advances, they were unable to understand Ramanujan’s enigmatic words, or find a finite formula for the partition numbers. “We were standing on some huge rocks, where we could see out over this valley and hear the falls, when we realized partition numbers are fractal,” Ono says. Photo by Zach Kent. Ono’s “dream team” wrestled with the problems for months. “Everything we tried didn’t work,” he says. A eureka moment happened in September, when Ono and Zach Kent were hiking to Tallulah Falls in northern Georgia. As they walked through the woods, noticing patterns in clumps of trees, Ono and Kent began thinking about what it would be like to “walk” amid partition numbers. “We were standing on some huge rocks, where we could see out over this valley and hear the falls, when we realized partition numbers are fractal,” Ono says. “We both just started laughing.” The term fractal was invented in 1980 by Benoit Mandelbrot, to describe what seem like irregularities in the geometry of natural forms. The more a viewer zooms into “rough” natural forms, the clearer it becomes that they actually consist of repeating patterns (see youtube video, below). Not only are fractals beautiful, they have immense practical value in fields as diverse as art to medicine. Their hike sparked a theory that reveals a new class of fractals, one that dispensed with the problem of infinity for partition numbers. “It’s as though we no longer needed to see all the stars in the universe, because the pattern that keeps repeating forever can be seen on a three-mile walk to Tallulah Falls,” Ono says. Ramanujan’s congruences are explained by their fractal theory. The team also demonstrated that the divisibility properties of partition numbers are “fractal” for every prime. “The sequences are all eventually periodic, and they repeat themselves over and over at precise intervals,” Ono says. “It’s like zooming in on the Mandelbrot set,” he adds, referring to the most famous fractal of them all. An Atlanta traffic jam played a role in the final breakthrough of a formula. But this extraordinary view into the superstructure of partition numbers was not enough. The team was determined to go beyond mere theories and hit upon a formula that could be implemented in the real world. The final eureka moment occurred near another Georgia landmark: Spaghetti Junction. Ono and Jan Bruinier were stuck in traffic near the notorious Atlanta interchange. While chatting in the car, they hit upon a way to overcome the infinite complexity of Rademacher’s method. They went on to prove a formula that requires only finitely many simple numbers. “We found a function, that we call P, that is like a magical oracle,” Ono says. “I can take any number, plug it into P, and instantly calculate the partitions of that number. P does not return gruesome numbers with infinitely many decimal places. It’s the finite, algebraic formula that we have all been looking for.” The work by Ono and his colleagues resulted in two papers that will be available soon on the AIM website Ken Ono's public lecture on the new theories How a hike in the woods led to math 'Aha!' 16 comments: 1. AWESOME WOW 2. Congratulations! This is great breakthrough in the number theory. Fractals are everywhere. 3. we are fractal 4. Right there with you, Anon. 5. This discovery is great news! My mind immediately wanders to the application of this discovery to a topic just over the horizon: quantum computing. It seems apparent that a quantum fractal with many dimensions could serve as a flawless model for an entire reality. Being able to perfectly simulate a reality is synonymous with actually creating one. I encourage every possible inference into that statement. There are profound innovations and advancements on the way, and this is the tip of the iceberg. Keep the discoveries coming! Great work! 6. This is awesome. A great write up Ms.Clark. Lucid and great explanation. Thanks. 7. Great Article!! We have to keep making breakthroughs like this one and keep in mind that everything is connected to nature. 8. i smell a nobel!... congratulations! 9. @Sergio: No nobel for mathematics, I'm afraid! The Fields Medal, perhaps. 10. Amazing. I can't wait to read the paper! 11. This is a well-written piece on a tough topic for us laymen to grasp. Nicely done. And kudos to editors for resisting the urge to tart up the article with lots of pretty fractal pictures, but to actually illustrate it with (gasp) real mathematicians and some mathematics! 12. any bearing on p=np? 13. aja¡...ruptura de la linealidad numérica de la partición por continuidad múltiple de la partición especifica del numero. en tanto que X es la potencialidad móvil de N =Uc.algoritm/X 14. and so onto fermat... 15. There appears to be some slight of hand in the proof 16. Emerson BiajotiJanuary 27, 2012 at 4:45PM A matemática ainda guarda grandes segredos. A Teoria dos Números é uma das mais belas partes da Matemática.	{"url":"https://esciencecommons.blogspot.com/2011/01/new-theories-reveal-nature-of-numbers.html","timestamp":"2024-11-03T09:25:05Z","content_type":"application/xhtml+xml","content_length":"224770","record_id":"<urn:uuid:4f410df1-2db0-4354-a79c-af7761692e00>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00382.warc.gz"}
Interagency Modeling and Analysis Group A compliant 1 compartment lung with resistance to air flow, driven by external positive pressure ventilator. The equations governing airflow in and out of a one compartment lung are given by the following analogy to electrical circuits: Airway pressure is analogous to voltage. Air flow is analogous to current flow. Volume is analogous to charge. Resistance to air flow is analogous to electrical resistance. Compliance, the relationship between pressure and volume, is analogous to capacitance, the relationship between charge and voltage. The model shows that various quantities are governed by exponential decay with time constant tau=RCom. The main assumption is that the human lungs can be approximated as a single compartment modeled by an RC circuit where the quantities of interest, air flow, volume of air, pressure, compliance, and resistance are analogous to current, charge, voltage, capacitance, and resistance respectively. GENERAL RESULTS: The ventilator, using a driving pressure of 10 mmHg gives an approximately normal tidal volume of 500 ml. Normally of course, the force is provided by expansion of the chest, creating a negative pressure in the intrapleural space, just the oppposite of this positive pressure ventilator. Figure: Pressure, volume, flow vs. time. The forcing ventilator pressure is Pmouth (black). The numerical solutions:Fair Flow at mouth (green dashes), Volume circles; Plung red dashes. TestF is black line (exp(-t/Rescom)), a verification test fitting Fair(t). where P[mouth] is the pressure at the mouth; P [atmos], Ref pressure, external to body; P[vent], the driving pressure from a ventilator; ScalPvent, scaler of amplitude of Pvent; P[lung] is the pressure in the lung; Fair is the air flow at the mouth; R is the resistance of the airway; Com is the compliance of the lung; V[FRC] is the functional residual capacity of the lung; and V[lung] is the volume of air in the lung. The equations for this model may be viewed by running the JSim model applet and clicking on the Source tab at the bottom left of JSim's Run Time graphical user interface. The equations are written in JSim's Mathematical Modeling Language (MML). See the Introduction to MML and the MML Reference Manual. Additional documentation for MML can be found by using the search option at the Physiome home Download JSim model project file • Download JSim model MML code (text): • Download translated SBML version of model (if available): Model Feedback We welcome comments and feedback for this model. Please use the button below to send comments: M.G. Levitsky, Pulmonary Physiology, Sixth Edition, McGraw Hill, 2003. Key terms lung compliance RC circuit lung mechanics airflow in trachea tidal volume positive pressure ventilation Please cite https://www.imagwiki.nibib.nih.gov/physiome in any publication for which this software is used and send one reprint to the address given below: The National Simulation Resource, Director J. B. Bassingthwaighte, Department of Bioengineering, University of Washington, Seattle WA 98195-5061. Model development and archiving support at https://www.imagwiki.nibib.nih.gov/physiome provided by the following grants: NIH U01HL122199 Analyzing the Cardiac Power Grid, 09/15/2015 - 05/31/2020, NIH /NIBIB BE08407 Software Integration, JSim and SBW 6/1/09-5/31/13; NIH/NHLBI T15 HL88516-01 Modeling for Heart, Lung and Blood: From Cell to Organ, 4/1/07-3/31/11; NSF BES-0506477 Adaptive Multi-Scale Model Simulation, 8/15/05-7/31/08; NIH/NHLBI R01 HL073598 Core 3: 3D Imaging and Computer Modeling of the Respiratory Tract, 9/1/04-8/31/09; as well as prior support from NIH/NCRR P41 RR01243 Simulation Resource in Circulatory Mass Transport and Exchange, 12/1/1980-11/30/01 and NIH/NIBIB R01 EB001973 JSim: A Simulation Analysis Platform, 3/1/02-2/28/07.	{"url":"https://www.imagwiki.nibib.nih.gov/physiome/jsim/models/webmodel/NSR/onealvlungassist","timestamp":"2024-11-06T21:34:49Z","content_type":"text/html","content_length":"62553","record_id":"<urn:uuid:0389cc95-4283-437d-b483-a6fea142c11b>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00303.warc.gz"}
Fitting Continuous Piecewise Linear Functions Continuous piecewise linear functions (CPWL) are an interesting class of functions. The attractive features include their efficiency and continuity. The CPWL function regression can be better behaving than the polynomial regression, and is often used for approximation of complex functions. The CPWL functions are quite efficient in terms of the computational and memory requirements, which allows demanding applications, such as resource-constrained microcontrollers or graphics processing. However, there seems to be a lack of methods of fitting the CPWL functions to other functions or experimental sets of data. In particular, when the x coordinates of the CPWL function segments are fixed and only the y coordinates are unknown. The following paper offers a solution to this problem. The paper describes an application of the least-squares method to fitting a continuous piecewise linear function. It shows that the solution is unique and the best fit can be found without resorting to iterative optimization techniques.	{"url":"https://www.golovchenko.org/home/pwl_fit","timestamp":"2024-11-08T16:09:21Z","content_type":"text/html","content_length":"73170","record_id":"<urn:uuid:17461612-c1ef-431e-82b4-3163e0f0b200>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00782.warc.gz"}
Convert between Tables, Graphs, Mappings, and Lists of Points \| sofatutor.com Convert between Tables, Graphs, Mappings, and Lists of Points Basics on the topic Convert between Tables, Graphs, Mappings, and Lists of Points A function is a particular kind of relation between sets. A function takes every element x in a starting set, called the domain, and tells us how to assign it to exactly one element y in an ending set, called the range. We can represent functions in different ways; let’s take a look at a few of them through the following example: At a stationary supply store, one pen costs $2.75. So the price the customer pays depends on how many pens they decide to buy. We can represent the amount the customer needs to pay with the function p(x)=2.75x. We have that the domain is the number of pens x the customer buys and the range is the amount the customer has to pay, 2.75x. Let’s look at some ways in which we can represent p(x): One way of representing p(x) is via a mapping diagram. A mapping shows how the elements of the domain and range are paired. It’s a flow chart consisting of two parallel columns, showing the input x (first column) and output y (second column) values. Lines or arrows are drawn from domain to range, to represent the relation between any two elements. So for our example we would map 1 to 2.75, 2 to 5.50, 3 to 8.25, 4 to 11.00, 5 to 13.75, and so on. We can also use tables, ordered pairs, and graphs to represent p(x). The easiest way to make a graph is to begin by making a table containing inputs and outputs. We would then call (x,p(x)) an ordered pair where x is the input and p(x) is the output. For our example, we can write the data from such a table as ordered pairs: (1, 2.75), (2, 5.50), (3, 8.25), (4,11.00), and (5,13.75). These ordered pairs can be plotted and lines can be drawn between them to get a graph. Interpret functions that arise in applications in terms of the context. Transcript Convert between Tables, Graphs, Mappings, and Lists of Points Emilia and Oscar are playing a board game called The Joyous Journey when Oscar gets a call. Seizing the opportunity, Emilia rolls the dice and deftly moves her game piece. When Oscar returns from his call, he notices something fishy SOMEHOW, Emilia ended up on a space where she gets a bonus card! Oscar asks Emilia where her turn started and what she rolled to land on this space. Conveniently, Emilia doesn't seem to remember those details anymore, but she does remember she rolled a 3. Oscar decides to examine the game board a bit more carefully. Mapping Diagram We can use a mapping diagram to help us visualize the situation. We know Emilia ended up on space 7 and that she moved forward. If Emilia started at the first space and rolled a 3, she would have ended up at space 4, but space 4 sends a player back to the start so she would still be at the first space. We can show this relation by drawing an arrow that starts at our input, 1, and ends at our output, 1. So Emilia didn't start at the first space, what about the second? Since she rolled a 3, she would have gone to space 5. But space 5 moves the player 3 additional steps forward. This brings us to space 8. Space 8 moves a player back 2 steps, which finally ends our journey at space 6 where you have to take a though-card. So an input of 2 gives us an output of 6. We can show this by drawing an arrow beginning at the input, 2, and ending at the output, 6. Starting at space 3 and rolling a 3 also leads to space 6. And since space 4 sends a player back to space 1, it’s not possible that this space was Emilia's starting point. Space 5 isn't a possibility because it sends a player 3 spaces forward to space 8, and space 8 sends a player 2 spaces back. Emilia's chances are running out. Oscar looks at the last possible starting point, space 6; rolling a 3 would lead to space 9. All input-values, also called x-values, are called the domain. And all output-values, also called y-values, are called the range. In this example, the domain is {1,2,3,6} and the range is {1,6,9}. Since there is exactly one output for every possible input this relation is a function. Putting the Data in Table and Graph Form This is not the only way to show these things, however. We can group these inputs and outputs into ordered pairs, which also allows us to put the data in table and graph form. For our first input, we have 1 and from our mapping diagram, we see that the output is also 1. The ordered pair for this is (1, 1). Our second input, 2, has an output of 6. This ordered pair is (2, 6). Think you're getting the hang of it? Our third input, 3, also has an output of 6, making its ordered pair (3, 6). And our final ordered pair is (6, 9). To show this relationship in a table, we write all the x-values to the left and the corresponding y-values to the right. Our graph, however, is arranged a bit differently. Our inputs are our x-values and our outputs are our y-values. This gives us our points on the graph. Oscar's upset because it's obvious Emilia cheated to get to the lucky space. But did Oscar play fair? Convert between Tables, Graphs, Mappings, and Lists of Points exercise Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Convert between Tables, Graphs, Mappings, and Lists of Points. • Determine to which positions Emilia could have started. Use the board game. An input can't have two different outputs, but it is possible for two different inputs to have the same output. If Emilia started at the first space and rolled a $3$, she would have ended up at space $4$, but space $4$ sends a player back to the start so she would still be at the first space. We can show this relation by drawing an arrow that starts at our input, $1$, and ends at our output, $1$. So Emilia didn't start at the first space...what about the second? Since she rolled a $3$, she would have gone to space $5$. But space $5$ moves the player $3$ additional steps forward. This brings us to space $8$. Space $8$ moves a player back $2$ steps, which finally ends our journey at space $6$...where you have to take a though-card. So an input of $2$ gives us an output of $6$. We can show this by drawing an arrow beginning at the input, $2$, and ending at the output, $6$. Starting at space $3$ and rolling a $3$ also leads to space $6$. And since space $4$ sends a player back to space $1$, it’s not possible that this space was Emilia's starting point. Space $5$ isn't a possibility because it sends a player $3$ spaces forward to space $8$, and space $8$ sends a player $2$ spaces back. Oscar looks at the last possible starting point, space $6$; rolling a $3$ would lead to space $9$. • Explain how to represent the input and output of a function with a graph. Keep the order of pairs in a coordinate system in mind. Here is a table representing the possible moves Emilia could have made in her board game with Oscar. From the table, we can see that starting at position $3$ leads to an ending position of $6$. An example of an ordered pair is $(3,6)$. For a graph we draw a coordinate system with a $x$- and a $y$-axis. You can see such a coordinate system in the picture. The input is represented by the $x$-values and the output is represented by the $y$-values. So we can draw each ordered pair $(x,y)$ into the coordinate system. • Plot the given points in a graph. Keep the order of the coordinates in mind. For example, with $(3,7)$, you have the line $x=3$ parallel to the $y$-axis and the line $y=7$ parallel to the $x$-axis. The intersection of those lines is the point $(3,7)$. The horizontal axis is the $x$-axis, while the vertical axis is the $y$-axis. The number of pairs given by the table is five. From the table, $\begin{array}{c\|c} x&y\\ \hline 1&1\\ 3&2\\ 5&3\\ 7&3\\ 8&4 \end{array}$ we get the ordered pairs $(1,1)$, $(3,2)$, $(5,3)$, $(7,3)$, and $(8,4)$. To graph these points, keep in mind that the first coordinate of an ordered pair is the $x$-coordinate. So for example, go to $x=3$ and then two steps up to get the point $(3,2)$. For instance, to differentiate between the points, or ordered pairs, $(5,3)$ and $(3,5)$: □ $(5,3)$ lies on a line through $x=5$ parallel to the $y$-axis. □ $(3,5)$ lies on a line through $x=3$ parallel to the $y$-axis. You can see the point from the table plotted in the graph above and to the right. • Decide which graph corresponds to which mapping diagram. You can write each pair of input $x$ and output $y$ as an ordered pair. The relation where each $x$ is assigned to the same $y$ is called a constant relation. The corresponding points lie on a line parallel to one coordinate axis. Keep the order of the coordinates in mind: The first coordinate (the input) corresponds to the $x$-axis, while the second corresponds to the $y$-axis. You can write down ordered pairs for each relation given by a mapping diagram. Let's start with the diagram on the right. You see the corresponding graph above and to the left, and can check that they match by checking the ordered pairs given by the diagram. Next, let's look at the relation where each input is assigned to the same output, namely $3$. The corresponding graph is the one where all the points lie on a line parallel to the $x$-axis. Now the first mapping diagram from the left gives us the pairs $(1,2)$, $(2,2)$, $(3,4)$, $(4,4)$, and $(5,6)$. The diagram where each input is assigned to itself gives us the graph where all points lie on a line from bottom left to top right. • Match the image with its corresponding definition. In a mapping diagram, the input is the domain and the output the range. For a graph, we use a coordinate system. An ordered pair in general is given by $(x,y)$. To represent any relation between an input and an output we can use different imagines. From left to the right we have: The mapping diagram On the left side we have the domain, the set of all inputs, and on the right the range, where we find all possible outputs. We use arrows to show the corresponding relationships. Ordered pairs Using the relations represented by the mapping diagram, we can also write ordered pairs: the first position is the input $x$ and the second the output $y$. $\begin{array}{c\|c} x&y\\ \text{input}&\text{output}\\ \hline 1&1\\ 2&6\\ 3&6\\ 6&9 \end{array}$ Here you see the corresponding table: row by row you see the ordered pairs. Again you can recognize the input on the left and the output on the right side. To represent the relation in a more visual way we use graphs: for this we draw each ordered pair $(x,y)$ into a coordinate system, where the $x$-axis stands for the input while the $y$-axis represents the output. • Determine the coordinates of the ordered pairs. Keep the order of the coordinates in mind. The first coordinate is the $x$-coordinate, which corresponds to the $x$-axis, or the input. For example, the order pairs of this graph, from left to right: □ $(1,5)$ □ $(3,4)$ □ $(5,3)$ □ $(7,2)$ Let's start with the topmost graph: □ For $y=3$, we get $x=2$. □ For $x=4$, we get $y=5$. □ For $x=5$, we get $y=5$. □ The last point is $(7,2)$ For the middle graph, we get the point : For the lower graph, we get: □ $(2,1)$ □ $(3,3)$ □ $(4,5)$ □ $(5,7)$	{"url":"https://us.sofatutor.com/math/videos/convert-between-tables-graphs-mappings-and-lists-of-points","timestamp":"2024-11-12T21:56:36Z","content_type":"text/html","content_length":"156557","record_id":"<urn:uuid:c484bf27-d649-4b6a-848b-755a25c480bf>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00359.warc.gz"}
< 1 2 3 4 5 6 7 8 > We find the number 2 so obvious that we sometimes omit it. You write the square root of 3 as and actually it is You never write it this way, because everyone knows of course that it is the 2^nd root if there is nothing indicated. If you put it in more detail, it becomes It is a form of exponentiation, because you can also write it as and, of course, as Sometimes this is more comfortable in a calculation. Deutsch Español Français Nederlands 中文 Русский	{"url":"http://www.maeckes.nl/Getal%20twee%20(weg)%20GB.html","timestamp":"2024-11-01T22:05:20Z","content_type":"application/xhtml+xml","content_length":"5046","record_id":"<urn:uuid:21b556ab-134f-4918-a113-fba187db319b>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00503.warc.gz"}
Correlation Tool \| Data Encyclopedia Correlation Coefficient A correlation coefficient is a number between -1 and 1 measuring the strength of a relationship between two variables. A positive correlation means that returns move together. If you plotted returns of two cryptoassets on a scatterplot, dots sloping upwards and right in a diagonal line would imply a positive correlation. Note that no causality is inferred. The actual derivation involves taking the covariance of two variables. Covariance tells you whether two variables move together or not, but it is unbounded and unstandardized. Finding a correlation involves standardizing those arbitrarily large figures by dividing by the sample standard deviations of both variables. This reduces covariance to a range between -1 and 1, and it is now informative as to the magnitude of moves between two assets. Put simply correlation tells you: whether two variables are related, and the magnitude of their inter-relatedness Tool Options The tool offers the user several options for determining the correlation coefficient: Log vs. Arithmetic Returns Timeframe: 30/60/90/180/360 days Handing of Data Gaps: Linearly Interpolate or Exclude Pearson and Spearman Correlation Our tool presents two options for correlation: Pearson and Spearman Correlation. The orthodox and conventional way of doing things in finance is to take the Pearson correlation of logarithmic daily asset returns, preferably over a long period. If you aren’t interested in getting into any additional complexity, you can stop there and use those settings on the charts. (A common mistake in correlating assets is to use raw prices rather than returns. This is a mistake, as prices are often trended and non-stationary, meaning that you often get spurious positive correlations. Our tool uses the asset returns.) Pearson correlation basically assumes that the relationship between the two variables is linear, and it measures it on this basis. If you have nonlinear yet meaningful relationships, Pearson will report a weak correlation, when in fact there may be something interesting going on behind the scenes. One solution to this problem is using log daily returns, which helps in processing data, and has some other useful properties if the data is normally distributed. To give our users an alternative, we introduced Spearman correlation. Spearman correlation takes a ranking of all the data points in the sample, and then it runs a Pearson correlation on the rankings data. It then compares the two variables based on how much their rankings move together. This enables the ability to capture co-movements in datasets that are nonlinear. Spearman would find the correlation between a perfect exponential relationship as 1, whereas Pearson would declare it positive but not perfectly correlated. For more on this, we suggest reading this excellent breakdown in the differences between the two. To put it simply: if you think your variables are linearly related, and you want to measure the strength of those co-movements, use Pearson correlation. If you want to capture variables which move together, but not at a constant rate (as with an exponential relationship), consider Spearman correlation. Spearman makes no assumptions as to linearity but simply assesses the relationship in terms of whether one variable increases when the other increases, and vice versa. Arithmetic vs Logarithmic Returns While Arithmetic Returns are easily understood, as mentioned above, there are some good reasons for using a Logarithmic Return in certain occasions. Our took allows you to choose Arithmetic or Logarithmic returns. The tool allows you to select the number of daily observations to consider when comparing returns to determine the data sets' correlation. If, for example, you picked 180 days, the 180 daily returns prior to the point/date on the chart were considered when determining the correlation coefficient returned for the two data series at that point/date on the chart. In other words, the number you see for today refers to a sample of the previous 180 days. Keep in mind that if you "exclude gaps" in the data set (see below), then the number of days selected will reflect only the number of days for which there are observations. Handling of Data Gaps On occasion, you may want to compare a continuous data series with one with gaps (such as when you compare cryptoasset returns with a traditional index like S&P500 that has no values on the weekends or holidays). The tool gives you two options for handling this: Linearly interpolate the data to derive a value for the data gap prior to calculating the returns Exclude the observations on the dates where one data series has a gap, but the other doesn't prior to calculating the returns (note: this results in fewer observations for the non-gapped series) The tool will default to "excluding" data where gaps are identified. You can adjust this setting from the settings menu on the right toolbar. Let’s say you select a Pearson correlation of the logarithmic returns for the previous 180 days. This number means “in the 180 days leading up to the point on the chart you’re looking at, the two assets that you have selected had co-movement in their daily returns of a magnitude corresponding to x.” If x is 1, their daily returns were perfectly positively correlated in the 180-day period leading up to the date where you observe the coefficient of 1. If x is 0.3, their returns moved together more often than not, but they weren’t particularly closely linked. Attempting to predict the future by looking at historical returns is always somewhat fraught. There’s no guarantee that a historical correlation will persist into the future. Because of this, we recommend looking for useful correlations over longer periods. Things that should grab your attention: Very positive correlations between two assets, especially over longer periods of time. Consistently high correlations deserve investigation. Persistent negative correlations of any sort are tremendously interesting, albeit rare. The strength of the signal increases with the quantity of historical evidence. Be aware that different market phases exist. Longstanding correlations can break down during certain market conditions. Try to discern when your correlation analysis holds little explanatory power. Avoid mistaking noise for signal. And use this tool with caution. It’s also worth reading more about the statistical underpinnings of correlation measures. \	{"url":"https://gitbook-docs.coinmetrics.io/data-visualization/charting-tool/correlation-tool","timestamp":"2024-11-07T22:14:35Z","content_type":"text/html","content_length":"654555","record_id":"<urn:uuid:60c55b52-ce1e-4285-89c2-eb73754328b5>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00075.warc.gz"}
Pick's Theorem This is a development version of this entry. It might change over time and is not stable. Please refer to release versions for citations. We formalize Pick's theorem for finding the area of a simple polygon whose vertices are integral lattice points. We are inspired by John Harrison's formalization of Pick's theorem in HOL Light, but tailor our proof approach to avoid a primary challenge point in his formalization, which is proving that any polygon with more than three vertices can be split (in its interior) by a line between some two vertices. Our formalization involves augmenting the existing geometry libraries in various foundational ways (e.g., by adding the definition of a polygon and formalizing some key properties Related publications • Grunbaum, B., & Shephard, G. C. (1993). Pick’s Theorem. The American Mathematical Monthly, 100(2), 150. https://doi.org/10.2307/2323771 • HARRISON, J. (2011). A formal proof of Pick’s Theorem. Mathematical Structures in Computer Science, 21(4), 715–729. https://doi.org/10.1017/s0960129511000089 Session Picks_Theorem	{"url":"https://devel.isa-afp.org/entries/Picks_Theorem.html","timestamp":"2024-11-01T22:47:29Z","content_type":"text/html","content_length":"10273","record_id":"<urn:uuid:aceece04-3248-4d43-9e03-a121a7133e5d>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00243.warc.gz"}
The Benefit of Online Math Tutoring - Online Tutoring - Online Tutors \| Growing StarsThe Benefit of Online Math Tutoring Good math skills are the key to future career success. The job opportunities of the future with the most openings and best potential salaries all require excellent math skills. Unfortunately, not everybody is naturally gifted in math and not every school system has the resources to give individual attention to those who need extra help. Online math tutoring can give struggling students the help they require in order to master this challenging subject. How Tutoring Can Help Students sometimes need extra assistance to be able to understand complex material. Math, in particular, requires the understanding of formulas that can be applied to equations. Those formulas are the building blocks for more advanced problems and equations. Once you have mastered one, you can move on to the next. While these formulas may not be complex, that doesn’t mean that comprehending them is always easy. Each teacher has an individual way of explaining material with only a very limited amount of time to cover each topic. If a student’s ability to understand doesn’t match the teacher’s style of explaining, a student may be left with an inability to grasp the material. Before you know it, the student is confused and lost, which only gets worse as the class moves on. Tutoring provides one-on-one attention to ensure that students learn crucial material in a manner they can best understand. Why Online Tutoring? Even when you know that tutoring is a great option for you or your student, finding time to get that needed help can be very challenging. Most families already lead very busy lives, and the thought of adding another time commitment into packed schedules can seem intimidating. Rather than trying to rearrange a family’s entire schedule to accommodate an extra commitment, online tutoring is flexible and can be done at a convenient time. Online tutoring is also individualized so that the material is presented in a way the student will easily understand. A student can even ask for explanations as many times as necessary and can continue practicing a skill until they’ve mastered it. This ensures that they will build a strong foundation and be well prepared for the next lesson. Learning math skills is no longer optional if you want to be successful. The job market is only expected to become more competitive in the coming years. Invest in the tutoring you or your child will need to make sure they’re ready for the future. Thanks to math tutoring online, getting that extra help is more accessible than ever. Look to the experts at Growing Stars when a little support is needed to succeed.	{"url":"https://www.growingstars.com/the-benefit-of-online-math-tutoring/","timestamp":"2024-11-13T12:53:10Z","content_type":"text/html","content_length":"339602","record_id":"<urn:uuid:e8170748-fcf5-4e8d-b42c-86942f61b5cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00447.warc.gz"}
IC Frequency Dividers and Counters, January 1969 Electronics World January 1969 Electronics World Table of Contents Wax nostalgic about and learn from the history of early electronics. See articles from Electronics World, published May 1959 - December 1971. All copyrights hereby acknowledged. The subject of IC-based digital circuits was relatively new in 1969 when this story was printed in Electronics World magazine and was probably considered expert knowledge. Note that it was written by a research engineer at the Lockheed Missiles and Space Company. Today, introductory digital logic courses begin with material presented here and advance rapidly into programmable logic, ASICs microprocessors, and beyond. Even if you have already taken courses on digital logic with counters and dividers, the knowledge might be in a dusty corner of the gray matter and could stand a refresh. One of my main motivations for studying for and obtaining my Amateur Radio licenses (recently promoted to Extra level), was to have an excuse to review and re-learn concepts first studied long ago. I have not yet acquired the issue of Electronics World in which Part 1 was published. IC Frequency Dividers & Counters By Donald L. Steinbach / Research Engineer Lockheed Missiles and Space Co. Complete frequency divider and counter systems. Synchronous dividers for division ratios from two through ten are given, along with a simple decade counter using inexpensive, readily available integrated circuits. Part 1 of this article discussed in some detail the characteristics of a family of IC flip-flops, gates, and buffers. In this part we extend this information to complete frequency divider and counter Logic Elements in Dividers and Counters The frequency dividers and counters described in this article are made up of one or more JK FF's (flip-flops) interconnected in such a way that as each CP (clock pulse) arrives at the divider input one or more of the FF's in the divider change state. This is accomplished by "forcing" the FF's to particular states by use of the S and C inputs and/or appropriate selection of the source for input T. In some cases gates are used to derive signals for S and C that do not already exist somewhere in the divider. Buffers are used as required to increase drive levels and/or provide isolation from external circuitry. The frequency divider has a division ration if its output waveform passes through one complete cycle as its input waveform passes through n complete cycles. The number of FF's required in a particular divider is determined by the desired division ratio. The highest possible division ratio for a given number of FF's is 2^x where x is the number of FF's in the divider. Thus, one FF is needed to divide by two; two FF's are needed to divide by three or four; three FF's are needed to divide by five, six, seven, or eight, etc. The signal input connected to T (pin 2 of the particular C discussed last month in Part 1) of each FF in the divider will be either the incoming CP or the output of a preceding FF. If T of each FF is connected to the incoming CP, the divider is a synchronous divider. If the incoming CP is connected to T of the first FF only, and T of the second FF is connected to the output (Q or Q) of the first FF, etc., then the divider is called an asynchronous divider. The FF propagation delays in the asynchronous divider are cumulative and the time between CP's must be sufficient to allow each FF in the divider string to change state. In general, no more than about six 9923 JK FF's should be used in anyone asynchronous divider intended to operate at an input frequency of 2 MHz. All FF's in the synchronous divider are triggered simultaneously and the time delay between the CP and the resulting change in divider output is equal to the propagation delay time of one FF rather than the combined propagation delays of a string of FF's. Synchronous dividers should always be used when the divider input and output waveforms must be in synchronism. The control inputs 5 and C (pins 1 and 3 respectively) of a particular FF are connected to ground (0 level), + V[CC] (1 level), or the outputs of other FF's either directly or via gates. The level applied to S, the level applied to C, and the "present" FF state determine the state of the FF after the next CP or 1-to-0 transition of a preceding FF. Fig. 1 is an expanded version of the truth table in Fig. 7 of Part 1. It lists all possible combinations of S, C, and JK FF states that might exist before the arrival of a CP and gives the FF state that will then result after the arrival of the CP. Keep in mind that a CP is simply a 1-to-0 transition at T (pin 2) and that the state of the FF is the level at Q (pin 7). If Preset (pin 6) of all the 9923 FF's in a divider are connected together, all of the FF's will be forced to the 0 state (output Q at the 0 level) when this "preset line" is momentarily connected to V[CC] (+3.6 V d.c.). This technique provides a convenient starting point for the dividing action both in the operating circuit and on paper. It is customary to define the instantaneous state of the divider as the states of the FF's in the divider written in some logical order. Thus, if the divider is made up of four FF's labeled FF1, FF2, FF3, and FF4, and FF states are 1, 0, 1, and 1, respectively, then the state of the divider is 1011. Since each FF has two states (1 and 0), the number of possible divider states is 2x where x is the number of FF's in the divider. All possible states of a divider having 1, 2, 3, or 4 FF's are tabulated in Fig. 2. Circuit waveforms for the more complex dividers are determined from a state table. The state table is a CP-by-CP tabulation of the levels at S, C, and Q of every FF in the divider. It is most convenient to assume that the divider starts from the Preset state (i.e., all Q's at 0). The levels of each FF S and C input are then determined from the divider schematic. Knowing S, C, and Q, the FF states after the first CP may then be determined from Fig. 1. The "new" S and C levels are determined and the FF states after the second CP are determined. This process is continued until the state table begins to repeat itself, indicating that one complete division cycle has occurred. The completed state table should be compared with Fig. 2 to determine which (if any) of the possible divider states in Fig. 2 do not appear in the state table. Additional state tables are then constructed using each of these "unused" divider states as the initial starting point in order to determine if the divider will recover and divide by the desired ratio. If it does not, two courses of action are available: redesign the divider, or make provisions for Presetting the divider. There may be more than one circuit that will yield a particular division ratio. The circuit finally chosen will usually be the one that uses the fewest components or provides the most desirable output waveform for the particular application. It frequently happens that more than one division ratio can be obtained from a single divider. For example, a divide-by-ten circuit may be able to simultaneously deliver a divide-by-two or divide-by-five output from some point in the circuit. The circuits to follow are drawn using the logic symbols of the IC devices. Refer to Fig. 8 in Part 1 for the actual Fairchild IC pin numbers. Although not shown, pin 4 of each IC is grounded and pin 8 of each IC is connected to V[CC] (+3.6 V d.c.). If the Preset feature of the JK FF's is employed, then connect pin 6 of each of the FF's together and connect this to V[CC] through a normally open momentary switch. Either Q or Q of any FF in the divider may be chosen as the divider output (s). For a given FF, the more lightly loaded of the two output terminals is usually used, although this is not mandatory as long as the output drive factor of the FF is not exceeded. In the figures that follow, the input signal is drawn as a square wave only for purposes of illustration. The input waveform may be of any shape as long as the fall-time is small enough to be accepted by the FF's as a clock pulse. The area marked "first complete division cycle" is the waveform that will repeat with every n clock pulse. Dividing by Two The simplest possible frequency divider is an n =2 divider made from a single JK FF. If the S and C inputs are both (permanently) at 0, the FF changes state with each CP. If the FF is initially in the 0 state, it will change to the 1 state when the first CP arrives. When the second CP arrives, the FF returns to the 0 state. On the third CP, the FF changes back to the 1 state, completing the output cycle. The FF state alternates with each consecutive CP and the output frequency is one-half the input frequency - or the output waveform period is twice the input waveform period. Dividing by Three Then n = 3 divider in Fig. 3 is a synchronous divider - the CP is applied simultaneously to the T input of both FF's. The input load factor is 10 since each FF has an input load factor of 5. The state table for the divider of Fig. 3 is constructed as follows: a. After Preset, Q1 (output Q of FF1) and Q2 (output Q of FF2) are both 0. These 0's are entered in the Q1 and Q2 columns on the Preset line of the table. b. Now that Q1 and Q2 are known, all of the S and C levels may be determined directly from the divider schematic: S1 = Q2 =1; C1 = 0; S2 = 0; and C2 = Q1 = 1. These levels are entered in their appropriate columns on the Preset line of the table. c. The levels entered on the Preset line of the table are the levels at S, C, and Q that now exist before the arrival of the first CP. The levels at Q1 and Q2 after the arrival of the first CP are obtained directly from Fig. 1: Q1 = 1 and Q2 = 0. These levels for Q1 and Q2 are entered on the CP1 line of the table. d. Now that Q1 and Q2 after CP1 are known, the S and C levels after CP1 are determined: S1 = Q2 = 1; C1 = 0; S2 = 0; and C2 = Q1 = 0. These levels are entered in their appropriate columns on the CP1 line of the table. e. Continuing in this manner, after CP2: Q1 = 1; and Q2 = 1. Also, S1 = 0; C1 = 0; S2 = 0; and C2 = 0. f. After CP3: Q1 = 0 and Q2 = 0. Also S1 = 1; C1 = 0; S2 = 0; and C2 = 1. This divider state is identical to the Preset state; therefore, the cycle will be repetitive. If a CP4 line were added to the table, it would look exactly the same as the CP1 line; a CP5 line would be the same as the CP2 line; a CP6 line would be the same as the CP3 line, a CP7 line would be the same as the CP1 line, etc. The waveforms in Fig. 3 are constructed directly from the information in the state table. The three states of the divider are 00, 10, and 11. Comparing these states with Fig. 2 reveals that the 01 state is missing from the divider operating sequence. A state table using an initial state (U1 ) of 01 demonstrates that the divider will perform exactly the same as when the initial state is 00; the divider state one CP after the initial state is the same in both cases and the waveforms are identical. Bear in mind that if the Preset function is used as explained earlier, this evaluation of the divider recovery from an "unused" state is unnecessary. It is explained in this section only to demonstrate the technique. Dividing by Four The simplest means of dividing by four is to divide by two twice. The output of the first divider is at one-half the input frequency and is connected to the input of the second divider. The second divider divides the output frequency of the first divider by two and the resulting output frequency is one-fourth the input frequency to the first divider. Simultaneous n = 2 and n = 4 outputs may be obtained from this divider. The n = 2 output is synchronous, but the n = 4 output is asynchronous since it is delayed from the input CP by the sum of the propagation delays of both flip-flops. The synchronous divider in Fig. 4 also provides simultaneous n = 2 and n = 4 outputs. The synchronous n = 4 output is obtained at the expense of slightly increased circuit complexity and a larger input load factor. Dividing by Five, Six, Seven, Eight & Nine Fig. 5 is a n = 5 divider. All outputs are synchronous and have the same waveshape, but are time-displaced from one another. The circuit will recover from its unused states so the use of the Preset function is optional. Fig. 6 is a simple synchronous n = 6 divider. In addition to the n = 6 output from FF3, n = 3 outputs are available from either FF1 or FF2. The divider has two unused states and will recover from either. A division ratio of six may also be obtained by dividing by two and then by three (or vice versa). A synchronous n = 7 divider appears in Fig. 7. The divider has one unused state and will recover on the next CP. An asynchronous n = 7 divider can be constructed and requires one less gate than the synchronous divider. An asynchronous n = 8 divider is most easily assembled by cascading three n = 2 dividers. A synchronous divider is shown in Fig. 8. Regardless of the method used, n = 2 and n = 4 outputs will also be available and there are no unused states. Fig. 9 is a synchronous n = 9 divider. It has seven unused states and will recover from each. Cascading two n = 3 dividers will provide an asynchronous n = 9 output and a synchronous n = 3 output. Dividing by Ten Many n = 10 dividers have been devised due to their popularity in dividing and counting applications. An asynchronous n = 10 divider could be built from an n = 2 divider and an n = 5 divider. A synchronous n = 2 or n = 5 output (depending on which divider is connected to the incoming CP) would also be available. The synchronous n = 10 divider in Fig. 10 operates in the so-called "shift mode." Although this divider requires one additional FF and provides only n = 10 outputs, it is particularly useful in counting applications as we shall see later. The divider has 22 unused states - an ideal opportunity to utilize the Preset function. Practical Systems Typical frequency-divider systems consist of one or more divider stages cascaded to provide the desired division ratios and outputs. A common application is that of cascading several n = 10 dividers to divide a 1-MHz or 100-kHz signal down to 10 kHz, 1 kHz, etc. Buffers are used between divider stages when an increase in drive level is required. They should also be provided on the output lines if external circuit loading is appreciable. Whatever the ultimate application, the first problem encountered is usually that of converting the input waveform to a fast-fall-time pulse to act as a clock pulse for the dividers. The circuit of Fig. 11 will accept any input waveform and has been used by the author to drive some of the dividers in this article at frequencies in excess of 10 MHz. The circuit functions like a low-hysteresis Schmitt trigger that switches at a threshold voltage of about 0.9 V d.c. D1 may be any signal diode - its only function is to protect IC1 from negative-going inputs. Naturally, the voltage at pin 1 of IC1 should not exceed 3.6 volts in the positive direction. The choice of C1 is based on the input signal amplitude and frequency. For inputs of 100 kHz and over, a 0.1-μF capacitor is adequate. If R4 is set midway between the two trigger levels, the circuit will operate reliably on a.c. inputs well under 100 m V peak-to-peak. If R3 and R4 are omitted, then the minimum a.c. input must be on the order of 2 volts peak-to-peak. The power supply requires no particular attention other than assuring that its output voltage is low in ripple and transient-free. A full-wave rectifier followed by a filter capacitance of 25,000 μF or more will be adequate on both counts. An allowance of about 25 mA d.c. per IC will suffice for estimating total d.c. current requirements. Fig. 11. Circuit for generating clock pulses from any input. Output can drive the equivalent of 16 flip-flop "T" inputs. Fig. 12. Decoder for the n = 10 divider shown in Fig. 10. Frequency Counters Although the term "frequency divider" has been used throughout this article, the frequency dividers are actually repetitive counters. When the input CP's to the counter are regular and periodic, frequency division is obtained as a by-product of the repetitive counting operation. In frequency-dividing applications, one is interested in the time-varying waveforms present at the FF outputs during the counting operation; in counting applications, one is interested in the states of the individual FF's at a particular instant of time. In order for the counter to be of any real value, information on the FF states must be presented in some usable form. Typically a lamp-driver/lamp circuit is used. The lamp driver is designed so that the lamp illuminates when the lamp driver input is at the 1 level and extinguishes when it is at the 0 level. Decade Counters The decade counter is designed to display the number of clock pulses counted in numbers from 0 to 9. On the 10th CP the counter resets to 0 and delivers an output pulse. If this pulse is connected to the input of a second decade counter, the display of the second counter advances one count for every ten counts of the first decade counter. Connected in this manner, one decade counter counts from 0 to 9 clock pulses, two decade counters count from 0 to 99 clock pulses, three decade counters can count up to 999 clock pulses, etc. A decade counter is designed around an n = 10 divider. Since the divider state is different for each successive CP, the lamp responses can be related to the divider states through appropriate decoding techniques. Any n = 10 divider can be decoded, but the divider of Fig. 10 is ideally suited since it can be completely decoded using only ten two-input gates. A decoder for the Fig. 10 divider is shown in Fig. 12. The inputs Q1 through Q5 are connected to the corresponding FF outputs of Fig 10. When the divider/counter is Preset (all FF outputs at the 0 level) only the "0" decoded output is at the 1 level. After the first CP only the ''1'' decoded output is at the 1 level. After the second CP only the "2" decoded output is at the 1 level. After the ninth CP only the "9" decoded output and output X are at the 1 level. Coincident with the tenth CP output X switches to the 0 level providing a CP to drive a second decade counter. Posted August 16, 2024 (updated from original post on 5/23/2017)	{"url":"https://rfcafe.com/references/electronics-world/ic-frequency-dividers-counters-electronics-world-january-1969.htm","timestamp":"2024-11-06T12:18:07Z","content_type":"text/html","content_length":"48590","record_id":"<urn:uuid:da835c6f-9dd2-49c7-911d-5ba2512d0ad9>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00638.warc.gz"}
2014 AMC 10B Problems/Problem 21 Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$? $\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26$ Solution 1 $[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D) /2,W); label("14",(B+C)/2,E); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("E",DD,N); label("F",CC,N); draw(C--CC); draw(D--DD); [/asy]$ In the diagram, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$. Denote $\overline{AE} = x$ and $\overline{DE} = h$. In right triangle $AED$, we have from the Pythagorean theorem: $x^2+h^2=100$. Note that since $EF = DC$, we have $BF = 33-DC-x = 12-x$. Using the Pythagorean theorem in right triangle $BFC$, we have $(12-x)^2 + h^2 = 196$. We isolate the $h^2$ term in both equations, getting $h^2= 100-x^2$ and $h^2 = 196-(12-x)^2$. Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$. Now, we can determine that $h^2 = 100-4 \implies h = 4\sqrt{6}$. $[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D) /2,W); label("14",(B+C)/2,E); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("D",D,SW); label("E",DD,SE); label("F",CC,SW); draw(C--CC); draw(D--DD); label("21",(CC+DD)/ 2,N); label("2",(A+DD)/2,N); label("10",(CC+B)/2,N); label("4\sqrt{6}",(C+CC)/2,W); label("4\sqrt{6}",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\overline{AC}$ and $\overline{BD}$. Using the Pythagorean theorem again on $\bigtriangleup AFC$ and $\bigtriangleup BED$, we can find these lengths to be $\sqrt{96+529} = 25$ and $\sqrt{96+961} = \sqrt{1057}$. Since $\sqrt{96+529}<\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\boxed{\textbf{(B) }25}$. • Or, alternatively, one can notice that the two triangles have the same height but $\bigtriangleup AFC$ has a shorter base than $\bigtriangleup BED$. Solution 2 $[asy] size(7cm); pair A,B,C,D,E; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(12,1),E); label("14",(2,1),E); label("12",(A+E)/2,N); label("21",(E+B)/2,N); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("D",D,SW); label("E",E,N); [/asy]$ The area of $\Delta AED$ is by Heron's, $4\sqrt{9(4)(3)(2)}=24\sqrt{6}$. This makes the length of the altitude from $D$ onto $\overline{AE}$ equal to $4\sqrt{6}$. One may now proceed as in Solution $1$ to obtain an answer of $\boxed{\textbf{(B) }25}$. Solution 3 Using the same way as Solution 1, obtain that AE=2. Let us assume that point G is point A moving in a straight line down until it touches DC if DC was extended both ways by infinity. We know that GC= 23. Thus the answer would be the square root of (23 square and (4square root of 6) square). Thus we get 625. Square root of 625 would be 25. -Reality Writes See Also The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.	{"url":"https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_21","timestamp":"2024-11-14T09:09:20Z","content_type":"text/html","content_length":"51704","record_id":"<urn:uuid:b4569fad-b476-4e2a-9eab-3d6f71a8715f>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00273.warc.gz"}
Least common multiple and greatest common factor worksheets least common multiple and greatest common factor worksheets Related topics: news in math iv in advance algebra and statisticst solve 4y-8+y+38=8y+30-6y mathematical equations 7-8 graders; free online Answers For Algebra Problems mme maths balance equation ti-83 plus equation solver free learn basic algebra simplifying algebraic expressions,2 model academic standards for mathematics,2 decimal to mixed number online calculator Author Message Liepecheep Posted: Friday 24th of Jan 09:24 I am in desperate need of help in completing a project in least common multiple and greatest common factor worksheets. I need to finish it by next week and am having a difficult time trying to figure out a couple of tricky problems. I tried some of the online help sites but have not gotten what I want so far. I would be really grateful if anyone can help me. Back to top kfir Posted: Friday 24th of Jan 11:40 I do understand what your problem is, but if you could explain in greater detail the areas in which you are facing struggling, then I might be in a better position to guide you. Anyhow I have a suggestion for you, try Algebrator. It can solve a wide range of questions, and it can do so within minutes. And there’s more to come, it also gives a detailed step-by-step description of how it arrived at a particular solution . That way you don’t just solve your problem but also get to understand how to go about solving it. I found this software to be particularly useful for solving questions on least common multiple and greatest common factor worksheets. But that’s just my experience, I’m sure it’ll be good no matter what the topic is Registered: . From: egypt Back to top DoniilT Posted: Saturday 25th of Jan 07:19 I have used various programs to grapple with my difficulties with conversion of units, trinomials and converting decimals. Of them, my experience with Algebrator has been the finest. All I needed to do was to merely key in the problem. Punch the solve key. The reply appeared almost instantly with an easy to understand steps indicating how to get the answer. It was just too effortless . Since then I have relied on this Algebrator for my problems with Intermediate algebra, Algebra 1 and Intermediate algebra. I would highly urge you to try out Algebrator. Back to top Simon N Posted: Monday 27th of Jan 12:35 Maths can be real fun if there actually is some program like this. Please send me the link to the software. From: c:\uk Back to top fveingal Posted: Monday 27th of Jan 19:48 A great piece of math software is Algebrator. Even I faced similar difficulties while solving monomials, factoring and slope. Just by typing in the problem workbookand clicking on Solve – and step by step solution to my algebra homework would be ready. I have used it through several math classes - College Algebra, Pre Algebra and Intermediate algebra. I highly recommend the program. From: Earth Back to top MoonBuggy Posted: Tuesday 28th of Jan 07:55 I think this is the best link: https://softmath.com/about-algebra-help.html. I think they give an absolute money back guarantee, so you have nothing to lose. Best of Luck! Leeds, UK Back to top	{"url":"https://www.softmath.com/algebra-software/subtracting-exponents/least-common-multiple-and.html","timestamp":"2024-11-12T00:43:53Z","content_type":"text/html","content_length":"43459","record_id":"<urn:uuid:8a9fb85f-85a3-4f22-91bb-242ee5a636f3>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00351.warc.gz"}
Which equations of 5 string theories show that elementary particles formed by strings? 1474 views It is commonly said that "elementary particles are indeed formed by strings." (from E Witten interview) Which equations of string theory show that elementary particles are indeed formed by strings (2d string worldsheets)? How to see elementary particles in equations of each version of string theory? • In Type I? • In Type IIA? • In Type IIB? • In Type SO(32)? • In Type $$E_8 \times E_8$$? Can we take electrons and $$u,d$$ quarks as examples? This post imported from StackExchange Physics at 2020-12-03 13:03 (UTC), posted by SE-user annie marie heart Well, particles arising in the spectrum of a string can be seen in even the simpler model of the bosonic string, if that general mechanism is what you are after. This post imported from StackExchange Physics at 2020-12-03 13:04 (UTC), posted by SE-user JamalS "that elementary particles formed by strings?" Elementary particles are described theoretically by the standard model a well developed and flexible quantum field theory, with the group structure of SU(3)xSU(2)xU(1). The interest in string theories arises because this group structure can be embedded in the group behavior of the vibrations of a string. Automatically all the success of the standard model can be reproduced by string theories in their mathematical construction, assuming that the elementary particles are vibrations on a string,( of any type as long as they can carry the group structure). In a sense , if strings had been studied before the experimental observation of SU(3)xSU(2)xU(1) , they could have predicted it. It is not in the equations , but in the solutions, elementary particles are identified with the vibrations of the string of any type. This post imported from StackExchange Physics at 2020-12-03 13:04 (UTC), posted by SE-user anna v But where do we see SU(3)xSU(2)xU(1) from string theory and where do we see quarks and leptons from string theory? This post imported from StackExchange Physics at 2020-12-03 13:04 (UTC), posted by SE-user annie marie heart excuse me, I am hoping more details to show the spectra of elementary particles. This post imported from StackExchange Physics at 2020-12-03 13:04 (UTC), posted by SE-user annie marie heart the groups structure exists in string theories, the spectra, if you mean the masses depend on the type of theory one chooses and the energy . If at very high energies all masses are zero (in specific theories for cosmology for example). we just try to map the theoretical map on the data, and it fits because the same groups appear. It isongoing research. justthe possibility exists. see these talks physics.ox.ac.uk/pp/seminars/String%20Phenomenology.pdf nikhef.nl/~t58/Presentations/VU_26_2.pdf for the progress This post imported from StackExchange Physics at 2020-12-03 13:04 (UTC), posted by SE-user anna v quarks and lepton spectra experimentally are organized in the symmetries of SU(3)xSU(2)xU(1) , those are the spectra at low energies the symmetries are broken and we see a different mass. the group structure of the spectra is expected to persist at high energies, and that is why any theory of everything has to embed SU(3)xSU(2)xU(1) This post imported from StackExchange Physics at 2020-12-03 13:04 (UTC), posted by SE-user anna v just saw this , and it might interest you to see how the particle spectrum appears in a specific model arxiv.org/abs/2007.13248 This post imported from StackExchange Physics at 2020-12-03 13:04 (UTC), posted by SE-user anna v	{"url":"https://www.physicsoverflow.org/43862/equations-string-theories-elementary-particles-formed-strings","timestamp":"2024-11-02T18:42:44Z","content_type":"text/html","content_length":"142623","record_id":"<urn:uuid:9a89c121-c7cf-4b74-8473-ce08ee89abac>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00560.warc.gz"}
Multiplying Whole Numbers By Fractions Word Problems Worksheet 2024 - NumbersWorksheets.com Multiplying Whole Numbers By Fractions Word Problems Worksheet Multiplying Whole Numbers By Fractions Word Problems Worksheet – This multiplication worksheet concentrates on teaching students how to psychologically flourish whole phone numbers. Students can use personalized grids to suit particularly 1 query. The worksheets also protectfractions and decimals, and exponents. You will even find multiplication worksheets by using a spread home. These worksheets can be a need to-have for your personal math concepts school. They can be employed in school to learn how to psychologically flourish entire line and numbers them up. Multiplying Whole Numbers By Fractions Word Problems Worksheet. Multiplication of complete phone numbers If you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets can help you master this basic strategy. You can opt for one digit multipliers or two-digit and three-digit multipliers. Power of 10 will also be an incredible alternative. These worksheets will help you to exercise long practice and multiplication looking at the phone numbers. They are also a great way to support your kids recognize the value of understanding the several types of total phone numbers. Multiplication of fractions Possessing multiplication of fractions on a worksheet might help educators prepare and get ready classes effectively. Employing fractions worksheets allows instructors to quickly examine students’ knowledge of fractions. Pupils can be questioned in order to complete the worksheet in a a number of some time and then mark their solutions to see exactly where they need more instruction. College students can benefit from expression issues that associate maths to actual-existence scenarios. Some fractions worksheets involve examples of contrasting and comparing numbers. Multiplication of decimals Whenever you grow two decimal amounts, make sure to team them vertically. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. As an example, 01 x (11.2) x 2 can be similar to 01 by 2.33 by 11.2 except when the item has decimal places of below two. Then, this product is circular towards the local complete Multiplication of exponents A arithmetic worksheet for Multiplication of exponents will allow you to practice multiplying and dividing figures with exponents. This worksheet will also give issues that will require pupils to flourish two distinct exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. Apart from, you can even key in special instructions around the worksheet alone. When you’re done, you are able to just click “Make” and the worksheet will be acquired. Division of exponents The essential principle for department of exponents when multiplying amounts is usually to subtract the exponent from the denominator in the exponent inside the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. For instance, $23 divided up by 4 will equal 27. This method is not always accurate, however. This procedure can result in uncertainty when multiplying phone numbers which can be too large or too small. Linear functions If you’ve ever rented a car, you’ve probably noticed that the cost was $320 x 10 days. So, the total rent would be $470. A linear purpose of this sort offers the type f(x), exactly where ‘x’ is the amount of times the automobile was hired. In addition, it provides the shape f(x) = ax b, where by ‘b’ and ‘a’ are actual amounts. Gallery of Multiplying Whole Numbers By Fractions Word Problems Worksheet Multiplication Fraction Word Problems Worksheet Multiplying Fractions Word Problems 2 Worksheets 99Worksheets Multiplying Fractions Word Problems Worksheet Dividing Whole Numbers Leave a Comment	{"url":"https://numbersworksheet.com/multiplying-whole-numbers-by-fractions-word-problems-worksheet/","timestamp":"2024-11-03T07:32:31Z","content_type":"text/html","content_length":"53701","record_id":"<urn:uuid:b99f74a8-6110-4fda-953e-9cad3e64e965>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00193.warc.gz"}
295.4 grams to ounces Convert 295.4 Grams to Ounces (gm to oz) with our conversion calculator. 295.4 grams to ounces equals 10.419927784 oz. Enter grams to convert to ounces. Formula for Converting Grams to Ounces: ounces = grams ÷ 28.3495 By dividing the number of grams by 28.3495, you can easily obtain the equivalent weight in ounces. Converting grams to ounces is a common task that many people encounter, especially when dealing with recipes, scientific measurements, or everyday items. Understanding how to perform this conversion can help bridge the gap between the metric and imperial systems, making it easier to communicate measurements across different contexts. The conversion factor between grams and ounces is essential for accurate measurement. One ounce is equivalent to approximately 28.3495 grams. This means that to convert grams to ounces, you need to divide the number of grams by this conversion factor. Knowing this allows you to easily switch between the two units of measurement. To convert 295.4 grams to ounces, you can use the following formula: Ounces = Grams ÷ 28.3495 Now, let’s break down the calculation step-by-step: 1. Start with the amount in grams: 295.4 grams. 2. Use the conversion factor: 28.3495. 3. Perform the division: 295.4 ÷ 28.3495. 4. The result is approximately 10.41 ounces. When rounded to two decimal places, 295.4 grams is equal to 10.41 ounces. This level of precision is particularly useful in various applications, ensuring that measurements are both accurate and The importance of converting grams to ounces cannot be overstated. In cooking, for instance, many recipes may list ingredients in ounces, while others use grams. Being able to convert between these units allows you to follow recipes accurately, ensuring that your dishes turn out as intended. Similarly, in scientific contexts, precise measurements are crucial for experiments and data collection, where the difference of a few grams or ounces can impact results significantly. Everyday scenarios also benefit from this conversion. Whether you’re weighing food items, measuring ingredients for a DIY project, or even calculating nutritional information, knowing how to convert grams to ounces can simplify your tasks and enhance your understanding of measurements. In conclusion, converting 295.4 grams to ounces is a straightforward process that can be accomplished with a simple formula and a clear understanding of the conversion factor. By mastering this skill, you can navigate between the metric and imperial systems with ease, making your cooking, scientific endeavors, and daily activities more efficient and accurate. Here are 10 items that weigh close to 295.4 grams to ounces – • Standard Baseball Shape: Spherical Dimensions: 9 inches in circumference Usage: Used in the sport of baseball for pitching, hitting, and fielding. Fact: A baseball is made of a cork center wrapped in layers of yarn and covered with leather. • Medium-Sized Pineapple Shape: Oval Dimensions: Approximately 12 inches tall and 6 inches wide Usage: Consumed fresh, juiced, or used in cooking and baking. Fact: Pineapples are a symbol of hospitality and are often used as a decorative centerpiece. • Small Watermelon Shape: Round Dimensions: About 8 inches in diameter Usage: Eaten fresh, in salads, or blended into smoothies. Fact: Watermelons are 92% water, making them a refreshing summer treat. • Large Avocado Shape: Pear-like Dimensions: Approximately 7 inches long and 4 inches wide Usage: Used in salads, spreads, and guacamole. Fact: Avocados are often referred to as “alligator pears” due to their bumpy skin. • Medium-Sized Bag of Flour Shape: Rectangular Dimensions: 10 inches tall and 5 inches wide Usage: Used in baking and cooking for various recipes. Fact: Flour is a staple ingredient in many cuisines around the world. • Standard Laptop Shape: Rectangular Dimensions: 15 inches wide, 10 inches deep, and 1 inch thick Usage: Used for computing tasks, browsing the internet, and multimedia consumption. Fact: The first laptop was introduced in 1981 and weighed over 5 kg! • Large Book Shape: Rectangular Dimensions: 9 inches wide and 12 inches tall Usage: Used for reading, studying, and reference. Fact: The largest book in the world is over 5 meters tall and weighs more than 1,400 kg! • Medium-Sized Pumpkin Shape: Round Dimensions: Approximately 10 inches in diameter Usage: Used for decoration, cooking, and making pumpkin pie. Fact: Pumpkins are a type of squash and are native to North America. • Standard Soccer Ball Shape: Spherical Dimensions: 8.6 inches in diameter Usage: Used in the sport of soccer for kicking and passing. Fact: A regulation soccer ball has 32 panels and is designed for optimal aerodynamics. • Medium-Sized Bag of Rice Shape: Rectangular Dimensions: 12 inches tall and 8 inches wide Usage: Used as a staple food in many cultures around the world. Fact: Rice is the primary food source for more than half of the world’s population. Other Oz <-> Gm Conversions –	{"url":"https://www.gptpromptshub.com/grams-ounce-converter/295-4-grams-to-ounces","timestamp":"2024-11-13T01:41:49Z","content_type":"text/html","content_length":"186201","record_id":"<urn:uuid:5357fad5-850b-4524-bd36-72907e13bc91>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00137.warc.gz"}
MP Board 10th math question paper 2022 : imp question, model paper - Physics Hindi MP Board 10th math question paper 2022 : imp question, model paper MP Board 10th math question paper 2022 : Hi buddy, what’s up ,how are you, hope all of you are fine, today I am going to describe about MP Board 10th math question paper 2022. Lovely students Madhya Pradesh Board of Secondary Education has declared final board exam of class 10th. Besides this MP Board issued sample papers of tenth math question paper. Dear students Madhya Pradesh Board of Secondary higher education has announced various sample of math model question paper 2022. Friends sample paper is a medium by which any student can make a good performance in MP Board 10th math exam MP Board 10th math question paper 2022 Dear candidates since last 3 years pandemic has been it’s top peak. Therefore students could not get sufficient times for their complete study almost every college and institutions were closed in MP Board last three years. Many students think that mathematics is a scare. But dear students you have no need to afraid with mathematics, according me mathematics is only a fun nothing more than that only you need some daily practice. Dear friends if you secure highest marks in mathematics then first of all you will have to learn and understand mathematics formulas and then by using formula focusing on mathematics sums. You can solve any math problem. MP Board 10th model paper 2022 Dear candidates as I referred in the above keyword, MP Board Department of Education notified that MP Board class 10th final exam will be held on 18th February and last on 14 March. Whereas there are many students Such as who did not prepare exam preparation properly, because there were no proper schools since last 3 years. So students are haunting an unknown fear that what types of class 10th math question paper 2022 comes and what was the pattern of class 10th math question paper 2022. So Students don’t worry, We are here to give you a perfect solution for your worries. Students first of all I tell you that MP Board Department of Education had redusced 30% of syllabus and after that MP Board Department of Education issued mp sample papers for easiness of the students. MP Sample papers 2022 tell bout the pattern of board exams 2022. After solve MP Board class tenth Math sample paper students should analysis For their weaknesses. And then make improvement of their studies. And students are advised that they should try to solve questions in standard time. MP Board 10th math exam paper 2022 overview In as much as MP board has already cleared that mp board 10th final exam paper will start on 18 february. And class 10th math paper is on 22 february which we are mentioning in this article. This year mp board 10 math modal paper 2022 are also issued by MPBSE to inform students for the blueprint of high school and higher secondary exams. Name of Board MPBSE Bhopal State Madhya Pradesh Grade Name HSC (10th) Subject Math Exam Date 18 Feb 2022 Exam paper MP Board 10th modal paper Exam period 18 Feb to 10 march 2022 Medium Hindi & English Category Boards Exam Official Website www.mpbse.nic.in MP Board 10th math question paper 2022 How to solve 10th math mp board exam paper 2022 Dear student In this keyword We shall discuss about how to solve MP board math 10th question exam paper 2022. Dear students maths solving is an art for fun. Math is fun on those who practice mathematics. And scare on those who dodge with it. There are giving some tips or hints to solve MP Board 10th math question paper 2022 – • Students first you gonna run out of mind that mathematics is scary, and have to bear in mind that mathematics is only fun nothing more than that. • After that you will have to understand with the base of mathematics such as BODMAS, fraction, common, etc. • Now you have to learn formula because without formula you can not move question one bit ahead. • Math problem can be solve by three simple step process, first you visualize the math problem, second approach to be followed for that problem, and at last you can solve anyhow math problem • One more other math solving four step process is that ,first of all you should carefully read ,understand, identify the problem, after that draw and review the problem in your mind, at third stage you should develop the plan for solving that problem, and at last you can solve the problem. • The difficulty question stage is three types. easy 40%,middle 45%,high level 15%. So do not be panic • Always touch with your teachers • never be gape school study except casuals. • make proper notes and make underline special formulae and questions. • make a proper complete study time schedule. MP Board 10th math exam 2022 Lovely students all of you you are familiar with that time schedule which announced by MP Board Department of Education now there is no time remain for students to complete their exams preparations. So students have need to prepare oneself. So there is only one way for preparation that which sample papers all guess papers which are issued by MP Board Department of Education. The sample paper are very important to understand students pattern of the final exam. MP Board class 10th final exams starting from 18 February2022 and lasting on 10 March 2022. All those students who want to boost their exam post paper then they should advise to download sample papers or guess papers buy the official website of MP Board Department of Education.mpbse.nic.in . Therefore students must complete the syllabus at time and start to solve MP Board 10th math question paper, to promote there accuracy. These Sample papers are the keyword or blueprint of the main exam papers. MP Board 10th math exam paper 2022 overview In as much as MP board has already cleared that mp board 10th final exam paper will start on 18 february. And class 10th math paper is on 22 february which we are mentioning in this article. This year mp board 10 math modal paper 2022 are also issued by MPBSE to inform students for the blueprint of high school and higher secondary exams. MP Board 10th math Imp question 2022 Some board 10th class imp question are following below • 50 मीटर ऊंची पहाड़ी के शिखर से किसी मीनार की चोटी और आधार के अवनमन कोण क्रमशः 30° और 45° हु \| मीनार की ऊंचाई ज्ञात कीजिए • एक त्रिभुज की भुजाएँ 4 सेमी, ६ सेमी और 8 सेमी है \| इसका परिगत वृत्त खींचिए तथा रचना के पद लिखिए • दो क्रमागत धनात्मक पूर्णांक ज्ञात कीजिये जिनके वर्गों का योग 365 हो। • एक घड़ी की मिनट की सुई जिसकी लंबाई 14 CM है \| इस सुई द्वारा 5 मिनट में रचित क्षेत्रफल ज्ञात कीजिये? • वृत के क्षेत्रफल का सूत्र लिखो। • किसी परिमेय संख्या हर में 5 जोड़ने और उसके अंश में से घटाने पर 1/7 प्राप्त होता है \| यदि उसके अंश में से 4 घटाया जाए तो 1/3 प्राप्त होता है \| वह संख्या ज्ञात कीजिए \| • यदि tanA= cotS, तो सिद्ध कीजिए कि A+B= 90° . • sin60°.cos30°+ sin30. cos60. मान ज्ञात करो। • बिंदुओं (2, 3) और (4, 1) के बीच की दूरियाँ ज्ञात कीजिए • 15,36,18,26,19,25,29,20,37 की माध्यका ज्ञात करो- • समरूपता का क्या है, लिखिए • समांतर श्रेढ़ी का व्यापक रूप क्या है, लिखिए • विलोपन विधि का प्रयोग करके निम्न रैखिक समीकरण युग्म को हल कीजिए – Hence students I hope now you will be all understandable by A single bit of mathematics. To get such type information daily and regularly, visit our website continuously. If you have any doubt related MP Board 10th math question paper 2022, fell freely and ask in comments, we will try to reach you soon. Official Website Click Here APSMHOW HOME PAGE Click Here Leave a Comment	{"url":"https://physicshindi.com/mp-board-10th-math-question-paper-2022/","timestamp":"2024-11-08T15:33:39Z","content_type":"text/html","content_length":"94027","record_id":"<urn:uuid:45e5062c-cfd5-4326-940d-7b1c2971b704>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00720.warc.gz"}
Linear Algebra Video-Part 4 This is a sneak peek into the video Linear Algebra Part 4 . Here, you will learn about linear transformations, linear independence, eigenvalues and eigenvectors, This is a part of College / Engineering Mathematics. To view the entire video, you can contact me . The easiest way to study Mathematics! Study at your own pace. Revise formulas everyday. Sign in for webinars. This is great for College students learning Math. You need to get good scores .	{"url":"https://www.mathmadeeasy.co/post/linear-algebra-video-part-4","timestamp":"2024-11-02T18:51:28Z","content_type":"text/html","content_length":"1050379","record_id":"<urn:uuid:cb5cd783-8c24-41dd-9e95-d36e08cc2bb4>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00185.warc.gz"}
Degree Course in Academic Year 2016/2017 - 2° Year Learning Objectives The student will continue the acquisition of Mathematical concepts. The themes will be linked to concepts learned in other disciplines. In particular, the course has the following objectives: Knowledge and understanding (knowledge and understanding): the student will be introduced in the study of metric spaces, normed spaces and Hilbert spaces. You will see how the results known from the course Analysis I are part of a much wider and more general context. The rigorous abstraction capacity and in the same time critical synthesis will be developed. The student will become familiar with differential calculus of several variables vector functions, with the study of differential equations of which already knows the solution methods but will be introduced to fundamental issues such as the theorems of existence, uniqueness of solutions. Expand the knowledge related to the measurability of sets in the plane and in space by studying the measure theory and Lebesgue Peano. So students will be introduced to the integral calculus of several variables and functions to the different applications. Some insights will be entrusted to the most willing students who, alone or in groups, will present them in short seminars. Applying knowledge and understanding (applying knowledge and understanding): the student will not only learn the individual concepts but fail to connect them and will be conducted, in particular, to reflect on the structural properties (eg topology) that form the basis of various topics studied. It will be used to reflect on an issue and build a mathematical model to be studied analytically. Students can also exercise their ability to use their knowledge in situations other than those in which they were submitted, for example they will be invited to demonstrate independently similar results to those studied, and to perform many exercises of application of the studied theorems. This will be done through guided exercises in the classroom and through exercises - both manipulative that demonstration - that will be proposed for self-study. Making judgments (Making judgments): students can study the topics of the lectures not to get used to deepen their own knowledge and will be encouraged to search for additional applications of the arguments. also he is able to critically engage with the other students during tutoring hours to identify the most appropriate solutions. Communication skills (communication skills): by listening to lectures and reading books recommended, the student will be helped to structure so more complex mathematical thinking and to better treat the mathematical language which was already introduced in the first year of course. Through guided exercises and seminars, learn how to communicate clearly and rigorous both orally and in writing. You will learn to use a proper language is one of the most important means to acquire the mentality mathematics. Learning skills (learning skills): the student will be led to acquire a study method that allows him to turn to a new topic now recognizing what are the necessary prerequisites. Develop, in addition, the computing capacity and handling of the studied mathematical objects. Detailed Course Content Metric spaces. Normed Banach and Hilbert spaces. Sequences and series of functions.. Didfferential calculus for functions of several variables. Implicit functions. Measure theory Integrals of functions of several variables. Curves. Integral of differential forms.Surfaces and surface Integrals. Systems of differential equations. Textbook Information Giovanni Emmanuele Analisi Matematica II Foxwell and Davies 2004	{"url":"https://web.dmi.unict.it/courses/l-35/course-units/?cod=4130","timestamp":"2024-11-06T23:51:56Z","content_type":"text/html","content_length":"24320","record_id":"<urn:uuid:0eb0a0ab-a326-4049-b5b0-b9b1801468cf>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00278.warc.gz"}
Basic Number Bar This page has moved to http://www.artofchording.com/layout/basic-number-bar.html. Click here if you are not redirected. Basic Number Bar The final key on the steno layout is also the biggest one. It lies on top of the whole layout, accessible with any finger from the top row. The number bar is used in steno to produce digits. The number bar is functionally just one key. On some steno machines it is split into many parts for comfort or aesthetics. Single Digits Pressing the number bar alone doesn't do anything. Instead, it behaves sort of like a shift key that turns keys into numbers. If you just press the bare number bar, unless you have it mapped to some translation, you will just get the pound symbol #. You need to hit the number bar with certain steno keys to get digits. The mapping from steno keys to numbers is as follows: S T P H A O F P L T Subsequent strokes made with the number bar will attach to each other. So if we stroke #S the output is "1". If we stroke #S #T #P the output is "123". Note that the order is pretty predictable, except the 0 which is sitting in the middle of the layout. It can be remembered by noticing that the O key looks like a zero. For fingering, you can hit the number bar with any finger you like. Some people hit it with the letter they are using (for example, #S would be the pinky in the crack between the number bar and the top S key), and others hit it with the same fingers all the time (such as the middle finger, which is the longest). Multiple Digits If you hit multiple steno keys at the same time as the number bar, you will get the equivalent of pressing them all in steno order. "123456" can be written in just a single stroke with #STPHAF. Using multiple digits, we can break down large numbers into just a couple strokes by writing down the parts that are in steno order. For example, "1384257" can be written in three strokes with #SP-L #H #TAP. The last tool we will learn about for the number bar in this introduction is reversal. By adding the chord EU to our number stroke, the steno order of the numbers is reversed. #TP is "23", #TPEU is In Plover's dictionary, only pairs of numbers (10-98) are reversible by default, but you can get the rest of the possible number chords by adding a supplementary dictionary. This will allow you to go up to #STPHAOEUFPLT: "9876054321". Now we can write complex numbers like "602208" in just two strokes: #TOEUF #TOL. 1. Translate Write the English sentence represented by these outlines, including punctuation. 1. E H #TO OR SO TP-PL 2. -F -T #SAO E KEPT #AEUP KW-PL OR HAF TP-PL 3. WHA R -T #ST STEPS H-F 2. Find Outlines Find steno outlines that will write these English sentences, including punctuation. 1. She is 40. 2. He had 500. 3. Step 1: start. 4. 90210. Can you write 90210 in just two strokes?	{"url":"https://morinted.gitbooks.io/plover-theory/content/layout/basic-number-bar.html","timestamp":"2024-11-04T05:17:56Z","content_type":"text/html","content_length":"39613","record_id":"<urn:uuid:8e4f7096-0961-4b4e-acbc-b195595d773d>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00260.warc.gz"}
Is it possible to find a solution for any $ae^{bx}+cx+d=0$? So I enjoy solving problems of the form$$ae^{bx}+cx+d=0$$$$a,b,c\ne0$$however what I don't particularly enjoy is having to solve that equation every single time I come across it. So what I want to know is: Is there a way to solve every non-algebraically solvable equation of that form? Here is my attempt:$$ae^{bx}+cx+d=0\implies a+(cx+d)e^{-bx}=0$$$$\implies(cx+d)e^{-bx}=-a\quad\text{rearranging terms}$$$$\implies((-c/b)(-bx-bd/c))e^{-bx}=-a$$$$\implies(-bx-bd/c)e^{-bx}=ab/c$$$$\ implies(-bx-bd/c)e^{-bx-bd/c}=(ab/c)e^{-bd/c}$$and now taking the Lambert W of both sides (note that the solution uses $W_k(z)$ since all branches of the product log function will yield a valid solution),$$-bx-bd/c=W_k((ab/c)e^{-bd/c})$$which leads to the solution$$x=(1/b)W_k((ab/c)e^{-bd/c})-d/c$$However, my question is: Is my solution valid, or how would one go about finding the solution for any $ae^{bx}+cx+d$ where $a,b,c\ne0$? 0 answers Sign up to answer this question »	{"url":"https://math.codidact.com/posts/291375","timestamp":"2024-11-03T15:05:51Z","content_type":"text/html","content_length":"41427","record_id":"<urn:uuid:b0242550-df7e-4714-812d-3226c5647ee9>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00674.warc.gz"}
Branch Circuit Design Calculations – Part Two In the previous article " Branch Circuit Design Calculations – Part One " in our new course " Course EE-3: Basic Electrical design course – Level II ", I explained some essential definitions and I listed different types of Branch circuits. Today I will list additional types of branch circuits and explain the basic features for any branch circuit as follows. 4.A Additional Branch circuits types 1- According to location A- Indoor branch circuit: It includes All branch circuits inside the premises for lighting, receptacles, equipments, HVAC or other outlets. B- Outdoor Branch Circuits It is the branch circuits run on or between buildings, structures, or poles on the premises; and electrical equipment and wiring for the supply of utilization equipment that is located on or attached to the outside of buildings, structures, or poles. Examples for Outdoor Branch Circuits are listed in table 225.3 in below. 2- According to purpose of the Load A- General-Purpose Branch Circuits It included in Chapters 1, 2, 3, and 4 and not listed under Specific-Purpose Branch Circuits tables 210.2, 220.3 and chapters 5, 6&7. B- Specific-Purpose Branch Circuits The provisions for branch circuits supplying equipment listed in Table 210.2 amend or supplement the provisions in this article and shall apply to branch circuits referred to therein. Also table 220.3 list the calculation of loads in specialized applications that are in addition to, or modifications of, those Of general rules. NEC code Chapters 5, 6, and 7 apply to special occupancies, special equipment, or other special conditions. These chapters supplement or modify the general rules. 3- According to type of building A- Dwelling Building A building providing complete and independent living facilities for one or more persons, including permanent provisions for living, sleeping, cooking, and sanitation. Dwelling buildings include the following types: • Dwelling, One-Family: A building that consists solely of one dwelling unit. • Dwelling, Two-Family: A building that consists solely of two dwelling units. • Dwelling, Multifamily: A building that contains three or more dwelling units. B- Non-dwelling buildings other buildings than dwelling like schools, hotels, factories, etc. 4- According to building’s construction status A- Branch circuits for Existing building. B- Branch circuits for New building. 5- According to Branch circuit voltage A- Branch Circuits Not More Than 600 Volts It is divided to (4) categories as follows: a- Branch Circuits not exceed 120 volts In residences and hotel rooms, circuits supplying lighting fixtures and small receptacle loads. b- Branch Circuits 120 volts and less it may be used to supply lampholders, auxiliary equipment of electric-discharge lamps, receptacles, and permanently wired equipment. c- Branch circuits 120 volts - 277 volts It may supply mogul-base screw-shell lampholders, ballasts for fluorescent lighting, ballasts for electric-discharge lighting, plug-connected appliances, and hard-wired appliances. Incandescent lighting operating over 150 volts is permitted in commercial construction. d- Branch circuits 277 volts - 600 volts it can supply mercury-vapor and fluorescent lighting, provided the lighting units are installed at heights not less that 22 feet above grade and in tunnels at heights no less than 18 feet. B- Branch Circuits over 600 Volts It used for some special uses in non-dwelling buildings. 6- According to Branch circuit rating: A- 15 and 20 Ampere Branch Circuits It shall be permitted to supply lighting units or other utilization equipment, or a combination of both. B- 30 Ampere Branch Circuits It shall be permitted to supply fixed lighting units with heavy-duty lampholders in other than a dwelling unit(s) or utilization equipment in any occupancy. C- 40 and 50 Ampere Branch Circuits It shall be permitted to supply cooking appliances that are fastened in place in any occupancy. In other than dwelling units, such circuits shall be permitted to supply fixed lighting units with heavy-duty lampholders, infrared heating units, or other utilization equipment. D- Branch Circuits Larger Than 50 Amperes It supply only non-lighting outlet loads. 5. Branch circuit features 5.1 Branch circuit Voltages • The voltage rating of electrical equipment shall not be less than the nominal voltage of a circuit to which it is connected. • Unless other voltages are specified, for purposes of calculating branch-circuit and feeder loads, nominal system voltages of 120, 120/240, 208Y/120, 240, 347, 480Y/277, 480, 600Y/347, and 600 volts shall be used. • Branch-circuit voltage limits are provided by NEC Code. These limits are based on the equipment being supplied by the circuit. 5.2 Branch Circuit Conductors Conductors normally used to carry current shall be of copper unless otherwise noted. Conductors size unit: The American Wire Gage (AWG) for size identification, which is the same as the Brown and Sharpe (BS) Gage. The notation circular mils is used for Conductors larger than 4/0 AWG which will be equal to 250,000 circular mils. the notation MCM is used instead of circular mils, where MCM = 1000 circular mils. so, a 250,000-circular-mil conductor was labeled 250 MCM. the notation kcmil is used by both UL standards and IEEE standards instead of MCM. where kcmil = MCM = 1000 circular mils. The circular mil area of a conductor is equal to its diameter in mils squared (1 in. = 1000 mils). What is the circular mil area of an 8 AWG solid conductor that has a 0.1285-in. diameter? 0.1285 in. x 1000 = 128.5 mils 128.5 x 128.5 = 16,512.25 circular mils 5.3 Branch Circuit Rating • The rating of any branch circuit will be the maximum permitted ampere rating or setting of the overcurrent device protecting this branch circuit. • Branch circuits serving only one device can have any rating, while a circuit supplying more that one load is limited to ratings of 15, 20, 30, 40, or 50 amps. A branch circuit wired with 10 AWG copper conductors has an allowable ampacity of at least 30 amperes, If this branch-circuit overcurrent protective device is a 20-ampere circuit breaker or fuse, what is the rating of this branch circuit?. the rating of this branch circuit is 20 amperes, based on the size or rating of the overcurrent protective device. 5.4 Number of Branch Circuits The minimum number of branch circuits shall be determined from the total calculated load and the size or rating of the circuits used. In all installations, the number of circuits shall be sufficient to supply the load served. I will explain the branch circuit design calculations as per the classification of branch circuit according to type of load mentioned in Previous Article " Branch Circuit Design Calculations – Part Two ". First: Lighting Branch circuits In the broad sense, lighting loads may be categorized as follows: 1. General lighting. 2. Show-window lighting. 3. Track lighting. 4. Sign and outline lighting. 5. Other lighting. Each lighting load is computed separately and then combined to determine the total lighting load. In the next Article, I will explain the design calculations for general lighting branch circuit. Please keep, following.	{"url":"http://www.electrical-knowhow.com/2013/02/branch-circuit-design-calculations-part_6.html","timestamp":"2024-11-03T14:09:05Z","content_type":"application/xhtml+xml","content_length":"96967","record_id":"<urn:uuid:f7a4033e-8e57-4a04-95a3-606a7c5cddd5>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00034.warc.gz"}
ACT Math Strategies \|\| Free ACT Training from AP Guru ACT Math Basics ACT Math Strategies Some strategies are so powerful that they can singlehandedly increase your ACT Math Test score by more than 100 points. At AP Guru, we absolutely recommend students use these three strategies time and again. If you learn just three strategies and build them into your routine, you’ll notice a ludicrous shift in your math performance. These strategies can be applied to the majority of your ACT Math Test 1. Stealing the Answers The ACT is a multiple-choice exam. On the Math portion, this has a profound implication: the correct answer is sitting right in front of you. Why would you ever go through the horrible work of actually solving a math problem or working harder than you need to when you can just cherry pick the right answer instead? All problems that allow you to steal answers have something in common: they point to the answer choices. Whenever you can steal answers, you’ll see language like this: 1. “What is a number that satisfies these conditions?” In other words, “there’s only one number, and it’s in the answer choices...so plug the choices in and figure it out...” 2. “Which of the following...” Sometimes, the question literally says, “Of these answers, one does X, Y, or Z.” In those cases, you know it’s stealing time! 3. “What is the largest/smallest possible value of...” The problem is saying, “there are a few values that work, but the largest/smallest is in the answer choices, and you can just plug them in to see if they work.” 4. “What could be a value of...” Most problems involving “possibility” that also contain the words “could” or “might” will allow you to steal answers. Just like any other strategy, this one takes a bit of practice before it’s totally useful. The more You practice, the better you’ll become at identifying when you can steal or eliminate. Examples are always the best way to learn, so let’s launch right into the good stuff Example 1 Andrea manages a company that currently has 116 customers, which is 8 more than twice the number of customers the company had 1 year ago. How many customers did the company have 1 year ago? A. 50 B. 54 C. 61 D. 66 Solution: The best way to approach this problem is to take each answer choice, double it, and add 8. Start at the top and see which one works. For example, option A would be 50 x 2 = 100. Add 8 to it that would be 108. Therefore, it is not the answer. Option B would be 54 x 2 = 108. Adding 8 gives you 116. There we have it. B is the correct answer (there can’t be two right answers to a math problem, so there’s no need to test options C or D). That was easy! Example 2 Which of the following (x,y) pairs is the solution for the system of equations x + 2y = 4 and -2x + y = 7 A. (-2.3) B. (-1.2,5) C. (1,1.5) D. (2,1) Solution: Like before, plug each answer in, and you’re good to go! One will work, the others won’t. Start with A. If we plug the X and Y values in, we get: -2 + 6 = 4 for the first equation. Check! For the second, we get 4+3 = 7. Check! It works! The answer is A. Boom. We’re done. Testing in the rest would be a waste of time. 2. Plugging In Numbers Many ACT Math Test problems don’t deal in concrete numbers - they deal in variables. FOR INSTANCE: "A certain table has a width four times as great as its length” or “The amount of strawberries sold, S, is twice the amount of blueberries sold, B.” And so on and so forth. In both cases, and all cases like them, you’re not dealing with numbers you’re dealing with the idea of numbers. That makes things unnecessarily difficult, and who likes when things are difficult? We don’t, and neither should you. Math is about numbers. When you do math without numbers, it’s hard. When you do math with numbers, it’s easy. With that in mind, our goal is to use real numbers whenever we possibly can and to make them up if they don’t exist. Always follow this Golden Rule: If you don’t need to know the value of anything in any given math problem, make the value up. It doesn’t matter whether it’s the value of x, the side length of a square, the number of students in a class, whatever. If you don’t need to know it specifically, then make it up and plug it in. Whenever you read an ACT Math Test problem, simply ask: is there a certain number I don’t need to know? If so, can I make it up? If you can, do it. Keep these things in mind: 1. If you don’t need to know the exact value of something, you should make it up 2. It’s most common in ratio and percentage problems 3. You never have to use this strategy but it’s extremely helpful. So if you ever get totally stuck on a problem, pull yourself back and see if you might be able to make up values for one of the variables or mystery numbers. You just might! Let's see the strategy in action using a few solved examples: Example 3 If the average of X and Y is m, and m doesn’t equal 0, then which of the following is the average of X, Y, and 2m? A. m B. 4m/3 C. 3m/21 D. 2m Solution: Now, you could solve this the difficult way using algebra and numbers that aren’t real. Or you could take the massive shortcut and simply make up the values of x and y!!! For this problem, you don’t need to know what the values of x and y actually are, so you can just make up any random values: Using the plugged in, made up numbers, you see that the average of x and y is 10. So that means that: M=10. The question asks: “what’s the average of x, y, and 2m?” The answer is the average of 7, 13 and 20. That is 40/3. By plugging m = 10 into the answer choices, we figure the right answer is B. Example 4 When the positive integer x is divided by 7, the remainder is 4. Which of the following expressions will yield a remainder of 3 when it is divided by 7? A. X + 2 B. X + 3 C. X + 4 D. X + 6 Solution: At first, this problem seems weird and confusing. Unless, that is, we make up a number to represent X. What if we make X = 4. Then 4 divided by 7 has a remainder of 4 (and a quotient of Now that we have an X, let’s test the answers: A. 4+2=6. 6 divided by 7 has a remainder of 6 (quotient 0, remainder 6). Doesn’t work B. 4+3=7. 7 divided by 7 has a remainder of 0 (quotient 1, remainder 0). Doesn’t work C. 4+4=8. 8 divided by 7 has a remainder of 1 (quotient 1, remainder 1). Doesn’t work D. 4+6=10. 10 divided by 7 has a remainder of 3 (quotient 1, remainder 3). Works Here’s the amazing thing: so long as you follow the rules of the problem, the plug in strategy always works! Following the rules of the problem is important, though. If I picked “x=5,” this wouldn’t have worked, since 5 divided by 7 doesn’t have a remainder of 4. But try ANY number that has a remainder of 4 when divided by 7 (try 11, 18, or 25), and you’ll get the same answer D. 3. Creating Variables So what happens if you can’t come up with real numbers? What if there are defined values in the problem you’re solving that you need to work with? And what if the answer choices don’t help you to solve the problem? That’s where the third and final strategy comes in. The hardest part of solving any ACT Math Test problem is figuring out what you’re trying to solve in the first place. Always ask yourself: can I create a variable to stand in for a certain value? As we’ve mentioned, math is a visual science and requires manipulation of values. All of algebra boils down to “manipulating equations” so that the thing you want isolated is on one side of the = sign, and everything else is on the other side. Almost every ACT Math Test problem boils down to an algebraic equation – once you have an equation in your hands, a problem is far less difficult to solve. Of course, to create an equation, you need one thing: A VARIABLE. Without variables, there are no equations. What would be the point of saying 9 = 9, or 4+5=9? These aren’t “solvable” because there’s nothing to figure out. When you are stuck on a problem, and you can’t figure out what to do, it’s usually because you don’t have any variables to manipulate, and without variables, no real mathematics can take place. Let’s walk you through this idea with an EXAMPLE.. Suppose you have a problem: 5 less than 2 times a number is the same as 4 times that number. "What is the number?” This problem may seem somewhat complicated without assigning the unknown number a variable. So, assign it one! Then all you need to do is convert the entire sentence into algebrese, the language of “The number” is now X. Let's try to solve it now. The equation translates to: 2X – 5 = 4X. The number is -5/2. By actively processing problems and forcing yourself to rewrite all the information you receive, you’ll create more useful, visual models of every single problem you have to solve.	{"url":"https://www.apguru.com/act-hub/act-math-strategies","timestamp":"2024-11-13T22:06:29Z","content_type":"text/html","content_length":"33349","record_id":"<urn:uuid:8411532d-41ca-4909-9862-8239a7099d18>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00605.warc.gz"}
Our users: What I like about this software is the simple way of explaning which anybody can understand. And, by 'anybody, I really mean it. D.H., Tennessee You've been extremely patient and helpful. I'm a "late bloomer" in the college scene, and attempting math classes online are quite challenging to say the least! Thank you! David Brown, CA As a single mom attending college, I found that I did not have much time for my daughter when I was struggling over my Algebra homework. I tried algebra help books, which only made me more confused. I considered a tutor, but they were just simply to expensive. The Algebrator software was far less expensive, and walked me through each problem step by step. Thank you for creating a great product. Maria Leblanc Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among Search phrases used on 2013-10-22: • differentiation formulaes • free printable school examination papers • find y value on ti 83 graphing calculator • solve richard equation by matlab soft ware • real world font download • mcdougal littell algebra 1 resource book answers • "math worksheets" fraction pie charts elementary students • great common divider • alegabra help • algebra complex equation worksheets free • free Algebra 1 worksheets • evaluate logarithmic expression square root • "aptitude in c"+ppt • Quetions on english apptitude • how to figure out common denominator • printable homework for 3rd graders • examples of a math trivia with answer • convert 3 3/4 to decimal • Aptitude test paper • solve for x calculator • mathematica tutor • what's 3x-6y=12 • verbal algebra 1 • "scientific computing " heath "solution manual" • subtracting integers worksheet • adding negatives calculator • "math problem solver" • Free First Grade Math Sheets • factor monomials calculator • real life application of completing square • radical equation solver free online • free answers for math books • graphing and solving radicals • calculator with simplifying function • interactive calculations I=PRT • mathematics for dummies • solving expressions calculator • variable worksheets • ti-83 plus solve systems of equations • 9th grade Star testing study guide • permutations and combinations book • holt algebra answers • sample multiple choice exams on polynomial • free worksheets for money computation • Linear combinations special types worksheet • what is the suare root of 777 • convert decimal to fraction calculator • worksheets by mcdougal littell • square route problems 7 grade free • patterns and algebra test • solve alegbra equation • rational exponents worksheet • GCSE Transformation • sample algebra problem • TI84 quadratic formula • solve nonhomogeneous wave equation • simplifying radicals calculator • mathamatical formulas • ti-89 quadratic • third grade math sheets to print • online algebra calculator • math pre algebra nc books prentice hall • implicit differentiation online solver • 20 EXAMPLES QUADRATIC ALGREBRA • "quadratic formula" + area + square • english : gerak parabola • linear equation and inequalities calculator • teach third grade permutations • Free Math Permutations Practce • completing the square calculator • 6th grade permutations and combinations • java code testing for prime numbers • free downloadable aptitude test • java time converter • FIFTH GRADE MATH WORKSHEETS • division worksheet w/ remainders • hands on equations math problem solver • first grade fraction printables • cramers rule online solver	{"url":"https://emathtutoring.com/factoring-algebraic/interval-notation/9th-math-worksheets-free.html","timestamp":"2024-11-10T04:54:46Z","content_type":"text/html","content_length":"86101","record_id":"<urn:uuid:e472f493-edc1-4e35-a1d5-cddeddbdf3ca>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00295.warc.gz"}
Feet and Inches Calculator Have you ever struggled with adding up feet and inches or figuring out the area of your room? Well, you're in luck! A Feet and Inches Calculator is here to save the day. This amazing tool helps you work with these measurements easily and quickly. Why is it useful? Imagine you're helping your parents measure for new curtains or trying to figure out if your new bed will fit in your room. This calculator can help you add, subtract, and even find areas without getting a headache from all the math! Understanding Feet and Inches Before we dive into using the calculator, let's talk about feet and inches. A foot is a unit of length, and it's about as long as an adult's foot (hence the name!). An inch is a smaller unit of length - in fact, there are 12 inches in one foot. We use feet and inches all the time in everyday life. For example: • Measuring how tall you are (like 5 feet 4 inches) • Figuring out the size of a TV screen (maybe 55 inches) • Measuring rooms in a house Area Calculator - Feet and Inches Now, let's talk about area. Area is the space inside a flat shape. If you're looking at your bedroom floor, the area is all the space you could cover with a giant rug. To use the calculator for area, you'll need to know the length and width of the space you're measuring. Here's how it works: 1. Measure the length in feet and inches 2. Measure the width in feet and inches 3. Enter these numbers into the calculator 4. The calculator will multiply length by width to get the area Let's try an example: Your room is 12 feet 6 inches long and 10 feet 3 inches wide. Here's how you'd calculate the area: Length: 12 feet 6 inches Width: 10 feet 3 inches Area = 12'6" × 10'3" = 128.125 square feet The calculator does all this math for you in seconds! You could use this to figure out how much carpet you need for your room or how much paint for the ceiling. How to Add Feet and Inches Adding feet and inches can be tricky because we need to remember that 12 inches make a foot. Here are the basic steps: 1. Add up all the feet 2. Add up all the inches 3. If the inches add up to 12 or more, convert them to feet 4. Add any extra feet from step 3 to your feet total Let's try an example: You want to know the total length of your hallway and living room. The hallway is 8 feet 9 inches long, and the living room is 15 feet 5 inches long. Hallway: 8 feet 9 inches Living room: 15 feet 5 inches Total: 23 feet 14 inches But remember, 14 inches is more than 12, so we need to convert that: 14 inches = 1 foot 2 inches So our real total is: 24 feet 2 inches The Feet and Inches Calculator does all of this for you automatically! How to Subtract Feet and Inches Subtracting feet and inches can be even trickier than adding them. Here's how it works: 1. Subtract the feet 2. Subtract the inches 3. If you can't subtract the inches (because the number is bigger than what you're subtracting from), you need to borrow a foot 4. Remember, borrowing a foot gives you 12 more inches to work with Let's look at an example: Your room is 15 feet 8 inches long. You want to put in a bookshelf that's 3 feet 10 inches wide. How much space will be left? Room length: 15 feet 8 inches Bookshelf width: 3 feet 10 inches Space left: 11 feet 10 inches Here's what happened: • 15 - 3 = 12 feet • We can't do 8 - 10 for the inches, so we borrow a foot • Now we have 11 feet and 20 inches • 20 - 10 = 10 inches • So we end up with 11 feet 10 inches Again, the calculator does all this hard work for you! Converting Between Feet and Inches Sometimes you might need to change feet to inches or the other way around. Here's how: To convert feet to inches: Multiply the number of feet by 12 Example: 5 feet = 5 × 12 = 60 inches To convert inches to feet: Divide the number of inches by 12 Example: 30 inches = 30 ÷ 12 = 2 feet 6 inches (because 6 is left over) Try this: How many inches are in 4 feet 7 inches? 4 feet = 4 × 12 = 48 inches 48 inches + 7 inches = 55 inches Comparing Feet and Inches Sometimes you need to know which measurement is bigger. The calculator can help with this too! It will convert everything to inches to compare. For example: Which is taller, 5 feet 11 inches or 6 feet 0 inches? 5 feet 11 inches = (5 × 12) + 11 = 71 inches 6 feet 0 inches = 6 × 12 = 72 inches So 6 feet 0 inches is taller by 1 inch! Feet and Inches in Construction Builders and carpenters use feet and inches all the time. They might use the calculator to: • Figure out how much wood they need for a project • Measure rooms for flooring • Make sure furniture will fit through doors For example, if a builder is making a fence that's 50 feet long with posts every 8 feet, they could use the calculator to figure out how many posts they need: Fence length: 50 feet Space between posts: 8 feet Number of spaces: 50 ÷ 8 = 6.25 Number of posts needed: 7 (we round up to make sure the fence is secure) Other Similar Calculators Check out other calculators that are similar to this one. FAQs (Frequently Asked Questions) Why do we use feet and inches? Feet and inches are part of the Imperial system of measurement. They're commonly used in the United States and a few other countries. Many people find them easy to estimate with their bodies (a foot is about the length of a foot, an inch is about the width of a thumb). How many inches are in a foot? There are 12 inches in a foot. This is why we "carry over" to feet when we get to 12 inches in our calculations. Can I use this calculator for other measurements? This specific calculator is designed for feet and inches. For other measurements like meters or centimeters, you'd need a different calculator. What if I get a result with more than 12 inches? If your result has more than 12 inches, you should convert those extra inches to feet. For example, if you get 5 feet 14 inches, you would convert that to 6 feet 2 inches. What's the proper way to write feet and inches? For feet, you have two options: • Use the abbreviation "ft" (e.g., 4 ft) • Use a single apostrophe (e.g., 4') For inches, you can: • Use the abbreviation "in" (e.g., 10 in) • Use a double apostrophe (e.g., 10") So, 5 feet 3 inches could be written as 5 ft 3 in, or 5' 3". How do I convert between feet and inches? Remember: 1 foot = 12 inches To convert feet to inches: Multiply the number of feet by 12 Example: 2 feet = 2 × 12 = 24 inches To convert inches to feet: Divide the number of inches by 12 Example: 30 inches = 30 ÷ 12 = 2 feet 6 inches (or 2.5 feet) How can I express a large number of inches in feet and inches? Let's use 62 inches as an example: 1. Divide by 12: 62 ÷ 12 = 5 remainder 2 2. The whole number (5) is the number of feet 3. The remainder (2) is the number of inches So, 62 inches = 5 feet 2 inches (or 5' 2") How do I calculate area when dealing with feet and inches? When calculating area, it's often easiest to convert everything to inches first. Here's an example: Let's find the area of a rectangle that's 4 feet 6 inches long and 3 feet wide. 1. Convert all measurements to inches: □ 4 feet 6 inches = (4 × 12) + 6 = 54 inches □ 3 feet = 3 × 12 = 36 inches 2. Multiply the length by the width: 54 × 36 = 1,944 square inches 3. If needed, convert back to square feet: 1,944 ÷ 144 = 13.5 square feet The area is 1,944 square inches or 13.5 square feet. Why do we use feet and inches? Feet and inches are part of the Imperial system of measurement. They're commonly used in the United States and a few other countries. Many people find them easy to estimate with their bodies (a foot is about the length of a foot, an inch is about the width of a thumb). Can I use this calculator for other measurements? This specific calculator is designed for feet and inches. For other measurements like meters or centimeters, you'd need a different calculator. Even with a calculator, sometimes things can go wrong. Here are some common errors to watch out for: • Typing mistakes: Always double-check the numbers you enter • Forgetting to carry over: Remember, when you have 12 or more inches, that equals an extra foot • Mixing up feet and inches: Make sure you're putting the right numbers in the right places To check your work, you can try doing the calculation backwards. For example, if you added two lengths, try subtracting one from your total to see if you get the other length back.	{"url":"https://thatcalculator.com/calculators/feet-inches/","timestamp":"2024-11-08T21:50:22Z","content_type":"text/html","content_length":"31228","record_id":"<urn:uuid:12fb0745-d21b-41c9-98ed-106cfc07ed14>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00245.warc.gz"}
[molpro-user] Ask help on molpro calculation as a new user Yin,Shi Shi.Yin at colostate.edu Wed Dec 7 17:32:47 CET 2016 Dear Molpro folks, I am, Shi Yin, working in Prof. Elliot Bernstein group, Chemistry department at Colorado State University. Recently, I start to use molpro to do calculation on iron sulfur negative ion clusters. I am trying to do calculations of their theoretical first vertical detachment energies (VDE = E neutral at optimized anion geometry - E optimized anion). I start with FeS- as a test firstly. The following is my input file: Fe -0.33994333 0.04249292 0.00000000 S 1.83005667 0.04249292 0.00000000 optg !geometry optimization of quartet anion FeS- using hf wf,spin=2,charge=0} !single point energy of triplet neutral FeS with above optimized geometry of quartet anion FeS- using hf wf,spin=4,charge=0} !single point energy of quintet neutral FeS with above optimized geometry of quartet anion FeS- using hf The result I obtained is: UHF-SCF UHF-SCF OPTG(UHF) UHF-SCF -1659.77314480 -1659.70612062 -1659.86274040 -1659.85196075 Thus, I can obtain the VDE = -1659.77314480 - (-1659.86274040) = 0.0895956 Hartree = 2.438 eV. And then, I did the same thing using B3lyp and TZVP. The input: Fe -0.33994333 0.04249292 0.00000000 S 1.83005667 0.04249292 0.00000000 The results: UKS-SCF UKS-SCF OPTG(UKS) UKS-SCF -1661.78663123 -1661.76503538 -1661.76845789 -1661.76799811 Thus, I can obtain the VDE = -1661.78663123 - (-1661.76845789) = -0.01817334 Hartree = -0.494 eV. Because I used Gaussian previously, I compared obtained results with Gaussian, and experimental results, as listed in the following table. Calculated VDE of quartet anion FeS- (ground state) 2.438 eV -0.494 eV 0.83 eV 1.43 eV Experimental VDE of anion FeS- 1.85 eV I am confused, because the results obtained by Molpro are very different with that obtained by Gaussian, and the Gaussian results seems more reasonable comparing with the experimental result. Since I am a new user of Molpro, I am not sure my Molpro calculations are right or not. Could someone help me to check my input file to see if there is any problem or mistake? Or please feel free to let me know any comments and suggestions to understand the above results listed in the table. I will appreciate your reply and help very much! Best regards, Shi Yin Shi Yin Web site: http://sites.chem.colostate.edu/bernsteinlab/SY.htm Department of Chemistry Colorado State University Fort Collins, CO 80523 Phone Numbers: Office 970-491-5741; Lab -5787 -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://www.molpro.net/pipermail/molpro-user/attachments/20161207/49e178e7/attachment.html> More information about the Molpro-user mailing list	{"url":"https://www.molpro.net/pipermail/molpro-user/2016-December/006828.html","timestamp":"2024-11-02T12:49:41Z","content_type":"text/html","content_length":"6055","record_id":"<urn:uuid:bf9bc282-fc18-416e-af8c-4f3e7a9991e4>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00637.warc.gz"}
Tutorial 3 - Scaling, Inverse Kinematics, and Inverse Dynamics The tutorial below is designed for use with OpenSim version 4.0 and later. A version of the tutorial compatible with OpenSim version 3.3 is available here. I. Objectives The purpose of this tutorial is to demonstrate how OpenSim solves an inverse kinematics and dynamics problem using experimental data. To diagnose movement disorders and study human movement, biomechanists frequently ask human subjects to perform movements in a motion capture laboratory and use computational tools to analyze these movements. A common step in analyzing a movement is to compute the joint angles and joint moments of the subject during movement. OpenSim has tools for computing these quantities: 1. Inverse kinematics is used to compute joint angles. 2. Inverse dynamics is used to compute net joint reaction forces and net joint moments. Inverse kinematics computes the joint angles for a musculoskeletal model that best reproduce the motion of a subject. Inverse dynamics then use joint angles, angular velocities, and angular accelerations of the model, together with the experimental ground reaction forces and moments, to solve for the net reaction forces and net moments at each of the joints. The schematic below shows an overview of the inverse kinematics and inverse dynamics problems. In this tutorial, you will: • Become familiar with OpenSim's Scale, Inverse Kinematics and Inverse Dynamics tools • Solve an inverse kinematics and inverse dynamics problem using experimental data • Interpret the results of the inverse dynamics solution • Investigate the dynamic inconsistencies that arise during inverse dynamics Each section of the tutorial guides you in using certain tools within and asks you to answer a few questions. The menu titles and option names you must select and any commands you must type to run OpenSim will appear in boldface. The questions can be answered based on information from OpenSim and basic knowledge of the human musculoskeletal system. After you complete the tutorial, feel free to explore OpenSim and the other analysis tools further on your own. Depending on the amount of exploration you do, this tutorial should take 1-2 hours to complete. II. Generic Musculoskeletal Model In this tutorial, you will be using a generic musculoskeletal model with 23 degrees of freedom and actuated by 54 muscles entitled 3DGaitModel2354. Note: Detailed information about the model can found on the Gait2392 and 2354 Models page To load the generic musculoskeletal model into OpenSim: • Click the File menu and select Open Model. • Find the Gait2354_Simbody folder in your default OpenSim resources directory— \Documents\OpenSim\Models for PC and Mac. Note: When you first launch OpenSim, you are prompted to provide a path to install the resources folder, the default is in your systems Documents folders. • Open the Gait2354_Simbody folder, select the file gait2354_simbody.osim, and click Open. III. Scaling A Musculoskeletal Model Subject-specific modeling involves (i) scaling a generic musculoskeletal model to modify the anthropometry, or physical dimensions, of the generic model so that it matches the anthropometry of a particular subject and (ii) registering the markers placed on the model to match the locations on the subject. Scaling and Registration are the most important steps in solving inverse kinematics and inverse dynamics problems because IK and ID solutions are sensitive to the accuracy of the scaling and registration. To scale the generic model and register the markers: • Click the Tools menu and select Scale Model. • At the bottom of the Scale Tool dialog, click Load to input a settings file. • In the file browser, ensure that you are in the Gait2354 folder, select the file subject01_Setup_Scale.xml and click Open. This Scale Setup file is an xml file that contains pre-configured settings to scale and register the generic gait2354 musculoskeletal model to the dimensions of a subject that we have experimental data for. A detailed explanation of the Scale Tool can be found on the Scaling page of the documentation. Model Scaling In OpenSim, the scaling step adjusts both the dimensions of the body segments, as well as the mass properties (mass and inertia tensor). Scaling can be performed using a combination of two methods: (1) Manual Scaling: Scaling that allows the user to scale a segment based on some predetermined scale factor. Manual scaling is sometimes necessary when suitable data are not available, or if the scale factors were determined using an alternative algorithm. (2) Measurement-based Scaling: Scaling that determines scale factors for a body segment by comparing distance measurements between specified landmarks on the model, known as model markers, and the corresponding experimental marker positions. Marker Registration In OpenSim, the registration step adjusts the location of model markers to match the location of markers on the subject. To do this, you must first estimate a pose for the model that closely resembles the pose of the subject during the experimental static trial. To complete the scale step: • In the Scale Tool dialog, click Run. • When complete, a new, scaled and registered model entitled subject01 will appear in Visualizer window. Notice the pink model markers around the new model. • To save the scaled model, either click File and select Save Model, or right-click on the model name, subject01, in the Navigator window, and select Save As. • Save the scaled model as gait2354_scaled.osim, and click Save. Note: ensure that you are in the Gait2354 folder. • Once you have answered Questions 1-5, below, close the Scale Tool Dialog by clicking Close. At this point, you may close the generic model (right-click the model name in the Navigator window, and select Close) or hide the model (right-click the model name, and select Display -> Hide). 1. Based on information in the Scale Tool dialog, what is the mass of the generic musculoskeletal model? What was the mass of the subject? 2. What frequency was the experimental motion data captured? Hint: Look for the box titled Marker Data. 3. Click on the Scale Factors tab. Which body segments were scaled manually? IV. Inverse Kinematics Kinematics is the study of motion without considering the forces and moments that produce that motion. The purpose of inverse kinematics (IK) is to estimate the joint angles of a particular subject from experimental data. In this section, you will estimate a subject's joint angles during walking by performing an IK analysis using the subject scaled model and experimentally collected walking For each time step of recorded motion data, IK computes a set of joint angles that put the model in a configuration that "best matches" the experimental kinematics. OpenSim determines this "best match" by solving a weighted least squares optimization problem with the goal of minimizing marker error. Marker error is defined as the distance between an experimental marker and the corresponding model marker. Each marker has an associated weighting value, specifying how strongly that marker's error term should be minimized in the least squares problem. For each time step, the inverse kinematics tool solves for a vector of generalized coordinates (e.g., joint angles), q, that minimizes the weighted sum of marker errors, which is expressed as where q is the vector of generalized coordinates (e.g., joint angles), x[i]^exp is the position of experimental marker i, x[i](q) is the position of the corresponding model marker i (which depends on q), and w[i] is the weight associated with marker i. To set up an inverse kinematics analysis: • Open the Inverse Kinematics Tool dialog window by clicking the Tools menu and selecting Inverse Kinematics. • Load an inverse kinematics tool setup file by clicking Load, selecting the file subject01_Setup_IK.xml, and clicking Open. Note: In the file browser, ensure that you are in the Gait2354_Simbody folder, subject01_Setup_IK.xml contains pre-configured settings for the inverse kinematics tool. Notice the text boxes in the dialog window are now filled with values. A detailed explanation of the Inverse Kinematics Tool can be found on the Inverse Kinematics page of the documentation. Navigate to the Weights tab. • View which markers are selected for use in the inverse kinematics analysis, and their corresponding weights. • Enable the tracking for the marker R.Knee.Lat. Notice the row turns red and the Run button is now greyed out. You will be unable to run the inverse kinematics tool because there is no experimental data found for the R.Knee.Lat marker in the subject_walk1.trc file. Disable the R.Knee.Lat marker and notice the Run button is now clickable. To perform inverse kinematics: • Click Run. The model will begin to move as the inverse kinematics problem is being solved for each frame of the experimental data. • Notice the progress bar in the lower right-hand corner of the program. Wait until the bar disappears before proceeding. Note: Closing the inverse kinematics tool dialog during the analysis doesn't affect the Inverse Kinematics tool running. • To visualize the inverse kinematics solution, animate the model by using the motion slider and video controls. The model should walk through one full gait cycle. Note: You can loopand control the speed of the animation. • The inverse kinematics solution is saved to subject01_walk1_ik.mot, as specified in the setup file. Note: Be sure to use the exact file name given, as this file is used later. • To compare experimental marker data with inverse kinematics results, in the Navigator panel, go to Motions and right-click on IKResults (which are what the Inverse Kinematics Tool just generated). Then choose Associate Motion Data... from the drop-down menu. Choose subject01_walk1.trc and click Open. Model markers are shown in pink and experimental markers are shown in blue. Hit play in the Motion Toolbar. The virtual markers should correspond closely to the experimental marker locations as the animation proceeds. • Click the Window menu and select Messages. The Messages window records details of all steps you have performed. Take a minute to explore the Messages window. Then, scroll to the very bottom. The line above InverseKinematicsTool completed... provides the markers errors and model coordinate errors (e.g., joint angle errors) associated with the last frame of the motion. Note: All marker errors have units in meters, and all coordinate errors have units in radians. • Once you have answered Questions 4-6, below, close the Inverse Kinematics Tool Dialog by clicking Close. 4. In the Inverse Kinematics Tool dialog window, click the Weights tab and scroll through the list of markers in the top half of the weights tab. Which markers have weighting values less than one? Hint: Think about joints that have not been modeled. 5. Based on information in the Messages window, what is the root-mean-squared (RMS) error of all the markers in the last frame of the motion? Include units. Does this seem reasonable? Explain. 6. What was the value of the maximum marker error in the last frame? Include units. Which marker had this maximum error, and why? Hint: Think about the weighted least squares problem. V. Inverse Dynamics Dynamics is the study of motion and the forces and moments that produce that motion. To perform inverse dynamics, estimation of mass and inertia is required. The purpose of inverse dynamics is to estimate the forces and moments that cause a particular motion, and its results can be used to infer how muscles are utilized in that motion. To determine these forces and moments, equations of motion for the system are solved iteratively [3]. The equations of motion are derived using the kinematic description and mass properties of a musculoskeletal model. Then, using the joint angles from inverse kinematics and experimental ground reaction force data, the net reaction forces and net moments at each of the joints are calculated such that the dynamic equilibrium conditions and boundary conditions are satisfied [3]. To setup an inverse dynamics analysis: • Open the inverse dynamics tool dialog window by clicking the Tools menu and selecting Inverse Dynamics. • Load an inverse dynamics tool setup file by clicking Load, selecting file subject01_Setup_InverseDynamics.xml, and clicking Open. Note: If the Motion From File textbox appears red, this means the textbox was filled with an inappropriate file name. Make sure the motion file was saved with the correct file name in the Inverse Kinematics section. • Note the folder listed in the Directory textbox, located in the Output section of the dialog. The storage file containing the inverse dynamics results will be saved in this folder: Documents\ A detailed explanation of the Inverse Dynamics Tool can be found on the Inverse Dynamics page of the documentation. To perform inverse dynamics: • Click Run. • Notice the progress bar in the lower right-hand corner of the program. Wait until the bar disappears before proceeding. Note: Closing the inverse dynamics tool dialog during the analysis doesn't affect the Inverse Dynamics tool running. • To visualize the inverse dynamics solution, animate the model by using the motion slider and video controls. The model should walk through one full gait cycle. Note: You can loopand control the speed of the animation. • It is often useful to view the ground reaction forces with the inverse dynamics results. in the Navigator panel, go to Motions and right-click on IDResults (which are what the Inverse Dynamics Tool). Then choose Associate Motion Data... from the drop-down menu. Choose subject01_walk1_grf.mot and click Open. Green arrows are now shown that represent ground reaction force vectors collected from a force plate. • Close the inverse dynamics tool dialog window. When completed, examine the results of the inverse dynamics solution by plotting the net moments at the left and right ankles: • Click Tools and select Plot. • In the Plotter window, click the Y-Quantity button and select Load File. • In the file browser, go to the ResultsInverseDynamics folder, select the file inverse_dynamics.sto, and click Open. • In the menu, select ankle_angle_r_moment and ankle_angle_l_moment by clicking the corresponding checkboxes, then click OK. Note: To quickly find these quantities, type ankle into the pattern text box. • Click the X-Quantity button, select time, and click OK. • Back in the Plotter window, click Add to add the moment curves to the plot. • Print your plot by right-clicking on the plot and selecting Print. Note: To export the plot as an image by right-clicking the plot and selecting Export Image. • After printing the plot and answering questions 7-8, Close the Plotter and inverse dynamics dialog window In solving the inverse dynamics problem, both kinematic data and force plate data were used, making this an over-determined problem. In other words, the problem has more equations than unknowns (i.e., degrees of freedom). Due to errors in the experimental motion data and inaccuracies in the musculoskeletal model, it turns out that Newton's second law is violated, or [3]. One method to handle this inconsistency is to compute and apply residual forces and moments to a particular body segment in the model, such that Newton's second law becomes: † An analogous equation relates the ground reaction moment, to the residual moment, . In this musculoskeletal model, the residuals are applied to the pelvis segment. To see the residuals from the inverse dynamics solution, in a new plot window, Plot pelvis_tx_force, pelvis_ty_force, and pelvis_tz_force versus time. Using this plot, answer question 9. While applying residual forces and moments makes the model's motion dynamically consistent with the external forces , this strategy is undesirable because the residuals can be large. More advanced strategies have been developed to deal with the problem of residuals and dynamic inconsistencies, such as least-squares optimization [3], the Residual Elimination Algorithm (REA) [5], and the Residual Reduction Algorithm (RRA) [6]. OpenSim implements a Residual Reduction Algorithm as part of its workflow for generating muscle-actuated simulations [6]. A detailed explanation of the Residual Reduction Algorithm (RRA) can be found on the Residual Reduction Algorithm page of the documentation. For additional information on these strategies, please also see [3], [5], [6], and [7]. 7. On your plot of the ankle moments, identify when heel strike, stance phase, toe off, and swing phase occur for each curve (i.e., left leg and right leg). 8. Based on your plot and the angle convention for the ankle, give an explanation of what is happening at the ankle just before toe-off. Hint: It may be useful to use the Coordinate sliders to understand the angle convention for the ankle. 9. What are the maximum magnitudes of the residual forces? Using the mass of the subject from Question 1, what fraction of body weight are the maximum residual forces? The experimental gait data were collected by Jill Higginson and Chand John in the Neuromuscular Biomechanics Lab at the University of Delaware [8]. The data include marker trajectories and ground reaction forces for an adult male walking at a self-selected speed on an instrumented split-belt treadmill. Please note that the data distributed with OpenSim is from a different subject than the one described in the paper. Data collection protocols were the same for both subjects. 1. Delp, S.L., Loan, J.P., Hoy, M.G., Zajac, F.E., Topp E.L., Rosen, J.M. An interactive graphics-based model of the lower extremity to study orthopaedic surgical procedures. IEEE Transactions on Biomedical Engineering, vol. 37, pp. 757-767, 1990. 2. Anderson, F.C., Pandy, M.G. A dynamic optimization solution for vertical jumping in three dimensions. Computer Methods in Biomechanical and Biomedical Engineering, vol. 2, pp. 201-231, 1999. 3. Kuo, A.D. A least squares estimation approach to improving the precision of inverse dynamics computations, Journal of Biomechanical Engineering, vol. 120, pp. 148-159, 1998. 4. Winter, D.A. Biomechanics and Motor Control of Human Movement, Wiley and Sons, pp. 77-79, 1990. 5. Thelen, D.G., Anderson, F.C. Using computed muscle control to generate forward dynamic simulations of human walking from experimental data, Journal of Biomechanics, vol. 39, pp. 1107-1115, 2006. 6. John, C.T., Anderson, F.C., Guendelman, E., Arnold, A.S., Delp, S.L. An algorithm for generating muscle-actuated simulations of long-duration movements, Biomedical Computation at Stanford (BCATS) Symposium, Stanford University, 21 October 2006, Poster Presentation. 7. Delp, S.L., Anderson, F.C., Arnold, A.S., Loan, P., Habib, A., John, C.T., Guendelman, E., Thelen, D.G. OpenSim: Open-source software to create and analyze dynamic simulations of movement. IEEE Transactions on Biomedical Engineering, vol. 55, pp. 1940-1950, 2007. 8. Chand T. John, Frank C. Anderson, Jill S. Higginson & Scott L. Delp (2012): Stabilisation of walking by intrinsic muscle properties revealed in a three-dimensional muscle-driven simulation, Computer Methods in Biomechanics and Biomedical Engineering. OpenSim is supported by the Mobilize Center , an NIH Biomedical Technology Resource Center (grant P41 EB027060); the Restore Center , an NIH-funded Medical Rehabilitation Research Resource Network Center (grant P2C HD101913); and the Wu Tsai Human Performance Alliance through the Joe and Clara Tsai Foundation. See the People page for a list of the many people who have contributed to the OpenSim project over the years. ©2010-2024 OpenSim. All rights reserved.	{"url":"https://opensimconfluence.atlassian.net/wiki/spaces/OpenSim/pages/53089741/Tutorial+3+-+Scaling+Inverse+Kinematics+and+Inverse+Dynamics?atl_f=content-tree","timestamp":"2024-11-04T14:33:45Z","content_type":"text/html","content_length":"1029191","record_id":"<urn:uuid:ed47516d-c339-4058-bc68-1e1c5fa8bc74>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00265.warc.gz"}
Program for Wednesday, September 18th previous day next day all days View: session overviewtalk overview 09:00-09:50 Session 17A: SYNASC Invited talk Multi-objective sequence learning for Chemistry and Computer games 09:00 ABSTRACT. Monte Carlo Tree Search is a popular method for dealing with complex, highly branched tree search problems that represent sequences, be it steps to win a game or reactions to make a molecule. However, nearly all available algorithm variants deal with one objective only. But what if we have multiple objectives? Up to now, there are very few methods for this and I report on our attempts to use these methods for retrosynthesis (find ways how to make a specific target molecule according to several criteria) and also in game AI. It seems clear that this is a research area with high potential but little activity as of now. 09:00-09:50 Session 17B: FROM Invited talk Symbolic Computation in Automated Program Reasoning ABSTRACT. We describe applications of symbolic computation towards automating the formal analysis of while-programs implementing polynomial arithmetic. We combine methods from static analysis, 09:00 symbolic summation and computer algebra to derive polynomial loop invariants, yielding a finite representation of all polynomial equations that are valid before and after each loop execution. While deriving polynomial invariants is in general undecidable, we identify classes of loops for which we automatically can solve the problem of invariant synthesis. We further generalize our work to the analysis of probabilistic program loops. Doing so, we compute higher-order statistical moments over (random) program variables, inferring this way quantitative invariants of probabilistic program loops. Our results yield computer-aided solutions in support of formal software verification, compiler optimization, and probabilistic reasoning. 10:10-11:00 Session 18A: SYNASC Invited talk An introduction to Gaussian processes applied to Bayesian regression 10:10 ABSTRACT. The talk will address the Gaussian Process Theory and its application to Global Illumination, rendering, regression, etc. Regression can be useful in many applications. For example it allows to perform a precise regression for a BRDF or an environment map for which the sampled data (incident radiance, incident-reflected radiances) are not the most significant. Note that the Gaussian Process theory can be applied to different research fields such as crowd simulation, computer vision, etc. 10:10-11:10 Session 18B: FROM Symposium Intuitionistic Propositional Logic in Lean 10:10 ABSTRACT. In this paper we present a formalization of Intuitionistic Propositional Logic in the Lean proof assistant. Our approach focuses on verifying two completeness proofs for the studied logical system, as well as exploring the relation between the two analyzed semantical paradigms - Kripke and algebraic. In addition, we prove a large number of theorems and derived deduction Unification in Matching Logic — Revisited ABSTRACT. Matching logic is a logical framework for specifying and reasoning about programs using pattern matching semantics. A pattern is made up of a number of structural components and 10:30 constraints. Structural components are syntactically matched, while constraints need to be satisfied. Having multiple structural patterns poses a practical problem as it requires multiple matching operations. This is easily remedied by unification, for which an algorithm has already been defined and proven correct in a sorted, polyadic variant of matching logic. This paper revisits the subject in the applicative variant of the language while generalising the unification problem and mechanizing a proven-sound solution in Coq. Optics, functorially: Extended abstract ABSTRACT. Optics are valuable categorical constructions encapsulating the general notion of bidirectional data accessors, and extensive research has been dedicated to comprehending these 10:50 structures in recent years. Profunctor representation is a successful framework for encompassing various types of optics, such as lenses, prisms or traversals, starting from a pair of actions of a monoidal category. We enhance the above framework as to include structure-preserving functors of the two monoidal actions involved. This has the advantage of unifying within a single 2-functorial framework not only optics (which usually are treated as arrows in categories), but also morphisms between them, and makes it convenient to manipulate and to reason about them in a uniform manner. From a more practical perspective, there is potential in developing better suited applications like complex data access schemes. 11:20-13:00 Session 19A: Workshop SegWEDA AR and VR for the Casa Romei Museum 11:20 ABSTRACT. The temporary exhibition "Across Your Senses" at the Casa Romei Museum in Ferrara offered, through a multisensory approach, a new visiting experience not only for the general tourist but also for those who are already familiar with and frequent this monumental building and its assets. This text describes some installations and activities designed in relation to virtual reality and augmented reality. A SERIOUS GAME TO LEARN DORIC GREEK ARCHITECTURE ABSTRACT. The present contribution aims to describe the development of a serious game designed to disseminate knowledge in the field of architecture and archeology. In recent decades, 11:40 educational video games, or serious games, have emerged as a significant means of engaging students across various educational disciplines. Notably, some of these games have shown considerable potential in conveying both the tangible and intangible aspects of cultural heritage, particularly within the teaching of historical subjects. The case study under consideration involves the initial phase of a game focused on the reconstruction of one corner of the Zeus Temple in the Archeological Park of Agrigento. Following an evaluative test will be conducted with high school and first year university students, in order to gather data generated by its use. Level Generation Using ChatGPT: A Case Study on the Science Birds Game 12:00 ABSTRACT. The increasing complexity of video games and development costs have necessitated innovative approaches to content creation. Procedural Content Generation (PCG) and AI tools like ChatGPT offer promising solutions. This paper explores the application of ChatGPT in Procedural Level Generation (PLG) for the game "Science Birds". Through a combination of theoretical exploration and practical experiments, we demonstrate how ChatGPT can be leveraged to automate level design, enhancing both efficiency and creativity in game development. Graphic design for architectural serious games 12:20 ABSTRACT. As part of the PhD program in History, Representation, and Restoration of Architecture, a serious game for children based on Palazzo Barberini in Rome has been designed, focusing on its graphic component. This study highlights the role of architects in the gamification of cultural heritage, ensuring accurate architectural representation and enhancing communication. Fostering Agentic Play Between Technology and Democracy 12:40 ABSTRACT. The Knowledge Technologies for Democracy (KT4D) project seeks to produce new mechanisms to support the protection of human agency in the context of tensions between democracy and emerging technologies. Addressing this challenge from the perspectives of education, regulation and innovation, the project has chosen to deliver a number of its interventions through the paradigm of serious games. The place of agency in gaming is contested, however, leading the project to weave multiple mechanisms to foster agency as a collective process across its approaches to critical digital literacy, narrative and co-creation. 11:20-12:40 Session 19B: Track: Artificial Intelligence (2) SUDS: A Simplified UNet with Depth-wise Separable Convolutions ABSTRACT. Medical image segmentation is one of the most developed part of image segmentation and it plays a crucial role in computer-aided diagnosis. U-Net paved the way for a series of 11:20 variants that took advantage of the key features of the network. In this article, different features proposed in the variants of U-Net are adapted and experimented on to create a new architecture that still leaves the idea of a U-shape structure unchanged. The proposed architecture takes advantage of the efficient depth-wise separable convolution, but with a twist. Instead of using the pointwise convolution as the last step in the depth-wise separable convolution, it uses the so called Ghost Module. This results in a very efficient network with a reduced complexity, while still having a great segmentation performance. We compared SUDS with U-Net and its variants across multiple segmentation tasks: skin lesion segmentation and colonscopy segmentation. Experiments demonstrate that SUDS has similar segmentation accuracy compared to U-Net and its variants, while the parameters and floating-point operations are greatly reduced. Ideal Centroid Striving: An unsupervised and prediction parameterized anomaly detection method ABSTRACT. Automated machines are widely used in industrial environments for the production of different items and generate lots of data following these processes. The proper execution of the machines influences the production output, thus the detection of anomalous behavior in machines' activity using the generated data must be considered to avoid unpleasant outcomes. This paper 11:40 presents an ensemble unsupervised anomaly detection method characterized by parameterized prediction. The proposed method consists of 2 stages - the first stage uses statistical-based methods to assign artificial labels to the input data. In the second stage, the artificially labeled instances and a feature bagging technique are employed to construct the model's estimators - each estimator calculates the percentiles of the distances computed between its centroid and each instance from the training subset. The model's prediction function is parameterized by a percentile rank. Each estimator computes the distance between its centroid and the evaluated instance: if this distance exceeds the specified percentile value, the estimator classifies the instance as anomaly. A majority vote is applied to determine the final outcome. Class-Incremental Learning Enhanced: Assessing New Exemplar Selection Strategies in the iCaRL Framework ABSTRACT. The field of machine learning has increasingly focused on incremental learning, enabling systems to continually adapt and improve by integrating new knowledge while retaining previously learned information. One particular area of interest is class-incremental learning, where the learning system sequentially acquires new classes without access to or with limited exposure to past data. The primary challenge in class-incremental learning is catastrophic forgetting, wherein the model tends to overlook previously learned classes when confronted with new 12:00 tasks. One of the class-incremental learning approaches to mitigate catastrophic forgetting is the Incremental Classifier and Representation Learning (iCaRL) framework. In this paper, we propose three new selection criteria for the iCaRL approach. Our best selection criterion, which uses the K-Means clustering algorithm to create diverse groups and then selects exemplars close to the centroids of the clusters, outperforms the original iCaRL selection criterion by over 16% for the MNIST dataset and by over 12% for the FashionMNIST dataset in terms of average accuracy. The full implementation of the iCaRL approach, along with the three proposed selection criteria and detailed experimental results logs, can be found in our publicly available GitHub repository. By contributing to the ongoing development of class-incremental learning techniques, we aim to support the creation of more effective and robust lifelong machine learning systems. MACE: Malware Analysis and Clustering Engine ABSTRACT. Analysing malicious samples is a lengthy task, and, given a set of hundreds, even thousands of files, a security researcher would have to spend multiple hours looking at each sample. In many cases, this is not a feasible solution, considering that new malware is created every day. The flow of such files will not stop, and therefore a way of greatly reducing analysis time is 12:20 needed. This paper presents the Malware Analysis and Clustering Engine (MACE), a proposed solution to address this problem. It offers the user a modular framework implementing multiple feature extraction methods and clustering algorithms, which can provide a multitude of clustering configurations. The user can experiment with different configurations and choose one that best suits their scenario, obtaining a clustering result that allows them to only analyse a single sample from each cluster and draw a conclusion both for it and all of its peers. This approach significantly accelerates the analysis process. The engine was tested on four different datasets, each trying to exemplify a scenario that is likely to be encountered when using the tool in real situations. Based on the results, it was concluded that MACE does achieve its goal, providing multiple configurations for each dataset that reach both great accuracies and a number of created clusters that is close to the one that was expected. 14:00-15:30 Session 21A: SYNASC & FROM Tutorial Induction in Saturation-based Proving ABSTRACT. Induction in saturation-based first-order theorem proving is a new exciting direction in automated reasoning, bringing together inductive reasoning and reasoning with full first-order logic extended with theories. In this tutorial, we dive into our recent results in this area. Traditional approaches to inductive theorem proving, based on goal-subgoal reduction, are incompatible with saturation algorithms where the search space can simultaneously contain hundreds of thousands of formulas, with no clear notion of a goal. Rather, our approach applies induction by theory lemma generation: from time to time we add to the search space instances of induction axioms, which are valid in the underlying theory but not valid in first-order predicate logic. To formalize this, we introduce new inference rules adding (clausal forms of) such induction axioms within saturation. Applications of these rules are triggered by recognition of formulas in the search space that can be considered as goals solvable by induction. We also propose additional reasoning methods for strengthening inductive reasoning, as well as for handling recursive function definitions. We implemented our work in the Vampire theorem prover 14:00 and will demonstrate the practical impact in experiments. The tutorial will consist of the following parts supported by live demonstrations: Introduction to saturation-based reasoning and superposition Integration of induction into saturation [1] Case studies: integer induction and recursive definitions [2, 3] How far can we go with induction in saturation? Future outlooks: using automated inductive proving in proof assistants and beyond The tutorial is based on work made with several co-authors and the tutorial content was jointly created with Petra Hozzova. References: [1] Induction in Saturation-Based Proof Search (2019), G. Reger and A. Voronkov, in Proc. of CADE https://doi.org/10.1007/978-3-030-29436-6_28 [2] Integer Induction in Saturation (2021), P. Hozzová, L. Kovács, and A. Voronkov, in Proc. of CADE https://doi.org/10.1007/978-3-030-79876-5_21 [3] Induction with Recursive Definitions in Superposition (2021), M. Hajdu, P. Hozzová, L. Kovács, and A. Voronkov, in Proc. of FMCAD https://doi.org/10.34727/2021/isbn.978-3-85448-046-4_34 For a survey, also see: Getting Saturated with Induction (2022), M. Hajdu, P. Hozzová, L. Kovács, G. Reger, and A. Voronkov, in Principles of Systems Design https://doi.org/10.1007/978-3-031-22337-2_15 14:00-15:20 Session 21B: Workshop IAFP (1) Recent developments in the fixed point theory of enriched contractive mappings. A survey 14:00 ABSTRACT. The aim of this paper is threefold: first, we present a few relevant facts about the way in which the technique of enriching contractive mappings was introduced; secondly, we expose the main contributions in the area of enriched mappings established by the authors and their collaborators by using this technique; and third, we survey some related developments in the very recent literature which were authored by other researchers. Fixed points for Feng-Liu multi-valued operators with an application 14:20 ABSTRACT. In this talk we will use the notion of nonlinear multi-value Feng-Liu operator in order to prove some fixed point theorems in the context of a set endowed with two metrics. An application to an integral inclusion is also given. Maia’s fixed point theorems in a space with distance 14:40 ABSTRACT. We establish new fixed point theorems for Maia’s fixed point theorem in the setting of a space with a distance, more precisely when one of the metrics is replaced with a distance. We also present some examples to illustrate the theoretical results. SOME FIXED POINT THEOREMS IN THE FRAMEWORK OF f -METRIC SPACES ABSTRACT. This paper deals with the notion of f-metric space, a genuine generalization of the concept of b-metric space. We present Matkowski, Kannan and Chatterjea type fixed point theorems in the context of f-metric spaces. 15:50-17:10 Session 22A: Workshop IAFP (2) Fixed points and coupled fixed points with applications ABSTRACT. Let $(X,d)$ be a metric space, $P(X)$ be the set of all nonempty subsets of $X$ and $T:X\to P(X)$ be a a multi-valued operator. Then, the pair $(x^,y^)\in X\times X$ is called a 15:50 coupled fixed point for $T$ if $$\label{ecfp} \left\{\begin{array}{lll} x^\in T(x^,y^)\\ y^\in T(y^,x^). \end{array}\right.$$ It is easy to observe that $z^:=(x^,y^)$ is a coupled fixed point of $T$ if and only if $z^$ is a fixed point of the multi-valued operator $$F_T:X\times X\to P(X\times X), \ F_T(x,y)=T(x,y)\times T(y,x).$$ Exploiting this remark, in this paper we will give some existence and approximations results for the coupled fixed point problem. Several extensions and some open questions are also considered. Existence and Approximation of Fixed Points of Enriched Nonexpansive Operators using Krasnoselskii-Mann type Algorithms ABSTRACT. In this paper, we present some results about the aproximation of fixed points of enriched nonexpansive operators. There are numerous works in this regard (for example [3], [4], [5] 16:10 [6], [26], [30] and references to them). Of course, the bibliografical references are extensive and they are mentioned at the end of this paper. In order to approximate the fixed points of enriched nonexpansive mappings, we use the Krasnoselskii-Mann iterative algorithm and a modified Krasnoselskii-Mann iterative algorithm for which we prove weak and strong convergence theorems. Also, in this paper, we make a comparative study about some classical convergence theorems from the literature in the class of enriched nonexpansive mappings using the two algorithms. Fixed Points of b-Enriched Multivalued Nonexpansive Mappings and *-b-enriched nonexpansive mappings in Hilbert Spaces 16:30 ABSTRACT. The main purpose of this paper is to extend some results of fixed point theorems from the general class of nonexpansive mappings, introduced by Vasile Berinde, denoted b-enriched nonexpansive mappings to multivalued mappings in Hilbert space Common fixed point theorems for enriched contraction under the assumption of R-weakly commuting condition 16:50 ABSTRACT. Abstract. We introduce common fixed point theorems using the technique of enriched Banach contractions for single-valued mappings under the assumption of Rweakly commuting condition, R-weakly commuting of type (Ag) and R-weakly commuting of type (Af): We give examples for each theorem in turn to suport our results. We obtained related results of Pant [3], H. K. Pathak, Y. J. Cho and S. M. Kang [2] and V. Berinde and M. Pacurar [1]. 15:50-17:10 Session 22B: Track: Artificial Intelligence (3) A comparative analysis of Genetic Algorithms and NSGA-II on the Portfolio Optimisation Problem ABSTRACT. This paper contains a comparison between a Genetic Algorithm (GA) and a Non-dominated Sorting Genetic Algorithm II (NSGA-II) on the Portfolio Optimisation Problem, based on the Modern Portfolio Theory proposed by Markowitz (1952, 1956). 15:50 We use the real-world data of the S&P 500 index (quarterly returns of top 200 stocks from 2019 to 2024, and top 442 stocks from 2004 to 2016), and show that both algorithms can yield returns above the index, in mixed bull and bear market conditions. N-point Crossover accelerates algorithm convergence, and by using the Sharpe Ratio, the NSGA-II outperformed most models based on Stochastic Dominance we tested, according to the metrics in Bruni el. al. (2017). Assessing Features Importance in the 15-Class Galaxy Classification Problem ABSTRACT. This study explores feature selection for classifying galaxy morphology using the extensive Galaxy Zoo 2 dataset. We investigate supervised and unsupervised learning methods to group 16:10 galaxies based on key features, aiming to replicate supervised learning results. We evaluate various feature selection methods and compare them to an existing classification approach. Our results demonstrate that a reduced set of features based on adjusted vote fractions improves classification accuracy and potentially reduces computational complexity. While unsupervised clustering partially groups galaxies by morphology, further optimization is required. This work suggests that feature selection and unsupervised learning are promising techniques for the efficient classification of large galaxy datasets in upcoming astronomical surveys. Enhanced Anomaly Detection in Automotive Systems Using SAAD: Statistical Aggregated Anomaly Detection Dacian Goina ABSTRACT. This paper presents a novel anomaly detection methodology termed Statistical Aggregated Anomaly Detection (SAAD). The SAAD approach integrates advanced statistical techniques with machine learning, and its efficacy is demonstrated through validation on real sensor data from a Hardware-in-the-Loop (HIL) environment within the automotive domain. The key innovation of SAAD lies in its ability to significantly enhance the accuracy and robustness of anomaly detection when combined with Fully Connected Networks (FCNs) augmented by dropout layers. Comprehensive experimental evaluations indicate that the standalone statistical method achieves an accuracy of 72.1%, whereas the deep learning model alone attains an accuracy of 71.5%. In contrast, the aggregated method achieves a superior accuracy of 88.3% and an F1 score of 0.921, thereby outperforming the individual models. These results underscore the effectiveness of SAAD, demonstrating its potential for broad application in various domains, including automotive systems. 17:30-19:00 Session 23A: Tutorial Generalized multisets over infinite alphabets with atoms ABSTRACT. A multiset over X is a function from a set X to the set N of positive integers, indicating that each element of X has associated a positive multiplicity. This notion was generalized by introducing the hybrid sets which also allow negative multiplicities. Since the set of all integers are denoted by Z, the hybrid sets can be named Z-multisets. The set Z of integers is a group under the operation of addition. Starting from this observation, we introduce a generalization of the hybrid sets by defining the group-valued multisets. These multisets over a set X are functions from X to an arbitrary group G, ensuring an inverse for each multiplicity (not necessarily a number) of the elements of X (together with other features derived from the group properties). We denote them by G-multisets. In particular, Z-multisets are G-multisets; in general, Z can be replaced by any group G, and this aspect allows to get deeper relationships and correlations among the quantitative attributes (multiplicities) for elements of X, useful for various models and optimizations (e.g., in economy). For instance, whenever we use elements of X having certain (quantitative) attributes, we can precisely describe and use the elements of X having the inverse (quantitative) attributes. In our books [Springer 2016, Springer 2020], we studied the multisets allowing negative multiplicities both in the Zermelo-Fraenkel framework and in the finitely supported framework (where only finitely supported sets are allowed), analyzing the correspondence between some properties of these generalized multisets obtained in finitely supported framework and those obtained in the classical ZermeloFraenkel framework. Finitely supported sets are related to the permutation models of Zermelo-Fraenkel set theory with atoms. These models were introduced in 1930s by Fraenkel, Lindenbaum and Mostowski to prove the independence of the axiom of choice from the other axioms of Zermelo-Fraenkel set theory with atoms (ZFA). More recently, finitely supported sets have been developed in Zermelo-Fraenkel (ZF) set theory by equipping ZF sets with actions of a group of one-to-one and onto transformations of some basic elements called atoms. Sets with permutation actions were used to investigate the variables binding, renaming and choosing fresh names in the theory of programming since the notions of structural recursion and structural induction can be adequately transferred into this new framework, as well as in describing automata, languages and Turing machines that operate over infinite alphabets. The notions of invariant set and finitely supported structure are introduced and described in previous papers of the authors. An invariant set is defined as a usual ZF set endowed with a group action of the group of all one-to-one and onto transformations of certain fixed infinite ZF set A of basic elements (called atoms) satisfying a finite support requirement. This requirement states that any element in an invariant set has to be finitely supported, i.e. for any such element x there should exist a finite set of atoms S_x such that any permutation of atoms fixing S_x pointwise also leaves the element x invariant under the related group action. A finitely supported set is defined as a finitely supported element in the powerset of an invariant set. A finitely supported structure is defined as a finitely supported set equipped with a finitely supported internal algebraic law or relation (which should be finitely supported as a subset of a Cartesian product of finitely supported sets). The theory of finitely supported structures allows a discrete representation of possibly infinite sets containing enough symmetries to be concisely handled. More specifically, this theory allows us to treat as equivalent the 17:30 objects that have a certain degree of similarity and to focus only on those objects that are “really different” by involving the notion of finite support. The framework of finitely supported structures contains both the family of non-atomic ZF structures which are proved to be trivially invariant (i.e. all their elements are empty supported) and the family of atomic structures with f inite (possibly non-empty) supports. We have to analyze whether a ZF result preserves its validity when reformulating it by replacing ‘non-atomic ZF structure’ with ‘atomic and finitely supported structure’. The meta-theoretical technique for the translation of ZF results into the framework of atomic finitely supported structures is based on a closure property for finite supports in an hierarchical construction, called ‘S-finite support principle’ claiming that “for any finite set S of atoms, anything that can be defined in higher-order logic from structures supported by S, by using each time only constructions supported by S, is also supported by S”. The formal involvement of the related S-finite support principle implies a step-by-step construction of the support of a structure by using, at every step, the previously constructed supports of the substructures of the related structure. However, there are ZF results that cannot be translated into an atomic framework as we proved in [Springer 2020]. In this tutorial, by extending the results in [Springer 2016] and [Springer 2020], we define and investigate the group-valued multisets, also in the framework of finitely supported sets. By involving the theory of finitely supported sets, we are able to study group-valued multisets over infinite sets X in a finitary manner. We introduce finitely supported groups and provided some relevant examples of these structures. The finitely supported groups are finitely supported sets equipped with finitely supported internal group laws. We prove that the set of all finitely supported bijections of a finitely supported set, the set of all finitely supported automorphisms of a finitely supported group, the set of all finitely supported inner automorphisms of a finitely supported group, and the set of all equivalence classes of finite words with letters belonging to a finitely supported set, all are finitely supported groups. We prove an isomorphism theorem for finitely supported groups and a result showing that the finitely supported group of all finitary permutations of atoms coincide with the finitely supported group of all bijections of atoms. We also prove some counting properties for the supports of free groups. Then we introduce and study finitely supported group-valued multisets (called G-multisets). For each G-multiset we provide a relationship result between its support and its algebraic support (formed by the family of elements with a non-empty multiplicity). The set of all G-multisets on a finitely supported universe of discourse X can be organized as a finitely supported group and satisfies some (Cayley-type) embedding results, as well as isomorphism and universality theorems. We provide some examples of infinite finitely supported groups that are Dedekind finite (i.e. they do not contain infinite, finitely supported countable subsets). Finally, we connect the concepts of G-multiset, free group and hybrid set (Z-multiset with finite algebraic support) via some universality properties. [Springer 2016] A. Alexandru, G. Ciobanu. Finitely Supported Mathematics: An Introduction, Springer-Nature, 185 pages, 2016. [Springer 2020] A. Alexandru, G. Ciobanu. Foundations of Finitely Supported Structures: A set theoretical viewpoint, Springer-Nature, 210 pages, 2020. 17:30-18:50 Session 23B: Workshop IAFP (3) Picard Operators: retraction-displacement condition and admissible perturbation of a multivalued operator 17:30 ABSTRACT. In this paper, there are studied some strict fixed-point results and stability properties for multi-valued operators satisfying some contraction type conditions. The main purpose of this work is to analyse under which conditions imposed on the admissible perturbation of a multivalued operator, the strict fixed-point results and stability properties still hold true. Examples of b-Metric Spaces Endowed with a Partial Order ABSTRACT. The purpose of this paper is to present some examples of ordered rectangular b-metric spaces. On some local versions of contraction mapping principle with applications to integral equations ABSTRACT. In this paper we discuss some local fixed point theorems of the Banach contraction principle which are adapted to provide local convergence theorems for Picard iteration in the class of Fredholm inegral equations of the second kind. Several examples are discussed. 18:10 References [1] Ezquerro J. A. and Hernandez-Veron, M. A. On the application of some fixed-point techniques to Fredholm integral equations of the second kind, J. Fixed Point Theory Appl. (2024) 26:29 https://doi.org/10.1007/s11784-024-01119-6 [2] Granas, A.; Dugundji, J. Fixed point theory. Springer Monographs in Mathematics. Springer-Verlag, New York, 2003. [3] Rus, I. A.; Petruşel, A.; Petruşel, G. Fixed point theory. Cluj University Press, Cluj-Napoca, 2008. Best proximity points and their applications ABSTRACT. Our aim is to present recent results regarding best proximity point theory. To arouse interest in this field we present recent applications of best proximity and we propose a new direction of study in the case of a cyclic operator.	{"url":"https://easychair.org/smart-program/SYNASC2024/2024-09-18.html","timestamp":"2024-11-12T19:09:35Z","content_type":"text/html","content_length":"51488","record_id":"<urn:uuid:928ea497-a80b-499b-a7e9-c04d4d73a8e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00250.warc.gz"}
CSCI 3104 Problem Set 2 1. (20 pts total) Solve the following recurrence relations using any of the following methods: unrolling, tail recursion, recurrence tree (include tree diagram), or expansion. Each each case, show your work. (a) T(n) = T(n − 2) + C n if n > 1, and T(n) = C otherwise (b) T(n) = 3T(n − 1) + 1 if n > 1, and T(1) = 3 (c) T(n) = T(n − 1) + 3n if n > 1, and T(1) = 3 (d) T(n) = T(n ) + 1 if n > 2 , and T(n) = 0 otherwise 2. (10 pts) Consider the following function: def foo(n) { if (n > 1) { print( ’’hello’’ ) In terms of the input n, determine how many times is “hello” printed. Write down a recurrence and solve using the Master method. 3. (30 pts) Professor McGonagall asks you to help her with some arrays that are raludominular. A raludominular array has the property that the subarray A[1..i] has the property that A[j] > A[j + 1] for 1 ≤ j < i, and the subarray A[i..n] has the property that A[j] < A[j + 1] for i ≤ j < n. Using her wand, McGonagall writes the following raludominular array on the board A = [7, 6, 4, −1, −2, −9, −5, −3, 10, 13], as an example. (a) Write a recursive algorithm that takes asymptotically sub-linear time to find the minimum element of A. 1 (b) Prove that your algorithm is correct. (Hint: prove that your algorithm’s correctness follows from the correctness of another correct algorithm we already know.) (c) Now consider the multi-raludominular generalization, in which the array contains k local minima, i.e., it contains k subarrays, each of which is itself a raludominular array. Let k = 2 and prove that your algorithm can fail on such an input. (d) Suppose that k = 2 and we can guarantee that neither local minimum is closer than n/3 positions to the middle of the array, and that the “joining point” of the two singly-raludominular subarrays lays in the middle third of the array. Now write an algorithm that returns the minimum element of A in sublinear time. Prove that your algorithm is correct, give a recurrence relation for its running time, and solve for its asymptotic behavior. 4. (15 pts extra credit) Asymptotic relations like O, Ω, and Θ represent relationships between functions, and these relationships are transitive. That is, if some f(n) = Ω(g(n)), and g(n) = Ω(h(n)), then it is also true that f(n) = Ω(h(n)). This means that we can sort functions by their asymptotic growth.1 Sort the following functions by order of asymptotic growth such that the final arrangement of functions g1, g2, . . . , g12 satisfies the ordering constraint g1 = Ω(g2), g2 = Ω(g3), . . . , g11 = Ω(g12). n n 1.5 8 lg n 4 lg∗ n n! (lg n)! 5 4 n n 1/ lg n n lg n lg(n!) e n 42 Give the final sorted list and identify which pair(s) functions f(n), g(n), if any, are in the same equivalence class, i.e., f(n) = Θ(g(n)). 1The notion of sorting is entirely general: so long as you can define a pairwise comparison operator for a set of objects S that is transitive, then you can sort the things in S. For instance, for strings, we use a comparison based on lexical ordering to sort them. Furthermore, we can use any sorting algorithm to sort S, by simply changing the comparison operators >, <, etc. to have a meaning appropriate for S. For instance, using Ω, O, and Θ, you could apply QuickSort or MergeSort to the functions here to obtain the sorted list.	{"url":"https://codingprolab.com/answer/csci-3104-problem-set-2/","timestamp":"2024-11-02T14:51:34Z","content_type":"text/html","content_length":"108231","record_id":"<urn:uuid:0685e703-aaa5-4563-986d-56268b908977>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00037.warc.gz"}
Introduction to Geometry Spot : Basic Geometry Fundamentals - trhicks.com Geometry Spot is an online educational platform designed to assist students in learning geometry and other mathematical concepts. Launched in October 2022, it features a variety of resources including over 30 articles, tutorials, and interactive activities that cover a wide range of geometry topics. The website aims to enhance understanding through engaging content and regular updates, making it a valuable tool for students looking to improve their geometry skills . Understanding Geometry Spot Geometry Spot is more than just a collection of shapes and theorems; it is a comprehensive approach to understanding the world through mathematical concepts. By delving into Geometry Spot, students can explore the relationship between different geometric figures and grasp the underlying principles that govern their interactions. This area of study forms the foundation for various advanced mathematical and scientific disciplines, making it essential for students to master its concepts early on. Historical Development of Geometry The history of geometry is rich and varied, dating back to ancient civilizations such as the Egyptians and Babylonians who used rudimentary geometric techniques for practical purposes like land surveying and construction. However, it was the Greeks, particularly Euclid, who laid the systematic foundations of geometry as we know it today. Euclid’s “Elements,” a compilation of all known work on geometry during his time, is still considered one of the most influential works in mathematics. As geometry evolved, it branched into various subfields, including coordinate geometry, differential geometry, and topology, each contributing to the broader understanding of the discipline. Fundamental Concepts in Geometry Spot At its core, geometry begins with understanding the basic building blocks: points, lines, and planes. A point represents a location in space without any dimensions, while a line is a one-dimensional figure extending infinitely in both directions, defined by two points. A plane, on the other hand, is a flat, two-dimensional surface that extends infinitely in all directions. These fundamental concepts are crucial for understanding more complex geometric figures and Types of Angles in Geometry Angles are a fundamental aspect of geometry, formed by two rays with a common endpoint. Understanding the different types of angles is essential for solving various geometric problems. The primary types include acute angles (less than 90 degrees), right angles (exactly 90 degrees), obtuse angles (greater than 90 degrees but less than 180 degrees), and straight angles (exactly 180 degrees). Each type of angle has unique properties and plays a critical role in the study of geometric figures. Triangles: The Foundation of Geometry Triangles are perhaps the most studied geometric figures due to their fundamental properties and versatility. A triangle consists of three sides and three angles, with the sum of the interior angles always equaling 180 degrees. There are various types of triangles, including equilateral (all sides and angles are equal), isosceles (two sides and two angles are equal), and scalene (all sides and angles are different). Understanding the properties of triangles is essential for exploring more complex geometric concepts. The Pythagorean Theorem One of the most famous theorems in geometry is the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem, attributed to the ancient Greek mathematician Pythagoras, is fundamental in various applications, from simple calculations to complex problem-solving in trigonometry and algebra. Polygons and Their Properties Polygons are multi-sided figures with varying properties depending on the number of sides. Common polygons include quadrilaterals, pentagons, hexagons, and octagons. Each polygon has specific properties, such as the sum of its interior angles, which can be calculated using the formula (n−2)×180(n-2) times 180 degrees, where nn is the number of sides. Understanding these properties is crucial for solving geometric problems involving polygons. Circles and Their Properties Circles are unique geometric figures defined by a single constant distance from a central point, known as the radius. Key properties of circles include the diameter (twice the radius), circumference (the distance around the circle), and area (the space enclosed by the circle). Theorems related to circles, such as the properties of chords, tangents, and arcs, play a significant role in geometry and its applications. Coordinate Geometry Spot Coordinate geometry, also known as analytic geometry, involves studying geometric figures using a coordinate system. By representing points, lines, and shapes with coordinates, this branch of geometry bridges algebra and geometry, providing a powerful tool for solving complex geometric problems. Coordinate geometry is essential for understanding graphs, plotting points, and analyzing geometric relationships algebraically. Transformations in Geometry Transformations involve changing the position or size of a geometric figure. The main types of transformations are translations (sliding a figure), rotations (turning a figure around a point), reflections (flipping a figure over a line), and dilations (resizing a figure). Understanding transformations is crucial for analyzing the properties of geometric figures and their relationships in different contexts. Symmetry in Geometry Spot Symmetry is a fundamental concept in geometry, referring to the balance and proportion of geometric figures. A figure is symmetric if it can be divided into identical parts. Types of symmetry include reflectional symmetry (mirror image), rotational symmetry (rotation around a central point), and translational symmetry (repeating patterns). Symmetry is not only aesthetically pleasing but also has practical applications in various fields, including architecture and design. Geometric Proofs and Theorems Geometric proofs are logical arguments that use deductive reasoning to establish the truth of geometric statements. These proofs often involve a series of statements and reasons, leading to a conclusion. Theorems, such as the Pythagorean theorem, theorems involving angles, and properties of polygons, are proven using these logical sequences. Mastering geometric proofs is essential for developing critical thinking and problem-solving skills. Solid Geometry Spot Solid geometry focuses on three-dimensional figures, such as polyhedra (solid figures with flat faces), spheres, cylinders, and cones. Understanding the properties of these solids, including volume and surface area, is crucial for solving real-world problems in fields like engineering and architecture. Solid geometry provides a deeper understanding of the spatial relationships and measurements of three-dimensional objects. Surface Area and Volume Calculating the surface area and volume of geometric solids is an essential skill in geometry. Surface area refers to the total area of all the faces of a solid, while volume measures the space enclosed within the solid. These calculations are vital for practical applications, such as determining the amount of material needed for construction or the capacity of a container. Real-World Applications of Geometry Geometry is not just an abstract mathematical discipline; it has numerous real-world applications. Engineers use geometric principles to design structures, while architects rely on geometry to create aesthetically pleasing and functional buildings. Artists and designers use geometric shapes and patterns to create visually striking works, and scientists employ geometric concepts to model natural Geometry in Technology The role of geometry in modern technology cannot be overstated. From computer graphics and animation to robotics and artificial intelligence, geometric principles are fundamental to various technological advancements. Understanding geometry is essential for developing algorithms, designing virtual environments, and creating realistic simulations in digital technology. Geometry Spot in Nature Geometry is inherently present in nature, with countless examples of geometric patterns and shapes found in plants, animals, and natural formations. The Fibonacci sequence, for instance, appears in the arrangement of leaves, flowers, and seeds, while hexagonal patterns are observed in honeycombs and snowflakes. Studying these natural geometries can provide insights into the fundamental principles that govern the natural world. Advanced Topics in Geometry For those looking to delve deeper into geometry, advanced topics such as non-Euclidean geometry and topology offer fascinating avenues of study. Non-Euclidean geometry explores geometric systems that differ from traditional Euclidean principles, while topology examines properties of shapes that remain constant under continuous transformations. These advanced topics expand the boundaries of geometric understanding and have applications in various scientific fields. Tips for Studying Geometry Spot Studying geometry effectively requires a combination of theoretical understanding and practical application. Here are some tips for mastering geometry: 1. Practice regularly by solving various geometric problems. 2. Visualize concepts using diagrams and models. 3. Understand the underlying principles rather than just memorizing formulas. 4. Use online resources and software to explore geometric concepts interactively. 5. Collaborate with peers to discuss and solve complex problems. Common Mistakes in Geometry and How to Avoid Them Learning geometry can be challenging, and students often make common mistakes. Here are some tips to avoid them: 1. Misinterpreting geometric figures: Always read and analyze the problem carefully. 2. Incorrectly applying formulas: Ensure you understand when and how to use specific formulas. 3. Neglecting units of measurement: Always include and convert units as necessary. 4. Overlooking special cases: Pay attention to unique properties of specific geometric figures. 5. Skipping steps in proofs: Follow a logical sequence and ensure each step is justified. Geometry Spot offers a comprehensive understanding of the fundamental and advanced concepts of geometry, highlighting its significance in various fields. By mastering geometric principles, students can enhance their analytical and problem-solving skills, opening doors to numerous academic and professional opportunities. As we continue to explore and innovate, the role of geometry in shaping our understanding of the world will remain ever essential. Also Read: Discover the Ancient Mysticism of Örviri What is Geometry Spot? Geometry Spot is an in-depth exploration of geometric concepts, making them accessible and engaging for learners. Why is geometry important? Geometry is fundamental for understanding spatial relationships and is crucial in various fields such as engineering, architecture, and technology. How can I improve my understanding of geometry? Regular practice, visualizing concepts, and using online resources can enhance your understanding of geometry. What are some common mistakes in learning geometry? Common mistakes include misinterpreting figures, incorrectly applying formulas, and overlooking units of measurement. What resources are available for learning geometry? Recommended resources include textbooks like “Elements” by Euclid, online platforms like Khan Academy, and interactive software like GeoGebra. What are some advanced topics in geometry? Advanced topics include non-Euclidean geometry, which explores systems differing from Euclidean principles, and topology, which examines properties of shapes under continuous transformations.	{"url":"https://trhicks.com/geometry-spot/","timestamp":"2024-11-03T23:15:19Z","content_type":"text/html","content_length":"83450","record_id":"<urn:uuid:4ef7c43a-7818-4c1a-bef3-f2fe3f85ceb6>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00390.warc.gz"}
How to Sum by Year in Google Sheets ( Quick & Easy Guide ) - 2024 In this tutorial, you will learn How to Sum by Year in Google Sheets • To Sum by Year in Google Sheets, you can first extract the year using the Year() function from your dataset then from the result identify the unique Year. • Once the Unique Years are identified. You can use the SUMIF function to get the SUM by Year in Google Sheets. How to Sum by Year in Google Sheets: Step by Step Guide Enclosed are the steps on How to Sum by Year in Google Sheets : Step 1: Prepare your dataset Below is the dataset which we will use to Sum by Year in Google Sheets. The following dataset consists of Dates and Visa Applications on that day Step 2: Use the YEAR() function to extract Year from dates Now we will use the Year() function to extract the Years from the available dates in our data set. Below is the formula which we will use in our example. We will type =Year(A2) in the cell D2 To get the result on the rest of the cells in Column D . Move the cursor on top of the blue pointer in Cell D2 until it turns into a plus sign and starts dragging it downwards to apply the formula to the rest of the cells. Step 3: Identify the Unique Year Now we will identify the Uniques Year from the previous result using the =UNIQUE() function We will type the formula =UNIQUE(D2:D11) in Cell E2 in our example Once we press enter we get a list of Unique Year. Step 4: Use the SUMIF function to find the Sum by Year Now to find the Sum by Year we will use the SUMIF(range, criterion, sum_range) function to calculate the sum of visa applications in each year. We will type the below formula in Cell F2 in our example In our example, we’ll type the following formula in cell F2: =SUMIF($D$2:$D$11, E2, $B$2:$B$11) Apply the formula downwards to the rest of the cells. Now we have the total number of applications as shown below The above result means : • The total number of Visa applications in the year 2019 was 28 • The total number of Visa applications in the year 2020 was 45 • The total number of Visa applications in the year 2021 was 27 • The total number of Visa applications in the year 2022 was 159 How to Sum by Year in Google Sheets(Conclusion) In the above tutorial, we have shown you steps using which you Sum by Year in Google Sheets. We hope this tutorial on How to Sum by Year in Google Sheets was useful. Related articles : How to Insert Spin Button in Google Sheets ( Easy Guide ) How to Freeze Rows in Google Sheets How to Calculate Square Root and Cube Root in Google Sheets(Quick & Easy Guide) How to use SUMSQ Function in Google Sheets(Quick & Easy Guide )	{"url":"https://addnewskills.com/how-to-sum-by-year-in-google-sheets/","timestamp":"2024-11-01T23:47:58Z","content_type":"text/html","content_length":"125230","record_id":"<urn:uuid:a0fe1bfd-9594-40e6-bfe7-f9991eaa7c08>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00861.warc.gz"}
Oxford Mathematics Public Lectures - Euler’s pioneering equation: "the most beautiful theorem in mathematics" - Robin Wilson Euler’s equation, the ‘most beautiful equation in mathematics’, startlingly connects the five most important constants in the subject: 1, 0, π, e and i. Central to both mathematics and physics. So what is this equation – and why is it pioneering? Robin Wilson is an Emeritus Professor of Pure Mathematics at the Open University, Emeritus Professor of Geometry at Gresham College, London, and a former Fellow of Keble College, Oxford.	{"url":"https://www.podcasts.ox.ac.uk/oxford-mathematics-public-lectures-eulers-pioneering-equation-most-beautiful-theorem-mathematics?video=1","timestamp":"2024-11-11T21:18:12Z","content_type":"text/html","content_length":"31232","record_id":"<urn:uuid:6352b8cc-f083-4647-a6ed-7eaf67ef084d>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00396.warc.gz"}
The case for WeBWorK WeBWorK is an open-source online homework system. I believe it is a very good idea to use WeBWorK in mathematics classes for the following reasons. It lends itself to active learning. When teaching by lecture or lecture-discussion, the teacher often has a coherent and insightful train of thought in mind, which, however, is often lost on the students, as any teacher knows. Indeed the lecture format is inherently problematic, and is based on an often naive and excessive faith in the power of passive osmosis. An alternative is to translate the same train of thought into a path of guided inquiry, where the students themselves have to supply various steps along the way. WeBWorK can serve as the infrastructure for creating an active classroom along these lines. Further thoughts on this here. It refocusses the teacher role from judge to guide. In a traditional course, when the student hands in a test or assignment, the teacher locks himself in a dark room for hours critically judging it, and a week later the student receives it back with red marks on it. Insofar as the teacher wrote feedback, the only grade-pragmatist use of this is to scan for point deductions that seem weakly motivated so that one can confront the teacher for a grade appeal. The teacher becomes an aloof judge, and his feedback becomes justifications for his harsh verdicts. With WeBWorK, the teacher does not spend his time judging, but guiding. Instead of spending his time judging the students’ work after the fact, he spends his time alongside them as they are doing the work. He assists students when they get stuck. His feedback is of the form “try this” or “look at it this way” instead of “here’s why you’re wrong” or “this is what you should have done.” It incentivises persistent work. In a traditional course, students are typically assigned many practice problems that are never graded, since hand-grading is unfeasible for the volume of practice problems that students need. Students must be driven by their own conscience and discipline to complete these assignments. They will eventually be held accountable in the form of quizzes and exams, but this does not directly incentivise them to complete all assignments and to complete them on time. Many students are natural procrastinators who seek the path of least resistance. If they will not be held accountable for this week’s material until a quiz three weeks down the road, many will neglect the assignments and hope to make up for it with some last-minute cramming before the test. In fact, since a test can only contain a few questions and could not possibly cover everything from the assignments, doing all assignments is guaranteed to mean much “wasted effort” as far as crude grade-pragmatism goes. With WeBWorK, all problems count toward the student’s grade. The students now have a direct incentive to complete all assignments and to complete them on time. Attending class, asking questions, discussing problems with classmates, using tutoring resources---all of these things are now more strongly incentivised, since the students can keep their WeBWorK assignments before them, on their smartphones for instance, and see concrete and direct contributions toward their final grade accumulate in proportion to the effort and time they put it. It incentivises learning from mistakes. In a traditional course, when students receive back a graded exam or assignment they have little incentive to do more than to check the final score. Ideally you would want them to learn from their mistakes, but since the grade has already been set anyway the only mechanism for this is a kind of self-shaming, which is not pleasant. Furthermore, even if they wanted to learn from their mistakes, the feedback comes much too late, several days after the students wrote it, meaning that the student will have to devote much effort to retracing their original reasoning before they could analyse their mistake and understand how to learn from it. Redoing the problem and relearning the material correctly in this way would cost much time, and since the grade is already set grade-pragmatism dictates that this time is better spent on the next assignment. With WeBWorK, students receive instant feedback and have the chance to correct their mistakes right away while the problem is still fresh in their minds. They thus have a direct incentive to try to figure out what went wrong and to correct their reasoning. This also eliminates the demoralising phenomena, unavoidable in one-off grading schemes, that a small computational error can have a heavy effect on the final score. It encourages collaborative work and structural thinking. WeBWorK problems incorporate randomisation, making them ideal for collaborative work without direct copying, as the underlying idea or strategy rather than the answer must be conveyed to help a It is suited for assigning a variety of conceptual questions. Traditional hand-grading tends to be focussed on a few quite substantial computational problems. The one-sided nature of these problems, however, can have a disastrous effects. Students can become masters in computational procedures while still understanding next to nothing of the visual or verbal meaning of what they are doing. It is crucial, therefore, to challenge them with conceptual problems that forces them to reflect and form a fuller mental image of the notions at hand. But this is not easily done in hand-graded homework assignments. Conceptual, explanation-type questions are much harder to grade fairly and objectively than clear-cut computational problems with a straightforward “right answer.” This difficulty can be avoided by focussing on true-false problems, multiple-choice verbal-interpretation questions, graph-matching problems, etc. But such questions are less suitable as hand-graded homework assignments since a large quantity of them are needed and the answers to such problems are easily copied. With WeBWorK, randomisation helps eliminate the threat of copying, and the volume of problems ceases to be problematic, thus making it possible to target conceptual understanding more efficiently. WeBWorK is free and open-source and idealistically developed by actual mathematics teachers. This is the exact opposite of the scammy, filthy-expensive online homework systems pushed by major commercial publishers, which are designed to sell, not to teach. WeBWorK runs beautifully on any smartphone, tablet, or laptop right in your browser without any installation or complications. It also generates beautiful PDF output for printing.	{"url":"https://intellectualmathematics.com/blog/the-case-for-webwork/","timestamp":"2024-11-12T23:13:35Z","content_type":"text/html","content_length":"45326","record_id":"<urn:uuid:3e001e81-cb5d-4310-a97b-3f2264d1f739>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00540.warc.gz"}
Multiple Choice - [PDF Document] (2024) 1. Evaluate the lim (x^2-16)/(x-4). a. 1 b. 8 c. 0 d. 16 2. Evaluate the limit (x-4)/(x^2-x-12) as x approaches 4. a. undefined b. 0 c. infinity d. 1/7 3. What is the limit of cos (1/y) as y approaches infinity? a. 0 b. -1 c. infinity d. 1 4. Evaluate the limits of lim (x^3-2x+9) /(2x^3-8). a. 0 b. -9/8 c. α d. ½ 5. Evaluate the limit of (x^3-2x^2-x+2) /(x^2-4) as x approaches 2. a. α b. ¾ c. 2/5 d. 4/7 6. Evaluate the limit of √(x-4)/√(x^2-16) as x approaches 4. a. 0.262 b. 0.354 c. 0 d. α 7. Evaluate the limit of (x^2-x-6)/(x^2-4x+3) as x approaches 3. a. 3/2 b. 3/5 c. 0 d. 5/2 8. Evaluate the limit of (4x^2-x)/ (2x^2+4) as x approaches α. a. 2 b. 4 c. α d. 0 9. Evaluate the limit of (x-2)/(x^3-8) as x approaches 2. a. α b. 1/12 c. 0 d. 2/3 10. Evaluate the limit of θ/(2 sinθ) as θ approaches 0. a. 2 b. ½ c. 0 d. α 11. Evaluate the limit of (1-sec^2 (x)/ cos (x)-1 as x approaches 0. a. -2 b. α c. 0 d. 1 12. Evaluate the limit (x^3-27)/(x-3) as x approaches to 3. a. 0 b. infinity c. 9 d. 27 13. Evaluate the limit (3x^3-4x^2-5x+2)/ (x^2-x-2) as x approaches to 2. a. α b. 5 c. 0 d. 7/3 14. Evaluate the limit of (4 tan^3 (x)/ 2sin(x)-x as x approaches 0. a. 1 b. 0 c. 2 d. α 15. Evaluate the limit of 8x/(2x-1) as x approaches α. a. 4 b. 3 c. 2 d. -1 16. Evaluate the limit of (x^2-1)/ (x^2+3x-4) as x approaches 1. a. 2/5 b. 1/5 c. 3/5 d. 4/5 17. Evaluate the limit of (x+2)/(x-2) as x approaches α. a. α b. -1 c. 1 d. 4 18. Evaluate the limit of (1-cosx)/(x^2) as x approaches 0. a. α b. ½ c. 1 d. 0 19. Find the limit of [sqrt(x+4)-2]/x as x approaches 0. a. α b. ¼ c. 0 d. ½ 20. Find the limit [sqrt(x+9)-3] /x as x approaches 0. a. α b. 1/6 c. 0 d. 1/3 21. Evaluate the limit (x^2+x-6)/(x^2-4) as x approaches to 2. a. 6/5 b. 5/4 c. 4/3 d. 3/2 22. Evaluate the limit (x^4-81)/(x-3) as x approaches to 3. a. 108 b. 110 c. 122 d. 100 23. Evaluate the limit (x+sin2x)/ (x-sin2x) as x approaches to 0. a. -5 b. -3 c. 4 d. -1 24. Evaluate the limit (ln sin x)/(ln tan x) as x approaches to 0. a. 1 b. 2 c. ½ d. α 25. Compute the equation of the vertical asymptote of the curve y=(2x-1)/(x+2). a. x+2=0 b. x-3=0 c. x+3=0 d. x-2=0 26. Compute the equation of the horizontal asymptote of the curve y=(2x-1)/(x+2). a. y=2 b. y=0 c. y-1=0 d. y-3=0 27. The function y=(x-4)/(x+2) is discontinuous at x equals? a. -2 b. 0 c. 1 d. 2 28. An elliptical plot of garden has a semi-major axis of 6m and a semi-minor axis of 4.8meters. If these are increased by 0.15m each, find by differential equations the increase in area of the garden in sq.m. a. 0.62π b. 1.62π c. 2.62π d. 2.62π 29. The diameter of a circle is to be measured and its area computed. If the diameter can be measured with a maximum error of 0.001cm and the area must be accurate to within 0.10sq.cm. Find the largest diameter for which the process can be used. a. 64 b. 16 c. 32 d. 48 30. The altitude of a right circular cylinder is twice the radius of the base. The altitude is measured as 12cm. With a possible error of 0.005cm, find the approximately error in the calculated volume of the cylinder. a. 0.188 cu cm b. 0.144 cu cm c. 0.104 cu cm d. 0.126 cu cm 31. What is the allowable error in measuring the edge of a cube that is intended to hold a cu m, if the error in the computed volume is not to exceed 0.03 cu m? a. 0.002 b. 0.0025 c. 0.003 d. 0.001 32. If y=x^(3/2) what is the approximate change in y when x changes from 9 to 9.01? a. 0.045 b. 0.068 c. 0.070 d. 0.023 33. The expression for the horsepower of an engine is P=0.4 n x^2 where n is the number of cylinders and x is the bore of cylinders. Determine the power differential added when four cylinder car has the cylinders rebored from 3.25cm to 3.265cm. a. 0.156 hp b. 0.210 hp c. 0.319 hp d. 0.180 hp 34. A surveying instrument is placed at a point 180m from the base of a bldg on a level ground. The angle of elevation of the top of a bldg is 30 degrees as measured by the instrument. What would be error in the height of the bldg due to an error of 15minutes in this measured angle by differential equation? a. 1.05m b. 1.09m c. 2.08m d. 1.05m 35. If y=3x^2-x+1, find the point x at which dy/dx assume its mean value in the interval x=2 and x=4. a. 3 b. 6 c. 4 d. 8 36. Find the approximate increase by the use of differentials, in the volume of the sphere if the radius increases from 2 to 2.05. a. 2.51 b. 2.25 c. 2.12 d. 2.86 37. If the area of a circle is 64π sq mm, compute the allowable error in the area of a circle if the allowable error in the radius is 0.02 mm. a. 1.01 sq mm b. 1.58 sq mm c. 2.32 sq mm d. 0.75 sq mm 38. If the volume of a sphere is 1000π/6 cu mm and the allowable error in the diameter of the sphere is 0.03 mm, compute the allowable error in the volume of a sphere. a. 6.72 cu mm b. 4.71 cu mm c. 5.53 cu mm d. 3.68 cu mm 39. A cube has a volume of 1728 cu mm. If the allowable error in the edge of a cube is 0.04 mm, compute the allowable error in the volume of the cube. a. 17.28 cu mm b. 16.88 cu mm c. 15.22 cu mm d. 20.59 cu mm 40. Find the derivative of y=2^(4x). a. 3^(4x+2) ln 2 b. 2^(4x+2) ln 2 c. 6^(3x+2) ln 2 d. 4^(4x+2) ln 2 41. Find the derivative of h with respect to u if h=π^ (2u). a. π^(2u) b. 2u ln π c. 2π^(2u) ln π d. 2π^(2u) 42. Find y’ if y=ln x a. 1/x b. ln x^2 c. 1/ln x d. x ln x 43. Find y’ if y=arc sin (x) a. √(1-x^2) b. 1/√(1-x^2) c. 1/(1+x^2) d. (1+x)/√(1-x^2) 44. Find the derivative of loga u with respect to x. a. log u du/dx b. u du/ln a c. loga e/u d. log a du/dx 45. Find the derivative of arc cos (2x). a. -2/√(1-4x^2) b. 2/√(1-4x^2) c. 2/(1+4x^2) d. 2/ √(2x^2-1) 46. Find the derivative of 4 arc tan (2x). a. 4/(1+x^2) b. 4/(4x^2+1) c. 8/(1+4x^2) d. 8/(4x^2+1) 47. Find the derivative of arc csc (3x). a. -1/[x√(9x^2-1)] b. 1/[3x√(9x^2-1)] c. 3/[x√(1-9x^2)] d. 3/[x√9x^2-1)] 48. Find the derivative of arc sec (2x) a. 1/[x√(4x^2-1)] b. 2/[x√(4x^2-1)] c. 1/[x√(1-4x^2)] d. 2/[x√(1-4x^2)] 49. If ln (ln y) + ln y = ln x, find y’. a. x/(x+y) b. x/(x-y) c. y/(x+y) d. y/(x-y) 50. Find y” if y=a^u. a. a^u ln a b. u ln a c. a^u/ln a d. a ln u 51. Find the derivative of y with respect to x if y = x ln x – x. a. x ln x b. ln x c. (ln x)/x d. x/ln x 52. If y=tanh x, find dy/dx. a. sech^2 (x) b. csch^2 (x) c. sinh^2 (x) d. tanh^2 (x) 53. Find the derivative of y=x^x. a. x^x (2+ln x) b. x^x (1+ln x) c. x^x (4-ln x) d. x^x (8+ln x) 54. Find the derivative of y=loga 4x. a. y’=(loga e)/x b. y’=(cos e)/x c. y’=(sin e)/x d. y’=(tan e)/x 55. What is the derivative with respect to x of (x+1)^3 – x^3. a. 3x+3 b. 3x-3 c. 6x-3 d. 6x+3 56. What is the derivative with respect to x of sec^2 (x)? a. 2x sec^2 (x) tan^2 (x) b. 2x sec (x) tan (x) c. sec^2 (x) tan^2 (x) d. 2 sec^2 (x) tan^2 (x) 57. The derivative with respect to x of 2cos^2 (x^2+2). a. 4 sin (x^2+2) cos (x^2+2) b. -4 sin (x^2+2) cos (x^2+2) c. 8x sin (x^2+2) cos (x^2+2) d. -8x sin (x^2+2) cos (x^2+2) 58. Find the derivative of [(x+1)^3]/x. a. [3(x+1)^2]/x – [(x+1)^3]/x^2 b. [2(x+1)^3]/x – [(x+1)^3]/x^3 c. [4(x+1)^2]/x – [2(x+1)^3]/x d. [(x+1)^2]/x – [(x+1)^3]/x 59. Determine the slope of the curve y=x^2-3x as it passes through the origin. a. -4 b. 2 c. -3 d. 0 60. If y1=2x+4 and y2=x^2+C, find the value of C such that y2 is tangent to y1. a. 6 b. 5 c. 7 d. 4 61. Find the slope of (x^2)y=8 at the point (2,2). a. 2 b. -1 c. -1/2 d. -2 62. What is the first derivative dy/dx of the expression (xy)^x=e. a. –y(1-ln xy)/x^2 b. –y(1+ln xy)/x c. 0 d. x/y 63. Find y’ in the following equation y=4x^2-3x-1. a. 8x-3 b. 4x-3 c. 2x-3 d. 8x-x 64. Differentiate the equation y=(x^2)/(x+1). a. (x^2+2x)/(x+1)^2 b. x/(x+1) c. 2x^2/(x+1) d. 1 65. If y=x/(x+1), find y’. a. 1/(x+1)^3 b. 1/(x+1)^2 c. x+1 d. (x+1)^2 66. Find dy/dx in the equation y=(x^6+3x^2+50)/(x^2+1) if x=1 a. -21 b. -18 c. 10 d. 16 67. Find the equation of the curve whose slope is (x+1)(x+2) and passes through point (- 3, -3/2). a. y=x^2+2x-4 b. y=(x^3)/3+(3x^2)/2+2x c. y= 3x^2+4x-8 d. y=(3x^2)/2+4x/3+2 68. Find the equation of the curve whose slope is 3x^4-x^2 and passes through point (0,1). a. y=(3x^5)/5-(x^3)/3+1 b. y=(x^4)/4-(x^3)+1 c. y=(2x^5)/5-2x+1 d. y=(3x^5)-(x^3)/3+1 69. What is the slope of the tangent to y=(x^2+1)(x^3-4x) at (1,-6)? a. -8 b. -4 c. 3 d. 5 70. Find the coordinate of the vertex of the parabola y=x^2-4x+1 by making use of the fact that at the vertex, the slope of the tangent is zero. a. (2,-3) b. (3,2) c. (-1,-3) d. (-2,-3) 71. Find the slope of the curve x^2+y^2-6x+10y+5=0 at point (1,0). a. 2/5 b. ¼ c. 2 d. 2 72. Find the slope of the ellipse x^2+4y^2-10x+16y+5=0 at the point where y=2+8^0.5 and x=7. a. -0.1654 b. -0.1538 c. -0.1768 d. -0.1463 73. Find the slope of the tangent to the curve y=2x-x^2+x^3 at (0,2). a. 2 b. 3 c. 4 d. 1 74. Find the equation of the tangent to the curve y=2e^x at (0,2). a. 2x-y+3=0 b. 2x-y+2=0 c. 3x+y+2=0 d. 2x+3y+2=0 75. Find the slope of the curve y=2(1+3x)^2 at point (0,3). a. 12 b. -9 c. 8 d. -16 76. Find the slope of the curve y=x^2(x+2)^3 at point (1,2). a. 81 b. 48 c. 64 d. 54 77. Find the slope of the curve y=[(4-x)^2]/x at point (2,2). a. -3 b. 2 c. -2 d. 3 78. If the slope of the curve y^2=12x is equal to 1 at point (x,y), find the value of x and y. a. x=3, y=6 b. x=4, y=5 c. x=2, y=7 d. x=5, y=6 79. If the slope of the curve x^2+y^2=25 is equal to -3/4 at point (x,y) find the value of x and y. a. 3,4 b. 2,3 c. 3,4.2 d. 3.5,4 80. If the slope of the curve 25x^2+4y^2=100 is equal to -15/8 at point (x,y), find the value of x and y. a. 1.2,4 b. 2,4 c. 1.2,3 d. 2,4.2 81. Determine the point on the curve x^3-9x-y=0 at which slope is 18. a. x=3, y=0 b. x=4, y=5 c. x=2, y=7 d. x=5, y=6 82. Find the second derivative of y=(2x+1)^2+x^3. a. 8+6x b. (2x+1)^3 c. x+1 d. 6+4x 83. Find the second derivative of y=(2x+4)^2 x^3. a. x^2(80x+192) b. 2x+4 c. x^3(2x+80) d. x^2(20x+60) 84. Find the second derivative of y=2x+3(4x+2)^3 when x=1. a. 1728 b. 1642 c. 1541 d. 1832 85. Find the second derivative of y=2x/[3(4x+2)^2] when x= 0. a. -1.33 b. 1.44 c. 2.16 d. -2.72 86. Find the second derivative of y=3/(4x^-3) when x=1. a. 4.5 b. -3.6 c. 2.4 d. -1.84 87. Find the second derivative of y=x^-2 when x=2. a. 0.375 b. 0.268 c. 0.148 d. 0.425 88. Find the first derivative of y=2cos(2+x^2). a. -4x sin (2+x^2) b. 4x cos (2+x^2) c. x sin (2+x^2) d. x cos (2+x^2) 89. Find the first derivative of y=2 sin^2 (3x^2-3). a. 24x sin (3x^2-3) cos (3x^2-3) b. 12 sin (3x^2-3) c. 6x cos (3x^2-3) d. 24x sin (3x^2-3) 90. Find the first derivative of y=tan^2 (3x^2-4). a. 12xtan(3x^2-4)sec^2(3x^2-4) b. x tan (3x^2-4) c. sec^2 (3x^2-4) d. 2 tan^2(3x^2-4)csc^2(3x^2-4) 91. Find the derivative of arc cos 4x a. -4/(1-16x^2)^0.5 b. 4/ (1-16x^2)^0.5 c. -4/(1-4x^2)^0.5 d. 4/(1-4x^2)^0.5 92. The equation y^2=cx is the general equation of. a. y’=2y/x b. y’=2x/y c. y’=y/2x d. y’=x/2y 93. Find the slope of the curve y=6(4+x)^1/2 at point (0,12). a. 1.5 b. 2.2 c. 1.8 d. 2.8 94. Find the coordinate of the vertex of the parabola y=x^2-4x+1 by making use of the fact that at the vertex, the slope of the tangent is zero. a. (2,-3) b. (3,2) c. (-1,-3) d. (-2,-3) 95. Find dy/dx by implicit differentiation at the point (3,4) when x^2+y^2=25. a. -3/4 b. ¾ c. 2/3 d. -2/3 96. Find dy/dx by implicit differentiation at point (0,0) if (x^ 3)(y^3)-y=x. a. -1 b. -2 c. 2 d. 1 97. Find dy/dx by implicit differentiation at point (0,-2) if x^3-xy+y^2=4. a. ½ b. -2 c. -2/3 d. ¾ 98. Find the point of inflection of f(x)=x^3-3x^2-x+7. a. 1,4 b. 1,2 c. 2,1 d. 3,1 99. Find the point of inflection of the curve y=(9x^2-x^3+6)/6. a. 3,10 b. 2,8 c. 3,8 d. 2,10 100. Find the point of inflection of the curve y=x^3-3x^2+6. a. 1,4 b. 1,3 c. 0,2 d. 2,1 101. Locate the point of inflection of the curve y=f(x)=(x square)(e exponent x). a. -2 plus or minus (sqrt of 3) b. 2 plus or minus (sqrt of 2) c. -2 plus or minus (sqrt of 2) d. 2 plus or minus (sqrt of 3) 102. The daily sales in thousands of pesos of a product is given by S=(x^2-x^3+6)/6 where x is the thousand of pesos spent on advertising. Find the point of diminishing returns for money spent on advertising. a. 5 b. 4 c. 3 d. 6 103. y=x to the 3rd power -3x. Find the maximum value of y. a. 2 b. 1 c. 0 d. 3 104. Find the curvature of the parabola y^2=12x at (3,6). a. -√2/24 b. √2/8 c. 3√2 d. 8√2/3 105. Locate the center of curvature of the parabola x^2=4y at point (2,2). a. (-2,6) b. (-3,6) c. (-2,4) d. (-3,7) 106. Compute the radius of curvature of the parabola x^2=4y at the point (4,4). a. 22.36 b. 24.94 c. 20.38 d. 18.42 107. Find the radius of curvature of the curve y=2x^3+3x^2 at (1,5). a. 97 b. 90 c. 101 d. 87 108. Compute the radius of curvature of the curve x=2y^3-3y^2 at (4,2). a. -97.15 b. -99.38 c. -95.11 d. -84.62 109. Find the radius of curvature of a parabola y^2-4x=0 at point (4,4). a. 22.36 b. 25.78 c. 20.33 d. 15.42 110. Find the radius of curvature of the curve x=y^3 at point (1,1). a. -1.76 b. -1.24 c. 2.19 d. 2.89 111. A cylindrical boiler is to have a volume of 1340 cu ft. The cost of the metal sheets to make the boiler should be minimum. What should be its diameter in feet? a. 7.08 b. 11.95 c. 8.08 d. 10.95 112. A rectangular corral is to be built with a required area. If an existing fence is to be used as one of the sides, determine the relation of the width and the length which would cost the least. a. width=twice the length b. width=1/2 length c. width=length d. width=3 times the length 113. Find the two numbers whose sum is 20, if the product of one by the cube of the other is to be minimum. a. 5 and 15 b. 10 and 10 c. 4 and 16 d. 8 and 12 114. The sum of two numbers is 12. Find the minimum value of the sum of their cubes. a. 432 b. 644 c. 346 d. 244 115. A printed page must contain 60 sq m of printed material. There are to be margins of 5cm on either side and margins of 3cm on top and bottom. How long should the printed lines be in order to minimize the amount of paper used? a. 10 b. 18 c. 12 d. 15 116. a school sponsored trip will cost each students 15 pesos if not more than 150 students make the trip, however the cost per student will reduced by 5 centavos for each student in excess of 150. How many students should make the trip in order for the school to receive the largest group income? a. 225 b. 250 c. 200 d. 195 117. A rectangular box with square base and open at the top is to have a capacity of 16823 cu cm. Find the height of the box that requires minimum amount of materials required. a. 16.14 cm b. 14.12 cm c. 12.13 cm d. 10.36 cm 118. A closed cylindrical tank has a capacity of 576.56 cu m. Find the minimum surface area of the tank. a. 383.40 cu m b. 412.60 cu m c. 516.32 cu m d. 218.60 cu m 119. A wall 2.245 m high is x meters away from a building. The shortest ladder that can reach the building with one end resting on the ground outside the wall is 6m. What is the value of x? a. 2m b. 2.6m c. 3.0m d. 4.0m 120. With only 381.7 sq m of materials, a closed cylindrical tank of maximum volume is to be the height of the tank, in m? a. 9m b. 7m c. 11m d. 13m 121. If the hypotenuse of a right triangle is known, what is the ratio of the base and the altitude of the right triangle when its are is maximum? a. 1:1 b. 1:2 c. 1:3 d. 1:4 122. The stiffness of a rectangular beam is proportional to the breadth and the cube of the depth. Find the shape of the stiffest beam that can be cut from a log of given size. a. depth=√3 breadth b. depth=breadth c. depth=√2 breadth d. depth=2√2 breadth 123. What is the maximum length of the perimeter if the hypotenuse of a right triangle is 5m long? a. 12.08 m b. 15.09 m c. 20.09 m d. 8.99 m 124. An open top rectangular tank with square s bases is to have a volume of 10 cu m. The material fir its bottom is to cost 15 cents per sq m and that for the sides 6 cents per sq m. Find the most economical dimensions for the tank. a. 2 x 2 x 2.5 b. 2 x 5 x 2.5 c. 2 x 3 x 2.5 d. 2 x 4 x 2.5 125. A trapezoidal gutter is to be made from a strip of metal 22m wide by bending up the sides. If the base is 14m, what width across the top gives the greatest carrying capacity? a. 16 b. 22 c. 10 d. 27 126. Divide the number 60 into two pats so that the product P of one part and the square of the other is maximum. Find the smallest part. a. 20 b. 22 c. 10 d. 27 127. The edges of a rectangular box are to be reinforced with a narrow metal strips. If the box will have a volume of 8 cu m, what would its dimensions be to require the least total length of strips? a. 2 x 2 x 2 b. 4 x 4 x 4 c. 3 x 3 x 3 d. 2 x 2 x 4 128. A rectangular window surmounted by a right isosceles triangle has a perimeter equal to 54.14m. Find the height of the rectangular window so that the window will admit the most light. a. 10 b. 22 c. 12 d. 27 129. A normal window is in the shape of a rectangle surrounded by a semi-circle. If the perimeter of the window is 71.416, what is its radius and the height of the rectangular portion so that it will yield a window admitting the most light? a. 10 b. 22 c. 12 d. 27 130. Find the radius of a right circular cone having a lateral area of 544.12 sq m to have a maximum volume. a. 10 b. 20 c. 17 d. 19 131. A gutter with trapezoidal cross section is to be made from a long sheet of tin that is 15cm wide by turning up one third of its width on each side. What width across the top that will give a maximum capacity? a. 10 b. 20 c. 15 d. 13 132. A piece of plywood for a billboard has an area of 24 sq ft. The margins at the top and bottom are 9 inches and at the sides are 6 in. Determine the size of plywood for maximum dimensions of the painted area. a. 4 x 6 b. 3 x 4 c. 4 x 8 d. 3 x 8 133. A manufacturer estimates that the cost of production of x units of a certain item is C=40x-0.02x^2-600. How many units should be produced for minimum cost? a. 1000 units b. 100 units c. 10 units d. 10000 units 134. If the sum of the two numbers is 4, find the minimum value of the um of their cubes. a. 16 b. 18 c. 10 d. 32 135. If x units of a certain item are manufactured, each unit can be sold for 200-0.01x pesos. How many units can be manufactured for maximum revenue? What is the corresponding unit price? a. 10000, P100 b. 10500, P300 c. 20000, P200 d. 15000, P400 136. A certain spare parts has a selling price of P150 if they would sell 8000 units per month. If for every P1.00 increase in selling price, 80 units less will be sold out pr month. If the production cost is P100 per unit, find the price per unit for maximum profit per month. a. P175 b. P250 c. P150 d. P225 137. The highway department is planning to build a picnic area for motorist along a major highway. It is to be rectangular with an area of 5000 sq m is to be fenced off on the three sides not adjacent to the highway. What is the least amount of fencing that ill be needed to complete the job? a. 200m b. 300m c. 400m d. 500m 138. A rectangular lot has an area of 1600 sq m. Find the least amount of fence that could be used to enclose the area. a. 160m b. 200m c. 100m d. 300m 139. A student club on a college campus charges annual membership due of P10, less 5 centavos for each member over 60. How many members would give the club the most revenue from annual dues? a. 130 members b. 420 members c. 240 members d. 650 members 140. A company estimates that it can sell 1000 units per weak if it sets the unit price at P3.00, but that its weekly sles will rise by 100 units for each P0.10 decrease in price. Find the number of units sold each week and its unit price per max revenue. a. 2000, P2.00 b. 1000, P3.00 c. 2500, P2.50 d. 1500, P1.50 141. In manufacturing and selling x units of a certain commodity, the selling price per unit is P=5-0.002x and the production cost in pesos is C=3+1.10x. Determine the production level that will produce the max profit and what would this profit be? a. 975, P1898.25 b. 800, P1750.75 c. 865, P1670.50 d. 785, P1920.60 142. ABC company manufactures computer spare parts. With its present machines, it has an output of 500 units annually. With the addition of the new machines the company could boosts its yearly production to 750 units. If it produces x parts it can set a price of P= 200-0.15x pesos per unit and will have a total yearly cost of C=6000+6x-0.003x in pesos. What production level maximizes total yearly profit? a. 660 units b. 237 units c. 560 units d. 243 units 143. The fixed monthly cost for operating a manufacturing plant that makes transformers is P8000 and there are direct costs of P110 for each unit produced. The manufacturer estimates that 100 units per month can be sold if the unit price is P250 and that sales will in crease by 20 units for each P10 decrease in price. Compute the number of units that must be sold per month to maximize the profit. Compute the unit price. a. 190, P205 b. 160, P185 c. 170, P205 d. 200, P220 144. The total cost of producing and marketing x units of a certain commodity is given as C=(80000x-400x^2+x^3)/40000. For what number x is the average cost a minimum? a. 200 units b. 100 units c. 300 units d. 400 units 145. A wall 2.245m high is 2m away from a bldg. Find the shortest ladder that can reach the bldg with one end resting on the ground outside the wall. a. 6m b. 9m c. 10m d. 4m 146. If the hypotenuse of a right triangle is known, what is the relation of the base and the altitude of the right triangle when its area is maximum? a. altitude=base b. altitude=√2 base c. altitude=√2 base d. altitude=2 base 147. The hypotenuse of a right triangle is 20cm. What is the max possible area of the triangle in sq cm? a. 100 b. 170 c. 120 d. 160 148. A rectangular field has an area of 10,000 sq m. What is the least amount of fencing meters to enclose it? a. 400 b. 370 c. 220 d. 560 149. A monthly overhead of a manufacturer of a certain commodity is P6000 and the cost of material is P1.0 per unit. If not more than 4500 units are manufactured per month, labor cost is P0.40 per unit, but for each unit over 4500, the manufacturer must pay P0.60 for labor per unit. The manufacturer can sell 4000 units per month at P7.0 per unit and estimates that monthly sales will rise by 100 for each P0.10 reduction in price. Find the number of units that should be produced each month for maximum profit. a. 4700 units b. 2600 units c. 6800 units d. 9900 units 150. Find two numbers whose product is 100m and whose sum is minimum. a. 10, 10 b. 12, 8 c. 5, 15 d. 9, 11 151. Find two numbers whose sum is 36 if the product of one by the square of the other is a maximum. a. 12, 24 b. 13, 23 c. 20, 16 d. 11, 25 152. Find the minimum amount of thin sheet that can be made into a closed cylinder having a volume of 108 cu in. in square inches. a. 125.5 b. 127.5 c. 123.5 d. 129.5 153. A buyer is to take a plot of land fronting street, the plot is to be rectangular and three times its frontage added to twice its depth is to be 96 meters. What is the greatest number of sq m be may take? a. 384 sq m b. 352 sq m c. 443 sq m d. 298 sq m 154. A company has determined that the marginal cost function for the production of a particular cost function for the production of a particular commodity is given as y”=125+10x-(x^2)/9 where y is the cost of producing x units of the commodity. If the fixed cost is 250 pesos, what is the cost of producing 15 units? a. 250 b. 225 c. 300 d. 200 155. A pig weighing 300lb gains 8 pounds per day and cost 6 pesos per day to maintain. The market price for the pig is seven pesos and fifty centavos per pound but is decreasing 10 centavos per day. When should the pig be sold? a. 15 days b. 18 days c. 20 days d. 10 days 156. It costs a bus company P125 to run a bus on a certain tour, plus P15 per passenger. The capacity of the bus is 20 persons and the company charges P35 per ticket if the bus is full. For each empty seat, however, the company increases the ticket price by P2.0. For maximum profit how many empty seats would the company like to see? a. 5 b. 3 c. 6 d. 4 157. A book publisher prints the pages of a certain book with 0.5 inch margins on the top, bottom and one side and a one inch margin on the other side to allow for the binding. Find the dimensions of the page that will maximize the printed area of the page if the area of the entire page is 96 sq inches. a. 8 inches b. 7 inches c. 9 inches d. 10 inches 158. The cost of manufacturing an engine parts is P300 and the number which can be sold varies inversely as the fourth power of the selling price. Find the selling price which will yield the greatest total net profit. a. 400 b. 350 c. 450 d. 375 159. The price of the product in a competitive market is P300. If the cost per unit of producing the product is 160+x where x is the number of units produced per month, how many units should the firm produce and sell to maximize its profit? a. 70 b. 80 c. 60 d. 50 160. If the cost per unit of producing a product by ABC company is 10+2x and if the price on the competitive market is P50, what is the maximum daily profit that the company can expect of this product? a. 200 b. 300 c. 400 d. 600 161. An entrepreneur starts new companies and sells them when their growth is maximized. Suppose the annual profit for a new company is given by P(x)=22-x/2- 18/(x+1) where P is in thousand of pesos and x is the number of years after the company is formed. If the entrepreneur wants to sell the company before profit begins to decline, after how many years would the company be sold? a. 5 b. 4 c. 6 d. 7 162. The profit function for a product is P(x)=5600x+85x^2-x^3-x-200000. How many items will produce a maximum profit? a. 80 b. 60 c. 70 d. 40 163. The following statistics of a manufacturing company shows the corresponding values for manufacturing x units. Production cost=60x+10000 pesos Selling price/unit=200-0.02x pesos How many units must be produced for max profit? a. 3500 b. 3300 c. 4000 d. 3800 164. The cost per unit of production is expressed as (4+3x) and the selling price on the competitive market is P100 per unit. What maximum daily profit that the company can expect of this product? a. P768 b. P876 c. P657 d. P678 165. A certain unit produced by the company can be sold for 400-0.02x pesos where x is the number of units manufactured. What would be the corresponding price per unit in order to have a max revenue? a. P200 b. P220 c. P150 d. P180 166. Given the cost equation of a certain product as follows C=50t^2-200t+10000 where t is in years. Find the maximum cost from the year 1995 to 2002. a. P9,800 b. P6,400 c. P7,200 d. P10,600 167. The total cost of production a shipment of a certain product is C=5000x+125000/x where x is the number of machines used in the production. How many machines will minimize the total cost? a. 5 b. 20 c. 10 d. 15 168. The demand x for a product is x=10000-100P where P is the market price in pesos per unit. The expenditure for the two product is E=Px. What market price will the expenditure be the greatest? a. 50 b. 60 c. 70 d. 100 169. Analysis of daily output of a factory shows that the hourly number of units y produced after t hours of production is y=70t+(t^2)/2-t^3. After how many hours will the hourly number of units be maximized? a. 5 b. 6 c. 7 d. 8 170. An inferior product with large advertising budget sells well when it is introduced, but sales fall as people discontinue use of the product. If the weekly sales are given by S=200t/(t+1)^2 where S is in millions of pesos and t in weeks. After how many weeks will the sales be maximized? a. 1 b. 2 c. 3 d. 4 171. In the coming presidential election of 1998, it is estimated that the proportions P of votes that recognizes a certain presidentiables name t months after the campaign is given by P=[7.2t/(t^2+16)]+0.20. After how many months is the proportional maximized? a. 4 b. 3 c. 5 d. 6 172. A car manufacturer estimates that the cost of production of x cars of a certain model is C= 20x-0.01x^2-800. How many cars should be produced for a minimum cost? a. 1000 b. 1200 c. 900 d. 1100 173. Analysis of daily output of a factory shows that the hourly number of units y produced after t hours of production is y=70t+(t^2)/2-t^3. After how many hours will the hourly number of units be maximized and what would be the maximum hourly output? a. 5hrs, 237.5 b. 4hrs, 273.6 c. 6hrs, 243.5 d. 3hrs, 223.6 174. A time study showed that on average, the productivity of a worker after t hours on the job can be modeled by the expression P=27+6t-t^3 where P is the number of units produced per hour. What is the maximum productivity expected? a. 36 b. 34 c. 44 d. 40 175. The sum of two numbers is equal to S. Find the minimum sum of the cube of the two numbers/ a. (S^3)/4 b. S/4 c. (S^2)/4 d. (S^3)/5 176. Given the cost equation of a certain product as follows: C=50t^2-200t+10000 where t is in years. Find the maximum cost from year 1995 to 2002. a. P9000 b. P9800 c. P8500 d. P7300 177. A manufacturer determines that the profit derived from selling x units of a certain item is given by P=0.003x^2+10x. Find the marginal profit for a production of 50 units. a. P10.30 b. P12.60 c. P15.40 d. P17.30 178. The total cost of production spare parts of computers is given as C=4000x- 100x^2+x^3 where x is the number of units of spare parts produced so that the average cost will be minimum? a. 50 b. 10 c. 20 d. 4 179. A viaduct is traversed by a truck running at 15mph at the same time that another truck traveling at a speed of 30mph on the street 22ft below and at right angle to the viaduct, approached the point directly below the viaduct from a distance of 55ft. Find the nearest distance between the trucks. a. 33 ft b. 44 ft c. 29 ft d. 39 ft 180. A sector is cut out of a circular disk of radius √3 and the remaining part of the disk I bent up so that the two edges join and a cone is formed. What is the largest volume for the cone? a. 2π/3 b. π/3 c. 3π/4 d. π/4 181. Four squares are cut out of a rectangular cardboard 50cm by 80 cm. in dimension and the remaining piece is folded into a closed, rectangular box with two extra flaps trucked in. What is the largest possible volume for such a box? a. 9000 b. 6000 c. 7000 d. 8000 182. An isosceles triangle with equal sides of 20cm has these sides at a variable equal angle with the base. Determine the max area of the triangle. a. 200 sq cm b. 250 sq cm c. 300 sq cm d. 280 sq cm 183. Formerly, for a package to go by parcel post, the sum of its length and girth could not exceed 120cm. Find the dimensions of the rectangular package of greatest volume that could be sent. a. 20 x 20 x 40 b. 20 x 20 x 20 c. 20 x 40 x 10 d. 40 x 20 x 30 184. The cross-section of a trough is an isosceles trapezoid. If the trough is made by bending up the sides of s strip of metal 12cm wide, what would be the angle of inclination of the sides and the width across the bottom if the cross-sectional area is to be a maximum? a. 60 degrees b. 120 degrees c. 45 degrees d. 75 degrees 185. Find the minimum amount of thin sheet that can be made into a closed cylinder having a volume of 108cu inches in square inches. a. 125.5 b. 127.5 c. 123.5 d. 129.5 186. Compute the abscissa of the min point of the curve y=x^3-12x-9. a. 2 b. -2 c. -1 d. 1 187. What value of x does a maximum of y=x^3-3x occur? a. -1 b. 1 c. 2 d. -2 188. Determine the point on the curve y^2=8x which is nearest to the external curve (4,2). a. (2,4) b. (4,3) c. (3,5) d. (6,8) 189. The LRT system runs from the Bonifacio Monument to Baclaran for a total distance of 15km. The cost of electric energy consumed by a train per hour is directly proportional to the cube of its speed and is P250 per hour at 50kph. Other expenses such as salaries, depreciation, overhead, etc. amounts to P1687.50 per hour. Find the most economical speed of the train in kph. a. 75 b. 80 c. 65 d. 60 190. A businessman found out that his profit varies as the product of the amount spent for production and the square root of the amount spent for advertisement. If his total budget for these expenses is P1.5 million, how much must be allocated for advertisement to maximize his profit? a. 0.5M b. 0.7M c. 0.8M d. 1.0M 191. A steel girder 16m long is moved on rollers along a passageway 8m wide and into a corridor at right angles with the passageway. Neglecting the width of thr girder, how wide must the corridor be? a. 3.6 m b. 1.4 m c. 1.8 m d. 2.8 m 192. A can manufacturer receives an order for milk cans having a capacity of 100 cu cm. Each can is made from a rectangular sheet of metal by rolling the sheet into a cylinder; the lids are stamped out from another rectangular sheet. What are the most economical proportions of the can? a. 2.55 b. 2.59 c. 2.53 d. 3.67 193. A triangle has a variable sides x, y and z subject to the constraint that the P is fixed to 18cm. What is the maximum possible area for the triangle? a. 15.59 sq cm b. 18.71 sq cm c. 14.03 sq cm d. 17.15 sq cm 194. Postal regulations require that a parcel post package shall be not greater than 600cm in the sum of its length and girth (perimeter of the cross-section). What is the volume in cu cm of the largest package allowed by the postal regulations if the package is to be rectangular in cu cm? a. 2 x 10^6 b. 3 x 10^6 c. 1.5 x 10^6 d. 4 x 10^6 195. Divide 60 into 3 parts so that the product of the three parts will be a maximum, find the product. a. 8000 b. 4000 c. 6000 d. 12000 196. Find the radius of the circle inscribe in a triangle having a max area of 173.205 sq cm. a. 3.45 cm b. 5.77 cm c. 4.96 cm d. 2.19 cm 197. The area of a circle inscribe in a triangle is equal to 113.10 sq cm. Find the max area of the triangle. a. 186.98 sq cm b. 156. 59 sq cm c. 175.80 sq cm d. 193. 49 sq cm 198. Find the perimeter of a triangle having a max area that is circumscribing a circle of radius 8cm. a. 83.13 cm b. 85.77 cm c. 84.96 cm d. 92.19 cm 199. Suppose y is the number of workers in the labor force neededtp produce x units of a certain commodity and x =4y^2. If the production of the commodity this year is 25000 units and the production is increasing at the rate and the production is increasing at the rate of 18000 units per year, what is the current rate at which the labor force should be increased? a. 9 b. 7 c. 10 d. 15 200. Sugar juice is filtering through a conical funnel 20cm, deep and 12cm across top, into a cylindrical container whose diameter is 10cm. When the depth of the juice in the funnel is 10cm, determine the rate at which its level in the cylinder is rising. a. 0.45 b. 1.25 c. 0.75 d. 0.15 201. An airplane, flying horizontally at an altitude of 1km, passes directly over an observer. If the constant speed of the plane is 240kph, how fast is its distance from the observer increasing 30seconds later? a. 214.66 kph b. 256.34 kph c. 324.57 kph d. 137.78 kph 202. A metal disk expands during heating. If its radius increases at the rate of 20 mm per second, how fast is the area of one of its faces increasing when its radius is 8.1 meters? a. 1.018 sq m per sec b. 1.337 sq m per sec c. 0.846 sq m per sec d. 1.632 sq m per sec 203. The structural steel work of a new office building is finished. Across the street 20m from the ground floor of the freight elevator shaft in the building, a spectator is standing and watching the freight elevator ascend at a constant rate of 5 meters per second. How fast is the angle of elevation of the spectator’s line of sight to the elevator increasing 6 seconds after his line of sight passes the horizontal? a. 1/13 b. 1/15 c. 1/10 d. 1/12 204. A boy rides a bicycle along the Quezon Bridge at a rate of 6m /s. 24m directly below the bridge and running at right angles to it is a highway along which an automobile is traveling at the rate of 80m/s. How far is the distance between the boy and the automobile changing when the boy is 6m, past the point directly over the path of the automobile and the automobile is 8m past the point directly under the path of the boy? a. 26 m/s b. 20 m/s c. 28 m/s d. 30 m/s 205. A point moves on the parabola y^2=8 in such a way that the rate of change of the ordinate is always 5 units per sec. How fast is the abscissa changing when the ordinate is 4? a. 5 b. 4 c. 3 d. 7 206. An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to one another. One plane is 150miles from the point and is moving at 450 mph. The other plane is 200 miles from the point and has the speed of 600 mph. How much time does the traffic controller have to get one of the planes on a different flight path? a. 20 min b. 25 min c. 30 min d. 15 min 207. An LRT train 6 m above the ground crosses a street at a speed of 9 m/s, at the instant that a car approaching at a speed of 4 m/s is 12 m up the street. Find the rate of the LRT train and the car are separating one second later. a. 3.64 m/s b. 4.34 m/s c. 6.43 m/s d. 4.63 m/s 208. A street light is 8m from a wall and 4m from a point along the path leading to the shadow of the man 1.8m tall shortening along the wall when he is 3m from the wall. The man walks towards the wall at the rate of 0.6m/s. a. -0.192 m/s b. -1.018 m/s c. -0.826 m/s d. -0.027 m/s 209. A mercury light hangs 12 ft above the island at the center of Ayala Avenue whish is 24 ft wide. A cigarette vendor 5ft tall walks along the curb of the street at a speed of 420 fpm. How fast is the tip of the shadow of the cigarette vendor moving at the same instant? a. 12 fps b. 15 fps c. 10 fps d. 14 fps 210. The sides of an equilateral triangle are increasing at the rate of 10m/s. What is the length of the sides at the instant when the area is increasing 100 sq m/sec? a. 20/√3 b. 22/√3 c. 25/√3 d. 15/√3 211. Water is the flowing into a conical vessel 15cm deep and having a radius of 3.75cm across the top. If the rate at which water is rising is 2cm/s, how fast is the water flowing into the conical vessel when the depth of water is 4cm? a. 6.28 cu m/min b. 4 cu m/min c. 2.5 cu m/ min d. 1.5 cu m/min 212. Two sides of a triangle are 5 and 8 units respectively. If the included angle is changing at the rate of one radian pr second, at what rate is the third side changing when the included angle is 60 degrees? a. 4.95 units/sec b. 5.55 units/sec c. 4.24 units/sec d. 3.87 units/sec 213. The two adjacent sides of a triangle are 5 and 8 meters respectively. If the included angle is changing at the rate of 2 rad/sec, at what rate is the area of the triangle changing if the included angle is 60 degrees? a. 20 sq m/sec b. 25 sq m/sec c. 15 sq m/sec d. 23 sq m/sec 214. A triangular trough is 12m long, 2m wide at the top and 2m deep. If water flows in at the rate of 12 cu m per min, find how fast the surface is rising when the water is 1m deep. a. 1 b. 2 c. 3 d. 4 215. A man starts from a point on a circular track of radius 100m and walks along the circumference at the rate of 40m/min. An observer is stationed at a point on the track directly opposite the starting point and collinear with the center of the circular track. How fast is the man’s distance form the observer changing after one minute? a. -7.95 m/min b. -6.48 m/min c. 8.62 m/min d. 9.82 m/min 216. A plane 3000ft from the earth is flying east at the rate of 120mph. It passes directly over a car also going east at 60mph. How fast are they separating when the distance between them is 5000ft? a. 70.4 ft/sec b. 84.3 ft/sec c. 76.2 ft/sec d. 63.7 ft/sec 217. A horseman gallops along the straight shore of a sea at the rate of 30mph. A battleship anchored 3 miles offshore keeps searchlight trained on him as he moved along. Find the rate of rotation of the light when the horseman is 2 miles down the beach? a. 6.92 rad/sec b. 4.67 rad/sec c. 5.53 rad/sec d. 6.15 rad/sec 218. Find the point in the parabola y^2=4x at which the rate of change of the ordinate and abscissa are equal. a. (1,2) b. (-1,4) c. (2,1) d. (4,4) 219. Water flows into a vertical cylindrical tank, at the rate of 1/5 cu ft/sec. The water surface is rising at the rate of 0.425ft/min. What is the diameter of the tank? a. 6 ft b. 10 ft c. 8 ft d. 4 ft 220. The radius of a sphere is changing at a rate of 2 cm/sec. Find the rate of change of the surface area when the radius is 6cm. a. 96π sq cm/sec b. 78π sq cm/sec c. 84π sq cm /sec d. 68π sq cm/sec 221. The radius of a circle is increasing at the rate of 2cm/min. Find the rate of change of the area when r=6cm. a. 24 π sq cm/sec b. 36 π sq cm/sec c. 18 π sq cm/sec d. 30 π sq cm/sec 222. All edges of a cube are expanding at the rate of 3cm/sec. How fast is the volume changing when each edge is 10cm long? a. 900 cu cm/sec b. 800 cu cm/sec c. 600 cu cm/sec d. 400 cu cm/ sec 223. A spherical balloon is inflated with gas at the rate of 20 cu m/min. How fast is the radius of the balloon changing at the instant the radius is 2cm? a. 0.398 b. 0.422 c. 0.388 d. 0.498 224. The base radius of a cone is changing at a rate of 3cm/sec. Find the rate of change of its volume when the radius is 4cm and its altitude is 6cm. a. 48 π cu cm/sec b. 24 π cu cm/sec c. 18 π cu cm/sec d. 36 π cu cm/sec 225. The edge of cube is changing at a rate of 2 cm/min. Find the rate of change of its diagonal when each edge is 10cm long. a. 3.464 cm/min b. 5.343 cm/min c. 2.128 cm/min d. 6.283 cm/min 226. The radius of a circle is changing at a rate of 4cm/sec. Determine the rate of change of the circumference when the radius is 6cm. a. 8 π cm/sec b. 6 π cm/sec c. 10 π cm/sec d. 4 π cm/sec 227. When a squares of side x are cut from the corners of a 12cm square piece of cardboard, an open top box can be formed by folding up the sides. The volume of this box is given by V=x(12-2x)^2. Find the rate of change of volume when x=1cm. a. 60 b. 40 c. 30 d. 20 228. As x increases uniformly at the rate of 0.002 ft/sec, at what rate is expression (1+x) to the third power increasing when x becomes 8ft? a. 0.486 cfs b. 0.430 cfs c. 0.300 cfs d. 0.346 cfs 229. A trough 10m long has as it ends isosceles trapezoids, altitude 2m, lower base, 2m upper base 3m. If water is let in at a rate of 3 cu m/min, how fast is the water level rising when the water is 1m deep? a. 0.12 b. 0.18 c.0.21 d. 0.28 230. a launch whose deck is 7 m below the level of a wharf is being pulled toward the wharf by a rope attached to a ring on the deck. If a winch pulls in the rope at the rate of 15 m/min, how fast is the launch moving through the water when there are 25m of rope out? a. -15.625 b. 14.525 c. -14.526 d. 15.148 231. An object is dropped freely from a bldg. having a height of 40m. An observer at a horizontal distance of 30m from a bldg is observing the object is it was dropped. Determine the rate at which the distance between the object and the observer is changing after 2sec. a. -11.025 b. 12.25 c. -10.85 d. 14.85 232. Car A moves due east at 30kph at the same instant car B is moving S 30deg E.with a speed of 30kph. The distance from A to B is 30km. Find how fast is the speed between them are separating after one hour. a. 45 kph b. 36 kph c. 40 kph d. 38 kph 233. Water is flowing into a frustum of a cone at a rate of 100 liter/min. The upper radius of the frustum of a cone is 1.5m while the lower radius is 1m and a height of 2m. If the water rises at the rate of 0.04916 cm/sec, find the depth of water. a. 15.5cm b. 10.3cm c. 13.6cm d. 18.9cm 234. Water is flowing into a conical vessel 18 cm deep and 10 cm across the top. If the rate at which the water surface is rising is 27.52 mm/ sec, how fast is the water flowing into the conical vessel when the depth of water is 12cm? a. 9.6 cu cm/sec b. 7.4 cu cm/sec c. 8.5 cu cm/sec d. 6.3 cu cm/sec 235. Sand is falling off a conveyor onto a conical pile at the rate of 15 cu cm/min. The base of the cone is approximately twice the altitude. Find the height of the pile if the height of he pile is changing at the rate of 0.047746 cm/ min. a. 10 cm b. 12 cm c. 8 cm d. 6 cm 236. A company is increasing its production of a certain product at the rate of 100 units per month. The monthly demand function is given by P=100-x/800. Find the rate of change of the revenue with respect to time in months when the monthly production is 4000. a. P9000/month b. P8000/month c. P6000/month d. P4000/month 237. A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at the rate of 0.05 cm/sec and the volume V is 128π cu cm. At what rate is the length h changing when the radius is 2.5cm? a. 0.8192 cm/sec b. 0.7652 cm/sec c. 0.6178 cm/sec d. 0.5214 cm/sec 238. Two sides of a triangle are 15cm and 20cm long respectively. How fast is the third side increasing if the angle between the given sides is 60 degrees and is increasing at the rate of 2deg/sec? a. 0.05 cm/s b. 2.70 cm/s c. 1.20 cm/s d. 3.60 cm/s 239. Two sides of a triangle are 30cm and 40cm respectively. How fast is the area of the triangle increasing if the angle between the sides is 60 degrees and is increasing at the rate of 4deg/sec? a. 20.94 b. 29.34 c. 14.68 d. 24.58 240. A man 6ft tall is walking toward a building at the rate of 5ft/sec. If there is a light on the ground 50ft from a bldg, how fast is the man’s shadow on the bldg growing shorter when he is 30ft from the bldg? a. -3.75 fps b. -7.35 fps c. -5.37 fps d. -4.86 fps 241. The volume of the sphere is increasing at the rate of 6 cu cm/hr. At what rate is its surface area increasing when the radius is 50 cm (in cu cm/hr)? a. 0.36 b. 0.50 c. 0.40 d. 0.24 242. A particle moves in a plane according to the parametric equations of motions: x=t^2, y=t^3. Find the magnitude of the acceleration when t=2/3. a. 4.47 b. 5.10 c. 4.90 d. 6.12 243. A particle moves along the right-hand part of the curve 4y^3=x^2 with a speed Vy=dy/dx=constant at 2. Find the speed of motion when y=4. a. 12.17 b. 14.10 c. 15.31 d. 16.40 244. The equations of motion of a particle moving in a plane are x=t^2, y=3t-1 when t is the time and x and y are rectangular coordinates. Find the speed of motion at the instant when t=2. a. 5 b. 7 c. 9 d. 10 245. A particle moves along the parabola y^2=4x with a constant horizontal component velocity of 2m/s. Find the vertical component of the velocity at the point (1,2). a. 2 m/s b. 7 m/s c. 5 m/s d. 4 m/s 246. The acceleration of the particle is given by a=2+12t in m/s^2 where t is the time in minutes. If the velocity of this particle is 11 m/s after 1min, find the velocity after 2minutes. a. 31 m/s b. 45 m/s c. 37 m/s d. 26 m/s 247. A particle moves along a path whose parametric equations are x=t^3 and y=2t^2. What is the acceleration when t=3sec. a. 18.44 m/sec^2 b. 15.93 m/sec^2 c. 23.36 m/sec^2 d. 10.59 m/sec^2 248. A vehicle moves along a trajectory having coordinates given as x=t^3 and y=1-t^2. The acceleration of the vehicle at any point of the trajectory is a vector having magnitude and direction. Find the acceleration when t=2. a. 12.17 b. 13.20 c. 15.32 d. 12.45 249. The search light of a lighthouse which is positioned 2km from the shoreline is tracking a car which is traveling at a constant speed along the shore. If the searchlight is rotating at the rate of 0.25 rev per hour, determine the speed of the car when it is 1km away from the point on the shore nearest to the lighthouse. a. 3.93 kph b. 4.16 kph c. 2.5 kph d. 1.8 kph 250. A light is at the top of a pole 80 ft high. A ball is dropped at the same height from a point 20 ft from the light. Assuming that the ball falls according to S=16t^2, how fast is the shadow of the ball moving along the ground 1 second later? a. -200 ft/sec b. -180 ft/sec c. -240 ft/sec d. -140 ft/sec 251. Water is poured at the rate of 8 cu ft/min into a conical shaped tank, 20 ft deep and 10 ft diameter at the top. If the tank has a leak in the bottom and the water level is rising at the rate of 1 inch/min, when the water is 16 ft deep, how fast is the water leaking? a. 3.81 cu ft/min b. 4.28 cu ft/min c. 2.96 cu ft/min d. 5.79 cu ft/min 252. An airplane is flying at a constant speed at an altitude of 10000ft on a line that will take it directly over an observer on the ground. At a given instant the observer notes that the angle of elevation of the airplane is π/3 radians and is increasing at the rate of 1/60 rad/sec. Find the speed of the airplane. a. -222.22 ft/sec b. -232.44 ft/sec c. -332.22 ft/sec d. -432.12 ft/sec 253. A horizontal trough is 16 m long and its ends are isosceles trapezoids with an altitude of 4m lower base of 4m and an upper base of 6m. If the water level is decreasing at the rate of 25 cm/min, when the water is 3m deep, at what rate is water being drawn from the trough? a. 22 cu m/min b. 25 cu m/min c. 20 cu m/min d. 30 cu m/min 254. The sides of an equilateral triangle is increasing at rate of 10 cm/min. What is the length of the sides if the area is increasing at the rate of 69.82 sq cm/min? a. 8 cm b. 10 cm c. 5 cm d. 15 cm 255. The two adjacent sides of a triangle are 6m and 8m respectively. If the included angle is changing at the rate of 3 rad/min, at what rate is the area of a triangle changing if the included angle is 30 degrees? a. 62.35 sq m b. 65.76 sq m c. 55.23 sq m d. 70.32 sq m 256. Water is pouring into a swimming pool. After t hours, there are t+√t gallons in the pool. At what rate is the water pouring into the pool when t=9hours? a. 7/6 gph b. 1/6 gph c. 3/2 gph d. ½ gph 257. A point on the rim of a flywheel of radius cm, has a vertical velocity of 50 cm/sec at a point P, 4cm above the x-axis. What is the angular velocity of the wheel? a. 16.67 rad/sec b. 14.35 rad/sec c. 19.95 rad/sec d. 10.22 rad/sec 258. A spherical balloon is filled with air at the rate of 2 cu cm/min. Compute the time rate of change of the surface are of the balloon at the instant when its volume is 32π/3 cu cm. a. 2 cu cm/min b. 3 cu cm/min c. 4 cu cm/min c. 5 cu cm/min 259. The coordinate (x,y) in ft of a moving particle P are given by x=cos(t)-1 and y=2sin(t)+1, where t is the time in seconds. At what extreme rates in fps is P moving along the curve? a. 2 and 1 b. 3 and 2 c. 2 and 0.5 d. 3 and 1 260. A bomber plane is flying horizontally at a velocity of 440 m/s and drops a bomb to a target h meters below the plane. At the instant the bomb was dropped, the angle of depression of the target is 45 degrees and is increasing at the rate of 0.05 rad/sec. Determine the value of h. a. 4400 m b. 2040 m c. 3500 m d. 6704 m 261. Glycerine is flowing into a conical vessel 18cm deep and 10 cm across the top at the rate of 4 cu cm per min. The deep of glyerine is h cm. If the rate which the surface is rising is 0.1146 cm/min, find the value of h. a. 12 cm b. 16 cm c. 20 cm d. 25 cm 262. Helium is escaping from a spherical balloon at the rate of 2 cu cm/min. When the surface area is shrinking at the rate of sq cm/min, find the radius of the spherical balloon. a. 12 b. 16 cm c. 20 cm d. 25 cm 263. Water is running into hemispherical bowl having a radius of 10 cm at a constant rate of 3 cu cm/min. When the water is h cm deep, the water level is rising at the rate of 0.0149 cm/min. What is the value of h? a. 4 cm b. 6 cm c. 2 cm d. 5 cm 264. A train, starting noon, travels north at 40 mph. Another train starting from the same pint at 2pm travels east at 50mph. How fast are the two trains separating at 3pm? a. 56.15 mph b. 98.65 mph c. 46.51 mph d. 34.15 mph 265. An automobile is traveling at 30 fps towards north is approaching an intersection. When the automobile is 120ft from the intersection, a truck traveling at 40fps towards east is 60ft from the same intersection. The automobile and the truck are on the roads that are at right angles to each other. How fast are they separating after 6 sec? a. 47.83 fps b. 87.34 fps c. 23.74 fps d. 56.47 fps 266. A train, starting noon, travels north at 40 mph. Another train starting from the same point at 2pm travels east at 50 mph. How fast are the trains separating after a long time? a. 64 mph b. 69 mph c. 46 mph d. 53 mph 267. At noon a car drives from A towards the east at 60mph. Another car starts from B towards A at 30 mph. B has a direction and distance of N 30 degrees east and 42m respectively from A. Find the time when the cars will be nearest each other. a. 24 min after noon b. 23 min after noon c. 25 min after noon d. 26 min after noon 268. A ferris wheel 15 m in diameter makes 1 rev every 2 min. If the center of the wheel is 9m above the ground, how many fast is a passenger in the wheel moving vertically when he is 12.5 above the ground? a. 20.84 m/min b. 24.08 m/min c. 22.34 m/min d. 25.67 m/min 269. A bomber plane, flying horizontally 3.2 km above the ground is sighting on at a target on the ground directly ahead. The angle between the line of sight and the pad of the plane is changing at the rate of 5/12 rad/min. When the angle is 30 degrees, what is the speed of the plane in mph? a. 200 b. 260 c. 220 d. 240 270. Two railroad tracks are perpendicular to each other. At 12pm there is a train at each track was approaching the crossing at 50kph, one being 100km the other 150km away from the crossing. How fast in kph is the distance between the two trains changing at 4pm? a. 67.08 kph b. 68.08 kph c. 69.08 kph d. 70.08 kph 271. a ball is thrown vertically upward and its distance from the ground is given as S=104t-16t^2. Find the maximum height to which the ball will rise if S is expressed in meters and t in seconds. a. 169m b. 190m c. 187m d. 169m 272. If f(x)=ax^3+bx^2+cx, determine the value of a so that the graph will have a point of inflection at (1,-1) and so that the slope of the inflection tangent there will be -3. a. 2 b. 5 c. 3 d. 4 273. If f(x)=ax^3+bx^2, determine the values of a and b so that the graph will have a point of inflection at (2,16). a. -1, 6 b. -2, 5 c. -1, 7 d. -2, 8 274. Under what condition is the inflection point of y=ax^3+bx^2+cx+d on the y-axis? a. b=0 b. b=1 c. b=3 d. b=4 275. Find the equation of the curve whose slope is 4x-5 and passing through (3,1). a. 2x^2-5x-2 b. 5x^2-9x-1 c. 5x^2+7x-2 d. 2x^ 2-8x+5 276. The point (3,2) is on a curve and at any point (x,y) on the curve the tangent line has a slope equal to 2x-3. Find the equation of the curve. a. y=x^2-3x-4 b. y=x^2-3x+2 c. y=x^2+8x+5 d. y=x^3+3x-3 277. If m is the slope of the tangent line to the curve y=x^2-2x^2+x at the point (x,y), find the instantaneous rate of change of the slope m per unit change in x at the point (2,2). a. 8 b. 9 c. 10 d. 11 278. Suppose the daily profit from the production and sale of x units of a product is given by P=180x-(x^2)/1000-2000. At what rate is the profit changing when the number of units produced and sold is 100 and is increasing at 10 units per day? a. P1798 b. P1932 c. P2942 d. P989 279. The population of a city was found to be given by P=40500e^(0.03t) where t is the number of years after 1990. At what rate is the population expected to be growing in 2000? a. 1640 b. 2120 c. 2930 d. 1893 280. A bridge is h meters above a river which lies perpendicular to the bridge. A motorboat going 3 m/s passes under the bridge at the same instant that a man walking 2 m/s reaches that point simultaneously. If the distance between them is changing, at the rate of 2.647 m/s after 3 seconds, find the value of h. a. 10 b. 12 c. 14 d. 8 281. What is the area bounded by the curve x^2=-9y and the line y+1=0. a. 6 b. 5 c. 4 d. 3 282. What is the area bounded by the curve y^2=x and the line x-4=0? a. 10 b. 32/3 c. 31/3 d. 11 283. What is the area bounded by the curve y^2=4x and x^2=4y. a. 6 b. 7.333 c. 6.666 d. 5.333 284. Find the area bounded by the curve y=9-x^2 and the x-axis. a. 25 sq units b. 36 sq units c. 18 sq units d. 30 sq units 285. Find the area bounded by the curve y^2=9x and its latus rectum. a. 10.5 b. 13.5 c. 11.5 d. 12.5 286. Find the area bounded by the curve 5y^2=164x and the curve y^2=8x-24. a. 30 b. 20 c. 16 d. 19 287. Find the area bounded by the curve y^2=4x and the line 2x+y=4. a. 10 b. 9 c. 7 d. 4 288. Find the area bounded by the curve y=1/x with and upper limit of y=2 and a lower limit of y=10. a. 1.61 b. 2.61 c. 1.81 d. 2.81 289. By integration, determine the area bounded by the curves y=6x-x^2 and y=x^2-2x. a. 25.60 sq units b. 21.33 sq units c. 17.78 sq units d. 30.72 sq units 290. What is the appropriate total area bounded by the curve y=sin x and y=0 over the interval 0≤x≤2π (in radians). a. π/2 b. 2 c. 4 d. 0 291. What is the area between y=0, y=3x^2, x=0 and x=2? a. 8 b. 24 c. 12 d. 6 292. Determine the tangent to the curve 3y^2=x^3 at (3,3) and calculate the area of the triangle bounded by the tangent line, the x-axis and the line x=3. a. 3.50 sq units b. 2.50 sq units c. 3.00 sq units d. 4.00 sq units 293. Find the areas bounded by the curve y=8-x^3 and the x-axis. a. 12 sq units b. 15 sq units c. 13 sq units d. 10 sq units 294. Find the area in the first quadrant bounded by the parabola, y^2=4x and the line x=3 and x=1. a. 9.535 b. 5.595 c. 5.955 d. 9.955 295. Find the area (in sq units) bounded by the parabola x^2-2y=0 and x^2=-2y+8. a. 11.7 b. 4.7 c. 9.7 d. 10.7 296. In x years from now, one investment plan will be generating profit at the rate of R1(x)=50+x^2 pesos per yr, while a second plan will be generating profit at the rate R2(x)=200+5x pesos per yr. For how many yrs will the second plan be more profitable one? Compute also the net excess profit if the second plan would be used instead of the first. a. 15yrs, P1687.50 b. 12yrs, P1450.25 c. 14yrs, P15640.25 d. 10yrs, P1360.25 297. An industrial machine generates revenue at the rate R(x)=5000-20x^2 pesos per yr and results in cost that accumulates at the rate of C(x)=2000+10x^2 pesos per yr. For how many yrs (x) is the use of this machine profitable? Compute also that net earnings generated by the machine at this period. a. 10yrs, P20000 b. 12yrs, P25000 c. 15yrs, P30000 d. 14yrs, P35000 298. Find the area under one arch of the curve y=sin(x/2). a. 4 b. 7 c. 3 d. 5 299. Find the area bounded by the curve y=arc sin x, x=1 and y=π/2 on the first quadrant. a. 0 b. 2 c. 1 d. 3 300. Find the area bounded by the curve y=8-x^3, x=0, y=0. a. 12 b. 11 c. 15 d. 13 301. Find the area bounded by the curve y=cos hx, x=0, x=1 and y=0. a. 1.175 b. 1.234 c. 1.354 d. 1.073 302. Find the area in the first quadrant under the curve y-sin hx from x=0 to x=1. a. 0.543 b. 0.453 c. 0.345 d. 0.623 303. Find the area of the region in the first quadrant bounded by the curves y=sin x, y=cos x and the y-axis. a. 0.414 b. 0.534 c. 0.356 d. 0.486 304. Find the area of the region bounded by the x-axis, the curve y=6x-x^2 and the vertical lines x=1 and x=4. a. 24 b. 23 c. 25 d. 22 305. Find the area bounded by the curve y=e^x, y=e^-x and x=1, by integration. a. [(e-1)^2]/e b. (e^2-1)/e c. (e-1)/e d. [(e-1)^2]/(e^2) 306. Suppose a company wants to introduce a new machine that will produce a rate of annual savings S(x)=150-x^2 where x is the number of yrs of operation of the machine, while producing a rate of annual costs of C(x)=(x^2)+(11x/4). For how many years will it be profitable to use this new machine? a. 7 yrs b. 6 yrs c. 8 yrs d. 10 yrs 307. Suppose a company wants to introduce a new machine that will produce a rate of annual savings S(x)=150-x^2 where x is the number of yrs of operation of the machine, while producing a rate of annual costs of C(x)=(x^2)+(11x/4). What are the net total savings during the first year of use of the machine? a. 122 b. 148 c. 257 d. 183 308. Suppose a company wants to introduce a new machine that will produce a rate of annual savings S(x)=150-x^2 where x is the number of yrs of operation of the machine, while producing a rate of annual costs of C(x)=(x^2)+(11x/4). What are the net total savings over the entire period of use of the machine? a. 771 b. 826 c. 653 d. 711 309. The price in pesos for a certain product is expressed as p(x)=900-80x-x^2 when the demand for the product is x units. Also the function p(x)=x^2+10x gives the price in pesos when the supply is x units. Find the consumer and producers surplus. a. P4500; P3375 b. P3400; P4422 c. P5420; P3200 d. P4000; P3585 310. A horse is tied ouside of a circular fence of radius 4m by a rope having a length of 4π m. Determine the area on which the horse can graze. a. 413.42 sq m b. 484.37 sq m c. 398.29 sq m d. 531.36 sq m 311. A dog is tied to an 8m circular tank by a 3m length of cord. The cord remains horizontal. Find the area over which the dog can move. a. 16.387 sq m b. 15.298 sq m c. 10.286 sq m d. 13.164 sq m 312. Find the area bounded by the curve y^2=8(x-4), the line y=4, y-axis and x-axis. a. 18.67 b. 14.67 c. 15.67 d. 17.67 313. Find the area enclosed by the parabola y^2=8x and the latus rectum. a. 32/3 sq units b. 29/4 sq units c. 41/2 sq units d. 33/2 sq units 314. What is the area bounded y the curve x^2=-9y and the line y+1=0 a. 6 sq units b. 5 sq units c. 2 sq units d. 4 sq units 315. What is the area bounded by the curve y^2=x and the line x-4=0. a. 23/4 sq units b. 32/3 sq units c. 54/4 sq units d. 13/5 sq units 316. Find the area bounded by The parabola x^2=4y and y=4. a. 21.33 sq units b. 33.21 sq units c. 31.32 sq units d. 13.23 sq units 317. What is the area bounded by the curve y^2=-2x and the line x=-2. a. 18/3 sq units b. 19/5 sq units c. 16/3 sq units d. 17/7 sq units 318. Find the area enclosed by the curve x^2+8y+16=0 the x-axis, y-axis and the line x- 4=0. a. 10.67 b. 9.67 c. 8.67 d. 7.67 319. Find the area bounded by the parabola y=6x-x (square) and y=x(square)-2x. Note, the parabola intersects at point (0,0) and (4,8). a. 44/3 b. 64/3 c. 74/3 d. 54/3 320. Find the area of the portion of the curve y=cos x from x=0 to x=π/2. a. 1 sq unit b. 2 sq units c. 3 sq units d. 4 sq units 321. Find the area of the portion of the curve y=sin x from x=0 to x=π. a. 2 sq units b. 3 sq units c. 1 sq unit d. 4 sq units 322. Find the area bounded by the curve r^2=4cos2φ. a. 8 sq units b. 2 sq units c. 4 sq units d. 6 sq units 323. Find the area enclosed by the curve r^2=4cosφ. a. 4 b. 8 c. 16 d. 2 324. Determine the period and amplitude of the function y=2sin5x. a. 2π/5, 2 b. 3π/2, 2 c. π/5, 2 d. 3π/10, 2 325. Determine the period and amplitude of the function y=5cos2x. a. π, 5 b. 3π/2, 2 c. π/5, 2 d. 3π/10, 2 326. Determine the period and amplitude of the function y=5sinx. a. 2π, 5 b. 3π/2, 5 c. π/2, 5 d. π, 5 327. Determine the period and amplitude of the function y=3 cos x. a. 2π, 3 b. π/2, 3 c. 3/2, 3 d. π, 3 328. Find the area of the curve r^2=a^2cosφ. a. a^2 b. a c. 2a d. a^3 329. Find the area of the region bounded by the curve r^2=16cosθ. a. 32 sq units b. 35 sq units c. 27 sq units d. 30 sq units 330. Find the area enclosed by the curve r=a (1-sinθ). a. (3a^2)π/2 b. (2a^2)π c. (3a^2)π d. (3a^2)π/5 331. Find the surface area of the portion of the curve x^2=y from y=1 to y=2 when it is revolved about the y-axis. a. 19.84 b. 17.86 c. 16.75 d. 18.94 332. Find the area of the surface generated by rotating the portion of the curve y=(x^3)/3 from x=0 to x=1 about the x-axis. a. 0.638 b. 0.542 c. 0.782 d. 0.486 333. Find the surface area of the portion of the curve x^2+y^2=4 from x=0 to x=2 when it is revolved about the y-axis. a. 8π b. 16π c. 4π d. 12π 334. Compute the surface area generated when the first quadrant portion if the curve x^2- 4y+8=0 from x=0 to x=2 is revolved about the y-axis. a. 30.64 b. 28.32 c. 26.42 d. 31.64 335. Find the total length of the curve r=4 (1-sin θ) from θ=90deg to θ=270deg and also the total perimeter of the curve. a. 16, 32 b. 18, 36 c. 12, 24 d. 15, 30 336. Find the length of the curve r=4sinθ from θ=0 to θ=90deg and also the total length of the curve. a. 2π; 4π b. 3π; 6π c. π; 2π d. 4π; 8π 337. Find the length of the curve r=a (1-cos θ) from θ=0 to θ=π and also the total length of curve. a. 4a; 8a b. 2a; 4a c. 3a; 6a d. 5a; 10a 338. Find the total length of the curve r= a cos θ. a. πa b. 2πa c. 3πa/2 d. 2πa/3 339. Find the length of the curve having a parametric equations of x= a cos^3 θ y=a sin^2 θ from θ=0 to θ=2π. a. 5a b. 6a c. 7a d. 8a 340. Find the centroid of the area bounded by the curve y=4-x^2 the line x=1 and the coordinate axes. a. 1.85 b. 0.46 c. 1.57 d. 2.16 341. Find the centroid of the area under y=4-x^2 in the first quadrant. a. 0.75 b. 0.25 c. 0.50 d. 1.15 342. Find the centroid of the area in first quadrant bounded by the curve y^2=4ax and latus rectum. a. 3a/5 b. 2a/5 c. 4a/5 d. 1a 343. A triangular section has coordinates of A(2,2), B(11,2) and C(5,8). Find the coordinates of the centroid of the triangular section. a. (7, 4) b. (6, 4) c. (8, 4) d. (9, 4) 344. The following cross section has the following given coordinates. Compute for the centroid of the given cross section A(2,2); B(5,8); C(7,2); D(2,0) and E(7,0). a. 4.6, 3.4 b. 4.8, 2.9 c. 5.2, 3.8 d. 5.3, 4.1 345. Sections ABCD is a quadrilateral having the given coordinates A(2,3); B(8,9); C(11,3); D(11,0). Compute the coordinates of the centroid of the quadrilateral. a. (7.33, 4) b. (7, 4) c. (6.22, 3.8) d. (7.8, 4.2) 346. A cross section consists of a triangle ABC and a semi circle with AC as its diameter. If the coordinates of A(2,6); B(11,9) and C(14,6), compute the coordinates of the centroid of the cross section. a. 4.6, 3.4 b. 4.8, 2.9 c. 5.2, 3.8 d. 5.3, 4.1 347. Locate the centroid of the area bounded by the parabola y^2=4x, the line y=4 ad the y-axis. a. 6/5, 3 b. 2/5, 3 c. 3/5, 3 d. 4/5, 3 348. Find the centroid of the area bounded by the curve x^2=-(y-4), the x-axis and the yaxis on the first quadrant. a. ¾, 8/5 b. 5/4, 7/5 c. 7/4, 6/5 d. ¼, 9/5 349. Locate the centroid of the area bounded by the curve y^2=-3(x-6)/2 the x-axis and the y-axis on the first quadrant. a. 12/5, 9/8 b. 13/5, 7/8 c. 14/5, 5/8 d. 11/5, 11/8 350. Locate the centroid of the area bounded by the curve 5y^2=16x and y^2=8x-24 on the first quadrant. a. x=2.20; y=1.51 b. x=1.50; y=0.25 c. x=2.78; y=1.39 d. x=1.64; y=0.26 351. Locate the centroid of the area bounded by the parabola x^2=8y and x^2=16(y-2) in the first quadrant. a. x=2.12; y=1.6 b. x=3.25; y=1.2 c. x=2.67; y=2.0 d. x=2; y=2.8 352. Given the area in the first quadrant bounded by x^2=8y, the line y-2 and the y-axis. What is the volume generated this area is revolved about the line y-2=0? a. 53.31 cu units b. 45.87 cu units c. 28.81 cu units d. 33.98 cu units 353. Given the area in the first quadrant bounded by x^2=8y, the line x=4 and the x-axis. What is the volume generated by revolving this area about y-axis? a. 78.987 cu units b. 50.265 cu units c. 61.523 cu units d. 82.285 cu units 354. Given the area in the first quadrant bounded by x^2=8y, the line y-2=0 and the yaxis. What is the volume generated when this area is revolved about the x-axis? a. 20.32 cu units b. 34.45 cu units c. 40.21 cu units d. 45.56 cu units 355. Find the volume formed by revolving the hyperbola xy=6 from x=2 to x=4 about the x-axis. a. 28.27 cu units b. 25.53 cu units c. 23.23 cu units d. 30.43 cu units 356. The region in the first quadrant under the curve y=sin h x from x=0 to x=1 is revolved about the x-axis. Compute the volume of solid generated. a. 1.278 cu units b. 2.123 cu units c. 3.156 cu units d. 1.849 cu units 357. A square hole of side 2cm is chiseled perpendicular to the side of a cylindrical post of radius 2cm. If the axis of the hole is going to be along the diameter of the circular section of the post, find the volume cut off. a. 15.3 cu cm b. 23.8 cu cm c. 43.7 cu cm d. 16.4 cu cm 358. A hole radius 1cm is bored through a sphere of radius 3cm, the axis of the hole being a diameter of a sphere. Find the volume of the sphere which remains. a. (64π√2)/3 cu cm b. (66π√3)/3 cu cm c. (70π√2)/3 cu cm d. (60π√2)/3 cu cm 359. Find the volume of common to the cylinders x^2+y^2=9 and y^2+z^2=9. a. 241 cu m b. 533 cu m c. 424 cu m d. 144 cu m 360. Given is the area in the first quadrant bounded by x^2= 8y, the line y-2=0 and the yaxis. What is the volume generated when this area is revolved about the line y-2=0. a. 28.41 b. 26.81 c. 27.32 d. 25.83 361. Given is the area in the first quadrant bounded by x^2=8y, the line x=4 and the xaxis. What is the volume generated when this area is revolved about the y-axis? a. 50.26 b. 52.26 c. 53.26 d. 51.26 362. The area bounded by the curve y^2=12 and the line x=3 is revolved about the line x=3. What is the volume generated? a. 185 b. 187 c. 181 d. 183 363. The area in the second quadrant of the circle x^2+y^2=36 is revolved about the line y+10=0. What is the volume generated? a. 2218.63 b. 2228.83 c. 2233.43 d. 2208.53 364. The area enclosed by the ellipse (x^2)/9 + (y^2)/4 = 1 is revolved about the line x=3, what is the volume generated? a. 370.3 b. 360.1 c. 355.3 d. 365.10 365. Find the volume of the solid formed if we rotate the ellipse (x^2)/9 + (y^2)/4 = 1 about the line 4x+3y=20. a. 48 π^2 cu units b. 45 π^2 cu units c. 40 π^2 cu units d. 53 π^2 cu units 366. The area on the first and second quadrant of the circle x^2+y^2=36 is revolved about the line x=6. What is the volume generated? a. 2131.83 b. 2242.46 c. 2421.36 d. 2342.38 367. The area on the first quadrant of the circle x^2+y^2=25 is revolved about the line x=5. What is the volume generated? a. 355.31 b. 365.44 c. 368.33 d. 370.32 368. The area on the second and third quadrant of the circle x^2+y^2=36 is revolved about the line x=4. What is the volume generated? a. 2320.30 b. 2545.34 c. 2327.25 d. 2520.40 369. The area on the first quadrant of the circle x^2+y^2=36 is revolved about the line y+10=0. What is the volume generated? a. 3924.60 b. 2229.54 c. 2593.45 d. 2696.50 370. The area enclosed by the ellipse (x^2)/16 + (y^2)/9 = 1 on the first and 2nd quadrant is revolved about the x-axis. What is the volume generated? a. 151.40 b. 155.39 c. 156.30 d. 150.41 371. The area enclosed by the ellipse 9x^2+16y^2=144 on the first quadrant is revolved about the y-axis. What is the volume generated? a. 100.67 b. 200.98 c. 98.60 d. 54.80 372. Find the volume of an ellipsoid having the equation (x^2)/25 + (y^2)/16 + (z^2)/4 = 1. a. 167.55 b. 178.40 c. 171.30 d. 210.20 373. Find the volume of a prolate spheroid having the equation (x^2)/25 + (y^2)/9 + (z^2)/9 = 1. a. 178.90 cu units b. 184.45 cu units c. 188.50 cu units d. 213.45 cu units 374. The region in the first quadrant which is bounded by the curve y^2=4x, and the lines x=4 and y=0, is revolved about the x-axis. Locate the centroid of the resulting solid of revolution. a. 8/3 b. 7/3 c. 10/3 d. 5/3 375. The region in the first quadrant which is bounded by the curve x^2=4y, and the line x=4, is revolved about the line x=4. Locate the centroid of the resulting solid of revolution. a. 0.8 b. 0.5 c. 1 d. 0.6 376. The area bounded by the curve x^3=y, the line y=8 and the y-axis is to be revolved about the y-axis. Determine the centroid of the volume generated. a. 5 b. 6 c. 4 d. 7 377. The area bounded by the curve x^3=y, and the x-axis is to be revolved about the xaxis. Determine the centroid of the volume generated. a. 7/4 b. 9/4 c. 5/4 d. ¾ 378. The region in the 2nd quadrant, which is bounded by the curve x^2=4y, and the line x=-4, is revolved about the x-axis. Locate the cenroid of the resulting solid of revolution. a. -4.28 b. -3.33 c. -5.35 d. -2.77 379. The region in the 1st quadrant, which is bounded by the curve y^2=4x, and the line x=-4, is revolved about the line x=4. Locate the cenroid of the resulting solid of revolution. a. 1.25 units b. 2 units c. 1.50 units d. 1 unit 380. Find the moment of inertia of the area bounded by the curve x^2=4y, the line y =1 and the y-axis on the first quadrant with respect to x-axis. a. 6/5 b. 7/2 c. 4/7 d. 8/7 381. Find the moment of inertia of the area bounded by the curve x^2=4y, the line y=1 and the y-axis on the first quadrant with respect to y-axis. a. 19/3 b. 16/15 c. 13/15 d. 15/16 382. Find the moment of inertia of the area bounded by the curve x^2=8y, the line x=4 and the x-axis on the first quadrant with respect to x-axis. a. 1.52 b. 2.61 c. 1.98 d. 2.36 383. Find the moment of inertia of the area bounded by the curve x^2=8y, the line x=4 and the x-axis on the first quadrant with respect to y-axis. a. 25.6 b. 21.8 c. 31.6 d. 36.4 384. Find the moment of inertia of the area bounded by the curve y^2=4x, the line x=1 and the x-axis on the first quadrant with respect to x-axis. a. 1.067 b. 1.142 c. 1.861 d. 1.232 385. Find the moment of inertia of the area bounded by the curve y^2=4x, the line x=1 and the x-axis on the first quadrant with respect to y-axis. a. 0.571 b. 0.682 c. 0.436 d. 0.716 386. Determine the moment of inertia with respect to x-axis of the region in the first quadrant which is bounded by the curve y^2=4x, the line y=2 and y-axis. a. 1.6 b. 2.3 c. 1.3 d. 1.9 387. Find the moment of inertia of the area bounded by the curve y^2=4x, the line y=2 and the y-axis on the first quadrant with respect to y-axis. a. 0.095 b. 0.064 c. 0.088 d. 0.076 388. Find the moment of inertia with respect to x-axis of the area bounded by the parabola y^2=4x and the line x=1. a. 2.35 b. 2.68 c. 2.13 d. 2.56 389. What is the integral of sin^6(φ)cos^4 (φ) dφ if the upper limit is π/2 and lower limit is 0? a. 0.0184 b. 1.0483 c. 0.1398 d. 0.9237 390. Evaluate the integral of cos^7 φ sin^5 φ dφ if the upper limit is 0. a. 0.1047 b. 0.0083 c. 1.0387 d. 1.3852 391. What is the integral of sin^4 x dx if the lower limit is 0 and the upper limit is π/2? a. 1.082 b. 0.927 c. 2.133 d. 0.589 392. Evaluate the integral of cos^5 φ dφ if the lower limit is 0 and the upper limit is π/2. a. 0.533 b. 0.084 c. 1.203 d. 1.027 393. Evaluate the integral (cos3A)^8 dA from 0 to π/6. a. 27π/363 b. 35π/768 c. 23π/765 d. 12π/81 394. What is the integral of sin^5 x cos^3 x dx if the lower limit is 0 and the upper limit is π/2? a. 0.0208 b. 0.0833 c. 0.0278 d. 0.0417 395. Evaluate the integral of 15sin^7 (x) dx from 0 to π/2. a. 6.857 b. 4.382 c. 5.394 d. 6.139 396. Evaluate the integral of 5 cos^6 x sin^2 x dx if the upper limit is π/2 and the lower limit is 0. a. 0.307 b. 0.294 c. 0.415 d. 0.186 397. Evaluate the integral of 3(sin x)^3 dx from 0 to π/2. a. 2 b. π c. 6 d. π/2 398. A rectangular plate is 4 feet long and 2 feet wide. It is submerged vertically in water with the upper 4 feet parallel and to 3 feet below the surface. Find the magnitude of the resultant force against one side of the plate. a. 38 w b. 32 w c. 27 w d. 25 w 399. Find the force on one face of a right triangle of sides 4 m, and altitude of 3m. The altitude is submerged vertically with the 4m side in the surface. a. 58.86 kN b. 53.22 kN c. 62.64 kN d. 66.27 kN 400. A plate in the form of a parabolic segment of base 12m and height of 4m is submerged in water so that the base is in the surface of the liquid. Find the force on the face of the plate. a. 502.2 kN b. 510.5 kN c. 520.6 kN d. 489.1 kN 401. A circular water main 4 meter in diam. is closed by a bulkhead whose center is 40 m below the surface of the water in the reservoir. Find the force on the bulkhead. a. 4931 kN b. 5028 kN c. 3419 kN d. 4319 kN 402. A plate in the form of parabolic segment is 12m in height and 4m deep and is partly submerged in water so that its axis is parallel to end 3m below the water surface. Find the force acting on the plate. a. 993.26 kN b. 939.46 kN c. 933.17 kN d. 899.21 kN 403. A cistern in the form of an inverted right circular cone is 20 m deep and 12 m diameter at the top. If the water is 16 m deep in the cistern, find the work done in Joules in pumping out the water. The water is raised to a point of discharge 10 m above the top cistern. a. 68166750 Joules b. 54883992 Joules c. 61772263 Joules d. 76177640 Joules 404. A bag containing originally 60 kg of flour is lifted through a vertical distance of 9m. While it is being lifted, flour is leaking from the bag at such rate that the number of pounds lost is proportional to the square root of the distance traversed. If the total loss of flour is 12 kg find the amount of work done in lifting the bag. a. 4591 Joules b. 4290 Joules c. 5338 Joules d. 6212 Joules 405. According to Hooke’s law, the force required to stretch a helical spring is proportional to the distance stretched. The natural length of a given spring is 8 cm. a force of 4kg will stretch it to a total length of 10 cm. Find the work done in stretching it from its natural length to a total length of 16 cm. a. 6.28 Joules b. 5.32 Joules c. 4.65 Joules d. 7.17 Joules 406. The top of an elliptical conical reservoir is an ellipse with major axis 6m and minor axis 4m. it is 6m deep and full of water. Find the work done in pumping the water to an outlet at the top of the reservoir. a. 554742 Joules b. 473725 Joules c. 493722 Joules d. 593722 Joules 407. A bag of sand originally weighing 144 kg is lifted at a rate of 3m/min. the sand leaks out uniformly at such rate that half of the sand is lost when the bag has been lifted 18m. find the work done in lifting the bag of sand at this distance. a. 6351 Joules b. 4591 Joules c. 5349 Joules d. 5017 Joules 408. A cylindrical tank having a radius of 2m and a height of 8m is filled with water at a depth of 6m. Compute the work done in pumping all the liquid out of the top of the container. a. 3 698 283 Joules b. 4 233 946 Joules c. 5 163 948 Joules d. 2 934 942 Joules 409. A right cylindrical tank of radius 2m and a height 8m is full of water. Find the work done in pumping the tank. Assume water to weigh 9810 N/m^3. a. 3945 kN . m b. 4136 kN . m c. 2846 kN . m d. 5237 kN . m 410. A conical vessel 12m across the top and 15m deep. If it contains water to a depth of 10m find the work done in pumping the liquid to the top of the vessel. a. 12 327.5 kN . m b. 24 216.2 kN . m c. 14 812.42 kN . m d. 31 621 kN . m 411. A hemispherical vessel of diameter 8m is full of water. Determine the work done in pumping out the top of the tank in Joules. a. 326 740 pi b. 627 840 pi c. 516 320 pi d. 418 640 pi 412. A spring with a natural length of 10cm is stretched by 1/2 cm by a Newton force. Find the work done in stretching from 10 cm to 18cm. Express your answer in joules. a. 7.68 Joules b. 8.38 Joules c. 7.13 Joules d. 6.29 Joules 413. A 5 lb. monkey is attached to a 20 ft hanging rope that weighs 0.3 lb/ft. the monkey climbs the rope up to the top. How much work has it done? a. 160 b. 170 c. 165 d. 180 414. A bucket weighing 10 Newton when empty is loaded with 90 Newton of sand and lifted at 10 cm at a constant speed. Sand leaks out of a hole in a bucket at a uniform rate and one third of sand is lost by the end of the lifting process in Joules. a. 850 Joules b. 900 Joules c. 950 Joules d. 800 Joules 415. A conical vessel is 12 m across the top and 15 m deep. If it contains water to a depth of 10m find the work done in pumping the liquid to a height 3m above the top of the vessel. a. 560pi w N.m b. 660 pi w N.m c. 520 pi w N.m d. 580 pi w N.m 416. A small in the sack of rice cause some rice to be wasted while the sack is being lifted vertically to a height of 30m. The weight lost is proportional to the cube root of distance traversed. If the total loss was 16 kg, find the work done in lifting the said sack of rice which weighs 110kg. a. 2940 kg.m b. 2369 kg.m c. 3108 kg.m d. 2409 kg.m 417. A hemispherical tank of diameter 20 ft is full of oil weighing 20pcf. The oil is pumped to a height of 10 ft, above the top of the tank by an engine of 1/2 horsepower. How long will it take the engine to empty the tank? a. 1 hr. 44.72 min b. 1 hr. 15.47 min c. 1 hr. 24.27 min d. 2 hrs. 418. A full tank consists of a hemisphere of radius 4m surmounted by a circular cylinder of the same radius of altitude 8m. Find the work done in pumping the water to an outlet of the top of the tank. a. (2752/3) pi w b. (2255/3) pi w c. (2527/3) pi w d. (5722/3) pi w 419. Determine the differential equation of a family of lines passing thru (h, k). a. (y-k) dx – (x-h) dy = 0 b. (x-h) + (y-k) = dy/dx c. (x-h) dx – (y-k) dy = 0 d. (x+h) dx – (y-k) dy = 0 420. What is the differential equation of the family of parabolas having their vertices at the origin and their foci on the x-axis a. 2x dy – y dx = 0 b. x dy + y dx = 0 c. 2y dx – x dy = 0 d. dy/dx – x = 0 421. Find the differential equations of the family of lines passing through the origin. a. ydx – xdy = 0 b. xdy – ydx = 0 c. xdx + ydy = 0 d. ydx + xdy = 0 422. The radius of the moon is 1080 miles. The gravitation acceleration of the moons surface is 0.165 miles the gravitational acceleration at the earth’s surface. What is the velocity of escape from the moon in miles per second? a. 2.38 b. 1.47 c. 3.52 d. 4.26 423. Find the equation of the curve at every point of which the tangent line has a slope of 2x. a. x = -y^2 + C b. y = -x^2 + C c. x = y^2 + C d. y = x^2 + C 424. The radius of the earth is 3960 miles. If the gravitational acceleration of earth surface is 31.16 ft/sec^2, what is the velocity of escape from the earth in miles/sec? a. 6.9455 b. 5.4244 c. 3.9266 d. 7.1842 425. Find the velocity of escape of the Apollo spaceship as it is projected from the earth’s surface that is the minimum velocity imparted to it so that it will never return. The radius of the earth is 400 miles and the acceleration of the spaceship is 32.2 ft/sec^2. a. 40478 kph b. 50236 kph c. 30426 kph d. 60426 kph 426. The rate of population growth of a country is proportional to the number of inhabitants. If a population of a country now is 40 million and expected to double in 25 years, in how many years is the population be 3 times the present? a. 39.62 yrs. b. 28.62 yrs. c. 18.64 yrs. d. 41.2 yrs. 427. From the given differential equation xdx+6y^5dy = 0 solve for the constant of integration when x = 0, y = 2. a. 27x dx + 4y^2 dy = 0 b. 58 c. 48 d. 64 428. Find the equation of the curve which passes through points (1, 4) and (0, 2) if d^2 y/ dx^2 = 1 a. 2y = x^2 + 3x + 4 b. 4y = 2x^2 + x + 4 c. 5y = x^2 + 2x + 2 d. 3y = x^2 + x + 4 429. The rate of population growth of a country is proportional to the number of inhabitants. If a population of a country now is 40 million and 50 million in 10 years time, what will be its population 20years from now? a. 56.19 b. 71.29 c. 62.18 d. 59.24 430. The Bureau of Census record in 1980 shows that the population in the country doubles compared to that of 1960. In what year will the population trebles assuming that the rate of increase in the population is proportional to the population? a. 34.60 b. 31.70 c. 45.65 d. 38.45 431. A tank contains 200 liters of fresh water. Brine containing 2 kg/liter of salt enters the tank at the rate of 4 liters per min, and the mixture kept uniform by stirring, runs out at 3 liters per min. Find the amount of salt in the tank after 30 min. a. 196.99 kg b. 186.50 kg c. 312.69 kg d. 234.28 kg 432. In a tank are 100 liters of brine containing 50 kg total of dissolved salt. Pure water is allowed to run into the tank at the rate of 3 liters per minute. Brine runs out of the tank at rate of 2 liters per minute. The instantaneous concentration in the tank is kept uniform by stirring. How much salt is in the tank at the end of 1 hour? a. 20.50 b. 18.63 c. 19.53 d. 22.40 433. Determine the general solution of xdy + ydx=0. a. xy = c b. ln xy = c c. ln x + ln y = c d. x + y = c 434. The inverse laplace transform of s/[(square) + (w square)] is: a. sin wt b. w c. (e exponent wt) d. cos st 435. The laplace transform of cos wt is: a. s/[(square) + (w square)] b. w/[(square) + (w square)] c. w/s + w d. s/s + w 436. K divided by [(s square) + (k square)] is inverse laplace transform of: a. cos kt b. sin kt c. (e exponent Ky) d. 1.0 437. Find the inverse transform of [2/(s+1)] – [(4/(s+3)] is equal to: a. [2 e (exp – t) – 4e (exp – 3t)] b. [e (exp – 2t) + e (exp – 3t)] c. [e (exp – 2t) – e (exp - 3t)] d. [2e (exp – t) – 2e (exp - 2t)] 438. What is the laplace transform of e^(-4t) a. 1/ (s + 1) b. 1/ (s + 4) c. 1/ (s – 4) d. 1/ (s + t) 439. Determine the laplace transform of I(S) = 200 / [(s^2) + 50s + 10625] a. I(S) = 2e^(-25t) sin100t b. I(S) = 2te^(-25t) sin100t c. I(S) = 2e^(-25t) cos100t d. I(S) = 2te^(-25t) cos100t 440. Determine the inverse laplace transform of (s+a) / [(s+a) ^2 + w^2] a. e^(-at) sin wt b. te^(-at) cos wt c. t sin wt d. e^(-at) cos wt 441. Determine the inverse laplace transform of 100/ [(S+10) (S+20)] a. 10e^(-10t) – 20e^(-20t) b. 10e^(-10t) + 20e^(-20t) c. 10e^(-10t) – 10e^(-20t) d. 20e^(-10t) + 10e^(-20t) 442. A thin heavy uniform iron rod 16m long is bent at the 10 m mark forming a right angle L – shaped piece 6m by 10m of bend. What angle does the 10m side make with the vertical when the system is in equilibrium? a. 28° 12’ b. 19° 48’ c. 24° 36’ d. 26° 14’ 443. Three men carry a uniform timber. One takes hold at one end and the other two carry by means of a crossbar placed underneath. At what point of timber must the bar be placed so that each man may carry one third of the weight of the weight of the timber? The timber has a length of 12 m. a. 4m b. 5m c. 2.5 m d. 3m 444. A painters scaffold 30m long and a mass of 300 kg, is supported in a horizontal position by a vertical ropes attached at equal distances from the ends of the scaffold. Find the greatest distance from the ends that the ropes may be attached so as to permit a 200 kg man to stand safely at one end of scaffold. a. 8m b. 7m c. 6m d. 9m 445. A cylindrical tank having a diameter of 16 cm weighing 100 kN is resting on a horizontal floor. A block having a height of 4 cm is placed on the side of the cylindrical tank to prevent it from rolling. What horizontal force must be applied at the top of the cylindrical tank so that it will start to roll over the block? Assume the block will not slide and is firmly attached to the horizontal floor. a. 68.36 kN b. 75.42 kN c. 58.36 kN d. 57.74 kN 446. Two identical sphere weighing 100 kN are each place in a container such that the lower sphere will be resting on a vertical wall and a horizontal wall and the other sphere will be resting on the lower sphere and a wall making an angle of 60 degrees with the horizontal. The line connecting the two centers of the spheres makes an angle of 30 degrees with the horizontal surface. Determine the reaction between the contact of the two spheres. Assume the walls to be frictionless. a. 150 b. 120 c. 180 d. 100 447. The 5 m uniform steel beam has a mass of 600 kg and is to be lifted from the ring B with two chains, AB of length 3m, and CB of length 4m. Determine the tension T in chain AB when the beam is clear of the platform. a. 2.47 kN b. 3.68 kN c. 5.42 kN d. 4.52 kN 448. A man attempts to support a stack of books horizontally by applying a compressive force of F=120 N to the ends of the stack with his hands, determine the number of books that can be supported in the stack if the coefficient of friction between any two books is 0.40. a. 15 books b. 20 books c. 10 books d. 12 books 449. Two men are just to lift a 300 kg weight of crowbar when the fulcrum for this lever is 0.3m from the weight and the man exerts their strengths at 0.9 m and 1.5 m respectively from the fulcrum. If the men interchange positions, they can raise a 340 kg weight. What force does each man exert? a. 25 kg, 40 kg b. 35 kg, 45 kg c. 40 kg, 50 kg d. 30 kg, 50 kg 450. A man exert a maximum pull of 1000 N but wishes to lift a new stone door for his cave weighing 20 000 N. if he uses lever how much closer must the fulcrum be to the stone than to his hand? a. 10 times nearer b. 20 times farther c. 10 times farther d. 20 times nearer 451. A simple beam having a span of 6m has a weight of 20 kN/m. It carries a concentrated load of 20 kN at the left end and 40 kN at 2m from the right end of the beam. If it is supported at 2m from the left end and the right end, compute the reaction at the right end of the beam. a. 40 kN b. 20 kN c. 50 kN d. 30 kN 452. When one boy is sitting 1.20 m from the center of a seesaw another boy must sit on the other side 1.50 m from the center to maintain an even balance. However, when the first boy carries an additional weight of 14 kg and sit 1.80 m from the center, the second boy must move 3m from the center to balance, Neglecting the weight of the see weight of the heaviest boy. a. 42 kg b. 35 kg c. 58 kg d. 29 kg 453. A wire connects a middle of links AC and AB compute the tension in the wire if AC carries a uniform load of 600 N/m. AC is 4.5 m long and BC is 7.5 m. point A is hinged on the wall and joint C is also hinged connecting the links AC and CB. AC is horizontal while B is supported by roller acting on the wall AB. a. 2700 N b. 3600 N c. 300 N d. 2200 N 454. An airtight closed box of weight P is suspended from a spring balance. A bird of weight W is place on the floor of the bow, and the balance reads W + P. If the bird flies without accelerating. What is the balance reading? a. P + W b. P c. P – W d. P + 2W 455. A tripod whose legs are each 4 meters long supports a load of 1000 kg. the feet of the tripod are at the vertices of a horizontal equilateral triangle whose side are 3.5 meters. Determine the load of each leg. a. 386.19 kg b. 347.29 kg c. 214.69 kg d. 446.27 kg 456. A uniform square table top ABCD having sides 4m long is supported by three vertical supports at A, E and F, E is midway n the side BC and F is 1m from D along the side DC. Determine the share of load in percent carried by supports at A, E and F. a. A = 29%, E = 42%, F = 29% b. A = 32%, E = 46%, F = 20% c. A = 28%, E = 40%, F = 32% d. A = 36%, E = 32%, F = 32% 457. The square steel plate has a mass of 1800 kg with mass at its center G. Calculate the tension at each of the three cables with which the plate is lifted while remaining horizontal. a. Ta = Tb = 6.23 kN, Tc = 10.47 kN b. Ta = Tb = 7.47 kN, Tc = 7.84 kN c. Ta = Tb = 5.41 kN, Tc = 9.87 kN d. Ta = Tb = 4.42 kN, Tc = 6.27 kN 458. A horizontal Circular platform of radius R is supported at three points A, B and C on its circumference. A and B are 90 degrees apart and C is 120 degrees from A. The platform carries a vertical load of 400 kN at its center and 100 kN at a point d on the circumference diametrically opposite A. Compute the reaction at C. a. 253.45 kN b. 321.23 kN c. 310.10 kN d. 287.67 kN 459. A ladder 4m long having a mass of 15kg is resting against a floor and an wall for which the coefficients of static friction are 0.30 for the floor to which a man having a mass of 70 kg can climb without causing the plank to slip if the plank makes an angle of 40 degrees with the horizontal. a. 2 b. 1 c. 2.5 d. 3 460. A homogenous block having dimension of 4cm by 8cm is resting on an inclined plane making an angle of θ with the horizontal. The block has a weight of 20 kN. If the coefficient of friction between the block and the inclined plane is 0.55, find the value of θ before the block starts to move. The 8cm side is perpendicular to the inclined plane. a. 26.57° b. 28.81° c. 27.7° d. 23.4° 461. A uniform ladder on a wall at A and at the floor at B. Point A is 3.6m above the floor and point B is 1.5m away from the wall. Determine the minimum coefficient of friction at B required for a mass weighing 65 kg to use the ladder assuming that there is no friction at A. a. 0.42 b. 0.50 c. 0.48 d. 0.54 462. A block having a mass of 250 kg is placed on top of an inclined plane having a slope of 3 vertical to 4 horizontal. If the coefficient of friction between the block and the inclined plane is 0.15, determine the force P that may be applied parallel to the inclined plane to keep block from sliding down the plane. a. 1177.2 N b. 1088.2 N c. 980.86 N d. 1205.30 N 463. A 3.6 m ladder weighing 180 N is resting on a horizontal floor at A and on the wall at B making an angle of 30 degrees from the vertical wall. When a man weighing 800 N reaches a point 2.4m from the lower end (point A), the ladder is just about to slip. Determine the coefficient of friction between the ladder and the floor if the coefficient of friction between the ladder and the wall is 0.20. a. 0.35 b. 0.42 c. 0.28 d. 0.56 464. A dockworker adjusts a spring line (rope) which keeps the ship from drifting along side a wharf. If he exerts a pull of 200N on the rope, which ahs 1 ¼ turns around the mooring bit, what force T can he support? The coefficient of friction between the rope and the cast-steel mooring bit is 0.30. a. 2110 N b. 1860 N c. 155 N d. 142 N 465. Determine the distance “x” to which the 90 kg painter can climb without causing the 4m ladder to slip at its lower end A. The top of the 15 kg ladder has a small roller, and the ground coefficient of static friction is 0.25. the lower end of the ladder is 1.5 m away from the wall. a. 2.55 m b. 3.17 m c. 1.58 m d. 0.1 m 466. The uniform pole of length 4m and mass 100kg is leaned against a vertical wall. If the coefficient of static friction between the supporting surfaces and the ends of the poles is 0.25, calculate the maximum angle θ at which the pole may be placed with the vertical wall before it starts to slip. a. 28.07° b. 26.57° c. 31.6° d. 33.5° 467. A horizontal force P acts on the top of a 30 kg block having a width of 25 cm, and a height of 50cm. if the coefficient of friction between the block and the plane is 0.33, what is the value of P for motion to impend? a. 7.5 kg b. 5.3 kg c. 6.6 kg d. 8.2 kg 468. A 600 N block rests on a 30° plane. If the coefficient of static friction is 0.30 and the coefficient of kinetic friction is 0.20, what is the value of P applied horizontally to prevent the block from sliding down the plane? a. 141.85 N b. 183.29 N c. 119.27 N d. 126.59 N 469. A 600 N block rests on a 30° plane. If the coefficient of static friction is 0.30 and the coefficient of kinetic friction is 0.20, what is the value of P applied horizontally to keep the block moving up the plane? a. 527.31 N b. 569.29 N c. 427.46 N d. 624.17 N 470. Solve for the force P to obtain equilibrium. Angle of friction is 25° between block and the inclined plane. a. 96.46 kg b. 77.65 kg c. 69.38 kg d. 84.22 kg 471. A 200 kg crate impends to slide down a ramp inclined at an angle of 19.29° with the horizontal. What is the frictional resistance? Use g = 9.81 m/s^2. a. 648.16 N b. 638.15 N c. 618.15 N d. 628.15 N 472. A 40kg block is resting on an inclined plane making an angle of 20° from the horizontal. If the coefficient of friction is 0.60, determine the force parallel to the incline that must be applied to cause impending motion down the plane. Use g = 9.81 a. 87 N b. 82 N c. 72 N d. 77 N 473. A 40 kg block is resting on an inclined plane making an angle of θ from the horizontal. Coefficient of friction is 0.60, find the value of θ when force P = 36.23 is applied to cause the motion upward along the plane. a. 20° b. 30° c. 28° d. 23° 474. A 40 kg block is resting on an inclined plane making an angle θ from the horizontal. The block is subjected to a force 87N parallel to the inclined plane which causes an impending motion down the plane. If the coefficient of motion is 0.60, compute the value of θ. a. 20° b. 30° c. 28° d. 23° 475. A rectangular block having a width of 8cm and height of 20 cm, is reating on a horizontal plane. If the coefficient of friction between he horizontal plane and the block is 0.40, at what point above the horizontal plane should horizontal force P will be applied at which tipping will occur? a. 10 cm b. 14 cm c. 12 cm d. 8 cm 476. A ladder is resting on a horizontal plane and a vertical wall. If the coefficient of friction between the ladder, the horizontal plane and the vertical wall is 0.40, determine the angle that the ladder makes with the horizontal at which it is about to slip. a. 46.4° b. 33.6° c. 53.13° d. 64.13° 477. Three identical blocks A, B and C are placed on top of each other are place on a horizontal plane with block B on top of A and C on top of B. The coefficient of friction between all surfaces is 0.20. if block C is prevented from moving by a horizontal cable attached to a vertical wall, find the horizontal force in Newton that must be applied to B without causing motion to impend. Each block has a mass of 50kg. a. 294.3 Newtons b. 274.7 Newtons c. 321.3 Newtons d. 280.5 Newtons 478. A car moving downward on an inclined plane which makes an angle of θ from the horizontal. The distance from the front wheel to the rear wheel is 400cm and its centroid is located at 50 cm from the surface of the plane. If only rear wheels provide breaking, what is the value of θ so that the car will start to slide if the coefficient of friction is 0.6? a. 15.6° b. 18.4° c. 16.8° d. 17.4° 479. A 40kg block is resting on an inclined plane making an angle of 20° from the horizontal. If the coefficient of friction is 0.60, determine the force parallel to the inclined plane that must be applied to cause impending motion up the plane. a. 355.42 N b. 354.65 N c. 439.35 N d. 433.23 N 480. A block weighing 40 kg is placed on an inclined plane making an angle of θ from th horizontal. If the coefficient of friction between the block and the inclined plane is 0.30, find the value of θ, when the block impends to slide downward. a. 16.70° b. 13.60° c. 15.80° d. 14.50° 481. A block having a weight W is resting on an inclined plane making an angle of 30° from the horizontal. If the coefficient of friction between the block and the inclined plane is 0.50. Determine the value of W is a force 300 N applied parallel to the inclined plane to cause an impending motion upward. a. 321.54 N b. 493.53 N c. 450.32 N d. 354.53 N 482. 40kg block is resting on an inclined plane making an angle of 20° from the horizontal. The block is subjected to a force 87 N parallel to the inclined plane which causes an impending motion down the plane. Compute the coefficient of friction between the block and the inclined plane. a. 0.60 b. 0.80 c. 0.70 d. 0.50 483. A 20kg cubical block is resting on an inclined plane making an angle of 30° with the horizontal. If the coefficient of friction between the block and the inclined plane is 0.30, what force applied at the uppermost section which is parallel to the inclined plane will cause the 20kg block to move up? a. 134 N b. 130 N c. 146 N d. 154 N 484. The coefficient of friction between the 60 kN block is to remain in equilibrium, what is the maximum allowable magnitude for the force P? a. 15 kN b. 12 kN c. 18 kN d. 24 kN 485. Find the value of P acting to the left that is required to pull the wedge out under the 500kg block. Angle of friction is 20° for all contact surfaces. a. 253.80 kg b. 242.49 kg c. 432.20 kg d. 120.50 kg 486. The accurate alignment of a heavy duty engine on its bed is accomplished by a screw adjusted wedge with a 20° taper as shown in the figure. Determine the horizontal thrust P in adjusting screw necessary to raise the mounting if the wedge supports one fourth of the total engine weight of 20 000N. The total coefficient of friction for all surfaces is 0.25. a. 4640 N b. 4550 N c. 5460 N d. 6540 N 487. Two blocks connected by a horizontal link AB are supported on two rough planes as shown. The coefficient of friction for block A on the horizontal plane is 0.40. the angle of friction for block B on the inclined plane is 15°. What is the smallest weight of block A for which equilibrium of the system can exists? a. 1000 kN b. 1500 kN c. 2000 kN d. 500 kN 488. Is the system in equilibrium? If not, find the force P so that the system will be in equilibrium. a. 80 kg b. 90 kg c. 100 kg d. 70 kg 489. A 12 kg block of steel is at rest on a horizontal cable. The coefficient of static friction between the block a table is 0.52. What is the magnitude of the force acting upward 62° from the horizontal that will just start the block moving? a. 65.9 N b. 78.1 N c. 70.2 N d. 72.4 N 490. The pull required to overcome the rolling resistance of a wheel is 90 N acting at the c enter of the wheels. If the weight of the wheel is 18 000 N and the diameter of the wheel is 300mm, determine the coefficient of rolling resistance. a. 0.60 mm b. 0.75 mm c. 0.50 mm d. 0.45 mm 491. A 1000 kN weight is to be moved by using 50 mm diameter rollers. If the coefficient of the rolling resistance for the rollers and floor is 0.08 mm and that for rollers and weight is 0.02 mm. determine the pull required. a. 2000 N b. 1500 N c. 2500 N d. 1000 N 492. A ball is thrown vertically upward with an initial velocity of 3m/sec from a window of a tall building. The ball strikes at the sidewalk at ground level 4 sec later. Determine the velocity with which the ball hits the ground. a. 30.86 m/sec b. 36.24 m/sec c. 42.68 m/sec d. 25.27 m/sec 493. A train starts from rest at station P and stops from station Q which is 10km from station P. the maximum possible acceleration of the train is 15km/hour/min. if the maximum allowable speed is 60 kph, what is the least time the train go from P to Q? a. 15 min b. 10 min c. 12 min d. 20 min 494. A car starting from rest picks up at a uniform rate and passes three electric post in succession. The posts are spaced 360 m apart along a straight road. The car takes 10sec to travel from first post to sec post and takes 6 sec to go from the second to the third post. Determine the distance from the starting point to the first post. a. 73.5 m b. 80.3 m c. 77.5 m d. 70.9 m 495. A stone is dropped from the deck of Mactan Bridge. The sound of the splash reaches the deck 3 seconds later. If sound travels 342 m/s in still air, how high is the deck of Mactan Bridge above the water? a. 40.6 m b. 45.2 m c. 57.3 m d. 33.1 m 496. At a uniform rate of 4 drops per second, water is dripping from a faucet. Assuming the acceleration of each drop to be 9.81 m/sec^2 and no air resistance, find the distance between two successive drops in mm if the upper drop has been in motion for 3/8 seconds. a. 1230 mm b. 2340 mm c. 2231 mm d. 1340 mm 497. A racing car during the Marlboro Championship starts from rest and has a constant acceleration of 4m/sec^2. What is its average velocity during the first 5 seconds of motion? a. 10 m/s b. 4 m/s c. 6 m/s d. 12 m/s 498. A train is to commute between Tutuban station and San Andres station with a top speed of 250 kph but can not accelerate nor decelerate faster than 4 m/s. What is its min. distance between the two stations in order for the train to be able to reach its top speed? a. 1106.24 b. 1205.48 c. 1309.26 d. 1026.42 499. A block having a weight of 400N rests on an inclined plane making an angle of 30° with the horizontal is initially at rest after it was released for 3 sec, find the distance the block has traveled assuming there is no friction between the block and plane. Determine the velocity after 3 seconds. a. 14.71 m/sec b. 15.39 m/sec c. 14.60 m/sec d. 13.68 m/sec 500. A car accelerates for 6 sec from an initial velocity of 10 m/s. the acceleration is increasing uniformly from zero to 8 m/s^2 in 6 sec. during the next 2 seconds, the car decelerates at a constant rate of m/s^2. Compute the total distance the car has traveled from the start after 8 sec. a. 169 m b. 172 m c. 180 m d. 200 m 501. A train passing at point A at a speed of 72 khp accelerates at 0.75 m/s^ 2 from one minute along a straight path then decelerates at 1.0 m/s^2. How far from point a will be 2min after passing point A. a. 6.49 km b. 7.30 km c. 4.65 km d. 3.60 km 502. A car accelerate 8 seconds from rest, the acceleration increasing uniformly from zero to 12 m/s^2. During the next 4 sec, the car decelerates at a constant rate of -11 m/s^2. Compute the distance the car has traveled after 12 sec from the start. a. 232 m b. 240 m c. 302 m d. 321 m 503. A car moving at 6 m/s accelerates at 1.5 m/s^2 for 15 sec, then decelerates at a rate of 1.2 m/s^2 for 12 sec. Determine the total distance traveled. a. 558.75 m b. 543.80 m c. 384.90 m d. 433.75 m 504. A train starting at initial velocity of 30 kph travels a distance 21 km in 8 min. determine the acceleration of the train at this instant. a. 0.0865 m/s^2 b. 0.0206 m/s^2 c. 0.3820 m/s^2 d. 0.0043 m/s^2 505. From a speed of 75 kph, a car decelerates at the rate of 500 m/min^2 along a straight path. How far in meters will it travel in 45 sec? a. 790.293 m b. 791.357 m c. 796.875 m d. 793.328 m 506. An object experiences rectilinear acceleration a(t)= 10 – 2t. How far does it travel in 6 sec if its initial velocity is 10 m/s? a. 182 b. 168 c. 174 d. 154 507. An object experiences the velocity as shown in the diagram. How far will it move in 6 seconds? a. 40 m b. 60 m c. 80 m d. 100 m 508. An object is accelerating to the right along a straight path at 2m/s. the object begins with a velocity 10 m/s to the left. How far does it travel in 15 seconds? a. 125 m b. 130 m c. 140 m d. 100 m 509. What is the acceleration of a body that increases in velocity from 20 m/s to 40 m/s in 3 sec? Answer in SI. a. 8.00 m/s^2 b. 6.67 m/s^2 c. 50. m/s^2 d. 7.0 m/s^2 510. A shell is fired vertically upward with an initial velocity of 2000 fps. It is timed to burst in 7 sec. Four seconds after firing the first shell, a second shell is fired with the same velocity. This shell is time to burst in 5 sec. An observer stationed in a captive balloon near the line of fire hears both burst. At the same instance what is the elevation or height of the balloon. Assume velocity of sound to be 1100 fps. a. 10 304 ft b. 18 930 ft c. 13 400 ft d. 14 030 ft 511. An object from a height of 92 m and strikes the ground with a speed of 19 m/s. Determine the height that the object must fall in order to strike with a speed of 24 m/s. a. 146.94 m b. 184.29 m c. 110.12 m d. 205.32 m 512. A ball is dropped from a balloon at a height of 195 m. if the balloon is rising 29.3 m/s. Find the highest point reached by the ball and the time of flight. a. 238.8 m b. 487.3 m c. 328.4 m d. 297.3 m 513. A ball is thrown vertically upward with an initial velocity of 3m/sec from a window of a tall building. The ball strikes at the sidewalk at ground level 4 sec later. Determine the velocity with which the ball hits the ground and the height of the window above the ground level. a. 36.2 m/s; 66.79 m b. 24.4 m/s; 81.3 m c. 42.3 m/s; 48.2 m d. 53.2 m/s; 36.8 m 514. A ball is dropped freely from a balloon at a height 195 m. If the balloon is rising 29.3 m/s. Find the highest point reached by the ball and the velocity of the ball as it strikes the ground. a. 43.76 m; 68.44 m/s b. 22.46 m; 71.66 m/s c. 36.24 m; 69.24m/s d. 12.8 m; 31.2 m/s 515. How far does the automobile move while its speed increases uniformly from 15 kph to 45 kph in 20 sec? a. 185 m b. 167 m c. 200 m d. 172 m 516. An automobile is moving at 20 kph and accelerates at 0.5 m/s^2 for a peroiud of 45 sec. Compute the distance traveled by the car at the end of 45 sec. a. 842.62 m b. 765.45 m c. 672.48 m d. 585.82 m 517. A ball is thrown vertically upward with an initial velocity of 3m/sec from a window of a tall building, which is 70 m above the ground level. How long will it take for the ball to hit the ground? a. 3.8 sec b. 4.1 sec c. 5.2 sec d. 6.1 sec 518. A ball is thrown vertically upward with an initial velocity of 3m/sec from a window of a tall building. The ball strikes the ground 4 sec later. Determine the height of the window above the ground. a. 66.331 m b. 67.239 m c. 54.346 m d. 72.354 m 519. A stone was dropped freely from a balloon at a height of 190m above the ground. The balloon is moving upward at a speed of 30 m/s. Determine the velocity of the stone at it hits the ground. a. 56.43 m/s b. 68.03 m/s c. 62.45 m/s d. 76.76 m/s 520. A ball is thrown vertically at a speed of 20 m/s from a building 100 m above the ground. Find the velocity and the position of the ball above the ground after 5 seconds. a. 3.34 m, 45.23 m/s b. 4.54 m, 47.68 m/s c. 5.67 m, 56.42 m/s d. 6.23 m, 34.76 m/s 521. A ball is thrown vertically at a speed of 30 m/s from a building 200 m above the ground. Determine the velocity and the time that it strikes the ground. a. 11.50 sec, 65.80 m/s b. 11.45 sec, 66.59 m/s c. 10.30 sec, 67.21 m/s d. 10.14 sec, 69.45 m/s 522. A ball is thrown vertically with a velocity of 20 m/s from the top of a building 100m high. Find the velocity of the ball at a height of 40 m above the ground. a. 39.71 m/s b. 40.23 m/s c. 39.88 m/s d. 39.68 m/s 523. A ball is shot at a ground level at an angle of 60 degrees with the horizontal with an initial velocity of 100 m/s. Determine the height of the ball after 2 seconds. a. 162.46 m b. 153.59 m c. 175.48 m d. 186.42 m 524. A ball is shot at an average speed of 200 m/s at an angle of 20° with the horizontal. What would be the velocity of the ball after 8 seconds? a. 188.21 m/s b. 154.34 m/s c. 215.53 m/s d. 198.37 m/s 525. A projectile has a velocity of 200 m/s acting at an angle 20 degrees with the horizontal. How long will it take for the projectile to hit the ground surface? a. 13.95 sec b. 15.75 sec c. 10.11 sec d. 24.23 sec 526. A solid homogenous circular cylinder and a solid homogenous sphere are placed at equal distances from the end of an inclined plane. Assuming that no slipping occurs as the two bodies roll down the plane, which of them will reach the end of the plane first? Assume that they have the same weight and radius. a. sphere b. cylinder c. both cylinder and sphere d. none of these 527. A homogenous sphere rolls down as inclined plane making an angle of 30° with the horizontal. Determine the minimum value of the coefficient of friction which will prevent slipping. a. 0.165 b. 0.362 c. 1.028 d. 0.625 528. At what weight “h” above the billiard table surface should a billiard ball of radius 3cm be struck by a horizontal impact in order that the ball will start moving with no friction between the ball and the table? a. 4.9 cm b. 3.4 cm c. 4.2 cm d. 5.5 cm 529. A common swing 7.5 m high is designed for a static load of 1500 N (tension in the rope is equal to 1500 N). Two boys each weighing 500 N are swinging on it. How much many degrees on each side of the vertical can they swing without exceeding the designed load? a. 41.41° b. 45.45° c. 30.35° d. 54.26° 530. A wooden block weighing 20N rests on a turn table having a radius of 2m at a distance on 1m from the center. The coefficient of friction between the block and the turn table is 0.30. The rotation of the table is governed by the equation Ø = 4t^2 where Ø is in radians and t in seconds. If the table starts rotating from rest at t=0, determine the time elapsed before the block will begin to slip. a. 0.21 sec b. 0.55 sec c. 1.05 sec d. 0.10 sec 531. A ball at the end of a cord 121 cm long is swinging with a complete vertical just enough velocity to keep it in the top. If the ball is released from the cord where it is at the top point of its path, where will it strike the ground 245 cm below the center of the circle. a. 297.61 cm b. 332.64 cm c. 258.37 cm d. 263.63 cm 532. At what RPM is the ferriswheel turning when the riders feel “weightless” or zero gravity every time the each rider is at the topmost part of the wheel 9m in radius? a. 9.97 rpm b. 8.58 rpm c. 10.73 rpm d. 9.15 rpm 533. A wooden block having a weight of 50 N is placed at a distance 1.5m from the center of a circular platform rotating at a speed of 2 radians per second. Determine the minimum coefficient of friction of the blocks so that it will not slide. Radius of circular platform is 3m. a. 0.61 b. 0.84 c. 0.21 d. 1.03 534. A 2N weight is swing in a vertical circle of 1m radius and the end of the cable will break if the tension exceeds 500 N. Which of the following most nearly gives the angular velocity of the weight when the cable breaks? a. 49.4 rad/sec b. 37.2 rad/sec c. 24.9 rad/sec d. 58.3 rad/sec 535. A weight is attached to a chord and forms a conical pendulum when it is rotated about the vertical axis. If the period of rotation is 0.2 sec, determine the velocity of the weight if the chord makes an angle of 25° with the vertical. a. 0.146 m/s b. 0.823 m/s c. 1.028 m/s d. 0.427 m/s 536. A ball having a weight of 4N is attached to a cord 1.2 m long and is revolving around a vertical axis so that the cord makes an angle of 20° with the vertical axis. Determine the rpm. a. 28.17 b. 24.16 c. 22.12 d. 25.18 537. A wheel is rotating at 4000 rpm. If it experience a deceleration of 20 rad/sec^2 through how many revolutions will it rotate before it stops? a. 400 b. 698 c. 520 d. 720 538. Find the maximum acceleration of a mass at the end of a 2m long string. It swing like a pendulum with a maximum angle of 30°. a. 4.91 m/s^2 b. 3.61 m/s^2 c. 6.21 m/s^2 d. 7.21 m/s^2 539. A turbine started from rest to 180 rpm in 6 min at a constant acceleration. Find the number of revolution that it makes within the elapsed time. a. 550 revolutions b. 540 revolutions c. 630 revolutions d. 500 revolutions 540. Traffic travels at 65 mph around banked highway curved with a radius of 3000 feet. What banking angle is necessary such that friction will not be required to resist the centrifugal force? a. 3.2° b. 2.5° c. 5.4° d. 18° 541. The rated speed of a highway curve of 60m radius of 50 kph. If the coefficient of friction between the tires and the road is 0.60, what is the maximum speed at which a car can round a curve without skidding? a. 93.6 kph b. 84.2 kph c. 80.5 kph d. 105.2 kph 542. A solid disk flywheel (I = 200 kg.m) is rotating with a speed of 900 rpm. What is the rotational kinetic energy? a. 730 x 10^3 J b. 680 x 10^3 J c. 888 x 10^3 J d. 1100 x 10^3 J 543. A cyclist on a circular track of radius r = 800ft travelling at 27 ft/s. His speed at the tangential direction increases at the rate of 3 ft/s^2. What is the cyclist’s total acceleration? a. 2.8 ft/s^2 b. -3.12 ft/s^2 c. -5.1 ft/s^2 d. 3.31 ft/s^2 544. An automobile travels on a perfectly horizontal, unbanked circular track of radius R. The coefficient of friction between the tires and the track is 0.3. If the car’s velocity is 15 m/s, what is the smallest radius it may travel without skidding? a. 68m b. 69.4 m c. 76.5 m d. 71.6 m 545. Determine the angle of super elevation for a highway curves of 183 m radius, so that there will be no “slide thrust” for a speed of 72 kilometer per hour. At what speed will skidding impend if the coefficient of friction is 0.3? a. 12.57°; 31.72 m/s b. 13.58°; 25.49 m/s c. 15.29°; 34.24 m/s d. 10.33°; 30.57 m/s 546. A child places a picnic basket on the outer rim of merry go round that has a radius of 4.6m and revolves once every 24 sec. How large must the coefficient of static friction be for the basket to stay on the merry go round? a. 0.032 b. 0.024 c. 0.045 d. 0.052 547. A driver’s manual that a driver traveling at 48kph and desiring to stop as quickly as possible travels 4m before the foot reaches the brake. The car travels and additional 21 m before coming to rest. What coefficient of friction is assumed in this calculation? a. 0.43 b. 0.34 c. 0.56 d. 0.51 548. A point on the rim of a rotating flywheel changes its speed its speed from 1.5m/s to 9 m/s while it moves 60 m. If the radius of the wheel is 1m, compute the normal acceleration at the instant when its speed is 6m/s. a. 36 m/s^2 b. 24 m/s^2 c. 18 m/s^2 d. 20 m/s^2 549. The angular speed of a rotating flywheel a radius of 0.5m, is 180/π rpm. Compute the value of its normal acceleration and the tangential speed. a. 16 m/s^2; 2 m/s b. 18 m/s^2; 3 m/s c. 14 m/s^2; 1.5 m/s d. 12 m/s^2; 1.0 m/s 550. A pulley has an angular velocity of 2 rad/sec, and a tangential speed of 4 m/s. Compute the normal acceleration. a. 8 m/sec^2 b. 6 m/sec^2 c. 4 m/sec^2 d. 3 m/sec^2 551. A 10.7 kN car travelling at 134 m/s attempts to round an unbanked curve with a radius of 61 m. What force of friction is required to keep the car on its circular path? a. 3211 N b. 3445 N c. 3123 N d. 4434 N 552. A rotating wheel has a radius of 2 feet and 6 inches. A point on the rim of the wheel moves 30ft in 2sec. Find the angular velocity of the wheel. a. 6 rad/sec b. 2 rad/sec c. 4 rad/sec d. 5 rad/sec 553. A prismatic AB bar 6m long has a weight of 500 N. It is pin connected at one end at A. If it is rotated about a vertical axis at Ai how fast would it be rotated when it makes an angle of 30° with the vertical? a. 1.68 rad/sec b. 2.58 rad/sec c. 1.22 rad/sec d. 2.21 rad/sec 554. A prismatic bar weighing 25 kg is rotated horizontally about one of its ends at a speed of 2.5 rad/sec. Compute the length of the prismatic bar when it makes an angle of 45° with the vertical. a. 6.5 m b. 3.33 m c. 6.20 m d. 7.35 m 555. A bullet enters a 50 mm plank with a speed of 600 m/s and leaves with a speed of 24 m/s. Determine the thickness of the plank that can be penetrated by the bullet. a. 55 mm b. 60 mm c. 65 mm d. 70 mm 556. A balikbayan box is placed on top on a flooring of a delivery truck with a coefficient of friction between the floor and the box equal to 0.40. If the truck moves at 60 kph, determine the distance that the truck will move before the box will stop slipping. The box weighs 200 N. a. 70.8 m b. 60.8 m c. 50.8 m d. 40.8 m 557. At what speed must a 10 kN car approach a ramp which makes an angle of 30° with the horizontal an 18m high at the top such that it will just stop as it reaches the top. Assume resisting force of friction, to be 0.60 kN. a. 71.57 kph b. 60.46 kph c. 54.46 kph d. 82.52 kph 558. A car weighing 10 kN approaches a ramp which makes a slope of 20° at the speed of 75 kph. At the foot of the ramp, the motor is turned off. How far does the car travel up the inclined before it stops? a. 64.57 m b. 46.74 m c. 74.84 m d. 54.84 m 559. A car is running up a grade of 1 in 250 at a speed of 28.8 kph when the engine conk out. Neglecting friction, how far will the car have gone after 3 minutes from the point where the engine conk out? a. 808.2 m b. 607.8 m c. 542.4 m d. 486.8 m 560. A 70 kg man stands on a spring scale on an elevator. During the first 2 seconds starting from rest, the scale reads 80 kg. Find the velocity of the elevator at the end of 2 seconds and the tension T in the supporting cable fro the during the acceleration period. The total weight of the elevator, man and scale is 7000N. a. 2.8 m/sec; 8000N b. 3.4 m/sec; 7000N c. 4.3 m/sec; 9000N d. 1.5 m/sec; 6000N 561. A cylinder having a mass of 40 kg with a radius of 0.5 m is pushed to the right without rotation and with acceleration 2 m/sec^2. Determine the magnitude and location of the horizontal force P if the coefficient of friction is 0.30. a. 198 N; 20.2 cm b. 200 N; 32.4 cm c. 183 N; 15.7 cm d. 232 N; 34.2 cm 562. A block having a weight of 400 N rests on an inclined plane making an angle of 30° with the horizontal is initially at rest. After it was released for 3sec, find the distance that the block has traveled assuming that there is no friction between the block and the plane. Compute also the velocity after 3 sec. a. 22.07 m, 14.71 m/s b. 27.39 m, 15.39 m/s c. 20.23 m, 14.60 m/ s d. 15.69 m, 13.68 m/s 563. A block having a weight of 200 N rests on an inclined plane making an angle of 30° with the horizontal is initially at rest. If the block is initially at rest and the coefficient of friction between the inclined plane and the block is 0.20, compute the time to travel a distance of 14.45m, and the velocity of the block after 3 sec. a. 3 sec, 9.63 m/s b. 2 sec, 10.12 m/s c. 4 sec, 12.20 m/s d. 5sec, 11.20 m/s 564. A 100kg block is released at the top of 30° incline 10 m above the ground. The slight melting of ice renders the surfaces frictionless; calculate the velocity of the foot of the incline. a. 20 m/s b. 15 m/s c. 25 m/s d. 22 m/s 565. Starting from rest, an elevator weighting 9000 N attains an upward velocity of 5 m/s in 4 sec. with uniform acceleration. Find the apparent weight of 600 N man standing inside the elevator during its ascent and calculate the tension in the supporting cable. a. 10 823 N b. 11 382 N c. 9 254 N d. 12 483 N 566. A body weighing 40 lb starts from rest and slides down a plane at an angle of 30° with the horizontal for which the coefficient of friction f = 0.30. How far will it move during the third second? a. 19.63 ft b. 19.33 ft c. 18.33 ft d. 19.99 ft 567. What force is necessary to accelerate a 3000 lbs railway electric car at the rate of 1.25 ft/sec^2, if the force required to overcome the frictional resistance is 400 lbs. a. 1564.596 lbs b. 1267.328 lbs c. 1653.397 lbs d. 1427.937 lbs 568. A freight car having a mass of 15 Mg is towed along the horizontal track. If the car starts from rest and attains a speed of 8 m/s after traveling a distance of 150m, determine the constant horizontal towing force applied to the car. Neglect friction and the mass of the wheels. a. 3.2 kN b. 2.2 kN c. 4.3 kN d. 4.1 kN 569. An elevator weighing 2000 lb attains an upward velocity of 16 fps in 4 sec with uniform acceleration. What is the tension in supporting= cables? a. 2250 lb b. 2495 lb c. 1950 lb d. 2150 lb 570. A block weighing 200N rests on a plane inclined upward to the right at slope 4 vertical to 3 horizontal. The block is connected by a cable initially parallel to the plane passing through a pulley which is connected to another block weighing 100 N moving vertically. The coefficient of kinetic friction between the 200N block and the inclined plane is 0.10, which of the following most nearly give the acceleration of the system. a. 2.93 m/sec^2 b. 0.37 m/sec^2 c. 1.57 m/sec^2 d. 3.74 m/sec^2 571. A pick-up truck is traveling forward a5 m/s the bed is loaded with boxes, whose coefficient of friction with the bed is 0.4. What is the shortest time that the truck can be bought to a stop such that the boxes do not shift? a. 4.75 b. 2.35 c. 5.45 d. 6.37 572. Two barges are weighing 40 kN and the other 80 kN are connected by a cable in quiet water. Initially the barges are 100 m apart. If friction is negligible calculate the distance moved by the 80 kN barge. a. 20 m b. 30 m c. 12 m d. 25 m 573. Two blocks A and B weighs 150 N and 200 N respectively is supported by a flexible cord which passes through a frictionless pulley which is supported by a rod attached to a ceiling. Neglecting the mass and friction of the pulley, compute the acceleration on the blocks and the tension on the rod supporting the frictionless pulley. a. 1.40 m/s^2, 342.92 N b. 1.50 m/s^2, 386.45 N c. 1.80 m/s^ 2, 421.42 N d. 2.2 m/s^2, 510.62 N 574. A pendulum with the concentrated mass “m” is suspended vertically inside a stationary railroad freight car by means of a rigid weightless connecting rod. If the connecting rod is pivoted where it attaches to the boxcar, compute the angle of that the rod makes with the vertical as a result of constant horizontal acceleration of 2m/s. a. 11°31’ b. 9°12’ c. 6°32’ d. 3°56’ 575. Two 15 N weights A and B are connected by a massless string hanging over a smooth frictionless peg. If a third weight of 15N is added to A and the system is released, by how much is the force on the peg increased? a. 10 kN b. 12 kN c. 15 kN d. 20 kN 576. Three crates with masses A = 45.2 kg, B = 22.8 kg, and C = 34.3 kg are placed with B besides A and C besides B along a horizontal frictionless surface. Find the force exerted by B and C by pushing to the right with an acceleration of 1.32 m/s^2. a. 45.3 kN b. 54.2 kN c. 43.2 kN d. 38.7 kN 577. Three blocks A, B and C are placed on a horizontal frictionless surface and are connected by chords between A, B and C. Determine the tension between block B and C when a horizontal tensile force is applied at C = 6.5 N. Masses of blocks are A = 1.2 kg, B 2.4 kg, and C = 3.1 kg. a. 3.50 N b. 4.21 N c. 3.89 N d. 4.65 N 578. A constant force P = 750 N acts on the body shown during only the first 6 m of its motion starting from rest. If u = 0.20, find the velocity of the body after it has moved a total distance of 9m. a. 3.93 m/s^2 b. 4.73 m/s^2 c. 2.32 m/s^2 d. 3.11 m/s^2 579. A weight 9 kN is initially suspended on a 150 m long cable. The cable weighs 0.002 kN/m. If the weight is then raised 100 m how much work is done in Joules. a. 915000 b. 938700 c. 951000 d. 905100 580. What is the kinetic energy of 4000 lb automobile which is moving at 44 fps? a. 1.2 x 10^5 ft-lb b. 2.1 x 10^5 ft-lb c. 1.8 x 10^5 ft-lb d. 3.1 x 10^5 ft-lb 581. A box slides from rest from a point A down a plane inclined 30° to the horizontal. After reaching the bottom of the plane, the box move at horizontal floor at distance 2m before coming to rest. If the coefficient of friction between the box and the plane and the box and the floor is 0.40, what is the distance of point “A” from the intersection of the plane and the floor? a. 7.24 m b. 5.21 m c. 4.75 m d. 9.52 m 582. A 400 N block slides on the horizontal plane by applying a horizontal force of 200 N and reaches a velocity of 20 m/s in a distance of 30m from rest. Compute the coefficient of friction between the floor and the block. a. 0.18 b. 0.24 c. 0.31 d. 0.40 583. A car weighing 40 tons is switched to a 2 percent of upgrade with a velocity of 30 mph. If the train resistance is 10 lb/ton, ho9w far up the grade will it go? a. 1124 ft on slope b. 2014 ft on slope c. 1203 ft on slope d. 1402 ft on slope 584. A car weighing 10 kN is towed along a horizontal surface at a uniform velocity of 80 kph. The towing cable is parallel with the road surface. When the car is at foot of an incline as shown having an elevation of 30m, the towing cable was suddenly cut. At what elevation in the inclined road will the car stop in its upward motion? a. 55.16 m b. 60.24 m c. 51.43 m d. 49.62 m 585. A wooden block starting from rest, slides 6 m down a 45° slope, then 3m along the level surface and then up 30° incline until it come to rest again. If the coefficient of friction is 0.15 for all surfaces in contact compute the total distance traveled. a. 20 m b. 11 m c. 14 m d. 18 m 586. The block shown starting from rest and moves towards the right. What is the velocity of block B as it touches the ground? How far will block A travel if the coefficient of friction between block A and the surface is 0.20? Assume pulley to be frictionless. a. 1.44 m b. 2.55 m c. 5.22 m d. 3.25 m 587. After the block in the figure has moved 3m from rest the constant force P=600N is removed find the velocity of the block when it is returned to its initial position. a. 8.6 m/s b. 5.6 m/s c. 6.4 m/s d. 7.1 m/s 588. A 10 kg block is raised vertically 3 meters. What is the change in potential energy? Answer in SI units closest to: a. 350 kg-m^2/ s b. 320 J c. 350N-m d. 294 J 589. A car weighing 40 tons is switched 2% upgrade with a velocity of 30 mph. If the car is allowed to run back what velocity will it have at the foot of the grade? a. 37 fps b. 31 fps c. 43 fps d. 34 fps 590. A 200 ton train is accelerated from rest to a velocity of 30 miles per hour on a level track. How much useful work was done? a. 12 024 845 b. 13 827 217 c. 11 038 738 d. 10 287 846 591. A drop hammer weighing 40 kN is dropped freely and drives a concrete pile 150 mm into the ground. The velocity of the drop hammer at impact is 6m/s. What is the average resistance of the soil in kN? a. 542.4 b. 489.3 c. 384.6 d. 248.7 592. A force of 200 lbf act on a block at an angle of 28° with respect to the horizontal. The block is pushed 2 feet horizontally. What is the work done by this force? a. 320 J b. 480 J c. 540 J d. 215 J 593. A small rocket propelled test vehicle with a total mass of 100 kg starts from rest at A and moves with negligible friction along the track in the vertical plane as shown. If the propelling rocket exerts a constant thrust T of 1.5 kN from A to position B. where it is shut off, determine the distance S that the vehicle rolls up the incline before stopping. The loss of mass due to the expulsion of gases by the rocket is small and may be neglected. a. 170 m b. 165 m c. 160 m d. 175 m 594. A body that weighs W Newton fall from rest from a height 600 mm and strikes a sping whose scale is 7 N/mm. If the maximum compression of the spring is 150 mm, what is the value of W? Disregard the mass of spring. a. 105 N b. 132 N c. 112 N c. 101 N 595. A 100 N weight falls from rest from a height of 500 mm and strikes a spring which compresses by 100 mm. Compute the value of the spring constant, neglecting the mass of the spring. a. 10 N/mm b. 15 N/mm c. 12 N/mm d. 8 N/mm 596. A 200 N weight falls from rest at height “h” and strikes spring having a spring constant of 10 N/mm. the maximum compression of spring is 100 mm, after the weight the weight stikes the spring. Compute the value of h is meter. a. 0.12 m b. 0.10 m c. 0.15 m d. 0.21 m 597. A block weighing 500 N is dropped from a height of 1.2 m upon a spring whose modulus is 20 n/mm. What velocity will the block have at the instant the spring is deformed 100 mm? a. 6.55 m/s^2 b. 5.43 m/s^2 c. 4.65 m/s^2 d. 3.45 m/s^2 598. A 50 kg object strikes the unstretched spring to a vertical wall having a spring constant of 20 kN/m. Find the maximum deflection of the spring. The velocity of the object before it strikes the spring is 40 m/s. a. 1 m b. 2 m c. 3 m d. 4 m 599. A large coil spring with a spring constant k=120 N/m is elongated, within its elastic range by 1m. Compute the store energy of the spring in N-m. a. 60 b. 40 c. 50 d. 120 600. To push a 25 kg crate up a 27° inclined plane, a worker exerts a force of 120 N, parallel to the incline. As the crate slides 3.6 m, how much is the work done by the worker and by the force of gravity. a. 400 Joules b. 420 Joules c. 380 Joules d. 350 Joules 601. A train weighing 12 000 kN is accelerate up a 2% grade with velocity increasing from 30 kph to 50 kph in a distance of 500 m. Determine the horse power developed by the train. a. 5.394 kW b. 5.120 kW c. 4.468 kW d. 4.591 kW 602. An elevator has an empty weight of 5160 N. It is designed to carry a maximum load of 20 passengers from the ground floor to the 25th floor of the building in a time of 18 seconds. Assuming the average weight of the passenger to be 710 N. and the distance between floors is 3.5 m, what is the minimum constant power needed for the elevator motor? a. 85.5 kW b. 97.4 kW c. 94.3 kW d. 77.6 kW 603. An engine hoist 1.50 m^3 of a concrete in a 2200 N bucket moves a distance of 12 m in 20 seconds. If concrete weighs 23.5 kN/m^3, determine the engine horsepower assuming 80 efficiency. a. 32.12 hp b. 42.23 hp c. 37.74 hp d. 28.87 hp 604. A train weighs 15 000 kN. The train’s resistance is 9 n per kilo-Newton. If 5000 is available to pull this train up to 2% grade, what will be its maximum speed in kph? a. 46.2 kph b. 50 kph c. 40 kph d. 35 kph 605. An engine having a 40 hp rating is used as an engine hoist to lift a certain weight of height 8 m. Determine the maximum weight it could lift in a period of 20 sec. a. 85.5 kW b. 97.4 kW c. 74.6 kW d. 77.6 kW 606. A 100 kg body moves to the right 5 m/s and another 140 kg body moves to the left at 3 m/s. They collided and after impact the 100 kg body rebounds to the left at 2 m/s. Compute the coefficient of restitution. a. 0.40 b. 0.50 c. 0.30 d. 0.60 607. A ball is dropped from an initial height of 6m above a solid floor, how high will the ball rebound if the coefficient of restitution is e = 0.92? a. 5.08 b. 5.52 c. 5.41 d. 5.12 608. A ball strikes the ground at an angle of 30° with the ground surface; the ball then rebounds at a certain angle θ with the ground surface. If the coefficient of restitution is 0.80, find the value of θ. a. 24.79° b. 18.48° c. 32.2° d. 26.7° 609. A ball is thrown with an initial horizontal velocity of 30m/s from a height of 3m above the ground and 40 m from a vertical wall. How high above the ground will the ball strike if the coefficient of restitution is 0.70? a. 1.46 m b. 2.52 m c. 1.11 m d. 0.89 m 610. Two cars having equal weights of 135 kN are traveling on a straight horizontal track with velocities of 3m/s to the right and 1.5 m/s to the left respectively. They collide and are coupled during impact. Neglecting friction due to skidding, determine their final common velocity and the gain or loss in kinetic energy after impact. a. 7.74 m/s; 69.67 kN-m b. 1.25 m/s; 66.35 kN-m c. 2.06 m/s; 57.25 kN-m d. 3.12 m/ s; 77.36 kN-m 611. A man weighing 68 kg jumps from a pier with horizontal velocity of 6 m/s onto a boat that is rest on the water. If the boat weighs 100 kg, what is the velocity of the boat when the man comes to rest relative to the boat? a. 2.43 m/s b. 3.53 m/s c. 2.88 m/s d. 1.42 m/s 612. A man weighing 68 kg jumps from a pier with horizontal velocity of 5 m/s onto a 100 kg boat moving towards the dock at 4m/s What would be the velocity of the boat after the man lands on it? a. -0.56 m/s b. -0.36 m/s c. -0.78 m/s d. -1.33 m/s 613. A ball is thrown at an angle of 40° from the horizontal toward a smooth foor and it rebounds at an angle of 25° with the horizontal floor. Compute the value of coefficient of restitution. a. 0.56 b. 0.66 c. 0.46 d. 0.76 614. Car B is moving at a speed of 12 m/s and is struck by car A which is moving at a speed of 20 m/s. The weight of car A is 14 tons and of car B is 10 tons. Determine the velocities of the car after impact assuming that the bumpers got locked after impact. Both cars are moving in the same direction to the right. a. 16.67 m/s b. 14.25 m/s c. 15.42 m/s d. 13.62 m/s 615. Two cars A and B have weights equal to 12 tons and 8 tons respectively are moving in opposite directions. The speed of car A is 22m/s to the right and that of car B is 18 m/s to the left. Two cars bumped each other. Determine the velocity of the cars after impact assuming the bumpers get locked. a. 6 m/s b. 8 m/s c. 4 m/s d. 3 m/s 616. A 6000 N drop hammer falling freely trough a height of 0.9 m drives a 3000N pile 150mm vertically to the ground. Assuming the hammer and the pile to cling together after the impact, determine the average resistance to penetration of the pile. a. 32 976 N b. 42 364 N c. 30 636 N d. 28 476 N 617. Two identical balls collide as shown. What is V2’ if the coefficient of restitution is 0.8? a. 4.8 m/s b. 3.6 m/s c. 5.6 m/s d. 2.4 m/s 618. A 16gm mass is moving at 30 cm/s while a 4gm mass is moving opposite direction at 50cm/sec. They collide head on and stick together. Their velocity after collision is: a. 0.14 m/s b. 0.21 m/s c. 0.07 m/s d. 0.28 m/s 619. A bullet weighing 0.014 kg and moving horizontally with a velocity of 610 m/s strikes centrally a block of wood having a mass of 4.45 kg which is suspended by a cord from a point 1.2 m above the center of the block. To what angle from the vertical will the block and embedded bullet swing? a. 31.79° b. 29.32° c. 30.12° d. 28.64° 620. A body having a mass of 100kg and having velocity of 10m/s to the right collides with an 80 kg mass having a velocity of 5 m/s to the left. If the coefficient of restitution is 0.5, determine the loss of kinetic energy after impact. a. 3750 N-m b. 4260 N-m c. 3640 N-m d. 4450 N-m 621. A 0.44N bullet is fired horizontally to an 89.18 N block of wood resting on a horizontal surface which the coefficient of friction is 0.30. If the block is moved a distance 375 mm along the surface, what was the velocity of the bullet before striking? a. 303.49 m/s b. 204.61 m/s c. 142.52 m/s d. 414.25 m/s 622. A 60 ton rail car moving at 1 mile per hour is instantaneously coupled to a stationary 40 to rail car. What is the speed of the coupled cars? a. 0.88 mi/hr b. 1.0 mi/hr c. 0.60 mi/hr d. 0.40 mi/hr 623. What momentum does a 40 lbm projectile posses if the projectile is moving 420 mph? a. 24 640 lbf-sec b. 16 860 lbf-sec c. 765 lbf-sec d. 523.6 lbf-sec 624. A 300 kg block is in contact with a level of coefficient of kinetic friction is 0.10. if the block is acted upon by a horizontal force of 50kg what time will elapse before the block reaches a velocity of 48.3 m/min from rest? If the 50 kg is then removed, hos much longer will the block continue to move? a. 12.08 sec; 8.05 sec b. 15.28 sec; 9.27 sec c. 10.42 sec; 7.64 sec d. 13.52 sec; 10.53 sec 625. A 100kg body moves to the right at 5m/s and another body of mass W is moves to the left 3m/s. they meet each other and after impact the 100kg body rebounds to the left at 2m/s. Determine the mass of the body if coefficient of restitution is 0.5. a. 140 kg b. 150 kg c. 100 kg d. 200 kg 626. A wood block weighing 44.75 N rests on a rough horizontal plane, the coefficient of friction being 0.40. if a bullet weighing 0.25 N is fired horizontally into the block with the velocity of 600m/s, how far will the block be displaced from its initial position? Assume that the bullet remains inside the block. a. 1.41 m b. 2.42 m c. 1.89 m d. 0.98 m 627. The system is used to determine experimentally the coefficient of restitution. If ball A is released from rest an ball B swings through θ=53.1°, after being struck, determine the coefficient of restitution. Weight of A is 150 N while that of b is 100 N. a. 0.537 b. 0.291 c. 1.083 d. 0.926 628. The ball A and B are attached to stiff rods of negligible weight. Ball A is released from rest and allowed to strike B. If the coefficient of restitution is 0.60, determine the angle θ through which ball B will swing. If the impact lasts for 0.01 sec, also find the average impact force. Mass of A is 15kg and that of B is 10kg. a. 64.85° ; 6720 N b. 60.58° ; 6270 N c. 57.63° ; 7660 N d. 73.32° ; 7670 N 629. A 1500N block is in contact with a level plane whose coefficient of kinetic friction is 0.10. If the block is acted upon a horizontal force of 250 N, at what time will elapse before the block reaches a velocity of 14.5m/s starting from rest? a. 22.17 sec b. 18.36 sec c. 21.12 sec d. 16.93 sec 630. A 1600N block is in contact with a level plane whose coefficient of kinetic friction is 0.20. If the block is acted upon a horizontal force of 300 N initially when the block is at rest and the force is removed when the velocity of the block reaches 16m/s. How much longer will the block continue to move? a. 8.15 sec b. 6.25 sec c. 4.36 sec d. 5.75 sec 631. A bullet weighing 0.30N is moving 660m/s penetrates an 50N body and emerged with a velocity of 180 m/s. For how long will the body moves before it stops? Coefficient of friction is 0.40. a. 7.34 sec b. 6.84 sec c. 5.24 sec d. 8.36 sec 632. A 1000N block is resting on an incline plane whose slope is 3 vertical to 4 horizontal. If a force of 1500 N acting parallel to the inclined plane pushes the block up the inclined plane, determine the time required to increase the velocity of the block from 3m/s to 15m/s. Coefficient of friction between the block and the plane is 0.20. a. 1.65 sec b. 1.86 sec c. 2.17 sec d. 3.64 sec 633. The 9kg block is moving to the right with a velocity of 0.6 m/s on the horizontal surface when a force P is applied to it at t=0. Calculate the velocity V2 of the block when the t=0.4 sec. the kinetic coefficient of friction is Mk = 0.30. a. 1.82 m/s b. 1.23 m/s c. 2.64 m/s d. 2.11 m/s 634. A 50kg block initially at rest is acted upon by a force P which varies as shown. Knowing the coefficient of kinetic friction between the block and the horizontal surface is 0.20, compute the velocity of the block after 5 s, and after 8s. a. 15.19 m/s; 16.804 m/s b. 13.23 m/s; 15.534 m/s c. 10.65 m/s; 17.705 m/s d. 17.46 m/s; 14.312 m/s 635. The motion of the block starting from rest is governed by the a-t curve shown. Determine the velocity and distance traveled after 9 sec. Neglecting friction. a. 66 m/s; 228 m b. 45 m/s; 233 m c. 57 m/s, 423 m d. 72 m/s; 326 m 636. From the V-t curve shown compute the distance traveled by a car starting form rest after 6 sec. a. 50m b. 60 m c. 70 m d. 80 m 637. A bullet weighing 50g is fired into a block of wood weighing 20 lbs on a top of a horizontal table. The block moves 45 cm. the coefficient of friction between the block and table is 0.30. What is the speed of the bullet in kph before hitting the block? Assume that the bullet is embedded of the block. a. 1068.77 kph b. 1843.53 kph c. 1144.38 kph d. 1683.78 kph 638. A block weighing 60 N is subjected to a horizontal force P = (10 + t^3) and a friction resisting equal to (6 + t^2). Compute the velocity of the block 2 s after it has started from rest. a. 2.4 m/s b. 1.8 m/s c. 2.8 m/s d. 1.4 m/s 639. A price tag of P1200 is specified if paid within 60 days but offers 3% discount for cash in 30 days. Find the rate of interest. a. 37.11% b. 38.51 % c. 40.21 % d. 39.31 % 640. It is the practice of almost all banks in the Philippines that when they grant a loan, the interest the interest for one year is automatically deducted from the principal amount upon release of money to a borrower. Let us that you applied for a loan with a bank and the P80 000 was approved with interest rate of 14% which P11 200was deducted and you were given a check of P68 800. Since you have to pay the amount of P80 000 in one year after, what then will be the effective interest rate? a. 16.28% b 16.18% c. 16.30% d. 16.20% 641. Mr. J. dela Cruz borrowed money from a bank. He received from the bank P1 340 and promised to pay P1 500 at the end of 9 months. Determine the simple interest rate and the corresponding discount rate or often referred to as “Banker’s Discount”. a. 15.92%; 13.73 % b. 18.28%; 13.12 % c. 12.95%; 17.33 % d. 19.25%; 13.33% 642. A man borrowed from a bank with a promissory note that he signed in the amount of P25000 for a period of one year. He received only the amount of P21, 915 after the bank collected the interest and additional amount of P85 00 for notarial and inspection fees. What was the rate of interest that the bank collected in advance? a. 13.64% b. 18.37% c. 16.43% d. 10.32% 643. Agnes Abadilla was granted a loan of P20 000 by her employer CPM Industrial Fabricator and Construction Corporation with an interest rate of 6% for 180 days on the principal collected in advance. The corporation will accept a promissory note for P20 000 non-interest for 180 days. If discounted at once, find the proceeds on the note. a. P18 800 b. P19 000 c. P18 000 d. P18 400 644. P400 is borrowed for 75 days at 16% per annum simple interes. How much wil be due at the end of 75 days? a. P4168.43 b. P5124.54 c. P4133.33 d. P5625.43 645. Mr. Almagro made a money market of P1 000 000 for 30 days at 7.5% per year. If the withholding tax is 20%, what is the net interest that he will receive at the end of the month? a. P3 000 b. P4 000 c. P6 000 d. P5 000 646. L for a motorboat specifies a cost of P1200 due at the end of 100 days but offers 4% discount for cash in 30 days. What is the highest rate, simple interest at which the buyer can afford to borrow money in order to take advantage of the discount? a. 18.4% b. 19.6% c. 20.9% d. 21.4% 647. In buying a computer disk, the buyer was offered the options of paying P250 cash at the end of 30 days or P270 at the end of 120 days. At what rate is the buyer paying simple interest if he agree to pay at the end of 120 days. a. 32% b. 40% c. 28% d. 25% 648. On March 1, 1996 Mr. Almagro obtains a loan of P1500 from Mr. Abella and signs a note promising to pay the principal and accumulated simple interest at rate of 5% at the end of 120 day. On March 15, 1996, Mr. Abella discounts the note at the bank whose discount rate is 6%. What does he receive? a. P 2 201.48 b. P1 123.29 c. P1 513.56 d. P938.20 649. A deposit of P110 000 was made for 31 days. The net interest after deducting 20% withholding tax is P890.36. Find the rate of return annually. a. 12.25 b. 11.75 c. 12.75 d. 11.95 650. If you borrowed money from your friend with simple interest of 12%, find the present worth of P50 000, which is due at the end of 7 months. a. P46 200 b. P44 893 c. P46 729 d. P45 789 651. A man borrowed P2000 from a bank and promises to pay the amount for one year. He received only the amount of P1920 after the bank collected an advance interest of P80. What was the rate of discount and the rate of interest that the bank collected in advance. a. 4%, 4.17% b. 3%, 3.17% c. 4%, 4.71% d. 3%, 3.71% 652. The amount of P12 800 in 4yrs. At 5% compounded quarterly is ______? a. P 14 785.34 b. P 15 614.59 c. P 16 311.26 d. P 15 847.33 653. A man borrows money from a bank which uses a simple discount rate of 14%. He signs a promissory note promising to pay P500 per month at the end of 4th, 6th, and 7th months respectively. Determine the amount of money that he received from the bank. a. P1403.68 b. P1340.38 c. P1102.37 d. P1030.28 654. A nominal interest of 3% compounded continuously is given on the account. What is the accumulated amount of P10 000 after 10 years? a. P13 610.10 b. P13 500.10 c. P13 498.60 d. P13 439.16 655. By the condition of a will, the sum of P2000 is left to a girl to be held in trust fund by her guardian until it amounts to P50 000. When will the girl receive the money if the fund is invested at 8% compounded quarterly? a. 7.98 years b. 10.34 years c. 11.57 years d. 10.45 years 656. A man expects to receive P25 000 in 8 years. How much is that worth now considering interest at 8% compounded quarterly? a. P13 859.12 b. P13 958.33 c. P13 675.23 d. P13 265.83 657. P500 000 was deposited at an interest of 6% compounded quarterly. Compute the compounded interest after 4 years and 9 months. a. 163 475.37 b. 178 362.37 c. 158 270.37 d. 183 327.37 658. If the nominal interest rate is 3%, how much is P5000 worth in 10 years in a continuous compounded account? a. P5750 b. P6750 c. P7500 d. P6350 659. P200 000 was deposited for a period of 4 years and 6 months and bears on interest of P85 649.25. What is the rate of interest if it is compounded quarterly? a. 8% b. 6% c. 7% d. 5% 660. How many years will P100 000 earned a compound interest o P50 000 if the interest rate is 9% compounded quarterly? a. 3.25 b. 4.55 c. 5.86 d. 2.11 661. A certain amount was deposited 5 years and 9months ago at an intrest 8% compounded quarterly. If the sum now is P315 379.85, how much was the amount deposited? a. P200 000 b. P180 000 c. P240 000 d. P260 000 662. Compute the effective annual interest rate which is equivalent to 5% nominal annual interest compounded continuously. a. 5.13% b. 4.94% c. 5.26% d. 4.9% 663. Find the time required fro a sum of money to triple itself at 5% per annum compounded continuously? a. 21.97 yrs. b. 25.34 yrs. c. 18.23 yrs. d. 23.36 yrs. 664. A man wishes to have P40 000 in a certain fund at the end of 8years. How much should he invest in a fund that will pay 6% compounded continuously? a. P24 751.34 b. P36 421.44 c. P28 864.36 d. P30 486.42 665. If the effective annual interest rate is 4%, compute the equivalent nominal annual interest compounded continuously. a. 3.92% b. 4.10% c. 3.80% d. 4.09% 666. What is the nominal rate of interest compounded continuously for 10 years if the compounded amount factor is equal to 1.34986? a. 3% b. 4% c. 5% d. 6% 667. American Express Corp. charges 1.5% interest per month, compounded continuously on the unpaid balance purchases made on this credit card. Compute the effective rate of interest. a. 19.72% b. 20.25% c. 21.20% d. 13.19% 668. If the nominal interest is 12% compounded continuously, compute the effective rate of annual interest. a. 12.75% b. 11.26% c. 12.40% d. 11.55% 669. Compute the difference in the future amount of P500 compounded annually at nominal rate of 5% and if it is compounded continuously for 5 years at the same rate. a. P3.87 b. P4.21 c. P5.48 d. P6.25 670. If the effective interest rate is 24%, what nominal rate of interest is charged for a continuously compounded loan? a. 21.51% b. 22.35% c. 23.25% d. 21.90% 671. What is the nominal rate of interest compounded continuously for 8 years if the pre4sent worth factor is equal to 0.6187835? a. 4% b. 5% c. 6% d. 7% 672. What is the difference of the amount 3 yrs. for now for a 10% simple interest and 10% compound interest per year? a. P155 b. P100 c. same d. P50 673. Find the discount is P2000 is discounted for 6 months and at 8% compounded quarterly. a. P76.92 b. P80.00 c. P77.66 d. P78.42 674. If a sum of money triples in a certain period of time at a given rate of interest, compute the value of the single payment present worth factor. a. 0.333 b. 3 c. 0.292 d. 1.962 675. If the single payment amount factor for a period of 5 years is 1.33822. What is the nearest value of the interest rate? a. 8% b. 7% c. 5.50% d. 6% 676. If the single payment present worth factor for a period of 8 years is 0.58201, compute the nearest value of the rate for that period. a. 6% b. 7% c. 6.5% d. 8% 677. If money is worth 8% compounded quarterly, compute the single payment amount factor for a period of 6 years. a. 1.60844 b. 0.62172 c. 1.70241 d. 0.53162 678. Which of these gives the lowest effective rate of interest? a. 12.35% compounded annually b. 11.9% compounded semi-annually c. 12.2% compounded quarterly d. 11.6% compounded monthly 679. It takes 20.15 years to quadruple your money if invested x% compounded semiannually. Find the value of x. a. 8% b. 6.5% c. 7% d. 5% 680. It takes 13.87 years to treble the money at the rate of x% compounded quarterly. Compute the value of x. a. 5% b. 6% c. 7% d. 8% 681. Money was invested at x% compounded quarterly. If it takes the money up to quadruple in 17.5 years, find the value of x. a. 8% b. 6% c. 7% d. 5% 682. Fifteen years ago P1000 was deposited in a bank account an today it is worth P2370. The bank pays semi-annually. What was the interest rate paid on this account? a. 4.9% b. 5.8 % c. 5.0% d. 3.8% 683. You borrow P3500 for one year from a friend at an interest rate of 1.5 per month instead of taking a loan from a bank at rate of 18% a year. Compare how much money you will save or lose on the transaction. a. You will pay P155 more than if you borrowed from the bank b. You will save P55 by borrowing from your friend c. You will pay P85 more that if you borrowed from the bank d. You will pay P55 less than if you borrowed from the bank 684. Find the present worth of a future payment of P100 000 to be made in 10 years with an interest of 12% compounded quarterly. a. P30444.44 b. P33000.00 c. P30655.68 d. P30546.01 685. An initial deposit of P80 000 in a certain bank earns 6% interest per annum compounded monthly. If the earnings from the deposit are subject to 20% tax what would be the net value of the deposit be after three quarters? a. P95324.95 b. P82938.28 c. P68743.24 d. P56244.75 686. The effective rate of interest of 14% compounded semi-annually is: a. 14.49% b. 14.36% c. 14.94% d. 14.88% 687. The amount of P50 000 was deposited in the bank earning an interest of 7.5% per annum. Determine the total amount at the end of % years, if the principal and interest were not withdrawn during the period? a. P71 781.47 b. P72 475.23 c. P70 374.90 d. P78 536.34 688. What is the effective rate corresponding to 18% compounded daily? Take 1 year is equal to 360 days. a. 18.35% b. 19.39% c. 18.1% d. 19.72% 689. If P1000 becomes P1126.49 after 4 years when invested at a certain nominal rate of interest compounded semi-annually determine the nominal rate and the corresponding effective rate. a. 3% and 3.02% b. 4.29% and 4.32% c. 2.30% and 2.76% d. 3.97% and 3.95% 690. Convert 12% semi-annually to compounded quarterly a. 19.23% compounded quarterly b. 23.56% compounded quarterly c. 14.67% compounded quarterly d. 11.83% compounded quarterly 691. What is the corresponding effective interest rate of 18% compounded semiquarterly? a. 19.25% b. 19.48% c. 18.46% d. 18.95% 692. If P5000 shall accumulate for 10 years at 8% compounded quarterly, find the compounded interest at the end of 10 years. a. P6005.30 b. P6000.00 c. P6040.20 d. P6010.20 693. A couple borrowed P4000 from a lending company for 6 years at 12%. At the end of 6 years, it renews the loan for the amount due plus P4000 more for 3 years at 12%. What is the lump sum due? a. P14 842.40 b. P16 712.03 c. P12 316.40 d. P15 382.60 694. How long (years) will it take money if it earns 7% compounded semi-annually? a. 26.30 b. 40.30 c. 33.15 d. 20.15 695. P200 000 was deposited on Jan. 1, 1988 at an interest rate of 24% compounded semi-annually. How much would the sum be on Jan. 1, 1993? a. P421 170 b. P521 170 c. P401 170 d. P621 170 696. If P500 000 is deposited at a rate of 11.25% compounded monthly; determine the compounded interest rate after 7 years and 9 months. a. 690 849 b. 670 258 c. 680 686 d. 660 592 697. P200 000 was deposited at an interest rate of 24% compounded semi-annually. After how many years will the sum be P621 170? a. 4 b. 3 c. 5 d. 6 698. A bank is advertising 9.5% accounts that yield 9.84 annually. How often is the interest compounded? a. monthly b. bi- monthly c. quarterly d. daily 699. Evaluate the integral of 2dx/(4x+3) if the upper limit is 5 and the lower limit is 1. a. 0.595 b. 0.675 c. 0.486 d. 0.387 700. Evaluate the integral of 2xdx/(2x^2+4) if the upper limit is 6 and the lower limit is 3. a. 0.620 b. 0.675 c. 0.486 d. 0.580	{"url":"https://flowerstlc.com/article/multiple-choice-pdf-document","timestamp":"2024-11-08T21:48:08Z","content_type":"text/html","content_length":"286342","record_id":"<urn:uuid:9c8bf09d-6cd6-4168-ad33-12ffe63bf884>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00079.warc.gz"}
The Stacks project Lemma 15.112.2. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension. Then there are $e \geq 1$ and $f \geq 1$ such that $e_ i = e$ and $f_ i = f$ for all $i$ (notation as in Remark 15.111.6). In particular $[L : K] = n e f$. Comments (0) Post a comment Your email address will not be published. Required fields are marked. 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). All contributions are licensed under the GNU Free Documentation License. In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09EB. Beware of the difference between the letter 'O' and the digit '0'. The tag you filled in for the captcha is wrong. You need to write 09EB, in case you are confused.	{"url":"https://stacks.math.columbia.edu/tag/09EB","timestamp":"2024-11-14T22:10:19Z","content_type":"text/html","content_length":"14232","record_id":"<urn:uuid:1cea864e-b22d-4558-8ad8-4556d3edcf33>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00305.warc.gz"}
As shown in figure, the left block rests on a table at distance-Turito Are you sure you want to logout? As shown in figure, the left block rests on a table at distance A. Left block reach the edge of the table B. Right block hit the table C. Both (A) and (B) happens at the same time D. Can’t say anything The correct answer is: Left block reach the edge of the table Get an Expert Advice From Turito.	{"url":"https://www.turito.com/ask-a-doubt/physics-as-shown-in-figure-the-left-block-rests-on-a-table-at-distance-from-the-edge-while-the-right-block-is-kept-qc91cbe","timestamp":"2024-11-06T11:09:41Z","content_type":"application/xhtml+xml","content_length":"562245","record_id":"<urn:uuid:8a27f65a-6816-47cd-b578-f8d549d46ac5>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00610.warc.gz"}
Let K, K′ be two centrally symmetric convex bodies in En, with their centres at the origin o. Let Vr denote the r-dimensional volume function. A problem of H. Busemann and C. M. Petty [1], see also, H. Busemann [2] asks:— “If, for each (n − 1)-dimensional subspace L of En, does it follow that If n = 2 or, if K is an ellipsoid, then Busemann [3] shows that it does follow. However we will show that, at least for n ≥ 12, the result does not hold for general centrally symmetric convex bodies K, even if K′ is an ellipsoid. We do not construct the counter example explicitly; instead we use a probabilistic argument to establish its existence.	{"url":"https://core-cms.prod.aop.cambridge.org/core/search?filters%5BauthorTerms%5D=D.%20G.%20Larman&eventCode=SE-AU","timestamp":"2024-11-10T08:59:07Z","content_type":"text/html","content_length":"1003556","record_id":"<urn:uuid:9fc17c8f-fe83-44f7-a145-827f89a0f9e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00349.warc.gz"}
Christoph Schär: Katalogdaten im Frühjahrssemester 2019 Name Herr Prof. em. Dr. Christoph Schär Lehrgebiet Hydrologie und Klimatologie Institut für Atmosphäre und Klima ETH Zürich, CHN M 12.2 Adresse Universitätstrasse 16 8092 Zürich Telefon +41 44 632 81 99 E-Mail schaer@env.ethz.ch URL http://www.iac.ethz.ch/people/schaer/ Departement Umweltsystemwissenschaften Beziehung Professor emeritus Nummer Titel ECTS Umfang Dozierende 401-5930-00L Seminar in Physics of the Atmosphere for CSE 4 KP 2S H. Joos, C. Schär Kurzbeschreibung In this seminar the knowledge exchange between you and the other students is promoted. You attend lectures on scientific writing and you train your scientific writing skills by writing a proposal for your MSc thesis. You receive critical and constructive feedback through an in-depth review process by scientific writing experts and your future supervisors. Lernziel In this seminar the knowledge exchange between you and the other students is promoted. You attend lectures on scientific writing and you train your scientific writing skills by writing a proposal for your MSc thesis. You receive critical and constructive feedback through an in-depth review process by scientific writing experts and your future supervisors. 651-4095-01L Colloquium Atmosphere and Climate 1 1 KP 1K C. Schär, H. Wernli, D. N. Bresch, D. Domeisen, N. Gruber, H. Joos, R. Knutti, U. Lohmann, T. Peter, S. I. Seneviratne, K. Steffen, M. Wild Kurzbeschreibung The colloquium is a series of scientific talks by prominent invited speakers assembling interested students and researchers from around Zürich. Students take part of the scientific Lernziel The colloquium is a series of scientific talks by prominent invited speakers assembling interested students and researchers from around Zürich. Students take part of the scientific Voraussetzungen To acquire credit points for this colloquium, please visit the course's web page and sign up for one of the groups. / Besonderes 651-4095-02L Colloquium Atmosphere and Climate 2 1 KP 1K C. Schär, H. Wernli, D. N. Bresch, D. Domeisen, N. Gruber, H. Joos, R. Knutti, U. Lohmann, T. Peter, S. I. Seneviratne, K. Steffen, M. Wild Kurzbeschreibung The colloquium is a series of scientific talks by prominent invited speakers assembling interested students and researchers from around Zürich. Students take part of the scientific Lernziel The colloquium is a series of scientific talks by prominent invited speakers assembling interested students and researchers from around Zürich. Students take part of the scientific Voraussetzungen To acquire credit points for this colloquium, please visit the course's web page and sign up for one of the groups. / Besonderes 651-4095-03L Colloquium Atmosphere and Climate 3 1 KP 1K C. Schär, H. Wernli, D. N. Bresch, D. Domeisen, N. Gruber, H. Joos, R. Knutti, U. Lohmann, T. Peter, S. I. Seneviratne, K. Steffen, M. Wild Kurzbeschreibung The colloquium is a series of scientific talks by prominent invited speakers assembling interested students and researchers from around Zürich. Students take part of the scientific Lernziel The colloquium is a series of scientific talks by prominent invited speakers assembling interested students and researchers from around Zürich. Students take part of the scientific Voraussetzungen To acquire credit points for this colloquium, please visit the course's web page and sign up for one of the groups. / Besonderes Numerical Methods in Environmental Sciences Belegung ist NUR erlaubt für MSc Studierende, die diese Lerneinheit als 701-0461-AAL Auflagenfach verfügt haben. 3 KP 6R C. Schär, O. Fuhrer Alle anderen Studierenden (u.a. auch Mobilitätsstudierende, Doktorierende) können diese Lerneinheit NICHT belegen. This lecture imparts the mathematical basis necessary for the development and application of Kurzbeschreibung numerical models in the field of Environmental Science. The lecture material includes an introduction into numerical techniques for solving ordinary and partial differential equations, as well as exercises aimed at the realization of simple models. This lecture imparts the mathematical basis necessary for the development and application of Lernziel numerical models in the field of Environmental Science. The lecture material includes an introduction into numerical techniques for solving ordinary and partial differential equations, as well as exercises aimed at the realization of simple models. Classification of numerical problems, introduction to finite-difference methods, time integration schemes, non-linearity, conservative numerical techniques, an overview of spectral and finite-element methods. Examples and exercises from a diverse cross-section of Environmental Science. Three obligatory exercises, each two hours in length, are integrated into the lecture. The implementation language is Matlab (previous experience not necessary: a Matlab introduction is given). Example programs and graphics tools are supplied. Literatur List of literature is provided. 701-1216-00L Numerical Modelling of Weather and Climate 4 KP 3G C. Schär, N. Ban The course provides an introduction to weather and climate models. It discusses how these models are built addressing both the dynamical core and the physical parameterizations, and Kurzbeschreibung it provides an overview of how these models are used in numerical weather prediction and climate research. As a tutorial, students conduct a term project and build a simple atmospheric model using the language PYTHON. Lernziel At the end of this course, students understand how weather and climate models are formulated from the governing physical principles, and how they are used for climate and weather prediction purposes. The course provides an introduction into the following themes: numerical methods (finite differences and spectral methods); adiabatic formulation of atmospheric models (vertical Inhalt coordinates, hydrostatic approximation); parameterization of physical processes (e.g. clouds, convection, boundary layer, radiation); atmospheric data assimilation and weather prediction; predictability (chaos-theory, ensemble methods); climate models (coupled atmospheric, oceanic and biogeochemical models); climate prediction. Hands-on experience with simple models will be acquired in the tutorials. Skript Slides and lecture notes will be made available at Literatur List of literature will be provided. Voraussetzungen Prerequisites: to follow this course, you need some basic background in atmospheric science, numerical methods (e.g., "Numerische Methoden in der Umweltphysik", 701-0461-00L) as well / Besonderes as experience in programming. Previous experience with PYTHON is useful but not required.	{"url":"https://www.vorlesungen.ethz.ch/Vorlesungsverzeichnis/dozent.view?lang=de&dozide=10002709&ansicht=2&semkez=2019S&","timestamp":"2024-11-02T11:15:59Z","content_type":"text/html","content_length":"22824","record_id":"<urn:uuid:e3f7d830-d120-4207-b518-a887bda17a33>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00602.warc.gz"}
quantum mechanics briefly explained - Quantum Physics Lady quantum mechanics « Back to Glossary Index In quantum mechanics, physicists study how tiny particles behave. These include atoms, the components of atoms like electrons and protons, and other tiny particles like photons of light. These tiny particles follow the laws of quantum mechanics and are called “quantum particles.” Carbon molecules emitting a photon, a tiny particle of light (in green). Yes, the photon looks like it’s moving towards the molecules—no accounting for the whims of artists. [Image source: Nancy Ambrosiano, Los Alamos National Laboratory, July 2017 News Release, “Single-photon emitter has promise for quantum info-processing,” (Public domain) Billiard balls epitomize objects that follow Newton’s Laws of Motion. In the early 1900’s, physicists started experimenting with atoms and molecules. They discovered that atoms and their components follow different laws from those of ordinary objects like tables and chairs, so-called “macroscopic” objects. The mathematical laws governing the movements and forces among tables and chairs, “classical mechanics,” were first articulated by Isaac Newton in the 1600’s. For example, Newton developed the mathematical law of the amount of force exerted by a chair when thrown at a table: Force = Mass times Acceleration. However, when physicists of the early 1900’s used the equations of classical mechanics to predict the results of experiments with electrons and tiny particles of light, their predictions turned out to be erroneous. After considerable dismay, confusion, and fumbling around, they came up with quantum theory. By the 1920’s and 30’s, they were able to articulate the laws of quantum theory with mathematical equations. The mathematical version of this field then became known as “quantum mechanics” on analogy with “classical mechanics.” Quantum mechanics is also often called “quantum physics.” a term which doesn’t emphasize the mathematical aspect of the field. The term “quantum physics” can also be applied to the early years before the mathematical equations of quantum mechanics were developed. The quantum sublevel of reality appears a red mesh wave. It forms a particle in our macroscopic level of reality (orange/yellow dot). [Image source: stills from Fermilab video by Dr. Don Lincoln, “Quantum Field Theory” (in the public domain) Jan. 14, 2016; https://www.youtube.com/watch?v=FBeALt3rxEA&feature=youtu.be.] Quantum mechanics describes a different reality from the one that we experience daily. It’s a sublevel of reality that co-exists with ours and underlies it. I’ll call it “Quantumland.” In Quantumland, an electron is a “wave” that instantaneously turns into a particle the moment that we detect it. “Wave” is in quotes because it’s not a wave like a water wave or other waves that we encounter at the macroscopic level. The mathematical description of a quantum wave is quite different from that of a water wave. The mathematical equations that describe quantum particles don’t fit how we think mathematics should describe things in the physical universe. Quantumland (left) and macroscopic world of objects that we perceive (right). The transformation from the wavy quantum state to particles that we detect is called the “Collapse of the Wave Function.” [Image source: David Chalmers and Kelvin McQueen, “Consciousness and the Collapse of the Wave Function” http://consc.net/slides/collapse.pdf In Quantumland, quantum particles behave in a number of odd ways: • As mentioned, electrons, photons, and other quantum particles can act like a kind of wave at one moment and a particle the next. This transition is demonstrated in the above video. • Quantum particles, in certain regards, appear to behave with true randomness. The properties of individual particles, such as their position when detected, do not appear to be determined by causal events. However, the probability that a collection of particles will manifest particular properties can be predicted. For example, the percentage of particles that will be detected in particular positions can be predicted. • Some of the properties of quantum particles, such as momentum, are not defined until the particles interact with some other part of the physical universe. For example, the direction of travel of a photon of light appears to be many directions at the same time, as if smelling all the possibilities. When detected, that is, absorbed by an electron, the photon will be found in a definite position, one in accord with the straight-line travel. • Quantum particles appear to cause effects prior in time (retro-causality). • Quantum particles such as electrons can disappear from one side of a barrier and instantaneously appear on the other. This, despite having insufficient energy for this maneuver and without actually passing through the barrier. This is “quantum tunneling.” • Particles which have interacted together are “entangled.” The behaviors of entangled particles are correlated despite any amount of distance separating them. For example, two entangled electrons sitting halfway across the universe from each other will always coordinate their spins. If one begins to spin clockwise, the other will begin to spin counterclockwise. According to the Theory of Special Relativity, which prohibits faster than light communication, this cannot be due to signaling each other. It appears to be due to a simultaneity of behavior which physicists can describe but not explain. • Particles, known as “virtual particles,” can appear with no material source nor energy source and disappear almost instantaneously. These particles might be thought to violate the Law of Conservation of Energy. But the violation occurs so briefly that the mathematics of quantum mechanics, specifically, the Heisenberg Uncertainty Principle, is adequate to describe this behavior. Interpretations of Quantum Mechanics Most physicists agree on the key mathematical equations of quantum mechanics. But physicists have not reached agreement on the implications of quantum mechanics for the nature of quantum particles and the nature of reality. Physicists have developed about 15 different interpretations which assign meaning to quantum physics, describing both how quantum particles behave and the nature of Quantumland is a term coined by Dr. Ruth E. Kastner, one of the developers of the Transactional Interpretation of quantum mechanics. Ruth E. Kastner, Understanding Our Unseen Reality, Solving Quantum Riddles; Imperial College Press, 2015, London. *The mathematical description of a quantum wave is called a “wave function.” A wave function can include imaginary numbers. Imaginary numbers are based on the square root of negative 1. You are right if you’re wondering what number squared (multiplied times itself) equals negative 1. There is no such number. The square root of negative 1 does not exist and is called an “imaginary number.” Nevertheless, imaginary numbers are needed when describing quantum waves. This is only one of the issues that besets our ability to conceptualize the mathematics of quantum waves and to consider these waves real. Another issue is that the quantum wave describes probabilities as to how the particle will behave rather than describing definite properties.	{"url":"https://quantumphysicslady.org/glossary/quantum-mechanics/","timestamp":"2024-11-09T00:02:50Z","content_type":"text/html","content_length":"87923","record_id":"<urn:uuid:c9eb7c04-3daf-407f-852b-38160806dc86>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00456.warc.gz"}
CS 61B: Data Structures (Spring 2014) Midterm II Problem 1. (10 points) Trees. a. Many students provided answers in which some of the leaves are at a greater depth than other leaves. If you learn only one thing about 2-3-4 trees, know that all the leaves must be at the same b. The smallest possible number of keys in a valid 2-3-4 tree of depth 3 is 15. d. The total running time for all n insertion operations together is in Θ(n), because each newly inserted key stays at the bottom level of the tree and does not bubble up (being the largest key in the tree), so each insertion operation takes constant time. e. The total running time for all n removal operations together is in Θ(n log n). f. The total running time for all n insertion operations together is in Θ(n^2). The tree looks like this. g. The total running time for all n removal operations together is in Θ(n), because each remove() operation removes a root that has only one child, which is done in constant time. Problem 2. (7 points) Asymptotic Analysis. a. O(9^x + log z + y √log z + y^2 / z^2). b. (i) False. A counterexample is f(n) = 2n; g(n) = n. Clearly f(n) ∈ O(g(n)), but 2^f(n) = 4^n ∉ O(2^n) = O(2^g(n)), because 4^n is larger than 2^n by a factor of 2^n (which is not a constant). b. (ii) True. We assume that f(n) ∈ O(g(n)), which means that there exist constants c and N such that for all n ≥ N, f(n) ≤ c g(n). It follows that for all n ≥ N, log f(n) ≤ log c + log g(n). Let N' be a number such that for all n ≥ N', g(n) ≥ c. Then for all n ≥ max{N, N'}, log f(n) ≤ 2 log g(n), which proves that log f(n) ∈ O (log g(n)). Problem 3. (8 points) Algorithms. a. The score of the root node is 5. b. The fastest possible algorithm to answer this question takes Θ(k^2) time with an adjacency list (using linked lists), and Θ(n) time with an adjacency matrix. c. Note: many students didn't read this question carefully enough. The graph does not come with color preassigned to the vertices; it's your job to pick a color for each vertex. Pick an arbitrary starting vertex and assign it an arbitrary color (red or blue). Perform a depth-first search or a breadth-first search from that vertex. Whenever you visit a new vertex for the first time, assign it the color opposite of the color of the last vertex you just came from. Whenever you follow an edge and find a vertex that has already been visited, check its color; if it has the same color as the last vertex you just came from, output a message saying that the graph is not bipartite and halt. If the DFS or BFS finishes without halting early, the graph is bipartite. d. Before answering the question, I want to comment on what makes this question hard. Many students claimed that it is necessary to check every edge of the graph, but that's not true. If the graph is not bipartite, you can stop checking edges as soon as you identify one cycle with an odd number of edges. Whether the graph is bipartite or not, here's a clever trick that allows you to skip looking at a few of the edges. Do a DFS, but stop it short as soon as you visit the last vertex, which we'll call v. At this point, you have assigned colors to all the vertices, but you've inspected only one of the edges adjoining v. You can avoid inspecting any other edge adjoining v as follows. Suppose that v is red. Then inspect all the edges out of all the red vertices except v. (Note: no need to re-inspect outbound edges already inspected by DFS.) If all those edges lead to blue vertices, then you know that v doesn't adjoin a red vertex either, even without inspecting any edge adjoining v (other than the edge through which we found v). If the total degree of all the blue vertices equals the total degree of all the red vertices, then you also know that no blue vertex adjoins a blue vertex, so the graph is bipartite. (Observe that the vertex degrees give us information that permits us to avoid looking at some of the edges.) Now let's answer the exam question. If it isn't necessary to inspect every edge, which edges must you inspect? Consider a graph G that is bipartite, and suppose that it is “sufficiently well-connected” that you cannot disconnect it by removing two edges. Consider an algorithm (any correct algorithm whatsoever) that determines whether G is bipartite. Suppose there are two edges that the algorithm has not looked at, (r[1], b[1]) and (r[2], b[2]), where r[1] and r[2] are two different red vertices, and b[1] and b[2] are two different blue vertices. If you remove those two edges and replace them with (r[1], r[2]) and (b[1], b[2]), you obtain a modified graph G' that is not bipartite. (Note that we cannot change the coloring of the vertices, because we assumed that removing two edges from G does not disconnect it.) However, observe that there is no other difference between the data structures representing G and G' besides four changed destination vertices in the adjacency list: two changed edges times two directions. In particular, no vertex degree has changed, and the adjacency list records the same edges in the same order except that four destination vertices have changed. Therefore, we absolutely cannot distinguish G from G' without inspecting one of the two edges. Every correct algorithm must inspect at least one of them. To be certain that a “sufficiently well-connected” graph is bipartite, we must keep inspecting edges until there are no two uninspected edges that don't share a common vertex. We can stop inspecting edges only when every edge still uninspected adjoins a common vertex v. Therefore, we must inspect at least e – n + 1 edges (because the degree of v is at most n). If e ≥ 2n, this is Ω(e) edges, so the algorithm takes Ω(e) time. If e < 2n, it takes Ω(n) time just to inspect each vertex, which is at least Ω(e) time. Now that you've seen a rigorous answer, you might wonder what answers actually received credit on the exam. (Obviously, they weren't this long.) We didn't give credit for claiming that it is necessary to inspect every edge of the graph—partly because that's wrong, but also because we only give credit for attempting to prove that we must inspect most of the edges. We looked for answers that point out that you can transform a bipartite graph to a non-bipartite graph by changing just an edge or two, and that there are many such transformations that can be made to the same bipartite Mail inquiries to cs61b@cory.eecs	{"url":"https://people.eecs.berkeley.edu/~jrs/61bs14/exam/14mid2.html","timestamp":"2024-11-14T01:59:21Z","content_type":"text/html","content_length":"8795","record_id":"<urn:uuid:74e28962-472f-4bbc-bda7-5f771454af03>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00371.warc.gz"}
Properties of a chaotic transition in ecosystem dynamics TYPE Statistical & Bio Seminar Speaker: Yanay Tovi Affiliation: Physics Department, Technion Date: 14.09.2021 Time: 13:30 - 14:30 Location: Lidow Nathan Rosen (300) Remark: Msc seminar Abstract: Natural ecosystems often harbor many interacting species, which suggests that they might be amenable to analysis using tools from statistical physics, with high diversity playing the role of the thermodynamic limit. In this limit, dynamical phase-transitions have been found in theoretical models. Here we study one such transition, from a fixed-point phase to a chaotic phase. We analyze the behavior near this continuous transition using a perturbation expansion, and derive a self-consistent equation for the size of the dynamical fluctuations there. We show evidence towards an intriguing conclusion: that this self-consistent equation might not admit a non-trivial solution corresponding to a chaotic phase.	{"url":"https://phys.technion.ac.il/en/events/future-events/mevent/2445:properties-of-a-chaotic-transition-in-ecosystem-dynamics?tmpl=component","timestamp":"2024-11-03T08:58:25Z","content_type":"application/xhtml+xml","content_length":"4715","record_id":"<urn:uuid:2e43cef5-89d0-4d05-8ff3-539f1e6164b6>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00014.warc.gz"}
sq m to sq yd and sq yd to m Easily convert square meters to square yards on Examples.com. Simply enter your area measurement for quick conversion outcomes. sq m to sq yd Formula: square yards (sq yd) = Length in square meter ( sq m ) x 1.195990 Square Meter: Square Yard: Square Meter Square Yard Exponential 1 1.1959900463 1.1959900463e+0 sq yd to sq m Formula: Length in square meter = Length in square yard (sq yd ) × 0.836127 Square Yard: Square Meter: Square Yard Square Meter Exponential 1 0.83612736 8.3612736e-1 Area Converters to Square Meter (sq m) Area Converters to Square Yard (sq yd) Conversion Factors • Square meters to square yards: 1 square meter = 1.19599 square yards. • Square yards to square meters: 1 square yard = 0.836127 square meters. How to Convert Square Meters to Square Yards To convert square meters to square yards, multiply the number of square meters by 1.19599. Square yards=Square meters×1.19599 Example: Convert 200 square meters to square yards. Square yards=200×1.19599=239.198square yards How to Convert Square Yards to Square Meters To convert square yards to square meters, multiply the number of square yards by 0.836127. Square meters=Square yards×0.836127 Example: Convert 300 square yards to square meters. Square meters=300×0.836127=250.8381square meters. Square Meters to Square Yards Conversion Table Square Meters (sq m) Square Yards (sq yd) 1 sq m 1.196 sq yd 2 sq m 2.392 sq yd 3 sq m 3.588 sq yd 4 sq m 4.784 sq yd 5 sq m 5.980 sq yd 6 sq m 7.176 sq yd 7 sq m 8.372 sq yd 8 sq m 9.568 sq yd 9 sq m 10.764 sq yd 10 sq m 11.960 sq yd 20 sq m 23.920 sq yd 30 sq m 35.880 sq yd 40 sq m 47.840 sq yd 50 sq m 59.800 sq yd 60 sq m 71.760 sq yd 70 sq m 83.720 sq yd 80 sq m 95.680 sq yd 90 sq m 107.640 sq yd 100 sq m 119.600 sq yd sq m to sq yd Conversion Chart Square Yards to Square Meters Conversion Table Square Yards (sq yd) Square Meters (sq m) 1 sq yd 0.836 sq m 2 sq yd 1.672 sq m 3 sq yd 2.508 sq m 4 sq yd 3.345 sq m 5 sq yd 4.181 sq m 6 sq yd 5.017 sq m 7 sq yd 5.853 sq m 8 sq yd 6.689 sq m 9 sq yd 7.525 sq m 10 sq yd 8.361 sq m 20 sq yd 16.723 sq m 30 sq yd 25.084 sq m 40 sq yd 33.446 sq m 50 sq yd 41.807 sq m 60 sq yd 50.169 sq m 70 sq yd 58.530 sq m 80 sq yd 66.892 sq m 90 sq yd 75.253 sq m 100 sq yd 83.615 sq m sq yd to sq m Conversion Chart Difference Between Square Meters to Square Yards Aspect Square Meters (sq m) Square Yards (sq yd) Unit System Metric system Imperial system Symbol sq m or m² sq yd Equivalent Size 1 sq m = 1.196 sq yd 1 sq yd = 0.836 sq m Usage Commonly used globally, especially in scientific contexts Mostly used in the U.S. and U.K. for real estate, land area, etc. Standard Conversion Factor 1 sq m = 10.7639 sq ft 1 sq yd = 9 sq ft Typical Application Floor plans, architecture, landscaping, and scientific areas Residential and commercial real estate transactions, construction, etc. Subunits Square centimeters (sq cm) Square feet (sq ft) Area Comparison A larger value numerically compared to the same area in sq yd A smaller value numerically compared to the same area in sq m 1. Solved Examples on Converting Square Meters to Square Yards Example 1: Convert 50 square meters to square yards. 50 square meters is approximately 59.80 square yards. Example 2: Convert 100 square meters to square yards. 100 square meters is approximately 119.60 square yards. Example 3: Convert 175 square meters to square yards. 175 square meters is approximately 209.30 square yards. Example 4: Convert 250 square meters to square yards. 250 square meters is approximately 299.00 square yards. Example 5: Convert 360 square meters to square yards. 360 square meters is approximately 430.56 square yards. 2. Solved Examples on Converting Square Yards to Square Meters Example 1: Convert 50 square yards to square meters. 50 square yards is approximately 41.81 square meters. Example 2: Convert 100 square yards to square meters. 100 square yards is approximately 83.61 square meters. Example 3: Convert 175 square yards to square meters. 175 square yards is approximately 146.32 square meters. Example 4: Convert 250 square yards to square meters. 250 square yards is approximately 209.03 square meters. Example 5: Convert 360 square yards to square meters. 360 square yards is approximately 301.01 square meters. Why would I need to convert square meters to square yards? Converting square meters to square yards is often necessary when working with measurements in both the metric and imperial systems, particularly in real estate, construction, or international research projects. Is there a quick way to estimate square meters to square yards without a calculator? While not entirely accurate, you can approximate the conversion by multiplying the square meters by roughly 1.2 to obtain the equivalent square yards. Is converting square meters to square yards linear or exponential? The relationship between square meters to square yards is linear because it involves a consistent conversion factor. Does converting square meters to square yards affect accuracy? Converting square meters to square yards using precise calculations ensures accurate conversions, especially when the original measurement is precise. Do different industries require unique conversion factors for square meters to square yards? No, the same factor of 1.19599 is used universally across different industries to convert square meters to square yards. What are the challenges when converting large areas from square meters to square yards? When converting large areas from square meters to square yards, maintaining consistent precision is crucial, especially with large datasets.	{"url":"https://www.examples.com/maths/sq-m-sq-yd","timestamp":"2024-11-02T11:57:30Z","content_type":"text/html","content_length":"112082","record_id":"<urn:uuid:9268bc76-de6b-4d2e-bb7c-fd9850366534>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00775.warc.gz"}
A Class of Nonhydrostatic Global Models 1. Introduction In this paper we describe the extension of a small- to mesoscale nonhydrostatic model to global scales. Traditionally, numerical models for simulating planetary-scale weather and climate employ the hydrostatic primitive equations—a truncated form of Navier–Stokes’ equations that neglects vertical accelerations and uses simplified Coriolis forces.^^1 Although there is no evidence so far that including nonhydrostatic effects in global models has any physical significance for large-scale solutions, there is an emerging trend in the community toward restoring the Navier–Stokes’ equations (or at least their less constrained forms) in global models of atmospheres and oceans. The primary motivation is that state-of-the-art computers already admit resolutions where local nonhydrostatic effects become noticeable. Other advantages include: the convenience of local mesh refinement, better overall accuracy while entailing insubstantial computational overhead relative to hydrostatic models, generality and the resulting convenience of maintaining a single large code, the theoretical well-posedness in the continuous limit (Oliger and Sundström 1978), conceptual simplicity and mathematical elegance—features important for education and dissemination. The few existing nonhydrostatic global models differ in their analytic formulation and numerical design, reflecting their different origins and purposes. The compressible atmospheric models of Semazzi et al. (1995) and Cullen et al. (1997) aim at high-resolution numerical weather prediction (NWP) and extend approximation methods traditional in NWP on nonhydrostatic dynamics. The incompressible Boussinesq model of Marshall et al. (1997a) extends a convection-scale stratified-fluid model to a flexible research tool for all-scale studies of oceanic circulations. The development in this paper is philosophically akin to that in Marshall et al. (1997a) but addresses atmospheric circulations. Much of our recent research (Anderson et al. 1997; Smolarkiewicz and Margolin 1997) has aimed to improve the design of our high-performance numerical solver for simulating flows of moist (and precipitating), rotating, stratified fluids past a specified time-dependent irregular lower boundary. This model is representative of a class of small-scale nonhydrostatic atmospheric codes that employ the anelastic equations of motion in a terrain-following curvilinear framework. A unique feature of our model is the parallel implementations of nonoscillatory forward-in-time (NFT) semi-Lagrangian and Eulerian approximations (Smolarkiewicz and Pudykiewicz 1992; Smolarkiewicz and Margolin 1993), options that are selectable by the user. This approach has been employed in a variety of small- and mesoscale geophysical applications and the quality of results suggest that NFT methods are superior to the more traditional centered-in-time-and-space schemes in terms of accuracy, computational efficiency, flexibility, and robustness (cf. Smolarkiewicz and Margolin 1997, 1998). The goal of this study is to prepare the ground for a future “global cloud model”—a research tool to investigate effects of small- and mesoscale phenomena on global flows and vice versa. In order to arrive at an optimal design, we explore several extensions of our small-scale NFT model to a mountainous sphere. Each of the resulting global models naturally dispenses with the traditional simplifications of hydrostaticity, gentle terrain slopes, and weak rotation. However, not all possible extensions are practical. For example, straightforward extensions that account only for the appropriate metric coefficients, forces, and boundary conditions inevitably lead to computationally intractable global models of the earth’s meteorology. The key issue is the formulation and efficient solution of the elliptic equations that arise in nonhydrostatic modeling. One consequence of nonhydrostatic modeling using either filtered equations or implicit discretization methods, is the necessity of solving a full 3D Poisson or Helmholtz equation, which typically does not possess the regularity properties that allow the use of fast standard methods. The condition number (of the elliptic operator) grows in proportion to the square of the horizontal scale of the model, and so does the degree of difficulty in solving the problem. All published works on existing nonhydrostatic global models (cf. Cullen et al. 1997; Marshall et al. 1997a; Semazzi et al. 1995) note the importance of this key issue. Because NFT methods are inherently two-time-level, an accurate (i.e., time-centered) integration of forces complicates the problem even further, leading ultimately to the inversion of a large nonsymmetric linear system that represents a complex nonself-adjoint 3D elliptic partial differential equation for pressure. For deep fluids, such a problem can be solved easily using standard Krylov subspace methods for nonsymmetric operators (Smolarkiewicz and Margolin 1994; Smolarkiewicz et al. 1997). For shallow fluids typical of the earth’s meteorology, however, the resulting elliptic operator is extremely stiff,^^2 necessitating additional enhancements to the Krylov solvers to assure convergence. We will compare two strategies in this paper. One is to use the hydrostatic solution for pressure as an initial guess for the iterative nonhydrostatic pressure solver. This approach naturally facilitates the optional implementation of hydrostatic or nonhydrostatic models (cf. Marshall et al. 1997a). The second strategy dispenses with the decomposition of the pressure into a hydrostatic and a nonhydrostatic part but relies on effective preconditioners for the Krylov solver. This results in a simpler, more general, and elegant variant of the model. Yet the most important benefit of the latter variant is the ease of further extension to the semi-implicit (with respect to the internal gravity waves)^^3 NFT scheme [see the appendix for a formal exposition;for a complete technical development, see appendix B in Smolarkiewicz et al. We present our nonhydrostatic anelastic semi-implicit global NFT model as it emerges from a series of sequential extensions of a small-scale dynamics model. Using benchmark “dynamical core” experiments that idealize weather and climate simulations (while excluding moist processes, radiation, and subgrid-scale turbulence schemes), we compare the relative accuracy and efficiency of hydrostatic, nonhydrostatic, implicit, explicit, semi-Lagrangian, Eulerian, elastic, anelastic, incompressible, etc., variants of the global model. We measure the differences due to analytic formulation of the governing equations against the truncation errors of legitimate optional second-order-accurate discretization approximations. This analysis builds our confidence that nonhydrostatic anelastic global models derived from small-scale codes (i.e., not relying on large-scale balances in their analytic/numerical design) adequately capture a broad range of planetary flows while requiring relatively minor overhead due to the nonhydrostatic formulation. The numerical models considered in this study have been developed over several years; some were abandoned as impractical, while others have been pursued and generalized. To aid the reader’s orientation in the discussions that follow, we list all models in a historical order and characterize them briefly in Table 1. 2. Model description In this paper, we focus on an inviscid, adiabatic, density-stratified fluid whose undisturbed, geostrophically balanced “ambient” (or “environmental”) state is described by the potential temperature = ϒ ) and the velocity ). The expanded form of the governing anelastic equations used in the model, which accounts for the coordinate transformations, has been given in Smolarkiewicz et al. (1998 closely following Clark et al. 1996 To facilitate further discussion on alternate model formulations, here we present only compact, symbolic forms of the governing equations. We start with the anelastic system of Lipps and Hemler 1982 (see also Lipps 1990 Here the operators Grad, and Div have their usual meaning of material derivative, gradient, and divergence; denotes the velocity vector; ^^5 M symbolizes appropriate metric forces [Christoffel terms proportional to products of velocity components; see Smolarkiewicz et al. (1998 ]; ϒ, denote potential temperature, density, and pressure; symbolize the vectors of the “Coriolis parameter” and gravity, respectively. Primes denote deviations from the ambient state, and overbars refer to the horizontally homogeneous hydrostatic reference state of the Boussinesq expansion around a constant stability profile [see section 2b in Clark and Farley (1984) for a discussion]. The anelastic equations (1)–(3) may be viewed as combining two distinct approximations in the compressible Euler equations: the Boussinesq type linearization of the pressure gradient forces and mass fluxes in, correspondingly, ; and the anelasticity per se equivalent to taking the limit of an infinite speed of sound. To assess the impact of both assumptions inherent in the anelastic system we employ, respectively, the fully incompressible Euler equations and a Boussinesq elastic system. Using the same notation as in , the incompressible Euler equations are written as follows The elastic Boussinesq model consists of the momentum and entropy equations (1) , formally identical with those of the anelastic system, and the mass continuity equation is the speed of sound (constant to the Boussinesq approximation, taken here to be 300 m s ). Technically, the conversion from the anelastic to incompressible equations is achieved easily within the framework of the explicit numerical model (entry 3 in Table 1 ), by redefining the pressure gradient forces and buoyancy, and by equivalencing the ϒ field to the fluid density ) (cf. Rotunno and Smolarkiewicz 1995 ). The conversion to the elastic Boussinesq system is a particularly simple alteration of the model (either explicit or implicit), merely requiring replacement of the anelastic mass continuity equation (3) . In effect, the Poisson equation for pressure is transformed into a slightly better-conditioned Helmholtz equation (A7) ; see the appendix for discussions. Our basic NFT approach for approximating integrals of the governing equations on a discrete mesh is second-order-accurate in space and time. The two optional model algorithms, semi-Lagrangian ( Smolarkiewicz and Pudykiewicz 1992) and Eulerian (Smolarkiewicz and Margolin 1993), correspond to the trajectory-wise and point-wise integrals of the evolution equations (1, 2) or (4, 5). All variables are defined at the same grid points x[i]—a choice important for the efficacy of the unified semi-Lagrangian–Eulerian NFT approach (Smolarkiewicz and Margolin 1997). The theory underlying the NFT numerical algorithms has been documented in earlier works, and their basic properties of computational stability, accuracy, and efficiency are already established. In this paper we focus on the overall design of nonhydrostatic NFT models and their relative performance in representative complex planetary flow problems. We write the resulting finite-difference approximations in the compact form Here, LE denotes either an advective semi-Lagrangian or a flux-form Eulerian NFT transport operator [sections 3.1 and 3.2 in Smolarkiewicz and Margolin (1997) , respectively]; ^^6 ψ̃ + 0.5Δ , where the indices denote the spatial and temporal location on a (logically) rectangular Cartesian mesh. Transporting the auxiliary field (rather than the fluid variable alone) has been shown to be important for the accuracy and stability of forward-in-time approximations. In the Eulerian algorithm, transporting is a consequence of compensating the first-order truncation error proportional to the divergence of the advective flux of , while in the semi-Lagrangian algorithm it derives straightforwardly from the trapezoidal-rule approximation for the integral on the right-hand side (rhs) of the evolution equations; see Smolarkiewicz and Margolin (1997 , for further discussions. Equation (8) represents a system implicit with respect to the dependent variables (e.g., ϒ and components of v). Completion of the model algorithm requires a straightforward algebraic inversion of (8), resulting in the formulation of the boundary value problem for pressure implied by the mass continuity constraint—for example, Eq. (3). In the appendix, we outline the essential steps of this fairly standard projection procedure and focus on key features of the different numerical models considered. The resulting elliptic equations are solved, subject to appropriate boundary conditions, using a conjugate-gradient method. We have found the generalized conjugate-residual (GCR) approach of Eisenstat et al. (1983)—a nonsymmetric Krylov solver akin to the celebrated GMRES algorithm (Saad and Schultz 1986)—especially convenient for geophysical fluid models. Further technical details of the projection procedure, an explicit form of the resulting elliptic operator for the anelastic system, and a description of the complete preconditioned Krylov solver can be found in Smolarkiewicz et al. (1999). 3. Results In this section we analyze in detail two examples of idealized global flows. These are representative of two distinct classes of application—basic geophysical fluid dynamics of a mechanically forced laminar system, and climate studies of a thermally forced turbulent system. These examples serve as convenient test beds for the overall accuracy (both analytic and numerical) and relative performance of our nonhydrostatic global models. Note that the overall intent of our analysis is not to demonstrate that our small-scale code works well at resolutions appropriate for nonhydrostatic dynamics (this is well documented in the literature already), but, to the contrary, to demonstrate that the globally extended small-scale models adequately capture a broad range of large-scale responses even though they do not exploit global balances in their analytic–numerical design. a. Orographic flow The first problem is an orographic flow of a shallow fluid on a rotating sphere. It was proposed by Williamson et al. (1992) for evaluating the accuracy and efficiency of numerical methods for global-scale dynamics and has become a benchmark in the field. Its original formulation is in terms of the shallow water equations. Here, we extend this problem to a continuously stratified 3D Boussinesq fluid while keeping other flow parameters consistent with the benchmark. We use the Boussinesq option of the anelastic model to enable meaningful comparisons of the three governing systems (1–3), (4–6), and (1, 2, 7). In this manner, we can quantify the effects of the non-Boussinesq baroclinicity and finite speed of sound relative to the truncation errors of finite-difference approximations. We anticipate these results would extrapolate to similar comparisons of the anelastic and fully compressible Euler equations. Figure 1 shows the pattern of vertical and meridional velocity components after 15 days of simulated zonal flow of a stratified Boussinesq fluid past a large hill (consistent with the linear solution of Grose and Hoskins 1979 ; cf. their Fig. 3a) using the semi-Lagrangian implicit variant of the model discussed above. The flow parameters are and the corresponding thermally balanced state = ϒ [1 + ], and the Brunt–Väisälä frequency = 10 . Here r, R, ϕ, and Ω denote, respectively, the radial component of the vector radius, sphere’s radius, latitude, and angular velocity of the planetary rotation. Also, Γ ≡ ′ ≡ The conical hill with height 2 × 10 m and base radius Π/9 (in terms of lat–long) is centered at ( ) = (3Π/2,Π/6), where is the longitude. The globe is covered with a uniform spherical mesh with = 128 × 64 grid intervals (no grid points at the poles) and the = 8 × 10 m deep atmosphere is resolved with = 20 uniform grid intervals. The time step Δ = 7.2 × 10 The simulation in Fig. 1 employs no viscous filters and remains, effectively, inviscid. The residual dissipation due to the nonoscillatory remapping algorithm (Smolarkiewicz and Grell 1992) suffices merely to preserve monotonicity of the transported variables [consistent with analytic properties of fundamental conservation laws; Smolarkiewicz and Pudykiewicz (1992), see also section 2d in Prusa et al. (1996), and Margolin et al. (1999), for further discussions]. For reference, the execution time on a single processor CRAY J-90 is ∼1.4 × 10^4 s. This value can be easily reduced by a factor of about 2 by using dissipative filters in the polar regions (a common practice in global circulation models). With the large time step employed here, most of the computational effort is already in the elliptic solver, so that overall model efficiency strongly depends on such technical issues as stopping criteria (Smolarkiewicz et al. 1997) and effective preconditioning (Skamarock et al. 1997). We believe that improvement in the latter will lead to further acceleration of the model. In order to establish the solution dependence on model design and on the numerical scheme employed, as well as on alternate formulations of the governing equations of motion, we have performed a series of simulations like that in Fig. 1. The results of this sensitivity study are summarized in Table 2. Table 2 collects various runs in a certain logical order as follows. Entries 1–5 are for the semi-Lagrangian and Eulerian runs of the semi-implicit anelastic model (entry 6 in Table 1) at different time steps; run 3b is identical to run 3 except for a tighter convergence criterion (one order of magnitude) in the GCR pressure solver. Entries 6–8 are for the hydrostatic and nonhydrostatic runs using explicit (with respect to internal gravity waves) variants of the semi-Lagrangian anelastic model (entries 2 and 3 in Table 1). Runs 6 and 7 are for the hydrostatic and nonhydrostatic options of the model based on the hydrostatic first guess, and run 8 is for the alternate nonhydrostatic model with an ADI-type preconditioner of the 3D elliptic pressure solver. Since the explicit model runs are prohibitively expensive, further applications of the explicit models (entries 9–12) incorporate a heavy dissipative filter in polar regions^^7 to allow a substantially larger time step. Thus, runs 9 and 10 were designed to make the explicit and semi-implicit nonhydrostatic semi-Lagrangian anelastic runs 8 and 4 directly comparable. Finally, entries 11 and 12 collect results of the simulations using alternate governing equations of motion (entries 5 and 4 in Table 1). Run 11 uses the fully nonlinear incompressible Euler equations (4)–(6), and so addresses the impact of the Boussinesq type linearization inherent in the anelastic model (1)–(3). Run 12, in turn, addresses the impact of incompressibility inherent in the anelastic model by admitting a finite speed of sound while retaining the Boussinesq approximation. For ease of comparison, the table lists L[∞] and L[2] norms of the meridional and vertical velocity fields—natural perturbation fields with respect to the ambient flow (9)—as well as the computational expense of the model measured by the CPU time (in minutes). As illustrated by the values collected in tables, all the listed solutions agree to about 10%. In fact, they are all similar to that shown in Fig. 1 and are hardly distinguishable in the figures. In order to quantify the differences between various experiments, we have performed analyses of the appropriate difference fields. The results are summarized in Table 3.^^8 Table 3 supports a number of interesting conclusions. We draw attention to a few points of special note. In general, the differences due to the higher-order truncation errors of legitimate modes of executing contemporary global models (i.e., various time steps, Eulerian, Lagrangian, various pressure solvers, etc.) overwhelm the differences due to analytic formulation of the governing equations [cf. Nance and Durran (1994) for a similar conclusion in the area of small-scale dynamics]. The largest differences observed are due to the six-fold difference in Δt (using the semi-Lagrangian semi-implicit model), while the smallest differences are between the hydrostatic and nonhydrostatic model formulations. For illustration, in Fig. 2, we show the respective δw difference fields (cf. plate a in Fig. 1).^^9 In particular, note the maximal vertical velocity δw\|[Δ(6,7)] ∼ 10^−3 m s^−1; that is, an order of magnitude less than the vertical velocities of the actual flow. The relative overhead of solving the nonhydrostatic problem is ∼90% in the model where the nonhydrostatic solution is sought as a perturbation to the hydrostatic result, but only ∼44% in the simpler model without the hydrostatic counterpart. For a sufficiently small Δt, the difference between the explicit and implicit Δ(9,10) model formulations is small. Note that it is the implicit model that is computationally more efficient, most likely due to the better conditioning of the elliptic pressure operator. The results collected in Tables 2 measure the proximity of various numerical solutions, but offer no information about their absolute accuracy (except for the correct magnitude of the vertical velocity). In order to assess the accuracy of approximate solutions (note the lack of the analytic solution), as well as to comment on relative accuracy of semi-Lagrangian and Eulerian versions of the model algorithm, we consider two separate measures. Table 4 lists values of the density-normalized inverse flow Jacobian is the Jacobian of the coordinate transformation ( Smolarkiewicz et al. 1998 ). This quantity is readily available only in semi-Lagrangian models. The Lagrangian form of the mass continuity equation, , implies that the exact value of should be 1. The quantity listed in the last column of Table 4 , is defined as where 〈〉 denotes the domain integral. Since the selected test problem is fully adiabatic, the exact value of is zero. Note, however, that neither semi-Lagrangian nor Eulerian model algorithms assure exact conservation. The semi-implicit design dictates that both models transport ϒ′ rather than full ϒ [cf. Eq. (A3) in the appendix], so even “conservative” flux-form Eulerian model contains a residual source of ϒ, whose magnitude depends on the accuracy specified for enforcing the anelastic incompressibility , via imposing a stopping criterion in the iterative elliptic solver ( Smolarkiewicz et al. 1997 ). In turn, the semi-Lagrangian model is not conservative (i.e., in advective form) by design. The results collected in Table 4 show, in general, that the errors in are small, both for semi-Lagrangian and Eulerian options of the model. Except for the semi-Lagrangian simulation with Δ = 7200 s, maximum errors of the flow Jacobian are less than 0.5%. Conservation errors are clearly smaller for the Eulerian option, and can be reduced even further by imposing a more stringent stopping criterion in the elliptic solver. From the practical viewpoint, however, both model options appear equally viable, at least for this smooth flow case. b. Idealized climate states The orographic planetary flow discussed in the preceding section is fairly laminar and deterministic. The equivalent results generated with many different variants of the model closely match each other, documenting both the hydrodynamic stability of the flow and robustness of the model design. The example considered in this section is very different in nature. Simulations of the idealized climates of Held and Suarez (1994), basically representing thermally forced baroclinic instability on the sphere, bear striking resemblance to large-eddy simulations of convective boundary layers ( Nieuwstadt et al. 1992), where small differences in model setups can lead to totally different instantaneous flow realizations, and where different model designs can lead to quite divergent integral flow characteristics. In other words, these simulated flows are both turbulent and stochastic [for illustration, see Fig. 3 in Smolarkiewicz et al. (1999)]. They typify the response of an initially stagnant and uniformly stratified fluid to a diabatic forcing that mimics the long-term thermal and frictional forcing in the earth atmosphere. This diabatic forcing attenuates ϒ and v to the specified equilibrium temperature ϒ[EQ](\|y\|,r′) and v\|[r′<z[i]] = 0, where z[i] represents a height of the boundary layer (see section 2 in Held and Suarez (1994), for details). The corresponding forcing functions augment the governing equations of motion (1)–(2) with appropriate Rayleigh friction and Newtonian cooling–heating terms. The original forcing functions of Held and Suarez are expressed in normalized pressure coordinates σ ≡ p/p[s] (where p[s] and p denote, respectively, the full thermodynamic pressures at the surface and in the atmosphere aloft), so their diabatic forcing may evolve in time. In the anelastic model, only gradients of the perturbation pressure are meaningful and the full thermodynamic pressure is, in essence, unavailable. In order to avoid cumbersome procedures attempting to recover the true σ coordinate in our anelastic model, we have simply assumed a standard atmosphere with the density scale of 7 km to evaluate the fixed forcing functions. The significance of such a simplification can be verified easily within the framework of a σ-coordinate model. In section 3a, we assumed a shallow fluid and used the Boussinesq approximation ρ(z[c]) = ρ[o] and ϒ = ϒ[o]. Here we consider a deep atmosphere and, therefore, solve the anelastic equations (1)–(3) routinely with a variable reference density and potential temperature implied by N = 10^−2 s^−1 Brunt–Väisällä frequency assumed for the reference state [cf. section 2b in Clark and Farley (1984)]. The implicit numerical model (entry 6 in Table 1) is employed, and the environmental profiles v[e] = 0 and ϒ[e] = ϒ[EQ](Π/2,r′) are assumed. The globe is covered by uniform spherical mesh with nx × ny = 64 × 32 grid intervals (no grid points at the poles) and the H = 32 × 10^3 m deep atmosphere is resolved with nz = 40 uniform grid intervals. The time step of integration is Δt = 900 s. The dissipative filter in the polar regions assumes a reciprocal of the attenuation timescale that increases linearly from 0 to 1/86400 s^−1 over the four grid intervals near each pole. Also, in lieu of the biharmonic diffusion used in the original Held–Suarez experiments, we exploit the implicit viscosity of the advection algorithms by employing the first-order upwind scheme at every sixth time step of the model (cf. Liska and Wendroff 1998) in both the Eulerian and the semi-Lagrangian simulations.^^12 The particular simulation depicted in Fig. 3 used the massively parallel version^^13 of the Eulerian model algorithm with the standard and linearized nonoscillatory MPDATA transport schemes for, respectively, ϒ′ and momenta (Smolarkiewicz and Margolin 1998). Figure 3 displays the resulting “climate”; that is, zonally averaged 3-yr means of u and ϒ—the data from the first 200 days is excluded. Figure 4 supplements Fig. 3 with equivalent displays of υ and w to complete the picture of the simulated climatic circulation. Figure 3 corresponds to the results in Figs. 1 and 2 of Held and Suarez (1994). (Note that their plots are drawn in the σ coordinate.) The agreement of the two solutions is qualitative, which should not be surprising considering the substantial differences between the two models. Our trade winds and equatorial easterlies aloft are somewhat weaker, but our subpolar easterlies are more pronounced. Our westerlies are about as strong but shifted toward the equator. A pronounced neutral stratification in polar regions is an artifact of our filter attenuating local solutions toward ϒ[EQ], well mixed near the poles, cf. Fig. 1b in Held and Suarez (1994); it is inconsequential for the discussions that follow. Although the Held–Suarez original solutions are for the primitive equations, we do not believe (based on the results of the preceding section) that either hydrostaticity, compressibility, or simplified Coriolis and metric forces are responsible for the differences. Among “physical” factors, perhaps the Boussinesq linearization of the pressure gradient terms is important, but even this seems unlikely in the light of the following results. We have performed numerous experiments to address the sensitivity of Held–Suarez climates to various aspects of the numerical model design. Among these, we have tested sensitivities to the initial and ambient conditions; to the selected reference state; to the definition of the σ-coordinate in the forcing functions; to spatial and temporal resolution; to model depth; to the design, strength, and spatial extent of polar filters; to the various viscosities in the model (in particular, the relative viscosity in the entropy and velocity equations, i.e., Prandtl number); to the flux versus advective (i.e., Eulerian versus semi-Lagrangian) model formulation; and to the linear z versus mass σ (recall δp ∼ −ρδz) vertical coordinate representation. The latter—see Fig. 5, for illustration—has been achieved by an exponential stretching of the vertical coordinate mimicking σ coordinate for the standard atmosphere with the density scale height of 7 km. We have found that the tropospheric climate is fairly robust, with fine details depending both on the model resolution and characteristics of polar filters. In contrast, the stratospheric solutions are quite sensitive even to the details of the model design.^^14 For example, the simulation identical to that summarized in Fig. 3 except for using a somewhat-less viscous option of MPDATA for ϒ′, results in weaker equatorial easterlies aloft and a slow O(year) meridional oscillation in the stratosphere (cf. Untch 1999). Also, a similar simulation but using semi-Lagrangian advection, produces westerlies whose magnitude increases monotonically with height (no closed jets). However, both models tend to reproduce the solution in Fig. 3 when the model depth is doubled and stretched “mass” coordinates are employed with a vertical gravity wave absorber layer at the top of the model. Experiments using enhanced viscosity in the vertical transport terms—a crude convection parameterization—demonstrate another strong sensitivity of the solutions. More such examples could be presented (cf. Boville and Cheng 1988; Boville and Baumhefner 1990;and the references therein). Tables 5 and 6 quantify and compare selected simulated climates. Table 5 lists extrema and standard deviations of zonal velocity (m s^−1) and potential temperature (K), and L[∞] and L[2] norms of meridional, and vertical velocities (m s^−1). Experiment EU-dfl is the default run discussed extensively earlier in this section. Run 1 (EU-rfs) assumes a twice-larger Brunt–Väisällä frequency for the reference state but is otherwise identical to the default run, and run 2 (EU-bou) tests the shallow-convection (classical incompressible Boussinesq) approximation with a constant density and potential temperature in the reference state. The latter two experiments address the adequacy of the anelastic approximation. Run 3 (EU-f4b) differs from EU-dfl only by a 10-times weaker polar filter. Run 4 (EU-adv) uses a slightly less viscous MPDATA option for the advection of ϒ′, and the run 5 (SL-dfl) is identical to EU-dfl but uses the semi-Lagrangian option of the model. When the difference fields are compared (Table 6), it is apparent that the solutions are more sensitive to details of the numerical schemes than to fairly substantial departures in specific realizations of the anelastic system of the equations. The Boussinesq linearization around the reference state with four-times larger stability—an anelastic model with the stratosphere in focus—affects the resulting climates much less (in terms of standard deviations) than, for example, switching between closely related second-order-accurate semi-Lagrangian and Eulerian advection algorithms. The latter has an even more pronounced effect than making the unjustifiable shallow-convection approximation within the anelastic system (thereby inhibiting the non-Boussinesq amplification of vertically propagating disturbances). The corresponding comparison of the dispersions of mean climatic states indicates a roughly constant range of weather variability for all climatic states (Table 7), whereas an analysis of the respective difference fields of σ^X (not shown) tends to correlate with the results of Table 6. The picture emerging from our sensitivity study—consistent with other reported results (Chen and Bates 1996; Untch 1999)—is that the simulated Held–Suarez climates depend strongly on the viscous properties of the numerical models employed (and on subgrid-scale transport, in general), regardless of whether those arise via explicit parameterizations or implicit effects of the truncation errors. In our experience, the smaller the viscosity of the model, the less robust are the climate simulations. The latter implies the dependence on details of convection parameterizations and, ultimately, the need for resolving convective scales in climate modeling. Direct modeling of clouds on the global scale is beyond the reach of the computational technology perhaps for decades. However, this is where nonhydrostatic global models may be helpful. Less constrained forms of the Navier–Stokes’ equations open new research possibilities that would be either cumbersome or even invalid in models based on the primitive equations. For example, the cloud-resolving convection parameterization (Grabowski and Smolarkiewicz 1999) embeds a 2D cloud-resolving model in each column of a large-scale model. There, the compatibility of the governing equations simplifies the formalism of the model coupling—a common issue in multilevel methods and mesh-refinement schemes. Another example is a series of simulations in Smolarkiewicz et al. (1999), where the Held–Suarez experiments were extended on abstract aquaplanets with radii decreasing successively by a factor of 10. Such rescaled globes cover range of scales unaccessible to primitive equations. In principle, they could be employed to investigate certain effects of small- and mesoscale phenomena on global flows and vice versa at a fundamental level. 4. Summary remarks In the global atmospheric–oceanic modeling community, there is a trend toward replacing the traditional hydrostatic primitive equations with less constrained nonhydrostatic forms of the Navier–Stokes’ equations (Semmazi et al. 1995; Marshall et al. 1997a; Cullen et al. 1997). The few existing nonhydrostatic global models vary in analytic formulation and numerical design, reflecting their different purposes and origins. Drawing from the philosophy of the “all-scale-research” ocean model of Marshall et al. (1997a), we have extended our anelastic NFT small-to-mesoscale atmospheric model (broadly documented in the literature) to a mountainous sphere and have dispensed naturally with the traditional meteorological simplifications of hydrostaticity, gentle terrain slopes, and weak rotation. To assess the impact of the assumptions inherent in the small-scale model on global flows, we have measured relative departures of various variants of the global model (hydrostatic, nonhydrostatic, elastic Boussinesq, anelastic, and incompressible Euler) against the truncation errors of optional (e.g., Eulerian, semi-Lagrangian, implicit, explicit) second-order-accurate discretization approximations. We have performed an extensive analysis for an idealized orographic planetary flow, representative of basic geophysical fluid dynamics of a mechanically forced laminar system. This analysis revealed that the differences due to the higher-order truncation errors of legitimate modes of executing contemporary global models overwhelm the differences due to analytic formulation of the governing equations. For an idealized climate problem, representative of a thermally forced turbulent system, a similar analysis was computationally affordable only within the context of the semi-implicit anelastic model. There, we measured departures of the climates simulated with different anelastic reference states (emphasizing either tropospheric or stratospheric circulations) against the truncation errors. Similar as in the laminar case, we have found that the solutions’ departures due to fairly substantial variations in the optional realizations of the anelastic system are smaller than the higher-order truncation errors. In general terms, our results document that nonhydrostatic anelastic global models derived from small-scale codes can capture adequately a broad range of planetary flows while requiring relatively minor overhead due to the nonhydrostatic formulation. Since our small-scale models secure no large-scale balances a priori in their analytic–numerical design, this is a success of an “atomistic” school of thought in numerical modeling of atmospheres and oceans. In addition to a purely theoretical interest, the adequacy of the anelastic approximation has important practical consequences. The Boussinesq linearization inherent in the anelastic system greatly simplifies the task of designing accurate, flexible, and computationally efficient all-scale-research models for atmospheric circulations. This is especially important within the class of NFT models, where two-time-level self-adaptive nonlinear numerics leads inevitably to difficult nonlinear elliptic problems for implicit discretizations of fully compressible Euler equations. The semi-implicit nonhydrostatic NFT model presented in this paper is only a first step toward a future“global cloud model” intended for studies of the role of small- and mesoscale phenomena in global flows and climate. Full potential of the nonhydrostatic modeling, in general, and of our semi-implicit model, in particuler, becomes more apparent once the moist processes are included.^^15 These developments will be reported in the future. Stimulating discussions with James Hack, Darryl Holm, and David Williamson are gratefully acknowledged. Special thanks go to Mitchell Moncrieff, Philip Rasch, and William Skamarock for their personal reviews of this paper. Comments of three anonymous referees helped to improve the manuscript. 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Proc. 1998 ECMWF Seminar on Recent Developments in Numerical Methods for Atmospheric Modelling, Reading, United Kingdom, ECMWF, • Williamson, D. L., J. B. Drake, J. J. Hack, R. Jakob, and P. N. Swarztrauber, 1992: A standard test set for numerical approximations to the shallow water equations on the sphere. J. Comput. Phys.,102, 211–224. Elliptic Pressure Equations In order to avoid solving a nonlinear elliptic equation for pressure, the metric forces M contributing to F^n+1 on the rhs of (8) are either approximated explicitly in the spirit of Adams–Bashfort schemes, or the entire subset of (8) corresponding to three components of the momentum equations is iterated with the metric terms lagged behind. In both cases the metric forces enter the rhs of the resulting linear elliptic pressure equation. Although the iterative approach requires solving the elliptic equation at each iteration, it is advantageous overall: in the limit, this approach converges to the trapezoidal-rule approximation, which preserves the neutral character of the metric force. With the first guess M^n+1\|0 = M^n, one iteration suffices for second-order accuracy. In typical applications for flows on the earth, the results are not sensitive to the number of iterations beyond 2, and the overhead associated with two passes through the pressure solver is insignificant, as the second pass requires only a few iterations of the Krylov solver to maintain the accuracy of the first pass. Keeping the above discussion in mind, we can now proceed with formulation of an elliptic pressure equation. First, let us rewrite in a simplified notation * is shorthand for the explicit part of the solution, 0.5Δ has been absorbed in the definition of , and all grid indices have been dropped as there is no ambiguity. Second, let us distinguish the pressure gradient force from other contributions to such as Coriolis force, buoyancy, and gravity wave absorbers [for the latter, see Smolarkiewicz et al. (1998 ]. For momenta, this leads to an alternate form of symbolizes a normalized finite-difference gradient operator, ∼ 0.5Δ ′ denotes a normalized pressure variable, and the linear operator combines the remaining contributions to . The explicit part of the solution * always includes the LE( ) transport term and the metric-force contribution ∼ . However, the specific finite-difference definitions of depend on both the governing system and the overall design (explicit or implicit) of the time-integration scheme. In models explicit with respect to internal gravity waves, the entropy equations (2) are integrated prior to the momentum equations. This allows the buoyancy term in to be evaluated explicitly at + 1 in —in the spirit of Runge–Kutta schemes—and to enter the * term on the rhs of . In the explicit incompressible Euler model, the finite-difference representation of is weighted by the reciprocal of = 0.5Δ ′. In the anelastic model implicit with respect to internal gravity waves, the entropy equation (2) is treated in an alternate form is the complete convective derivative operator (see Smolarkiewicz et al. 1998 for details). must be solved simultaneously with the momentum equations. The buoyancy term in is evaluated implicitly at + 1 in —in the spirit of the trapezoidal-rule scheme—and it enters both the * term and the coefficients of the linear operator in Regardless of the specific definitions employed in , the straightforward algebraic inversion of with respect to components of —a benefit of the colocated grid—is common to all models: For all anelastic models, the Poisson equation for follows the mass continuity constraint denotes the finite difference divergence, ** and * are shorthand for the modified elements of , and the whole equation has been premultiplied by −Δ In the incompressible Euler system, the elliptic pressure equation follows similarly from , whereas in the Boussinesq elastic model, is inserted into the discretized form of leading to the Helmholtz equation , and ≡ 2/(Δ ). Although might be manipulated to more standard forms (e.g., to a linear problem ), we purposely emphasize the one adopted here. It exposes links to standard projection procedures ( Chorin 1968 Clark 1977 Kapitza and Eppel 1992 ) and to relaxation methods via physical analogies ( Birkhoff and Lynch 1984 , chapter 4), it simplifies the design of Neuman boundary conditions for implied by Dirichlet boundary conditions for Smolarkiewicz and Margolin 1994 ), and it suggests physically motivated stopping criteria for iterative solvers ( Smolarkiewicz et al. 1997 Fig. 1. Planetary wave propagation on a sphere. Contours show patterns of (a) vertical and (b) meridional velocity components in m s^−1, with imposed flow vectors, at 4 km after 15 days of simulation. Contour maxima (cmx), minima (cmn), and intervals (cnt) are shown in the upper-left corner of each panel. Negative values are dashed, and zero contours are omitted. The mountain is illustrated with thick solid circles. Maximum vector lengths (here identical) are shown in the upper-right corner of each plate. Citation: Journal of the Atmospheric Sciences 58, 4; 10.1175/1520-0469(2001)058<0349:ACONGM>2.0.CO;2 Fig. 2. The vertical velocity difference field δw, with imposed flow difference vectors (δu,δυ), for (a) the first Δ(3,4) and (b) the last Δ(6,7) entry in Table 3. The contouring convention is the same as in Fig. 1. Citation: Journal of the Atmospheric Sciences 58, 4; 10.1175/1520-0469(2001)058<0349:ACONGM>2.0.CO;2 Fig. 3. The zonally averaged 3-yr means of (a) potential temperature and (b) zonal velocity for the simulation of the Held–Suarez idealized climate. Contour extrema and intervals are shown in the upper-left corner of each plate. Negative values are dashed. Maximum vector length is shown in the upper-right corner of (b). Citation: Journal of the Atmospheric Sciences 58, 4; 10.1175/1520-0469(2001)058<0349:ACONGM>2.0.CO;2 Fig. 4. The zonally averaged 3-yr means of (a) meridional and (b) vertical velocity for the simulation of the Held–Suarez idealized climate. Contouring convention is the same as in Fig. 3. Citation: Journal of the Atmospheric Sciences 58, 4; 10.1175/1520-0469(2001)058<0349:ACONGM>2.0.CO;2 Fig. 5. The zonally averaged 3-yr means of zonal velocity, linear z vs mass σ vertical coordinate representation (note stretched z labels in the lower row). Contouring convention is similar to that used in Fig. 3. The z-coordinate solution displayed in (a) z and (d) σ coordinates. The σ-coordinate solution displayed in (c) σ and (b) z coordinates. The z-coordinate solution uses nz = 81 grid points in the vertical with uniform interval Δz = 800 m. The σ-coordinate solution covers the same model depth with nz = 41 stretched levels (see the text for further details). Citation: Journal of the Atmospheric Sciences 58, 4; 10.1175/1520-0469(2001)058<0349:ACONGM>2.0.CO;2 Table 1. Models considered in this study. The first column refers to the governing fluid equations. The second column distinguishes between explicit or implicit treatment of internal gravity waves. The third column distinguishes between semi-Lagrangian (SL) and Eulerian (EU) options of the model algorithm. The fourth column identifies the means of accelerating the Krylov solver for the elliptic pressure Table 2. Comparison of various simulations of the 15-day evolution of the orographic flow on the sphere. The first column numbers runs for later reference. The second column lists the solver (e.g., Eulerian vs semi-Lagrangian, hydrostatic vs nonhydrostatic, explicit vs implicit, etc.; see the text for details) and the time step (Δt in s) employed. Columns 3–6 provide norms of the meridional and vertical velocity components (in m s^−1), respectively. The last column lists single-processor Cray J90 CPU times (min). Table 3. Difference analysis of various runs collected in Table 2. Table 4. Density normalized inverse flow Jacobians and the entropy conservation measure for the semi-implicit semi-Lagrangian and Eulerian experiments from Table 2. Table 5. Comparison of selected executions of the semi-implicit anelastic model for simulating the idealized Held–Suarez climate. Eulerian model runs 1–4 differ from the default run 0 by, respectively, a modified reference state, Boussinesq approximation, modified polar filter, and altered advection algorithm; run 5 is for the default semi-Lagrangian variant of the model. Table 6. Difference analysis of runs collected in Table 5. Table 7. Variability of instantaneous Held–Suarez flows for corresponding mean states collected in Table 5, as measured with σ^X ≡ [(X − [X])^2]^1/2, where [] refers to zonal and temporal averaging. A fair estimate of the spectral condition number is κ ∼ O(10^10). In all models discussed in this paper, the external mode is eliminated by assuming a rigid-lid upper boundary. The anelastic model on a mountainous globe results from a composition of two mappings: it can be derived by transforming the anelastic variant of the Navier–Stokes’ equations on a rotating sphere ( Gill 1982; Dutton 1986) to a standard terrain-following coordinates (Gal-Chen and Somerville 1975). We employ the standard representation where components of v are defined in terms of a local tangent Cartesian framework aligned with standard geographical coordinates. Specifically, the semi-Lagrangian algorithm remaps transported fields to the departure points of flow trajectories arriving at grid points (x[t],t^n+1) (Smolarkiewicz and Grell 1992), while the Eulerian scheme integrates the homogeneous transport equation ρψ̃[t], + Div(ρvψ̃) = 0 (Smolarkiewicz and Clark 1986); both methods preserve sign and monotonicity of the transported variables. The Rayleigh friction and Newtonian cooling–heating terms are included, respectively, in the momentum and entropy equations, with the reciprocal of the attenuation timescale increasing linearly from 0 to 1/150 s^−1 over the five grid intervals near each pole. Our intention was to order the analyses of the difference fields in the decreasing magnitude of the differences. In some cases, this required a subjective judgement, since not all the norms used decrease at the same rate. The corresponding δυ fields are not shown because: (i) they mostly mimic the wave field in Fig. 1b, so the numeric values of the δυ norms in Table 3 adequately describe the differences; and (ii) the vectors of the flow difference already give a sense of δυ field. In general, for a flow to be realizable (topological), 0 < J < ∞ [see chapter 2 in Ottino (1989) for a discussion]. The parallelization strategy adopted in the global model (a single program multiple data message-passing approach with an explicit 2D horizontal grid decomposition) closely follows that used in the small-scale anelastic model (Anderson and Smolarkiewicz 1997; Anderson et al. 1997), a precursor of the present code. Note that comparisons in z coordinates clearly expose the differences in stratospheric climate, see Fig. 5 for illustration. In contrast to more traditional semi-implicit schemes, we implicitly center in time the entire ẋ∇ϒ[e] term on the rhs of (A3), to assure stable integrations even for rapidly rotating small globes with steep orography—naturally, this is done at the cost of complicating the resulting elliptic pressure equation [see appendix B in Smolarkiewicz et al. (1999) for a complete technical development]. The factor (−1) assures the formal negative-definiteness of the elliptic operator on the lhs of (A5); further normalization by Δt/(ρ) gives the residual errors of (A5) the sense of the divergence of a dimensionless velocity on the grid.	{"url":"https://journals.ametsoc.org/view/journals/atsc/58/4/1520-0469_2001_058_0349_acongm_2.0.co_2.xml","timestamp":"2024-11-02T21:41:35Z","content_type":"text/html","content_length":"589278","record_id":"<urn:uuid:1c9c9c68-e85f-4903-a374-6ff264d46311>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00667.warc.gz"}
How Can I Use the Basic Skills Module? There are several ways to use the Basic Skills module. Below are the three most common approaches: Situation 1: Self-Study and Portfolio Development After completing the activation questions, students' starting levels become clear. This forms the basis for further training in the programme. The goal is for students to achieve 100% in all Basic Skills at the desired target level. You can generate a report of the student's results yourself in the student monitoring system. For more information, see this article: Where can I find the practice results of my students? How to get started: 1. Let the students first choose the desired target level (3F or 4F for Dutch, or B1, B2, or C1 for English). 2. Let the students complete the activation questions. 3. They can then work towards 100% in each subject. 4. Over time, review the results in the student tracking system. 5. Optional: Let the students write a reflection report. 6. Once the student has achieved 100% for all subjects (and optionally a passing grade for their reflection report), the module is successfully completed. Situation 2: Self-Study and Summative test After completing the activation questions, students' starting levels become clear. This forms the basis for further training in the programme. It is up to the student to prepare thoroughly for the summative assessment. The institution determines which level will be assessed and the grading criteria, with Hogeschooltaal providing a standard. The programme offers summative assessments at all mentioned levels and also provides practice tests for these levels, available in the practice environment. These tests are based on the same test matrices as the official assessments, allowing students to use them as a measuring tool before taking the summative assessment. Once students pass the summative assessment according to Hogeschooltaalâ€™s standards, you can award an official Hogeschooltaal certificate. The standard for Dutch assessments is: at least 80% for the verb spelling component and at least 80% average across the four Basic Skills components. For English, a minimum of 80% overall is required at all levels. How to get started: 1. Schedule the summative assessment in advance. 2. Schedule an intake test and let students complete it. 3. Let students complete the activation questions and prepare for the test by working in the practice environment. 4. Monitor studentsâ€™ practice behaviour and encourage them to keep practising for the summative test. 5. Students can prepare by taking practice tests in the programme or you can schedule a formative test. 6. Students take the summative test on the scheduled date and time. After the test period, students have 24 hours to review their results. 7. You can print the results immediately after the test from the student monitoring system. Situation 3: Self-Study, Lesson Integration, Portfolio Development, Summative test The method of implementation is a key indicator for determining academic success, where good guidance is essential. Situation 3 is therefore the most effective approach, leading to significantly higher results. For self-study, portfolio development, and summative test, the previously described approach applies. Hogeschooltaal can be integrated into lessons by having students work independently and providing additional explanations on the components they struggle with. Classroom time can also be well spent on conducting formative tests and evaluating them. How to get started: 1. Schedule the summative test in advance. 2. Schedule an intake test and let students complete it. 3. Let students complete the activation questions and prepare for the test by working in the practice environment. 4. In lessons, students work independently in the programme for half of the time. The remaining time can be used to answer submitted questions, let students complete a practice test, or teach about theoretical issues. 5. Monitor studentsâ€™ practice behaviour and encourage them to keep practising for the summative test. 6. Students can prepare by taking practice tests in the programme or you can schedule a formative test. 7. Students take the summative test on the scheduled date and time. After the assessment period, students have 24 hours to review their results. 8. You can print the results immediately after the test from the student monitoring system. Did you know we also have a comprehensive Writing Skills module? For more information on using this module, see the article: How can I use the Writing Skills module? 0 comments Please sign in to leave a comment.	{"url":"https://hogeschooltaal.zendesk.com/hc/en-gb/articles/4408149415575-How-Can-I-Use-the-Basic-Skills-Module","timestamp":"2024-11-09T06:27:16Z","content_type":"text/html","content_length":"32722","record_id":"<urn:uuid:ba700df0-f93a-4d35-bd61-8dc82fc0355a>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00529.warc.gz"}
Shaded plots python shaded plots python compareplot. It is quite easy to do that in basic python plotting using matplotlib library. Seaborn distplot lets you show a histogram with a line on it. Apart from Datashader itself the code relies on other Python packages from the HoloViz project that are each designed to make it simple to lay out plots and widgets into an app or dashboard in a notebook or for serving separately build interactive web based plots without writing JavaScript The usual. gradient elev slope nbsp import cartopy. Otherwise the details Kdeplot is a Kernel Distribution Estimation Plot which depicts the probability density function of the continuous or non parametric data variables i. Difficulty Hard. Which results in the python stacked bar chart with legend as shown below. 7410 0 0 1 0. using nbsp 4 Apr 2019 In addition you may add text labels lines and shading to the graph. frame this can be as simply as plot x y providing 2 columns variables in the data. Plot of data. It is with the plot function that we specify the transparency of the plot. Feb 13 2016 Figure 2 Transformed Data Plot with Projected Discriminant Functions. This can take a string such as quot quot quot . 5 and 12. We can save a plot as an image easily by following the steps mentioned in this article. Plotting a time series helps us actually see if there is a trend a seasonal cycle outliers and more. So here is a plot of the instant probability of recidivism. Violin plots have many of the same summary statistics as box plots the white dot represents the median the thick gray bar in the center represents the interquartile range The main aim of this plot is to find whether the feature is important or not. Improve this page. Click Output tabPlot panelPage Setup Manager. Note RGB RGB Red Green Blue describes what kind of light needs to be emitted to produce a given color. 2 days ago Plotly is a famous library used for creating interactive plotting and dashboards in Python. It can be done by using nbsp import numpy as np import matplotlib. The first Jun 25 2019 Matplotlib is a Python 2D plotting library used to create 2D graphs and plots by using python scripts. YMMV View my complete profile Sep 26 2020 Here is a traditional KM plot based on the exploded discrete time training dataset. numpy and matplotlib . show always blocks the execution of python script Code for reproduction I try plt. 2 if the non overlapping area gt 0. Find resources and tutorials that will have you coding in no time. The size and shade of each circle represents the strength of each relationship while the color represents the direction either negative or positive. Python provides a large number of libraries to work with. arange 0 5 0. frame elements If the rst argument to plot is a data. Example usage python pyephem_example. In response to the standard polygon approach I wrote a function called shadenorm that will produce a plot of a normal density with whatever area you want shaded. plot x y When 39 gouraud 39 each quad will be Gouraud shaded. bar X A color 39 w 39 hatch nbsp Prepare the data Basic area plots Change line types and colors Change colors by groups library ggplot2 p lt ggplot df aes x weight Basic area plot p Johns Hopkins University Specialization Python for Everybody by University nbsp 21 Nov 2017 plot the shaded range of the confidence intervals. 40 182 44 3 Recently I stumb Python is one of the most powerful and popular dynamic languages in use today. Set Shaded Viewport Options Click the layout tab for which you want to set shaded viewport options. plot_cohorts G B T model 39 kaplan meier 39 ci 0. For example in the image code below I would like the plot to be filled between 20 40 and 50 60 rather than 20 30 and a spike under 40 . 3 Enhancement When plotting the python generated data we set the seperator to for easy using it in csv The GUI python program 3dgraph. My shapefile is from a much bigger place than my data and I want to zoom in. boxplot data column by ax fontsize Make a box plot from DataFrame columns. What I am trying to do is set some threshold such as 0. In addition to these basic options the errorbar function has many options to fine tune the outputs. So you can embed the panel into any wx. plot linewidth 0 19 Plot the graph Here we can plot the graph for survival probability. How do I plot a step function with Matplotlib in Python The problem is that right now it is only shading directly under where the signal 1 rather than to the next change to signal 0 step function . If required these data can also be moved in time. Aug 13 2019 a Pass numeric type data as a Series 1d array or list to plot histogram. Cartesian Plots Parametric Plots Polar Plots Plotting Data Points Contour Plots things like coloring fills and shading to give you a sense of the possibilities. A time series dataset does not make sense to us until we plot it. pyplot as plt import matplotlib. I would be grateful to know if anyone of you have an idea of a concise way of describing the timespan between 16 00pm and 8 00am and how to pass it in a recursive way to Nov 01 2018 Bug report matplotlib. 5 0 0. In this step by step tutorial you 39 ll learn how to handle spreadsheets in Python using the openpyxl package. Python Machine learning Scikit learn Exercises Practice and Solution Write a Python program using seaborne to create a kde Kernel Density Estimate plot of two shaded bivariate densities of Sepal Width and Sepal Length. I am happy that I got this far but I nbsp 25 Feb 2014 Subplot 23 Shaded Relief Map plt. The complete example is listed below. plot x y 39 39 plt Apr 10 2020 Experience in Python MATLAB or R is not necessary but preferred. from osgeo import gdal import numpy as np import matplotlib. 0 shaded our second x y combination contained values up to 1 and 1. I 39 m giving away a few eBook versions if you 39 re willing to write a review on Amazon. Nov 01 2018 Bug report matplotlib. plot_data and gnuplot. Until then Nov 24 2017 Sometimes we need to plot multiple lines on one chart using different styles such as dot line dash or maybe with different colour as well. For plots pymc3. forestplot. comdataviz caveats by Data to Viz. python How do you make an errorbar plot in matplotlib using linestyle None in rcParams When plotting errorbar plots matplotlib is not following the rcParams of no linestyle. matplotlib s stateful functional interface is discouraged by matplotlib. A kernel density estimate KDE plot is a method for visualizing the distribution of observations in a dataset analagous to a histogram. pyplot as plt Re Model hoc file plotting output in Python Post by uri. Peak Temperature nbsp . Apr 13 2016 The last thing I want to change is a smaller detail which is the actual size of the plot. In this case we set the transparency equal to a very low value 0. Matplotlib. Plots with shaded standard deviation. Feb 26 2020 Python Math Exercise 77 with Solution. Jun 14 2020 On this collection of articles on Python based plotting libraries we ll have a conceptual have a look at plots utilizing pandas the massively standard Python knowledge manipulation library. Jun 05 2020 Plotting Geographical Data in Python Published by A choropleth map is a type of thematic map in which areas are shaded or patterned in proportion to a statistical Jul 02 2019 Made a scatter plot of our data and shaded or changed the icon of the data according to cluster. pyplot This plot displays a histogram of lidar dem elevation values with 3 bins. Note that passing in both an ax and sharex True will alter all x axis labels for all subplots in a figure. Nov 25 2017 Create a line graph by clicking on the Charts tab in the Excel ribbon clicking the Line icon under the Insert Chart area and selecting the Marked Line plot. Interpret autocorrelation plots If autocorrelation values are close to 0 then values between consecutive observations are not correlated with one another. The following are 30 code examples for showing how to use seaborn. I am assuming you have python if not click here. As bbum says it 39 s so quot google can organize my head. DISLIN is a high level plotting library that contains subroutines and functions for displaying data graph icallly as curves bar graphs pie charts 3 D colour plots surfaces contours and maps. FacetGrid food_consumption row quot food_category quot hue quot food_category quot aspect 5 height 1 In Python recent modules and technics have made DEM rasters very simple to process and that 39 s what we are going to integrate and explore in this post. The most basic skew T can be plotted with only five lines of Python. Instead it 39 s plotting all of the points connected with a line. Today I ll be ta Python s shelve module is a powerful way to include persistence and to save objects in an easy access database format. Matplotlib is a nbsp Matplotlib is a multiplatform data visualization library built on NumPy arrays In 1 import matplotlib as mpl import matplotlib. Lines of Code 40. for a pandas DataFrame. So let 39 s go with the code Basic plot customizations with a focus on plot legends and text titles axes labels and plot layout. set_extent 12 13 47 48 plt. 2 draw a shaded relief image m . In 2 . The darker the shade the denser it is. You may modify the appearance of a graph using dialogs or via the set of nbsp 12 May 2016 Furthermore it is possible to perform solar shading simulations and timeseries data It is built on top of numpy scipy matplotlib and shapely. The axes to plot the histogram on. Plot univariate or bivariate distributions using kernel density estimation. Previous to Ferret v6. Finally plot the DataFrame by adding the following syntax df. When you shelve an object you must assign a key by which the object value is known. 1415927 180. To set custom RGB color in grace look my previous post here . 20 Confidence interval The confidence interval gives us the range of values we are fairly sure our true values lie in. Pandas is a regular instrument in Python for scalably remodeling knowledge and it has additionally change into a well liked option to import and export from CSV and Excel Sep 26 2020 mantid. The . There are many different variations of bar charts. Matplotlib is a python library used to create 2D graphs and plots by using python scripts. A script illustrating how to use the confband method below is available. Then we also use map to create a horizontal line using plt. The plotting extension is based on the data plotting library DISLIN that is available for several C Fortran 77 and Fortran 90 compilers. plots The functions in this module are intended to be used with matplotlib s object oriented abstract program interface API . In this article we will be using offline plotly to visually represent data in the form of different geographical maps. histogram function is from easyGgplot2 R package. Visvis can be used in Python scripts interactive Python sessions as with IPython or IEP and can be embedded in applications. Apr 03 2011 It works like a charm but it is not the most intuitive way to let users produce plots of normal densities. There are 18 categorical features in the dataset. cohen Thu May 10 2012 10 31 pm PyNEURON installed from pypi or bitbucket does not include the NEURON GUI and hence cannot be used to visualize cells of plot simulation result. Explained in simplified parts so you gain the knowledge and a clear understanding of how to add modify and layout the various components in a plot. Each filled area corresponds to one value of the column given by the line_group parameter. com May 09 2014 series and plot it as a column chart. I never had to plot something like this before but recently a friend asked me how to this with python and I got interested. The second option is an object oriented interface which is much more powerful. We performed PCA via the pccomp function that is built into R. 2 To plot python generated data we use gnuplot. Bar charts is one of the type of charts it can be plot. Feb 04 2019 Please do note that Joint plot is a figure level function so it can t coexist in a figure with other plots. . Once you understood how to build a basic density plot with seaborn it is really easy to add a shade under the line library amp dataset import seaborn as sns df sns. I was thinking on using the axvspan function to achieve this. import pandas as pd. Any column row The easiest way to get started with plotting using matplotlib is often to use the MATLAB like API provided by matplotlib. This site is open source. axhline with the goal to highlight the x axis line for each facet. how to shade regions between contours ContourStyle Automatic the style for contour lines EvaluationMonitor None expression to evaluate at every function evaluation Exclusions Automatic x y curves to exclude ExclusionsStyle None what to draw at excluded curves Frame True whether to put a frame around the plot FrameTicks The function returns a Matplotlib container object with all bars. set_xticks and ax. Aug 18 2019 Bar graph or Bar Plot Bar Plot is a visualization of x and y numeric and categorical dataset variable in a graph to find the relationship between them. It is easy to learn because its syntax emphasizes readability whic Want to seriously level up your coding game Check out the Complete 2020 Python Programming Certification Bundle currently 97 off for Android Authority readers. In total there are 154 observation days . In the box plot a box is created from the first quartile to the third quartile a verticle line is also there which goes through the box at the median. library a general purpose library for exploratory analysis of Bayesian models. A basic understanding of any o 1 499 4 1 Python programming language Th Python doesn t come prepackaged with Windows but that doesn t mean Windows users won t find the flexible programming language useful. Programming Arrays and lists are some of the most useful data structures in programming although few people really use them to their full potential. This next plot is simple but has many customization options that you can view here An Introduction to corrplot Package. contourf method. You might like the Matplotlib gallery. Note SHADE OVERLAY with time axes Previous to Ferret v6. We can plot the data easily in Pandas by calling the plot function on the DataFrame. barplot function helps to visualize dataset in a bar graph. random. Alternatively you can specify specific break points that you want Python to use when it bins the data. 2018 12 29T00 02 24 05 30 2018 12 29T00 02 24 05 30 Amit Arora Amit Arora Python Programming Tutorial Python Practical Solution Share on Facebook Share on Twitter Aug 21 2018 Applied Plotting Charting amp Data Representation in Python W2 Assignment 2 Applied Plotting Charting amp Data Representation in Python When we plot a line with slope and intercept we usually traditionally position the axes at the middle of the graph. Next let 39 s draw a preliminary graph of the recession periods. Write a Python program to convert RGB color to HSV color. plotting. I am happy that I got this far but I wonder why it is not rendered as 2 lines but rahter individual points. IDL Python Description Save plot to a graphics file. The computation intensive steps in this process are written in Python but With code like the above you can plot 300 million points of data one per person in nbsp 4 Sep 2020 These values of x variable are placed as vertical lines on the plot and the area between these lines is shaded. Dec 28 2019 The plot shows customer counts of over 5000 No Churn and close to 2000 Yes Churn. colors as cs m. Jan 06 2017 POST OUTLINE Motivation Get Data Default Plot with Recession Shading Add Chart Titles Axis Labels Fancy Legend Horizontal Line Format X and Y Axis Tick Labels Change Font and Add Data Markers Add Annotations Add Logo Watermarks Rich Shepard was interested in plotting quot S curves quot and quot Z curves quot and a little bit of googling suggests that the S curve is a sigmoid and the Z curve is simply 1. Matplotlib allows you to specify the color of the graph plot. For the default plot the line width is in pixels so you will typically use 1 for a thin line 2 for a medium line 4 for a thick line or more if you want a really thick line. Jul 29 2020 we use the pandas df. 0980 0 0. Here you can see in the above graph the light blue color shade represents the confidence interval of survival. quot Interpret the results. This lesson of the Python Tutorial for Data Analysis covers plotting histograms and box plots with pandas . 1. 2018 12 29T00 02 24 05 30 2018 12 29T00 02 24 05 30 Amit Arora Amit Arora Python Programming Tutorial Python Practical Solution Share on Facebook Share on Twitter Here the canvas is implemented on the panel self. Introduction. base. 3D surface and 3D wireframe plots are graphs that you can use to explore the potential relationship between three variables. Create snake add food increase snake size score etc. import matplotlib. You can set the width of the plot line using the linewidth parameter. random N X np. . Read documentation for more techniques. pyplot as interesting orography. splot_data . Matplotlib consists of several plots like line bar scatter histogram etc. The ultimate guide to the ggplot histogram SHARP SIGHT density plot is just a variation of the histogram but instead of the y axis showing the number of Oct 27 2014 Previous Post How to change size of Matplotlib plot Next Post How to install Eclipse Luna on Ubuntu 2 thoughts on How to get and set rotation angle of 3D plot in Matplotlib The Y axis of the spike raster plot can represent either a neuron or a trial number of the experiment on a specific neuron. a Sage Python data type called a dictionary and turns out to be useful for more nbsp 2 Feb 2019 This is the type of curve we are going to plot with Matplotlib. 0 sigmoid. Legend is plotted on the top left corner. Contributors VictoriaLynn matthewjwoodruff and jdherman See example_images for the outputs from these files PNG . gist_stern_r nbsp 18 Jul 2019 Python is known to be good for data visualization. python scatter. But we do have our kde plot function which can draw a 2 d KDE onto specific Axes. By using pyplot we can create plotting easily and control font properties line controls formatting axes etc Python Training https www. kdeplot df 39 sepal_width 39 shade True sns. The plot needs to contain data. 4 like this. distplot tips_df quot total_bill quot bins 55 Output gt gt gt Jun 05 2019 A Computer Science portal for geeks. It is redundant for me to say that there are various articles which show the utilisation of Plotly for Arduino Raspberry Pi etcetera. we can plot for the univariate or multiple variables altogether. Find In the Page Setup Manager Page Setups area select the page setup that you want to modify. In this article we show how to change the color of a graph plot in matplotlib with Python. This one liner hides the fact that a plot is really a hierarchy of nested Python objects. plot to plot the data you defined. 25 then the probability that a randomly chosen can of soda has a fill weight that is between 11. For clarification we 39 ll revist our grid. subplot 111 projection ccrs. Shades contour regions given low and or high values using colors or patterns. Hundreds of charts are present always realised with the python programming language. It is important that you use the maskout function when plotting the sea ice otherwise you will overwrite all of the SST data with zeros. show block False it failed and appear in a small moment then close itself. title name quot Shaded nbsp Kdeplot is a Kernel Distribution Estimation Plot which. All video and text tutorials are free. So we can make two sets of a 3 3 count plots for each categorical feature. lt function gt are now aliases for ArviZ functions. io import srtm import matplotlib. Feb 02 2010 This blog started as a record of my adventures learning bioinformatics and using Python. If we shade the rectangle that defines each pair of categories we end up with a Categorical Heatmap. I would like to understand why the shaded region of my plot takes a curved shape My understanding so far is that the confidence band changes because the variance is estimated by Bartlett s formula however I would like to understand this formula and why it s used on a ACF Panel plot with subplot shaded contour data and colour bar Passing list to function Python module for reading obstore file of Unified Model Data Assimilation System In this post we will build a bar plot using Python and atplotlib. To overlay a raster you will plot two different raster datasets in the same plot in matplotlib such as a Digital Terrain Model DTM and a hillshade raster. Excel has a built in capability to add nbsp 28 May 2019 Note SHADE OVERLAY with time axes . Matplotlib WXAgg For more professional plot you can use matplotlib more specifically matplotlib WXAgg backend where almost all the matplotlib features are available to wx. quot The programs here are developed on OS X using R and Python plus other software as noted. 4 using y Exp x 2 2 2 . 1 y np. pyplot as plot. It contains well written well thought and well explained computer science and programming articles quizzes and practice competitive programming company interview Questions. imshow for showing images. The eventplot function matplotlib. Switch the rows amp columns of the chart by clicking the column button the icon with the table and highlighted column in the data section of the Charts ribbon. Other tools that may be useful in panel data analysis include xarray a python package that extends pandas to N dimensional data structures. Even though the PV cell is the primary power generation unit PV modeling is often done at the module level for simplicity because module level parameters are much more available and it significantly reduces the computational scope of the simulation. The issue is that the grid is not uniform ie there is not a z component for every Related course The course below is all about data visualization Data Visualization with Matplotlib and Python. Jan 14 2012 In this post we will see how to visualize a function of two variables in two ways. 9545. Here is the complete Python code Kdeplot is a Kernel Distribution Estimation Plot which depicts the probability density function of the continuous or non parametric data variables i. You will need to set the maximum value of the Y axis and use the same value for your column series data. KDE represents the data using a continuous probability density curve in one or more dimensions. Modern society is built on the use of computers and programming languages are what make any computer tick. px. ncl Shows how to overlay a shaded contour plot on a filled contour plot and get labelbars for nbsp 28 Nov 2018 A compilation of the Top 50 matplotlib plots most useful in data analysis and The blue shaded region in the plot is the significance level. pyplot Seaborn distplot lets you show a histogram with a line on it. pyplot. py Download Jupyter notebook fill_between_demo. frame airquality which measured the 6 air quality in New York on a daily basis between May to September 1973. plot_posterior. Following is a simple example of the Matplotlib bar plot. And I 39 d like to do this using python. Arrays and lists are some of the most useful data structures in programming although few people use them to their full potential. etopo or map. The data range goes from about 3 to a little over 10 . Window object. MATLAB is a commercial platform. 2 then assert that the feature is important otherwise not. There are many simple forms for sigmoids eg the hill boltzman and arc tangent functions. For example let s plot the cosine function from 2 to 1. bins 1600 1800 2000 2100 In this case Python will count the number of pixels that occur within each value range as follows Mar 02 2020 Step 3 Plot the DataFrame using pandas. plot x y If the first argument is a vector and the second is a matrix the vector is plotted versus the columns or rows of the matrix. Matplotlib may be used to create bar charts. It has expanded to include Cocoa R simple math and assorted topics. Jul 22 2018 sns. This is a good opportunity to get inspired with new dataviz techniques that you could apply on your data. Here 39 s a minimum working example import matpl Load and Plot Dataset. Overlay Rasters in Python. shadedrelief scale scale lats and longs are nbsp 4 Jun 2018 I recently created a code for plotting shaded dials figures that look like gauges or speedometers in python and I thought I 39 d share my code nbsp 15 Mar 2017 Dear Python Experts I have to plot some data as a line chart but somehow it looks more like a scatter plot. Place the value against each data point you want shading. 3250 0. 178848 degrees Motivation . A hierarchy here means that there is a tree like structure of matplotlib objects underlying each plot. Learn how. In the previous article Line Chart Plotting in Python using Matplotlib we have seen the following plot. In the below code we move the left and bottom spines to the center of the graph applying set_position 39 center 39 while the right and top spines are hidden by setting their colours to none with set_color 39 none 39 . show This should show something similar to this The fact that some complaints take 30 years to resolve is pretty baffling. It shows the number of students enrolled for various courses offered at an institute. For example if we want to shade the area under a normal distribution up to the z score 1. shade is a NumPy array of RGBA values for each data point. express Plotly Express is the easy to use high level interface to Plotly which operates on a variety of types of data and produces easy to style figures. For example even after 2 years this article is one of the top posts that lead people to this site. We set a colormap with the cmap argument. Code import numpy as np import matplotlib. Click Modify. Apr 30 2020 A Box Plot is also known as Whisker plot is created to display the summary of the set of data values having properties like minimum first quartile median third quartile and maximum. Apr 16 2020 2D plots plot y If a single data argument is supplied it is taken as the set of Y coordinates and the X coordinates are taken to be the indices of the elements starting with 1. pyplot as plt x np. It is built on top of the lower level CartoPy covered in a separate section of this tutorial and is designed to work with GeoPandas input. area creates a stacked area plot. histogram is an easy to use function for plotting histograms using ggplot2 package and R statistical software. plot x y1 x y2 color Jun 28 2014 More Python plotting libraries In this tutorial I focused on making data visualizations with only Python s basic matplotlib library. Guido van Rossum developed Plotly With Python Recently I stumbled upon Plotly a beautiful online Data Visualization system by virtue of a MAKE article. The method bar creates a bar chart. Python. But really what we are modeling in this set up is the instant hazard not the cumulative hazard. Use FacetGrid to create the facet with one column ridge_plot sns. The number MUST have a three in front but the others don t matter. These examples are extracted from open source projects. Est. Dec 15 2015 Polar graphs can be a good way to represent cyclical data such as traffic by day of week. You 39 ll learn how to manipulate Excel spreadsheets extract information from spreadsheets create simple or more complex spreadsheets including adding styles charts and so on. Matplotlib is a Python library used for plotting. By default the kernel used is Gaussian this produces a Gaussian bell curve . In this post I take stack overflow data and plot the usage of tags throughout the week. There are three Matplotlib functions that can be helpful for this task plt. Create a SkewT object. How can I shade a region under a curve in matplotlib from x 1 to x 1 given the following plot im matplotlib documentation Shaded Plots Dear Python Experts I have to plot some data as a line chart but somehow it looks more like a scatter plot. plot to visualize the distribution of a dataset. Is it possible to make the area between the two lines slightly gray Any help is much appreciated. com Apr 20 2018 plt. Filling within a single trace In this example we show how to construct a trace that goes from low to high X values along the upper Y edge of a region and then from high to low X values along the lower Y edge of the region. How to create a crime heatmap in R SHARP SIGHT More recently I recommended learning and mastering the 2 density plot. Matplotlib is an amazing visualization library in Python for 2D plots of arrays. cm. Scroll down and look for python 3. For this we need some basic concept of two popular modules of Python in the world of plotting figure or any diagram i. Related course Matplotlib Examples and Video Course. May 16 2017 Python Recipes for CDFs May 16 2017 As a researcher in computer systems I find myself one too many times googling code snippets to represent cumulative distribution functions CDFs derived from data points. The function then fills the areas between the curves based on the shape of Y If Y is a vector the plot contains one curve. I would like to know if i can easely set something like most recent 10 dotes in B 3. sharex bool default True if ax is None else False. Now we ll see how to save this plot. Note the plt. Chapter 11 Python and External Hardware Chapter 11 Python and External Hardware Introduction PySerial Bytes and Unicode Strings Controlling an LED with Python Reading a Sensor with Python Summary Project Ideas Chapter 12 MicroPython Chapter 12 MicroPython Introduction What is MicroPython 1 Line plots The basic syntax for creating line plots is plt. Data is defined after the imports. Dec 17 2019 Using Periscope Data 39 s Python Integration we have built a function that allows users to create Gantt charts with the flexibility to alter a few parameters gantt_chart df show_today True groupby 39 team 39 df a dataframe object containing the following 4 columns project team start_date and end_date convoys. Ok so I ll leave this post here as we have covered most of the distribution plot capabilities next post I will move on to categorical plots and see what Seaborn can offer there. Note that the inequality y lt V22 x holds true for positions inside the semicircle of radius 2 centered at the origin. Jul 21 2020 Snake Game in Python using Pygame which is free and open source Python library used to create games. You will need OpenGL to get transparency to work. Figure 2 shows the projected data along with the three projected discriminant functions corresponding to the three wine cultivars. plot_surface X Y Z args kwargs Create a surface plot. 0 and 1. Below is a code for a 3 3 count plot visualization for the first set of nine categorical features. 8500 0. Use plt. In the Feb 13 2020 This initial graph simply plots the unemployment rate with all default graph options. The discriminant functions in this case are lines in but are projected into in the plot. contour for contour plots plt. You should first reshape the data using the tidyr package Collapse psavert and uempmed values in the same column new column . Nov 07 2016 Step 6 Saving a Plot. The method also adds errors to the matplotlib polar plot as a shaded region to help understand the variability in the data. Mar 09 2020 We use the shade True to fill the density plot with color. shadedrelief we can not zoom in to a smaller region since it will generate a blur image. Mesh plot shade_surf z loadct 3 Surface plot Scatter cloud plots. Then open IDLE pytho 2 187 12 3 Today i will show you h Python s string module provides a template class to process strings powerfully and en masse. The next code section builds a shaded contour plotting using Matplotlib 39 s ax. Datasets used in Plotly examples and documentation. Ask Question Asked 2 years 1 month ago. plot is called. Panel plot with subplot shaded contour data and colour bar Passing list to function Python module for reading obstore file of Unified Model Data Assimilation System The X Y axis from the quot base quot plot is the one that gets used for both plots which may cause your quot overlay quot plot to be cut off if its X and or Y axis outside the range of the X Y axis of the quot base quot plot. and at the same time increase the quality and the possiblities of pyny3d plots. Geoplot is a Python library providing a selection of easy to use geospatial visualizations. pyplot as plt to show quot shaded quot areas from a specific illumination source 39 s zenith and azimuth. hi all is there a way in matplotlib to plot lines with errorbars e. unset for i 1 200 label i 1. Simple Plotting Matplotlib shaded regions y x Plot junk and then a filled region plot x y Make a blue box that is somewhat see through and has a I have a set of latitude longitude and elevation pairs roughly a grid of such values though it is not uniform and I 39 d like to be able to plot an elevation map and perhaps also a shaded relief image for this data. This lecture has provided an introduction to some of pandas more advanced features including multiindices merging grouping and plotting. I want to use matplotlib to illustrate the definite integral between two regions x_0 and x_1. Here the canvas is implemented on the panel self. Aug 01 2015 The object returned by light. fill_between may be used to add shaded areas to charts. In this example the X axis of the second plot is longer than the first and the Y axis of the first plot is longer than the second. While it is easy to generate a plot using a few lines of code it Most of other python packages used for plotting spatio temporal data are based on matplotlib. displot penguins x quot flipper_length_mm quot hue quot species quot multiple quot stack quot The stacked histogram emphasizes the part whole relationship between the variables but it can obscure other features for example it is difficult to contourf Z creates a filled contour plot containing the isolines of matrix Z where Z contains height values on the x y plane. A distplot plots a univariate distribution of observations. color color_shading alpha . Notice the regions where there are gaps in the COADS data set Feb 09 2019 Example of python code to plot a normal distribution with matplotlib How to plot a normal distribution with matplotlib in python NCL gsn functions color routines NCL Home gt Documentation gt Graphics gt Graphical Interfaces gsn_contour_shade. You can plot a drawing that contains shaded 3D solids as it is displayed in wireframe with hidden lines removed or as rendered. Plotly also makes Dash a framework for building interactive web based applications with Python code . Plots Plots are delegated to the ArviZ. contourf for filled contour plots and plt. You can set the line style using the linestyle parameter. Line number 11 bar function plots the Happiness_Index_Female on top of Happiness_Index_Male with the help of argument bottom Happiness_Index_Male. The above plot is much too wide for a single column plot so I need to resize it. Surface plots Axes3D. shade_lowest is an optional advanced visualisation pattern. However I do see it becoming a popular supplement to the Power BI platform. Save figure Matplotlib can save plots directly to a file using savefig . This is the PART 3 of a series of posts called Integrating amp Exploring . in the given specification between the upper and lower bounds is shaded. Plotly is also a company that allows us to host both online and offline data visualisatoins. This repository is inspired by ICEbox. kml 39 39 39 from datetime import datetime timedelta from math import pi degrees radians from operator import mod import ephem from lxml import etree from pykml. Shaded KDE plot for tenure. In this ggplot2 tutorial we will see how to make a histogram and to customize the graphical parameters including main title axis labels legend background and colors. 3 days ago Choropleth maps are popular thematic maps used to represent statistical data through various shading patterns or symbols on predetermined nbsp 25 Nov 2017 This tutorial describes how to create error bands or confidence intervals in line graphs using Excel for Mac. The main principle of matplotlib. pyplot as plt N 8 A np. The goal is to visualize the impact of certain features towards model prediction for any supervised learning algorithm using partial dependence plots . makes a plot showing the three dimensional region in which pred is True. shade returns a 4 element array of the Red Green Blue and Alpha value for that point. Jun 30 2014 This is a quick post to test Cartopy 39 s shaded relief plots. PlateCarree x y np. pyplot as plt y1 60 65 65 70 75 y1_max This tutorial explains matplotlib 39 s way of making python plot like scatterplots bar charts and customize th components like figure subplots legend title. edureka. import numpy as np import matplotlib. ipynb Keywords matplotlib code example codex python plot pyplot Gallery generated by Sphinx Gallery Dec 15 2018 Introduction to Plotly. It s also easy to learn. Plot the pressure and temperature note that the pressure the independent variable is first even though it is plotted on the y axis . Light is added together to create form from darkness. Python is one of the most powerful and popular dynamic languages in use today. The predictor variables are displayed on the x and y scales and the response z variable is represented by a smooth surface 3D surface plot or a grid 3D wireframe plot . If partial autocorrelation values are beyond this confidence interval regions then you can assume that the observed partial autocorrelation values are statistically significant. Picking what size in inches you want for your plot is a bit of just trial and error but I typically find that for single column plots a width of 6 inches typically works well Plot multiple time series data. Due to the small number of samples this interval is large. A Figure object is the outermost container for a matplotlib graphic which can contain multiple Axes objects. These lines perform the following tasks Create a Figure object and set the size of the figure. pyplot as plt x 1 10 y 3 6 plt. It s a high level open source and general purpose programming language that s easy to learn and it features a broad standard library. overlay_10. RGB stores individual values for red green and blue. g. Matplotlib aims to have a Python object representing everything that appears on the plot for example recall that the figure is the bounding box within which plot elements appear. If you want to make the graph plot have a very low transparency you would give the alpha attribute a very high value. Search for Geoprocessing in Python Sponsored Link Plot Polygon Edges. The anatomy of a violin plot. To do so we need to provide a discretization grid of the values along the x axis and evaluate the function on each x Aug 01 2015 The object returned by light. 11 Jan 2016 On this tutorial we cover the basics of 3D line scatter wire frames surface and contour plots. 826165 degrees latitude 51. Line number 10 bar functions plots the Happiness_Index_Male first. We do this with the alpha attribute. After a quick Google search I found several blogs post about this. Here we ll plot the variables psavert and uempmed by dates. py is a 3D plot package for graphically displaying image or specturm data in a flexible and comprehensible fashion. The Python example code draws overlapped stacked and percentage based area plots. 3. Dec 14 2011 Using this method you can get plots like this where the shaded area corresponds to the 2sigma confidence band of the linear fit shown in green. So please suggest me further if I am missing any hidden concepts here. frame . sin x plt. With Python we used the PCA class in the scikit learn library. plot outlined in Simple Line Plots and Simple Scatter Plots. So let us begin. For each point in the input array light. traceplot. 39 set gxout shaded 39 39 set dfile 2 39 Sets Default file to file 2 Sea ice I am trying to create a 3D surface energy diagram where an x y position on a grid contains an associated z level. If you only want to plot the edges of the polygon things are quite simple. The following are 15 code examples for showing how to use matplotlib. The dark dot below is Melbourne and it has some very interesting data plotted on it. If you don t feel like tweaking the plots yourself and want the library to produce better looking plots on its own check out the following libraries. But the colors are hard to see. By default it will be colored in shades of a solid color but it also supports color mapping by supplying the cmap argument. In Python s Matplotlib the x tick and y tick marks of the plot can be changed using functions ax. Using the Python Seaborn module we can build the Kdeplot with various functionality added to it. Result Plot Same plot with overlap region with different color chosen between red and black color Same getOverlap function useful to plot the shaded overlap region also. It has a module named pyplot which makes things easy for plotting by providing the feature to control line styles font properties formatting axes etc. I would like to add a dark shaded background every day between 16 00pm and 8 00am to represent the quot night quot time. In this article we are going to learn how to fill the area of any figure with color in matplotlib using Python. Jun 08 2016 When using python Basemap to plot maps a nice background would be a big plus. You can vote up the ones you like or vote down the ones you don 39 t like and go to the original project or source file by following the links above each example. plot x y where x and y are arrays of the same length that specify the x y pairs that form the line. Matplotlib Scatter Colormap. 97 A restriction in PPLUS requires that if time is an axis of the shaded plot the nbsp 21 Apr 2016 import pygrib import matplotlib. Hi To make it simple i have three variables X Y Z I have set up a scatter plot which represent X on axe X Y on axe Y and Z as a dotes as Legend . Installing Python Imaging Library PIL for image processing plot and fill between y1 and y2 where a logical condition is met ax. Other plotting tools can use this data to draw a shaded surface. Plotly is a company that makes visualization tools including a Python API library. Plots enable us to visualize data in a pictorial or graphical representation. See full list on towardsdatascience. The people from the Tango project Wikimedia Commons Python is an interpreted object oriented high level programming language. You can see that in the first week out almost 1. tsplot Nov 24 2017 The appearance of the plot can be modified via input argumets and or the handles of the returned plot objects. 9290 0. plot x 39 Year 39 y 39 Unemployment_Rate 39 kind 39 line 39 You ll notice that the kind is now set to line in order to plot the line chart. Sometimes it is useful to display three dimensional data in two dimensions using contours or color coded regions. It is possible to have plots with two categorical axes. The gray function takes a number between 0 and 1 that specifies a shade of gray between black 0 and white 1 This plot also shows the statistical background inherent in Seaborn plots. x1 range n y1 range n x2 range m y2 range m Jun 17 2011 I have been dabbling with the TikZ package to create some diagrams relevant to a first year microeconomics course. kdeplot function to plot a density plot . Matplotlib is more flexible and capable for plotting. e. There is a TARDIS in the Earth Fleet scene of Iron Sky. But when using map. A box plot a histogram a scatter plot or something else That will depend on the purpose of the plot is it for performing an inspection on your data EDA or for showing your results conclusions to people and the number variables that you want to plot. First we will create an intensity image of the function and second we will use the 3D plotting capabilities of matplotlib to create a shaded surface plot. Bar Chart. In this series of articles I 39 m focusing on plotting with Python libraries. IDL Python Description set_plot 39 PS Python Data Visualisation Code Snippets. Feb 25 2020 Now lets use the SHADE command to make a color shaded plot of the sea surface temperature at the first time step and overlay quot filled quot continents instead of an outline. plot x y pch 15 col rgb 1 4 4 0 0 z When we have to print in grayscale R also supplies a function for building shades of gray which is called unsurprisingly gray . Example of modeling cell to cell mismatch loss from partial module shading. 2020 June 6 2020 by Stefan. plot line needs to be called before any other plot details are specified. Multiple bivariate KDE plots In 41 import seaborn as sns import matplotlib. Generate a surface plot of the function sin x cos y on the black shaded domain illustrated below. Shelve is a powerful Python module for object persistence. set_yticks . shape 0 ub lb . Hence it is not free. IPython Notebook nbsp Example Python program to draw an overlapped area plot. Bootstrap plot on mean median and mid range statistics. Each Matplotlib object can also act as a container of sub objects for example each figure can contain one or more axes objects each of which in turn contain other Useful GRAPHs cont 21. regplot . bins If the dataset contains data from range 1 to 55 and your requirement to show data step of 5 in each bar. The twoway area plot shades the area underneath a curve so using it on the recession indicator variable shades the recessions. pcolormesh x y data shading 39 flat 39 cmap plt. Plotting is comparatively not as flexible and capable as Python plotting. ax. Great now we have a plot with two different colors in 2 lines of code. A colormap is a range of colors matplotlib uses to shade your plots. These functions accepts an array of values representing tick mark positions. plt. Example Bar chart. In R the clusplot function was used which is part of the cluster library. Many features like shade type of distribution etc can be set using the parameters available in the functions. Plotting with Geoplot and GeoPandas . It gives us a feel for the data. The object oriented API allow for customization as well. RegionPlot3D pred 1 pred 2 plots several regions corresponding to the pred i . The shaded areas are confidence intervals which basically show the range in which our true value lies. The X axis of the spike raster plot represents the spike. Regards Jul 25 2019 Plot a dashed line. Use the seaborn plotting library for python specifically seaborn. You should have a basic understanding of Computer Programming terminologies. Format the column series to have a Gap Width of zero this option is on the Series Option of the format dialog. Related course The course below is all about data visualization Data Visualization with Matplotlib and Python May 08 2018 Trackbacks Pingbacks. As you can see each dates has a different color. If the population of fill weights follows a normal distribution and has a mean of 12 and a standard deviation of 0. Matplotlib is a widely used Python based library it is used to create 2d Plots and graphs easily through Python script it got another name as a pyplot. They can be implemented in a manner similar to filled area plots using scatter traces with the fill attribute. To draw an area plot method area on DataFrame. Sep 10 2020 Python Programming MATLAB It is an open source programming language free to use. plot_pacf function also returns confidence intervals which are represented as blue shaded regions. The plot below shows such a plot where the x axis categories are a list of years from 1948 to 2016 and the y axis categories are the months of the years. R function gather tidyr Create a grouping variable that with levels psavert and uempmed Course Description. Plotly has three different Python APIs giving you a choice of how to drive it guage Python. tripcolor . The code is available here May 28 2020 Plotly is a plotting ecosystem that allows you to make plots in Python as well as JavaScript and R. This can be shown in all kinds of variations. Python Programming tutorials from beginner to advanced on a massive variety of topics. 5 . bluemarble map. Plot a horizontal bar chart of the data contained in the df Pandas DataFrame Now that we are done with the quot SST quot and the land mask we are going to open up the sea ice file and plot the sea ice. plt. However you may have a certain color you want the plot to be. The plot will show the coefficient of thermal expansion CTE of three different materials based on a small data set. Time series data is omnipresent in the field of Data Science. Current color Old color 0 0. By using the alpha transparency argument shaded areas may overlap. Creating Subplots Your plot should then look similar to the one below. The idea of 3D scatter plots is that you can compare 3 characteristics of a data set instead of two. crs as ccrs from cartopy. In July 2014 the Monty Python comedy troupe opened their reunion show Monty Python Live Mostly with a trademark animation featuring the Tardis dubbed the quot retardis quot flying through space before the Pythons came on stage. pymc3. pyplot module marks lines at specified locations. show area X Y plots the values in Y against the x coordinates X. 1. We use seaborn in combination with matplotlib the Python plotting module. 5 ounces is 0. load_dataset 39 iris 39 density plot with shade sns. Plotting Examples Some examples for plotting different types of data in Matlab Python and R. How to avoid overplotting. Z are dates. 4 of the individuals reactivate. then we will first generate a list of x y coordinate pairs along the curve of the distribution from x 4 to x 1. The python seaborn library use for data visualization so it has sns. Last Updated 19 04 2020. SHADE OVERLAY Causes the indicated shaded plot to be overlaid on the existing plot. I certainly don t expect Python to replace DAX the Query Editor or Power BI s built in visuals nor would I want it to. Since we wanted the region between 1. There are many tools in Python enabling it to do so matplotlib pygal Seaborn Plotly etc. plot. 6. All possible colormaps are listed here. Mar 11 2015 The process to plot polygons in python can be different depending on whether you are happy to plot just the edges of the polygon or you would also like to plot the area enclosed by the polygon. Mar 26 2019 This method can be extended to draw filled curves or shade the area under a curve. Alpha controls the transparency of the point. figure figsize 10 10 ax plt. Examples Here the fmt is a format code controlling the appearance of lines and points and has the same syntax as the shorthand used in plt. kdeplot x shade True Great we can see that the two plots are the same and we have created our KDE plot correctly. Thus the links below will redirect you to ArviZ docs pymc3. In case subplots True share x axis and set some x axis labels to invisible defaults to True if ax is None otherwise False if an ax is passed in. pyplot as plt scale 0. Lets look at the data in the data. fill_between in Python. There are many different options and choosing the right one is a challenge. 95 groups groups pyplot. How to save a matplotlib plot as an image in Python. Sep 01 2014 Now see the below plot and compare with the above one you can see different color in the overlap regions. pyplot as plt if nbsp import numpy as np import matplotlib. We ll choose bwr which stands for blue white red. When we plot a line with slope and intercept we usually traditionally position the axes at the middle of the graph. ggplot2. To plot a dashed line a solution is to add 39 39 39 39 39 or 39 39 example import matplotlib. In that article I threw some shade at matplotlib and dismissed it during the analysis. Oct 26 2016 A violin plot is a hybrid of a box plot and a kernel density plot which shows peaks in the data. A collection of common dataviz caveats by Data to Viz. Typically data for plots is contained in Python lists NumPy arrays or Pandas dataframes. Now that we have finished our code let s run it to see our new customized plot. We used matplotlib to create the plot. Filled area plot with plotly. One such language is Python. In this way the shelve file becomes In matplotlib how do I plot error as a shaded region rather than error bars For example rather than. It has a module named pyplot which makes things easy for plotting by providing feature to control line styles font properties formatting axes etc. A Python version of this projection is available here. You have a lot of tools for plotting in Python. The following diagram of the probability density function pdf of a normal distri May 25 2020 These world objects can be anything from plots lines with markers to images 3D rendered volumes shaded meshes or you can program your own world object class. py A window should now open displaying our plot Next save the plot by clicking on the save button which is the disk icon located on the bottom toolbar. First matplotlib has two user interfaces The first mimics MATLAB plotting and uses the pylab interface. If the orientation of the plot is horizontal the lines drawn are vertical lines. So how do you use it The program below creates a bar chart. So when you create a plot of a graph by default matplotlib will choose a color for you. autocorrplot Please see ACF plot attached Python . 97 A restriction in PPLUS requires that if time is an axis of the shaded plot the overlaid variable must share the same time axis encoding as the base plot This page displays all the charts currently present in the python graph gallery. Using these additional options you can easily Jul 10 2019 Matplotlib is a huge library which can be a bit overwhelming for a beginner even if one is fairly comfortable with Python. arange N plt. Download Python source code fill_between_demo. 6940 0. This is done with the color attribute. 3D Scatter Plot with Python and Matplotlib Besides 3D wires and planes one of the most popular 3 dimensional graph types is 3D scatter plots. 1 giving the graph plot a lot of transparency. Examples showed above. May 28 2019 SHADE NOLABELS Suppresses all plot labels. co data science python certification course This Edureka Python Matplotlib tutorial Python Tutorial Blog https Calculating power loss from partial module shading . Vertical axis 5 Using color palettes within a Seaborn Kdeplot 6 Plotting two shaded Using the Python Seaborn module we can build the Kdeplot with various functionality added to it. MATLAB automatically selects the contour lines to display. factory import KML_ElementMaker as KML year 2011 month 12 day 22 longitude 1. Plot Histogram of quot total_bill quot with bins parameters sns. Plot data including options. 1250 Matplotlib is a Python module that lets you plot all kinds of charts. matplotlib documentation Scatter Plots Aug 15 2011 Use geom_rect to add recession bars to your time series plots rstats ggplot Posted on August 15 2011 by Jeffrey Breen in R bloggers 0 Comments This article was first published on Things I tend to forget R and kindly contributed to R bloggers . The kind of things I do with Python are data analysis numerical analysis programming structured and object oriented scripting plotting with publication quality handling of ASCII FITS tables interfacing with Fortran code etc. py gt test. 2 Enhancement If it s multiplot mode automatically call the following Gnuplot to unset the label g. The DataFrame class of Python pandas library has a plot member using which diagrams for visualizing the DataFrame are drawn. Aug 24 2018 As we have just seen Python is a powerful tool for data analysis and visualization that can be utilized to extend reporting in Power BI. Whether it is analyzing business trends forecasting company revenue or exploring customer behavior every data scientist is likely to encounter time series data at some point during their work. random N B np. Find resources and tutori Python Python programming language This tutorial is designed for software programmers who need to learn Python programming language from scratch. Polar Plots Python usr bin env python import math import dislin n 300 m 10 f 3. plotting lines with shaded transparent region for standard deviation. pnlPlot. Time 1hr but this varies by student What should I know before this project How to create a dataframe using the python package pandas The python visualization world can be a frustrating place for a new user. In this lesson you will learn about overlaying rasters on top of a hillshade for nicer looking plots in Python. It is designed to be compatible with MATLAB 39 s plotting functions so it is easy to get started with if you are familiar with MATLAB. . Inversely autocorrelations values close to 1 or 1 indicate that there exists strong positive or negative correlations between consecutive observations respectively. In this plot the outline of the full histogram will match the plot with only a single variable sns . 4470 0. Core concepts Nested loops Using Dataframes Dictionaries Lists Creating Scatter Plots. The following python code computes the projected Third Edition is on the shelves Geospatial concepts Geo python universe and pound for pound still the most pure python and minimal dependency examples you ll find anywhere so somebody somewhere out there will still be able to do the math. legend pyplot. plot function built over matplotlib or the seaborn library s sns. fill_between range mean. Saving showing clearing your plots show the plot save one or more figures to for example pdf files clear the axes clear the figure or close the plot etc. It s not quite a simple as installing the newest version however so let s make sure you get the right tools for the task at hand. Plotting a Gaussian normal curve with Python and Matplotlib Date Sat 02 February 2019 Tags python engineering statistics matplotlib scipy In the previous post we calculated the area under the standard normal curve using Python and the erf function from the math module in Python 39 s Standard Library. Join 250 000 subscribers and get a Python Coding Today i will show you how to make a simple ghost game in python. Aug 25 2020 Load and Plot Dataset. shaded plots python	{"url":"https://ebrflooring.co.uk/jordan-peterson/shaded-plots-python.php","timestamp":"2024-11-04T11:08:56Z","content_type":"text/html","content_length":"76424","record_id":"<urn:uuid:3609224b-b9a6-41a8-ac9a-8fd9ce3335d3>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00705.warc.gz"}
coefficient of skewness calculator Step 1 - Select type of frequency distribution Discrete or continuous Step 2 - Enter the Range or classes X. Now click the button Solve to get the. However, it is worth noting that the formula used for kurtosis in these programs actually calculates what is sometimes called "excess kurtosis" - put simply, the formula includes an adjustment so that a normal distribution has a kurtosis of zero. Yule, George Udny. WebPearson's Mode Skewness and Pearson's Median Skewness: Two calculations. Step 1: Create the Dataset First, lets create the following dataset in Excel: Step 2: Calculate the Pearson Coefficient of Skewness (Using the Mode) Next, we can use the following formula to calculate the Pearson Coefficient of Skewness using the mode: The skewness turns out to be 1.295. As you can see, for the denominator of this formula to not be equal to zero, you need at least three observations in the data sample. Skewness is a commonly used measure of the symmetry of a statistical distribution. If the two are equal, it has zero skewness. This adjusted FisherPearson standardized moment coefficient It indicates that there are observations at one of the extreme ends of the distribution, but that theyre relatively infrequent. Duncan Cramer (1997) Fundamental Statistics for Social Research. First off, since the data set is now a sample the standard deviation becomes 2, and is represented by s. 1 to 255 arguments for which you want to calculate skewness. With this test grade calculator, you'll quickly determine the test percentage score and grade. Karl Pearson Coefficient of Skewness Calculation Formula: Sk = 3 (m- Md) SD Md = median, Sk = Karl Pearson coefficient of Skewness F1 Score False Acceptance Rate (FAR) False Negative Rate (FNR) Statistics Calculators Defects per Million Opportunities E Power X Equally Likely Outcomes Probability Gamma Distribution Gamma Distribution Mean Formula: (Mean Median)/Standard Deviation. Skewness is a measure of the symmetry, or lack thereof, of a distribution. As such, kurtosis recognizes whether the tails of given dissemination contain extraordinary qualities. / 1 You might want to calculate the skewness of a distribution to: Describe the distribution of a variable alongside other descriptive statistics; Determine if a variable is normally distributed. A perfect normal distribution has a kurtosis coefficient equal to zero. Q WebHow to calculate Pearson's coefficient of skewness for grouped data? Algebra Applied Mathematics Calculus and Analysis Discrete Mathematics Foundations of Mathematics Geometry History and Terminology Number Theory Probability and Statistics Recreational Mathematics Topology Alphabetical Index New in b Sk = [(Q3 Q2) (Q2 Q1)]/(Q3 Q1) (Q3 2Q2 + Q1)/(Q3 Q1) The value of Sk would be zero if it is asymmetrical distribution. Skewness is a measure of the symmetry, or lack thereof, of a distribution. 1 c j 1 to 255 arguments for which you want to calculate skewness. A fundamental task in many statistical analyses is to characterize the location and variability of a data set. {\displaystyle b_{1}} for sufficiently large samples. The absolute value of kurtosis is high, suggesting that the height is more or less uniform in this class. Themeanor average of a given data is defined as the sum of all observations divided by the number of observations. i Premaratne, G., Bera, A. K. (2000). Revised on For non-normal distributions, WebSkewness Calculator Skewness Formula: g = n i = 1 ( x xi ) 3 (n 1)s3 Enter Inputs (separated by comma): Skewness (g) = Mean ( ) = Variance= = Standard Deviation ( ) = Sample Number = Skewness Calculator is a free online tool that displays the skewness in the statistical distributions. the sample variance). Note, however, that the converse is not true in general, i.e. To use this calculator, simply enter your data into the text box below, either one score per line or as a comma delimited list, and then press the "Calculate" button. To perform the calculation, enter a series of numbers. In fact you can compute the statistic by hand without a calculator for small data sets. A tail is a long, tapering end of a distribution. Since this value is negative, we interpret this to mean that the distribution is left-skewed, which means the tail extends to the left side of the distribution. Kurtosis = Fourth Moment / (Second Moment)2. Step 1 - Select type of frequency distribution (Discrete or continuous) Step 2 - Enter the Range or classes (X) seperated by comma (,) Step 3 - Enter the Frequencies (f) seperated by comma Step 4 - Click on "Calculate" for Pearson's coefficient of skewness calculation These other measures are: The Pearson mode skewness,[11] or first skewness coefficient, is defined as, The Pearson median skewness, or second skewness coefficient,[12][13] is defined as. Step 1 - Select type of frequency distribution (Discrete or continuous) Step 2 - Enter the Range or classes (X) seperated by comma (,) Step 3 - Enter the Frequencies (f) seperated by comma Step 4 - Click on "Calculate" for Pearson's coefficient of skewness calculation {\displaystyle \ nu } Hinkley DV (1975) "On power transformations to symmetry". Algebra Applied Mathematics Calculus and Analysis Discrete Mathematics Foundations of Mathematics Geometry History and Terminology Number Theory Probability and Statistics Recreational Mathematics Topology Alphabetical Index New in ( Sk = 3(Mean Median) sd = x M sx. Use this tool to calculate the sampling error incurred when inferring from a population. It can either be positive or negative, irrespective of the signs. {\displaystyle (x_{i},x_{j})} x A right-skewed distribution has a long tail on its right side. If the distribution is symmetric, then the mean is equal to the median, and the distribution has zero skewness. WebPearson Correlation Coefficient Calculator The Pearson correlation coefficient is used to measure the strength of a linear association between two variables, where the value r= 1 means a perfect positive correlation and the value r= -1 means a perfect negataive correlation. It is useful when data has a more repetitive number in the data set. D'Agostino's K-squared test is a goodness-of-fit normality test based on sample skewness and sample kurtosis. This definition leads to a corresponding overall measure of skewness[23] defined as the supremum of this over the range 1/2u<1. So, for example, you could Webscipy.stats.skew# scipy.stats. The value of skewness can be positive or negative, or even undefined. Measures of Skewness and Kurtosis. , For normally distributed data, the skewness should be about zero. OR. TOPICS. In the last step, we input these intermediate results into the skewness and kurtosis formulas. Step 1: Calculate the t value. x WebWhy Bowley Skewness works. Web How to Use Skewness Calculator. Using Median Step 1: Subtract the median from the mean. The long tail on its left represents the small proportion of students who received very low scores. For example, a zero value means that the tails on both sides of the mean balance out overall; this is the case for a symmetric distribution, but can also be true for an asymmetric distribution where one tail is long and thin, and the other is short but fat. {\displaystyle \\|\cdot \\|} Left skew is also referred to as negative skew. However, if a distribution is close to being symmetrical, it usually is considered to have zero skew for practical purposes, such as verifying model assumptions. You could also ignore the skew, since linear regression isnt very sensitive to skew. {\displaystyle x_{i}\geq x_{m}\geq x_{j}} Skew is a common way that a distribution can differ from a normal distribution. Step 3: Click on the "Reset" button to clear the field and enter the new values. Step 1 - Select type of frequency distribution (Discrete or continuous) Step 2 - Enter the Range or classes (X) seperated by comma (,) Step 3 - Enter the Frequencies (f) seperated by comma Step 4 - Click on "Calculate" for Pearson's coefficient of skewness calculation has an expected value of about 0.32, since usually all three samples are in the positive-valued part of the distribution, which is skewed the other way. If the reverse is true, it has positive skewness. If the value is less than zero, where is the mean, is the standard deviation, E is the expectation operator, 3 is the third central moment, and t are the t-th cumulants. Skewness can be calculated using various methods, whereas the most commonly used method is Pearsons coefficient. Sometimes, though, the distribution doesn't look as ideal as in the handbooks. However, the modern definition of skewness and the traditional nonparametric definition do not always have the same sign: while they agree for some families of distributions, they differ in some of the cases, and conflating them is misleading. 4 OR. A fundamental task in many statistical analyses is to characterize the location and variability of a data set. Kurtosis measures the tail-heaviness of the distribution. {\displaystyle \mu } Examples of distributions with finite skewness include the following. Since the number of sunspots observed per year is right-skewed, you can try to address the issue by transforming the variable. X & Frequency Class & Frequency. 0 WebStatistical Operations. Calculate the t value (a test statistic) using this formula: Example: Calculating the t value. , where Type your data in either horizontal or verical format, for seperator you can use '-' or ',' or ';' or space or tab. However, since the majority of cases is less than or equal to the mode, which is also the median, the mean sits in the heavier left tail. Get the result! Type the following formula into a second empty cell. Karl-Pearsons Method (Personian Coefficient of Skewness) Karl Pearson has suggested two formulas, where the relationship of mean and mode is established; where the relationship between mean and median is not established. Once you know what the skewness and kurtosis of a given data sample are, you need to interpret this value in a certain way. In the last step, we input these intermediate results into the skewness and kurtosis formulas. Skewness is a measure of the degree of asymmetry of a distribution. 1 One of the simplest is Pearsons median skewness. WebNow, the skewness of the salary distribution can be calculated by using the above formula, Skewness = = (- 6114.1 * 4 568.8 * 10 + 5.0 * 12 + 1607.5 * 6 + 10238.5 * 3) / [ (35 1) * (11.08) 3 ] = 0.22 Therefore, the salary distribution table exhibits symmetry as indicated by a skewness value of fairly close to zero. m {\ displaystyle g_{1}} g As such, kurtosis recognizes whether the tails of given dissemination contain extraordinary qualities. WebSkewness Calculator. You can use the Excel functions AVERAGE , MEDIAN and is a method of moments estimator. We obtain the following values: skewness = -0.3212 kurtosis = -0.7195 The negative value of the coefficient of skewness implies a slight skew to the left. ( Skewness indicates the direction and relative magnitude of a distribution's deviation from the normal distribution. Pearsons first coefficient of skewness To calculate skewness values, subtract a mode from a mean, and then divide the difference by standard deviation. The distribution is approximately symmetrical, with the observations distributed similarly on the left and right sides of its peak. Many statistical procedures assume that variables or residuals are normally distributed. Scribbr. and WebSkewness is a measure used in statistics that helps reveal the asymmetry of a probability distribution. Skewness is a measure of the symmetry, or lack thereof, of a distribution. Although a theoretical distribution (e.g., the z distribution) can have zero skew, real data almost always have at least a bit of skew. You are probably familiar with the normal distribution (if not, we have a normal distribution caculator). Since this value is negative, we interpret this to mean that the distribution is left-skewed, which means the tail extends to the left side of the distribution. Web Method 2. WebTo start, enter the above values in the Kurtosis calculator, and then press on the 'Calculate Kurtosis' button: You'll then get the Kurtosis of 1.85954: How to Manually Calculate the Kurtosis. In cases where one tail is long but the other tail is fat, skewness does not obey a simple rule. WebSkewness Calculator is an online statistics tool for data analysis programmed to find out the asymmetry of the probability distribution of a real-valued random variable. In other words, a left-skewed distribution has a long tail on its left side. Calculate Pearsons Skewness in Excel by Using Descriptive Statistics 1.1 Use AVERAGE, MODE.SNGL, STDEV.P Functions 1.2 Apply AVERAGE, MEDIAN, STDEV.P Functions 2. If the coefficient of skewness of a distribution is 032 the standard deviation is 65 and the mean is 296. WebIn probability theory and statistics, skewness is a measure of the asymmetry of the probability distribution of a real-valued random variable about its mean. skew (a, axis = 0, bias = True, nan_policy = 'propagate', *, keepdims = False) [source] # Compute the sample skewness of a data set. An introduction to the theory of statistics. , G WebThe steps to calculate the coefficient of skewness are as follows: Using Mode Step 1: Subtract the mode from the mean. For example, the mean zoology test score was 53.7, which is less than the median of 55. There is a long tail on the right, meaning that every few decades there is a year when the number of sunspots observed is a lot higher than average. 60.5 - 70 = -9.5 Step 7 - Gives output as Bowley's Coefficient of Skewness. 2. [28] It is the median of the values of the kernel function. Karl Pearson Coefficient of Skewness Calculation Formula: Sk = 3 (m- Md) SD Md = median, Sk = Karl Pearson coefficient of Skewness F1 Score False Acceptance Rate (FAR) False Negative Rate (FNR) Statistics Calculators Defects per Million Opportunities E Power X Equally Likely Outcomes Probability Gamma Distribution Gamma Distribution Mean Assume a data collection has a mean of 60, a mode of 70, a median of 75, and a standard deviation of 10. For a unimodal distribution, negative skew commonly indicates that the tail is on the left side of the distribution, and positive skew indicates that the tail is on the right. Thanks to our calculator, you will quickly test your dataset for normality and immediately identify any skew in the results. Skewness is close to zero, so the distribution is relatively symmetric. Modeling Asymmetry and Excess Kurtosis in Stock Return Data. WebHow to Use Skewness Calculator? ; their expected values can even have the opposite sign from the true skewness. WebSteps to calculate the coefficient of skewness: The coefficient of skewness can be calculated using either of the two formulas, depending on the data supplied. ) Arguments can either be numbers or names, arrays, or references that contain numbers. s So, how to calculate skewness? {\displaystyle G_{1}} Step 1: Calculate the t value. The numerator is difference between the average of the upper and lower quartiles (a measure of location) and the median (another measure of location), while the denominator is the semi-interquartile range 60.5 - 70 = -9.5 ) with probability one. Other measures of skewness have been used, including simpler calculations suggested by Karl Pearson[10] (not to be confused with Pearson's moment coefficient of skewness, see above). Thus there is a need for another measure of asymmetry that has this property: such a measure was introduced in 2000. It takes advantage of the fact that the mean and median are unequal in a skewed distribution. However, a symmetric unimodal or multimodal distribution always has zero skewness. Calculate the t value (a test statistic) using this formula: Example: Calculating the t value. This calculation computes the output values of skewness, mean and standard deviation according to the input values of data set. which can be calculated in Excel via the formula. {\displaystyle x_{m}} Skewness can be calculated using various methods, whereas the most commonly used method is Pearsons coefficient. Bowleys Coefficient of Skewness for grouped data. {\displaystyle {\overline {x}}} The weight and length of 10 newborns has a Pearson correlation coefficient of .47. WebWeb Karl Pearson Coefficient of Skewness Calculation Formula. On this Wikipedia the language links are at the top of the page across from the article title. WebHow to Use Skewness Calculator? To calculate the skewness, we have to first find the mean and variance of the given data. This calculator calculates the pearson median skewness using mean, mode, Skewness is a commonly used measure of the symmetry of a statistical distribution. Themean is calculated using the formula: Mean or Average =(x1+x2+x3+xn) / n, Wheren = total number of terms, x1,x2,x3, . Get the result! For example, the weights of six-week-old chicks are shown in the histogram below. Luckily, the skewness calculator finds this value by itself it is equal to s = 8.833. [6] is the standard deviation, the skewness is defined in terms of this relationship: positive/right nonparametric skew means the mean is greater than (to the right of) the median, while negative/left nonparametric skew means the mean is less than (to the left of) the median. 60.5 - 70 = -9.5 {\displaystyle k_{2}=s^{2}} This calculation computes the output values of skewness, mean and standard deviation according to the input values of data set. If the left tail (tail at small end of the distribution) is more pronounced than the right tail (tail at the large end of the distribution), the function is said to have negative skewness. x Skewness is a measure of symmetry, or more precisely, the lack of symmetry. 3 Easy Methods to Calculate Coefficient of Skewness in Excel 1. k WebSkewness Calculator Skewness Formula: g = n i = 1 ( x xi ) 3 (n 1)s3 Enter Inputs (separated by comma): Skewness (g) = Mean ( ) = Variance= = Standard Deviation ( ) = Sample Number = Skewness Calculator is a free online tool that displays the skewness in the statistical distributions. {\displaystyle \sigma } Check out 39 similar descriptive statistics calculators . WebIf the co-efficient of skewness is a positive value then the distribution is positively skewed and when it is a negative value, then the distribution is negatively skewed. Bowley's measure of skewness (from 1901),[14][15] also called Yule's coefficient (from 1912)[16][17] is defined as: where Q is the quantile function (i.e., the inverse of the cumulative distribution function). , This is closely related in form to Pearson's second skewness coefficient. We also need to find the standard deviation of this sample. The Karl Pearsons coefficient skewness for grouped data is given by. Webin statistics the skewness is a measure of the asymmetry of the probability distribution of a real-valued random variable about its mean. You can also use a single array or a reference to an array instead of arguments separated by commas. WebStep 2 - Enter the Range or classes (X) seperated by comma (,) Step 3 - Enter the Frequencies (f) seperated by comma. 1 WebHow to Use Skewness Calculator? Q Luckily, the skewness calculator finds this value by itself it is equal to s = 8.833. Forthcoming in Comm in Statistics, Simulation and Computation. and dSkew(X):=0 for X= (with probability 1). A left-skewed distribution is longer on the left side of its peak than on its right. WebPearsons second coefficient (median skewness): It is on the distributions mean, median, and standard deviation. Web Method 2. Measures of Skewness and Kurtosis. One reason you might check if a distribution is skewed is to verify whether your data is appropriate for a certain statistical procedure. , Sk = [(Q3 Q2) (Q2 Q1)]/(Q3 Q1) (Q3 2Q2 + Q1)/(Q3 Q1) The value of Sk would be zero if it is asymmetrical distribution. Bowleys Coefficient of Skewness for grouped data. Choose a distribution. Many textbooks teach a rule of thumb stating that the mean is right of the median under right skew, and left of the median under left skew. b A positive skew specifies that the tail on the right side is longer than the left side and the size of the values lie to the left of the mean. Measures of Skewness and Kurtosis. WebKurtosis is calculated using the formula given below. We obtain the following values: skewness = -0.3212 kurtosis = -0.7195 The negative value of the coefficient of skewness implies a slight skew to the left. WebPearson's Mode Skewness and Pearson's Median Skewness: Two calculations. is the median of the sample {\displaystyle \gamma _{1}=0} Several types of Step 1: Enter the numbers separated by a comma in the given input box. Published on As you can see above, Pearsons first coefficient of skewness has a mode as its one variable to calculate it. Most values cluster around a central region, with values tapering off as they go further away from the center. Insert Excel SKEW Function to Estimate Skewness Coefficient 3. = First off, since the data set is now a sample the standard deviation becomes 2, and is represented by s. This calculator computes the skewness and kurtosis of a distribution or data set. Alongside skewness, kurtosis is a significant unmistakable measurement of information dispersion. You can use the Excel functions AVERAGE , MEDIAN and OR. "Elements of Statistics, 4th Edn (New York, Charles Scribner)."(1920). Sk = Mean Mode) sd = x Mode sx. Suppose we calculate the skewness for this distribution and find that it is -1.3225. WebSkewness is a measure used in statistics that helps reveal the asymmetry of a probability distribution. A normal distribution has zero skew and is an assumption of many statistical procedures. With Cuemath, find solutions in simple and easy steps. Shaun Turney. Pearsons first coefficient of skewness To calculate skewness values, subtract a mode from a mean, and then divide the difference by standard deviation. . It may seem a bit skewed, with a tendency to the left or the right. , defined as:[4][5]. WebWhy Bowley Skewness works. WebKurtosis is calculated using the formula given below. This calculator calculates the pearson median skewness using mean, mode, You can find many different skewness formulas, each with a slightly different interpretation. x The result is divided by the standard deviation . Since we know that n = 10 and r = .47, we can calculate the t value: Insert Excel SKEW Function to Estimate Skewness Coefficient 3. Define the random variable. G Therefore, the distribution has approximately zero skew. Step 7 - Gives output as Bowley's Coefficient of Skewness. Pearsons first coefficient of skewness, or Pearson mode skewness, subtracts the mode from the mean and divides the difference by the standard deviation. This skewness and kurtosis calculator is a statistical tool for the advanced analysis of datasets. Since the kurtosis of the distribution is more than 3, it means it is a leptokurtic distribution. Skewness is a measure of symmetry, or more precisely, the lack of symmetry. WebFind Sample Skewness, Kurtosis for grouped data. If the two are equal, it has zero skewness. How to calculate skewness: an example using the skewness calculator. WebWhy Bowley Skewness works. Distance skewness is always between 0 and 1, equals 0 if and only if X is diagonally symmetric with respect to (X and 2X have the same probability distribution) and equals 1 if and only if X is a constant c ( / They arent perfectly equal because the sample distribution has a very small skew. where, x is the sample mean, M is the median, sx is the sample standard deviation. Alongside skewness, kurtosis is a significant unmistakable measurement of information dispersion. This page was last edited on 21 February 2023, at 08:26. A normal distribution has zero skew and is an assumption of many statistical procedures. {\displaystyle \{x_{1},x_{2},\ldots ,x_ {n}\}} x If your data has a value close to 0, you can consider it to have zero skew. Kurtosis ignores the central values, so it only analyzes what the tails of the distribution look like. [25], A value of skewness equal to zero does not imply that the probability distribution is symmetric. WebThe skewness equation is calculated based on the mean of the distribution, the number of variables, and the standard deviation of the distribution. For example, the mean number of sunspots observed per year was 48.6, which is greater than the median of 39. 3. WebPearsons coefficient of skewness (second method) is calculated by multiplying the difference between the mean and median, multiplied by three.	{"url":"https://curtisstone.com/x5sauxx/coefficient-of-skewness-calculator","timestamp":"2024-11-14T14:36:05Z","content_type":"text/html","content_length":"35267","record_id":"<urn:uuid:a16d20b1-f1e0-43b3-a457-800aeae7e756>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00654.warc.gz"}
Demographics & eGFR Accuracy - NIDDK Demographics & eGFR Accuracy Estimated glomerular filtration rate (eGFR) equations incorporate demographic variables to adjust for variation in concentrations of filtration markers across population subgroups. Adjusting for these variations helps to minimize error in patient subgroups. The recommended eGFR equations for adults adjust for age and sex. Previous eGFR equations also adjusted for race. However, these equations are no longer recommended and are provided for reference. To improve eGFR accuracy, adult equations include age to adjust for the observed decline in serum creatinine with increasing age. Kidney function declines with age in adults (see Table 1). Table 1. Population mean eGFRs from the National Health and Nutrition Examination Survey (NHANES) III^1 Age (Years) Mean eGFR (mL/min/1.73 m^2) 20–29 116 30–39 107 40–49 99 50–59 93 60–69 85 70+ 75 Current eGFR equations are based on sex assigned at birth and include different coefficients for female and male. Race Coefficient Research has demonstrated that race does not reliably account for genetic diversity.^2 As such, use of race in estimating equations is no longer recommended. Chronic Kidney Disease Epidemiology Collaboration (CKD-EPI) developed a new equation that does not include race [2021 CKD-EPI eGFRcr (age+sex)]. Elimination of race introduced a small bias but at levels considered acceptable for routine clinical decision making. Consistent with this, results from the Chronic Renal Insufficiency Cohort (CRIC) and other studies suggested that the use of serum creatinine to estimate GFR without race resulted in systematic misclassification. Data from both CKD-EPI and CRIC demonstrate that this error is mitigated with the use of cystatin C.^3 Thus, NIDDK supports the National Kidney Foundation–American Society of Nephrology and Kidney Disease Improving Global Outcomes (KDIGO) recommendation to assess eGFR using both cystatin C and serum creatinine, when possible, especially when eGFR results are within range of key clinical decision cut points.^4,5 However, a single parameter equation may be preferred when important non-GFR determinants are present for either creatinine or cystatin C. Tables 2 and 3 summarize the systematic error (bias) and accuracy, respectively, of the previous and updated CKD-EPI equations. Table 2. Systematic error (bias) in previous vs updated CKD equations for creatinine, cystatin C, and combined in the CKD-EPI 2021 validation dataset Filtration marker Equation Black individuals non-Black individuals Difference between Black and non-Black individuals (95% CI) creatinine 2009 eGFRcr (age+sex+race) -3.7 (-5.4, -1.8) -0.5 (-0.9, 0.0) -3.2 (-5.0, -1.3) creatinine 2021 eGFRcr (age+sex) 3.6 (1.8, 5.5) -3.9 (-4.4, -3.4) 7.6 (5.6, 9.5) cystatin C 2012 eGFRcys (age+sex) -0.1 (-1.5, 1.6) 0.7 (0.2, 1.2) -0.8 (-2.5, 0.8) creatinine-cystatin C 2012 eGFRcr-cys (age+sex+race) -2.5 (-3.7, -1.2) -0.6 (-0.9, -0.2) -1.9 (-3.2, -0.6) creatinine-cystatin C 2021 eGFRcr-cys (age+sex) 0.1 (-0.9, 1.6) -2.9 (-3.3, -2.5) 3.0 (1.6, 4.4) Table 3. Accuracy (P-30 in percent) of previous vs updated CKD equations for creatinine, cystatin C, and combined in the CKD-EPI 2021 validation dataset Filtration marker Equation Black individuals non-Black individuals Difference between Black and non-Black individuals (95% CI) creatinine 2009 eGFRcr (age+sex+race) 85.1 (82.2, 87.9) 89.5 (88.5, 90.4) -4.4 (-7.6, -1.2) creatinine 2021 eGFRcr (age+sex) 87.2 (84.5, 90.0) 86.5 (85.4, 87.6) 0.7 (-2.4, 3.8) cystatin C 2012 eGFRcys (age+sex) 84.6 (81.7, 87.6) 88.9 (87.9, 89.9) -4.3 (-7.5, -1.1) creatinine-cystatin C 2012 eGFRcr-cys (age+sex+race) 88.6 (85.8, 91.2) 92.4 (91.5, 93.2) -3.8 (-6.7, -0.9) creatinine-cystatin C 2021 eGFRcr-cys (age+sex) 90.5 (88.1, 92.9) 90.8 (89.9, 91.8) -0.3 (-3.0, 2.4) Comparing the Race-based 2009 and Race-free 2021 CKD-EPI Equations During the transition from the race-based GFR estimating equations to the race-free equations, it is important that laboratories clearly identify which equation was used to report eGFR values. Clinicians may need to continue to compare a patient’s eGFR values to results from older equations so as not to miss changes in kidney function that could be obscured by the use of a new equation with a different estimation output. The following figures show the differences in eGFR values using the previous CKD-EPI equations with age, sex, and race compared to the new CKD-EPI equations with only age and sex coefficients. As demonstrated in Figure 1, the 2009 CKD-EPI creatinine equation that uses age, sex, and race may overestimate GFR in Black individuals and to a lesser degree in non-Black individuals. The 2021 CKD-EPI creatinine equation that uses age and sex, and omits race, may underestimate measured GFR (mGFR) in Black individuals and overestimate mGFR in non-Black individuals. Therefore, using the 2021 CKD-EPI creatinine equation without the race coefficient may increase differential bias between Black and non-Black participants compared to the 2009 race-based equation. The 2021 CKD-EPI creatinine-cystatin C equation was more accurate than the 2021 CKD-EPI creatinine equation, with smaller differences between race groups. Figure 1. Comparison of measured vs estimated GFR by race groups across previous and updated CKD-EPI eGFR equations View full-sized image Figure 1: Comparison of measured vs estimated GFR by race groups across previous and updated CKD-EPI eGFR equations. Shows five graphs comparing mGFR and eGFR for Black individuals and non-Black individuals using previous and updated CKD-EPI eGFR equations in the validation data set for the 2021 CKD-EPI equations.^2 The first graph uses the previous 2009 eGFR creatinine equation. The second graph uses the updated 2021 eGFR creatinine equation. The third graph uses the previous 2012 eGFR creatinine-cystatin C equation. The fourth graph uses the updated 2021 creatinine-cystatin C equation. The final fifth graph uses the 2012 eGFR cystatin C equation. Each graph contains bias, P30, and correct classification data. The equations are referred to by the filtration marker or markers (creatinine [eGFRcr], cystatin C [eGFRcys], or creatinine–cystatin C [eGFRcr-cys]), the demographic factors (age, sex, and race or age and sex) that were used in their development, and the year in which the equations were developed (2009, 2012, or 2021). Bias is defined as the median difference between mGFR and eGFR. A positive sign indicates underestimation of mGFR, and a negative sign overestimation of mGFR. P30 is the proportion of eGFR within 30% of mGFR. Correct classification refers to agreement between mGFR and eGFR categories of more than 90, 60-89, 45-59, 30-44, 15-29, and less than 15 ml per minute per 1.73 m^2. Figures 2 and 3 demonstrate the differences in mean bias and P30 using the old and new equations. Figure 2. Systematic error (bias) and precision of the 2009 CKD-EPI creatinine equation vs 2021 CKD-EPI creatinine equation across eGFR subgroups View full-sized image Figure 2: Systematic error (bias) and precision of the 2009 CKD-EPI creatinine equation vs 2021 CKD-EPI creatinine equation across eGFR subgroups. Shows two graphs comparing bias and precision of the previous and updated CKD-EPI eGFR creatinine equations. The upper graph shows bias, measured as the difference between measured and estimated GFR, for the previous 2009 CKD-EPI creatinine equation with race coefficient versus the updated 2021 CKD-EPI creatine equation without race coefficient. The bottom graphs show accuracy, as measured by 1 - P30 or the percentage of estimates within 30% of mGFR, of the previous 2009 CKD-EPI creatinine equation with race coefficient versus the updated 2021 CKD-EPI creatine equation without race coefficient. The vertical bars indicate 95% confidence intervals. The dotted black line represents the difference in the GFR equation performance by race. eGFR stages include < 30, 30-59, 60-89, and > 90 ml/min/ 1.73m^2. eGFR was defined separately for each equation. The equations are referred to by the demographic factors (age, sex, and race or age and sex) that were used in their development and the year in which the equations were developed (2009 or 2021). The units for GFR are ml/min per 1.73 m^2. Figure 3. Systematic error (bias) and precision of the 2012 CKD-EPI creatinine-cystatin C equation vs the 2021 CKD-EPI creatinine-cystatin C equation across eGFR subgroups View full-sized image Figure 3: Systematic error (bias) and precision of the 2012 CKD-EPI creatinine-cystatin C equation vs the 2021 CKD-EPI creatinine-cystatin C equation across eGFR subgroups. Shows two graphs comparing bias and precision of the previous and updated CKD-EPI eGFR creatinine-cystatin C equations. The upper graph shows bias, as measured as the difference between measured and estimated GFR, of the previous 2012 CKD-EPI creatinine-cystatin C equation with race coefficient versus the updated 2021 CKD-EPI creatine-cystatin C equation without race coefficient. The bottom graphs show accuracy, as measured by 1 - P30 or the percentage of estimates within 30% of mGFR, of the previous 2012 CKD-EPI creatinine-cystatin C equation with race coefficient versus the updated 2021 CKD-EPI creatine-cystatin C equation without race coefficient. The vertical bars indicate 95% confidence intervals. The dotted black line represents the difference in the GFR equation performance by race. eGFR stages include < 30, 30-59, 60-89, and > 90 ml/min/ 1.73m^2. eGFR was defined separately for each equation. The equations are referred to by the demographic factors (age, sex, and race or age and sex) that were used in their development and the year in which the equations were developed (2012 or 2021). The units for GFR are ml/min per 1.73 m^2. Last Reviewed May 2024	{"url":"https://www.niddk.nih.gov/research-funding/research-programs/kidney-clinical-research-epidemiology/laboratory/factors-affecting-egfr-accuracy/demographics","timestamp":"2024-11-10T02:10:08Z","content_type":"text/html","content_length":"62701","record_id":"<urn:uuid:2758c46f-a399-4fcb-b684-516269210a2d>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00358.warc.gz"}
A Quantum Description of Physical Systems with Non-real Energies - JPS Hot Topics A Quantum Description of Physical Systems with Non-real Energies JPS Hot Topics 1, 044 © The Physical Society of Japan This article is on Editorial: Non-Hermitian quantum mechanics Prog. Theor. Exp. Phys. 2020, 12A001 (2020) . While quantum systems are traditionally described by Hermitian Hamiltonians, the formalism is extendable to a non-Hermitian description for systems that are dissipative or obey parity-time symmetry. The quantum mechanical worldview is, in many ways, strikingly different from our intuitions about the real world. One such conspicuous difference is that measurable physical quantities in quantum mechanics cannot be represented simply by numbers, but instead need to be considered as mathematical operators, or functions over a space of physical states onto another space of physical states. However, we usually insist that these operators give back real values, since, in the real world, we associate measurable quantities with real numbers. Consequently, physical operators are normally restricted to being what are called “Hermitian operators”. However, in my article I have explored two scenarios where this dogma can be questioned. One of the most important operators in quantum mechanics is the Hamiltonian operator, which corresponds to physical energy. It may seem intuitively obvious that the Hamiltonian operator must be Hermitian, but this is, in fact, only true so long as energy is conserved. Now while the energy of the entire universe is certainly conserved, it may not be necessarily so for a part of it, such as a radioactive nuclide. This system loses energy to the outside world whenever it emits an alpha or a beta particle and therefore, does not conserve energy. Such an open system can be described by an effective non-Hermitian Hamiltonian that yields complex energy values corresponding to short-lived “resonant states.” While the eigenfunctions of these states are not normalizable, it simply means that the states extend over the entire universe. Another instance is where the theory of parity-time (PT) symmetry (symmetry under the combined operations of spatial inversion and time reversal) considers the Hamiltonian of the entire universe as non-Hermitian but operating in a parameter space where its eigenvalues are exclusively real. For example, a system of two closed-shell atoms can be described using a non-Hermitian Hamiltonian that is PT symmetric and yields energy eigenvalues that can be rendered real by parameter tuning. This shows that Hermiticity is only a sufficient condition for real values, not a necessary one. The non-Hermitian formulation has since been picked up and applied by researchers in various fields, most notably in topological matter and many-body systems. Just as generalizing to complex numbers can often help describe physical systems more elegantly, a non-Hermitian generalization of quantum mechanics can help broaden our perspective of quantum systems. Editorial: Non-Hermitian quantum mechanics Prog. Theor. Exp. Phys. 2020, 12A001 (2020) . Share this topic	{"url":"https://jpsht.jps.jp/article/1-044/","timestamp":"2024-11-02T05:15:23Z","content_type":"text/html","content_length":"70919","record_id":"<urn:uuid:6c99a261-63bb-4fba-89ee-4edc4d461a6d>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00291.warc.gz"}
34500 in Words - How to Write 34500 in Words? \| Brighterly 34500 in Words Updated on January 3, 2024 The number 34500 is expressed in words as “thirty-four thousand five hundred”. It follows thirty-four thousand. For example, if there are thirty-four thousand five hundred apples in an orchard, it means the orchard has thirty-four thousand apples plus five hundred more. Thousands Hundreds Tens Ones How to Write 34500 in Words? The number 34500 is expressed in words as ‘Thirty-Four Thousand Five Hundred’. It has a ‘3’ in the ten-thousands place, a ‘4’ in the thousands place, a ‘5’ in the hundreds place, and ‘0’ in both the tens and ones places. Like counting thirty-four thousand five hundred items, you say, “I have thirty-four thousand five hundred.” Therefore, 34500 is written in words as ‘Thirty-Four Thousand Five 1. Place Value Chart: Thousands: 34, Hundreds: 5, Tens: 0, Ones: 0 2. Writing it down: 34500 = Thirty-Four Thousand Five Hundred This method is fundamental in learning how to convert larger numbers to words. FAQ on 34500 in Words Can you write the number 34500 using words? The number 34500 is written as ‘Thirty-four thousand five hundred’. What is the word form for the number 34500? Thirty-four thousand five hundred’ is the word form for 34500. If you count to 34500, how do you spell the number? Counting to 34500, spell it as ‘Thirty-four thousand five hundred’. Other Numbers in the Words: 58000 in Words 99 in Words 83000 in Words 1111 in Words 160000 in Words 15000 in Words 1200000 in Words Poor Level Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence. Mediocre Level Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence. Needs Improvement Start practicing math regularly to avoid your child`s math scores dropping to C or even D. High Potential It's important to continue building math proficiency to make sure your child outperforms peers at school.	{"url":"https://brighterly.com/math/34500-in-words/","timestamp":"2024-11-04T17:54:23Z","content_type":"text/html","content_length":"85217","record_id":"<urn:uuid:773541fd-a623-4610-ae94-8e548886c501>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00574.warc.gz"}
Transactions Online Liming ZHENG, Jooin WOO, Kazuhiko FUKAWA, Hiroshi SUZUKI, Satoshi SUYAMA, "Low-Complexity Algorithm for Log Likelihood Ratios in Coded MIMO Communications" in IEICE TRANSACTIONS on Communications, vol. E94-B, no. 1, pp. 183-193, January 2011, doi: 10.1587/transcom.E94.B.183. Abstract: This paper proposes a low-complexity algorithm to calculate log likelihood ratios (LLRs) of coded bits, which is necessary for channel decoding in coded MIMO mobile communications. An approximate LLR needs to find a pair of transmitted signal candidates that can maximize the log likelihood function under a constraint that a coded bit is equal to either one or zero. The proposed algorithm can find such a pair simultaneously, whereas conventional ones find them individually. Specifically, the proposed method searches for such candidates in directions of the noise enhancement using the MMSE detection as a starting point. First, an inverse matrix which the MMSE weight matrix includes is obtained and then the power method derives eigenvectors of the inverse matrix as the directions of the noise enhancement. With some eigenvectors, one-dimensional search and hard decision are performed. From the resultant signals, the transmitted signal candidates to be required are selected on the basis of the log likelihood function. Computer simulations with 44 MIMO-OFDM, 16QAM, and convolutional codes (rate =1/2, 2/3) demonstrate that the proposed algorithm requires only 1.0 dB more E[b]/N[0] than that of the maximum likelihood detection (MLD) in order to achieve packet error rate of 10^-3, while reducing the complexity to about 0.2% of that of MLD. URL: https://global.ieice.org/en_transactions/communications/10.1587/transcom.E94.B.183/_p author={Liming ZHENG, Jooin WOO, Kazuhiko FUKAWA, Hiroshi SUZUKI, Satoshi SUYAMA, }, journal={IEICE TRANSACTIONS on Communications}, title={Low-Complexity Algorithm for Log Likelihood Ratios in Coded MIMO Communications}, abstract={This paper proposes a low-complexity algorithm to calculate log likelihood ratios (LLRs) of coded bits, which is necessary for channel decoding in coded MIMO mobile communications. An approximate LLR needs to find a pair of transmitted signal candidates that can maximize the log likelihood function under a constraint that a coded bit is equal to either one or zero. The proposed algorithm can find such a pair simultaneously, whereas conventional ones find them individually. Specifically, the proposed method searches for such candidates in directions of the noise enhancement using the MMSE detection as a starting point. First, an inverse matrix which the MMSE weight matrix includes is obtained and then the power method derives eigenvectors of the inverse matrix as the directions of the noise enhancement. With some eigenvectors, one-dimensional search and hard decision are performed. From the resultant signals, the transmitted signal candidates to be required are selected on the basis of the log likelihood function. Computer simulations with 44 MIMO-OFDM, 16QAM, and convolutional codes (rate =1/2, 2/3) demonstrate that the proposed algorithm requires only 1.0 dB more E[b]/N[0] than that of the maximum likelihood detection (MLD) in order to achieve packet error rate of 10^-3, while reducing the complexity to about 0.2% of that of MLD.}, TY - JOUR TI - Low-Complexity Algorithm for Log Likelihood Ratios in Coded MIMO Communications T2 - IEICE TRANSACTIONS on Communications SP - 183 EP - 193 AU - Liming ZHENG AU - Jooin WOO AU - Kazuhiko FUKAWA AU - Hiroshi SUZUKI AU - Satoshi SUYAMA PY - 2011 DO - 10.1587/transcom.E94.B.183 JO - IEICE TRANSACTIONS on Communications SN - 1745-1345 VL - E94-B IS - 1 JA - IEICE TRANSACTIONS on Communications Y1 - January 2011 AB - This paper proposes a low-complexity algorithm to calculate log likelihood ratios (LLRs) of coded bits, which is necessary for channel decoding in coded MIMO mobile communications. An approximate LLR needs to find a pair of transmitted signal candidates that can maximize the log likelihood function under a constraint that a coded bit is equal to either one or zero. The proposed algorithm can find such a pair simultaneously, whereas conventional ones find them individually. Specifically, the proposed method searches for such candidates in directions of the noise enhancement using the MMSE detection as a starting point. First, an inverse matrix which the MMSE weight matrix includes is obtained and then the power method derives eigenvectors of the inverse matrix as the directions of the noise enhancement. With some eigenvectors, one-dimensional search and hard decision are performed. From the resultant signals, the transmitted signal candidates to be required are selected on the basis of the log likelihood function. Computer simulations with 44 MIMO-OFDM, 16QAM, and convolutional codes (rate =1/2, 2/3) demonstrate that the proposed algorithm requires only 1.0 dB more E[b]/N[0] than that of the maximum likelihood detection (MLD) in order to achieve packet error rate of 10^-3, while reducing the complexity to about 0.2% of that of MLD. ER -	{"url":"https://global.ieice.org/en_transactions/communications/10.1587/transcom.E94.B.183/_p","timestamp":"2024-11-10T02:06:48Z","content_type":"text/html","content_length":"65418","record_id":"<urn:uuid:6534e469-3238-444d-a2c8-254aa0e88540>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00729.warc.gz"}
1quad is an esolang based on Conway's Game of Life created by User:Yayimhere. it may seem random but it was made such that 1 was a quad a program is made up of a single unary number where the unary symbol is .. to add a decimal behind it you seperate the decimal part with a = now we have a unary number. lets call it n. now do this process to turn the program into conway's game of life: • set a variable P to ${\displaystyle abs({\frac {1}{n}})+(55385795187-n)}$ • the area(called the rectangle) for the GoL pattern is ${\displaystyle n\times (5+n)}$ by ${\displaystyle (n2)\times (n4)}$ • fit the binary of P into the rectangle(not caring about what goes outside the rectangle) where 1 is a live cell and 0 is a dead cell, and if there aren't enough digits we repeat the binary until there is. we ignore the decimal if there is any • run the GoL program so the program: returns this GoL layout: becomes the layout:	{"url":"https://esolangs.org/wiki/1quad","timestamp":"2024-11-14T17:43:06Z","content_type":"text/html","content_length":"21689","record_id":"<urn:uuid:d2da864a-3a2a-45c4-8e74-63beb87711f4>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00422.warc.gz"}
Re: An example LL(K) language that is not LL(K-1) ? Chris F Clark <cfc@shell01.TheWorld.com> Fri, 05 Feb 2010 18:12:01 -0500 From comp.compilers \| List of all articles for this month \| From: Chris F Clark <cfc@shell01.TheWorld.com> Newsgroups: comp.compilers Date: Fri, 05 Feb 2010 18:12:01 -0500 Organization: The World Public Access UNIX, Brookline, MA References: 10-02-020 Keywords: LL(1) Posted-Date: 06 Feb 2010 09:56:12 EST SLK Mail <slkpg@cox.net> writes: > Example: > S -> a A a > S -> b A b a > A -> b > A -> > Can you convert this to LL(1)? Yes, there is no recursion in this grammar, therefore the language is trivially regular and thus trivally, LL(1). The language is: S: aa \| aba \| bba \| bbba; Now, if you replaced all A's in the grammar with S's (or add the rule A -> S), making it recursive. I'm not so quick to jump to such a conclusion. The language wouldn't be left-factored in that case, so there would be work to do generally. Note, you only get a non-regular context free language when you intersect a regular language (languages with fixed strings, and left, and right recursion only) with a Dyck language (a language that describes balanced parethesized expressions, i.e. uses what I call center recursion). Thus, any LL(2) language will have at least one rule that has central recursion. Otherwise, if there is no recursion, or it is all left or right recursion (either direct or indirect), the language will be regular and thus LL(1). Hope this helps, Chris Clark email: christopher.f.clark@compiler-resources.com Compiler Resources, Inc. Web Site: http://world.std.com/~compres 23 Bailey Rd voice: (508) 435-5016 Berlin, MA 01503 USA twitter: @intel_chris Post a followup to this message Return to the comp.compilers page. Search the comp.compilers archives again.	{"url":"https://www.compilers.iecc.com/comparch/article/10-02-022","timestamp":"2024-11-11T11:13:53Z","content_type":"text/html","content_length":"7043","record_id":"<urn:uuid:c5269aed-5d9f-4a9b-b1f3-7c09fbac5b69>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00743.warc.gz"}
Microsoft Excel question - need help with totalling different items! Hey guys, question about Microsoft Excel here. I could really use an answer... I've solved a lot of tricky stuff in Excel before, but I'm baffled that I can't find a function for simply adding quantities of different items. For example: Fruit - Quantity Apples - 2 Bananas - 3 Coconuts - 8 Bananas - 1 Coconuts - 7 Apples - 4 Apples - 6 Coconuts - 5 Bananas - 9 Let's say I have the above, with the fruit names in one column and the quantity of them in the adjacent column. I want to be able to total each fruit type. Now here's the catch: I can't rearrange the rows by sorting, and I don't want to have to do any manual work after figuring this out the first time, at least no more than just inserting a column or another worksheet every time I get the data dump of columns similar to the above every month or whatever, and have all the totals I need. I know how to use macros, but I'm not sure that they can work with Auto-filter or PivotTables without still some manual involvement. I can think of two ways, but I'm hoping someone can come up with a better suggestion... 1. One way involves making separate columns with headers for each of the fruits. Using a simple IF/THEN function comparing the fruit name column with each header, I can say that if they equal, to transpose the quantity into that column. Then it's a simple sum of each column for each fruit. However, this requires a column for each fruit, and in my real scenario, I have dozens, if not hundreds of different items to tally, so this is not ideal. 2. The other way involves the DSUM function. But the weird thing is that I have to have a separate column reiterating what I want to total, so in addition to the example table above, let's say I have this column somewhere on the spreadsheet: The DSUM function requires a range (not a simple string of text or a single cell) for its criteria for some stupid reason, so I can select "Fruit" down to "Apples" to total just the Apples. I can also select "Fruit" down to "Bananas" to total the Apples and the Bananas together. But if I want to total just the Bananas? Can't. Not unless I have another column. So then I'd have to do something that looks like this: Fruit - Fruit Apples - Bananas So then I end up with a separate column for each fruit again (it can't be a row for each, for some stupid reason), and then it's just like in my first scenario. Actually, it's worse because now I have to waste the whole top row repeating "Fruit". I feel like I'm overlooking a function that's incredibly easy, because I don't see why something like Excel wouldn't have something as basic as adding certain numbers based on the text in an adjacent Any help would be appreciated.	{"url":"https://www.vgmaps.com/forums/index.php?topic=1367.0;prev_next=prev","timestamp":"2024-11-13T03:01:40Z","content_type":"application/xhtml+xml","content_length":"31467","record_id":"<urn:uuid:880ee81c-e922-45e0-b6e8-a42364a9eb4e>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00802.warc.gz"}
A363363 - OEIS Stephan Brandt, Ralph Faudree, and Wayne Goddard, Weakly pancyclic graphs , Journal of Graph Theory 27 (1998), 141-176. a(n) = 0 for n <= 5, because all graphs on at most 5 nodes are weakly pancyclic. There are a(6) = 4 not weakly pancyclic graphs on 6 nodes (all of them connected): a cycle of length 6 with one additional edge (two different graphs); the complete bipartite graph K_{3,3} with one edge removed;	{"url":"https://oeis.org/A363363","timestamp":"2024-11-10T09:02:15Z","content_type":"text/html","content_length":"13608","record_id":"<urn:uuid:043bec3a-1d25-4a5d-a659-91a4356f6a0f>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00636.warc.gz"}
Ks3 algebra worksheet ks3 algebra worksheet Related topics: algebra 1free perimeter and area worksheets Graphing Linear Equalities free work sheet for linear equation geometry linear equations in two variables multistep equations worksheet description of mathematics & statistics,2 algebra and trigonometry with subtracting 2 numbers programme in c language pre algebra cheats Author Message sdivem Posted: Saturday 11th of Apr 07:43 Can someone help me with my assignment questions ? Most of them are based on ks3 algebra worksheet. I have read a few articles on adding exponents and system of equations but that didn’t go a long way helping me in solving the questions on my assignment. I didn’t sleep last night since I have to turn in pretty soon. But the problem is no matter how much time I put in, I just don’t seem to be getting the hang of it. Every question poses a new challenge, one which seems to be tougher than climbing Mt.Everest! I need some help as soon as possible. Somebody please guide me. Registered: 25.08.2002 Back to top Vofj Timidrov Posted: Sunday 12th of Apr 08:08 Oh boy! You seem to be one of the top students in your class. Well, use Algebrator to solve those questions. The software will give you a comprehensive step by step solution. You can read the explanation and understand the problems. Hopefully your ks3 algebra worksheet class will be the best one. Registered: 06.07.2001 From: Bulgaria Back to top Xane Posted: Tuesday 14th of Apr 11:49 Even I’ve been through times when I was trying to figure out a way to solve certain type of questions pertaining to algebra formulas and exponent rules. But then I came across this piece of software and it was almost like I found a magic wand. In a flash it would solve even the most difficult problems for you. And the fact that it gives a detailed and elaborate explanation makes it even more handy. It’s a must buy for every math student. Registered: 16.04.2003 From: the wastelands between insomnia and Back to top Doldx Womj Posted: Tuesday 14th of Apr 20:02 This sounds very good . Do you know where I can buy the program ? Registered: 03.07.2002 From: Toronto, Canada Back to top molbheus2matlih Posted: Wednesday 15th of Apr 13:55 A truly piece of algebra software is Algebrator. Even I faced similar problems while solving unlike denominators, solving a triangle and graphing function. Just by typing in the problem workbookand clicking on Solve – and step by step solution to my algebra homework would be ready. I have used it through several math classes - Algebra 2, Algebra 2 and Algebra 1. I highly recommend the program. Registered: 10.04.2002 From: France Back to top Mov Posted: Thursday 16th of Apr 08:44 It’s right here: https://softmath.com/algebra-software-guarantee.html. Buy it and try it, if you don’t like it ( which is highly improbable) then they even have an unquestionable money back guarantee. Try using it and good luck with your project . Registered: 15.05.2002 Back to top	{"url":"https://softmath.com/algebra-software-2/ks3-algebra-worksheet.html","timestamp":"2024-11-14T14:55:54Z","content_type":"text/html","content_length":"43026","record_id":"<urn:uuid:3599325f-70e7-4dc0-89d5-ba9e40027612>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00086.warc.gz"}
Cite as Artur Czumaj, Guichen Gao, Shaofeng H.-C. Jiang, Robert Krauthgamer, and Pavel Veselý. Fully-Scalable MPC Algorithms for Clustering in High Dimension. In 51st International Colloquium on Automata, Languages, and Programming (ICALP 2024). Leibniz International Proceedings in Informatics (LIPIcs), Volume 297, pp. 50:1-50:20, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2024) Copy BibTex To Clipboard author = {Czumaj, Artur and Gao, Guichen and Jiang, Shaofeng H.-C. and Krauthgamer, Robert and Vesel\'{y}, Pavel}, title = {{Fully-Scalable MPC Algorithms for Clustering in High Dimension}}, booktitle = {51st International Colloquium on Automata, Languages, and Programming (ICALP 2024)}, pages = {50:1--50:20}, series = {Leibniz International Proceedings in Informatics (LIPIcs)}, ISBN = {978-3-95977-322-5}, ISSN = {1868-8969}, year = {2024}, volume = {297}, editor = {Bringmann, Karl and Grohe, Martin and Puppis, Gabriele and Svensson, Ola}, publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik}, address = {Dagstuhl, Germany}, URL = {https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.ICALP.2024.50}, URN = {urn:nbn:de:0030-drops-201938}, doi = {10.4230/LIPIcs.ICALP.2024.50}, annote = {Keywords: Massively parallel computing, high dimension, facility location, k-median, k-means}	{"url":"https://drops.dagstuhl.de/search?term=Guha%2C%20Sudipto","timestamp":"2024-11-06T00:49:30Z","content_type":"text/html","content_length":"183025","record_id":"<urn:uuid:9a99201f-d2ba-4a08-8410-b4ce516d2ba6>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00558.warc.gz"}
Number Properties Problem Solving : Factors \| Wizako GMAT Prep Blog Here is an interesting GMAT number property problem solving question. Concept tested : number of factors. Integer x has n factors; 3x has 3 factors; Which of the following values can n take? I. 1 II. 2 III. 3 A. I only B. II only C. I or II D. II or III E. I or III 1. Miguel Costa says 2 factors right? They must be 1 and 3. 2. Miguel Costa says Could you please help me out on this exercise? My thinking below: If 3x has exactly 3 factor (we dont know if they are distinct factors) it has necessarily 1;x;3;3x has factors, which implies that x=3. Considering this, 3 has 2 factors : 1 and 3, itself. So the answer must be 2 factors. right? Queries, answers, comments welcomeCancel reply	{"url":"https://gmat-prep-blog.wizako.com/gmat-quant-practice/arithmetic/number-properties-problem-solving-factors/","timestamp":"2024-11-06T15:17:41Z","content_type":"text/html","content_length":"68928","record_id":"<urn:uuid:666f4d0f-21df-48d9-8647-808aead850aa>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00472.warc.gz"}
slahrd: reduces the first NB columns of a real general n-by-(n-k+1) matrix A so that elements below the k-th subdiagonal are zero - Linux Manuals (l) slahrd (l) - Linux Manuals slahrd: reduces the first NB columns of a real general n-by-(n-k+1) matrix A so that elements below the k-th subdiagonal are zero SLAHRD - reduces the first NB columns of a real general n-by-(n-k+1) matrix A so that elements below the k-th subdiagonal are zero N, K, NB, A, LDA, TAU, T, LDT, Y, LDY ) INTEGER K, LDA, LDT, LDY, N, NB REAL A( LDA, * ), T( LDT, NB ), TAU( NB ), Y( LDY, NB ) SLAHRD reduces the first NB columns of a real general n-by-(n-k+1) matrix A so that elements below the k-th subdiagonal are zero. The reduction is performed by an orthogonal similarity transformation Qaq * A * Q. The routine returns the matrices V and T which determine Q as a block reflector I - VTVaq, and also the matrix Y = A * V * T. This is an OBSOLETE auxiliary routine. This routine will be aqdeprecatedaq in a future release. Please use the new routine SLAHR2 instead. N (input) INTEGER The order of the matrix A. K (input) INTEGER The offset for the reduction. Elements below the k-th subdiagonal in the first NB columns are reduced to zero. NB (input) INTEGER The number of columns to be reduced. A (input/output) REAL array, dimension (LDA,N-K+1) On entry, the n-by-(n-k+1) general matrix A. On exit, the elements on and above the k-th subdiagonal in the first NB columns are overwritten with the corresponding elements of the reduced matrix; the elements below the k-th subdiagonal, with the array TAU, represent the matrix Q as a product of elementary reflectors. The other columns of A are unchanged. See Further Details. LDA (input) INTEGER The leading dimension of the array A. LDA >= max(1,N). TAU (output) REAL array, dimension (NB) The scalar factors of the elementary reflectors. See Further Details. T (output) REAL array, dimension (LDT,NB) The upper triangular matrix T. LDT (input) INTEGER The leading dimension of the array T. LDT >= NB. Y (output) REAL array, dimension (LDY,NB) The n-by-nb matrix Y. LDY (input) INTEGER The leading dimension of the array Y. LDY >= N. The matrix Q is represented as a product of nb elementary reflectors = H(1) H(2) . . . H(nb). Each H(i) has the form H(i) = I - tau * v * vaq where tau is a real scalar, and v is a real vector with v(1:i+k-1) = 0, v(i+k) = 1; v(i+k+1:n) is stored on exit in A(i+k+1:n,i), and tau in TAU(i). The elements of the vectors v together form the (n-k+1)-by-nb matrix V which is needed, with T and Y, to apply the transformation to the unreduced part of the matrix, using an update of the form: A : = (I - VTVaq) * (A - Y*Vaq). The contents of A on exit are illustrated by the following example with n = 7, k = 3 and nb = 2: ( a h a a a ) ( a h a a a ) ( a h a a a ) ( h h a a a ) ( v1 h a a a ) ( v1 v2 a a a ) ( v1 v2 a a a ) where a denotes an element of the original matrix A, h denotes a modified element of the upper Hessenberg matrix H, and vi denotes an element of the vector defining H(i).	{"url":"https://www.systutorials.com/docs/linux/man/l-slahrd/","timestamp":"2024-11-11T20:28:02Z","content_type":"text/html","content_length":"12269","record_id":"<urn:uuid:c27fb630-68a6-4b7c-8546-cc10cba35ea8>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00697.warc.gz"}
Some Circuits in Graph or Network Theory A graph is a mathematical object made up of points (sometimes called nodes, see below) with lines joining some or all of the points. You can think of the world wide web as a graph. The points and lines are called vertices and edges just like the vertices and edges of polyhedra. Here is a graph representing a cube. Graph Theory is a whole mathematical subject in its own right, many books and papers are written on it and it is still an active research area with new discoveries still being made. Graphs are very useful in designing, representing and planning the use of networks (for example airline routes, electricity and water supply networks, delivery routes for goods, postal services etc.) Graphs are also used to solve problems in coding, telecommunications and parallel programming. In some of these applications the actual distances and the geometrical shape of the graph is not important, simply which vertices in the system are linked, and these applications come into the branch of maths known as topology. In other applications distances between the vertices, the direction of flow and the capacity of the 'pipes' are significant. Rather confusingly there are two different languages used by mathematicians. In this article we use the graph theory language. Graph Theory Network Applications Graph Network Vertex Node Edge Arc Degree Order Path Route You can trace a path in the graph by taking a pencil, starting at one of the vertices and drawing some of the edges of the graph without lifting your pencil off the paper. A path is simply a sequence of vertices where each vertex is connected by a line to the next one in the sequence. A circuit is any path in the graph which begins and ends at the same vertex. Two special types of circuits are Eulerian circuits, named after Leonard Euler (1707 to 1783), and Hamiltonian circuits named after William Rowan Hamilton (1805 to 1865). The whole subject of graph theory started with Euler and the famous Konisberg Bridge Problem. An Eulerian circuit passes along each edge once and only once, and a Hamiltonian circuit visits each vertex once and only once. Both are useful in applications; the Hamiltonian circuits when it is required to visit each vertex (say every customer, every supply depot or every town) and the Eulerian circuits when it is required to travel along all the connecting edges (say all the streets in a town to collect the garbage). Note that for a Hamiltonian circuit it is not necessary to travel along each edge. Here are two graphs, the first contains an Eulerian circuit but no Hamiltonian circuits and the second contains a Hamiltonian circuit but no Eulerian circuits. If you find it difficult to remember which is which just think E for edge and E for Euler. Can you draw for yourself other simple graphs which have one sort of circuit in them and not the other? Conditions for there to be Eulerian circuits are well know but in general it is a difficult problem to decide when a given graph has a Hamiltonian circuit. Finding conditions for the existence of Hamiltonian circuits is an unsolved problem. The degree of a vertex is the number of edges joining onto that vertex, and vertices are said to be odd or even according to whether the degree is odd or even. Euler circuits exist only in networks where there are no odd vertices, that is where all the vertices have an even number of edges ending there. Two edges are used each time the path visits and leaves a vertex because the circuit must use each edge only once. It follows that if the graph has an odd vertex then that vertex must be the start or end of the path and, as a circuit starts and ends at the same vertex, for a circuit to exist all the vertices must be even. When there are two odd vertices a walk can take place that traverses each edge exactly once but this will not be a circuit. Can you think why it is impossible to draw any graph with an odd number of odd vertices (e.g. one odd vertex)? Well the reason is that each edge has two ends so the total number of endings is even, so the sum of the degrees of all the vertices in a graph must be even, so there cannot be an odd number of odd vertices. Here is a simple puzzle, which we call the Prime Puzzle, for you to solve that uses and illustrates Hamiltonian circuits. Each of the following numbers is the product of exactly three prime factors and you have to arrange them in a sequence so that any two successive numbers in the sequence have exactly one common factor. The numbers are $222$, $255$, $385$, $874$, $2821$, $4199$, $11803$ and $20677$ and we have used only the first twelve prime numbers. First factorize the numbers, next start to draw the graph which will have $8$ vertices, one for each number. Take one number on a vertex and draw three edges from it and label them, one for each factor. Now attach the appropriate numbers at the ends of these edges. Repeat the procedure until the graph is complete. Any two vertices are joined by an edge if and only if they have a common factor. You should have eight vertices and twelve edges and this should suggest a neat way to draw the graph. You may wish to re-draw the graph so that the edges do not cross except at the eight vertices. The graph will be one where it is easy to find a Hamiltonian circuit and this circuit gives you the solution to the problem. Here is a similar but well known puzzle invented by Peterson where you have to arrange the ten cards in a loop so that each card has exactly one letter in common with each adjacent card. The words are HUT, WIT, SAW, CAR, CUB, MOB, DIM, RED, SON, HEN. The arrangement shown in the diagram looks very nearly correct but the words SON and RED do not match. If you try to solve the puzzle by re-arranging the cards you will not succeed because it is impossible. Now replace SON by SUN and HUT by HOT and the puzzle can be solved. The explanation is contained in the following two graphs. In the Peterson graph there are no Hamiltonian circuits so, unlike the Primes Puzzle above there is no way to put the cards into the required circuit. Changing two of the cards to SON and HUT makes it possible to find a Hamiltonian circuit and solve the problem. On the NRICH website you will find a lot of problems on graphs and networks which you might like to try. Also why not do some research on the web and find out about Euler and Hamilton, both giants in the mathematical world.	{"url":"http://nrich.maths.org/articles/some-circuits-graph-or-network-theory","timestamp":"2024-11-05T13:49:33Z","content_type":"text/html","content_length":"43136","record_id":"<urn:uuid:9e04a671-6ea4-43d6-8484-b408dfc8b370>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00150.warc.gz"}
Coq devs & plugin devs @Pierre-Marie Pédrot Is asking about why induction is so slow (https://github.com/coq/coq/pull/14174#discussion_r627670736) a good topic to add to the Coq call this week? Oh, sorry I missed that ping. I can have a look quickly though. It's not super clear that there is a single source of slowdown. It looks spread around, although mainly inside unification. Do you happen to have a slightly more extreme example? I tried nesting more records but I didn't get slower. It seems that the associativity of the records matter. @Pierre-Marie Pédrot Here's a slightly more extreme example (should work on Coq master too) Set Implicit Arguments. Definition ex_proj1 := fun {A : Prop} {P : A -> Prop} (x : exists y, P y) => match x with ex_intro _ a _ => a end. Definition ex_proj2 := fun {A : Prop} {P : A -> Prop} (x : exists y, P y) => match x return P (ex_proj1 x) with ex_intro _ _ a => a end. Axiom eq_sigT : forall {A : Type} {P : A -> Type} (u v : {a : A & P a}) (p : projT1 u = projT1 v), eq_rect (projT1 u) (fun a : A => P a) (projT2 u) (projT1 v) p = projT2 v -> u = v. Axiom eq_sigT_rect_uncurried : forall {A : Type} {P : A -> Type} {u v : {a : A & P a}} (Q : u = v -> Type), (forall pq : exists p : projT1 u = projT1 v, eq_rect (projT1 u) (fun a : A => P a) (projT2 u) (projT1 v) p = projT2 v, Q (eq_sigT u v (ex_proj1 pq) (ex_proj2 pq))) -> forall p : u = v, Q p. Axiom ex_rect : forall [A : Prop] [P : A -> Prop] (P0 : (exists y, P y) -> Type), (forall (x : A) (p : P x), P0 (ex_intro P x p)) -> forall e : exists y, P y, P0 e. Section inversion_sigma. Unset Implicit Arguments. Context A (B B' : A -> Prop) (C C' : forall a, B a -> Prop) (D : forall a b, C a b -> Prop) (E : forall a b c, D a b c -> Prop) (F : forall a b c d, E a b c d -> Prop) (G : forall a b c d e, F a b c d e -> Prop) (H : forall a b c d e f, G a b c d e f -> Prop) (I : forall a b c d e f g, H a b c d e f g -> Prop). Goal forall (x y : { a : A & { b : { b : B a & C a b } & { d : D a (projT1 b) (projT2 b) & { e : E _ _ _ d & { f : F _ _ _ _ e & { g : G _ _ _ _ _ f & { h : H _ _ _ _ _ _ g & I _ _ _ _ _ _ _ h } } } } } } }) (p : x = y), p = p. intros x y p; repeat match goal with H : sigT _ \|- _ => destruct H end; cbn [projT1 projT2] in . Time induction p as [p] using (@eq_sigT_rect_uncurried _ _ _ _); cbn beta iota delta [proj1_sig proj2_sig proj3_sig projT1 projT2 projT3 ex_proj1 ex_proj2 sig_of_sig2 sigT_of_sigT2 sig_of_sigT sigT_of_sig sig2_of_sigT2 sigT2_of_sig2] in p. Time induction p using ex_rect. ( Finished transaction in 2.99 secs (2.986u,0.003s) (successful) ) Indeed associativity seems to matter, but you should be able to keep adding sigmas on the right here. Strangely enough, if I do cbn beta match delta [projT1 projT2] before the second induction p, then the second induction gets much faster. Is the issue really that unification is slow because it spends a lot of time not realizing it needs to unfold projT1 and projT2? So, there seems to be a problem with the conclusion. The time is linear in the number of occurrences of p in the conclusion. If you replace p = p with a dummy predicate Blah (p, p, p, p ...) you can observe it easily. I still think it's kind-of atrocious that the induction p as [p] using (@eq_sigT_rect_uncurried _ _ _ _) takes .2 seconds, especially since adding match type of p with @existT ?A ?P ?x ?y=@existT _ _ ?x' ?y' => set (Y := y) in ; set (Y' := y') in ; set (PV := P) in end, which takes only 0.075s, drops the time of Time induction p as [p] using (@eq_sigT_rect_uncurried _ _ _ _). down to 0.005s, and then cbn [projT1 projT2] in *; induction p using ex_rect. takes only 0.006s rather than 0.157s Yeah, I think that this is slow because of second-order induction. Induction is trying to find your subterm in the conclusion to abstract it away. What do you mean by second-order induction? it tries to craft a predicate so for this it needs to abstract the stuff you're inducting on the pattern here is a quite big term if you know upfront what this is going to be you should pass it manually and here you do because you've just substituted a variable In general I don't, because I'm writing a general tactic to handle any equality of the form existT _ _ _ = existT _ _ _ which may be mentioned anywhere in the context and the goal But I'm a bit confused. It seems to me the unification problem should have a fast solution If instead of invoking induction I do revert p and then Time refine (@eq_sigT_rect_uncurried _ _ _ _ _ _)., this still takes 0.133s, but the unification problem is unifying something of the shape forall p : ?u = ?v, ?Q p with the goal, which has exactly the same shape modulo beta. Because p is not in the context of ?Q, this falls in the pattern fragment, and solving this should be as fast as just copying the goal. (So long as the goal syntactically has the form forall p : @eq (@sigT _ _) _ _, _, the instantiation is well-typed.) But Time match goal with \|- ?G => pose G end. is instantaneous, and so is cbv beta, so this seems much slower than copying the goal. I am not sure how useful this is but Set Debug Tactic Unification showed more failed unification attempts than my emacs buffer could handle. apply: (ex_rect _ _ p) (using Unicoq because evarconv cannot infer the predicate) doesn't show any failed unification whatsoever and takes a bit less than half the time that induction takes. (It's probably not doing the same thing at all.. I was just curious about what unification was doing here.) @Janno That is interesting, but I'm not sure what to make of it. My claim is that both calls to induction should be instantaneous, and hence it sounds like even the Unicoq one is too slow. My reasoning is that, so long as the hypothesis I'm inducting on is not a section variable mentioned in the body of some definition used in the context or the goal (i.e., so long as revert dependent p succeeds), then both calls to induction are guaranteed to succeed and I can, at least in theory, craft the partial proof term and the resulting goal and context without ever invoking unification. So unification should not be doing anything hard. I agree.. it is a bit baffling. And even the unicoq variant is definitely too slow. (I think unicoq spends a big chunk of its time instantiating evars but I am not sure about that.) I've made a parameterized version at https://github.com/coq/coq/issues/14301#issuecomment-837653213 where you can fill in the nat argument to make refine as slow as you'd like. The refine with holes is over 100x slower than the one where I solve the unification problem manually, despite the fact that the entire problem is in the pattern fragment I suspect we don't have a shortcut for the case what we're abstracting is actually a local variable... Hm, but it seems that the slowness is not pattern but apply (and to a lesser extent beta-iota normalization of intermediate and remaining goals)? Last updated: Oct 13 2024 at 01:02 UTC	{"url":"https://coq.gitlab.io/zulip-archive/stream/237656-Coq-devs-.26-plugin-devs/topic/Why.20induction.20is.20so.20slow.html","timestamp":"2024-11-12T06:55:25Z","content_type":"text/html","content_length":"40776","record_id":"<urn:uuid:0bceba64-c6c5-430d-ac88-2e3db7a95798>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00388.warc.gz"}
s a circle Why is a circle a circle? Why is a circle a circle? A circle is a plane figure bounded by one curved line, and such that all straight lines drawn from a certain point within it to the bounding line, are equal. The bounding line is called its circumference and the point, its centre. What is vertical bar in math? The vertical bar is used as a mathematical symbol in numerous ways: absolute value: , read “the absolute value of x” cardinality: , read “the cardinality of the set S” conditional probability: , reads “the probability of X given Y” determinant: , read “the determinant of the matrix A”. How do I insert a vertical bar in Word? Use a Bar Tab to Add a Vertical Line 1. Select the paragraph where you want to add the vertical line. 2. Go to Ribbon > Home. 3. Click the Tabs button at the bottom of the dialog. 4. In the Tab stop position box, enter the position where you want the vertical line to appear. 5. Click the Bar button in the Alignment section. Are perfect circles possible? No, it is not possible to make a perfect circle. At it’s most fundamental level, the universe has a minimum length of space called the ‘Planck Length’. Any and all circles that exist in nature therefore do not have infinite sides, but a number of sides equal to their circumference split into planck lengths. What does it mean to pass the vertical line test? In mathematics, the vertical line test is a visual way to determine if a curve is a graph of a function or not. A function can only have one output, y, for each unique input, x. If the vertical line you drew intersects the graph more than once for any value of x then the graph is not the graph of a function. Why do we get less sun in winter? During winter, the Northern Hemisphere leans away from the sun, there are fewer daylight hours, and the sun hits us at an angle; this makes it appear lower in the sky. There is less heating because the angled sun’s rays are “spread out” rather than direct. (Shadows are longer because of the lower angle of the sun.) What is the only type of straight line that would fail the vertical line test? Therefore, the vertical line test concludes that a curve in the plane represents the graph of a function if and only if no vertical line intersects it more than once. A plane curve which doesn’t represent the graph of a function is sometimes said to have failed the vertical line test. How close is the Sun to Earth right now? What happens to Earth’s orbit every 100 000 years? Currently, Earth’s eccentricity is near its least elliptic (most circular) and is very slowly decreasing, in a cycle that spans about 100,000 years. The total change in global annual insolation due to the eccentricity cycle is very small. How do you type a vertical bar? “\|”, How can I type it? 1. Shift-\ (“backslash”). 2. German keyboard it is on the left together with < and > and the Alt Gr modifier key must be pressed to get the pipe. 3. Note that depending on the font used, this vertical bar can be displayed as a consecutive line or by a line with a small gap in the middle. Is there a perfect circle? Perfect circles do not exist in nature, but you can see some close approximations around CMU’s main campus in Pittsburgh. What is the horizontal line test used for? In mathematics, the horizontal line test is a test used to determine whether a function is injective (i.e., one-to-one). Is a circle ever a function? Now, you ask whether a circle is a function, and the answer is clearly no, since a circle is typically a set of points. What does a circle symbolize in the Bible? It represents the notions of totality, wholeness, original perfection, the Self, the infinite, eternity, timelessness, all cyclic movement, God (‘God is a circle whose centre is everywhere and whose circumference is nowhere’ (Hermes Trismegistus)). Circle of Necessity: birth, growth, decline, death. What is the symbol for vertical? The vertical line, also called the vertical slash or upright slash ( \| ), is used in mathematical notation in place of the expression “such that” or “it is true that.” This symbol is commonly encountered in statements involving logic and sets. Also see Mathematical Symbols. How do you tell if a circle is a function? If you are looking at a function that describes a set of points in Cartesian space by mapping each x-coordinate to a y-coordinate, then a circle cannot be described by a function because it fails what is known in High School as the vertical line test. A function, by definition, has a unique output for every input. Is vertical up and down? Vertical describes something that rises straight up from a horizontal line or plane. The terms vertical and horizontal often describe directions: a vertical line goes up and down, and a horizontal line goes across. You can remember which direction is vertical by the letter, “v,” which points down. Why is Earth closest to the sun in January? Answer. Because the earth’s axis is tilted. In fact, the Earth is farthest from the sun in July and is closest to the sun in January! During the summer, the sun’s rays hit the Earth at a steep angle. Is every non vertical line a function? No, every straight line is not a graph of a function. Nearly all linear equations are functions because they pass the vertical line test. The exceptions are relations that fail the vertical line What do 2 vertical lines mean? Every number on the number line also has an absolute value, which simply means how far that number is from zero. The symbol for absolute value is two vertical lines. Since opposites are the same distance from the origin, they have the same absolute value. Is a vertical line a function? For a relation to be a function, use the Vertical Line Test: Draw a vertical line anywhere on the graph, and if it never hits the graph more than once, it is a function. If your vertical line hits twice or more, it’s not a function. Is a vertical line a linear relationship? A linear function is a function whose graph is a straight line. The line can’t be vertical, since then we wouldn’t have a function, but any other sort of straight line is fine. This graph shows a vertical line, which isn’t a function. This graph shows two lines, rather than one straight line. What letter will pass the vertical line test? Can a human draw a perfect circle? You can not draw a circle, you can only draw a picture of one. There are no “real” circles in nature. That is to say, an actual circle is an idea rather than a physical thing. According to Plato, the idea of a perfect circle is the Form of a circle,[1] which is to say, it’s a representation of a perfect circle. What are the 8 parts of a circle? The following figures show the different parts of a circle: tangent, chord, radius, diameter, minor arc, major arc, minor segment, major segment, minor sector, major sector. Why does vertical line test work? The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. Is the earth a perfect shape? Even though our planet is a sphere, it is not a perfect sphere. Because of the force caused when Earth rotates, the North and South Poles are slightly flat. Earth’s rotation, wobbly motion and other forces are making the planet change shape very slowly, but it is still round.	{"url":"https://lynniezulu.com/why-is-a-circle-a-circle/","timestamp":"2024-11-13T09:07:23Z","content_type":"text/html","content_length":"52037","record_id":"<urn:uuid:3bed5407-7d75-4ae0-be98-a33728033e28>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00613.warc.gz"}
Really Fast Tether Cars-How Long a Tether if the Car is Going the Speed of Light? I’m holding on to the tether for one of these really fast tether cars. Mine’s so fast, it goes the speed of light (ignore the physical problems with this; maybe my car’s composed of a single photon). How long must the tether be so that my revolution only takes, say, 2 seconds? I’m holding my end of the tether at arm’s length, about three feet. You want the circumference of your circle to be two light-seconds. Dividing by 2pi, we find that the radius is over 95 million meters. I think we can safely ignore the length of your arm. The strength of your arm, however, might be of some importance. My first thought was that it was that easy (the radius of a 2 light-second circle) but for some reason I got hung up on the length of my arm, and ended up asking here. So the tether’d have to be about 59,300 miles long. Thanks! Except, the energy required to rotate the tether goes up as the speed increases. Plus, the apparent mass gets heavier. Plus, at relativistic speeds there’s no such thing as a straight rope. Basically, the factor is IIRC 1/SQRT(1-(v^2)/(c^2)) so as V increases, it becomes 1 over a smaller and smaller number, until at C (which you’ll never reach) energy required is like 1/0 (work it out - 1, 2, 3, 4, …1000, …1000000, etc. The closer V gets to C, the huger the factor by which apparent weight goes up, the huger the energy needed to accelerate a bit more, etc. Plain old Newton a=vt and F=ma stop working at those speeds, they’re approximations that work well with speed dencetly below c. Making the photon change direction to swing in a circle is going to take energy, so it is probably going to start spitting out gamma radiation (which you are going to supply the energy for via the tether). Better wear lead lined underpants. So lets be somewhat more practical - outside of the event horizon of a black hole is a photon sphere, a zone that allows suitable photons to orbit the black hole forever (or at least until the black hole evaporates). So how overweight do you have to be to allow photons to orbit you at 53,900 miles? And lest you think I am joking, they have created artificial black hole analogues using refractive materials to illustrate the concept Wild. Tether cars and black holes. I love the Dope. So much force must our hammer-thrower exert to keep the photon in spin at the end of the tether? It depends on his/the black hole’s spin relative to the photon’s right? If he had 32,000 Solar Masses and were spinning at near c, it would be require less force to whip it, right? High school physics problem formulation required ASAP, please… That’s way more complicated than it needs to be. Just make a big hoop with a mirrored inner surface, and let the photon bounce around inside that. No gamma rays, no gamma factor. I’m not going near the relativistic implications of a rapidly spinning black hole. The OP specified a 30rpm (2 light second diameter) spin rate, is all. To be honest, my math is pretty much a ballpark guess - there is a bunch of relativity (frame dragging and space-time distortion) to account for, and I didn’t do any of that. And concepts like force have no real meaning to a photon, because it has no mass. Which is why we end up talking about curved space-time, cause it avoids all those nasty divide-by-zero/NaN/infinity issues. The amount of force required depends on the wavelength The concept of the gravitational force on a photon AFAIC see has zero practical meaning. However the spin of the black hole does affect where the circular orbits of photons exist (it’s worth pointing out though that I believe all orbits of photons around a black hole are unstable and certainly the circular ones). For a (non-rotating) Schwarzschild black hole there exists circular orbits in all planes at 1.5 times the Schwarzschild radius, but for a (rotating) Kerr black hole there exists two circular orbits for photons, both in the equatorial plane: one between 0.5-1.5 times (depending on the BH’s angular momentum parameter) the Schwarzschild radius for photons orbiting in the same direction as the black hole’s rotation and one at 1.5-2 times (ditto) the Schwarzschild radius for photons orbiting in the opposite direction to the black hole’s rotation. A near-extremal Kerr black hole (one which is spinning so fast it is almost a naked singularity) will have the orbits at just above 0.5 times the Schwarzschild radius and just below 2 times the Schwarzschild radius. on the off chance that it’s not clear, the car in the video is tethered to the central pole, and not the man.	{"url":"https://boards.straightdope.com/t/really-fast-tether-cars-how-long-a-tether-if-the-car-is-going-the-speed-of-light/670825","timestamp":"2024-11-03T19:18:53Z","content_type":"text/html","content_length":"48772","record_id":"<urn:uuid:b8a37083-5bf7-4a2e-a541-e91bccf287dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00839.warc.gz"}
Beat the Streak: Day Eight In this blog post, we will explore three factors that influence the probability of correctly selecting a player to get a hit on a given day. These are: 1. Individual batter strength, as measured by the proportion of plate appearances that resulted in a hit. 2. Team offensive strength, as measured by the average number of plate appearances per game by the batting team. 3. The position in the batting order. We plot the distribution of these statistics over (batter, year) pairs and (team, year) pairs. The plots below reveal that the best batters get a hit in about 30% of plate appearances, and the strongest offensive teams average 39 plate appearances per game. The tables below show the top-performing batters and teams: ┌────────────────┬────┬─────┐ ┌───────────┬────┬──────┐ │ batter │year│ │ │batter_team│year│ │ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Josh Hamilton │2010│0.326│ │ Mets │2011│39.000│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Trea Turner │2016│0.324│ │ Yankees │2017│39.006│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Jose Altuve │2014│0.319│ │ Braves │2018│39.051│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Daniel Murphy │2016│0.316│ │ Red Sox │2010│39.093│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Melky Cabrera │2012│0.315│ │ Reds │2018│39.143│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Dee Gordon │2015│0.315│ │ Red Sox │2013│39.253│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Hanley Ramirez │2013│0.312│ │ Yankees │2010│39.253│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │ Michael Young │2011│0.310│ │ Tigers │2013│39.278│ ├────────────────┼────┼─────┤ ├───────────┼────┼──────┤ │Carlos Gonzalez │2010│0.310│ │ Red Sox │2011│39.389│ └────────────────┴────┴─────┘ └───────────┴────┴──────┘ Note that the data I am working with currently covers years 2010 - 2018, which is somewhat stale, but should be sufficient for the purposes of understanding these general factors. For an individual team, we can effectively model the number of plate appearances with a negative binomial distribution. This is a distribution that counts the number of successes (i.e., non-outs) before a pre-specified number of failures (i.e., outs). We set the number of failures, $r$ to be $27$. Note that this model is simply an approximation to the true distribution. In home games, $24$ outs might be sufficient, and in extra-innings games more would be needed. Moreover, this model does not account for the possibility of double and triple plays, nor the fact that different players in the lineup have difference chances of success. Nevertheless, it serves as a simple and reasonable model for this quantity. We can fit the model to data using the moment matching method. Doing this on the 2016 Red Sox data yields the following plot: This demonstrates that the fit is reasonable. Now, we will analytically try to determine the probability that an individual player will get a hit as a function of the three factors mentioned at the beginning of this post. Let's assume that each player has a unique single-plate-appearance hit probability, which we will denote $p$. Now if that players has $n$ plate appearances, the probability that they will get a hit in any one plate appearance is simply $ 1 - (1 - p)^n $. However, the number of plate appearances for an individual player is a random quantity that depends on the number of plate appearances for the team. Let $N$ denote the number of plate appearances for the team, which we assumed was sampled from a negative binomial distribution. Then we have: $$ n = \lfloor 1 + \frac{N - order}{9} \rfloor $$ Here, "order" is the position in the lineup of the player. We can calculate the probability of each $n$ using the probability mass function of the negative binomial, to estimate the overall probability of a hit: $$ \sum_{N=27}^{\infty} f(N) (1 - (1 - p)^{ \lfloor 1 + \frac{N - order}{9} \rfloor }) $$ $f(N)$ above denotes the PMF of the negative binomial. Plotting this function for a set of reasonable inputs yields the following contour plot: This plot was constructed under the assumption of a lead-off batter. The color indicates the probability of at least one hit in the game, whereas the x-axis indicates the probability of a hit in a single plate appearance. This plot shows that, roughly speaking, picking a batter with 1% better chance of a hit in one plate appearance is equivalent to picking a team that gets 2 more plate appearances on average. This plot was for the lead-off batter. The plot below shows what the probability of a hit is for batters in different positions in the lineup, assuming $0.3$ hit probability per plate appearances and an average of 38 team plate appearances under the negative binomial model. This plot demonstrates that, all else equal, each position in the lineup costs about 0.8% to the probability of at least one hit. This blog post suggests that an 80% success rate in BTS should be attainable, by carefully considering the batter, the team, and the position in the lineup. For some reason that number has been elusive to me so far in my efforts to beat the streak. Even though as we saw above that some hitters were able to get hits in 30% of plate appearances in a given season, it's entirely possible that this impressive success rate was due in part by luck, and the true success rate (which is unknown) could be below 30%. This could partially explain why an 80% success rate has been so hard to achieve in BTS.	{"url":"http://www.ryanhmckenna.com/2022/01/beat-stt.html","timestamp":"2024-11-06T18:48:19Z","content_type":"application/xhtml+xml","content_length":"121423","record_id":"<urn:uuid:d063d2ae-bcf2-4bea-a53a-601af1cd9471>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00377.warc.gz"}
Discorsi e dimostrazioni matematiche : intorno à due nuoue scienze, attenenti alla mecanica & i movimenti locali del signor Galileo Galilei Linceo, filosofoe matematico primario del serenissimo grand duca di Toscana, Con una appendice del centro di grauità d’alcuni solidi - Rare & Special e-Zone Rare Books on History of Science Discorsi e dimostrazioni matematiche : intorno à due nuoue scienze, attenenti alla mecanica & i movimenti locali del signor Galileo Galilei Linceo, filosofoe matematico primario del serenissimo grand duca di Toscana, Con una appendice del centro di grauità d'alcuni solidi Galileo Galilei BOOK Appresso gli Elsevirii, 1638 306 (i.e. 314) p. : ill. ; 21 cm First edition of Galileo's last and greatest work; it is the first modern textbook of physics and the foundation of modern mechanics. "The two sciences with which the book principally deals are the engineering science of strength of materials and the mathematical science of kinematics… Galileo's Two New Sciences underlies modern physics not only because it contains the elements of the mathematical treatment of motion, but also because most of the problems that came rather quickly to be seen as problems amenable to physical experiment and mathematical analysis were gathered together in this book with suggestive discussions of their possible solution." --Dictionary of Scientific Biography , V, p.245. "Mathematicians and physicists of the later seventeenth century, Isaac Newton among the...[ Read more First edition of Galileo's last and greatest work; it is the first modern textbook of physics and the foundation of modern mechanics. "The two sciences with which the book principally deals are the engineering science of strength of materials and the mathematical science of kinematics… Galileo's Two New Sciences underlies modern physics not only because it contains the elements of the mathematical treatment of motion, but also because most of the problems that came rather quickly to be seen as problems amenable to physical experiment and mathematical analysis were gathered together in this book with suggestive discussions of their possible solution." --Dictionary of Scientific Biography , V, p.245. "Mathematicians and physicists of the later seventeenth century, Isaac Newton among them, rightly supposed that Galileo had begun a new era in the science of mechanics. It was upon his foundation that Huyens, Newton and others were able to erect the frame of the science of dynamics, and to extend its range to the heavenly bodies." --Printing And The Mind Of Man Close Viewer • Rare Books on History of Science • Galilei, Galileo, 1564-1642 • Mathematics Additional titles • Dialogues concerning two new sciences • 關於兩門新科學的對話 • Title vignette (printer's device) • Numerous errors in paging • Italian Call number • QA33 .G28 1638 • 10.14711/spcol/b523121 Permanent URL for this record: https://lbezone.hkust.edu.hk/bib/b523121	{"url":"https://lbezone.hkust.edu.hk/bib/b523121","timestamp":"2024-11-12T00:56:06Z","content_type":"text/html","content_length":"57509","record_id":"<urn:uuid:1ef153f2-e02e-4fb4-b317-4880fb93edef>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00310.warc.gz"}
operating speed of ball mill formula Typically R = 8. Rod Mill Charge: Typically 45% of internal volume; 35% 65% range. Bed porosity typically 40%. Height of bed measured in the same way as ball mills. Bulk density of rods = tons/ m3. In wet grinding, the solids concentration 1s typically 60% 75% by mass. A rod in situ and a cutaway of a rod mill interior. WhatsApp: +86 18838072829 For R = 1000 mm and r = 50 mm, = rpm But the mill is operated at a speed of 15 rpm. Therefore, the mill is operated at 100 x 15/ = % of critical speed. If 100 mm dia balls are replaced by 50 mm dia balls, and the other conditions are remaining the same, Speed of ball mill = [/ (2π)] x [/ (1 )] = rpm WhatsApp: +86 18838072829 How to Measure Grinding Efficiency. The first two Grinding Efficiency Measurement examples are given to show how to calculate Wio and Wioc for single stage ball mills. Figure 1. The first example is a comparison of two parallel mills from a daily operating report. Mill size x (′ x 20′ with a ID of 16′). WhatsApp: +86 18838072829 Ball Mill Power/Design Price Example #2 In Example this was determined that adenine 1400 HP wet grinder ball mill was required to grind 100 TPH of matter with an Bond Works Catalog of 15 ( guess that mineral type it is ) from 80% passing ¼ inch to 80% passing 100 mesh in closed circuit. WhatsApp: +86 18838072829 In total, 165 scenarios were simulated. When the mills charge comprising 60% of small balls and 40% of big balls, mill speed has the greatest influence on power consumption. When the mill charge is more homogeneous size, the effect of ball segregation is less and so the power consumption of the mill will be less affected. WhatsApp: +86 18838072829 There is a specific operating speed for most efficient . ... to determine the critical speed for any given size Mill, we use the following formula: divided by the square root of the radius ... WhatsApp: +86 18838072829 You've already forked sbm 0 Code Issues Pull Requests Packages Projects Releases Wiki Activity WhatsApp: +86 18838072829 An increase of over 10% in mill throughput was achieved by removing the ball scats from a single stage SAG mill. These scats are non spherical ball fragments resulting from uneven wear of balls WhatsApp: +86 18838072829 The critical speed of ball mill is given by, where R = radius of ball mill; r = radius of ball. But the mill is operated at a speed of 15 rpm. Therefore, the mill is operated at 100 x 15/ = % of critical speed. WhatsApp: +86 18838072829 Development of operation strategies for variable speed ball mills. Creator. Liu, Sijia. Publisher. University of British Columbia. Date Issued. 2018. Description. Mineral processing productivity relates to a range of operating parameters, including production rate, product grind size, and energy efficiency. WhatsApp: +86 18838072829 The invention belongs to the technical field of mineral processing, and particularly relates to a ball mill power calculation method, which is characterized by applying the following formula (9) as shown in the figure to obtain ball mill power, wherein in the formula, psi means media rotating speed (%); phi means media filling rate (%); delta means media loose density (t/m<3>); D means ball ... WhatsApp: +86 18838072829 The details of the ball mill motor are as follows. Power = kW or HP and the speed is 343 rpm. Load calculations (prior to failure analysis) The ball mill can experience failure based on the maximum normal stress theory as the working loads acting in the ball mill is concentrated across the seam of the mill periphery. WhatsApp: +86 18838072829 Figures in this table are generally slightly low compared to some reported plant data or other manufacturers' estimates, and are based on an empirical formula initially proposed by Bond (1961) designed to cover a wide range of mill dimensions, and the normal operating range of mill load (Vp = .35 to .50) and speed (Cs = .50 to .80). WhatsApp: +86 18838072829 Critical speed formula of ball mill. Nc = 1/2π √g/R r The operating speed/optimum speed of the ball mill is between 50 and 75% of the critical speed. Also Read: Hammer Mill Construction and Wroking Principal. Take these Notes is, Orginal Sources: Unit OperationsII, KA Gavhane WhatsApp: +86 18838072829 The formula for calculating critical mill of speed: N c = / √ (D d) Where: N c = Critical Speed of Mill D = Mill Diameter d = Diameter of Balls Let's solve an example; Find the critical speed of mill when the mill diameter is 12 and the diameter of balls is 6. This implies that; D = Mill Diameter = 12 d = Diameter of Balls = 6 WhatsApp: +86 18838072829 Show that the critical speed of rotation for a ball mill, defined as the speed required to take the balls just to the apex of revolution is equal to (g/a) 1/2 /2π revolutions per second, where a is the radius of the mill and g is the... WhatsApp: +86 18838072829 The video contain definition, concept of Critical speed of ball mill and step wise derivation of mathematical expression for determining critical speed of b... WhatsApp: +86 18838072829 Current operational results show that the SAG mill is operating at 5% ball charge level by volume and is delivering a K80 800 micron product as predicted. Power drawn at the pinion is 448 kW, kWh /tonne (SAG mill) and 570 kW, kWh/tonne (ball mill) when processing mtph, for a total of kWh/tonne or % above the ... WhatsApp: +86 18838072829 The idea of using a mixture of balls and pebbles, at a mill speed suitable for ballmilling, was revisited in this investigation, using a normal spectrum of pebble sizes (1975 mm). Batch tests in a pilotscale mill ( m diameter) were used to compare ballmilling to various ball/pebble mixtures. WhatsApp: +86 18838072829 The ultimate crystalline size of graphite, estimated by the Raman intensity ratio, of nm for the agate ballmill is smaller than that of nm for the stainless ballmill, while the milling ... WhatsApp: +86 18838072829 However, in combination with traditional optimization methods, ball mill grinding speed can be used to control energy input and offset the influences of ore variability. Optimum ball mill operating conditions can be determined based on circuit design and operating dynamics for any given runofmine ore. WhatsApp: +86 18838072829 Based on the calculations, the estimated mill capacity is 8,366,400 tonnes per year and the estimated mill shaft power is 1901 kW for a x wet ball mill operating in closed circuit with a 20:1 reduction ratio, 80% of the critical speed, and a bed porosity of 40%. WhatsApp: +86 18838072829 Mill Speed. Speed of ball mill is expressed as percentage of critical speed. Critical speed is the speed at which the centrifugal force is high enough that all media sticks to mill wall during rotation of the mill. Normal operating speed is about 75 % of critical speed WhatsApp: +86 18838072829 Figure 5. Particle size distribution of the ball mill feed and ball mill product with respect to feed rate Downloaded by [Dr Ahmad Hassanzadeh] at 04:04 15 October 2017 WhatsApp: +86 18838072829 e. Rotation speed of the cylinder. Several types of ball mills exist. They differ to an extent in their operating principle. They also differ in their maximum capacity of the milling vessel, ranging from liters for planetary ball mills, mixer mills, or vibration ball mills to several 100 liters for horizontal rolling ball mills. WhatsApp: +86 18838072829 The Formula derivation ends up as follow: Critical Speed is: Nc = () where: Nc is the critical speed,in revolutions per minute, D is the mill effective inside diameter, in feet. Example: a mill measuring 11'0" diameter inside of new shell liners operates at rpm. Critical speed is = / 11^ = rpm WhatsApp: +86 18838072829 Operating Speed of Rotation of a Ball MillIntroductionA ball mill is a cylindrical device used to grind or mix materials like ores, chemicals, ceramic raw materials, and paints. It rotates around a horizontal axis and partially filled with the material to be ground plus the grinding medium (balls). The diameter of the ball mill and the size of the balls can significantly affect the grinding ... WhatsApp: +86 18838072829 Ball Milling Theory Introduction: Figure 1: Ball milling terminology. I was first given the formula for gunpowder by my Uncle at age 14, after he had observed my apparent obsession with class C fireworks. Being a scientist who had experimented with the ancient recipe himself during his youth, he thought I should try making my own fireworks. WhatsApp: +86 18838072829 In recent research done by AmanNejad and Barani [93] using DEM to investigate the effect of ball size distribution on ball milling, charging the mill speed with 40% small balls and 60% big balls WhatsApp: +86 18838072829 Crushed ore is fed to the ball mill through the inlet; a scoop (small screw conveyor) ensures the feed is constant. For both wet and dry ball mills, the ball mill is charged to approximately 33% with balls (range 3045%). Pulp (crushed ore and water) fills another 15% of the drum's volume so that the total volume of the drum is 50% charged. WhatsApp: +86 18838072829 The formula for critical speed is CS = 1/2π √ (g/ (Rr) where g is the gravitational constant, R is the inside diameter of the mill and r is the diameter of one piece of media. This reduced to CS = /√ (Rr). Dry mills typically operate in the range of 50%70% of CS and most often between 60%65% of CS. WhatsApp: +86 18838072829 Critical speed on a ball mill Manufacturer Of Highend . Ball Mill Critical Speed Mill Grinding Cement. This paper presents a parison of the breakage parameters with fraction of mill critical speed under the standard conditions in a small laboratory ball mill of clinker and limestone samples, which are ground at the condition 70% of critical speed of cement ball mill in Glta cement factory ... 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Mean Absolute Error: A Comprehensive Guide - Navigating Complexity with Precision and Grace Mean Absolute Error: A Comprehensive Guide Jessy 08 mins The Mean Absolute Error (MAE) is a fundamental metric in the world of statistics and data science. It quantifies the average magnitude of errors between predicted and actual values, providing a clear measure of accuracy. In this blog post, we will explore the concept of Mean Absolute Error, its significance, calculation methods, and applications across various domains. What is Mean Absolute Error? Mean Absolute Error is a statistical measure that evaluates the average absolute differences between predicted values and actual observations. Unlike other metrics that can be influenced by the direction of errors, MAE focuses solely on the magnitude of errors, making it a robust measure of accuracy. Importance of Mean Absolute Error The importance of Mean Absolute Error lies in its simplicity and interpretability. By providing a straightforward average of errors, MAE helps analysts understand how well their models are performing. This makes it an essential tool for model validation and comparison in various fields, including finance, healthcare, and machine learning. How to Calculate Mean Absolute Error Calculating Mean Absolute Error is straightforward. The formula for MAE is: MAE=1n∑i=1n∣yi−y^i∣\text{MAE} = \frac{1}{n} \sum_{i=1}^{n} \| y_i – \hat{y}_i \| Here, yiy_i represents the actual values, y^i\hat{y}_i represents the predicted values, and nn is the number of observations. This formula ensures that all errors are treated equally, providing an unbiased measure of model accuracy. Advantages of Mean Absolute Error One of the key advantages of Mean Absolute Error is its interpretability. Unlike other metrics such as Mean Squared Error (MSE), MAE is expressed in the same units as the data, making it easier to understand. Additionally, MAE is less sensitive to outliers compared to MSE, which squares the errors, amplifying the impact of large deviations. Applications of Mean Absolute Error in Machine Learning In machine learning, Mean Absolute Error is widely used to evaluate regression models. It helps in assessing the accuracy of predictive models by providing a clear measure of how close predictions are to actual outcomes. Whether in predicting housing prices or stock market trends, MAE serves as a reliable metric for model performance. Mean Absolute Error vs. Other Error Metrics When comparing Mean Absolute Error with other error metrics like Mean Squared Error (MSE) or Root Mean Squared Error (RMSE), it’s essential to consider the context. While MSE and RMSE penalize larger errors more heavily due to squaring, MAE treats all errors equally. This makes MAE a preferred choice in scenarios where the magnitude of errors matters more than their direction. Limitations of Mean Absolute Error Despite its advantages, Mean Absolute Error has some limitations. One of the primary drawbacks is that it does not provide information about the direction of errors. This can be a disadvantage in scenarios where understanding whether predictions are consistently over or underestimating is crucial. Additionally, MAE may not be as sensitive to large errors as MSE. Improving Model Accuracy with Mean Absolute Error To improve model accuracy using Mean Absolute Error, it is essential to focus on reducing the magnitude of prediction errors. Techniques such as cross-validation, hyperparameter tuning, and feature engineering can help enhance model performance. By continuously monitoring MAE, analysts can make informed decisions to refine their models. Real-World Examples of Mean Absolute Error Mean Absolute Error is applied across various real-world scenarios. In finance, it helps in evaluating the accuracy of economic forecasts. In healthcare, MAE is used to assess predictive models for patient outcomes. These examples highlight the versatility and importance of MAE in different industries. Mean Absolute Error is a crucial metric for evaluating model accuracy. Its simplicity, interpretability, and robustness make it a valuable tool in data analysis and machine learning. By understanding and effectively utilizing MAE, analysts can enhance their models and make more accurate predictions. 1. What is the difference between Mean Absolute Error and Mean Squared Error? Mean Absolute Error measures the average absolute differences between predicted and actual values, while Mean Squared Error squares these differences before averaging. This makes MSE more sensitive to large errors compared to MAE. 2. Why is Mean Absolute Error important in machine learning? Mean Absolute Error is important in machine learning because it provides a clear measure of how close predictions are to actual outcomes, helping to evaluate and improve model performance. 3. Can Mean Absolute Error be used for classification problems? Mean Absolute Error is typically used for regression problems. For classification problems, metrics like accuracy, precision, and recall are more appropriate. 4. How can I reduce Mean Absolute Error in my models? Reducing Mean Absolute Error involves techniques such as cross-validation, hyperparameter tuning, and feature engineering to improve model accuracy and minimize prediction errors. 5. Is Mean Absolute Error sensitive to outliers? Mean Absolute Error is less sensitive to outliers compared to Mean Squared Error, as it does not square the errors, thus reducing the impact of large Leave a Reply Cancel reply	{"url":"https://errordomain.net/mean-absolute-error-a-comprehensive-guide/","timestamp":"2024-11-09T02:54:28Z","content_type":"text/html","content_length":"73482","record_id":"<urn:uuid:0e416a37-8f7c-4de3-898b-5d8016a4d8ed>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00291.warc.gz"}
Cluster Sampl How to Estimate a Population Total from a Cluster Sample This lesson describes how to estimate a population total, given survey data from a cluster sample. A good analysis should provide two outputs: First, we describe how to conduct a good analysis step-by-step. Then, we will illustrate the analysis with a sample problem. How to Analyze Survey Data A good analysis of survey data from a cluster sample includes eight steps: • Estimate a population parameter (in this case, the population total). • Compute sample variance within each cluster. • Compute sample variance between each cluster. • Compute standard error. • Specify a confidence level. • Find the critical value (often a z-score or a t-score). • Compute margin of error. • Define confidence interval. Let's look a little bit closer at each step - what we do in each step and why we do it. When you understand what is really going on, it will be easier for you to apply formulas correctly and to interpret analytical findings. Note: The formulas presented below are only appropriate for cluster sampling. Estimating a Population Total The first step in the analysis is to develop a point estimate for the population total. Before we can accomplish this objective, we need to compute a mean score or a proportion for each sampled Use this formula to compute the sample means: Sample mean in cluster h = x[h] = Σx[h] / m[h] where Σx[h] is the sum of all the sample observations in cluster h, and m[h] is the number of sample observations in cluster h. Once we know the sample mean in each cluster, we can estimate the population total (t) from the following formula: Population total = t = N/n * ΣM[h] * x[h] where N is the number of clusters in the population, n is the number of clusters in the sample, M[h] is the number of observations in the population from cluster h, and x[h] is the sample mean from cluster h. A proportion is a special case of the mean. It represents the number of observations that have a particular attribute divided by the total number of observations in the group. Use this formula to compute the proportion for each sampled cluster: p[h] = m'[h] / m[h] where p[h] is a sample estimate of the population proportion for cluster h, m'[h] is the number of sample observations from cluster h that have the attribute, and m[h] is the total number of sample observations from cluster h. Once we have estimated a sample proportion for each cluster, we can estimate a population total: Population total = t = N/n * ΣM[h] * p[h] where t is an estimate of the number of elements in the population that have a specified attribute, N is the number of clusters in the population, n is the number of clusters in the sample, M[h] is the number of observations from cluster h in the population, and p[h] is the sample proportion from cluster h. Because different samples can produce different point estimates, you can be fairly sure that the estimate from your sample does not equal the true value of the population parameter exactly. Therefore, you need a way to express the uncertainty inherent in your estimate. The remaining six steps in the analysis are geared toward quantifying the uncertainty in your estimate. Computing Variance Within Clusters If you are using one-stage cluster sampling, you can skip this step. But if you are using two-stage cluster sampling, you will need to compute the variance within each sampled cluster. For a mean score, the variance within each cluster can be estimated from sample data as: s^2[h] = Σ ( x[i][h] - x[h] )^2 / ( m[h] - 1 ) where s^2[h] is a sample estimate of population variance in cluster h, x[i][h] is the value of the ith element from cluster h, x[h] is the sample mean from cluster h, and m[h] is the number of observations sampled from cluster h. For a proportion, the variance within each cluster can be estimated as: s^2[h] = [ m[h] / (m[h] - 1) ] * p[h] * (1 - p[h]) where s^2[h] is a sample estimate of the variance within cluster h, m[h] is the number of observations sampled from cluster h, and p[h] is a sample estimate of the proportion in cluster h. Computing Variance Between Clusters Use the following formula to estimate the variance of total scores between sampled clusters (s^2[b]): s^2[b] = Σ ( t[h] - t/N )^2 / ( n - 1 ) where s^2[b] is a sample estimate of the variance between sampled clusters, t[h] is the total from cluster h, t is the sample estimate of the population total, N is the number of clusters in the population, and n is the number of clusters in the sample. Note: If you are working with proportions, t[h] is: t[h] = M[h] * p[h] where M[h] is the number of population elements in cluster h, and p[h] is the observed proportion in cluster h. Computing Standard Error The standard error is possibly the most important output from our analysis. It allows us to compute the margin of error and the confidence interval. When we estimate a population total from a cluster sample, the standard error (SE) of the estimate is: SE = N * sqrt { [ ( 1 - n/N ) / n ] * s^2[b]/n + N/n * Σ ( 1 - m[h]/M[h] ) * M^2[h] * s^2[h]/m[h] ) } where N is the number of clusters in the population, n is the number of clusters in the sample, s^2[b] is a sample estimate of the variance between clusters, m[h] is the number of elements from cluster h in the sample, M[h] is the number of elements from cluster h in the population, and s^2[h] is a sample estimate of the population variance in cluster h. With one-stage cluster sampling, the formula for the standard error reduces to: SE = N * sqrt { [ ( 1 - n/N ) / n ] * s^2[b]/n } Think of the standard error as the standard deviation of a sample statistic. In survey sampling, there are usually many different subsets of the population that we might choose for analysis. Each different sample might produce a different estimate of the value of a population parameter. The standard error provides a quantitative measure of the variability of those estimates. Specifying Confidence Level In survey sampling, different samples can be randomly selected from the same population; and each sample can often produce a different confidence interval. Some confidence intervals include the true population parameter; others do not. A confidence level refers to the percentage of all possible samples that produce confidence intervals that include the true population parameter. For example, suppose all possible samples were selected from the same population, and a confidence interval were computed for each sample. A 95% confidence level implies that 95% of the confidence intervals would include the true population As part of the analysis, survey researchers choose a confidence level. Probably, the most frequently chosen confidence level is 95%. Finding Critical Value Often expressed as a t-score or a z-score, the critical value is a factor used to compute the margin of error. To find the critical value, follow these steps: • Compute alpha (α): α = 1 - (confidence level / 100) • Find the critical probability (p): p = 1 - α/2 • To express the critical value as a z-score, find the z-score having a cumulative probability equal to the critical probability (p). • To express the critical value as a t-score, follow these steps: □ Find the degrees of freedom (df). To compute degrees of freedom for a cluster sample, use this equation: df = Σ ( m[h] - 1 ) where m[h] is the number of sample observations from cluster h. □ The critical t-score is the t statistic having degrees of freedom equal to df and a cumulative probability equal to the critical probability (p). Researchers use a t-score when sample size is small; a z-score when it is large (at least 30). You can use the Normal Distribution Calculator to find the critical z-score, and the t Distribution Calculator to find the critical t statistic. Computing Margin of Error The margin of error expresses the maximum expected difference between the true population parameter and a sample estimate of that parameter. Here is the formula for computing margin of error (ME): ME = SE * CV where SE is standard error, and CV is the critical value. Defining Confidence Interval Statisticians use a confidence interval to express the degree of uncertainty associated with a sample statistic. A confidence interval is an interval estimate combined with a probability statement. Here is how to compute the minimum and maximum values for a confidence interval. Mean Proportion CI[min] = x - SE * CV CI[min] = p - SE * CV CI[max] = x + SE * CV CI[max] = p + SE * CV In the table above, x is the sample estimate of the population mean, p is the sample estimate of the population proportion, SE is the standard error, and CV is the critical value (either a z-score or a t-score). And, the confidence interval is an interval estimate that ranges between CI[min] and CI[max]. Sample Problem This section presents a sample problem that illustrates how to estimate a population total when the sampling method is one-stage cluster sampling. Sample Size Calculator The analysis of data collected via cluster sampling can be complex and time-consuming. Stat Trek's Sample Size Calculator can help. The calculator computes standard error, margin of error, and confidence intervals. It assesses sample size requirements, estimates population parameters, and tests hypotheses. The calculator is free. You can find the Sample Size Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below. Sample Size Calculator Example 1 A botanist divides a field into 1000 equal-size plots. In each plot, he plants 100 clover seeds; and ultimately, each seed sprouts. The botanist randomly selects 20 plots and counts the number of four-leaf clovers in each sampled plot. His findings appear below: 0, 1, 1, 2, 2, 2, 3, 3, 3, 3 3, 3, 3, 3, 4, 4, 4, 4, 4, 8 Using sample data, estimate the total number of four-leaf clovers in the field. Find the margin of error and the confidence interval. Assume a 95% confidence level. Solution: To solve this problem, we follow the seven-step process described above. • Estimate the population total. Before we can estimate the population total, we need to first estimate the sample mean for each cluster. The formula for a cluster mean is: Sample mean in cluster h = x[h] = Σx[h] / m[h] where Σx[h] is the sum of all the sample observations in cluster h, and m[h] is the number of sample observations in cluster h. Using the above formula, we can compute a sample mean for each of the 20 sampled plots: Mean[plot 1] = x[1] = Σx[1] / n[1] = 0/100 = 0 Mean[plot 2] = x[2] = Σx[2] / n[2] = 1/100 = 0.01 . . . Mean[plot 19] = x[19] = Σx[19] / n[19] = 4/100 = 0.04 Mean[plot 20] = x[20] = Σx[20] / n[20] = 8/100 = 0.08 Given the sample means within strata, we can estimate the population total (t) from the following formula: t = N/n * ΣM[h] * x[h] t = 1000/20 * ( 100 * 0 + 100 * 0.01 + ... + 100 * 0.04 + 100 * 0.08 ) t = 50 * 60 = 3000 Therefore, based on sampled data, we estimate that there are 3000 four-leaf clovers in the field. • Compute sample variance within each cluster. If our problem involved two-stage cluster sampling, we would need to compute sample variance within each cluster. But since our problem uses one-stage cluster sampling, we don't need to compute variance within clusters. • Compute sample variance between clusters. We use the following formula to estimate the variance of total scores between sampled clusters (s^2[b]): s^2[b] = Σ ( t[h] - t/N )^2 / ( n - 1 ) s^2[b] = 1/19 * Σ ( t[h] - 3000/1000 )^2 = 0.05263 * Σ ( t[h] - 3 )^2 s^2[b] = 0.05263 * [(0-3)^2 + (1-3)^2 + ... + (4-3)^2 + (8-3)^2] s^2[b] = 0.05263 * [9 + 4 + 4 + 1 + 1 + 1 + . . . + 1 + 1 + 1 + 1 + 1 + 25] s^2[b] = 0.05263 * 50 = 2.63 • Compute standard error. With one-stage cluster sampling, the standard error (SE) of the estimate is: SE = N * sqrt [ ( 1 - n/N ) * s^2[b]/n ] SE = 1000 * sqrt [ ( 1 - 20/1000 ) * 2.63/20 ] SE = 1000 * sqrt ( 0.098 * 0.1315 ) SE = 1000 * sqrt (0.12887) = 1000 * 0.359 = 359 Thus, the standard error of the sampling distribution of the total is 359. • Select a confidence level. In this analysis, the confidence level is defined for us in the problem. We are working with a 95% confidence level. < • Find the critical value. The critical value is a factor used to compute the margin of error. To find the critical value, we take these steps. □ Compute alpha (α): α = 1 - (confidence level / 100) α = 1 - 95/100 = 0.05 □ Find the critical probability (p): p = 1 - α/2 = 1 - 0.05/2 = 0.975 □ Since the sample size (n = 20) is less than 30, we will express the critical value as a t-score with degrees of freedom (df) equal to: df = n - 1 = 20 - 1 = 19 Thus, the critical value is the t-score with 19 degrees of freedom that has a cumulative probability equal to 0.975. From the t-Distribution Calculator, we find that the critical value is about 2.09. • Compute the margin of error (ME): ME = critical value * standard error ME = 2.09 * 359 = 750 • Specify the confidence interval. The minimum and maximum values of the confidence interval are: CI[min] = x - SE * CV = 3000 - 750 = 2250 CI[max] = x + SE * CV = 3000 + 750 = 3750 In summary, here are the results of our analysis. Based on sample data, we estimate that there are 3000 four-leaf clovers in the field. Given a 95% confidence level, the margin of error around that estimate is 750; and the 95% confidence interval is 2250 to 3750.	{"url":"https://stattrek.com/survey-research/cluster-sampling-total","timestamp":"2024-11-07T23:32:17Z","content_type":"text/html","content_length":"66360","record_id":"<urn:uuid:7c8a6e0f-6484-4336-bba9-58589509f994>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00350.warc.gz"}
Workshop \| Advances in Homotopy Theory VI Advances in Homotopy Theory VI Workshop This is the sixth edition of a twice-yearly workshop alternating between the Southampton Centre for Geometry, Topology and Applications (CGTA) and the Beijing Institute of Mathematical Sciences and Applications (BIMSA). The aims are to promote exciting new work in homotopy theory, with an emphasis on younger mathematicians, and to showcase the subject’s wide relevance to other areas of mathematics and science. Daisuke Kishimoto ( Kyushu University ) Vector fields on noncompact manifolds I will present the Poincare-Hopf theorem for a bounded vector field on a connected noncompact manifold having a cocompact and properly discontinuous action of a discrete group. As a corollary, we will see that every bounded vector field on such a noncompact manifold has infinitely many zeros whenever the orbit manifold has nontrivial Euler characteristic and the acting group is amenable. Shuang Wu ( 吴双 , Beijing Forestry University & BIMSA ) Applications of GLMY theory in metabolomic networks of complex diseases Human diseases involve metabolic alterations. Metabolomic profiles have served as a biomarker for the early identification of high-risk individuals and disease prevention. However, current approaches can only characterize individual key metabolites, without taking into account their interactions. This work have leveraged a statistical physics model to combine all metabolites into bDSW networks and implement GLMY homology theory to analyze and interpret the topological change of health state from symbiosis to dysbiosis. The application of this model to real data allows us to identify several hub metabolites and their interaction webs, which play a part in the formation of inflammatory bowel diseases. Sandip Samanta ( Indian Institute of Science Education and Research Kolkata ) On Generalized Brace Product In the 1970s, James introduced the brace product for fibrations admitting a section to study the decomposability of certain fibrations. We have generalized this notion to investigate the H-splitting of based loop space fibrations of a given fibration. First, we will demonstrate that the generalized brace product is the sole obstruction to such splitting. Then, we will explore the connection between the generalized brace product and a generalized notion of Whitehead's J-homomorphism. Additionally, we will present examples in rationalized spaces, where the equivalence of the vanishing of the generalized brace product and the standard brace product simplifies computations. This work is based on our recent preprint, available at the link: https://arxiv.org/abs/2401.16206, co-authored with Dr. Somnath Basu and Dr. Aritra Bhowmick. Sadok Kallel ( American University of Sharjah ) The topology of spaces of maps from a Riemann surface to complex projective space The space of continuous maps Map(M,N) between two Riemannian manifolds M and N is a fundamental object of study in algebraic topology, more particularly when the source space M is a sphere. We will address the case when M=C is a Riemann surface of positive genus and N is a complex projective n-space. This mapping space has received considerable attention in the literature, by physicists and mathematicians alike. It breaks down into connected components indexed by an integer (the "charge"). We give an overview of the relevant results, and then describe the homology of these components. This is ongoing work with Paolo Salvatore (Rome). Fedor Vylegzhanin ( Moscow State University ) Loop homology of moment-angle complexes in the flag case Moment-angle complexes $\mathcal{Z}_{K}$, an important class of CW-spaces with torus action, are parametrized by simplicial complexes $K$. We study their homotopy invariants in the case of flag simplicial complexes. For arbitrary coefficient ring $k$ we describe a presentation of the Pontryagin algebra $H_*(\Omega\mathcal{Z}_K;k)$ by multiplicative generators and relations. Proof uses the connection between presentations of connected graded algebras and the Tor functor. Applying recent results by Huang, Berglund, Stanton, we prove that such moment-angle complexes are rationally coformal, give a necessary condition for their formality, and compute their homotopy groups in terms of homotopy groups of spheres. If time permits, I will outline similar results for homotopy quotients of moment-angle complexes, including quasitoric manifolds (work in progress). Ilya Alekseev; Vasiliy Ionin ( PDMI RAS & Saint Petersburg University; Chebyshev Laboratory, Saint Petersburg University ) Mixing braids, automorphisms, simplicial methods, and homotopy groups of spheres We discuss some fundamental connections between low-dimensional topology and homotopy theory. The first part of the talk is devoted to the interplay between braid groups and homotopy groups of spheres. We begin by investigating the impact of some geometric transformations on Brunnian braids and highlighting their non-preserving nature. This analysis leads us to new simplicial structures in braid theory and an action of the automorphism groups $\mathrm{Aut}(P_n)$ of the pure braid groups $P_n$ on homotopy groups of the two-sphere $S^2$. In particular, in the second part of the talk, we construct a simplicial group built on commutator subgroups $[P_n, P_n]$ of the pure braid groups, which relies on the Fulton-MacPherson compactifications of configuration spaces. By inspecting the derived subgroup of J. Milnor's free group construction, we prove that this simplicial group is homotopy equivalent to the three-sphere $S^3$. As an application, we show how this economical model for the three-sphere leads to some interesting Wu-type formulas for the homotopy groups $\pi_n(S^3)$. Juxin Yang ( 杨聚鑫 , BIMSA ) The extension problems for three far-unstable 33-stem homotopy groups and Toda brackets of diverse shapes Firstly, we will introduce our methods to tackle the extension problems for the homotopy groups $\pi_{39}(S^6)$, $\pi_{40}(S^7)$ and $\pi_{41}(S^8)$ localized at 2, the puzzles having unsolved for forty-five years. Our ability to address these extension problems is largely attributed to the utilization of a rectangular Toda bracket indexed by 2, a Toda bracket of new shape defined by us. Then we will introduce the Toda bracket (Tbr) in the spirit of Toda's 1962 monograph and its developments, including the 3-fold Tbr, 4-fold Tbr, left and right matrix Tbr, rectangular Tbr and Z-shape Tbr. We shall also introduce the applications we got in recent years, namely, the determinations of some homotopy groups of spheres and SO(n). It is worth mentioning that “the Toda bracket is an art of constructing homotopy liftings and homotopy extensions of maps, it plays a fundamental role in dealing with composition relations of homotopy classes” Briony Eldridge ( University of Southampton ) Loop spaces of polyhedral products associated with substitution complexes Polyhedral products are a topological space formed by gluing together ingredient spaces in a manner governed by a simplicial complex. They appear in many areas of study, including toric topology, combinatorics, commutative algebra, complex geometry and geometric group theory. A fundamental problem is to determine how operations on simplicial complexes change the topology of the polyhedral product. In this talk, we consider the substitution complex operation, a special case of the polyhedral join operation. We obtain a description of the loop space associated with some substitution complexes, and build a new family of simplicial complexes such that the homotopy type of the loop space of the moment angle complex is a product of spheres and loops on spheres. Antonio Viruel ( Universidad de Malaga ) On the group of self homotopy equivalences of polyhedral product of BG's In the context of the Kahn realization problem, we study the group of self-homotopy equivalences of polyhedral products of classifying spaces of simply connected simple compact Lie groups. Specifically, for a given polyhedron K and a simply connected simple compact Lie group G, we describe the group of self-homotopy equivalences of (BG)^K. We demonstrate that this group fits into a short exact sequence involving Aut(K) and Out(G). This is a joint work with Cristina Costoya (USC).	{"url":"https://qzc.tsinghua.edu.cn/en/info/1126/3707.htm","timestamp":"2024-11-09T23:44:01Z","content_type":"text/html","content_length":"35469","record_id":"<urn:uuid:cb1f484e-d008-4c57-9c3b-57d44ad4a064>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00611.warc.gz"}
I am research engineer at ICube, working within the Mimesis team at Inria. In the context of real-time simulation for pre-operative guidance, I am focusing on questions related to stochastic filtering, parameter estimation and system observability. My past research lies in the area of Complex Material Systems. My PhD (2009-2012), at the theory and simulation group of the Charles Sadron Institute in Strasbourg focused on spontaneous formation of structures and universal scaling behaviors in polymer and fiber systems. As a post-doctoral fellow at the Weizmann Institute of Science (2013-2015), I participated in a science-educational project targeting the fundamentals of Statistical Physics. My research interests include physics-based simulation, stochastic filtering, statistical physics of complex systems and multi-agent modeling. Langbeheim, S. Livne, N. Schulman, R. Chabays, S. Safran, E. Yerushalmi, Introductory-Level Course on Randomness and Order in Soft and Biological Matter. Biophysical Journal. 106:217A (2014). Schulmann, H. Meyer, T. Kreer, A. Cavallo, A. Johner, J. Baschnagel, J.P. Wittmer, Strictly two-dimensional self-avoiding walks: Density crossover scaling. Polymer Science Series C, 55, 990 (2013). Schulmann, H. Meyer, J.P. Wittmer, A. Johner, J. Baschnagel, Interchain Monomer Contact Probability in Two-Dimensional Polymer Solutions. Macromolecules, 45(3), 1646–1651 (2012). Schulmann, H. Xu, H. Meyer, P. Polinska, J. Baschnagel, J.P. Wittmer, Strictly two-dimensional self-avoiding walks: Thermodynamic properties revisited. Eur. Phys. J. E, 35(9), 16 pp (2012). Meyer, N. Schulmann, J.E. Zabel, J.P. Wittmer, The structure factor of dense two-dimensional polymer solutions. Computer Physics Communications, 183, 1949–1953 (2011). J.P. Wittmer, A. Cavallo, H. Xu, J.E. Zabel, P. Polinska, N. Schulmann, et al, Scale-Free Static and Dynamical Correlations in Melts of Monodisperse and Flory-Distributed A Review of Recent Bond-Fluctuation Model Studies. J. Stat. Phys., 145(4), 1017–1126 (2011). J.P. Wittmer, N. Schulmann, P. Polinska, J. Baschnagel, Note: Scale-free center-of-mass displacement correlations in polymer films without topological constraints and momentum conservation. J. Chem. Phys., 135(18), 2 pp, (2011).	{"url":"https://mimesis.inria.fr/members/nava-schulmann/","timestamp":"2024-11-09T04:20:18Z","content_type":"text/html","content_length":"173040","record_id":"<urn:uuid:296d1228-6681-4401-b9bd-2ceb34e4537f>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00680.warc.gz"}
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Information centre allocation A large number of potential sites are given and we have to choose k sites in order to set up information centres, where each centre is able to serve a limited number ofclients. The price a client pays for accessing a centre is proportional to the distance between the client and the centre. This problem belongs to a class of problems for which most theoretical computer scientists believe that there is no fast algorithm for finding an optimal solution. We therefore look for algorithms that produce an approximate solution. In this paper we present a fast algorithm that chooses k sites and assigns the clients to the centres in such a way that the maximum price a client pays is at most nine times the maximum price in an optimal solution. This algorithm works under the assumption that the number of chosen sites is small in comparison to the number of possible sites. Dive into the research topics of 'Information centre allocation'. Together they form a unique fingerprint.	{"url":"https://cris.biu.ac.il/en/publications/information-centre-allocation","timestamp":"2024-11-11T18:11:41Z","content_type":"text/html","content_length":"49366","record_id":"<urn:uuid:5051a447-b2b3-47be-b3fa-aec4f415e735>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00570.warc.gz"}
Probability and Ergodic Theory Karine Bertin, Sebastián Donoso, Joaquín Fontbona, Raúl Gouet, Rodolfo Gutiérrez, Alejandro Maass, Servet Martínez, Daniel Remenik, Jaime San Martín, Avelio Sepúlveda, Soledad Torres Coordinators: Alejandro Maass and Daniel Remenik About the research group This group puts together two research lines with a long-shared history within CMM, based on common interests and activities. The group works on a variety of fundamental topics in probability theory, stochastic processes, dynamical systems and ergodic theory, and in applications to other branches of pure and applied mathematics such as number theory, geometry, mathematical physics and mathematical biology. At the same time, members of the group develop theory and applications in a diverse range of subjects where randomness and information theory play a key role, including data science, systems biology and bioinformatics, astroinformatics, mining and geophysics. An important focus of work is stochastic models in mathematical physics. A highlight of the group’s research in this direction has been KPZ fixed point, the universal Markov process describing the asymptotic spatial fluctuations of all models in the KPZ universality class, and its surprising connections with dispersive PDEs. Another focus of interest is the long time behavior of interacting particle systems, where the group has made important contributions to the development of probabilistic techniques for propagation of chaos in mean field models in kinetic theory, such Kaç particle systems associated with the Boltzmann equation. A more recent focus of interest is two-dimensional random geometry. Here members of the group have made key contributions to the study of the Gaussian free field, which has been key to further understand properties of fundamental models in two-dimensional statistical physics including random curves such as the Schramm-Loewner evolution and random metric spaces such as Liouville quantum gravity; one highlight is the discovery of a topological phase transition for the Gaussian free field. Quasi-stationary distributions and M-matrices are two areas of long-standing interest for the group, which have led to two monographs describing the state of the art of these theories. Other work by the group focuses on stochastic modeling in genetics and population dynamics, on finite potential theory, and on stochastic analysis, particularly the characterization of multi-dimensional local martingales and its relation to the phenomenon of bubbles in finance. A recent focus of interest is the application of ideas from optimal transport in data science. In ergodic theory and dynamical systems, contributions were made in several directions, with connections to number theory, combinatorics, analysis, group theory and geometry. Deep results were obtained in Teichmüller dynamics, such as the resolution of the famous Kontsevich-Zorich conjecture on the simplicity of the Lyapunov spectrum for quadratic differentials. In addition, a program in symbolic dynamics was developed attacking fundamental problems for minimal Cantor systems, establishing a correspondence between finite rank systems and S-adic subshifts, and providing new tools to tackle several problems related to the complexity of minimal subshifts. Other important results deal with recurrence problems both in ergodic theory and topological dynamics, such as the analysis of multicorrelation sequences for Z^d actions and the study of sets of recurrence for actions of the multiplicative semigroup of integers.	{"url":"https://www.cmm.uchile.cl/?page_id=62","timestamp":"2024-11-08T05:18:39Z","content_type":"text/html","content_length":"51297","record_id":"<urn:uuid:9ef52b78-9763-4ab8-9e3b-8ad3d00611cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00347.warc.gz"}
Comparing Internal Flow in Freezing and Evaporating Water Droplets Using PIV Division of Fluid and Experimental Mechanics, Luleå University of Technology, SE-97187 Luleå, Sweden Author to whom correspondence should be addressed. These authors contributed equally to this work. Submission received: 27 April 2020 / Revised: 19 May 2020 / Accepted: 21 May 2020 / Published: 23 May 2020 The study of evaporation and freezing of droplets is important in, e.g., spray cooling, surface coating, ink-jet printing, and when dealing with icing on wind turbines, airplane wings, and roads. Due to the complex nature of the flow within droplets, a wide range of temperatures, from freezing temperatures to heating temperatures, have to be taken into account in order to increase the understanding of the flow behavior. This study aimed to reveal if natural convection and/or Marangoni convection influence the flow in freezing and evaporating droplets. Droplets were released on cold and warm surfaces using similar experimental techniques and setups, and the internal flow within freezing and evaporating water droplets were then investigated and compared to one another using Particle Image Velocimetry. It was shown that, for both freezing and evaporating droplets, a shift in flow direction occurs early in the processes. For the freezing droplets, this effect could be traced to the Marangoni convection, but this could not be concluded for the evaporating droplets. For both evaporating and freezing droplets, after the shift in flow direction, natural convection dominates the flow. In the end of the freezing process, conduction seems to be the only contributing factor for the flow. 1. Introduction Evaporation and freezing of droplets are two interesting areas with many applications. Evaporation of droplets is important in, e.g., spray cooling [ ], surface coating [ ], ink-jet printing [ ], and droplet based biosensing [ ]. For freezing droplets, the applications are found mainly when dealing with icing on wind turbines [ ], airplane wings [ ], and roads [ ]. Since icing is a big problem for these areas, effective icing prevention methods have been developed, but a lot is still unknown about the ice itself. One way in improving these existing icing prevention systems is to go back a few steps and start with a single freezing water droplet. Due to the complex nature of the flow within droplets, a wide range of temperatures, from freezing temperatures to heating temperatures, have to considered to increase the understanding of the flow behavior. This is the motivation for performing this comparative study between freezing and evaporating droplets. To the author’s best knowledge, not many studies have been performed studying the internal flow within the droplets for freezing of water droplets. Kavanami et al. [ ] used a numerical model considering both surface tension and the density maximum at 277 K. The results were also validated with experiments. They found that both natural and Marangoni convection are important mechanisms for the internal flow. Karlsson et al. [ ] modeled the inner flow when internal natural convection was included and excluded (i.e., only conduction from the surface) in the model and the result was validated with existing experimental data. Karlsson et al. [ ] performed experiments by releasing droplets on a cold surface and the inner flow was captured by using Particle Image Velocimetry (PIV). The flow field in the center of the droplet was visualized and the magnitude and direction of the velocities were derived. The evaporation of sessile droplets has been studied thoroughly both experimentally and numerically [ ]. A similar setup as in the experiments performed in this study was done by Erkan and Okamoto [ ] and Erkan [ ], where liquid droplets were released with different impact velocities onto a dry (and later, also heated) sapphire plate. The flow inside the droplet was investigated using time-resolved PIV, and a laser created a laser sheet illuminating a plane parallel to the surface. The radial velocity distribution during the early phase of spreading inside the droplets was compared to an analytical as well as a numerical model and both were proven to show similar results. However, none of the models could reproduce all of the experimental results in detail. An interesting study of the internal flow within droplets impacting on a surface was performed by Kumar et al. [ ] using the same ray-tracing algorithm as in this work proposed by Kang et al. [ ] to account for the image distortion at the droplet boundary. The vortices caused by the impact to the surface were traced and the vortex strength was computed. The internal flow was also studied by Thokchom et al. [ ] using PIV in externally heated sessile droplets confined in a narrow gap between two glass plates. The benefit of using a “Hele–Shaw” droplet is in the optics since no image correction has to be done and the surface flow is seen. The experiments yielded that an alteration in the free surface temperature changes the fluid flow profile inside the droplet and consequently also the deposition pattern of solute particles on the substrate. The heat sources were an IR light and a heating element that was moved along the droplet surface to create different effects. The measured velocity profiles were in qualitatively agreement with numerical simulations. In a recent study by Zhao et al. [ ], the full velocity field in evaporating sessile droplets was captured. Instead of correcting the raw PIV data as done in this work, the Scheimpflug principle was applied. This is a mapping method used to eliminate the perspective effects and therefore the flow near the droplet surface can be studied. This method is very useful when studying evaporating droplets, but when working with freezing droplets this method will not work properly because of the positioning of the camera. Since the camera is placed underneath the droplet, the frost layer and the created ice during freezing will make it impossible for the camera to retrieve good images of the freezing process. This is one of the reasons the method by Kang et al. [ ] was used in the freezing experiments in the current study. For easier comparison, the same method was also applied in the evaporation experiments. In this work, the effects of Marangoni convection on evaporating and freezing droplets were studied and the results from the two different processes were compared to each other. The effects of Marangoni convection on evaporating droplets has been studied before. To exemplify, Hu and Larson [ ] concluded from numerical simulations that a thermally driven Marangoni flow should be visible in an evaporating water droplet, but this is not as easy to visualize in experiments. One possible reason for this can be surfactant contaminants [ ]. Xu and Luo [ ] showed that, even though this is partially true, the surface tension and the surface temperature change nonmonotonously along the liquid surface due to a stagnation point at the droplet surface. Kita et al. [ ] induced a Marangoni flow in pure water drops by creating a temperature gradient on the drop using infrared thermography. They found that, to initiate a Marangoni flow as two similar vortices, a temperature gradient along the liquid–air interface of about 2.5 K ( $2.5 ∘$ C) is necessary and to maintain them the temperature difference has to be about 1.5 K ( $1.5 ∘$ C). The main consensus is that Marangoni convection is small for water droplets [ ], but can be induced [ ]. This motivates also a comparison between heated evaporating droplets and freezing droplets since it has been shown that the Marangoni convection plays an important role for the flow in the beginning of the freezing process, but after a while natural convection takes over [ ]. This is not as clear when studying evaporation of droplets, and this paper discusses the different effects in the droplets. Hence, in this study, the aim was to investigate and compare the inner flow when droplets are released on a cold and warm surface and to reveal if natural convection and/or Marangoni convection have a noticeable influence on the flow within the droplets. 2. Method To facilitate the presentation of the results, a schematic image of the direction of Marangoni flow for a constant gradient is shown in Figure 1 . The Marangoni flow when $T 1$ $T 2$ is seen to the left in the figure, i.e., the flow within an evaporating droplet, and the Marangoni flow when $T 1$ $T 2$ is seen to the right, i.e., the flow within a freezing water droplet. A similar setup and experimental procedure were used for the freezing and evaporating droplets. When these differ from one another, it is clearly stated in the text. 2.1. Experimental Setup Droplets were gently deposited on a sapphire plate (aluminum oxide, Al ) using a pipette. The sapphire plate measured 0.0508 m (2 ) in diameter, 0.005 m thick, and was chosen due to its high thermal conductivity in combination with its transparency. The pipette was a ThermoFisher Finnpipette F1 1–10 L, and it was kept in place by an optical rail to create repeatable deposition of the droplets in each experiment. The sapphire plate was resting on an aluminum holder that was heated using a Peltier element. The Peltier element was placed with its warm side up during the heating experiments and its cold side up during the freezing experiments. It was submerged in a box with either warm or cold circulating water depending on which experiment was performed (closed system, connected to a tank water held at = 295.85 K ± 0.98 K and = 277.05 K ± 1.3 K, respectively). The laser light was guided underneath the droplet using a prism that was put inside a central hole in the aluminum holder and a tunnel inside the holder allowed the light to pass through to the droplet. To protect the setup from disturbances in the room, a PMMA (plexiglass) chamber was put around the sapphire plate. Four holes was made in the chamber; one on top for the pipette to be able to release the droplets, one on the right side to allow the laser light to pass through to the prism, one on the left side to enable the camera to record the experiments and one on the back where the hygrometer was mounted. The sapphire plate was cleaned using Propanol ( $C 3 H 7 O H$ ) and deionized (DI) water using lens paper before each experiment. A schematic diagram of the experimental setup can be seen in Figure 2 . In both heating and freezing experiments, the liquid used was DI water kept at room temperature, $T w a t e r$ = 294.25 K ± 1.4 K. The seeding particles in the freezing experiments were chosen due to its good qualities when working with water. However, these particles did not work well in the heating experiments since the droplets collapsed as they reached the surface after the release from the pipette. Therefore, new particles were chosen for the heating experiments. Table 1 presents details of the seeding particles for each setup. The Stokes number, $S t k$ , was calculated for a “worst-case scenario” for both setups to determine if the particles was suitable for these experiments (see Table 1 ). Since both have a $S t k$ « 1, the conclusion is that the particles follow the flow well. Based on the guidelines, the amount of particles in the water was continuously evaluated to reach the recommended particles per interrogation area. The resulting concentration of particles for both setups can be found in Table 1 The temperature of the air and the humidity inside the chamber were monitored using a hygrometer and the temperature of the sapphire plate was measured using a thermocouple of K-type. The laser was a continuous 50 mW 532 nm Nd:YAG (Altechna Co Ltd., Vilnius, Lithuania) connected to a half wave-plate, a polarizing beam splitter (cube) and a beam dump. These components were used to adjust the amount of light transmitted through the droplet. A cylinder lens assembly from Dantec Dynamics created and focused the light sheet to a thickness of <0.0004 m. A 0.012-m-thick window placed on a rotation table was used to fine tune the position of the sheet to the center of the droplet (up and down to be able to adjust the light sheet to move sideways in the droplet) (see Figure 2 ). The motivation for guiding the laser light from underneath the droplet was the advantage of the plane surface reducing the light scatter. This arrangement also allows a good view of the symmetry line of the droplet. For the freezing setup, a CMOS camera (IDS Eye) with a spatial resolution of 1280 × 1024 pixels and pixel size 5.3·10 $− 6$ × 5.3·10 $− 6$$m 2$ together with a Navitar long distance microscope captured images of the particles. For the heating setup a similar CMOS camera (IDS UI-3140CP-M-GL Rev.2) was used with the same spatial resolution and microscope mounted, but with pixel size 4.8·10 $− 6$ × 4.8·10 $− 6$ . Details about the specific sampling rates can be found in Table 2 . The height from which the droplets were released was measured to be 0.0021 m and the velocity as the droplet hit the surface was calculated to be 0.052 m/s for the heating experiments and 0.0039 m and 0.077 m/s, respectively, for the freezing experiments. 2.2. Experimental Procedures The following statements can briefly summarize the experimental procedures: • The heating of the surface started. • At $T p l a t e$ = 313.15, 323.15 or 333.15 K (visually determined from the computer screen): The pipette was filled with the DI water and seeding particle suspension. The camera and the laser were switched on. The laser sheet was fine tuned to the center of the droplet (while the droplet was still hanging from the pipette). Finally, the droplet was released. • The camera and laser light were turned off after about 60 s after the droplet has hit the surface. • The surface was cleaned and a new experiment could begin. • The cooling of the surface started. • At $T p l a t e$ = 261.15 K or 265.15 (visually determined from the computer screen), the pressurized air was switched on and turned off again when RH was around 50%. This took about 60 s. • The pipette was filled with the DI water and seeding particle suspension. • The camera and the laser were switched on. The position of the light sheet was fined tuned to the center of the droplet (while the droplet was still hanging from the pipette). • The droplet was released when $T p l a t e$ reached 265.15 K (or 261.15 K) again (visually determined from the computer screen), which occurred approximately 30 s from when the pressurized air was switched off. • The cooling was turned off when the droplet was completely frozen. • The surface was cleaned and dried when $T p l a t e$ > 273.15 K and at $T p l a t e$ = 277.15 K a new experiment could begin. In the freezing experiments, it was found that a roughness on the surface was necessary to initialize the freezing process without delay. Due to the smoothness of the sapphire plate, this roughness was added by producing frost on the surface. This layer was generated by letting pressurized air pass through a container filled with water and this humidified air was then guided into the closed chamber surrounding the experimental setup. The temperature of the surface and air and the relative humidity inside the chamber were monitored to produce a layer of frost as similar as possible during each freezing (see Table 2 for details about the conditions for the frost). A longer discussion of how this frost layer can be found in the work of Karlsson et al. [ The result of the evaporation experiments is based on six evaporating droplets with a similar geometry, i.e., droplets with approximately same heights and radii, when $T p l a t e$ = 313.15, 323.15, and 333.15 K. These are presented in Table 3 . Two freezing droplets (when $T p l a t e$ = 261.15 and 265.15 K) are also included in this table. 2.3. Uncertainty Analysis The uncertainties introduced during the measurements can be divided into two categories: systematic (bias) and random errors. The systematic errors usually arise from the measuring equipment and the random errors are usually due to unknown or unpredictable changes in the experiments [ Similar to the experiments performed by Karlsson et al. [ ], the main sources of the systematic errors can be found in the pipette technique, i.e., the mixing of seeding particles in the water, the reading of the measurements instruments, and the positioning of the camera. Since these errors are introduced in each experiment, they may be difficult to detect and therefore have a large impact on the result. However, careful planning and execution of experiments can minimize these errors. Normally, the correlation error is 0.1 pixel [ The random errors can mainly be found in the release of the droplet, resulting in droplets with different geometries and different initial internal flows ( < 5 s). To get an estimate of the random errors, a repeatability test can be performed (see [ ]). Ten experiments for $T p l a t e$ = 333.15 K were considered (see Table 4 for details about the droplets), where the magnitude of the corrected velocities in the y-direction along the symmetry line between bottom and apex of the droplet was studied. Since the droplets varied in height, the interesting points is what happened at just above the heated surface, i.e., the lowest value of the corrected dataset and at the top of the corrected data in each case. In addition, the points at 25%, 50%, and 75% above the heated surface were determined. To get an estimate of the variations in the velocities along the symmetry line in the beginning of evaporation and as the flow settled in the droplet, the times studied were = 5 and 15 s. The specific time = 5 s was chosen because the shift in flow direction (see Section 3.2.1 ) had occurred for all 10 cases in this investigation; before this time the flow would be more difficult to evaluate. In Table 5 , the precision errors with a 95% confidence interval for the five points chosen are shown and it can be seen that the errors are below, or mostly well below 5.5%, suggesting that the random errors are relatively small regarding the velocities on the symmetry line, especially for when = 15 s. This can also be seen in Figure 3 , where the mean velocity in the five points is shown together with the precision error presented using error bars. This means that the velocities in the droplet are in fact comparable in each case despite their differences in geometry, and the six selected droplets in Table 3 can be used in the further study. 3. Results and Discussion 3.1. Impact of Release When a droplet of equal temperature as the sapphire plate, held at T = 293.15 K, is released and reaches the surface, all movement stops completely within less than a second. This could suggest that the movement inside the droplet after t = 1 s does not have to do with the release of the droplet. However, since the viscosity of the water decrease with an increase in temperature (drops about 50% from 293.15 to 333.15 K), any movement created by the release of the droplet should be more visible at higher temperatures. This indicates that depending on how the droplet impact the surface, different types of flow will be seen inside the droplets and there will be a seemingly random motion in each droplet directly after impact. This can be compared to the freezing droplet where the viscosity in the water is high as it reaches the surface and will only get larger as the temperature inside the droplet decreases. This means that the movement inside of the freezing droplet is not induced by the release of the droplet, but for the heated droplet the release may have a larger impact on the flow inside the droplet. 3.2. Evaporation After the droplet is released on the surface the flow is fluctuating in velocity and for at least the next 5 s; see Figure 4 where the magnitude of the mean velocity along the the symmetry line for $T p l a t e$ = 313.15, 323.15 and 333.15 K when = 1–50 s is plotted. The size of the fluctuations and how long these occur, increase with increasing temperature, i.e., when $T p l a t e$ = 313.15 K, the velocity settles more quickly than for when $T p l a t e$ = 333.15 K. After this initial time period, i.e., when = 5–20 s, the flow settles and the velocity decreases for the higher temperatures and is fairly constant for the lowest temperature. A form of “steady state” is reached, since the conditions in and around the droplet are not changing significantly. The only varying parameter is the drying of the droplet, which reduces the size and thereby also the velocities inside the droplet. This section is divided in two parts: evaporation until “steady state” and evaporation after “steady state”. 3.2.1. Evaporation until “Steady State” For all cases studied, during the first second or the first few seconds for some cases, the flow is moving down the symmetry line of the droplet and up along the curved surface. This is exemplified Figure 5 a for $T p l a t e$ = 333.15 K (Case 1). Note that the velocity vectors are normalized with the magnitude of the velocity of each vector. Within 15–20 s, the flow has settled and the flow is moving in the opposite direction up along the symmetry line. For $T p l a t e$ = 333.15 K (Case 1), this takes place already within 3 s. After this, the flow is moving upwards along the symmetry line and down along the curved surface. The initial direction of the flow may indicate that Marangoni convection can actually be seen in heated water droplets directly after the impact on the surface. After this initial time period, natural convection takes over changing the direction of the flow. This is interesting since the effects of Marangoni convection has also been shown to be the dominant for freezing droplets, up to about 15% of the total freezing process [ ]. However, even though the direction corresponds to Marangoni flow, it is difficult to conclude that this is the cause. Another explanation for this flow pattern is due to the release of the droplet. The force created as the droplet impact the heated surface will be directed upwards in the direction away from this surface forcing the water upwards along the curved surface in the droplet. Since the viscosity is decreasing with increasing temperature, the water moves more easily when a force is applied. Finally, a third possibility for this type of initial flow pattern might be due to the temperature differences, , caused by the heating of the surface. When = 0 (at room-temperature), no flow is seen inside the droplet and this could indicate that the droplets initial change in form do not induce a flow. However, when ≠ 0, this initial change in form might actually affect the flow and this could cause the flow moving in this initial direction, as exemplified in Figure 5 . All three causes are plausible and there might also be an interaction among all three contributing to the flow. The direction of the flow is altered within seconds after the release from the pipette. This can be seen in Table 6 , where the time for when the flow shifts in direction can be found for all cases. As the temperature increases, the time for when the flow is shifting is decreasing. Since the viscosity decreases with higher temperatures, the impact from the release might be seen for a longer time in the flow or it could also be that the effects from Marangoni has larger impact on the flow when the temperature increases. It should be noted that the flow seen in Figure 5 is similar, but not identical, in each of the six cases studied. Even though the goal for the release of the droplet on the surface is to be identical in each experiment, there are still differences, and therefore different types of flow emerges inside the droplet. Vortices are created at different locations in each droplet, but, as the flow settles (at = 5–20 s), two vortices are seen or are discernible at either side of the symmetry line in the droplet (see Figure 5 ). The flow direction, as discussed above, is however the same in all droplets. When the turn in flow direction occur the velocities drops considerably, as seen in Figure 6 = 3 s, where the magnitude of the velocity is exemplified for $T p l a t e$ = 333.15 K (Case 1). After the shift, the velocities recover and reach a similar level as before. Towards the end of the measuring period ( = 13 s), the flow slows down. When the temperature of the heated surface is higher, the velocities are also higher inside the droplet; see Figure 7 , where the spread (standard deviation) in velocity along the symmetry line in the droplet at = 1–15 s can be seen for all temperatures. Here, the mean velocity is seen as a solid and a dotted line representing the two cases for each temperature. As the temperature of the heated surface increases, the spread in velocity becomes larger in magnitude. This is due to both the release of the droplet and the large temperature differences in the droplet compared to the heated surface and is therefore expected. Some variations in velocity between the two cases compared at each temperature can be seen, but the magnitude of velocity is similar, suggesting that they are in fact comparable and repeatable as long as the droplet geometries are similar. 3.2.2. Evaporation after “Steady state” After the initial time period, the flow inside the evaporating droplets begins to settle. The velocity starts to decrease around = 15–20 s and the decrease in mean velocity is more rapid when $T p l a t e$ = 333.15 K (see Figure 4 ). To exemplify, if studying the velocity vectors (normalized with the magnitude of the velocity of each vector) for $T p l a t e$ = 333.15 K (Case 1) in Figure 8 = 20–50 s, it can be seen that the flow is approximately similar during this time. The vortices move somewhat between the time frames, probably due to a decrease in velocity and contact angle, but a type of “steady-state” has occurred. This “steady-state” is also reflected in Figure 9 , where the magnitude of the velocity along the symmetry line for the same case and corresponding times is seen and here the velocities are approximately similar with a slight decrease with time. In Figure 4 , this decrease is seen for all $T p l a t e$ temperatures and cases. Scrutinizing the full time period, i.e., = 1–50 s, yields that the decrease in velocity is more rapid in the beginning of the evaporation compared to the end. When the “steady-state” period is reached, the temperature differences between the water and the heated surface have evened out and the heat exchange with the surroundings is constant. In addition, the velocity differences increases with the increase in surface temperature of the heating surface if the full time period, = 1–50 s is considered. In Figure 10 , the spread (standard deviation) in velocity along the symmetry line in the droplet at = 20–50 s can be seen for all temperatures and here the mean velocity is seen as a solid and a dotted line representing the two cases for each temperature. Note that the maximum spread in velocity is still larger (if comparing to Figure 7 ) when the temperature of the heated surface is higher compared to a lower temperature. 3.3. Freezing The internal flow patterns with normalized velocity vectors for when $T p l a t e$ = 261.15 K at = 1–5 s are shown in Figure 11 . Note that the velocity vectors are normalized with the magnitude of the velocity of each vector. Here, the flow is moving up along the symmetry line and down along the curved surface up to when = 0.19 $t t o t a l$ , after which the flow changes direction and at = 0.22 $t t o t a l$ the flow is moving in the opposite direction—down along the symmetry line and up again along the curved surface. This flow behavior is in agreement with the flow behavior seen for $T p l a t e$ = 265.15 K by Karlsson et al. [ ], but the time when the shift is slightly different. The shift occur at approximately 16% of the full freezing time when $T p l a t e$ = 265.15 K [ ] compared to approximately 19% when $T p l a t e$ = 261.15 K (current study). In Figure 12 , the magnitude of the velocity along the symmetry line are presented for the corresponding times in Figure 11 . As for $T p l a t e$ = 265.15 K, the highest velocities are found in the beginning of freezing and the large drop in velocity that occur before the shift in the flow is also seen when $T p l a t e$ = 261.15 K. However, the increase in velocity after the shift is not as clear when $T p l a t e$ = 261.15 K as for $T p l a t e$ = 265.15 K [ ], but a similar flow behavior is seen for both temperatures. The spread (standard deviation) in velocities along the symmetry line for when $T p l a t e$ = 265.15 K and 261.15 K at = 1–4 s and = 1–5 s, respectively, is found in Figure 13 , where it can be seen that the velocities in the droplet varies significantly. In the beginning of freezing, the velocities are high, but shortly after the shift in flow direction the velocities are almost zero, resulting in a very large spread in velocities with respect to time. 3.4. Comparing the Flows within Freezing and Evaporating Droplets $T p l a t e$ temperatures result in higher velocities, but the highest velocities are found when $T p l a t e$ < 0. The highest velocities in this study are found when $T p l a t e$ = 261.15 K, suggesting that the velocity is increasing with decreasing $T p l a t e$ temperature for temperatures when $T p l a t e$ < 0. The maximum spread in velocities is also seen for when $T p l a t e$ = 261.15 K, suggesting that the highest velocities may be found here; however, when = 0.19 $t t o t a l$ , the velocities are almost zero, as for when $T p l a t e$ = 265.15 K [ ]. Regarding evaporation, the highest velocities are found the first 15 s in all cases (see Figure 7 Figure 10 ). The highest velocities for when $T p l a t e$ = 333.15 K is comparable to the highest velocities when $T p l a t e$ = 265.15 K, suggesting that Marangoni convection might actually affect the flow, at least for the first second or seconds. For freezing droplets, Marangoni convection seems to affect the flow due to the higher velocities in comparison to evaporation. This can also be compared to the theoretical value for the Marangoni ( $M a$ ), which is well above 80 for all cases studied. This means that the Marangoni effects should be present in both the freezing and evaporating droplets. Finally, when the flow shifts in direction the velocity approaches zero for both evaporation and freezing and after the shift the velocities increases again, but never to the highest levels as seen before the shift in flow. 4. Conclusions In this study, the internal flow in freezing and evaporating water droplets was investigated using Particle Image Velocimetry and a comparison between the two cases was performed. Three heating and two cooling temperatures were studied where the aim was to reveal if natural convection and/or if Marangoni effects influences the flow within the droplets. Due to the low viscosity in heated droplets, the early flow within these droplets might be influenced by the release of the droplet. For freezing droplets, the viscosity is high in the water and therefore the internal flow is not influenced by the release of the droplet, the movement is only due to freezing and $Δ$T. During evaporation, within 15 s after the release of the droplet to the surface, a form of “steady-state” occurs since the conditions in and around the droplet are not changing significantly. As for the freezing droplets, a shift in flow direction occurs early in evaporation. This might be due to the Marangoni effects, but it could not be concluded in this work. Two other explantations could be due to the temperature differences caused by the heated surface or because of the force created as the droplets impact the heated surface, pushing the water up along the curved surface. In the freezing droplets, the flow is similar between the two temperatures and shows a typical “Marangoni flow”-behavior, suggesting that the Marangoni effects influence the early flow within freezing droplets. However, the shift in direction of the flow and the magnitude of the velocities differ between the two cases. When comparing the velocities during freezing and evaporation, it was found that a warmer heating surface yields higher velocities and a larger spread in velocities compared to a colder heating temperature, although the highest velocities and the largest spread in velocities was found when the temperature of the cooling surface was below zero. Author Contributions Conceptualization, L.K., A.-L.L., and S.L.; methodology, L.K.; software, L.K.; validation, L.K. and A.-L.L.; formal analysis, L.K., A.-L.L., and S.L.; investigation, L.K.; resources, S.L.; data curation, L.K.; writing—original draft preparation, L.K.; writing—review and editing, L.K., A.-L.L., and S.L.; visualization, L.K.; supervision, A.-L.L., and S.L.; project administration, L.K.; and funding acquisition, A.-L.L. and S.L. All authors have read and agreed to the published version of the manuscript. This research received no external funding. Conflicts of Interest The authors declare no conflict of interest. A surface contact area (m$2$) d diameter (m) h height (m) r radius (m) t time (s) T temperature (K) RH relative humidity (%) $μ$ viscosity (kg/ms) $M a$ Marangoni number $S t k$ Stokes number Figure 1. Direction of Marangoni driven flow for a constant gradient: (Left) $T 1$ > $T 2$, evaporation; and (Right) $T 1$ < $T 2$, freezing. Figure 3. The mean velocity in the five points used in the repeatability study at t = 5 and 15 s for 10 experiments when $T p l a t e$ = 333.15 K. The precision error with a 95% confidence interval in each point is shown with error bars. Point 1 is located at the heated surface and Point 5 is located at the top of the corrected data. Points 2–4 are found in between Points 1 and 5. Figure 4. The magnitude of the mean velocity along the symmetry line for $T p l a t e$ = 313.15, 323.15, and 333.15 K when t = 1–50 s. Figure 5. Internal flow patterns with normalized velocity vectors for the corrected data when $T p l a t e$ = 333.15 K (Case 1) at t = 1–15 s. Figure 6. The magnitude of the velocity along the symmetry line for $T p l a t e$ = 333.15 K (Case 1) when t = 1–15 s and t = 7–15 s. Figure 7. The spread (standard deviation) in velocity along the symmetry line during evaporation for $T p l a t e$ = 313.15, 323.15 and 333.15 K (all case) when t = 1–15 s shown using error bars. The solid line is the mean velocity of all times. Figure 8. Internal flow patterns with normalized velocity vectors for the corrected data when $T p l a t e$ = 333.15 K (Case 1) at t = 20–50 s. Figure 9. The magnitude of the velocity along the symmetry line for $T p l a t e$ = 333.15 K (Case 1) when t = 20–50 s. Figure 10. The spread (standard deviation) in velocity along the symmetry line during evaporation for $T p l a t e$ = 313.15, 323.15 and 333.15 K (all case) when t = 20–50 s shown using error bars. The solid line is the mean velocity of all times. Figure 11. Internal flow patterns with normalized velocity vectors for the corrected data when $T p l a t e$ = 261.15 K at t = 1–5 s. Figure 12. The magnitude of the velocity along the symmetry line for $T p l a t e$ = 261.15 K when t = 1–5 s. Figure 13. The spread (standard deviation) in velocity along the symmetry line during freezing for $T p l a t e$ = 265.15 and 261.15 K when t = 1–4 s and t = 1–5 s, respectively, shown using error bars. The solid line is the mean velocity of all times. Case Particles Diameter, d (m) $Stk$ Conc. of Particles Heating Latex (Magsphere Inc., Pasadena, CA, USA 5.1 × 10$− 6$ <5 × 10$− 4$ 9.9 × 10$− 7$ m$3$ DI water and PS Red Fluorescent, 1055 kg/m$3$) 0.1 × 10$− 7$ m$3$ seeding particles Freezing Rhodamine B (microParticles 3.16 × 10$− 6$ <5 × 10$− 7$ 9.8 × 10$− 7$ m$3$ DI water and GmbH PS-FluoRed, 1050 kg/m$3$) 0.2·10$− 7$ m$3$ seeding particles Case Plate Temperature, $T p l a t e$ RH in Chamber Temperature in Chamber, $T c h a m b e r$ Sampling Rate Recording Times Heating 313.15 K ± 0.22 K, 323.15 K ± 0.11 K, 333.15 K ± 0.05 K 47.2% ± 1.2% 298.55 K ± 0.77 K 50–54 Hz 60 s Freezing 265.07 K ± 0.12 K, 261.11 K ± 0.06 K 50.4% ± 4.5% 289.85 K ± 1.7 K 50, 54 Hz Dependent on the freezing times of the droplets Case ($T p l a t e$-Temperature) Droplet Height, h (m) Droplet Radius, r (m) Contact Area at Surface, A (m$2$) 313.15 K: Case 1 0.00137 0.00188 11.1 × 10$− 6$ 313.15 K: Case 2 0.00143 0.00186 10.9 × 10$− 6$ 323.15 K: Case 1 0.00145 0.00186 10.9 × 10$− 6$ 323.15 K: Case 2 0.00147 0.00184 10.6 × 10$− 6$ 333.15 K: Case 1 0.00148 0.00192 11.6 × 10$− 6$ 333.15 K: Case 2 0.00145 0.00185 10.7 × 10$− 6$ 261.15 K 0.00178 0.00156 7.65 × 10$− 6$ 265.15 K 0.00142 0.00171 9.15 × 10$− 6$ Table 4. Values of the droplets geometry in the repeatability study at t = 5 and 15 s when $T p l a t e$ = 333.15 K. Note that the radius of the droplets and contact areas at the heated surface are the same for both times. Case Droplet Radius, r (m) Contact Area at Surface, A (m$2$) Droplet Height, h at t = 5 s (m) Droplet Height, h at t = 15 s (m) 1 0.00180 10.1 × 10$− 6$ 0.00143 0.00139 2 0.00186 10.9 × 10$− 6$ 0.00137 0.00133 3 0.00191 11.5 × 10$− 6$ 0.00149 0.00144 4 0.00178 9.98 × 10$− 6$ 0.00157 0.00152 5 0.00189 11.2 × 10$− 6$ 0.00143 0.00138 6 0.00186 10.8 × 10$− 6$ 0.00145 0.00142 7 0.00187 11.0 × 10$− 6$ 0.00144 0.00140 8 0.00177 9.80 × 10$− 6$ 0.00145 0.00142 9 0.00192 11.6 × 10$− 6$ 0.00147 0.00142 10 0.00184 10.6 × 10$− 6$ 0.00148 0.00143 Position t = 5 s t = 15 s At heated surface 1.00% 0.81% 25% 0.25% 1.78% 50% 5.37% 0.42% 75% 2.30% 0.22% Top 2.68% 0.02% Case Time (s) 313.15 K: Case 1 6 313.15 K: Case 2 6 323.15 K: Case 1 4 323.15 K: Case 2 4 333.15 K: Case 1 3 333.15 K: Case 2 2 © 2020 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:/ Share and Cite MDPI and ACS Style Karlsson, L.; Ljung, A.-L.; Lundström, T.S. Comparing Internal Flow in Freezing and Evaporating Water Droplets Using PIV. Water 2020, 12, 1489. https://doi.org/10.3390/w12051489 AMA Style Karlsson L, Ljung A-L, Lundström TS. Comparing Internal Flow in Freezing and Evaporating Water Droplets Using PIV. Water. 2020; 12(5):1489. https://doi.org/10.3390/w12051489 Chicago/Turabian Style Karlsson, Linn, Anna-Lena Ljung, and T. Staffan Lundström. 2020. "Comparing Internal Flow in Freezing and Evaporating Water Droplets Using PIV" Water 12, no. 5: 1489. https://doi.org/10.3390/ Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. 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Cost-Volume-Profit CVP Analysis: What It Is and the Formula for Calculating It - DEER Jeans Thus each unit sold contributes $100 to covering fixed costs and increasing profit. Segregation of total costs into its fixed and variable components is always a daunting task to do. This visual line chart tells your story clearly outlining revenue, fixed costs, and total expenses, and the breakeven point. To find out the number of units that need to be sold to break even, the fixed cost is divided by the contribution margin per unit. The simplest form of the break-even chart, wherein total profits are plotted on the vertical axis while units sold are plotted on the horizontal axis. Fixed costs are unlikely to stay constant as output increases beyond a certain range of activity. CVP simplifies the computation of breakeven in break-even analysis, and more generally allows simple computation of target income sales. It simplifies analysis of short run trade-offs in operational decisions. But we more than likely need to put a figure of sales dollars that we must ring up on the register (rather than the number of units sold). This involves dividing the fixed costs by the contribution margin ratio. The break-even point is reached when total costs and total revenues are equal, generating no gain or loss (Operating Income of $0). 1. Many might think that the higher the DOL, the better for companies. 2. Using the base case as an example, sales of $175,000 (cell D14) are calculated by multiplying the $250 sales price per unit (cell D5) by 700 units (cell D8). 3. The contribution margin indicates the amount of money remaining after the company covers its variable costs. Being plugged into your financial reports ensures this valuable data is updated in real-time. Computing the break-even point is equivalent to finding the sales that yield a targeted profit of zero. The first step required to perform a CVP analysis is to display the revenue and expense line items in a Contribution Margin Income Statement and compute the Contribution Margin Ratio. Each of these three examples could be illustrated with a change in the opposite direction. A decrease in sales quantity would not impact the contribution margin ratio. A decrease in unit selling price would also decrease this ratio, and a decrease in unit variable cost would increase it. Any change in fixed costs, although not illustrated in the examples, would not affect the contribution margin ratio. Alternatively, if the selling price per unit increases from $25 to $30 per unit, both operating income and the contribution margin ratio increase as well. Variable cost per unit remains at $10 and fixed costs are still $8,000. Understanding the basics of CVP analysis With $20,000 fixed costs/divided by the contribution margin ratio (.4) we arrive at $50,000 in sales. Therefore, if we ring up $50,000 in sales this will allow us to break even. CVP analysis is conducted to determine a revenue level required to achieve a specified profit. 2: Cost Volume Profit Analysis (CVP) CVP analysis makes several assumptions, including that the sales price, fixed and variable costs per unit are constant. Running a CVP analysis involves using several equations for price, cost, and other variables, which it then plots out on an economic graph. Before creating https://www.wave-accounting.net/ the graph, it’s important to have the necessary data ready. This typically includes the total fixed costs, the variable cost per unit, the selling price per unit, and the total volume or quantity. Once you have this data, you can move on to creating the graph. Difference Between CVP Analysis and Break Even Analysis The light green line represents the total sales of the company. This line assumes that as more units are produced more units are sold. The point where the total costs line crosses the total sales line represents the breakeven point. This point is where revenues from sales equal the total expenses. In other words, this is the point of production where sales revenue will cover the costs of This video will give you an example of the why andhow to do a contribution margin income statement. The contribution margin can be stated on a gross or per-unit basis. It represents the incremental money generated for each product/unit sold after deducting the variable portion of the firm’s costs. What limitations exist when using either graph? We calculate it by subtracting variable costs per unit (V) from the selling price per unit (S). The total revenue line shows how revenueincreases as volume increases. Total revenue is $ 120,000 for salesof 6,000 tapes ($ 20 per unit X 6,000 units sold). In the chart, wedemonstrate the effect of volume on revenue, costs, and net income,for a particular price, variable cost per unit, and fixed cost perperiod. Finally, if the selling price per unit remains at $25 and fixed costs remain the same, but unit variable cost increases from $10 to $15, total variable cost increases. In either case, the assumed cost relationships would no longer be valid. The contribution margin ratio with the unit variable cost increase is 40%. The additional $5 per unit in the variable cost lowers the contribution margin ratio 20%. Because a brewpub does not sell “units” of a specific product, the owners found the break-even point in sales dollars. The owners knew the contribution margin forced resignation letter ratio and all fixed costs from the financial model. With this information, they were able to calculate the break-even point and margin of safety. You most commonly see CVP analyses explained through graphs like the one below. While fixed costs remain constant at $33,050, total costs increase in proportion to units. Once sales and total costs intersect at the break-even point, all you see is profit. For accrual method businesses, depreciation and amortization count as fixed costs because they don’t change with the number of units your company sells. Since they’re non-cash expenses that don’t affect your business’s cash profits, you might choose to leave depreciation and amortization off your CVP calculation.	{"url":"https://deerjeans.id/cost-volume-profit-cvp-analysis-what-it-is-and-the-2/","timestamp":"2024-11-08T04:15:04Z","content_type":"text/html","content_length":"74770","record_id":"<urn:uuid:6bbe7eb3-8b0d-4132-bb44-f07047e751b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00891.warc.gz"}
T<sub>c</sub> enhancement in multiphonon-mediated multiband superconductivity We study a model of a multiband Fermi-surface structure to investigate its effect on the superconducting transition temperature in the limit of a high number of bands. We consider a simple limit consisting of an infinite number of identical locally pairwise coupled bands with intraband and interband hopping and a multiband generalized Bardeen-Cooper-Schrieffer Hamiltonian. The self-consistent mean-field system of equations which determines the intraband and interband order parameters decouples to two independent equations, unless the interband hopping integral is non-zero, in which case an energetically stable superconducting phase appears, where both the intraband and the interband gaps are non-zero. We demonstrate that for all values of the interband coupling constant the critical transition temperature is enhanced compared with the pure intraband critical transition temperature. The model is equivalent to a multiple momentum exchange originating from the interband coupling and thus modelling a highly anisotropic gap structure. Dive into the research topics of 'T[c] enhancement in multiphonon-mediated multiband superconductivity'. Together they form a unique fingerprint.	{"url":"https://researchportalplus.anu.edu.au/en/publications/tsubcsub-enhancement-in-multiphonon-mediated-multiband-supercondu","timestamp":"2024-11-04T23:41:58Z","content_type":"text/html","content_length":"53793","record_id":"<urn:uuid:a40a41ad-fc4b-4dc4-90b8-90a52a442265>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00822.warc.gz"}
Course - USC Schedule of Classes History of Mathematics (4.0 units) Evolution of mathematical ideas and techniques as seen through a study of the contributions of eminent mathematicians to the formulation and solution of celebrated problems. Recommended preparation: upper division MATH course. Section Session Type Time Days Registered Instructor Location Syllabus Info 39670D 001 Lecture 9:00-9:50am MWF 24 of 25 Nathaniel Emerson LVL13	{"url":"https://classes.usc.edu/term-20191/course/math-450/","timestamp":"2024-11-05T22:32:02Z","content_type":"text/html","content_length":"94576","record_id":"<urn:uuid:b38073c8-e8d0-448e-a894-46313fab3466>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00240.warc.gz"}
The measure of reciprocity defines the proportion of mutual connections, in a directed graph. It is most commonly defined as the probability that the opposite counterpart of a directed edge is also included in the graph. Or in adjacency matrix notation: $\sum_{ij} (A\cdot A')_{ij}$, where $A\cdot A'$ is the element-wise product of matrix $A$ and its transpose. This measure is calculated if the mode argument is default. Prior to igraph version 0.6, another measure was implemented, defined as the probability of mutual connection between a vertex pair, if we know that there is a (possibly non-mutual) connection between them. In other words, (unordered) vertex pairs are classified into three groups: (1) not-connected, (2) non-reciprocaly connected, (3) reciprocally connected. The result is the size of group (3), divided by the sum of group sizes (2)+(3). This measure is calculated if mode is ratio.	{"url":"https://www.rdocumentation.org/packages/igraph/versions/1.2.5/topics/reciprocity","timestamp":"2024-11-02T08:38:08Z","content_type":"text/html","content_length":"58213","record_id":"<urn:uuid:db94b3a3-850e-447c-837b-67f21cd0c3d2>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00449.warc.gz"}
Sigma DP1x Sigma DP1x Brand: Sigma Model: DP1x Megapixels: 4.70 Sensor: 20.7 x 13.8 mm Price: check here » Sensor info Sigma DP1x comes with a 20.7 x 13.8 mm Foveon sensor, which has a diagonal of 24.88 mm (0.98") and a surface area of 285.66 mm². Note: Sigma DP1x uses Foveon X3 sensor, which has 3 layers of photoelements stacked together in 1 pixel location. Traditional CCD/CMOS sensors have 1 pixel for 1 color, whereas Foveon sensor captures all 3 colors (blue, green, and red) at every pixel. Pixel density 1.65 MP/cm² If you want to know about the accuracy of these numbers, click here Actual sensor size Actual size is set to screen → change » This is the actual size of the DP1x sensor: 20.7 x 13.8 mm The sensor has a surface area of mm². There are approx. 4,700,000 photosites (pixels) on this area. Pixel pitch, which is a measure of the distance between pixels, is µm. Pixel pitch tells you the distance from the center of one pixel (photosite) to the center of the next. Pixel or photosite area is µm². The larger the photosite, the more light it can capture and the more information can be recorded. Pixel density tells you how many million pixels fit or would fit in one square cm of the sensor. Sigma DP1x has a pixel density of These numbers are important in terms of assessing the overall quality of a digital camera. Generally, the bigger (and newer) the sensor, pixel pitch and photosite area, and the smaller the pixel density, the better the camera. If you want to see how DP1x compares to other cameras, Brand: Sigma Model: DP1x Effective megapixels: 4.70 Total megapixels: 4.70 Sensor size: 20.7 x 13.8 mm Sensor type: Foveon Sensor resolution: 2655 x 1770 Max. image resolution: 2640 x 1760 x 3 Crop factor: 1.74 Optical zoom: 1x Digital zoom: No ISO: Auto, 50, 100, 200, 400, 800, 1600, 3200 RAW support: Manual focus: Normal focus range: 30 cm Macro focus range: Focal length (35mm equiv.): 28 mm Aperture priority: Yes Max aperture: Max. aperture (35mm equiv.): n/a Depth of field: simulate → Metering: Centre weighted, Evaluative, Spot Exposure Compensation: ±3 EV (in 1/3 EV steps) Shutter priority: Yes Min. shutter speed: 30 sec Max. shutter speed: 1/4000 sec Built-in flash: External flash: Viewfinder: None White balance presets: 6 Screen size: 2.5" Screen resolution: 230,000 dots Video capture: Storage types: SDHC, Secure Digital USB: USB 1.0 Battery: Lithium-Ion rechargeable battery Weight: 250 g Dimensions: 113.3 x 59.5 x 50.5 mm Year: 2010 Compare DP1x with another camera Diagonal is calculated by the use of Pythagorean theorem: = sensor width and = sensor height Sigma DP1x diagonal: = 20.70 mm = 13.80 mm Diagonal = √ 20.70² + 13.80² = 24.88 mm Surface area Surface area is calculated by multiplying the width and the height of a sensor. Width = 20.70 mm Height = 13.80 mm Surface area = 20.70 × 13.80 = 285.66 mm² Pixel pitch Pixel pitch is the distance from the center of one pixel to the center of the next measured in micrometers (µm). It can be calculated with the following formula: Pixel pitch = sensor width in mm × 1000 sensor resolution width in pixels Sigma DP1x pixel pitch: Sensor width = 20.70 mm Sensor resolution width = 2655 pixels Pixel pitch = 20.70 × 1000 = 7.8 µm Pixel area The area of one pixel can be calculated by simply squaring the pixel pitch: Pixel area = pixel pitch² You could also divide sensor surface area with effective megapixels: Pixel area = sensor surface area in mm² effective megapixels Sigma DP1x pixel area: Pixel pitch = 7.8 µm Pixel area = 7.8² = 60.84 µm² Pixel density Pixel density can be calculated with the following formula: Pixel density = ( sensor resolution width in pixels )² / 1000000 sensor width in cm You could also use this formula: Pixel density = effective megapixels × 1000000 / 10000 sensor surface area in mm² Sigma DP1x pixel density: Sensor resolution width = 2655 pixels Sensor width = 2.07 cm Pixel density = (2655 / 2.07)² / 1000000 = 1.65 MP/cm² Sensor resolution Sensor resolution is calculated from sensor size and effective megapixels. It's slightly higher than maximum (not interpolated) image resolution which is usually stated on camera specifications. Sensor resolution is used in pixel pitch, pixel area, and pixel density formula. For sake of simplicity, we're going to calculate it in 3 stages. 1. First we need to find the ratio between horizontal and vertical length by dividing the former with the latter (aspect ratio). It's usually 1.33 (4:3) or 1.5 (3:2), but not always. 2. With the ratio ( ) known we can calculate the from the formula below, where is a vertical number of pixels: (X × r) × X = effective megapixels × 1000000 → X = √ effective megapixels × 1000000 3. To get sensor resolution we then multiply with the corresponding ratio: Resolution horizontal: Resolution vertical: Sigma DP1x sensor resolution: Sensor width = 20.70 mm Sensor height = 13.80 mm Effective megapixels = 4.70 r = 20.70/13.80 = 1.5 X = √ 4.70 × 1000000 = 1770 Resolution horizontal: X × r = 1770 × 1.5 = 2655 Resolution vertical: X = 1770 Sensor resolution = 2655 x 1770 Crop factor Crop factor or focal length multiplier is calculated by dividing the diagonal of 35 mm film (43.27 mm) with the diagonal of the sensor. Crop factor = 43.27 mm sensor diagonal in mm Sigma DP1x crop factor: Sensor diagonal = 24.88 mm Crop factor = 43.27 = 1.74 35 mm equivalent aperture Equivalent aperture (in 135 film terms) is calculated by multiplying lens aperture with crop factor (a.k.a. focal length multiplier). Sigma DP1x equivalent aperture: Aperture is a lens characteristic, so it's calculated only for fixed lens cameras. If you want to know the equivalent aperture for Sigma DP1x, take the aperture of the lens you're using and multiply it with crop factor. Crop factor for Sigma DP1x is 1.74 Enter your screen size (diagonal) My screen size is inches Actual size is currently adjusted to screen. If your screen (phone, tablet, or monitor) is not in diagonal, then the actual size of a sensor won't be shown correctly.	{"url":"https://www.digicamdb.com/specs/sigma_dp1x/","timestamp":"2024-11-05T16:09:57Z","content_type":"text/html","content_length":"29880","record_id":"<urn:uuid:64f8ce43-5d5e-4c23-a539-1bfcec145363>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00702.warc.gz"}