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PHY/ECE 624, Open Quantum Systems, PHY/ECE 624, Open Quantum Systems, Spring 2023 Instructor: Prof. Thomas Barthel Class Sessions: Tuesday+Thursday 17:00-18:15 in Hudson Hall 208 Office hours: Thursday 11:15-12:30 in Physics 287 and upon request In several experimental frameworks, a high level of control on quantum systems has been accomplished. Due to practical constraints and our aim of manipulating these systems efficiently, they are inevitably open in the sense that they are coupled to the environment. This generally leads to dissipation and decoherence, which pose challenges for modern quantum technology. On the other hand, one can design environment couplings to achieve novel effects and, e.g., to stabilize useful entangled states for quantum computation and simulation. The description of open systems goes beyond the unitary dynamics covered in introductory quantum mechanics courses; it involves intriguing new mathematical aspects and physical phenomena. This course provides an introduction to open quantum systems. We will start by discussing quantum mechanics of composite systems, which leads us from pure states to density operators and from unitary dynamics to quantum channels. At this stage, we can already gain an understanding of decoherence, dephasing, and error correction. We will then derive and discuss the Lindblad master equation, which describes the evolution of Markovian systems. It covers, for example, systems weakly coupled to large baths or closed quantum systems with external noise. As we will see in applications for specific models, it can explain dissipation, decoherence, and thermalization. We will talk about prominent experimental platforms for quantum computation and simulation from this viewpoint. The analog of Hamiltonians for closed systems are Liouville super-operators for open systems. As they are non-Hermitian, interesting mathematical aspects arise. We will discuss fundamental properties like the spectrum and their connection to phase transitions in the nonequilibrium steady states. We will close with a summary of analytical and numerical methods for the investigation of open many-body systems, addressing third quantization, perturbation theory, exact diagonalization, quantum trajectories, tensor networks, and the Keldysh formalism. Depending on the available time, we may also address examples for non-Markovian dynamics and corresponding techniques. The course is intended for students from physics, quantum engineering, quantum chemistry, and math. We expect basic knowledge of quantum mechanics, e.g., on the level of PHY 464 or ECE 521 (Schrödinger equation, bra-ket notation, spin, tensor product). Lecture Notes [Are provided on the Sakai site PHYSICS.624.02.Sp23] Recommended reading for large parts of the course: • Breuer, Petruccione "Theory of Open Quantum Systems", Oxford University Press (2002), • Rivas, Huelga "Open Quantum Systems", Springer (2012), • Nielsen, Chuang "Quantum Computation and Quantum Information", Cambridge University Press (2000). The more advanced topics will be based on current research literature.	{"url":"http://webhome.phy.duke.edu/~barthel/L2023-01_OpenQuantumSystems_phyece624/","timestamp":"2024-11-03T22:53:39Z","content_type":"text/html","content_length":"5944","record_id":"<urn:uuid:9e89e174-9e8d-4091-bc8d-4f304b237b54>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00337.warc.gz"}
Feature Column from the AMS Bin Packing and Machine Scheduling 5. The list processing algorithm for machine scheduling A very simplified approach to machine scheduling will now be given. The scheduling problems we consider are completely deterministic. This means that the times for completing every task making up a complex job are known in advance. Furthermore, we are given a directed graph which is called a task-analysis digraph, which indicates which tasks come immediately before other tasks. An arrow from task T[i] to T[j] means that task T[i] must be completed before task T[j] can begin. Such restrictions are very common in carrying out the individual tasks that make up a complex job. We all know that in building a new home, the laying of the foundation must occur before the roof is put on. Note that this digraph will have no directed circuit (e.g a way to follow edges and start at a vertex and return there) because that would lead to a contradiction in the precedence relations between the tasks. Also, in the task-analysis digraph below, each vertex, which represents a task, has the time for the task inside the circle corresponding to that vertex. Figure 1 We want to schedule these six tasks on some fixed number of identical machines in such a manner that the tasks are completed by the earliest possible time. This is sometimes referred to as finding the makespan for the tasks that make up the job. We also want to be specific about which tasks should be scheduled on which machines during a given time period. We will assume that the scheduling is carried out so that no machine will remain voluntarily idle, and that once a machine begins working on a task it will continue to work on it without interruption until the task is completed. There is an appealing heuristic with which to approach this problem. The heuristic has the advantage of being relatively simple to program on a computer, and when carried out by hand by different people, leads to the same schedule for the tasks (because the ways ties can be broken is specified). Typically, the heuristic gives a good approximation to an optimal schedule. This algorithm (heuristic) is known as the list processing algorithm. The algorithm works by coordinating the scheduling of the tasks on the machines and taking account of a "priority list" and then coordinating this list with the demands imposed by the task analysis digraph. You can think of the given list as a kind of priority list which is independent of the scheduling requirements imposed by the task analysis digraph. The tasks are given in the list so that when read from left to right, tasks of higher priority are listed first. For example, the list may reflect an ordering of the tasks based on the size of cash payments that will be made when the tasks are completed. Alternatively, the list may have been chosen with the specific goal of trying to minimize the completion time for the tasks. With respect to constructing a schedule, a task is called ready at a time t if it has not been already assigned to a processor and all the tasks which precede it in the task analysis digraph have been completed by time t. For the task analysis digraph in Figure 1 the tasks ready at time 0 are T[1], T[2], and T[3]. Remember that we are assuming that machines do not stay voluntarily idle. This means that as soon as one processor's task is completed, it will look for a new task on which to work. In determining what this next task should be, one takes into account where on the priority list the task appears, as well as any constraints imposed by the task analysis digraph. A machine will stay idle for a period of time only if there are no ready tasks (unassigned to other machines) that are ready at the given time. The list processing heuristic assigns at a time t a ready task (reading from left to right) that has not already been assigned to the lowest-numbered processor which is not currently working on a As an example, consider trying to schedule the tasks in Figure 1 on two machines. I will refer to the two machines which must be scheduled as Machines 1 and 2. Suppose we are given the list L = T[1], T[2], T[3], T[4], T[5], T[6]. At time 0, Machine 1 being idle, and T[1] being ready and first in the list, we can schedule T[1] on Machine 1, which will keep that machine busy until time 8. Now Machine 2 is free at time 0 so it also seeks a task to work on at time 0. The next task in the list, T[2], is ready at 0 so Machine 2 can start at time 0 on task T[2]. Both machines are now "happily" working until time 8 when Machine 1 becomes free. Since T[3] is next in the priority list, and its predecessors (there are none) are done at time 8, Machine 1 can work on this task because it is ready at time 8. However when time 13 comes, Machine 2, just finishing T[2], would like to begin the next task in the priority list which has not yet been assigned to a machine. This would be T[4] but this task is not ready at time 13 because T[3] has not been completed. So Machine 2 tries the next task in the priority list, T[5], and this task being ready at 13 (both T[1] and T[2] are done) can be assigned to Machine 2. You can keep track of what is going on in Figure 2 below where the tasks assigned to Machine 1 are represented in the top row and the tasks assigned to Machine 2 are shown in the bottom row. Idle time on a machine is represented in blue. Figure 2 Continuing our analysis of how Figure 2 is generated, what happens when time 19 arrives, and Machine 2 tries to find an unassigned task in the priority list? Since both task T[4] and T[6] are not ready at time 19 (because they can only begin when T[3] is done), Machine 2 will stay idle until time 22. At time 22, both machines are free, and in accordance with our tie-breaking rule, T[4] gets assigned to Machine 1 while T[6] gets assigned to Machine 2. The completion time for this list is 34. Is this the best we can do? One way to check, when task times are integers, is that we know that a lower bound for completing the tasks is the ceiling function applied to the following quotient: the sum of all task times divided by the number of machines. In this case, we get a lower bound of 30, which means that there is perhaps some hope of finding a better schedule, one that finishes by time In addition to 30 there is potentially another independent lower bound that we can take into account. Suppose we find the length of all the paths in the task analysis digraph. In this example, two typical such paths are T[2], T[5] and T[3], T[4]. The lengths of these two paths, computed by summing the numbers inside the vertices making up the paths, are respectively, 19 and 26. How are these numbers related to the completion time for all of the tasks? Clearly, since we are working with directed paths, the early tasks in the paths must be completed before the later ones. Thus, the completion time for all the tasks is at least as long as the time necessary to complete all of the tasks in any of the paths in the task analysis digraph. In particular, the earliest completion time is at least as long as the length of the longest path in this digraph. In this example, the length of this longest path is 26 and the path involved is T[3], T[4]. The path in the task analysis digraph which has the longest length is known as a critical path. A given task analysis digraph has one or more critical paths. (This is true even if the digraph has no directed edges. In this case the length of the critical path is the amount of time it takes to complete the longest task.) Speeding up tasks that are not on the critical path does not affect this lower bound, regardless of the number of machines available. The earliest completion time must still be at least as long as the critical path, despite having a lot of processors to do the tasks not on the critical path(s). Is it possible to improve the schedule that is displayed in Figure 2? One idea is to use a list to prevent the difficulties when lengthy tasks appear late in a list or when tasks on the critical path (s) are given "high priority." Thus, one can construct the list which orders the tasks by decreasing time (ties broken in favor of tasks with a lower number). The decreasing time list in this case would be: T[3], T[2], T[4, ]T[1], T[6], T[5]. This list leads to a schedule with completion time 32. You can practice trying to use the list processing algorithm on this list with 2 processors. The result is a schedule that finishes at time 32 with only 4 time units of idle time on the second processor. Here is the way the tasks are scheduled: Machine 1: T[3], T[4], T[5]; Machine 2:[ ]T[2], T [1], T[6], idle from 28-32. Although this schedule is better than the one in Figure 2, it may not be the optimal one, because there is still hope that a schedule which uses 30 time units on each machine with no idle time is possible. Is there another list we could try that might achieve a better completion time? We have mentioned that tasks on the critical path are "bottlenecks" in the sense that when they are delayed, the time to complete all the tasks grows. This suggests the idea of a critical path list. Begin by putting the first task on a longest path (breaking ties with the task of lowest number) at the start of the list. Now remove this task and edges that come out of it from the task analysis digraph and repeat the process by finding a new task to add to the list that is at the start of a longest path. Doing this gives rise to the list T[3], T[2], T[1, ]T[4], T[6], T[5]. This list, though different from the decreasing time list actually, gives rise to exactly the same schedule we had before that finished at time 32. There are 6! = 720 different lists that can arise with six tasks, but the schedules that these lists give rise to need not be different, as we see in this case. We have tried three lists and they each finish later than the theoretical, but a priori possible, optimum time of 30. There is also the possibility that there is a schedule that completes at time 31, with 2 units of idle time. You can check that no sum of two sets of task times yields a value of 30. You can also check that although there are two sets of task times (e.g. 13, 12, and 6; 14, 8 7) that sum to 31 and 29, no schedule based on the assignment of the associated tasks in order to schedule two machines obeys the restrictions imposed by the task analysis digraph. Thus, with a bit of effort one sees that the optimal schedule in this case completes at time 32. The analysis of this small example mirrors the fact that for large versions of the machine scheduling problem, there is no known polynomial time procedure that will locate what the optimal schedule might be. This point brings us full circle to why the list processing heuristic was explored as a way of trying to find good approximate schedules. More is known for the case where the tasks making up the job can be done in any order, so-called independent tasks. This is the case where the task analysis digraph has no edges. Ronald Graham has shown that for independent tasks the list algorithm finds a completion time using the decreasing time list which is never more than where T is the optimal time to schedule the tasks and m (at least 2) is the number of machines the tasks are scheduled on. This result is a classic example of the interaction of theoretical and applied mathematics. Bin packing, an applied problem, led to many application insights as well as tools for solving a variety of theoretical problems. Bin packing relates to some machine scheduling problems, which in turn have rich connections with both pure and applied problems. Next month, I will explore some of these connections.	{"url":"http://www.ams.org/publicoutreach/feature-column/fcarc-packings5","timestamp":"2024-11-10T09:42:41Z","content_type":"text/html","content_length":"58395","record_id":"<urn:uuid:e6ccdb86-cdb3-4620-9b25-f0491f815dd1>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00092.warc.gz"}
2.3: Percents Last updated Page ID $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$ $ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$ $ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$ $ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vectorC}[1]{\textbf{#1}} $ $ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $ $ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $ $ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $ $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\ evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\ newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y} $ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real} {\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec} [3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array} {r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\ wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\ newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var} {\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left\|#1\right\|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\ bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\ widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$ The final two sections of this part will focus on a very specific type of ratio: a percent. In this chapter we'll focus on the one basic equation that describes all problems involving basic percents, and practice using it in context. • Understand and use the word "percent" in context • Recognize and convert between various representations of percents • Use the Basic Percent Equation to solve problems involving percents Definition and Representation of Percents We start with the main definition: A percent is a ratio with a denominator of 100. The percent sign $\%$ is equivalent to the fraction $\frac{n}{100}$. Recall that the denominator is the bottom of the fraction; that is, the second number in the ratio. The word percent literally means "per $100$." That is, a percent expressed as a portion of a whole when that whole is divided into hundredths. A quick note about language: some people use the words percent and percentage interchangeably, but technically they don't mean the same thing. A percentage is a relative amount (for example, "a large percentage") and a percent is a specific amount (for example, "$20$ percent"). This isn't a point we'll belabor, but this book will specifically use the word percent when referring to a specific Before we solve problems involving percents, it's important to know that a percent can be expressed in three ways: using a percent sign, as a decimal, or as a fraction. It is helpful to know how to convert between these. We'll do a couple examples in detail, and then provide a table of other examples for reference. • Express $96\%$ as a decimal and as a fraction. • Express $\frac{3}{20}$ as a decimal and as a percent. • The percent denoted $96\%$ is equivalent to the fraction $\frac{96}{100}$ by definition of percent. The equivalent decimal is $.96$. • Calculating $\frac{3}{20} = 3 \div 20$, we see that the fraction $\frac{3}{20}$ is equivalent to the decimal $.15$. This decimal is $15$ hundredths, or $\frac{15}{100}$ so the corresponding percent is $15\%$. Note in this example we see that multiple fractions can correspond to the same percent. In the previous example, you may have noticed a quick way to convert between decimals and percents: To convert a decimal to a percent, move the decimal point two places right, and then put a percent sign at the end. For example: \[.127 = 12.7\%\] To convert a percent to a decimal, move the decimal point two places left, and then drop the percent sign. For example: \[56.3\% = .563\] Additional examples are shown below to make these rules more clear: Percent Decimal Fraction $14\%$ $.14$ $\frac{14}{100}$ $8\%$ $.08$ $\frac{8}{100}$ $.5\%$ or "half a percent" $.005$ $\frac{.5}{100}$ or $\frac{5}{1000}$ $25\%$ $.25$ $\frac{25}{100}$ or $\frac{1}{4}$ $123\%$ $1.23$ $\frac{123}{100}$ The Basic Percent Equation There is one equation that governs all percent relationships. The equation that describes all relationships involving percents is \[\text{percent } \times \text{ whole } = \text{ part}\] We will call this the Basic Percent Equation. It will be used to solve all basic percent problems. The Basic Percent Equation can be rearranged using Division Undoes Multiplication in the following way: \[\text{percent } = \frac{\text{part}}{\text{whole}}\] or as \[\text{whole } = \frac{\text{part}}{\text{percent}}\] There is one important caveat to using this equation: To use the Basic Percent Equation, your percent must be expressed as a decimal. Let's see some examples of the Basic Percent Equation in action. $91%$ of people in the world are right-handed. In a randomly selected group of $745$ people, how many do you expect to be right-handed? We need to identify the percent, whole, and part in this equation. The percent, if given, will always have a percent sign next to it. So the percent in this case is $91\%$, which we need to express as the decimal, $.91$. In this problem, the number $745$ represents the whole, because that is the total number of people. We are being asked to find the part of that whole that is right-handed. So we will be finding the We use the Basic Percent Equation: \[\text{percent} \times \text{whole} = \text{part}\] and fill in what we know, which is $\text{percent } =.91$ and whole $=745$. We now have \[.91 \times 745= \ text{part}\] Once we compute the left side, we find that \[ 677.95 = \text{ part}\] We will round this to the nearest whole since we're talking about a number of people. Thus, $678$ people will be Here's another example: According to the WOU website, $34\%$ of WOU undergraduates are men. Suppose there are currently $1123$ undergraduate men at WOU. How many total undergraduates are there at WOU? Once again, we start by identifying the percent, whole, and part in this equation. The percent is $34\%$, which we express as the decimal $.34$. In this case, we are being asked to find the total, so the "whole" is the unknown quantity. We are told that the part — the number of undergraduate men — is equal to $1123$. Using the Basic Percent Equation, \[\text{percent} \times \text{ whole } = \text{part}\] we substitute what we are given: \[\text{percent} \times \text{whole } = \text{part}\] \[.34 \times \text{ whole } = 1123\] Now, since Division undoes Multiplication, we have that \[\text{whole } = \frac{1123}{.34} \approx 3306\] Since this is a number of people, we rounded to the nearest whole. This means that there are approximately $3306$ WOU undergraduates. In the next section, we'll see some particular applications of percents. Remember to read carefully and answer the question that is being asked! 1. Fill in the missing spots in this table. Copy the entire table in your answer. Percent Decimal Fraction 2. Twelve percent of Polk County residents speak Spanish fluently. There are $9048$ fluent Spanish speakers in Polk County. How many total residents are there in Polk County? 3. In a certain dorm on campus, $13$ people are social science majors, $12$ people are natural science majors, $17$ people are education majors, and $9$ people have other majors. What percentage of people in the dorm are natural science majors? Round to the nearest tenth of a percent. 4. In a certain acre of forest, there are $457$ deciduous trees and $1035$ trees in total. What percentage of the trees in this acre of forest are non-deciduous? Round to the nearest tenth of a 5. You're trying to save up to put a $10\%$ down payment on a house. You hope to purchase a $\$315,000$ house. Your plan to save equal amounts of money each month for four years to reach your goal. How much will you need to save each month? (Assume the saved money earns no interest.) 6. Read this article. After reading the article, answer the following questions: 1. Oregon's population in $2020$, when this article was written, was estimated to be $4,455,920$ people. According to the statistics given in this article, what number of people in Oregon are non-Hispanic white people? Make sure to show your calculation. 2. Assume there are $135,438$ people of color under the age of $15$ living in Oregon. According to a statistic given in this article (in the second half), calculate how many total people under the age of $15$ live in Oregon. Make sure to show your calculation. 3. Answer at least one of the following questions, writing at least $2$ sentences. ☆ Did anything surprise you when reading this article? If so, what was it? ☆ Are you curious about other statistics relating to the population of Oregon? If so, what would you ask?	{"url":"https://math.libretexts.org/Courses/Western_Oregon_University/Math_110%3A_Applied_College_Mathematics/02%3A_Numbers_in_Context/2.03%3A_Percents","timestamp":"2024-11-12T02:55:12Z","content_type":"text/html","content_length":"137646","record_id":"<urn:uuid:96e94396-fa08-4e1e-a6fe-d900c961c72c>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00501.warc.gz"}
Write your model¶ scikit-fdiff use Sympy to process the mathematical model given by the user. It should be able to solve every systems that can be written as \[\frac{\partial U}{\partial t} = F(U, \partial_{x,y,z...} U)\] U being a single dependent variable or a list of dependent variables. The time t is a mandatory independent variable (and is hard-coded), but you can have any other spatial independent variables (x, y, z, r…), as long as they are written with only one character. We can use the 2D heat diffusion as example: \[\frac{\partial T(x,y)}{\partial t} = k\,(\partial_{xx} T + \partial_{yy} T)\] You only have to write the right-hand side (rhs) : the python code will look like >>> from skfdiff import Model >>> model = Model("k * (dxxT + dyyT)", ... "T(x, y)", parameters="k") Be careful, the (x, y) after the dependent variable is mandatory. If not set, scikit-fdiff will consider a one-dimension problem (T(x)). You can also set a variable parameter that can depend on coordinate by specifying its coordinate dependency (as k(x, y)). You will then be able to use its derivative in the equation (aka, use dxk, dyk… in the evolution equations). from skfdiff import Model model = Model("k * (dxxT + dyyT) + dxk", "T(x, y)", parameters="k(x)") As sympy does the symbolic processing, all the function in the sympy namespace should be available if the backend allow it. The default numpy backend allow all the function available with sympy.lambdify using the modules="scipy" argument (all function available in numpy, scipy and :py:mod:scipy.special). Be aware that if sympy cannot derive them, the exact Jacobian matrix computation will not be available anymore, as well as the implicit solvers. Write the boundary conditions¶ The boundary conditions (bc) is written as a dictionary of dictionary. They follow this structure: bc ={("dependent_variable_1", "coordinate_1"): ("left_bc", "right_bc"), ("dependent_variable_1", "coordinate_2"): ("left_bc", "right_bc"), ("dependent_variable_2", "coordinate_1"): ("left_bc", "right_bc"),} For example, you can implement a mixed dirichlet (fixed temperature) and Neumann (fixed flux) for the 2D heat equation with >>> bc = {("T", "x"): ("dirichlet", "dirichlet"), ... ("T", "y"): ("dxT - phi_left", "dxT + phi_right")} You can then use it in the model definition model = Model("(dxxT + dyyT)", "T(x, y)", You can also use a periodic boundary by replacing a tuple (“left_bc”, “right_bc”) by “periodic” for periodic boundary conditions: >>> bc = {("T", "x"): ("dirichlet", "dirichlet"), ... ("T", "y"): "periodic"} And you use the alias “periodic” or “noflux” without dictionary for periodic or adiabatic boundaries everywhere. That latter is the default behaviour if no boundary condition is specified. >>> model = Model("(dxxT + dyyT)", "T(x, y)", ... boundary_conditions="periodic") >>> model = Model("(dxxT + dyyT)", "T(x, y)", ... boundary_conditions="noflux") is the same as >>> model = Model("(dxxT + dyyT)", "T(x, y)") Use upwind scheme to deal with advective problems¶ When a problem have a strong advective part and not enough diffusion to counter the advective part, the centred finite difference can not maintain stiff shapes. This can be dealt with different strategy. You can add some artificial diffusion at risk to loose the solution accuracy, use a more advanced filter (the diffusion can be seen as a simple filter that will smooth high frequencies) or use a strongly implicit temporal scheme that will inject numerical diffusion (as backward Euler with large time-step…). An other solution is to use upwind schemes, a dedicated discretization method that will de-centred the scheme in direction of the flow. That method will still add some numerical diffusion, especially at low order, but is more accurate that the other method mentioned earlier. It can be used with the custom function "c * dxU" ~= "upwind(c, U, x, n)", n being the upwind scheme order (between 1 and 3). You can have more info in the dedicated advection page. As small example, you can compare the inviscid Burger equation without and with upwind scheme in the cookbook. Don’t repeat yourself (DRY) with auxiliary definitions¶ You can make your model more concise by specifying one or more auxiliary definitions. It’s a dictionary written as {“alias”: “skfdiff expression”}. This avoid repetition, or can make a model easier to write and to read. For example, you can write a moisture flow in a wall with something like model = Model(["-u * dxCg + D * dxxCg - r", "r"], ["Cg(x)", "Cs(x)"], parameters=["d", "Pext", "Pint", "D", "k1", "k2"], subs=dict(u="d * (Pext - Pint)", r="k1 * Cg - k2 * Cs")) A word on backends: You can choose the backend when you write your model. For now, there is only two of them available: the default numpy backend and the numba one. The first has reasonable speed and is “easy” to install in every hardware. It’s also the one that have been most tested. But the parallisation only occur for big multidimensional domain. The latter is faster (~30% than numpy), but is linked to some hard dependencies as LLVM and relie to Just-In-Time (JIT) compilation. The first computation will be long to initiate (warm-up effect), and that warm-up can be very long for complex models. That being said, it allows full parallisation, leading to huge speed-up for explicit solvers. >>> from skfdiff import Model >>> numpy_model = Model("k * (dxxT + dyyT)", "T(x, y)", ... parameters="k", backend="numpy") >>> numba_model = Model("k * (dxxT + dyyT)", "T(x, y)", ... parameters="k", backend="numba") Initialize your fields¶ The Fields are the data (dependent variable and coordinates as well as parameters) that your simulation will need. They are stored in a xarray.Dataset, a smart container that allow coordinate aware arrays (see the Xarray documentation ). The model give you access to a dedicated builder that will raise an error if you forgot some data. Xarray Datasets come with nice features as built-in plots, interpolation… which is then available in skfdiff. >>> import pylab as pl >>> import numpy as np >>> from skfdiff import Model >>> model = Model("k * (dxxT + dyyT)", "T(x, y)", parameters="k") >>> x = np.linspace(-np.pi, np.pi, 56) >>> y = np.linspace(-np.pi, np.pi, 56) >>> xx, yy = np.meshgrid(x, y, indexing="ij") >>> T = np.cos(xx) * np.sin(yy) >>> initial_fields = model.Fields(x=x, y=y, T=T, k=1) >>> initial_fields["T"].plot() <matplotlib.collections.QuadMesh object ...> (png, hires.png, pdf) High level simulation handler¶ You have the model as well as the initial condition : you are ready to launch the simulation. The high level object in scikit-fdiff is the Simulation. It takes the model, the initial fields (as Fields or as dictionary), a time-step and some optional parameters that will feed the temporal scheme. Most of them takes at least a parameter time_stepping: bool that default to True and a parameter time_stepping: float that will drive the time stepping. For the other solver parameters, refer to the temporal scheme page. >>> import pylab as pl >>> import numpy as np >>> from skfdiff import Model, Simulation >>> model = Model("k * (dxxT + dyyT)", "T(x, y)", parameters="k") >>> x = np.linspace(-np.pi, np.pi, 56) >>> y = np.linspace(-np.pi, np.pi, 56) >>> xx, yy = np.meshgrid(x, y, indexing="ij") >>> T = np.cos(xx) * np.sin(yy) >>> initial_fields = model.Fields(x=x, y=y, T=T, k=1) >>> simulation = Simulation(model, initial_fields, dt=0.1, tmax=1) >>> tmax, last_fields = simulation.run() >>> last_fields["T"].plot() <matplotlib.collections.QuadMesh object ...> (png, hires.png, pdf) For better control of the simulation process, you can iterate on each time-step instead of running the simulation until you reach tmax. The Simulation object is an iterator itself that yield the tuple (t, fields) at each time step. You can then print the result, plot the fields, or do some control (stopping the simulation if something wrong happened for example). >>> simulation = Simulation(model, initial_fields, dt=0.1, tmax=10) >>> for t, fields in simulation: ... print("time: %g, T mean value: %g" % ... (t, fields["T"].mean())) time: ..., T mean value: ... See also How to use the Simulation.stream: Streamz.Stream interface to automate such process. Most of the daily process are already available in the skfdiff.plugins or implemented in the Simulation: • Post-process on the fields to compute derived fields (as flux, mean values…). • Keep the results for some or each time-steps. • Real time display in a jupyter notebook. Modify the fields on the fly with the hook¶ You can provide a hook to the simulation (in fact, you can provide a hook the same way to the lower-level skfdiff objects as the Temporal schemes and the Routines). This hook is a callable that take the actual time and fields, and return a modified fields. This hook will be called every time the evolution vector F or the Jacobian matrix J has to be evaluated. This can be useful to include strongly non-linear effect that would have been hard or impossible to formulate directly in the model. Be careful with the usage of the hook: as it allows to implement complex non-linear effect, they could be source of instabilities. It is a way to do some handy hack that overcome numerical difficulties but the resulting solution will be hardly trust-able. >>> def hook(t, fields): ... T = fields["T"].values ... # We bound the value of T, to be sure it does not go over 0.5 ... fields["T"] = ("x", "y"), np.where(T > 0.5, 0.5, T) ... return fields >>> simulation = Simulation(model, initial_fields, ... dt=0.1, tmax=10, hook=hook) Keep the results in a container¶ A container is an object that will save the fields for each or some time-steps. It can be in-memory container (by default) or save the data in a persistent way on disk. In that case, a netCDF file will be created on disk. The skfdiff.Container is a thin wrapper around the xarray library that will concatenate the fields of each time-steps and save it on disk if requested by the user. It has a main attributes, Container.data that return the underlying xarray.Dataset. >>> simulation = Simulation(model, initial_fields, dt=0.1, tmax=1) >>> container = simulation.attach_container() # in-memory container >>> t, fields = simulation.run() >>> container.data["T"].isel(t=slice(0, 10, 3)).plot(col="t") <xarray.plot.facetgrid.FacetGrid object ...> (png, hires.png, pdf) For persistent container, you can retrieve the container data with the function method retrieve_container(path). >>> from skfdiff import retrieve_container >>> simulation = Simulation(model, initial_fields, ... dt=0.1, tmax=1, ... id="heat_diff") >>> simulation.attach_container("/tmp/skfdiff_output/", ... force=True) # on-disk container path: ... >>> t, fields = simulation.run() >>> data = retrieve_container("/tmp/skfdiff_output/heat_diff") >>> data["T"].isel(t=slice(0, 10, 3)).plot(col="t") <xarray.plot.facetgrid.FacetGrid object ...> To avoid to balance the memory usage and I/O call, the data are not stored on-disk every time-steps. The buffer size can be set on container creation, and is equal to 50 by default. The container will create a new file each time it store the data. These file are merged at the end of the simulation. That strategy allow the user to manipulate the results already on disk in an other script or notebook, even with a very long simulation. If something have gone wrong and the data are not stored, you can force that with an instantiated Container container.merge(), or, using the path of the container with the static method Extend your simulation with post-processes¶ You can add extra step after each time-steps with the post-process interface. It allows to plug a callable that take the simulation as arguments to modify the simulation state itself, or to modify the actual fields. You can use it to add some metrics to the fields (as the running time, some probes as mean values or other). >>> import numpy as np >>> from skfdiff import Model, Simulation >>> model = Model("k * (dxxT + dyyT)", "T(x, y)", parameters="k") >>> x = np.linspace(-np.pi, np.pi, 56) >>> y = np.linspace(-np.pi, np.pi, 56) >>> xx, yy = np.meshgrid(x, y, indexing="ij") >>> T = np.sqrt(xx 2 + yy 2) >>> initial_fields = model.Fields(x=x, y=y, T=T, k=1) >>> simulation = Simulation(model, initial_fields, dt=.25, tmax=2) >>> def post_process(simul): ... dxT, dyT = np.gradient(simul.fields["T"].values) ... dx = np.gradient(simul.fields["x"].values) ... dy = np.gradient(simul.fields["y"].values) ... simul.fields["Tstd"] = (), simul.fields["T"].std().data ... simul.fields["dxT"] = ("x", "y"), (dxT / dx).data ... simul.fields["dyT"] = ("x", "y"), (dyT / dy).data ... simul.fields["magn"] = ("x", "y"), np.sqrt(simul.fields["dxT"] 2 + ... simul.fields["dyT"] 2).data >>> simulation.add_post_process("post_process", post_process) >>> print("Initial standard deviation: %g" % simulation.fields["Tstd"]) Initial standard deviation: ... >>> t, fields = simulation.run() >>> fig = pl.figure() >>> simulation.fields["T"].plot() <matplotlib.collections.QuadMesh object ...> >>> fig = pl.figure() >>> simulation.fields["magn"].plot() <matplotlib.collections.QuadMesh object ...> >>> print("Final standard deviation: %g" % simulation.fields["Tstd"]) Final standard deviation: ... Display the result in real-time¶ The displays are tools that allow real-time visualisation of the data, during the simulation. For now, we only provide visualisation for the simulation running in a jupyter notebook. You can plot 1D and 2D in a straightforward manner with display_fields(), and display 0D data (often obtained via a post-process) with display_probe(). You have to run the following example in a notebook. >>> from skfdiff import display_fields, enable_notebook >>> enable_notebook() >>> simulation = Simulation(model, initial_fields, ... dt=0.1, tmax=1) >>> simulation.add_post_process("post_process", post_process) >>> display_fields(simulation) <skfdiff.plugins.displays.Display object ...> >>> t, fields = simulation.run() We plan to provide some other, as real-time visualisation with matplotlib animation (that you can use everywhere a graphical server is available), and a cli tool that display the data from a on-disk skfdiff container. A word on the stream interface: The simulation object as a Simulation.stream: Streamz.Stream attribute that will be fed on the simulation itself at every time step. The post-process, display and container interfaces are built on it. You can easily build complex post-process analysis by using this stream. For that, refer to the Streamz documentation.	{"url":"https://scikit-fdiff.readthedocs.io/en/latest/overview.html","timestamp":"2024-11-12T05:13:14Z","content_type":"text/html","content_length":"68448","record_id":"<urn:uuid:fda6f0f0-44e0-4df3-8edd-4b9ebac0871b>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00260.warc.gz"}
We prove that if f is a reduced homogeneous polynomial of degree d, then its F-pure threshold at the unique homogeneous maximal ideal is at 1 least d−1^. We show, furthermore, that itsF-pure threshold equals 1 if and d−1 only if f ∈ m^[q] and d = q +1, where q is a power of p. Up to linear changes of coordinates (over a fixed algebraically closed field), we classify such “extremal singularities”, and show that there is at most one with isolated singularity. Finally, we indicate several ways in which the projective hypersurfaces defined by such forms are “extremal”, for example, in terms of the configurations of lines they can contain. ASJC Scopus subject areas • Mathematics (miscellaneous) Dive into the research topics of 'LOWER BOUNDS ON THE F-PURE THRESHOLD AND EXTREMAL SINGULARITIES'. Together they form a unique fingerprint.	{"url":"https://research.nu.edu.kz/en/publications/lower-bounds-on-the-f-pure-threshold-and-extremal-singularities","timestamp":"2024-11-04T18:56:41Z","content_type":"text/html","content_length":"52944","record_id":"<urn:uuid:8af005e4-83f5-48aa-8f90-2d37c2e06dc1>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00554.warc.gz"}
Use Elimination to solve 10c + 3s = 82 and 5c + 8s = 67 Use the elimination method to solve: 10c + 3s = 82 5c + 8s = 67 Check Format Equation 1 is in the correct format. Check Format Equation 2 is in the correct format. Step 1: Multiply Equation 1 by 5: 5 * (10c+3s=82) --> 50c + 15s = 410 Step 2: Multiply Equation 2 by 10: 10 * (5c+8s=67) --> 50c + 80s = 670 Step 3: Equation 1 - Equation 2: 50c + 15s = 410 - (50c + 80s = 670) -(50c + 80s = 670) 15s - 80s = 410 - 670 Step 4: simplify and solve for s: -65s = -260 s = 4 Step 5: Rearrange Equation 1 to solve for c: 10c = 82 - 3s Divide each side by 10 Step 6: Plug s = 4 into equation 1: c = 7 Test Your Knowledge? 1. What are the 3 methods to solve a system of 2 equations? 2. With the elimination method, what are you trying to eliminate? Common Core State Standards In This Lesson How does the Simultaneous Equations Calculator work? Free Simultaneous Equations Calculator - Solves a system of simultaneous equations with 2 unknowns using the following 3 methods: 1) Substitution Method (Direct Substitution) 2) Elimination Method 3) Cramers Method or Cramers Rule Pick any 3 of the methods to solve the systems of equations 2 equations 2 unknowns This calculator has 2 inputs. What 1 formula is used for the Simultaneous Equations Calculator? What 7 concepts are covered in the Simultaneous Equations Calculator? cramers rule an explicit formula for the solution of a system of linear equations with as many equations as unknowns to remove, to get rid of or put an end to a statement declaring two mathematical expressions are equal simultaneous equations two or more algebraic equations that share variables to put in the place of another. To replace one value with another a number or value we do not know Alphabetic character representing a number Math is always fascinating! Have you noticed? On the path to discovering the ultimate mysteries of math, we move forward with small steps. Many such discoveries of mathematical formulas validate the research efforts of those who came before us. If you want to inspire yourself and those around you, try making your own Custom Enamel Badges of achievement for small accomplishments. These vivid pin badges can be a constant reminder that numbers aren't exactly cold and that they can always be accompanied by fun and hope. Add This Calculator To Your Website	{"url":"https://www.mathcelebrity.com/simultaneous-equations.php?term1=10c+%2B+3s+%3D+82&term2=5c+%2B+8s+%3D+67&pl=Elimination","timestamp":"2024-11-12T22:18:38Z","content_type":"text/html","content_length":"49447","record_id":"<urn:uuid:92d42657-6220-4e04-a3c2-9ff2b9ea729f>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00795.warc.gz"}
RSA decryption failed, this is the code Topic: RSA decryption failed, this is the code WOLFSSL_RSA* ppri_key = NULL; int nfLen = 128; unsigned char uszDeData[1024] = {0}; ppri_key = wolfSSL_RSA_new(); ppri_key->e = wolfSSL_BN_bin2bn(PUBLIC_EXPONENT_HEX,3,NULL); ppri_key->d = wolfSSL_BN_bin2bn(PRIVATE_EXPONENT_HEX,128,NULL); ppri_key->n = wolfSSL_BN_bin2bn(MODULES_HEX,128,NULL); int nRet = wolfSSL_RSA_private_decrypt(nfLen,upszEnData,uszDeData,ppri_key,RSA_PKCS1_PADDING); It returns - 1,and uszDeData is empty. How to use large number to construct private key to decrypt？Please give me an example. Thanks! If the upszEnData length exceeds 2048,how to do it? Re: RSA decryption failed, this is the code Hi juhuaguai, I do not believe we have any examples showing the use of the WOLFSSL_RSA data structure being used in this way. One could use the internal RsaKey structure and possibly wc_RsaPublicKeyDecode to accomplish what you are attempting but let me check with the team tomorrow in a meeting and see if any of our other engineers have an example of using the WOLFSSL_RSA data structure in this way. Warm Regards, Re: RSA decryption failed, this is the code I went over this with the team and unfortunately we don't have any examples lying around, can you send a short complete test app that can be compiled and run with the commands: $ gcc test.c -o run-test -lwolfssl $ ./run-test Also note that you are only setting the e,d, and n, you're missing some elements of the private key also (p, q, dP, dQ, u). Warm Regards, Re: RSA decryption failed, this is the code I have uploaded the test code to GitHub.https://github.com/juhuaguai/temptest It is an MFC project of vs2019. The code shows the encryption and decryption of OpenSSL and wolfssl respectively. RSA of OpenSSL works normally. Wolfssl can't decrypt normally, but its encrypted string can be decrypted by OpenSSL. Please tell me how to modify the code. Thank you very much. Re: RSA decryption failed, this is the code Can you tell us a bit about the background of this project using .NET framework and solution setup? Warm Regards, Re: RSA decryption failed, this is the code This is a C + + Project and it didn't use c#,it doesn't need .Net framework。 Try it, you need to install Microsoft Visual Studio 2019, and choose C + + and MFC when installing. Do you have any mistakes? Can you send an error message? You can see the relevant code directly in the RsaTestDlg.cpp file. thank you. Re: RSA decryption failed, this is the code We would very much appreciate knowing if the project is commercial, open source, hobby, personal research etc to better classify the inquiry. Can you share a bit about the background of the project and what drove this work? Warm Regards, Re: RSA decryption failed, this is the code Don't worry. This is the temporary code I wrote during my personal research. The private key and so on are generated temporarily, without any risk. It doesn't involve any copyright and confidentiality. It's just an example for testing. All the codes in it can be shared, disclosed and modified at will.	{"url":"https://www.wolfssl.com/forums/post5188.html","timestamp":"2024-11-03T11:56:50Z","content_type":"text/html","content_length":"33084","record_id":"<urn:uuid:a2360868-6fba-4c08-8701-ecf26452abc7>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00789.warc.gz"}
What are Compound Angles\|Learn and Solve Questions Simply put, trigonometric functions—also called circular functions—are the functions of a triangle's angle. This means that these trigonometric functions provide the relationship between the angles and sides of a triangle. The fundamental trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. We need to know trigonometric ratios to get a clear view of the compound angles. An angle is created when two rays are united at a common point. The two rays are referred to as the arms of the angle, while the common point is the node or vertex. The symbol "$\pi$” stands for the angle. The Latin word "Angulus" is where the word "angle" originated. Typically, an angle is measured with a protractor and expressed in degrees. Here, several angles are shown at 30 degrees, 45 degrees, 60 degrees, 90 degrees, and 180 degrees. The angles' degrees values determine the many types of angles. Angles can be expressed in terms of pi ($\pi$) or radians. Trigonometric Ratios The ratios of the triangle's side lengths are known as trigonometric ratios. These ratios explain how the ratio of a right triangle's sides to each angle works in trigonometry. The sin, cos, and tan functions can obtain the other significant trigonometric ratios, cosec, sec, and cot. Trigon, which means "triangle," and metrôn, which means "to measure," are the roots of the term "trigonometry." It is a field of mathematics that examines how a right-angled triangle's angles and sides relate to one another. Trigonometric Functions Trigonometric Identities Trigonometric Identities come in handy whenever trigonometric functions are incorporated into an expression or equation. For every value of a variable appearing on both sides of an equation, a trigonometric identity is true. These trigonometric identities involve one or more angles' sine, cosine, and tangent. What are Compound Angles? An algebraic sum of two or more angles can define the compound angle. Compound angles represent different trigonometric identities. Compound angles are denoted using trigonometric identities. Compound angles can be used to compute the fundamental operations of finding the sum and difference of functions. Formula of Compound Angles The formula for trigonometric ratios of compound angles is given below: • $\sin {\rm{ }}(A + B){\rm{ }} = \sin {\rm{ }}A{\rm{ }}\cos {\rm{ }}B{\rm{ }} + \cos {\rm{ }}A{\rm{ }}\sin {\rm{ }}B$ • $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ • $\cos {\rm{ }}(A + B){\rm{ }} = \cos {\rm{ }}A{\rm{ }}\cos {\rm{ }}B - \sin {\rm{ }}A{\rm{ }}\sin {\rm{ }}B{\rm{ }}$ • $\cos {\rm{ }}(A - B){\rm{ }} = \cos {\rm{ }}A{\rm{ }}\cos {\rm{ }}B{\rm{ }} + \sin {\rm{ }}A{\rm{ }}\cos {\rm{ }}B{\rm{ }}$ • $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ • $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ • $\sin (A + B)\sin \left( {A - B} \right) = {\sin ^2}A - {\sin ^2}B = {\cos ^2}B - {\cos ^2}A$ • $\cos \left( {A + B} \right)\cos \left( {A - B} \right) = {\cos ^2}A - {\sin ^2}A - {\sin ^2}B = {\cos ^2}B - {\sin ^2}A$ Questions on Compound Angles Question 1: If $\cos A = \dfrac{4}{5}$ and $\cos B = \dfrac{{12}}{{13}},\dfrac{{3\pi }}{2} < A,B < 2\pi$, find the value of $\cos (A+B) $. According to the given question, $\cos A = \dfrac{4}{5}$ and $\cos B = \dfrac{{12}}{{13}}$. Also, $\dfrac{{3\pi }}{2} < A,B < 2\pi$, both A and B lie in the fourth quadrant. Then we can say that both sin A and sin B are negative. First we need to find out $\sin A$, $\begin{array}{c}\sin A = - \sqrt {1 + {{\cos }^2}A} \\ = - \sqrt {1 - \dfrac{{16}}{{25}}} \\ = - \dfrac{3}{5}\end{array}$ We need to find out $\sin B$, $\begin{array}{c}\sin B = - \sqrt {1 + {{\cos }^2}B} \\ = - \sqrt {1 - \dfrac{{144}}{{169}}} \\ = - \dfrac{5}{{13}}\end{array}$ Now, we can find out $\cos (A+B)$, $\begin{array}{c}\cos (A + B) = \cos A\cos B - \sin A\sin B\\ = \dfrac{4}{5} \times \dfrac{{12}}{{13}} - \left( {\dfrac{{ - 3}}{5}} \right) \times \left( {\dfrac{{ - 5}}{{13}}} \right)\\ = \dfrac {{48}}{{65}} - \dfrac{{15}}{{65}}\\ = \dfrac{{33}}{{65}}\end{array}$ Question 2: Determine the value of $\sin {15^\circ }$. According to the given question, we need to find the value of $\sin {15^\circ }$. $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ $\sin \left( {45 - 30} \right) = \sin 45\cos 30 - \cos 45\sin 30$ $\begin{array}{c}\sin {15^ \circ } = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}\\ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\end{array}$ Question 3: If A, B, and C are angles of a triangle, then prove that $\dfrac{A}{2} = \cot \left( {B + C} \right)/2$. According to the given question, we know A, B, and C are the angles of a triangle, $\begin{array}{l}A + B + C = n\\B + C = n - A\\\dfrac{{\left( {B + C} \right)}}{2} = \dfrac{n}{2} - \dfrac{A}{2}\end{array}$ $\begin{array}{l}\cot \dfrac{{\left( {B + C} \right)}}{2} = \cot \left( {\dfrac{n}{2} - \dfrac{A}{2}} \right)\\ = \tan \dfrac{A}{2}\end{array}$ Hence, proved. Trigonometry was developed to solve problems and take measurements involving triangles. Trigonometry is a tool used by meteorologists, engineers, scientists, engineers, engineers, engineers, and navigators. It is referred to as sum and difference formulas in American mathematics, but it is known as compound angles in British Mathematics. Trigonometric functions are utilised to obtain consistent results, including compound angles, trigonometric ratios, and multiple angles. FAQs on What are Compound Angles? Ques1: What are the different ways of naming an angle? An angle can be named in one of three ways: by its vertex, three points, or a letter or number written inside the angle's opening. Ques2: Who discovered the angle? Eudemus used the first idea, viewing an angle as a departure from a straight line; Carpus of Antioch used the second idea, viewing an angle as the distance or space between intersecting lines; and Euclid used the third idea. Ques3: Why are angles important in the real world? They are used by carpenters to make exact measurements for building doors, seats, tables, etc. Athletes use them to measure throw lengths and improve their athletic performance. Engineers use angle measurement to build structures such as buildings, bridges, homes, monuments, etc.	{"url":"https://www.vedantu.com/maths/what-are-compound-angles","timestamp":"2024-11-10T11:02:05Z","content_type":"text/html","content_length":"203804","record_id":"<urn:uuid:31587a55-93c8-4ce8-98fa-dbaacf28f30b>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00208.warc.gz"}
Benefits of Using Renewable Energy - Green Energy TipBenefits of Using Renewable Energy \| Green Energy Tip Benefits of Using Renewable Energy There are a lot of benefits of using renewable energy, in this article you will learn about them. Renewable energy is simply energy generated from complementary sources such as sun, wind, water, biological processes, and geothermal energy. Compared to fossil fuels and coal, these natural resources are usually referred to as clean forms of energy because they do not produce harmful emissions and pollutants in the atmosphere, and therefore have a little environmental impact during the production process, which contributes well to the conservation of our environment. There is no better way to reduce electricity bill charges than using renewable energy. However, some people may need more persuasion to switch to alternative sources of renewable energy. Some benefits of using renewable energy. Moreover, renewable energy is one that comes from natural resources that are consistently replenished. Examples include: • Geothermal Heat, from Earth’s interior heat • Solar Energy, from sunlight • Hydraulic Energy, from water Not every energy source is renewable, however. Non-renewable resources like coal and petroleum are limited and will go extinct one day. During this era of much-heated debate about environmental issues, renewable energy is quickly gaining strength and adherence. However, these alternative and clean means of moving the world are not always welcomed as old prejudice and ignorance remain. Nonetheless, the benefits that renewable energy can bring are many and unquestionable. Renewable energy helps protect the environment. If you are an ecologist, you should immediately switch to renewal. In a way, this is not about reducing energy bill fees. The environmental part of your care helps protect the world from pollution and global warming. The use of renewable energy can be very beneficial for the environment when it comes to reducing pollution, which is already a major threat to the entire population of the planet. Environmental impacts from the use of green energy are relatively small and more local than the widespread and widespread impacts of fossil fuels such as natural gas, coal, and oil. The use of many modern energy sources, such as coal, encourages the release of more greenhouse gases that contribute to global warming. If you use renewable energy sources, you will not pollute or heat the world. You can reduce your energy bill by 80%. Solar and wind energy can help reduce your electricity bill by up to 80%. This is a lot of savings for you. Also, some households that consume less energy may use lagging meters. This means that the power company may eventually have to pay for extra energy rather than what you do not use. reduces life cycle emissions and greenhouse gases. The major health problems caused by air pollution from fossil fuel production are of great concern. Air pollution, which worsens the respiratory function of children, has become the biggest factor in the death of many young children. Small particles caused air pollution in many deaths because the risk causes heart and lung disease. Green energy sources have very few products or are not released into the air, and are therefore the safest energy source. Economically, green energy has proven to be inexpensive in the energy-generating industry saving billions of dollars, resulting in large energy sources that will sustain many people for longer and The policies needed in many countries to create renewable energy will be restructured by many who want the next generation to be healthier and less concerned about pollution and other costly waste management issues. This should be more than sufficient to explain the benefits of renewable energy. Why Renewable Energy? There are several reasons why renewable energy is a superior choice for humanity to use in the future, for both the environment and the people living in it. Here are some reasons to consider when comparing it to the use of fossil fuels. Renewable Energy Helps the Fight Against Climate Change Over recent years, the conversation about climate change has come into the public consciousness. Many individuals are now concerned about their carbon footprint. Scientists warn that we will reach a ‘point of no return if we don’t act soon.. A big contributor to helping combat this is renewable A renewable resource such as wind or solar power is cleaner for the planet because it does not rely on emitting harmful greenhouse gases such as carbon dioxide or methane into the atmosphere. This, in turn, helps prevent damage to the ozone layer, which is the layer of molecules that protects the earth from the sun’s ultraviolet radiation. Renewable Energy Doesn’t Run Out Fossil fuels such as oil and natural gas are finite resources. Even if they were not harmful to the Earth, humanity will still eventually run out of these fuels. Renewable energy is an indispensable part of developing a self-sustaining energy system for the entire planet. As the name implies, renewable energy sources will naturally replenish themselves, or there is so much of it (i.e. water, solar power) that we can never possibly use it up. Switching to using renewable energy is an essential part of securing the future of this world. They never run out of supply The word renewable says it all. As long as the sun shines or the wind blows, you will have a reliable energy source that you can rely on. Of course, common sense will tell you that the sun and wind will never disappear until the end of the world. On the other hand, current energy sources such as coal and natural gas have limited resources that will eventually run out in the future. Renewable Energy Improves Public Health Air pollution can have a major effect on public health, especially in developing nations such as China and India. The microscopic airborne pollutants caused by fossil fuels when inhaled over the long term can lead to many negative effects on the body. When they enter the circulatory or respiratory system, they can result in a person’s lungs aging prematurely, as well as other impacts on the heart and brain, The UN considers air pollution to be a human rights violation. As a clean source of power, this is another reason why renewable energy can help contribute to a better quality of life for people across the globe and also lift the burden on national healthcare Improved Health One of the lesser-known advantages of using renewable energy is that it improves people’s health. While power plants like the ones based on coal, release toxic compounds in great quantities into the air, eventually finding its way into the bodies and lungs of people. But renewable power plants based on renewable energy don’t contain said compounds in high concentrations, significantly reducing cancer, breathing-related problems, heart attacks, and a myriad of other issues. Public health improvements Emission of carbon IV oxide harms our environment. They can also result in significant health problems. Switching to renewable energy sources improves our health; Solar, wind and hydro energy all produce fewer emissions. Also, geothermal and biomass are cleaner alternatives compared to fossil fuels. How Animals Benefit From Renewable Energy Renewable energy can help combat the effects on the lives of both wild and domesticated animals. With non-renewable sources having such a negative effect on the climate, extremely hot and cold conditions can cause heat stress for both pets and livestock. For wild animals, the pollutants from non-renewable sources can destroy their natural habitats and make it hard to access clean food and water. Oil spills can also have a disastrous effect on sea creatures. Renewable energy helps reduce the suffering of all sentient beings on the planet, not just humans. Renewable Energy is Becoming The Cheaper Option As well as being environmentally friendly, economic benefits are increasingly becoming one of the reasons why renewable energy is the best option to use. Because of the increasing environmental concern, investments are steadily increasing in renewable energy resources. Solar, wind, and hydropower are rapidly becoming competitively priced compared to their fossil fuel alternatives, while hydroelectricity leading to the cheapest energy source for electricity in the world today. In conclusion, there are many different reasons why renewable energy is the best choice for everyone. Whether the concern is being healthy, animal welfare, or the future of the planet. Lower Prices for Electricity Utilizing natural renewable resources is, by far, the smartest choice to make in the future as our electricity providers. In the first place, sources like sunlight or wind are abundant. On average, a house could get 5 hours of sunlight a day, enough for reducing the regular electricity supply and its bill in a large way. In addition, renewable resources cannot be commercialized by specific companies, therefore, they cannot be sold on fluctuating prices. And something even better is that if we generate any kind of renewable energy and don’t use it, that energy is sent back to the electricity grid and we get rewarded. Long-Term Profits and Economic Development Since you don’t have to pay for refuel, costs of maintenance and operation are significantly reduced for renewable energy, allowing countries to save millions of dollars every year. Not only that, but it also generates new jobs related to manufacturing, management, engineering, advertisement, and other fields. Another point is that regions that contain non-renewable resources, including raw materials for nuclear power production, are often places of constant armed conflict that end up interfering with its costs to governments, businesses, and consumers. This leads to another advantage of using renewable resources: self-sufficiency. Countries that can domestically provide all of its necessary energy won’t have to worry about the fluctuating prices of non-renewable resources like oil, and will be secure against both global crisis and resource conflicts. It can boost the economy The economy can benefit from renewable sources of energy in several ways. Solar power will need research and development. Manufacturing and construction will follow. Renewable energy also requires transportation and installation. There are more jobs created in line with the adoption of renewable energy. If 25 percent of our energy is from renewable energy, jobs will increase threefold, unlike when non-renewable energy sources are used. Consistent energy prices In almost all the countries that have adopted renewable sources, electricity is the most affordable and this helps in stabilizing energy prices. The materials that provide this energy need investment. Their operating costs are lower as the fuel in most cases is free and over time, the energy prices will become stable. Another benefit of renewable energy technology is that it is less costly. Projection forecast reduces cost as it becomes more mainstream. On the other hand, the prices of fossil fuel are volatile and can fluctuate. Prices can fall rapidly or rise rapidly as the market shifts. Benefits of Renewable Energy Resources Benefits of Renewable Energy Resources are huge, you will learn some of them here. Time To Take A Step The use of fossil fuels like oil, gas or coal as energy sources has left a mark on our planet. The global warming rises constantly and it’s about time to start taking measures to fight against this and keeping the Earth alive. A renewable energy source is the one that is almost impossible to ever run out. These sources naturally renew with time. In recent years several ways to generate electricity from renewable energy sources have been developed. The most popular ones are: • Solar: Storing the sunlight’s heat through solar panels to generate electricity. • Wind: Electricity is generated from wind motion. • Hydropower: The energy from water motion produces hydroelectricity. Many countries have hydropower plants, which provide an important share of the energy used. • Ocean: The electricity is generated from the rise and fall of the ocean tides. • Geothermal: This one is referred to the deep underground’s natural heat. The heat is stored and used to generate electricity. • Biomass: Consists of storing organic materials coming from plants and animals. The biomass is burned and the chemical energy in it is released as heat. Taking Advantage From Renewable Energy Several benefits come with switching to renewable energy, both for the planet, ¡and for us! Here are some of the most popular benefits: Slows Down Global Warming One of the biggest promoters of global warming is the use of fossil fuels. And renewable energy sources contribute to reducing dependence on fossil fuels, preventing global warming, a climate phenomenon caused by above-normal heat retention on the earth’s surface and in the The consequences of global warming can be catastrophic for the ecosystem: extinction of flora and fauna, droughts, floods and interference with agriculture. Renewable energy helps to reduce the impacts of production on the environment, reducing pollution and the emission of toxic chemicals. Reduces Impact on Global Warming The electricity sector represents one of the largest problems in terms of climate change. There’s an urgent need to decrease the electricity consumption coming from common fossil fuels and replace it with renewable and more “green” sources. Renewable energy sources help to keep our environment cleaner for a long time, producing almost none harmful gasses to the atmosphere. Moreover, we must do our best to control the effects of climate change. We can witness the increased temperatures across the globe, storms, and rising sea levels. Changes to food sources and habitats have a negative impact on wildlife as most species can’t adapt enough and the threat of extinction increases. Renewable Energy and Cleaner Air Renewable energy represents an improvement in public health. With lesser emission of gasses, the air we breathe becomes cleaner, preventing entire cities and countries people to acquire lung or heart diseases often related to pollution. Renewable Energy is the Present and the Future Given all the benefits that working with renewable energy brings, several companies have invested in it since now. It is presumed that for the year 2030, there will be over 24 million people working in this industry worldwide. It’s almost impossible thinking to replace electricity coming from fossil fuels, as every renewable source has a deficiency, like nights for solar energy, but what it is possible is to start reducing the negative impact that is causing to our planet the excessive consumption of electricity from non-renewable resources. Preservation of Natural Resources When it comes to renewable energy, preserving natural resources is the first benefit that comes to people’s minds, as most polluting energy sources are exhaustible. Oil is the classic example: Day after day, experts say that reserves will eventually run out, which would cause a global collapse. The same applies to gas and coal. Renewable energy, by contrast, does not abuse of natural resources in this way. It takes advantage of inexhaustible sources that can be tapped without impact. Currently, the main ones are solar energy and wind energy. You may wonder what are the types of Renewable Energy? There you will know some of them in this article. Renewable energy has been highly considered and in demand in many countries around the world. They are produced and restored naturally, such as the wind and sunlight. Unlike other sources of energy, renewable energies are inexhaustible. They are readily available and accessible to anybody. Limitless energy Renewable energy provides near limitless energy, unlike fossil fuels. They rely abundantly on sun, wind and plant matters. Furthermore, renewables include fast-flowing water and heat of the earth. Many countries are adopting renewable energy. Renewable energy is reliable We can occasionally lose electricity in our homes due to power outage. This affects many houses in one hit. However, renewable sources of energy, such as solar and wind, fail less on a large scale. Systems are spread over a large area. Therefore, if one area suffers from extreme weather, another area will generate energy. The modular aspects also assist with resilience and reliability. Many nonrenewable power stations rely on water for cooling. However, severe droughts or scarcity of water can affect energy generation. In contrast, wind and solar need no water to generate electricity. This makes them less susceptible to changes in water availability. In conclusion, the benefits of renewable energy are clear to see. It is time to make a change now. With technology continually evolving, it has never been a good time to turn to renewable energy. The benefits associated with our health, the health of our planet and the economy are evident. Renewable energy is readily available and we need to reduce our resilience on fossil fuels. The faster we adapt, the more we reduce pollution from fossil fuels. Types of Renewable Energy 1. Solar The head of all renewable energies. It is generated by converting the direct heat of the sunlight into electricity or power. It has a lot of potential, much more than other resources. Studies have proven that it can last for the next 5 billion years. To contain the solar energy produced by the sun, using a solar panel is the most common and most convenient way. In larger areas, concentrated solar power is more applicable. Thousands of glasses reflect heat onto a tower that later change into steam then electricity using turbines. It has special materials like silicon that transforms heat into usable electricity. 2. Wind It is an air in motion. It possesses kinetic energy that allows it to move something. Wind energy has been used right before the discovery of modern technologies. It was used to power sailboats and windmills by our ancestors. To this day, several countries in the world mounted windmills around rural areas to sustain their electricity needs. Aside from this, it is known to be clean and carbon dioxide-free which helps to protect the environment from air pollution. Although it is limited to some areas, wind energy is highly beneficial to a lot of people as it is cheaper and useful. 3. Hydroelectric It is the conversion of flowing water into electricity. It is continuously renewed by the sun, which is commonly known as the water cycle. There are two types of hydroelectricity. The most common one is the Dam. It works as the water flows down the turbines which produce electricity. Another type is the run of the river. Same as the Dam, it uses turbines and generators. Their only difference is that the Dam can store a large amount of water whereas the run of the river doesn’t. Both are environmentally friendly and reliable as to the amount of electricity being produced. 4. Geothermal Geothermal is acquired by utilizing the internal heat of the earth using geothermal wells dug approximately 3-10 km underground. The heat extracted below has 50,000 times energy than oils or any natural gas resources. These wells steam the water then pumped up by the turbines through the generators to produce electricity. It is efficiently effective, although it is less common than other renewable energy. Geothermal is pollution-free, but it is highly expensive to produce. 5. Biomass Biomass is the energy produced from organic matters. It is found by recently living plants and animals. Biomass can be transformed into several energy. Examples are burned woods and garbage into heat energy. Decomposed food crops, paper, animal manure, and yard waste into biogas. Biogas is a good alternative for fossil fuels such as coal and oil. It is one of the most sustainable sources of energy. With the use of technology, burning biomass emits less air pollution than non-renewable energies. Imagine the world if people will maximize using all the listed types of renewable energies. It will be a better place for us to live in as it promotes clean and safe energy. Moreover, it can create a great impact on combating climate change and decreasing pollution that threatens the present and future generations. What Are The Economic Benefits Of A Renewable Energy Source You will find The Economic Benefits Of A Renewable Energy Source in this post. The environment is going through considerable changes that are impacting life as we know it. Humans, animals, and plants are finding it harder to breath, dealing with extreme temperatures, and depleted resources. Sustainable or green energy can protect the environment if the right initiatives are put in place. For example, renewable energy sources are a great way to utilize the natural resources on the Earth while protecting its inhabitants. What are renewable energy sources and what kind of economic advantages does it have for the Earth? This blog defines renewable energy sources and their benefits. What Is A Renewable Energy Source? Renewable energy sources are taken from natural resources to provide sustainability. The sun, wind, rain, and even waves can be harnessed for their energy. Providing electricity or other resources with renewable energy can reduce carbon emissions and greenhouse gases by at least 67 percent, according to the Environmental Protection Agency. In fact, renewable energy drives down the costs and delivers clean energy which has a tremendous economic impact. When natural processes are used to create energy, they are constantly replenishing themselves without a man-made effort. Resources that are replenished on a human timescale have the potential to sustain life as we know it including animals and plants. What Are The Economic Advantages Of Renewable Energy? A renewable energy revolution is taking place in the U.S. renewable energy sector. As global warming and major ice cap melting threatens the Earth, a need for renewable energy resources has increased. Economically, using renewable energy can help Americans reduce their energy costs and encourage the deployment of the nation’s rich energy resources. In fact, it’s a great time for investors to take advantage of renewable energy stock. The future calls for a more sustainable energy resource. What are the other advantages of renewable energy? Reducing air pollution is another economical benefit of using renewable energy. Air pollution occurs when gases are released in the air. These gases can come from cars, factories, and exhaust. When we actively reduce the amount of man-made resources and rely on renewable energy resources, we’re creating these types of economical advantages. Plus, using renewable energy sources reduces the carbon monoxide being released by every home by 26 to 32 percent, according to the cleanerandgreener.org online. Carbon monoxide reduces the ability of the blood to bring oxygen to body cells and tissue. Humans must incorporate renewable energy into electricity and fueling their cars. We must stop burning fossil fuels, use alternative transportation, and take advantage of renewable energy. If we want to reduce the toxic compounds of air pollution and waste, it’s important to get more facts on renewable energy sources and it’s economic advantages. Every day millions of harmful pollutants are being released in the environment. These pollutants can also impact our water and soil and calls for a greater need for renewable energy. Remember, the long-term impact of not using renewable energy can impact human health, the economy, and the environment. Learn more about the advantages of renewable energy sources by becoming a part of the clean air movement today! Climate Change. Global Warming. Renewable Energy. These are just some of the hot keywords in today’s society related to the environment. And as we can see in this article, there’s a good reason for that, as the advantages of using renewable are many. Hopefully, this article has helped you better understand some of the reasons for using renewable energy and moving away from non-renewable resources.	{"url":"https://greenenergytip.com/benefits-renewable-energy-2/","timestamp":"2024-11-10T12:19:16Z","content_type":"text/html","content_length":"118261","record_id":"<urn:uuid:36f51554-58d3-4985-87c1-5ab05294b7cc>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00623.warc.gz"}
LeetCode – Max Chunks To Make Sorted (Java) Given an array arr that is a permutation of [0, 1, …, arr.length – 1], we split the array into some number of “chunks” (partitions), and individually sort each chunk. After concatenating them, the result equals the sorted array. What is the most number of chunks we could have made? For example, given [2,0,1], the method returns 0, as there can only be one chunk. The key to solve this problem is using a stack to track the existing chunk. Each chunk is represented a min and max number. Each chunk is essentially an interval and the interval can not overlap. Java Solution public int maxChunksToSorted(int[] arr) { return 0; // use [min,max] for each chunk Stack<int[]> stack = new Stack<int[]>(); for(int i=0; i<arr.length; i++){ int min=arr[i]; int max=arr[i]; int[] top = stack.peek(); if(arr[i] < top[1]){ min=Math.min(top[0], min); max=Math.max(max, top[1]); stack.push(new int[]{min,max}); return stack.size(); Time complexity is O(n). 3 thoughts on “LeetCode – Max Chunks To Make Sorted (Java)” 1. excellent post. Full Stack Training In Pune 2. Greetings. I know this is somewhat off-topic, but I was wondering if you knew where I could get a captcha plugin for my comment form? Iâ€™m using the same blog platform like yours, and Iâ€™m having difficulty finding one? Thanks a lot. 3. why maintain the minimum value? Leave a Comment	{"url":"https://www.programcreek.com/2016/01/leetcode-max-chunks-to-make-sorted-java/","timestamp":"2024-11-08T16:03:03Z","content_type":"text/html","content_length":"60235","record_id":"<urn:uuid:4717f561-33fd-471c-939d-54d5d7bf87cb>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00823.warc.gz"}
Price per Employee Calculator The Price per Employee Calculator can calculate the price for each employee based on the total price of all the employees. To calculate the price per employee, we divide the total price of all the employees by the number of employees. Please enter the total price of all the employees and the number of employees so we can calculate the price per employee: Price per Follower Calculator Here is a similar calculator you may find interesting.	{"url":"https://pricecalculator.org/per/price-per-employee-calculator.html","timestamp":"2024-11-12T05:08:38Z","content_type":"text/html","content_length":"6534","record_id":"<urn:uuid:565ebfc6-925c-4028-a186-7437a22a229e>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00224.warc.gz"}
Multiple Predictor Linear Model Return to Course Materials Lead Author(s): David Glidden, PhD Start presentation Slide 1: Multiple predictor linear regression • Models dependence of the mean of continuous outcome on multiple predictors simultaneously • By including multiple predictors we can try to □ control confounding of treatment effects by indication, risk factor effects by demographics, other covariates □ examine mediation of treatment, risk factor effects □ assess interaction of treatment effects or exposure with sex, race/ethnicity, genotype, other effect modifiers □ get at causal mechanisms in observational data □ also: account for stratified or multi-center design of RCT, increase precision of estimates Slide 2: Components of the Linear Model • Systematic: □ how does the average value of outcome y depend on values of the predictors? • Random: □ at each observed value of the predictors, values of y are distributed about the predicted average □ assumed distribution of deviations underlies hypothesis tests, p-values, and confidence intervals Slide 3: Systematic part of the model • In abstract terms, model written as □ Σ[y\|x]= \xDF0 + \xDF1x1 + \xDF2x2 + \xB7\xB7\xB7 + \xDFpxp • Σ[y\|x]: Expected or average value of y for a given set of predictors x = x1,x2,\xB7\xB7\xB7 ,xp • \xDFj: change in average value of outcome y per unit increase in predictor xj, holding all other predictors • \xDF0 (the intercept): average value of the outcomey when all predictors = 0 • "Linear predictor" common to linear, logistic, Cox, and longitudinal models Slide 4: Interpretation of regression coefficients • \xDFj: change in average value of outcome y per unit increase in predictor xj, holding all other predictors • Hold x2,...,xp constant, and let x1= k: □ Σ[y\|x]= \xDF0 + \xDF1k+ \xDF2x2 + \xB7\xB7\xB7 + \xDFpxp (1) • Now increase x1 by one unit to k+1: □ Σ[y\|x]= \xDF0 + \xDF1(k +1)+ \xDF2x2 + \xB7\xB7\xB7 + \xDFpxp (2) • Subtracting(1) from(2) gives \xDF1, for every value of k as well as x2,...xp • Note: assumes x1 does not interact with x2,...xp Slide 5: Interpretation of regression coefficients • \xDF0: average value of outcome y when all predictors = 0 • Let x1 = x2 = \xB7\xB7\xB7 = xp 0. Then E[y\|x] \xDF0 + \xDF1x1 + \xDF2x2 + \xB7\xB7\xB7 + \xDFpxp = \xDF0 • Intercept: where the regression line meets the y-axis in single-predictor models Slide 6: Review: centering predictors • Same as in single-predictor model • For many continuous predictors like age, SBP, LDL, no one has value 0 • Solution: center them on their sample means, so new variable has value 0 for observations at the mean • For binary predictors, 0 is the usual coding for the reference group, so not a problem for interpretation • With centering, \xDF0 estimates expected value of y for participant at reference level of binary predictors, mean of centered continuous predictors • Values and interpretation of other coefficients unaffected Slide 7: Review: rescaling predictors • Same as in single-predictor model • Rescaled variable Xrs = X/k • Coefficient for Xrs interpretable as increase in mean of outcome for a k-unit increase in X • If k = SD(X), coefficient forXrs interpretable as increase in mean of outcome for a 1 SD increase in X • \xDFˆ(Xrs)= k\xDFˆ(X);SE(\xDFˆ), 95% CI for \xDFˆalso rescaled • P-value for \xDF, intercept coefficient unaffected • Can accomplish the same thing using lincom Slide 8: Random part of the model yi =Σ[y\|xi]+εi • Outcome yi varies from the average at xi by an amount oi • ε represents unmeasured sources of variation, error • As in single-predictor model, four assumptions about o: 1. Normally distributed 2. mean zero at every value of x 3. constant variance 4. statistically independent • These assumptions underlie hypothesis tests, confidence intervals, p-values, also model checking Slide 9: Assumptions about the predictors • Nodistributional assumptions(e.g. Normality) □ predictorscanbecontinuous,discrete(e.g. counts), categorical(dichotomous, nominal, ordinal) • Linear regression works better if □ predictors are relatively variable □ there are no excessively "influential" points • Assumed measured without error(otherwise "regression dilution bias" and residual confounding) Slide 10: Update of two details • Fitted value:ˆyi = \xDFˆ0 + \xDFˆ1xi1 + \xB7\xB7\xB7 + \xDFˆpxip - estimated average or expected value of outcome y when x = xi, the predictor values for observation i □ now depends on multiple predictors instead of just one • Residual: ri = yi -yˆi =ˆoi □ difference between datapoint and fitted value □ sample analogue of oi, used in checking model fit □ not obvious what "vertical" means with multiple predictors Slide 11: Ordinaryleast squares(OLS) • Method for fitting linear regression models • OLS finds values of regression coefficients which minimize residual sumof squares(RSS; i.e. sumof squared residuals) • Good statistical properties: unbiased, efficient, easy to compute, but sensitive to outliers • For normally distributed outcomes, OLS is equivalent to "maximumlikelihood" (methodusedforlogistic,Cox, some repeated measures, many other models) Slide 12: Multi-predictor linear model for glucose Multi-predictor linear model for glucose • Upper left(ANOVA table) □ Total SS =Σn/i=1(yi-\xAFy)2: variability of outcome yi=1(yi - about the sample average \xAFy n □ Total MS =(yi -y\xAF)2/(n -1): sample variance i=1 of outcome y n □ Model SS = (ˆyi -y\xAF)2: variability of outcome i=1 accounted for by predictors included in model □ Model MS: numerator of model F-statistic n □ Residual SS =(yi -yˆi)2: residual variability i=1 not accounted for by predictors, what OLS minimizes □ Residual MS = yi)2/(n -p): sample i=1(yi - ˆvariance of residuals Slide 13: Interpreting Stata regression output Interpreting STATA regression output Slide 14: Summary of model • Multipredictor linear regression is a tool for estimating how the average value of a continuous outcome depends on multiple predictors simultaneously • Inferential machinery evaluates precision of estimates and whether sampling error can account for findings • Coefficients generally interpretable as the change in theaverage value of the outcome per unit increase in the predictor, holding all other predictors constant • Power helped by effect size, sample size, variability of predictor; hurt by correlation with other predictors, variability left unexplained Slide 15: Confounding • Can account for the some or all of the unadjusted association between a predictor and an outcome • Controlling confounding the primary reason for doing multi-predictor regression • Confounders must be associated with predictor and independently with outcome • Only an association adjusted for confounders can be viewed as possibly causal Slide 16: Unadjusted waist/glucose association Unadjusted waist/glucose association Slide 17: Adjusted waist/glucose association Adjusted waist/glucose association Slide 18: Primary predictor, confounder, and outcome Primary predictor, confounder and outcome Adjusting for a confounder • Primary predictor and confounder are correlated: □ values of primary predictor larger in subgroup 2 than subgroup 1 □ conversely, those with larger values of primary predictor more likely in subgroup 2 • Both continuousprimarypredictor andbinary confounder independently predict higher values of outcome • Unadjusted effect of primary predictor partly reflects effect of being in subgroup 2 • Adjustment for the confounder fixes the problem Slide 19: Interpretation of results • Unadjusted estimateforprimarypredictor(6.2) □ Estimates an observable trend in whole population □ Causal interpretation misleading in most contexts • Adjusted estimate(3.3) may have a causalinterpretation, because the effect of the confounder is not ignored • Regression lines for subgroups 1 and 2: □ slopes estimate predictor/outcome association within each subgroup("holding subgroup constant") □ assumedparallel(nointeraction - sameeffectinboth Behavior of regression coefficients for this case • When the primary predictor and confounder are positively correlated, both predict higher(or lower) • Values of the outcome adjusted coefficient for primary predictor is attenuated: that is, closer to zero than unadjusted coefficient in this case, still non-zero and signficant • Typical pattern for confounding Slide 20: Another case: so-called negative confounding • Confounding can also "mask" an independent association • Example: needlestick injuries and HIV-seroconversion □ overall, AZT prophylaxis does not predict seroconversion, but* use of AZT associated with severity of injury * severity of injury predicts seroconversion □ protective effect of AZT unmasked after controlling for severity of injury Slide 21: Negative confounding: two scenarios Negative confounding may arise between predictors that are 1. Positively correlated, with opposite effects on outcome: Example: injury severity, AZT, and seroconversion 2. Negatively correlated, with similar effects on outcome: Example: average BMI decreases with age in HERS cohort, but both predict increased SBP Slide 22: Summary: negative confounding • Average BMI decreases with age in HERS cohort, but both predict increased SBP • Adjustment for age increases BMI slope estimate from .21 to .30 mmHg per kg/m2 • Negative confounding is not all that uncommon • Implications for predictor selection: univariate screening, "forward" selection procedures may miss some negatively confounded predictors Slide 23: Confounding is difficult to rule out • Were all important confounders adjusted for? • Were they measured accurately? • Were their effects modeled adequately? □ modeled non-linearities in response to continuous predictors(Session 6) □ no omittedinteractions(Session5) □ no gross extrapolations • Modeling difficulties used to argue for propensity scores Slide 24: Summary • Confounders must be associated with predictor and independently with outcome • Unadjusted, adjusted coefficients estimate different things • Unadjusted association may be partly or completely explained or, conversely, unmasked after adjustment • Regression controlsfor confounding byjointly modeling effects ofpredictor and confounders(VGSMSect. 4.4) • Bigger samples don't help, except by making it easier to adjust • Controlling for covariates is easy enough, but residual confounding is difficult to rule out • Confounders are thought to cause the primary predictor, or are correlates of such a cause • In contrast, mediators are on the causal pathway from primary predictor to the outcome • In models, mediation and confounding behave alike and must be distinguished on substantive grounds • Example: to what extent is effect of BMI on SBP mediated by its effects on glucose levels? • Use a series of models to show that: □ primary predictor independently predicts mediator □ mediator predicts outcome independently of primary predictor □ adjustment for mediator attenuates estimate for primary predictor • The models: □ regress mediator on predictor and confounders □ regress outcome on predictor and confounders □ regress outcome on predictor, mediator, and confounder • Interpretation of coefficient estimates for primary predictor: □ before adjustment for mediator: overall effect □ after adjustment: effect, if any, via pathways other than the mediator • Assess mediation by difference between coefficients for primary predictor before and after adjustment for mediator • Hypothesis tests, CIs for difference and proportion of effect explained abitharder(seebookfor references) • Example: is association of BMI with SBP mediated by glucose levels? • BMI independently predicts higher glucose: 1.7 mg/dL (95% CI 1.4-1.9) for each kg/m2 increase in BMI • A 10 mg/dL increase in glucose levels is independently associated withhigherSBP:0.5 mmHg(95%CI0.3-0.7) • Overall BMI effect: before adjustment for glucose levels, each additional kg/m2 predicts an increase of .25 mmHg (95% CI 0.12-0.38) in average SBP • Direct BMI effect via other pathways: after adjustment for glucose levels, each kg/m2 predicts an increase of only .16 mmHg(95%CI0.03-0.30) • Degree of attenuation(PTE):glucoselevels explain (.25-.16)/.25*100 = 34% of the effect of BMI on SBP • An observational analysis even when the primary predictor is treatment in RCT; must control for confounding of mediator effects. • Evidence for mediation potentially stronger in longitudinal data □ but when predictor is both a mediator and a confounder, fancier methods required: e.g., "marginal structural models" • "Negative" mediation is possible: glitazones, weight, bone loss; HT, statin use, CHD events • TZDs cause bone loss in mouse models. • In HABC, TZD use not associated with bone loss, after controlling for confounders by indication • TZDs also cause weight gain, which is protective against bone loss • TZDs do predict bone loss, after controlling for weight gain: adverse effect emerges after controlling for beneficial effect via weight gain • In HERS, statin use differentially increased in placebo group, and controlling for this makes HT look a bit protective • Regression coefficients change when either a confounder or a mediator is added to the model; which is which depends on how you draw the causal arrows(statistics not informative) • Negative mediation is possible • Must control for confounders of mediator • Estimated independent effect of primary predictor □ before adjustment for mediator: overall effect □ after adjustment: direct effect via other pathways (assumingboth models adjust for confounders) Slide 32: Interpreting results for log-transformed variables • Positive continuous variables commonly log-transformed outcomes: normalize and equalize variance □ predictors: get rid of non-linearity, interaction □ more about this is session 6 • Bothlog-10(HIV viralload) and natural log transformations used • How does this affect interpretation of regression coefficients Slide 33: Log-transformed predictors • For natural-log or log-10 transformed predictor xj, \xDFˆj estimates the increase in the mean of the outcome for each 1-log increase in log-transformed xj - equivalently a 2.7-fold or 10-fold increase in untransformed value of xj. • \xDFˆjln(1+k/100) estimates the change in the mean of the outcome for each k% increase in untransformed xj. • Note: p-value for test of \xDFj =0 unaffected by choice of k • Use \xDFˆjlog10(1+k/100) if xj is log10 • Use nlcom to get interpretable estimates with confidence interval(lincom does not allow log() as argument)	{"url":"https://ctspedia.org/ctspedia/multipredictorlinearmodel","timestamp":"2024-11-11T20:56:25Z","content_type":"text/html","content_length":"24162","record_id":"<urn:uuid:32ace23a-73eb-40c0-903d-cbe10c7e0659>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00571.warc.gz"}
Draws an ellipse. draw_ellipse(x1, y1, x2, y2, outline); 参数描述 x1 The x coordinate of the left of the ellipse. y1 The y coordinate of the top of the ellipse. x2 The x coordinate of the right of the ellipse. y2 The y coordinate of the bottom of the ellipse. outline Whether the ellipse is drawn filled (false) or as a one pixel wide outline (true). 返回： N/A（无返回值） With this function you can draw either an outline of an ellipse or a filled ellipse by defining a rectangular area that will then have the ellipse created to fit. You can define how precise the drawing is with the function draw_set_circle_precision. NOTE: If you are wanting to draw a shape using a shader, you should be aware that most shaders expect the following inputs: vertex, texture, Colour. However, when using this function, only vertex and colour data are being passed in, and so the shader may not draw anything (or draw something but not correctly). If you need to draw shapes in this way then the shader should be customised with this in mind. draw_ellipse(100, 100, 300, 200, false); This will draw a filled ellipse within the defined rectangular area. © Copyright YoYo Games Ltd. 2018 All Rights Reserved	{"url":"https://docs.gamebar.me/gmsDocs/source/_build/3_scripting/4_gml_reference/drawing/forms/draw_ellipse.html","timestamp":"2024-11-07T23:17:19Z","content_type":"text/html","content_length":"6617","record_id":"<urn:uuid:4aecf55b-e5a4-4790-ab6d-2930917410f0>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00331.warc.gz"}
Algebraic Variables Understanding Algebraic Variables I'm sure you're wondering, "What is a variable and why are letters used in math?" If this hasn't crossed your mind, I'm sure it will as you begin your study of Pre-Algebra. Understanding algebraic variables is the very first component for understanding Algebra. So, let's get started... So, What is a Variable? A variable is a symbol, most often a letter, that represents a number in math. So, you say, "That's silly, why use a letter when you can just write the number? And... that's a very good question! The term variable means "to change". This is the reason why we use algebraic variables. If a number can change based on the situation, then we would use a variable in it's place. This is better represented when we talk about formulas. Take a look.... Formulas and Algebraic Variables Think about all of the math formulas that you have used in the past. I'm sure you are familiar with the area formula: A = lw or the perimeter formula: P= 2l+2w. Notice how these formulas use variables. We use variables in our formulas because these numbers change based on what you are measuring. Let's say that I'm calculating the money that I can make from my lawn service business. I might start by creating a chart: Since I multiply the number of lawns by $7 each time, I can write a formula (with a variable) to make this process easier. Using Variables to Make Calculations Easier In this way, I can always change "m" and substitute the number of lawns that I mowed that week and easily find my earnings. Do you see how the variables, E and m change based on the number of lawns mowed? This is an easy way to calculate your earnings without the added work of a chart. When working with variables, you must remember one thing: No matter how many times a variable is used in an expression, the value is always the same. For example take a look at the following expression: 2s + 6 - 3s Notice how the variable "s" is used twice in this equation. I CANNOT substitute two different numbers for "s". If s = 5, then 5 must be substituted for s in both places in the equation. For example: 2s + 6 - 3s where s = 5 This also means: 2(5) + 6 - 3(5) Notice how the number 5 takes the place of "s" in both places. If you have two different variables in an expression, then most likely they will represent different values. You will find that using variables in Algebra is actually an easy process. It will make some of the long labor intensive work in the past seem much easier! This leads us right into our next lesson on evaluating algebraic expressions. We would love to hear what you have to say about this page! Need More Help With Your Algebra Studies? Get access to hundreds of video examples and practice problems with your subscription! Click here for more information on our affordable subscription options. Not ready to subscribe? Register for our FREE Pre-Algebra Refresher course.	{"url":"https://www.algebra-class.com/algebraic-variables.html","timestamp":"2024-11-13T15:44:23Z","content_type":"text/html","content_length":"24880","record_id":"<urn:uuid:e1b1415e-daf5-4978-b8a9-8c410b089609>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00885.warc.gz"}
Remarks on Landau–Siegel zeros Speaker: Debmalya Basak Date: Tue, Jun 18, 2024 Location: PIMS, University of British Columbia Conference: Comparative Prime Number Theory Subject: Mathematics, Number Theory Class: Scientific CRG: L-Functions in Analytic Number Theory One of the central problems in comparative prime number theory involves understanding primes in arithmetic progressions. The distribution of primes in arithmetic progressions are sensitive to real zeros near $s = 1$ of L-functions associated to primitive real Dirichlet characters. The Generalized Riemann Hypothesis implies that such L-functions have no zeros near $s = 1$. In 1935, Siegel proved the strongest known upper bound for the largest such real zero, but his result is vastly inferior to what is known unconditionally for other L-functions. We exponentially improve Siegel’s bound under a mild hypothesis that permits real zeros to lie close to $s = 1$. Our hypothesis can be verified for almost all primitive real characters. Our work extends to other families of L-functions. This is joint work with Jesse Thorner and Alexandru Zaharescu. Additional Files:	{"url":"https://mathtube.org/lecture/video/remarks-landau%E2%80%93siegel-zeros","timestamp":"2024-11-08T11:09:55Z","content_type":"application/xhtml+xml","content_length":"27206","record_id":"<urn:uuid:25989c57-952e-4473-b516-7af8bd5c77d9>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00878.warc.gz"}
Games for adding integers :: Algebra Helper Our users: I want to thank you for all you help. Your spport in resolving how do a problem has helped me understand how to do the problems, and actually get the right result. Thanks So Much. Paul D'Souza, NC The Algebra Helper is my algebra doctor. Equations and inequalities were the two topics I used to struggle on but using the software wiped of my problems with the subject. Mr. Tom Carol, NY Excellent software, explains not only which rule to use, but how to use it. Oscar Peterman, NJ 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. 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ORDER NOW! Algebra Helper Download (and optional CD) Only $39.99 Click to Buy Now: 2Checkout.com is an authorized reseller of goods provided by Sofmath "It really helped me with my homework. I was stuck on some problems and your software walked me step by step through the process..." C. Sievert, KY 19179 Blanco #105-234 San Antonio, TX 78258 Phone: (512) 788-5675 Fax: (512) 519-1805	{"url":"https://algebra-answer.com/algebra-answer-book/equation-properties/games-for-adding-integers.html","timestamp":"2024-11-10T22:12:20Z","content_type":"text/html","content_length":"25083","record_id":"<urn:uuid:a17c9eb4-c594-4433-83d0-52f12c522460>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00048.warc.gz"}
About · solids4foam Why does solids4foam exist? • Desire to solve fluid-solid interactions in OpenFOAM • Desire to run solid mechanics cases natively in OpenFOAM • A modular approach for coupling different solid and fluid procedures • Research into finite volume methods for solid mechanics What is the solids4foam implementation philosophy? • If you can use OpenFOAM, you can use solids4foam • Support for main OpenFOAM forks • Easy to install • Emphasis on code design and style • Single executable design A brief history The finite volume solid mechanics procedures implemented in solids4foam can trace their routes to Demirdzic, Martinovic and Ivankovic (1988) and the subsequent developments of Demirdzic and co-workers. See the recent review article by Cardiff and Demirdzic (2021) for more details. The seminal Weller et al. (1998) FOAM/OpenFOAM paper demonstrated a simple small-strain linear elasticity solid mechanics solver, closely based on the methods of Demirdzic and co-workers. The commercial predecssor of OpenFOAM, called FOAM, presented a solid mechanics section on the Nabla Ltd website (courtesy of the way back machine). solids4foam builds on and generalises the Extend Bazaar FSI toolbox and the solidMechanics codes from foam-extend. How to cite If you use solids4foam for a publication, please cite the following references: P. Cardiff, A Karac, P. De Jaeger, H. Jasak, J. Nagy, A. Ivanković, Ž. Tuković: An open-source finite volume toolbox for solid mechanics and fluid-solid interaction simulations. 2018, 10.48550/ arXiv.1808.10736, available at https://arxiv.org/abs/1808.10736. Ž. Tuković, A. Karač, P. Cardiff, H. Jasak, A. Ivanković: OpenFOAM finite volume solver for fluid-solid interaction. Transactions of Famena, 42 (3), pp. 1-31, 2018, 10.21278/TOF.42301. The corresponding BibTeX entries are author = {P. Cardiff and A. Karac and P. De Jaeger and H. Jasak and J. Nagy and A. Ivankovi\'{c} and \v{Z}. Tukovi\'{c}}, title = {An open-source finite volume toolbox for solid mechanics and fluid-solid interaction simulations}, year = {2018}, doi = {10.48550/arXiv.1808.10736}, note = {\url{https://arxiv.org/abs/1808.10736}} author = {\v{Z}. Tukovi\'{c} and A. Kara\v{c} and P. Cardiff and H. Jasak and A. Ivankovi\'{c}}, title = {OpenFOAM finite volume solver for fluid-solid interaction}, year = {2018}, volume = {42}, number = {3}, pages = {1-31}, journal = {Transactions of FAMENA}, doi = {10.21278/TOF.42301} In addition, depending on the functionality selected, there may be additional relevant references (the solver output will let you know!). solids4foam is primarily developed by researchers at University College Dublin and the University of Zagreb with contributions from researchers across the OpenFOAM community. In particular, Philip Cardiff (Dublin) is the principal toolbox architect, with significant scientific and implementation contributions from Željko Tuković (Zagreb). A full list of contributors can be found in the contributors file.	{"url":"https://www.solids4foam.com/about/","timestamp":"2024-11-12T00:21:18Z","content_type":"text/html","content_length":"24695","record_id":"<urn:uuid:a443a4b4-6203-4a94-a725-6b7eb54f890e>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00488.warc.gz"}
Why do we subtract two when using Descartes rule of signs? - Cracking Cheats Why do we subtract two when using Descartes rule of signs? Descartes’ rule of sign is used to investigate the variety of real zeros of a polynomial function. It tells us that the number of effective real zeroes in a polynomial operate f(x) is an identical or less than through an even numbers as the variety of adjustments within the sign of the coefficients. Since we have four sign changes with f(x), then there is a possibility of four or four – 2 = 2 or four – 4 = zero effective real zeros. Notice how there aren’t any signal adjustments between successive terms. This implies there aren’t any negative real zeros. Also, what’s a real zero? Real Zeros. Don’t forget that a real 0 is where a graph crosses or touches the x-axis. Imagine of a few aspects alongside the x-axis. Correspondingly, why does Descartes rule of signs and symptoms work? Descartes’ rule of sign. Descartes’ rule of sign is used to check the number of genuine zeros of a polynomial function. It tells us that the variety of successful real zeroes in a polynomial function f(x) is a similar or less than with the aid of an even numbers because the variety of adjustments within the sign of the coefficients. How many real roots does the equation have? Total Number of Roots On the web page Essential Theorem of Algebra we clarify that a polynomial will have exactly as many roots as its measure (the degree is the maximum exponent of the polynomial). So we know a further thing: the degree is 5 so there are 5 roots in total. How many roots does a polynomial have? If we count roots per their multiplicity (see The Factor Theorem), then: A polynomial of degree n may have in simple terms an excellent wide variety fewer than n genuine roots. Thus, once we count multiplicity, a cubic polynomial could have only three roots or one root; a quadratic polynomial could have in simple terms two roots or zero roots. How many roots genuine or complicated does the polynomial 7 5x four 3x 2 have in all? Answer: All four roots are complex. How do you know how many zeros a function has? Finding the zero of a function ability to find the point (a,0) in which the graph of the operate and the y-intercept intersect. To locate the cost of a from the point (a,0) set the function equivalent to 0 and then resolve for x. What is I squared in algebra? An imaginary quantity is a posh wide variety that can be written as a genuine number multiplied through the imaginary unit i, that is defined by means of its property i2 = −1. The square of an imaginary wide variety bi is −b2. For example, 5i is an imaginary number, and its square is −25. What is a good root? The product of 2 numbers is effective if the two numbers have the same sign as is the case with squares and square roots. a2=a⋅a=(−a)⋅(−a) A square root is written with an intensive symbol √ and the quantity or expression contained in the radical symbol, lower than denoted a, is known as the radicand.	{"url":"https://crackingcheats.com/public-question/why-do-we-subtract-two-when-using-descartes-rule-of-signs/","timestamp":"2024-11-10T00:14:25Z","content_type":"text/html","content_length":"51263","record_id":"<urn:uuid:273d3f77-b242-4d34-950c-6603b850b661>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00880.warc.gz"}
matematicasVisuales \| Plane developments of geometric bodies (2): Prisms cut by an oblique plane In Plane developments of geometric bodies (1): Nets of prisms we can see how regular and non-regular prisms can be developed into a plane. Now we are going to study nets of prisms cut by an oblique This is one example: In the examples above a base was a regular polygons. But we can consider prisms whose bases are not regular polygons. In the next mathlet, bases are non-regular polygons (although they are inscribed in a circle and they are convex polygons). Each time we change the number of sides of the base a new prism is generated with sides randomly drawn: This is one example of a non-regular transparent prism cut by an oblique plane: Two examples of nets of non-regular prisms cut by an oblique plane: We study different cylinders cut by an oblique plane. The section that we get is an ellipse. Plane net of pyramids and pyramidal frustrum. How to calculate the lateral surface area. Plane net of pyramids cut by an oblique plane. Plane developments of cones and conical frustum. How to calculate the lateral surface area. Plane developments of cones cut by an oblique plane. The section is an ellipse. The first drawing of a plane net of a regular dodecahedron was published by Dürer in his book 'Underweysung der Messung' ('Four Books of Measurement'), published in 1525 . The first drawing of a plane net of a regular octahedron was published by Dürer in his book 'Underweysung der Messung' ('Four Books of Measurement'), published in 1525 . We can cut in half a cube by a plane and get a section that is a regular hexagon. Using eight of this pieces we can made a truncated octahedron. Using eight half cubes we can make a truncated octahedron. The cube tesselate the space an so do the truncated octahedron. We can calculate the volume of a truncated octahedron. Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the truncated octahedron. The truncated octahedron is an Archimedean solid. It has 8 regular hexagonal faces and 6 square faces. Its volume can be calculated knowing the volume of an octahedron. The volume of a tetrahedron is one third of the prism that contains it. The first drawing of a plane net of a regular tetrahedron was published by Dürer in his book 'Underweysung der Messung' ('Four Books of Measurement'), published in 1525 . The volume of an octahedron is four times the volume of a tetrahedron. It is easy to calculate and then we can get the volume of a tetrahedron. You can chamfer a cube and then you get a polyhedron similar (but not equal) to a truncated octahedron. You can get also a rhombic dodecahedron. A very simple technique to build complex and colorful polyhedra.	{"url":"http://matematicasvisuales.com/english/html/geometry/planenets/prismasobliq.html","timestamp":"2024-11-05T10:32:52Z","content_type":"text/html","content_length":"20723","record_id":"<urn:uuid:23af79d3-0afc-466c-9c3e-b8babdea93af>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00286.warc.gz"}
OptiVec: MF_solve DescriptionThese functions solve the system MA * X = B of simultaneous linear equations. In the general case, MF_solve, LU decomposition is used. For symmetric matrices, assumed to be positive-definite, MFsym_solve provides a roughly two times faster way, employing Cholesky decomposition. First, MF_solve is described. MFsym_solve follows at the end. MF_solve works well in all cases where there is one unique solution. If successful, it returns FALSE (0). If, on the other hand, the system is ill-determined, which happens in all cases where one or more of the equations are linear combinations of other equations of the same system, the resulting matrix becomes singular and the function fails with an error message. To avoid outright failure in an application where ill-determined matrices might occur, you can define a minimum pivot for the decomposition. If you wish to do so for all calls to MF_LUdecompose and to the functions based upon it, namely MF_inv and MF_solve, you can do so by calling MF_LUDsetEdit. However, as this method is not thread-safe, you cannot use it in order to set different thresholds for different calls to the functions mentioned. Instead of defining a default editing threshold then, use their "wEdit" variants, i.e. MF_LUdecomposewEdit, MF_invwEdit or MF_solvewEdit. They take the desired threshold as the additional argument thresh. Note that thresh is always real, also in the complex versions. The return value of MF_solve and MF_solvewEdit indicates if the linear system could successfully be solved: │Return value│Meaning │ │ 0 │Matrix MA is regular; linear system successfully solved │ │ 1 │Under-determined system; matrix MA is singular; result X contains no useful information │ │ 2 │Under-determined system; matrix MA is (nearly) singular; solution was achieved only by pivot editing; it depends on the specific application, if the result is useful or not.│ To check if MF_solve was successful, in single-thread programs, you may also call MF_LUDresult, whose return value will be FALSE (0), if the system could be solved without problems (and without pivot editing), and TRUE (1) for singular MA. In multi-thread programs, on the other hand, it would not be clear wich instance of MF_solve the call to MF_LUDresult would refer to. So, here, inspection of the return value of MF_solve is the only option. As an often preferrable alternative to pivot editing, you might switch to MF_safeSolve or MF_solveBySVD. For symmetric matrices, we already noticed that MFsym_solve provides a potentially much faster way. This routine first tries Cholesky decomposition. If the matrix turns out not to be positive-definite, LU decomposition is used. The return value of MFsym_solve indicates if the linear system could successfully be solved: │Return value│Meaning │ │ 0 │Linear system successfully solved, either by Cholesky or by LUD │ │ 1 │Under-determined system; both Cholesky and LUD failed; matrix MA is singular; result X contains no useful information │ │ 2 │Under-determined system; matrix MA is (nearly) singular; solution was achieved only by LUD with pivot editing; it depends on the specific application, if the result is partially useful│ │ │or not. │	{"url":"http://www.optivec.com/matfuncs/solve.htm","timestamp":"2024-11-12T06:22:03Z","content_type":"text/html","content_length":"10381","record_id":"<urn:uuid:21b4e420-2d95-4c57-bef1-614b91a8151f>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00368.warc.gz"}
Is mean absolute deviation biased? The mean absolute deviation of a sample is a biased estimator of the mean absolute deviation of the population. Therefore, the absolute deviation is a biased estimator. However, this argument is based on the notion of mean-unbiasedness. What is the mean absolute value deviation? Mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean. Mean absolute deviation is a way to describe variation in a data set. How do you calculate the mean absolute deviation? To find the mean absolute deviation of the data, start by finding the mean of the data set. Find the sum of the data values, and divide the sum by the number of data values. Find the absolute value of the difference between each data value and the mean: \|data value – mean\|. Why don’t we use mean absolute deviation? This gets at a pretty important point: unlike standard deviation, mean absolute deviation does not uniquely characterize the dispersion of a distribution. In statistics, we work with samples and thus don’t really know the true population mean. Is a higher or lower mean absolute deviation better? The mean absolute deviation is the “average” of the “positive distances” of each point from the mean. The larger the MAD, the greater variability there is in the data (the data is more spread out). The MAD helps determine whether the set’s mean is a useful indicator of the values within the set. Does mean absolute deviation have units? Mean absolute deviation describes the average distance between the values in a data set and the mean of the set. For example, a data set with a mean average deviation of 3.2 has values that are on average 3.2 units away from the mean. What is the difference between mean absolute deviation and standard deviation? Both measure the dispersion of your data by computing the distance of the data to its mean. The difference between the two norms is that the standard deviation is calculating the square of the difference whereas the mean absolute deviation is only looking at the absolute difference. What is mean absolute deviation quizlet? mean absolute deviation. one measure of variability; the average of how much the individual scores of a data set differ from the mean of the set. – abbreviation: MAD. How do you calculate mean deviation? Calculating the mean average helps you determine the deviation from the mean by calculating the difference between the mean and each value. Next, divide the sum of all previously calculated values by the number of deviations added together and the result is the average deviation from the mean. Which is better mean deviation or standard deviation? Standard deviation is considered the most appropriate measure of variability when using a population sample, when the mean is the best measure of center, and when the distribution of data is normal. What is the difference between mean and mean absolute deviation? – the mean (average) of all deviations in a set equals zero. The Mean Absolute Deviation (MAD) of a set of data is the average distance between each data value and the mean. The mean absolute deviation is the “average” of the “positive distances” of each point from the mean. Is the mean absolute deviation of a sample unbiased? Estimation. The mean absolute deviation of a sample is a biased estimator of the mean absolute deviation of the population. In order for the absolute deviation to be an unbiased estimator, the expected value (average) of all the sample absolute deviations must equal the population absolute deviation. However, it does not. How to calculate the mean deviation from the mean? Mean absolute deviation. Here’s how to calculate the mean absolute deviation. Step 1: Calculate the mean. Step 2: Calculate how far away each data point is from the mean using positive distances. These are called absolute deviations. Step 3: Add those deviations together. Step 4: Divide the sum by the number of data points. Following these… Is the absolute deviation of a location a biased estimator? Therefore, the absolute deviation is a biased estimator. However, this argument is based on the notion of mean-unbiasedness. Each measure of location has its own form of unbiasedness (see entry on biased estimator ). The relevant form of unbiasedness here is median unbiasedness. What is the average absolute deviation of size 3? The average of all the sample absolute deviations about the mean of size 3 that can be drawn from the population is 44/81, while the average of all the sample absolute deviations about the median is	{"url":"https://greatgreenwedding.com/is-mean-absolute-deviation-biased/","timestamp":"2024-11-02T01:36:43Z","content_type":"text/html","content_length":"45338","record_id":"<urn:uuid:c1bf9dbc-9787-4800-aaa1-22d1f7f75129>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00744.warc.gz"}
CSS3 Transitions: Bezier Timing Functions — SitePoint In the second part of this series we looked at the CSS3 property which controls how an animation varies in speed throughout the duration of the transition. This accepts keyword values such as which are normally enough for the most demanding CSS developer. However, you can define your own timing functions using a value. It sounds and looks complicated but can be explained with some simple diagrams. so, in effect, the proportion of the animation completed matches the time, e.g. 50% of the animation is complete half-way through the duration. • It starts slowly; approximately 12% of the animation is completed in the first 25% of time. • It ends slowly; the last 12% of the animation occurs in the last 25% of time. • Therefore, the middle 76% of the animation must occur during 50% of the time; it’ll be faster. In essence, the steeper the curve tangent, the faster the animation will occur at that time. If the line was vertical, the animation would be instantaneous at that point. This is demonstrated in the following diagram: What’s a Bézier Curve? ) and end point (P ) of a line, a bézier curve defines two control points for each end (P and P ). I won’t even begin to explain the mathematics but, if you’re interested, head over to for the stomach-turning equations. Luckily, we don’t need to worry about the complexities. Since our animation line starts at 0,0 and ends at 1,1, we just need to define points P and P in the value, e.g. /* cubic-bezier(p1x, p1y, p2x, p2y) / / identical to linear / transition-timing-function: cubic-bezier(0.25,0.25,0.75,0.75); / identical to ease-in-out */ transition-timing-function: cubic-bezier(0.420, 0.000, 0.580, 1.000); Note that the the x co-ordinates of P and P denote time and must be between 0 and 1 (inclusive). You couldn’t set a negative value since it would start earlier than it was triggered! Similarly, you couldn’t set a value greater than one since time cannot proceed to, say, 120% then reverse back to 100% (unless you have a TARDIS or flux capacitor to hand). However, the y co-ordinates denote the animation completed and can be any value less than zero or greater than one, e.g. transition-timing-function: cubic-bezier(0.5, -0.5, 0.5, 1.5); At approximately 15% of the duration, the animation is -10% complete! Therefore, if we were moving an element from 0px to 100px, it would be at -10px at that time. In other words, we have a bounce effect; head over to and click GO to see it in action. Let the Tools do the Work Defining bézier curves can involve trail and error to achieve the effect you want. Fortunately, there are a number of great tools to help you experiment and produce the correct code: In the final part of this series, we’ll look at a couple of advanced transition techniques. Frequently Asked Questions about CSS3 Transitions and Cubic Bezier Timing Function What is the cubic-bezier function in CSS3? The cubic-bezier function is a specific type of timing function in CSS3. It allows you to create custom easing effects for animations and transitions. The function takes four arguments, each representing two control points of a cubic Bézier curve. By manipulating these points, you can create a wide variety of easing effects, from simple linear transitions to complex, multi-stage How do I use the cubic-bezier function in CSS3? To use the cubic-bezier function, you need to specify it as the value of the ‘transition-timing-function’ or ‘animation-timing-function’ property in your CSS code. The function takes four arguments, which represent the coordinates of two control points of a cubic Bézier curve. For example, the code ‘transition-timing-function: cubic-bezier(0.1, 0.7, 1.0, 0.1);’ will create a custom easing effect for a transition. What are the default control points for the cubic-bezier function? The default control points for the cubic-bezier function are (0,0) and (1,1). These points create a linear transition, where the animation progresses at a constant speed from start to finish. By changing these points, you can create a variety of different easing effects. Can I use the cubic-bezier function to create a bounce effect? Yes, you can use the cubic-bezier function to create a bounce effect. This involves setting the control points to values that cause the animation to move back and forth along the transition path. However, creating a bounce effect with the cubic-bezier function can be complex and requires a good understanding of how Bézier curves work. How can I visualize the effect of different control points in the cubic-bezier function? There are several online tools that allow you to visualize the effect of different control points in the cubic-bezier function. These tools typically provide an interactive graph where you can manipulate the control points and see the resulting Bézier curve and easing effect. What is the difference between the cubic-bezier function and the steps function in CSS3? The cubic-bezier function and the steps function are both timing functions in CSS3, but they work in different ways. The cubic-bezier function creates a smooth transition that can be customized with control points, while the steps function creates a stepped transition with a fixed number of intervals. Can I use the cubic-bezier function with the ‘transform’ property in CSS3? Yes, you can use the cubic-bezier function with the ‘transform’ property in CSS3. This allows you to create complex animations where the element transforms over time according to a custom easing How does the cubic-bezier function relate to animation performance? The cubic-bezier function can have an impact on animation performance. Complex easing functions that require a lot of computation can slow down animations, especially on lower-powered devices. However, in most cases, the performance impact of the cubic-bezier function is minimal. Can I use the cubic-bezier function in all web browsers? The cubic-bezier function is supported in all modern web browsers, including Chrome, Firefox, Safari, and Edge. However, it may not work in older browsers or those that do not fully support CSS3. What are some common uses of the cubic-bezier function in web design? The cubic-bezier function is commonly used in web design to create custom animations and transitions. This can include things like hover effects, loading animations, slide-out menus, and much more. By using the cubic-bezier function, designers can create unique and engaging user experiences. Craig is a freelance UK web consultant who built his first page for IE2.0 in 1995. Since that time he's been advocating standards, accessibility, and best-practice HTML5 techniques. He's created enterprise specifications, websites and online applications for companies and organisations including the UK Parliament, the European Parliament, the Department of Energy & Climate Change, Microsoft, and more. He's written more than 1,000 articles for SitePoint and you can find him @craigbuckler. CSS3HTML5 Dev CenterHTML5 Tutorials & Articleslearn-advanced-csstransitions	{"url":"https://www.sitepoint.com/css3-transitions-cubic-bezier-timing-function/","timestamp":"2024-11-03T09:54:44Z","content_type":"text/html","content_length":"220615","record_id":"<urn:uuid:35671add-71a1-4847-abb2-7f681e0ec4bc>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00185.warc.gz"}
Gallery \| paradigm shift in physics Physics beyond Einstein's Relativity The theory of gravity "Newtonian quantum gravity" (NQG) is simple theory, because it precisely predicts so-called "general relativistic phenomena," as, for example, that observed at the binary pulsar PSR B1913 + 16, by just applying Kepler's second law on quantized gravitational fields. It is an irony of fate that the unsuspecting relativistic physicists still have to effort with the tensor calculations of an imaginary four-dimensional space-time. Everybody can understand that a mass that moves through space must meet more "gravitational quanta" emitted by a certain mass, if it moves faster than if it moves slower or rests against a certain mass, which must cause additional gravitational effects that must be added to the results of Newton's theory of gravity. However, today's physicists cannot recognize this because they are caught in Einstein's relativistic thinking and as general relativity can coincidentally also predict these quantum effects by a mathematically defined four-dimensional curvature of space-time.	{"url":"https://www.quantum-gravity.de/gallery/","timestamp":"2024-11-04T14:16:06Z","content_type":"text/html","content_length":"47090","record_id":"<urn:uuid:2dcf176c-7087-4f24-9e7c-2402a9e8807d>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00842.warc.gz"}
Good evening, hackers. Today’s missive is more of a massive, in the sense that it’s another presentation transcript-alike; these things always translate to many vertical pixels. In my defense, I hardly ever give a presentation twice, so not only do I miss out on the usual per-presentation cost amortization and on the incremental improvements of repetition, the more dire error is that whatever message I might have can only ever reach a subset of those that it might interest; here at least I can be more or less sure that if the presentation would interest someone, that they will find it. So for the time being I will try to share presentations here, in the spirit of, well, why the hell not. CPS Soup A functional intermediate language 10 May 2023 – Spritely Andy Wingo Igalia, S.L. Last week I gave a training talk to Spritely Institute collaborators on the intermediate representation used by Guile‘s compiler. CPS Soup Compiler: Front-end to Middle-end to Back-end Middle-end spans gap between high-level source code (AST) and low-level machine code Programs in middle-end expressed in intermediate language CPS Soup is the language of Guile’s middle-end An intermediate representation (IR) (or intermediate language, IL) is just another way to express a computer program. Specifically it’s the kind of language that is appropriate for the middle-end of a compiler, and by “appropriate” I meant that an IR serves a purpose: there has to be a straightforward transformation to the IR from high-level abstract syntax trees (ASTs) from the front-end, and there has to be a straightforward translation from IR to machine code. There are also usually a set of necessary source-to-source transformations on IR to “lower” it, meaning to make it closer to the back-end than to the front-end. There are usually a set of optional transformations to the IR to make the program run faster or allocate less memory or be more simple: these are the optimizations. “CPS soup” is Guile’s IR. This talk presents the essentials of CPS soup in the context of more traditional IRs. How to lower? (+ 1 (if x 42 69)) cmpi $x, #f je L1 movi $t, 42 j L2 movi $t, 69 addi $t, 1 How to get from here to there? Before we dive in, consider what we might call the dynamic range of an intermediate representation: we start with what is usually an algebraic formulation of a program and we need to get down to a specific sequence of instructions operating on registers (unlimited in number, at this stage; allocating to a fixed set of registers is a back-end concern), with explicit control flow between them. What kind of a language might be good for this? Let’s attempt to answer the question by looking into what the standard solutions are for this problem domain. Control-flow graph (CFG) graph := array<block> block := tuple<preds, succs, insts> inst := goto B \| if x then BT else BF \| z = const C \| z = add x, y BB0: if x then BB1 else BB2 BB1: t = const 42; goto BB3 BB2: t = const 69; goto BB3 BB3: t2 = addi t, 1; ret t2 Assignment, not definition Of course in the early days, there was no intermediate language; compilers translated ASTs directly to machine code. It’s been a while since I dove into all this but the milestone I have in my head is that it’s the 70s when compiler middle-ends come into their own right, with Fran Allen’s work on flow analysis and optimization. In those days the intermediate representation for a compiler was a graph of basic blocks, but unlike today the paradigm was assignment to locations rather than definition of values. By that I mean that in our example program, we get t assigned to in two places (BB1 and BB2); the actual definition of t is implicit, as a storage location, and our graph consists of assignments to the set of storage locations in the program. Static single assignment (SSA) CFG graph := array<block> block := tuple<preds, succs, phis, insts> phi := z := φ(x, y, ...) inst := z := const C \| z := add x, y BB0: if x then BB1 else BB2 BB1: v0 := const 42; goto BB3 BB2: v1 := const 69; goto BB3 BB3: v2 := φ(v0,v1); v3:=addi t,1; ret v3 Phi is phony function: v2 is v0 if coming from first predecessor, or v1 from second predecessor These days we still live in Fran Allen’s world, but with a twist: we no longer model programs as graphs of assignments, but rather graphs of definitions. The introduction in the mid-80s of so-called “static single-assignment” (SSA) form graphs mean that instead of having two assignments to t, we would define two different values v0 and v1. Then later instead of reading the value of the storage location associated with t, we define v2 to be either v0 or v1: the former if we reach the use of t in BB3 from BB1, the latter if we are coming from BB2. If you think on the machine level, in terms of what the resulting machine code will be, this either function isn’t a real operation; probably register allocation will put v0, v1, and v2 in the same place, say $rax. The function linking the definition of v2 to the inputs v0 and v1 is purely notational; in a way, you could say that it is phony, or not real. But when the creators of SSA went to submit this notation for publication they knew that they would need something that sounded more rigorous than “phony function”, so they instead called it a “phi” (φ) function. Really. 2003: MLton Refinement: phi variables are basic block args graph := array<block> block := tuple<preds, succs, args, insts> Inputs of phis implicitly computed from preds BB0(a0): if a0 then BB1() else BB2() BB1(): v0 := const 42; BB3(v0) BB2(): v1 := const 69; BB3(v1) BB3(v2): v3 := addi v2, 1; ret v3 SSA is still where it’s at, as a conventional solution to the IR problem. There have been some refinements, though. I learned of one of them from MLton; I don’t know if they were first but they had the idea of interpreting phi variables as arguments to basic blocks. In this formulation, you don’t have explicit phi instructions; rather the “v2 is either v1 or v0” property is expressed by v2 being a parameter of a block which is “called” with either v0 or v1 as an argument. It’s the same semantics, but an interesting notational change. Refinement: Control tail Often nice to know how a block ends (e.g. to compute phi input vars) graph := array<block> block := tuple<preds, succs, args, insts, control := if v then L1 else L2 \| L(v, ...) \| switch(v, L1, L2, ...) \| ret v One other refinement to SSA is to note that basic blocks consist of some number of instructions that can define values or have side effects but which otherwise exhibit fall-through control flow, followed by a single instruction that transfers control to another block. We might as well store that control instruction separately; this would let us easily know how a block ends, and in the case of phi block arguments, easily say what values are the inputs of a phi variable. So let’s do that. Refinement: DRY Block successors directly computable from control Predecessors graph is inverse of successors graph graph := array<block> block := tuple<args, insts, control> Can we simplify further? At this point we notice that we are repeating ourselves; the successors of a block can be computed directly from the block’s terminal control instruction. Let’s drop those as a distinct part of a block, because when you transform a program it’s unpleasant to have to needlessly update something in two places. While we’re doing that, we note that the predecessors array is also redundant, as it can be computed from the graph of block successors. Here we start to wonder: am I simpliying or am I removing something that is fundamental to the algorithmic complexity of the various graph transformations that I need to do? We press on, though, hoping we will get somewhere interesting. Basic blocks are annoying Ceremony about managing insts; array or doubly-linked list? Nonuniformity: “local” vs ‘`global’' transformations Optimizations transform graph A to graph B; mutability complicates this task • Desire to keep A in mind while making B • Bugs because of spooky action at a distance Recall that the context for this meander is Guile’s compiler, which is written in Scheme. Scheme doesn’t have expandable arrays built-in. You can build them, of course, but it is annoying. Also, in Scheme-land, functions with side-effects are conventionally suffixed with an exclamation mark; after too many of them, both the writer and the reader get fatigued. I know it’s a silly argument but it’s one of the things that made me grumpy about basic blocks. If you permit me to continue with this introspection, I find there is an uneasy relationship between instructions and locations in an IR that is structured around basic blocks. Do instructions live in a function-level array and a basic block is an array of instruction indices? How do you get from instruction to basic block? How would you hoist an instruction to another basic block, might you need to reallocate the block itself? And when you go to transform a graph of blocks... well how do you do that? Is it in-place? That would be efficient; but what if you need to refer to the original program during the transformation? Might you risk reading a stale graph? It seems to me that there are too many concepts, that in the same way that SSA itself moved away from assignment to a more declarative language, that perhaps there is something else here that might be more appropriate to the task of a middle-end. Basic blocks, phi vars redundant Blocks: label with args sufficient; “containing” multiple instructions is superfluous Unify the two ways of naming values: every var is a phi graph := array<block> block := tuple<args, inst> inst := L(expr) \| if v then L1() else L2() expr := const C \| add x, y I took a number of tacks here, but the one I ended up on was to declare that basic blocks themselves are redundant. Instead of containing an array of instructions with fallthrough control-flow, why not just make every instruction a control instruction? (Yes, there are arguments against this, but do come along for the ride, we get to a funny place.) While you are doing that, you might as well unify the two ways in which values are named in a MLton-style compiler: instead of distinguishing between basic block arguments and values defined within a basic block, we might as well make all names into basic block arguments. Arrays annoying Array of blocks implicitly associates a label with each block Optimizations add and remove blocks; annoying to have dead array entries Keep labels as small integers, but use a map instead of an array graph := map<label, block> In the traditional SSA CFG IR, a graph transformation would often not touch the structure of the graph of blocks. But now having given each instruction its own basic block, we find that transformations of the program necessarily change the graph. Consider an instruction that we elide; before, we would just remove it from its basic block, or replace it with a no-op. Now, we have to find its predecessor(s), and forward them to the instruction’s successor. It would be useful to have a more capable data structure to represent this graph. We might as well keep labels as being small integers, but allow for sparse maps and growth by using an integer-specialized map instead of an array. This is CPS soup graph := map<label, cont> cont := tuple<args, term> term := continue to L with values from expr \| if v then L1() else L2() expr := const C \| add x, y SSA is CPS This is exactly what CPS soup is! We came at it “from below”, so to speak; instead of the heady fumes of the lambda calculus, we get here from down-to-earth basic blocks. (If you prefer the other way around, you might enjoy this article from a long time ago.) The remainder of this presentation goes deeper into what it is like to work with CPS soup in practice. Scope and dominators BB0(a0): if a0 then BB1() else BB2() BB1(): v0 := const 42; BB3(v0) BB2(): v1 := const 69; BB3(v1) BB3(v2): v3 := addi v2, 1; ret v3 What vars are “in scope” at BB3? a0 and v2. Not v0; not all paths from BB0 to BB3 define v0. a0 always defined: its definition dominates all uses. BB0 dominates BB3: All paths to BB3 go through BB0. Before moving on, though, we should discuss what it means in an SSA-style IR that variables are defined rather than assigned. If you consider variables as locations to which values can be assigned and which initially hold garbage, you can read them at any point in your program. You might get garbage, though, if the variable wasn’t assigned something sensible on the path that led to reading the location’s value. It sounds bonkers but it is still the C and C++ semantic model. If we switch instead to a definition-oriented IR, then a variable never has garbage; the single definition always precedes any uses of the variable. That is to say that all paths from the function entry to the use of a variable must pass through the variable’s definition, or, in the jargon, that definitions dominate uses. This is an invariant of an SSA-style IR, that all variable uses be dominated by their associated definition. You can flip the question around to ask what variables are available for use at a given program point, which might be read equivalently as which variables are in scope; the answer is, all definitions from all program points that dominate the use site. The “CPS” in “CPS soup” stands for continuation-passing style, a dialect of the lambda calculus, which has also has a history of use as a compiler intermediate representation. But it turns out that if we use the lambda calculus in its conventional form, we end up needing to maintain a lexical scope nesting at the same time that we maintain the control-flow graph, and the lexical scope tree can fail to reflect the dominator tree. I go into this topic in more detail in an old article, and if it interests you, please do go deep. CPS soup in Guile Compilation unit is intmap of label to cont cont := $kargs names vars term \| ... term := $continue k src expr \| ... expr := $const C \| $primcall ’add #f (a b) \| ... Conventionally, entry point is lowest-numbered label Anyway! In Guile, the concrete form that CPS soup takes is that a program is an intmap of label to cont. A cont is the smallest labellable unit of code. You can call them blocks if that makes you feel better. One kind of cont, $kargs, binds incoming values to variables. It has a list of variables, vars, and also has an associated list of human-readable names, names, for debugging purposes. A $kargs contains a term, which is like a control instruction. One kind of term is $continue, which passes control to a continuation k. Using our earlier language, this is just goto k, with values, as in MLton. (The src is a source location for the term.) The values come from the term’s expr, of which there are a dozen kinds or so, for example $const which passes a literal constant, or $primcall, which invokes some kind of primitive operation, which above is add. The primcall may have an immediate operand, in this case #f, and some variables that it uses, in this case a and b. The number and type of the produced values is a property of the primcall; some are just for effect, some produce one value, some more. CPS soup term := $continue k src expr \| $branch kf kt src op param args \| $switch kf kt* src arg \| $prompt k kh src escape? tag \| $throw src op param args Expressions can have effects, produce values expr := $const val \| $primcall name param args \| $values args \| $call proc args \| ... There are other kinds of terms besides $continue: there is $branch, which proceeds either to the false continuation kf or the true continuation kt depending on the result of performing op on the variables args, with immediate operand param. In our running example, we might have made the initial term via: ($branch BB1 BB2 'false? #f (a0))) The definition of build-term (and build-cont and build-exp) is in the (language cps) module. There is also $switch, which takes an unboxed unsigned integer arg and performs an array dispatch to the continuations in the list kt, or kf otherwise. There is $prompt which continues to its k, having pushed on a new continuation delimiter associated with the var tag; if code aborts to tag before the prompt exits via an unwind primcall, the stack will be unwound and control passed to the handler continuation kh. If escape? is true, the continuation is escape-only and aborting to the prompt doesn’t need to capture the suspended continuation. Finally there is $throw, which doesn’t continue at all, because it causes a non-resumable exception to be thrown. And that’s it; it’s just a handful of kinds of term, determined by the different shapes of control-flow (how many continuations the term has). When it comes to values, we have about a dozen expression kinds. We saw $const and $primcall, but I want to explicitly mention $values, which simply passes on some number of values. Often a $values expression corresponds to passing an input to a phi variable, though $kargs vars can get their definitions from any expression that produces the right number of values. Kinds of continuations Guile functions untyped, can multiple return values Error if too few values, possibly truncate too many values, possibly cons as rest arg... Calling convention: contract between val producer & consumer • both on call and return side Continuation of $call unlike that of $const When a $continue term continues to a $kargs with a $const 42 expression, there are a number of invariants that the compiler can ensure: that the $kargs continuation is always passed the expected number of values, that the vars that it binds can be allocated to specific locations (e.g. registers), and that because all predecessors of the $kargs are known, that those predecessors can place their values directly into the variable’s storage locations. Effectively, the compiler determines a custom calling convention between each $kargs and its predecessors. Consider the $call expression, though; in general you don’t know what the callee will do to produce its values. You don’t even generally know that it will produce the right number of values. Therefore $call can’t (in general) continue to $kargs; instead it continues to $kreceive, which expects the return values in well-known places. $kreceive will check that it is getting the right number of values and then continue to a $kargs, shuffling those values into place. A standard calling convention defines how functions return values to callers. The conts cont := $kfun src meta self ktail kentry \| $kclause arity kbody kalternate \| $kargs names syms term \| $kreceive arity kbody \| $ktail $kclause, $kreceive very similar Continue to $ktail: return $call and return (and $throw, $prompt) exit first-order flow graph Of course, a $call expression could be a tail-call, in which case it would continue instead to $ktail, indicating an exit from the first-order function-local control-flow graph. The calling convention also specifies how to pass arguments to callees, and likewise those continuations have a fixed calling convention; in Guile we start functions with $kfun, which has some metadata attached, and then proceed to $kclause which bridges the boundary between the standard calling convention and the specialized graph of $kargs continuations. (Many details of this could be tweaked, for example that the case-lambda dispatch built-in to $kclause could instead dispatch to distinct functions instead of to different places in the same function; historical accidents abound.) As a detail, if a function is well-known, in that all its callers are known, then we can lighten the calling convention, moving the argument-count check to callees. In that case $kfun continues directly to $kargs. Similarly for return values, optimizations can make $call continue to $kargs, though there is still some value-shuffling to do. High and low CPS bridges AST (Tree-IL) and target code High-level: vars in outer functions in scope Closure conversion between high and low Low-level: Explicit closure representations; access free vars through closure CPS soup is the bridge between parsed Scheme and machine code. It starts out quite high-level, notably allowing for nested scope, in which expressions can directly refer to free variables. Variables are small integers, and for high-level CPS, variable indices have to be unique across all functions in a program. CPS gets lowered via closure conversion, which chooses specific representations for each closure that remains after optimization. After closure conversion, all variable access is local to the function; free variables are accessed via explicit loads from a function’s closure. Optimizations at all levels Optimizations before and after lowering Some exprs only present in one level Some high-level optimizations can merge functions (higher-order to first-order) Because of the broad remit of CPS, the language itself has two dialects, high and low. The high level dialect has cross-function variable references, first-class abstract functions (whose representation hasn’t been chosen), and recursive function binding. The low-level dialect has only specific ways to refer to functions: labels and specific closure representations. It also includes calls to function labels instead of just function values. But these are minor variations; some optimization and transformation passes can work on either dialect. Intmap, intset: Clojure-style persistent functional data structures Program: intmap<label,cont> Optimization: program→program Identify functions: (program,label)→intset<label> Edges: intmap<label,intset<label>> Compute succs: (program,label)→edges Compute preds: edges→edges I mentioned that programs were intmaps, and specifically in Guile they are Clojure/Bagwell-style persistent functional data structures. By functional I mean that intmaps (and intsets) are values that can’t be mutated in place (though we do have the transient optimization). I find that immutability has the effect of deploying a sense of calm to the compiler hacker – I don’t need to worry about data structures changing out from under me; instead I just structure all the transformations that you need to do as functions. An optimization is just a function that takes an intmap and produces another intmap. An analysis associating some data with each program label is just a function that computes an intmap, given a program; that analysis will never be invalidated by subsequent transformations, because the program to which it applies will never be mutated. This pervasive feeling of calm allows me to tackle problems that I wouldn’t have otherwise been able to fit into my head. One example is the novel online CSE pass; one day I’ll either wrap that up as a paper or just capitulate and blog it instead. Flow analysis A[k] = meet(A[p] for p in preds[k]) - kill[k] + gen[k] Compute available values at labels: • A: intmap<label,intset<val>> • meet: intmap-intersect<intset-intersect> • -, +: intset-subtract, intset-union • kill[k]: values invalidated by cont because of side effects • gen[k]: values defined at k But to keep it concrete, let’s take the example of flow analysis. For example, you might want to compute “available values” at a given label: these are the values that are candidates for common subexpression elimination. For example if a term is dominated by a car x primcall whose value is bound to v, and there is no path from the definition of V to a subsequent car x primcall, we can replace that second duplicate operation with $values (v) instead. There is a standard solution for this problem, which is to solve the flow equation above. I wrote about this at length ages ago, but looking back on it, the thing that pleases me is how easy it is to decompose the task of flow analysis into manageable parts, and how the types tell you exactly what you need to do. It’s easy to compute an initial analysis A, easy to define your meet function when your maps and sets have built-in intersect and union operators, easy to define what addition and subtraction mean over sets, and so on. Persistent data structures FTW • meet: intmap-intersect<intset-intersect> • -, +: intset-subtract, intset-union Naïve: O(nconts * nvals) Structure-sharing: O(nconts * log(nvals)) Computing an analysis isn’t free, but it is manageable in cost: the structure-sharing means that meet is usually trivial (for fallthrough control flow) and the cost of + and - is proportional to the log of the problem size. CPS soup: strengths Relatively uniform, orthogonal Facilitates functional transformations and analyses, lowering mental load: “I just have to write a function from foo to bar; I can do that” Encourages global optimizations Some kinds of bugs prevented by construction (unintended shared mutable state) We get the SSA optimization literature Well, we’re getting to the end here, and I want to take a step back. Guile has used CPS soup as its middle-end IR for about 8 years now, enough time to appreciate its fine points while also understanding its weaknesses. On the plus side, it has what to me is a kind of low cognitive overhead, and I say that not just because I came up with it: Guile’s development team is small and not particularly well-resourced, and we can’t afford complicated things. The simplicity of CPS soup works well for our development process (flawed though that process may be!). I also like how by having every variable be potentially a phi, that any optimization that we implement will be global (i.e. not local to a basic block) by default. Perhaps best of all, we get these benefits while also being able to use the existing SSA transformation literature. Because CPS is SSA, the lessons learned in SSA (e.g. loop peeling) apply directly. CPS soup: weaknesses Pointer-chasing, indirection through intmaps Heavier than basic blocks: more control-flow edges Names bound at continuation only; phi predecessors share a name Over-linearizes control, relative to sea-of-nodes Overhead of re-computation of analyses CPS soup is not without its drawbacks, though. It’s not suitable for JIT compilers, because it imposes some significant constant-factor (and sometimes algorithmic) overheads. You are always indirecting through intmaps and intsets, and these data structures involve significant pointer-chasing. Also, there are some forms of lightweight flow analysis that can be performed naturally on a graph of basic blocks without looking too much at the contents of the blocks; for example in our available variables analysis you could run it over blocks instead of individual instructions. In these cases, basic blocks themselves are an optimization, as they can reduce the size of the problem space, with corresponding reductions in time and memory use for analyses and transformations. Of course you could overlay a basic block graph on top of CPS soup, but it’s not a well-worn path. There is a little detail that not all phi predecessor values have names, since names are bound at successors (continuations). But this is a detail; if these names are important, little $values trampolines can be inserted. Probably the main drawback as an IR is that the graph of conts in CPS soup over-linearizes the program. There are other intermediate representations that don’t encode ordering constraints where there are none; perhaps it would be useful to marry CPS soup with sea-of-nodes, at least during some transformations. Finally, CPS soup does not encourage a style of programming where an analysis is incrementally kept up to date as a program is transformed in small ways. The result is that we end up performing much redundant computation within each individual optimization pass. CPS soup is SSA, distilled Labels and vars are small integers Programs map labels to conts Conts are the smallest labellable unit of code Conts can have terms that continue to other conts Compilation simplifies and lowers programs Wasm vs VM backend: a question for another day :) But all in all, CPS soup has been good for Guile. It’s just SSA by another name, in a simpler form, with a functional flavor. Or, it’s just CPS, but first-order only, without lambda. In the near future, I am interested in seeing what a new GC will do for CPS soup; will bump-pointer allocation palliate some of the costs of pointer-chasing? We’ll see. A tricky thing about CPS soup is that I don’t think that anyone else has tried it in other languages, so it’s hard to objectively understand its characteristics independent of Guile itself. Finally, it would be nice to engage in the academic conversation by publishing a paper somewhere; I would like to see interesting criticism, and blog posts don’t really participate in the citation graph. But in the limited time available to me, faced with the choice between hacking on something and writing a paper, it’s always been hacking, so far :) Speaking of limited time, I probably need to hit publish on this one and move on. Happy hacking to all, and until next time.	{"url":"http://www.wingolog.org/feed/atom?with=flow%20analysis","timestamp":"2024-11-03T17:15:18Z","content_type":"application/atom+xml","content_length":"114031","record_id":"<urn:uuid:87c19ba4-2ea2-4cde-880f-2db94dde41d2>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00068.warc.gz"}
Barry Martin's Hopalong Orbits Visualizer - WebVR // WebGL stats tracking Barry Martin's Hopalong Orbits Visualizer - Created using [Up-Down] Control speed - [Left-Right] Control rotation Barry Martin's Hopalong Orbits Visualizer These orbits are generated iterating this simple algorithm: (x, y) -> (y - sign(x)sqrt(abs(bx - c)), a -x ) where a, b, c are random parameters. This is known as the 'Hopalong Attractor'. 3D rendering done with three.js and vrrenderer.js Original version by Iacopo Sassarini	{"url":"http://hopalongvr.dghost.net/","timestamp":"2024-11-02T09:07:12Z","content_type":"text/html","content_length":"23931","record_id":"<urn:uuid:0fa72e91-9ed1-4a32-a0b5-d86a283d3ba3>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00636.warc.gz"}
Wrangling data with feature discretization, standardization \| TechTarget Wrangling data with feature discretization, standardization A variety of techniques help make data useful in machine learning algorithms. This article looks into two such data-wrangling techniques: discretization and standardization. This article is excerpted from the course "Fundamental Machine Learning," part of the Machine Learning Specialist certification program from Arcitura Education. It is the eighth part of the 13-part series, "Using machine learning algorithms, practices and patterns." This article continues the discussion begun in Part 7 on how machine learning data-wrangling techniques help prepare data to be used as input for a machine learning algorithm. This article focuses on two specific data-wrangling techniques: feature discretization and feature standardization, both of which are documented in a standard pattern profile format. Feature discretization: Overview • How can continuous features be used for model development when the underlying machine learning algorithm only supports discrete/nominal features? Or: How can the range of values that a continuous feature can take on be reduced in order to lower model complexity? • Some machine learning algorithms can only accept discrete or nominal features as input, which makes the inclusion of valuable continuous features impossible as an input for model development. Or: Using numerical features with a very wide range of continuous values makes the model complicated with further implications of overfitting and longer training and prediction times. • A limited number of discrete sets of values are derived from continuous features by employing statistical or machine learning techniques. • The continuous features are subjected to techniques such as binning and clustering that group continuous values into discrete bins, thereby discretizing continuous features into discrete ones. Feature discretization: Explained Some non-distance-based machine learning algorithms -- in other words, those that do not use distance measure for classification or clustering such as Naïve Bayes -- normally require input to comprise only categorical or discrete values. Even if an algorithm is able to work with continuous values, with the possibility of a very large number of feature values such as slight variations in decimal values of temperature readings with four decimal places, the dimension of the feature space becomes unwieldly. This makes the underlying mathematical operations too expensive. For example, with Naïve Bayes, the algorithm needs to calculate the probability of each unique value, which can be far too many in cases with large data sets with decimal values. This results in long training and prediction times (Figure 1). Figure 1: A training data set contains Feature B, which consists of various values. A probabilistic model that works best with discrete values needs to be trained using this data set (1). However, the training process runs for a very long time and in the end it errors out (2, 3). The continuous values are reduced to a manageable set of discrete values, such as ordinal or categorical values. The reduction is either achieved by a simple binning strategy, such that values in a certain range get replaced by the label of the corresponding bin, or by a more complex operation involving the use of other features such that the values grouped closely in n-dimensions are allocated to a single ordinal value. A number of techniques can be applied to achieve discretization, including binning and clustering. Binning is where ordered attribute values are grouped into intervals or bins, which can be created using either the equal-frequency or equal-width methods. The bin labels can be used in place of the original attribute values. This technique is affected by outliers and skewed data. In the equal-width method, the interval of values is the same, with each bin generally containing a different number of values. In the equal-frequency method, each bin contains the same number of values so the interval of values is generally not the same. To determine which type of binning to use, it is recommended to look at the shape of the distribution. With a Gaussian distribution, equal-frequency binning is normally used, whereas equal-width binning is normally used for a uniform distribution. Clustering involves the use of a clustering algorithm to divide a continuous attribute into a set of groups, based on the closeness and distribution makeup of the attribute values. Using clustering, outliers can be detected to prevent adverse effects of outliers on data discretization (Figure 2). Figure 2: A training data set contains Feature B, which consists of various values. A probabilistic model that works best with discrete values needs to be trained using this data set (1). The binning technique is applied to Feature C. However, before a binning strategy is chosen, the distribution of Feature B is examined (2). It is determined that the distribution is normal, and the equal-frequency binning strategy is consequently applied (3). This results in a data set where all feature values are discrete in nature (4). The model is then successfully trained using this data set (5, 6). Feature standardization: Overview • How can it be ensured that features with wide-ranging values do not overshadow other features carrying a smaller range of values? • Features whose values exist over a widescale carry the possibility of reducing the predictive potential of features whose values exist over a narrow scale, thereby resulting in the development of a less accurate model. • All numerical features in a data set are brought within the same scale so that the magnitude of each feature carries the same predictive potential. • Statistical techniques, such as min-max scaling, mean normalization and z-score standardization, are applied to convert the features' values in such a way that the values always exist within a known set of upper and lower bounds. Feature standardization: Explained A data set may contain a mixture of numerical features where some features operate at a different scale, such as one feature with values between -100 to +100 and another feature with values between 32 and 212. There can also be a difference in the units used by different features, such as one feature using meters while another uses kilometers. These differences incorrectly give more importance to features with higher magnitudes than lower ones. As a result, algorithms using distance calculation, such as Euclidean distance, focus on features with higher magnitudes and ignore the contribution of smaller magnitude features. In addition, a unit change in a feature using smaller units, such as yards, is treated the same as the feature using bigger units, such as miles, which should not be the case (Figure 3). Figure 3: A training data set comprises features with varying scales. Feature A's values are between 0 and 5000, and Feature B's values are between -15 and +15. Feature C's values are between 30 and 60 (1). The data set is used to train a model (2, 3). The resulting model has very low accuracy (4). All numerical feature values are transformed so that there is no difference in magnitude and they operate within the same scale. Two different techniques exist for standardization: scaling and normalization. Scaling modifies the range of the data and is used when the distribution is non-Gaussian or when there is uncertainty about the type of distribution. Normalization modifies the shape of the data (distribution) and transforms it into a normal distribution. It is used when the data is Gaussian in nature and the algorithm for model training requires input data to be in normal form. Both scaling and normalization are generally applied using data preprocessing functions available either within the model development software or as a separate library that can be imported. For scaling, a popular algorithm is min-max scaling, which brings the values within 0 and 1. However, based on requirements, another practice is to scale data between -1 and +1. For normalization, either z-score standardization or mean normalization can be used. Z-score standardization transforms the values into a distribution with zero mean and unit standard deviation, while mean normalization transforms the values into a distribution with zero mean and a range from -1 to +1 (Figure 4). Figure 4: A training data set comprises features with varying scales. Feature A's values are between 0 and 5000, and Feature B's values are between -15 and +15. Feature C's values are between 30 and 60 (1). The data set is exposed to the min-max scaling technique (2). The resulting standardized data set contains values such that all feature values are now between 0 and 1 (3). The standardized data set is then used to train a model (4, 5). The resulting model has very high accuracy (6). What's next The next article covers two supervised learning patterns: numerical prediction and category prediction.	{"url":"https://www.techtarget.com/searchenterpriseai/post/Wrangling-data-with-feature-discretization-standardization","timestamp":"2024-11-03T19:40:19Z","content_type":"text/html","content_length":"318007","record_id":"<urn:uuid:948ea236-25e3-47bb-a4f7-eaabbb31d58d>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00508.warc.gz"}
Lesson 14 Defining Rotations Problem 1 Draw the image of quadrilateral $ABCD$ when rotated $120^\circ$ counterclockwise around the point $D$. Problem 2 There is an equilateral triangle, $ABC$, inscribed in a circle with center $D$. What is the smallest angle you can rotate triangle $ABC$ around $D$ so that the image of $A$ is $B$? Problem 3 Which segment is the image of $AB$ when rotated $90^\circ$ counterclockwise around point $P$? Problem 4 The semaphore alphabet is a way to use flags to signal messages. Here's how to signal the letter Q. Describe a transformation that would take the right hand flag to the left hand flag. Problem 5 Here are 2 polygons: Select all sequences of translations, rotations, and reflections below that would take polygon $P$ to polygon $Q$. Rotate $180^\circ$ around point $A$. Translate so that $A$ is taken to $J$. Then reflect over line $BA$. Rotate $60^\circ$ counterclockwise around point $A$ and then reflect over the line $FA$. Reflect over the line $BA$ and then rotate $60^\circ$ counterclockwise around point $A$. Reflect over line $BA$ and then translate by directed line segment $BA$. Problem 6 1. Draw the image of figure $ABC$ when translated by directed line segment $u$. Label the image of $A$ as $A’$, the image of $B$ as $B’$, and the image of $C$ as $C’$. 2. Explain why the line containing $AB$ is parallel to the line containing $A’B’$. Problem 7 There is a sequence of rigid transformations that takes $A$ to $A’$, $B$ to $B’$, and $C$ to $C’$. The same sequence takes $D$ to $D’$. Draw and label $D’$:	{"url":"https://im-beta.kendallhunt.com/HS/teachers/2/1/14/practice.html","timestamp":"2024-11-03T07:12:02Z","content_type":"text/html","content_length":"93781","record_id":"<urn:uuid:3bc14848-845b-4856-bddf-a8f5e93ed51e>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00320.warc.gz"}
SM Toolbox The SM Toolbox has been developed by Meinard Müller, Nanzhu Jiang, Peter Grosche, and Harald G. Grohganz. It contains MATLAB implementations for computing and enhancing similarity matrices in various ways. Furthermore, the toolbox includes a number of additional tools for parsing, navigation, and visualization synchronized with audio playback. Also, it contains code for a recently proposed audio thumbnailing procedure that demonstrates the applicability and importance of enhancement concepts. The MATLAB implementations provided on this website are published under the terms of the General Public License (GPL). A general overview of the SM Toolbox is given in [1]. If you publish results obtained using these implementations, please cite [1]. For technical details, applications, or data please cite [2], [3], [4], [5], [6], [7]. MATLAB Code The MATLAB implementations provided on this website are published under the terms of the General Public License (GPL), version 2 or later. If you publish results obtained using these implementations, please cite the references below. Download SM Toolbox (Version 1.0. Last update: 2013-07-01): [zip] Computation and Visualization of SSMs Thumbnailing application The following demo files are provided. These demo files allow you to try out the code and give you a first overview of the toolbox. The necessary audio files to run the demos are also provided by the • demoSMtoolbox.m Demo showing various enhancement functionalities as described in [1]. • demoSMtoolbox_thumbnailing.m Demo for thumbnailing application as described in [1]. • demoSMtoolbox_thumbnailing_otherSettings.m Demo for thumbnailing application with other settings for various music recordings. Important Notes • For the SM Toolbox the MATLAB Signal Processing Toolbox is required. • For the feature computation the Chromagram Toolbox is required. For convenience, the Chroma Toolbox has been included in the folder MATLAB-Chroma-Toolbox_2.0 of the zip-file provided above. The feature extraction step may replaced using feature extraction functions supplied by other toolboxes. • The implementations have been tested using MATLAB 2012b or newer. • For questions, please contact Meinard Müller, Nanzhu Jiang or Harald G. Grohganz. Similarity Matrix The concept of similarity matrices (SMs) has been widely used for a multitude of music analysis and retrieval tasks including audio structure analysis or version identification. For such tasks, the improvement of structural properties of the similarity matrix at an early state of the processing pipeline has turned out to be of crucial importance. The SM toolbox contains MATLAB implementations for computing and enhancing similarity matrices in various ways. Original SSM Diagonal smoothing Tempo-invariant smoothing Forward-backward smoothing Tranposition-invariant SSM Tranposition index matrix Binary thresholding Thresholding with penalty • SM (Similarity Matrix): Given a audio recording, we first extract audio features such as chroma features from the recording, this is done by our Chroma Toolbox. Then a similarity measure between pair of features is specified. In our case, we use cosine similarity. Finally, we compute the similarity matrix with each element encodes the similarity between a certain pair of features. MATLAB function: features_to_SM.m • SM with diagonal smoothing: One important property of similarity matrices is the appearance of paths which represents high similarity of a pair of segments (the segments can be obtained by projecting the path on the vertical and horizontal axis respectively). One main task is to extract and identify such paths and using them to identify the similar pairs of segments. Due to musical and acoustic variations, there are noises around path structure which make the extraction and identification difficult. To further enhance the path structure, one general strategy is to apply some kind of smoothing filter along the direction of the main diagonal, resulting in an emphasis of diagonal information in S and a denoising of other structures. Controlling parameter: paramSM.smoothLenSM • SM with tempo-invariance: One of the main enhancement for a similarity matrix. In order to judge whether two segments are similar or not, a simple diagonal smoothing in the similarity matrix is usually not enough since music may have repeated parts with a faster or slower tempo. To deal with such tempo difference, we implement in our tool box that a similarity matrix is smoothed along various directions, each such direction corresponds to a tempo difference. Controlling parameters: paramSM.tempoRelMin, paramSM.tempoRelMax, paramSM.tempoNum • SM with forward and backward smoothing: By default, the implemented smoothing filter is realized to smooth in forward direction only. This results in a fading out of the paths in particular when using a large length parameter. To avoid this fading out, one can use a forward-backward option, which applies the filter also in backward direction. Controlling parameter: paramSM.forwardBackward • SM with transposition-invariance: It is often the case that certain musical parts are repeated in a transposed form. Such transpositions can be simulated by cyclically shifting chroma vectors. In our toolbox, we construct transposition-invariant similarity matrices by keeping one chroma feature sequence unaltered whereas the other chroma feature sequence cyclically shifted along the chroma dimension. Then, for each shifted version, a similarity matrix is computed, and the final similarity matrix is obtained by taking the cell-wise maximum over the twelve matrices. In this way, the repetitive structure is revealed even in the presence of key transpositions. Furthermore, storing the maximizing shift index for each cell results in another matrix referred to as transposition index matrix, which displays the harmonic relations within the music recording. Controlling parameter: paramSM.circShift • SM thresholded: In many music analysis applications, similarity matrices are further processed by suppressing all values that fall below a given threshold. On the one hand, such a step often leads to a substantial reduction of the noise while leaving only the most significant structures. On the other hand, weaker but still relevant information may be lost. Actually, the thresholding strategy may have a significant impact on the final results and has to be carefully chosen in the context of the considered application. In our toolbox, we offer some post-processing techniques such as thresholding, scaling, binarization or penalizing. . MATLAB functions and controlling parameters: threshSM.m, paramThres.threshTechnique, paramThres.threshValue, paramThres.applyBinarize, paramThres.applyScale, paramThres.penalty Thumbnailing Application As an illustrating application, our toolbox also contains the MATLAB code for a recently proposed audio thumbnailing procedure. For this task, the goal is to find the the most representative and repetitive segment of a given audio recording. Based on a suitable self-similarity matrix, the procedure in [4] computes for each audio segment a fitness value that expresses how well the given segment explains other related segments (also called induced segments) in the audio recording. These relations are expressed by a so-called path family over the given segment. The thumbnail is then defined as the fitness-maximizing segment. Furthermore, a triangular scape plot representation is computed, which shows the fitness of all segments and yields a compact high-level view on the structural properties of the entire audio recording. Fitness scape plot • Fitness scape plot: Starting with a self-similarity matrix, we derive the fitness scape plot where it encodes all fitness values representing the repetiveness for all possible segments of the given recording. In the computation, various step size and weighting parameters can be used to adjust the procedure. The resulting fitness scape plot can also be visualized. MATLAB functions: SSM_to_scapePlotFitness.m, visualizeScapePlot.m • Derive thumbnail: Using the fitness scape plot as input, we select the point which encodes the maximum fitness, and its corresponding segment is considered as the thumbnail. MATLAB function: scapePlotFitness_to_thumbnail.m • Induced segment family: Taking the thumbnail segment and the SSM, we compute the path family for the thumbnail and find all its repetition segments, these segments formed a induced segment family. The path family computation as well as the induced segment family can be visualizled. MATLAB functions: thumbnailSSM_to_pathFamily.m, visualizePathFamilySSM.m Further Functions • Parser of annotation file: As an assistant function, we provide a parser of annotation text file in our tool box. This parser can handle most of popular used structure annotating format. MATLAB function: parseAnnotationFile.m • Visualization with playback function: In order to get intuitive understanding of the relation between visualized phenomena and underlying music, we implemented a function which adds playback functionality to a given plot or image object. With the playback of sound file, one can easily inspect the figure for an audible analysis of certain points of interest. MATLAB function: makePlotPlayable.m 1. Meinard Müller, Nanzhu Jiang, and Harald G. Grohganz SM Toolbox: MATLAB Implementations for Computing and Enhancing Similarity Matrices In Proceedings of 53rd Audio Engineering Society (AES), 2014. author = {Meinard M{\"u}ller and Nanzhu Jiang and Harald G. Grohganz}, title = {SM Toolbox: MATLAB Implementations for Computing and Enhancing Similarity Matrices}, booktitle = {Proceedings of 53rd Audio Engineering Society ({AES})}, address = {London, UK}, year = {2014}, 2. Meinard Müller and Frank Kurth Enhancing Similarity Matrices for Music Audio Analysis In Proceedings of the International Conference on Acoustics, Speech and Signal Processing (ICASSP): 437–440, 2006. author = {Meinard M{\"u}ller and Frank Kurth}, title = {Enhancing Similarity Matrices for Music Audio Analysis}, booktitle = {Proceedings of the International Conference on Acoustics, Speech and Signal Processing ({ICASSP})}, address = {Toulouse, France}, year = {2006}, pages = {437--440}, 3. Meinard Müller and Michael Clausen Transposition-Invariant Self-Similarity Matrices In Proceedings of the International Conference on Music Information Retrieval (ISMIR): 47–50, 2007. author = {Meinard M{\"u}ller and Michael Clausen}, title = {Transposition-Invariant Self-Similarity Matrices}, booktitle = {Proceedings of the International Conference on Music Information Retrieval ({ISMIR})}, address = {Vienna, Austria}, year = {2007}, pages = {47--50}, 4. Meinard Müller, Nanzhu Jiang, and Peter Grosche A Robust Fitness Measure for Capturing Repetitions in Music Recordings With Applications to Audio Thumbnailing IEEE Transactions on Audio, Speech & Language Processing, 21(3): 531–543, 2013. author = {Meinard M{\"u}ller and Nanzhu Jiang and Peter Grosche}, title = {A Robust Fitness Measure for Capturing Repetitions in Music Recordings With Applications to Audio Thumbnailing}, journal = {IEEE Transactions on Audio, Speech {\&} Language Processing}, volume = {21}, number = {3}, year = {2013}, pages = {531-543}, 5. Meinard Müller Information Retrieval for Music and Motion Springer Verlag, ISBN: 3540740473, 2007. author = {Meinard M{\"u}ller}, title = {Information Retrieval for Music and Motion}, type = {Monograph}, year = {2007}, isbn = {3540740473}, publisher = {Springer Verlag} 6. Meinard Müller, Frank Kurth, and Michael Clausen Audio Matching via Chroma-Based Statistical Features In Proceedings of the 12th International Conference on Music Information Retrieval (ISMIR), 2011. author = {Meinard M{\"u}ller and Frank Kurth and Michael Clausen}, title = {Audio Matching via Chroma-Based Statistical Features}, booktitle = {Proceedings of the 12th International Conference on Music Information Retrieval ({ISMIR})}, year = {2011}, pages = {}, 7. Meinard Müller, Verena Konz, Wolfgang Bogler, and Vlora Arifi-Müller Saarland Music Data (SMD) In Proceedings of the International Society for Music Information Retrieval Conference (ISMIR): Late Breaking session, 2011. author = {Meinard M{\"u}ller and Verena Konz and Wolfgang Bogler and Vlora Arifi-M{\"u}ller}, title = {Saarland Music Data ({SMD})}, booktitle = {Proceedings of the International Society for Music Information Retrieval Conference ({ISMIR}): Late Breaking session}, year = {2011},	{"url":"https://www.audiolabs-erlangen.de/resources/MIR/SMtoolbox","timestamp":"2024-11-06T14:55:27Z","content_type":"text/html","content_length":"27311","record_id":"<urn:uuid:91c75541-8dd9-4bd7-a839-e78bcbd43e0c>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00358.warc.gz"}
Nanro ("Number Road") is a logic puzzle published by Nikoli. The task consists of a rectangular or square grid divided into regions. The goal is to fill in some cells with numbers. All numbers in a region must be the same. The given number in a region denotes how many cells in this region contain a number (all regions must have at least one number). When two numbers are orthogonally adjacent across a region boundary, the numbers must be different. Numbered cells must not cover an area of size 2 x 2 or larger. All cells with numbers must be interconnected. Cross+A can solve puzzles from 3 x 3 to 30 x 30.	{"url":"https://cross-plus-a.com/html/cros7nnr.htm","timestamp":"2024-11-07T01:34:09Z","content_type":"text/html","content_length":"1539","record_id":"<urn:uuid:c6df0c6f-a1da-4e43-bd9d-a29016586bc3>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00392.warc.gz"}
Fold Change Calculator \| Online Calculators Fold Change Calculator Do you want to compare two values to see how much one has changed relative to the other? The following Fold Change Calculator can helps you do just that. Enter the Initial Value and Final Value into the calculator to find the required change. What is Fold Change? Fold change is an easy way to see how much something has gone up or down compared to before. It shows how many times bigger or smaller the new amount is than the original one. For example, if the new number doubles the old one, the fold change would be 2. And if the new number is half the size of the old one, the fold change would be 0.5. Knowing the fold change tells the whole story simply. How to Use Calculator? 1. Basic Calculator: Enter any two values to find the missing one. For example, if you know the Initial Value and Final Value, the calculator will give you the Fold Change. Example: If the initial value is 10 and the final value is 30, the fold change will be 3. 2. Advanced Calculator: Enter the Initial Value and Final Value to calculate the Log2 Fold Change. Example: If the initial value is 5 and the final value is 40, the calculator will determine the Log2 Fold Change for you. The formula for calculating Fold Change is simple: $\text{Fold Change (FC)}=\frac{\text{Final Value}}{\text{Initial Value}}$ Variable Description Initial Value The starting value Final Value The ending value Fold Change The ratio of final to initial value How To calculate Fold Change Example 1: Let’s say the initial value is 20, and the final value is 60. Plug in the values: The Fold Change is 3, meaning the final value is 3 times the initial value. Example 2: Consider an initial value of 50 and a final value of 25. Plug in the values: The Fold Change is 0.5, meaning the final value is half the initial value. Input Guide 1. Initial Value: Enter the starting value of your measurement. 2. Final Value: Enter the ending value after a change has occurred. 3. Fold Change: The calculator will compute this automatically if you provide the other two values. How do I calculate fold change in R? To calculate fold change in R, you can use the formula fold_change = new_value / old_value or apply log transformation with log2FoldChange = log2(new_value / old_value). What does a fold change greater than 1 mean? A fold change greater than 1 indicates an increase. For example, a fold change of 2 means the new value is twice the original value. How do you calculate fold change in qPCR? In qPCR, fold change is calculated using the ΔΔCt method, which compares the target gene expression to a reference gene, and then calculates the fold difference between conditions. What does a fold change of 4 mean? A fold change of 4 means that the new value is four times larger than the original value. It signifies a significant increase. How do you handle fold change with zero values? Handling zero values in fold change calculations typically involves adding a small constant to avoid division by zero, or using log2 transformations for better comparison. Final Words I hope you found our Fold Change Calculator is a quick and easy way to understand how much a value has changed. Please let us know if you are facing any error using our tool. Leave a Comment	{"url":"https://lengthcalculators.com/fold-change-calculator/","timestamp":"2024-11-06T18:14:05Z","content_type":"text/html","content_length":"68054","record_id":"<urn:uuid:97348c1d-46be-4326-8044-4347ab6d8abd>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00507.warc.gz"}
Geometry Worksheets - 15 Worksheets.com Geometry Worksheets Related Worksheets About These Worksheets Geometry worksheets are essential educational resources designed to help students understand, practice, and master various geometric concepts. They provide structured opportunities for learners to engage with geometric principles through a variety of problems and exercises. These worksheets can be tailored to different educational levels, from elementary to high school and beyond, ensuring that the content matches the students’ proficiency and curriculum standards. What is Geometry? Geometry is a branch of mathematics that deals with the properties, relationships, and measurements of points, lines, angles, surfaces, and solids. It is one of the oldest fields of study in mathematics, with origins dating back to ancient civilizations such as the Egyptians, Babylonians, and Greeks. The term “geometry” comes from the Greek words “geo,” meaning earth, and “metron,” meaning measure, reflecting its historical role in surveying land and understanding spatial relationships. Geometry is profoundly embedded in various facets of the modern world, playing a critical role in fields like architecture, engineering, and computer science. In architecture, geometric principles guide the design and construction of buildings and structures, ensuring aesthetic appeal, stability, and functionality. Architects use geometry to create detailed blueprints and plans that translate abstract concepts into tangible forms, considering angles, symmetry, and spatial relationships to maximize space efficiency and structural integrity. Similarly, in engineering, geometry is essential for designing and analyzing components and systems. Engineers rely on geometric calculations to determine the dimensions, shapes, and tolerances necessary for machinery, vehicles, and infrastructure, ensuring that each part fits and functions correctly within the larger system. Beyond physical structures, geometry is indispensable in the digital realm, particularly in computer graphics, robotics, and geographic information systems (GIS). In computer graphics, geometric algorithms generate and manipulate visual images, enabling the creation of realistic 3D models, animations, and virtual environments used in video games, simulations, and films. Robotics employs geometric principles to navigate and interact with physical spaces, allowing robots to perform tasks with precision and efficiency. In GIS, geometry helps in mapping and analyzing spatial data, facilitating urban planning, environmental monitoring, and navigation systems. These applications highlight how geometry underpins technological advancements and contributes to solving complex problems in our increasingly interconnected world. Types of Problems Geometry worksheets encompass a wide range of topics, each addressing different geometric principles. These can be broadly categorized to help students systematically learn and apply various aspects of geometry. Basic Shapes and Properties – One fundamental type of problem in geometry worksheets involves identifying and naming basic geometric shapes. Students might be asked to recognize and label circles, squares, triangles, rectangles, and polygons. This foundational knowledge sets the stage for more complex geometric tasks. Additionally, exercises focused on the properties of these shapes delve into their defining characteristics, such as the number of sides, vertices, and angles. Understanding these properties is crucial for students to progress in their geometric studies. Angles – Worksheets addressing angles help students become familiar with the different types of angles and their properties. Problems might require students to identify and classify angles as acute, obtuse, right, or straight, sharpening their ability to distinguish between these fundamental geometric elements. Measuring angles with a protractor is another common exercise, teaching practical skills for precision. Furthermore, problems involving angle relationships-such as complementary, supplementary, and vertical angles-along with angles formed by parallel lines cut by a transversal, help students understand how angles interact in various configurations. Perimeter, Area, and Volume – Calculating perimeter, area, and volume is a key aspect of geometry that has practical applications in numerous fields. Perimeter exercises involve finding the total length around various shapes, both regular and irregular. Area problems require students to compute the space within basic shapes like squares, rectangles, triangles, and circles, as well as more complex figures such as trapezoids and parallelograms. For three-dimensional shapes, worksheets might include calculating surface area and volume, helping students understand the extent and capacity of objects like cubes, cylinders, spheres, and prisms. Coordinate Geometry – Coordinate geometry introduces students to the Cartesian plane and the concept of plotting points. Simple problems might involve placing points on a coordinate grid, while more advanced exercises include graphing lines and shapes based on given equations or coordinates. Distance and midpoint problems help students develop skills in calculating the distance between two points and finding the center point of a line segment. These tasks are foundational for higher-level mathematics and practical applications in various fields, including engineering and computer Transformations – Transformations are a significant area in geometry that involves moving shapes within the coordinate plane. Worksheets might include problems on translations, which shift shapes without altering their size or orientation. Rotations involve turning shapes around a fixed point, requiring students to understand angles and rotational symmetry. Reflections focus on flipping shapes over a line, while dilations involve resizing shapes proportionally. These exercises help students understand how shapes can change position and size while maintaining their essential Congruence and Similarity – Problems related to congruence and similarity help students understand the relationships between shapes. Identifying congruent shapes involves recognizing figures that are identical in shape and size. Proving congruence through postulates and theorems (such as SSS, SAS, ASA, AAS) deepens students’ understanding of geometric proofs. Similarity problems require students to identify shapes that are the same in form but different in size, often using proportions to solve for missing lengths. These concepts are critical for higher-level geometry and real-world Triangles – Triangles are a fundamental shape in geometry, and worksheets often include problems on classifying triangles based on their sides (equilateral, isosceles, scalene) and angles (acute, obtuse, right). The triangle inequality theorem, which determines if given lengths can form a triangle, is another key concept. Problems applying the Pythagorean theorem help students find missing sides in right triangles, a critical skill for various applications. Special right triangles, such as 30-60-90 and 45-45-90 triangles, are also a focus, with exercises designed to reinforce these specific geometric properties. Circles – Understanding circles is essential in geometry, and worksheets often include problems identifying parts of a circle such as the radius, diameter, chord, arc, and sector. Calculating the circumference and area of circles is a common exercise, teaching students practical measurement skills. Problems involving arc lengths and sector areas extend this understanding, requiring more complex calculations and a deeper comprehension of circular geometry. Geometric Constructions – Geometric constructions involve creating precise figures using a compass and straightedge, a skill that is both foundational and practical. Worksheets guide students through constructing geometric figures, such as bisectors and perpendicular lines, fostering a hands-on understanding of geometric principles. These basic constructions are essential for developing spatial reasoning and geometric intuition. Logical Reasoning and Proofs – Logical reasoning and proofs are at the heart of geometry, developing students’ critical thinking and problem-solving skills. Basic proofs introduce students to the structure and format of geometric arguments, including two-column proofs that logically deduce properties from given information. Problems that involve applying geometric theorems and postulates to prove statements about shapes and their properties deepen students’ understanding and ability to reason deductively, which is crucial for advanced mathematical thinking. Benefits of Geometry Worksheets Reinforce Learning By providing additional practice outside of classroom lessons, geometry worksheets help students reinforce and solidify their understanding of geometric concepts. These worksheets offer a variety of problems that allow students to apply what they have learned in different contexts, thus deepening their comprehension. Regular practice with these problems ensures that students retain the information and can recall it when needed, reducing the risk of forgetting key concepts over time. Additionally, worksheets can bridge the gap between classroom learning and homework, making it easier for students to stay on track with their studies. Assess Understanding Teachers can use geometry worksheets to assess how well students have grasped specific topics. These assessments can take the form of quizzes, homework assignments, or in-class activities, providing teachers with valuable insights into their students’ strengths and weaknesses. By analyzing worksheet results, teachers can identify which concepts need further clarification and tailor their instruction accordingly. This targeted approach helps ensure that all students achieve a solid understanding of geometry, enabling them to progress confidently through more advanced topics. Develop Problem-Solving Skills Regular practice with a variety of problems helps students develop critical thinking and problem-solving skills. Geometry worksheets often present challenges that require students to think logically and creatively to find solutions. By working through these problems, students learn to approach complex tasks systematically, breaking them down into manageable steps. This skill set is not only essential for success in geometry but also valuable in other areas of mathematics and in everyday life, where problem-solving abilities are crucial. Introduce New Concepts Worksheets can introduce new geometric concepts in a gradual and structured manner. By presenting new material incrementally, worksheets help prevent students from feeling overwhelmed by too much information at once. This step-by-step approach allows students to build on their existing knowledge, making it easier to understand and retain new concepts. Additionally, worksheets often include visual aids and guided practice problems that provide clear examples and reinforce the learning process, facilitating a smoother transition to more complex topics. Enhance Engagement Interactive and visually appealing worksheets can increase student engagement and interest in geometry. Worksheets that incorporate colorful diagrams, real-world applications, and hands-on activities make learning more enjoyable and relatable for students. Engaging content helps maintain students’ attention and motivation, encouraging them to invest more effort in their studies. Furthermore, interactive elements such as puzzles, games, and collaborative exercises can make learning geometry feel less like a chore and more like an exciting challenge, fostering a positive attitude toward the subject.	{"url":"https://15worksheets.com/worksheet-category/geometry/","timestamp":"2024-11-14T15:39:53Z","content_type":"text/html","content_length":"141596","record_id":"<urn:uuid:5abaef25-5b29-44ea-a146-d62e962512c9>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00746.warc.gz"}
Common Core 6th Grade Common Core: 6.EE.5 Common Core Identifier: 6.EE.5 / Grade: 6 Curriculum: Expressions And Equations: Reason About And Solve One-Variable Equations And Inequalities. Detail: Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. 34 Common Core State Standards (CCSS) aligned worksheets found: A set of single-variable, basic inequalities for students to graph on a number line. All the inequalities feature multiplication or division on one side of the inequality. Logged in members can use the Super Teacher Worksheets filing cabinet to save their favorite worksheets. Quickly access your most used files AND your custom generated worksheets! Please login to your account or become a member and join our community today to utilize this helpful feature. Find the value of the variable in each equation. Intermediate, one-step inequalities are graphed on a number line. This worksheet includes only addition or subtraction on the same side of the inequality as the variable. Negative numbers, decimals, and fractions are included On this page, students will solve the 2-step inequalities and then circle the possible values that satisfy each inequality. Rewrite each phrase as an inequality in standard algebraic form. (example: The product of 6 and x is greater than or equal to 48.) Solve the inequality and circle the numbers that are in the solution set. This version includes only whole numbers. Write an inequality to represent each of the situations described on the page. They are all single-variable inequalities. This page features two solve and graph problems, as well as a word problem. With the word problem, students must write and solve an inequality based on the situation described. They will graph the inequality too. Students must find and graph solutions to single variable inequalities. Uses negative numbers, fractions, decimals, and operators. Find the value of the variable in each algebraic equation. Each problem includes the number 24 somewhere in it. With this worksheet, students will practice a variety of skills involving two-step inequalities. They'll solve and graph, circle values that satisfy the inequality, and complete a word problem. Rewrite each sentence as an inequality in standard mathematical form. (example: x is at least 20.) Determine which numbers are in the solution set for each inequality. Circle the correct choices. This version includes only positive numbers. This printable file has an explanation and example of how to solve and graph an inequality with negative numbers, fractions, or decimals. Students can refer back to the example when solving the eight problems on their own. Like the Introduction worksheet, students graph inequalities that use negative numbers, decimals, fractions, and operators, but with less examples. A page of single variable, intermediate-level inequalities with a multiplication or division fact on one side of the inequality. The multiplication and division include decimals, fractions, and negative numbers. Why did the scarecrow win the Nobel Prize? Solve basic algebraic equations to find out. (example: 27 + x = 39) Solve each inequality and select the answers that are in the solution set. This version includes fractions and decimals. With this worksheet, students will solve the two-step inequalities and then graph them on the number line to the right. Solve and graph these 2-step inequalities. As intermediate level problems, they include negative numbers, decimals, and fractions. Students solve and graph basic, single variable inequalities. All problems have only positive, whole numbers. Less examples than the Introduction worksheet. Intermediate, single-variable inequalities are graphed on a number line. This worksheet includes only addition or subtraction on one side of the inequality. Negative numbers, decimals, and fractions are included. Another worksheet in which students are required to find the value of the variables. A page of one-step intermediate-level inequalities with a multiplication or division fact on the same side of the inequality as the variable. The multiplication and division include decimals, fractions, and negative numbers. Practice solving two-step inequalities with this printout. It includes an example and 10 problems. Rewrite each inequality in standard form. (example: The sum of -7.8 and x is less than 42.) On this worksheet, students will find and graph solutions to basic, single variable inequalities. All problems have only positive, whole numbers. Determine the value of the variable in each equation. The equations all include the number 2,024. On this worksheet, students will solve and graph two inequalities with negative numbers or decimals. They will also write and solve an inequality based on a word problem and then graph the solution. Rewrite each statement as an inequality with a variable. (example: z is not more than 3.14.) Decide which answers fit into the solution set for each number sentence. At the top of the page is an example and explanation of how to solve two-step inequalities and graph them. Reference this while solving and graphic the other eight problems. Students graph the basic, single-variable inequalities. All inequalities on this worksheet have addition or subtraction on one side of the inequality. A set of one-step, basic inequalities for students to graph on a number line. All the inequalities feature multiplication or division on the same side of the inequality as the variable.	{"url":"https://www.superteacherworksheets.com/common-core/6.ee.5.html","timestamp":"2024-11-12T03:31:15Z","content_type":"text/html","content_length":"138929","record_id":"<urn:uuid:fd9645c8-c154-4828-afaf-bfd91edd3b1b>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00495.warc.gz"}
Optics For Dummies Cheat Sheet Optics covers the study of light. Three phenomena — reflection, refraction, and diffraction — help you predict where a ray or rays of light will go. Study up on other important optics topics, too, including interference, polarization, and fiber optics. Reflection and refraction equations for predicting light's direction Reflection and refraction are two processes that change the direction light travels. Using the equations for calculating reflection and refraction, you can predict where rays encountering a surface will go — whether they reflect or refract (bounce off the surface or bend through it) — which is an important concept in the study of optics. The following equations help you determine reflection and refraction angles: • The law of reflection: The law of reflection shows the relationship between the incident angle and the reflected angle for a ray of light incident on a surface. The angles are measured relative to the surface normal (a line that is perpendicular to the surface), not relative to the surface itself. Here’s the formula: • The index of refraction: This quantity describes the effect of atoms and molecules on the light as it travels through a transparent material. Use this basic formula for the index of refraction: • Snell’s law or the law of refraction: Snell‘s law shows the relationship between the incident angle and the transmitted angle (refracted angle) for a ray of light incident on a surface of a transparent material. You can see how Snell’s law works in the following formula: • The critical angle for total internal reflection: Total internal reflection is the situation where light hits and reflects off the surface of a transparent material without transmitting through the surface. It utilizes the critical angle (the minimum angle of incidence where total internal reflection takes place.). For total internal reflection to occur, the light must start in the material with the higher index. Here’s the formula: Equations for optical imaging Imaging is a key function of optics. Specific optics equations can help you determine the basic characteristics of an image and predict where it will form. Use the following optics equations for your imaging needs: • Lateral magnification: Lateral magnification is one way you can describe how big the image is compared to the original object. Here are the equations: • Locating images formed by mirrors: An object placed a certain distance away from a mirror will produce an image at a certain distance from the mirror. In some cases where the mirrors are curved, you may be given the focal length of a mirror. Use these equations: • Location of images formed by a refracting surface: An object placed a certain distance away from a refracting surface will produce an image at a certain distance from the surface. The equation for this is • The lens maker’s formula: This equation allows you to calculate the focal length of a lens if all you know is the curvature of the two surfaces. Here’s the lens maker’s formula: • The thin lens equation: An object placed a certain distance away from a lens will produce an image at a certain distance from the lens, and the thin lens equation relates the image location to the object distance and focal length. The following is the thin lens equation: Optical polarization equations Optical polarization is the orientation of the planes of oscillation of the electric field vectors for many light waves. Optical polarization is often a major consideration in the construction of many optical systems, so equations for working with polarization come in handy. The following equations highlight some important polarization concepts. The equations listed here allow you to calculate how to make polarized light by reflection and to determine how much light passes through multiple polarizers: • Polarizing angle or Brewster’s angle: This angle is the angle of incidence where the reflected light is linearly polarized. Here’s the equation: • Malus’ law: This equation allows you to calculate how much polarized light passes through a linear polarizer. The equation for Malus’ law is • Phase retardation in a birefringent material: A birefringent material has two indexes of refraction. When you send polarized light into a birefringent material, the two components travel through the material with different speeds. This discrepancy can result in a change in the polarization state or simply rotate the polarization state. Use this equation: Optical interference equations Optical interference is just the interaction of two or more light waves. Optical interference is useful in many applications, so you need to understand some basic equations related to this optical phenomenon. The following equations allow you to calculate various quantities related to optical interference in the two most common interference arrangements. • The location of the bright and dark fringes in Young’s two-slit interference arrangement: The following equations allow you to calculate the location of the bright fringes (where constructive interference occurs) and dark fringes (where destructive interference occurs): • The phase shift due to the film thickness in thin film interference: When light is incident straight onto a thin film (such as an oil slick on the surface of a pool of water), light rays reflecting from the top and the bottom of the film interfere (either constructively or destructively depending on the film thickness and the wavelength of the light). The following equations determine constructive or destructive interference depending on whether the phase shift produced by reflection needs to be shifted by half the wavelength (the first equation) or maintained (the second equation): Optical diffraction equations Diffraction is light’s response to having something mess with its path, so diffraction occurs only when something blocks part of the wavefront. Diffraction is the phenomenon where light bends around an obstacle (this bending is not due to refraction, because the material doesn’t change as refraction requires). The following equations cover the most common situations involving diffraction, including resolution. • Resolution: Resolution is the minimum angular separation between two objects such that you can tell that there are two distinct objects. Here’s the equation for determining resolution: The location of the dark fringes produced by diffraction through a single slit: Because a slit has a width larger than the wavelength, light rays from different parts of the slit interfere with each other, creating a fringe pattern. You can relatively easily locate the points where the light destructively interferes by using the following equation: • The location of the different diffraction orders from a diffraction grating: A diffraction grating has a very large number of slits spaced closely together, such that the light from each of these slits interferes with the light from the others. You can pretty easily identify where the light constructively interferes by using the following equation: Equations for the characteristics of fiber-optics fibers Besides imaging, fiber-optic networks are probably the largest application of optics. Fiber optics are very long, thin glass fibers that transfer information-bearing light from one place to another, but that may not be in direct sight of each other. You need to be aware of a few characteristics of the particular fiber you’re using so that you can ensure the information is accurately transmitted from one end of the fiber to the other. The following equations cover three of the basic parameters necessary for proper use of optical fibers. • The maximum acceptance angle for a fiber: This angle is the largest angle of incidence at which light can enter the end of the fiber and be totally internally reflected inside the fiber. Angles of incidence larger than this angle will transmit through the sides of the fiber and not make it to the other end. The equation for this angle is • The numerical aperture for a fiber: The numerical aperture is a measure of the light-gathering power of the fiber. It has a maximum value of 1 (all the light remains trapped inside the fiber) and a minimum value of 0 (only light incident at an angle of 0 degrees on the end of the fiber remains trapped in the fiber). Use this equation: • Intermodal dispersion in a fiber: This characteristic measures the difference in time that different fiber modes take to reach the end of the fiber. The larger this time difference, the shorter the fiber has to be so that the information on this light doesn’t turn into junk. Here’s the equation: About This Article This article is from the book: This article can be found in the category:	{"url":"https://www.dummies.com/article/academics-the-arts/science/physics/optics-for-dummies-cheat-sheet-208578/","timestamp":"2024-11-02T01:12:52Z","content_type":"text/html","content_length":"96557","record_id":"<urn:uuid:2971b0dd-053a-4f3d-ac7f-52df487699a6>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00260.warc.gz"}
Software & Systems Journal influence Higher Attestation Commission (VAK) - К1 quartile Russian Science Citation Index (RSCI) Next issue Publication date: 09 September 2024 The Editorial Board / Serov Valeriy Sergejevich Serov V.S. was born on 4th of November 1952 in Pavlovo, Gorkovskaya obl., USSR. Department of Mathematical Sciences, University of Oulu, Finland. Associate professor of Moscow State University, Department of History. Sc.Dr., 1997, Moscow State University, Moscow, Russia. Thesis title: Some problems of the spectral theory of the elliptic differential operators with singularities. PhD, 1979, Moscow State University, Moscow, Russia. Thesis title: Absolute convergence of spectral expansions and fractional degrees of elliptic operators. Diploma, Applied Mathematics, 1975, Moscow State University, Faculty of Computational Mathematics and Cybernetics, Moscow, Russia. 1977–2002 Serov V.S. have been teaching at Moscow Lomonosov State University. Among the lectures courses are calculus, complex analysis, generalized functions, Fourier transforms, spectral theory of self–adjoint operators. The list of practical courses: calculus, analysis II, complex analysis, functional analysis. 2002–2009 Serov V.S. have been teaching at the University of Oulu three new courses: "Fourier transform, distributions and applications to the Schrödinger operator", "Spectral theory for elliptic dierential operators" and "Introduction to partial dierential equations". Serov V.S. have supervised 12 PhD and 12 Diploma works. In 1998 he was appointed the award of the Goerge Soros Honorary Docent. In 2002 he was accepted as the member of the Finnish Inverse Problems Society. Since 2006 Serov V.S. is a Senior Researcher of Finnish Centre of Excellence in Inverse Problems granted by Academy of Finland for the period 2006–2011. Serov V.S. also has experience as a visiting professor in different universities in Greece, the USA, Germany, Spain, Finland, Sweden and Italy. Among research interests: 1) Inverse Problems, 2) Spectral Theory, 3) Nonlinear Equations. Since 1990 Serov V.S. participated in more than 80 International conferences, seminars and symposiums. His list of publications includes more than 110 items; among them, three textbooks. 1. On a spectrum of the Schrödinger operator with Kato potential (with A.G. Razborov). Diff. Uravn. 2000, vol. 36 (5), pp. 689– 693. 2. About convergence of Fourier series on the eigenfunctions of Schrödinger operator with Kato potential. Matem. Zametki [Mathematical Notes]. 2000, vol. 67(5), pp. 755–763 (in Russ.) 3. Criteria for existence and stability of soliton solutions of the cubic–quintic nonlinear Schrödinger equation (with H.W. Schurmann). Physical Review E. 2000, vol. 62 (2), pp. 2821–2826. 4. Recovering singularities from backscattering in two dimensions (with P. Ola and L. Paivarinta). Comm.PDE. 2001, vol. 26 (3–4), pp. 697–715. 5. New mapping properties for the resolvent of the Laplacian and recovery of singularities of a multidimensional scattering potential (with L. Paivarinta). Inverse Problems. 2001, vol. 17 (5), pp. 6. Some inverse problems for the Schrödinger operator with Kato potential (with A.G. Razborov and M.K. Sagyndykov). Ill–Posed and Inverse Problems. 2002, vol. 10 (4), pp. 395–411. 7. Solutions to the Helmholtz equation on the line describing guided waves in a nonlinear three–layer structure (with Yu.V. Shestopalov and H.W. Schurmann). Journ. of Physics A: Mathematical and General. 2002, vol. 35 (50), pp. 10789–10801. 8. Reflection and transmission of a plane TE–wave at a lossless nonlinear dielectric film (with Yu.V. Shestopalov and H.W. Schurmann). Physica D. 2001, vol. 158, pp. 197–215. 9. On the theory of TE polarized waves guided by a lossless nonlinear three–layer structure (with Yu.V. Shestopalov and H.W. Schurmann). Proc. Progress in Electromagnetic Research Symp. Osaka, Japan, 2001, p. 632. 10. An n–dimensional Borg–Levinson theorem for singular potentials (with L. Paivarinta). Adv. Appl. Math. 2002, vol. 29 (4), pp. 509–520. 11. Recovery of the singularities of a potential in two dimensional Schrödinger operator from fixed angle scattering data. Russian Math. Dokl. 2002, vol. 358 (2), pp. 160–162. 12. Some inverse problems for two dimensional Schrödinger operators with singular potential. Born approximation. Proc. int. conf. KROMSH'2001. 2002, Simpheropol's Univ. Press, pp. 24–30. 13. Waves in three–layer structures with Kerr–type nonlinearity and variable permittivity (with Yu.V. Shestopalov and H.W. Schurmann). Abstracts of int. conf. Math. Modeling of Wave Phenomena. Växjö, Sweden, 2002, pp. 22–23. 14. Integral equation approach to reflection and transmission of a plane TE–wave at a (linear/nonlinear) dielectric film with spatially varying permittivity (with H.W. Schurmann and E.D. Svetogorova). Journ. of Physics A: Mathematical and General. 2004, vol. 37 (9), pp. 3489–3500. 15. Waves in three–layer structures with Kerr–type nonlinearity and variable permittivity (with Yu.V. Shestopalov and H.W. Schurmann). Proc. of int. conf. Math. Modeling of Wave Phenomena. Växjö, Sweden, 2004, vol. 7, pp. 217–226. 16. Reconstruction of singularities in two dimensional Schrödinger operator with fixed energy (with A.D. Chernova). Inverse and Ill–posed Problems. 2004, vol. 12 (4), pp. 413–421. 17. Traveling wave solutions of a generalized modified Kadomtsev–Petviashvili equation (with H.W. Schurmann). Journ. of Math. Phys. 2004, vol. 45 (6), pp. 2181–2187. 18. New estimates of Green–Faddeev function and recovering of singularities in two dimensional Schrödinger operator with fixed energy (with L. Paivarinta). Inverse Problems. 2005, vol. 21 (4), pp.1291–1301. 19. Integral equation approach to reflection and transmission of a plane TE–wave at a (linear, nonlinear, absorbing) dielectric film with spatially varying permittivity (with H.W. Schurmann and E.D. Svetogorova). PIERS 2004: electromagnetic research symposium. Pisa, 2004, vol. 2, pp. 121–124. 20. Weierstrass' solutions to certain nonlinear wave and evolution equations (with H.W. Schurmann). PIERS 2004: electromagnetic research symposium. Pisa, 2004, vol. 2, pp. 651–654. 21. Some inverse scattering problems for two–dimensional Schrödinger operator. Proc. of the 5th int. conf. on Inverse Problems in Engineering: Theory and Practice. Edited by D. Lesnic, Leeds Univ. Press, 2005, vol. 3, Leeds, UK. 22. Problems on theory of functions of complex variable with solutions (with T.A. Leontyeva and V.S. Panferov). Moscow, Mir, 2005, 360 p. 23. Inverse scattering problem for two–dimensional Schrödinger operator (with L. Paivarinta). Journ. Inverse and Ill–posed Problems. 2006, vol. 14 (3), pp. 295–305. 24. Some elliptic traveling wave solutions to the Novikov–Veselov equation (with H.W. Schurmann and J. Nickel). PIER. 2006, vol. 61, pp. 323–331. 25. Superposition in nonlinear wave and evolution equation (with H.W. Schurmann and J. Nickel). Int. Journ.of Theoretical Physics. 2006, vol. 45 (6), pp. 1057–1073. 26. Some elliptic traveling wave solutions to the Novikov–Veselov equation (with H.W. Schurmann and J. Nickel). Proc. PIERS. 2006, Cambridge/MA, USA, pp. 519–523. 27. Recovery of jumps and singularities of an unknown potential from limited data in dimension 1 (with M. Harju). Journ. of Physics A: Mathematical and General. 2006, vol. 39, pp. 4207–4217. 28. Fundamental solution and Fourier series in eigenfunctions of degenerate elliptic operator. Journ.of Mathematical Analysis and Applications. 2007, vol. 329 (1), pp. 132–144. 29. Reconstruction of discontinuities in one–dimensional nonlinear Schrödinger operator from limited data (with M. Harju). Inverse Problems. 2007, vol. 23 (2), pp. 493–506. 30. Recovery of jumps and singularities in the multidimensional Schrödinger operator from limited data (with L. Paivarinta). Inverse Problems and Imaging. 2007, vol. 1 (3), pp. 525–535. 31. Inverse Born approximation for the nonlinear two–dimensional Schrödinger operator. Inverse Problems. 2007, vol. 23 (3), pp. 1259–1270. 32. Partial recovery of the potentials in generalized nonlinear Schrödinger equation on the line (with M. Harju). Journ. of Mathematical Physics. 2007, vol. 48 (8), p.18. 33. On the theory of TM electromagnetic guided waves in a nonlinear three–layer structures (with H.W. Schurmann, Yu.V. Shestopalov and Yu.G. Smirnov). Proc. PIERS. Prague, Czech Republic, 2007, p. 34. A uniqueness theorem and reconstruction of singularities for a two–dimensional nonlinear Schrödinger equation (with M. Harju). Nonlinearity. 2008, vol. 21, pp. 1323–1337. 35. Inverse Born approximation for the generalized nonlinear Schrödinger operator in two dimensions. Modern Physics Letters B (MPLB). 2008, vol. 22 (23), pp. 2257–2275. 36. Inverse Born approximation for the generalized nonlinear Schrödinger operator in two dimensions. Journ. of Physics: Conference Series. 2008, vol. 35, 012092. 37. Inverse fixed angle scattering and backscattering problems in two dimensions. Inverse Problems. 2008, vol. 24, 065002. 38. TM electromagnetic guided waves in a (Kerr-) nonlinear three–layer structure (with H.W. Schurmann and K.A. Yuskaeva). PIERS Proc. Moscow, Russia, 2009, pp. 364–369. 39. Integral equation approach to TM electromagnetic waves guided by a (linear/nonlinear) dielectric film with a spatially varying permittivity (with H.W. Schurmann and K.A. Yuskaeva). PIERS Proc. Moscow, Russia, 2009, pp. 1915–1919. 40. The fundamental solution and Fourier series in eigenfunctions of the magnetic Schrödinger operator. Journ. of Physics A: Mathematical and Theoretical. 2009, vol. 42, no. 22, 225205. 41. An inverse Born approximation for the general nonlinear Schrödinger operator on the line. Journ. of Physics A: Mathematical and Theoretical. 2009, vol. 42, no. 33, 332002. 42. A domain description and Green's function for elliptic differential operator with singular potential (with U. Kyllonen). Journ. of Mathematical Analysis and Applications. 2010, vol. 366, iss. 1, pp. 11–23. 43. Application of the Banach fixed–point theorem to the scattering problem at a nonlinear three–layer structure with absorption (with H.W. Schurmann and E.D. Svetogorova). Fixed Point Theory and Applications. 2010. Available at: http://www.fixedpointtheoryandapplications.com/content/2010/1/439682 44. Green's function and convergence of Fourier series for elliptic differential operators with potential from Kato space. Abstract and Applied Analysis. Submitted, 2010. Available at: http:// 45. Inverse backscattering for the nonlinear Schrödinger operator in two dimensions (with J. Sandhu). Journ. Phys. A: Math. Theor. 2010, vol. 43, no. 32, 325206. 46. Inverse fixed energy scattering problem for the generalized nonlinear Schrödinger operator. Inverse Probl. 2012, vol. 28, no. 2. 47. Transmission eigenvalues for degenerate and singular cases. (with J. Sylvester). Inverse Probl. 2012, vol. 28, no. 6. 48. Inverse backscattering Born approximation for a two-dimensional magnetic Schrödinger operator. Inverse Probl. 2013, vol. 29, no. 7. 49. Inverse fixed angle scattering and backscattering for a nonlinear Schrödinger equation in 2D (with Fotopoulos G. and Harju M.). Inverse Probl. Imaging. 2013, vol. 7, no. 1, 183-197. 50. Transmission eigenvalues for non-regular cases. Commun. Math. Anal. 2013, vol. 14, no. 2, 129–142, (electronic only). 51. Inverse backscattering Born approximation for the magnetic Schrödinger equation in three dimensions. (with J.Sandhu) Math. Methods Appl. Sci. 2014, vol. 37, no. 2, pp. 265-269.	{"url":"https://swsys.ru/index.php?page=42&lang=en","timestamp":"2024-11-14T22:17:32Z","content_type":"application/xhtml+xml","content_length":"31335","record_id":"<urn:uuid:0ed18d6e-0197-40b8-ae39-50bafbaf219c>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00788.warc.gz"}
03/01/1995 09:00 AM Computer Engineering This paper describes a novel approach to find a tighter bound of the transformation of the Min-Max problems into the one of Least-Square Estimation. It is well known that the above transformation of one problem to the other can lead to the proof that their target functions linearly bound each other. However, this linear bound is not a tight one. In this paper, we prove that if we transform the Min- Max problem into two Least-Square Estimation problems, where one minimizes the Root-Mean-Square (RMS) of the original function and the other one minimizes the RMS of the difference between the original function and an arbitrary constant, one can obtain a tighter bound between their target functions. The tighter bound given by this novel approach depends on the outcome of the second Least-Square Estimation problem, so there is a great incentive to choose the arbitrary constant which gives the smallest RMS of the second Least-Square Estimation problem. For a problem with a large number of variables, this novel tighter bound can be two to three order tighter than the old one.	{"url":"https://tr.soe.ucsc.edu/research/technical-reports/UCSC-CRL-95-03","timestamp":"2024-11-13T11:47:52Z","content_type":"application/xhtml+xml","content_length":"13159","record_id":"<urn:uuid:2923d5e2-9865-4187-9df6-4ea247b2ec74>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00505.warc.gz"}
Conductor Fill Calculator - Calculator Doc Conductor Fill Calculator In electrical installations, understanding how much space conductors occupy within a conduit is essential for ensuring safety and compliance with standards. The Conductor Fill Calculator simplifies this task by allowing you to calculate the percentage of space occupied by conductors relative to the total available area. This helps in planning and designing efficient and safe electrical systems. To calculate the conductor fill percentage, use the formula: CF = (A * N) / T, where CF represents the conductor fill percentage, A is the area occupied by each conductor, N is the number of conductors, and T is the total area of the conduit. How to Use 1. Enter the Area (A): Input the area occupied by each conductor into the corresponding field. 2. Input the Number of Conductors (N): Enter the total number of conductors. 3. Specify the Total Area (T): Provide the total area of the conduit or space. 4. Calculate: Click the “Calculate” button to determine the conductor fill percentage. 5. View Result: The result will appear in the result field, showing the percentage of space occupied by the conductors. Suppose you have an area occupied by each conductor of 5 square units, with 10 conductors, and the total conduit area is 100 square units. To find the conductor fill percentage, enter these values into the calculator. After clicking “Calculate,” the result will be 5.00, indicating that the conductors occupy 5% of the total conduit area. 1. What is conductor fill? Conductor fill refers to the percentage of space within a conduit that is occupied by conductors. It is important for ensuring proper spacing and safety. 2. Why is conductor fill calculation important? It helps in adhering to safety standards and regulations, preventing overheating, and ensuring efficient operation of electrical systems. 3. Can the calculator handle decimal values? Yes, the calculator supports decimal values for precise results. 4. What happens if the total area (T) is zero? The calculator will display an error message since division by zero is not possible. Ensure the total area is a non-zero value. 5. Are there maximum input limits for area and number of conductors? There are no specific maximum limits, but extremely large values might affect the accuracy or performance. 6. How often should conductor fill be checked? It should be checked whenever designing or modifying conduit systems to ensure compliance with safety standards. 7. Can this calculator be used for different types of conduits? Yes, it is versatile and can be applied to various types of conduits. 8. What does the result represent? The result shows the percentage of conduit space occupied by conductors, which is crucial for proper installation. 9. Can I use this calculator for multiple conduits? You would need to adjust the input fields for each conduit separately. 10. How can I improve conductor spacing based on the result? If the fill percentage is too high, consider increasing conduit size or reducing the number of conductors to comply with safety standards. 11. Does the calculator support historical data comparison? No, it calculates the fill percentage for the current input values only. Historical data needs to be managed separately. 12. Can I print the results? Yes, you can manually record the results or use your browser’s print functionality. 13. Is this calculator free to use? Yes, the calculator is available for free use. 14. How accurate is the calculator? It provides results up to two decimal places, based on the precision of input values. 15. Can this calculator be integrated into a website? Yes, the HTML and JavaScript code can be embedded into your website for easy access. The Conductor Fill Calculator is a practical tool for accurately measuring the space occupied by conductors within a conduit. Its straightforward design ensures ease of use and responsiveness across devices, making it an essential resource for electrical engineers and contractors. By providing precise calculations, this tool helps maintain safety standards and optimize conduit usage.	{"url":"https://calculatordoc.com/conductor-fill-calculator/","timestamp":"2024-11-02T10:41:08Z","content_type":"text/html","content_length":"86358","record_id":"<urn:uuid:5c2f4f3d-1bc7-4ad2-9fdc-741f89d2d6a6>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00103.warc.gz"}
Understanding Mathematical Functions: Is This A Function Or Not Introduction: Understanding the Basics of Mathematical Functions Mathematical functions are a fundamental concept in mathematics, with diverse applications in various fields such as science, engineering, and economics. In this chapter, we will delve into the essence of mathematical functions, the significance of distinguishing between functions and non-functions, and the criteria for identifying a function. A. Define what a mathematical function is At its core, a mathematical function is a relation between a set of inputs (called the domain) and a set of outputs (called the codomain) with the property that each input is related to exactly one output. In simpler terms, a function assigns a unique output value to each input value. For instance, consider the function f(x) = 2x, where for every input value x, there is a unique output value 2x. This concept can be extended to more complex functions involving multiple variables and operations. B. Explain the importance of distinguishing between functions and non-functions The ability to distinguish between functions and non-functions is crucial in various mathematical and real-world contexts. In mathematics, functions serve as the basis for calculus, algebra, and other advanced topics. Furthermore, in fields such as computer science and data analysis, functions are used to model relationships and make predictions. Distinguishing a function from a non-function helps in accurately representing and analyzing these relationships. Furthermore, in real-world scenarios, such as financial modeling, physics equations, and computer programming, the correct identification of functions is essential for accurate predictions and C. Outline the criteria for identifying a function To determine whether a relation is a function, certain criteria must be fulfilled. The fundamental criterion is the requirement of each input having exactly one output. This can be assessed through methods such as the vertical line test, where a vertical line is drawn through the graph of the relation, and if it intersects the graph at more than one point, the relation is not a function. Additionally, another criterion is the absence of ambiguity, meaning that each input must lead to a unique output without any uncertainty or multiple possible values. • Each input has exactly one output • Absence of ambiguity in the output for each input • Adherence to the vertical line test By adhering to these criteria, one can accurately identify whether a given relation qualifies as a mathematical function. Key Takeaways • Functions have only one output for each input. • Check for repeating inputs with different outputs. • Graph the relationship to see if it passes the vertical line test. • Use algebraic methods to determine if it's a function. • Understand the concept of domain and range. The Concept of Mapping in Functions When it comes to understanding mathematical functions, the concept of mapping is essential. Mapping refers to the process of associating each element of a set of inputs with exactly one element of a set of outputs. This association forms the basis of functions in mathematics. A. Describe the idea of mapping from a set of inputs to a set of outputs In the context of functions, mapping involves taking an input value, applying a specific rule or operation to it, and obtaining an output value. This process allows us to establish a relationship between the input and output values, which is fundamental to understanding functions. B. Discuss the concept of domain and range In the context of mapping, the domain of a function refers to the set of all possible input values that can be used with the function. On the other hand, the range of a function represents the set of all possible output values that the function can produce. Understanding the domain and range of a function is crucial in determining its behavior and characteristics. C. Use examples to illustrate one-to-one and many-to-one mappings One-to-one mapping occurs when each element in the domain is associated with exactly one element in the range, and no two different elements in the domain are associated with the same element in the range. On the other hand, many-to-one mapping occurs when multiple elements in the domain are associated with the same element in the range. • One-to-one mapping example: Consider the function f(x) = 2x. For every input value of x, there is a unique output value of 2x. No two different input values produce the same output value, making it a one-to-one mapping. • Many-to-one mapping example: The function g(x) = x^2 represents a many-to-one mapping, as different input values can produce the same output value. For instance, g(2) = 4 and g(-2) = 4, demonstrating that multiple input values can result in the same output value. The Vertical Line Test When it comes to understanding mathematical functions, one important tool for identifying functions graphically is the vertical line test. This test provides a simple and visual way to determine whether a given graph represents a function or not. Introduce the vertical line test as a tool for identifying functions graphically The vertical line test is a method used to determine if a graph represents a function. It involves visually inspecting the graph and checking whether any vertical line intersects the graph at more than one point. If a vertical line intersects the graph at only one point for every possible x-value, then the graph represents a function. If the vertical line intersects the graph at more than one point for any x-value, then the graph does not represent a function. Show how to apply the vertical line test with illustrations Let's consider the graph of a simple linear function, y = 2x + 3. When we plot this graph on a coordinate plane, we can see that for every x-value, there is only one corresponding y-value. If we were to draw a vertical line at any point on the graph, it would only intersect the graph at one point, confirming that this graph represents a function. On the other hand, if we consider the graph of a circle, we can see that a vertical line drawn through the circle will intersect the graph at two points for certain x-values. This means that the graph of a circle does not represent a function, as it fails the vertical line test. Explain the reasoning behind the vertical line test and its implications for different types of relations The reasoning behind the vertical line test lies in the definition of a function. A function is a relation in which each input (x-value) is associated with exactly one output (y-value). When we apply the vertical line test, we are essentially checking whether each x-value has a unique corresponding y-value on the graph. If the test fails, it indicates that the graph does not meet the criteria of a function. Understanding the implications of the vertical line test is crucial when dealing with different types of relations. For example, when working with real-world data or mathematical models, it is important to know whether a given graph represents a function in order to make accurate predictions and interpretations. Function Notation and Representation Understanding mathematical functions involves being able to interpret and work with different representations of functions. Function notation and representation are essential concepts in this regard, as they provide a way to express and understand the behavior of functions. A. Standard Function Notation Standard function notation, such as f(x), is used to represent a function. The letter f represents the name of the function, while x is the input variable. This notation indicates that the function f operates on the input x to produce an output. B. Different Ways Functions Can Be Represented Functions can be represented in various ways, including equations, graphs, and tables of values. • Equations: Functions can be represented using algebraic equations, such as y = 2x + 3. This equation shows the relationship between the input variable x and the output variable y. • Graphs: Graphical representation of functions provides a visual way to understand the behavior of a function. The graph of a function shows how the output varies with changes in the input. • Tables of Values: Functions can also be represented using tables that list input-output pairs. This tabular representation provides a systematic way to organize and analyze the function's C. Interpreting and Translating Among Representations It is important to be able to interpret and translate among different representations of functions. For example, given an equation of a function, one should be able to sketch its graph or create a table of values to understand its behavior. Similarly, given a graph or a table of values, one should be able to write an equation that represents the function. Translating among representations involves understanding how changes in one representation affect the others. For instance, shifting a graph horizontally or vertically corresponds to specific changes in the equation of the function. Being able to make these connections is crucial for a comprehensive understanding of functions. Common Misunderstandings and Pitfalls When it comes to understanding mathematical functions, there are several common misunderstandings and pitfalls that many students and even some professionals encounter. In this chapter, we will identify these misconceptions, point out common errors when determining if a relation is a function, and provide strategies to avoid these mistakes. A. Identify frequent misconceptions about functions One frequent misconception about functions is that they are always expressed as equations. While many functions can be represented by equations, it's important to understand that a function is a relation between a set of inputs and a set of possible outputs where each input is related to exactly one output. This means that functions can also be represented as tables, graphs, or even verbal Another common misunderstanding is the belief that all relations are functions. In reality, not all relations are functions. A relation is only a function if each input is related to exactly one output. If there is an input that is related to multiple outputs, then the relation is not a function. B. Point out common errors when determining if a relation is a function One common error when determining if a relation is a function is failing to check for multiple outputs for the same input. It's important to carefully examine each input and ensure that it is related to only one output. If there are multiple outputs for the same input, then the relation is not a function. Another common error is assuming that a graph represents a function without verifying that the vertical line test is satisfied. The vertical line test states that if a vertical line intersects the graph of a relation in more than one point, then the relation is not a function. Failing to apply this test can lead to the misidentification of a relation as a function. C. Provide strategies to avoid these mistakes To avoid the misconception that all functions are expressed as equations, it's important to expose students to various representations of functions, such as tables, graphs, and verbal descriptions. This can help them understand that functions can take different forms and are not limited to equations. To prevent the error of failing to check for multiple outputs for the same input, students should be encouraged to systematically analyze each input and its corresponding output. Emphasizing the importance of precision and thoroughness in determining if a relation is a function can help avoid this mistake. Finally, to avoid the error of assuming that a graph represents a function without applying the vertical line test, students should be taught to always verify the criteria for a relation to be a function. This includes checking for multiple outputs for the same input and applying the vertical line test when dealing with graphs. Real-world Examples and Applications Understanding mathematical functions is crucial in various real-world scenarios and applications. Whether it's in the field of economics, engineering, or data science, the ability to identify and work with functions is essential for problem-solving and decision-making. A Showcase practical scenarios where identifying functions is crucial In the field of finance, understanding functions is crucial for analyzing and predicting market trends. For example, stock prices can be modeled using mathematical functions to understand their behavior over time. Similarly, in the field of biology, functions are used to model population growth and decay, which is essential for understanding ecological systems. Discuss functions in various fields, such as economics, engineering, and data science In economics, functions are used to model relationships between variables such as supply and demand, production costs, and consumer behavior. Engineers use functions to design and analyze systems, such as electrical circuits, mechanical structures, and chemical processes. In data science, functions are used to analyze and interpret large datasets, making it possible to extract valuable insights and make data-driven decisions. Offer insights on how understanding functions can lead to better problem-solving skills Understanding functions not only allows us to model and analyze real-world phenomena but also enhances our problem-solving skills. By being able to identify and work with functions, individuals can approach complex problems with a structured and analytical mindset. This can lead to more effective problem-solving and decision-making in various fields, ultimately contributing to innovation and Conclusion & Best Practices for Function Identification A Recap the significance of recognizing functions in mathematical analysis Understanding mathematical functions is crucial in mathematical analysis as it helps in modeling real-world phenomena, making predictions, and solving problems. Recognizing functions allows us to understand the relationship between variables and make informed decisions based on data and patterns. Summarize the key points from the post • Definition of a Function: A function is a relation between a set of inputs and a set of possible outputs where each input is related to exactly one output. • Function Notation: Functions are often represented using function notation, such as f(x), where 'x' is the input and 'f(x)' is the output. • Vertical Line Test: The vertical line test is a method used to determine if a graph represents a function. If any vertical line intersects the graph at more than one point, the graph does not represent a function. • Best Practices for Function Identification: It is important to carefully analyze the given data or graph to determine if it represents a function. Critical thinking and verification are essential in accurately identifying functions. Offer best practices and tips for accurate function identification, with an emphasis on critical thinking and verification When identifying functions, it is important to follow best practices to ensure accuracy. Here are some tips for accurate function identification: • Understand the Definition: Familiarize yourself with the definition of a function and the criteria that must be met for a relation to be considered a function. • Use Function Notation: Representing functions using function notation can help in clearly defining the input-output relationship. • Apply the Vertical Line Test: When dealing with graphs, use the vertical line test to determine if the graph represents a function. • Verify the Relationship: Verify that each input is related to exactly one output. If there are multiple outputs for a single input, it is not a function. • Think Critically: Analyze the given data or graph critically, considering all possible scenarios and relationships between variables. • Seek Confirmation: If in doubt, seek confirmation from a peer, instructor, or reliable source to ensure accurate function identification.	{"url":"https://dashboardsexcel.com/blogs/blog/understanding-mathematical-functions-is-this-a-function-or-not","timestamp":"2024-11-15T00:16:39Z","content_type":"text/html","content_length":"226439","record_id":"<urn:uuid:0abf50d6-bc77-4e1b-85ac-7f3d97a961a2>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00555.warc.gz"}
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My portfolio on Coingecko is having lots of errors on profit/loss calculations. I have posted an example screenshot of a small trade I did today - showing a A Profit Calculator to calculate the profit or loss value in money and pips CoinBSVBitcoin SVBTCBitcoinBTCBBitcoin BEP2BUSDBinance podarokb2b.ru PaymentLens's Crypto Currency Coin Profit / Loss Calculator lets you calculate profits or loss for your crypto currency investments. Crypto Profit Calculator. Select a coin and see how much return you'll get. Bitcoin. Ethereum. XRP. Solana. Entry amount. Exit amount. USD0. Profit / Loss. Universal Corp Stock \| Website Servers	{"url":"https://podarokb2b.ru/learn/coin-profit-loss-calculator.php","timestamp":"2024-11-11T23:25:03Z","content_type":"text/html","content_length":"13014","record_id":"<urn:uuid:13eab870-b8d6-4246-9e51-b68a733fd984>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00884.warc.gz"}
Choosing the Correct Average – CSS Wizardry More and more frequently I’m finding myself presenting data to clients. Wether it’s through code reviews or performance audits, taking a statistical look at code is an important part of my job, and it needs presenting to my clients in a way that is representative and honest. This has led me more and more into looking at different averages, and knowing when to use the correct one for each scenario. Interestingly, I have found that the mean, the average most commonly referred to as simply the average, is usually the least useful. I want to step through the relative merits of each, using some real life examples. Disclaimers: Firstly, I’m by no means a statistician or data scientist; if anything in here isn’t quite correct, please let me know. Secondly, I deal with relatively small data sets, so we don’t need to worry about more complex measurements (e.g. standard deviation or interquartile ranges); sticking with simplified concepts is fine. Finally, all of the data in this article is fictional, please do not cite any figures in any other articles or papers. The mean is the most common average we’re likely to be aware of. It works well when trying to get a representative overview of a data set with a small range; it is much less representative if we have extremities in our data. If you’ve ever split a bill at a restaurant, you’ve used a mean average. We arrive at the mean by adding all of our data points together and then dividing by the number of data points there were, e.g.: Person Cost Harry £42 Stephanie £39 Amit £41 Anna £47 Laura £39 If we’re going to split this bill, then we’d work out the mean: (42 + 39 + 41 + 47 + 39) ÷ 5 = 41.6 I’m sure you’d all agree that £41.60 is a pretty fair price to pay for your meal if we’re splitting the bill. However, if Chad comes along and orders the wagyu steak at £125, we’re going to have a different outcome altogether: Person Cost Harry £42 Stephanie £39 Amit £41 Anna £47 Laura £39 Chad £125 (42 + 39 + 41 + 47 + 39 + 125) ÷ 6 = 55.5 Paying £55.50 each to subsidise Chad’s expensive taste isn’t quite so fair. I’ve mentioned issues with the mean before, in my previous post about Parker. Sometimes it’s nicer to know either a more representative single number, or to know that most values are x. Use the mean if you want a fair or representative take on a data set with a very small range. Honestly, I find that the mean is seldom useful in the work I do, so let’s leave it there. The median is great for working out a good representation of a data set that contains statistical outliers, or data with a large range. The median is simply the middlemost data point in a set that has been arranged in ascending order; finding the middle point helps us to trim off any statistical outliers and anomalies. Here’s an actual use case from just yesterday. I was doing a code review for a client and was concerned that their Gulp task took a long time to complete. I took a measurement of five runs and ascertained the median value: Run Duration (s) So you take your data points: 74, 68, 70, 138, 69 And pick the centre point: 68, 69, 70, 74, 138 Tip: Whenever you run tests like this one, run an odd number of them (e.g. 5) so that the median is easier to find. As you can see here, run 4 took an unusually long time to complete. This would skew the data if we were to take the mean: (74 + 68 + 70 + 138 + 69) ÷ 5 = 83.8 83.8 seconds isn’t a very fair reflection when most runs were in the high sixties/low seventies, so the mean is not appropriate here. It turns out that the median (at exactly 70) is a very good representation. This means that I will be telling the client that compilation took an average of 70 seconds to complete (median run of five runs). compilation took an average of 70 seconds to complete (median run of five runs) Another good use of the median is in representing page load times: load the same page five times from an empty cache, record your outcomes, find the median. This means that you won’t get skewed results from things like DNS latency that might be out of your control. Again, this happened to me just the other day, and the mean would have been an unfair and inappropriate representation of the page’s general performance. Take a look at the DNS delays on this waterfall: This page took over 20 seconds to load, which is not at all representative. Unusual DNS slowdowns were causing unexpected delays, so the median measurement makes allowances for this. Use the median if you want to get a good view of a data set that has a large range and/or statistical outliers. The mode is a little harder to explain, but it works best with grouped data sets. With means and medians, the data is usually whatever-it-ends-up-being. That is to say, if run 1 takes 62s then the data point is 62; if it takes 93s then it is 93. However, if we were to make our own silos of data points, we can begin looking at finding a mode: instead of representing each data point individually, we put in into a pre-defined bin, e.g. ≤60s, > 60s, ≤90s, etc. Now our data isn’t whatever-it-ends-up-being, it’s actually inside a category we’re already aware of. Let’s look at a better scenario. If a client wants to know how well their site performs in general, I could do something like this: There are a couple of issues here: firstly, I’m going to end up with a very long x axis if I have profiled each individual page, meaning I’m having to work over a lot of data; secondly, this still only tells me how each page performs, and doesn’t give me a very good overview of the site’s performance like we wanted. This has told me little of any value. What would be better would be to chunk the data points into bins like ‘under 3s’, ‘between 3s and 5s’, ‘over 10s’, etc., and then represent it as a histogram. A histogram plots the frequency of a data point, not the data point itself. This allows us to very easily identify the mode: Now I have a much better of how the site performs in general, noting that most pages on the site load between 3 and 5 seconds. Most pages aren’t terrible, but we do also have a few outliers that we should probably focus on first. Here I can get a good idea of the distribution in general, and see a good holistic snapshot of the current site. Use the mode if you want to get a feel for what most things are like. E.g. ‘most images on the product page are under 75KB’. If you’re going to be measuring, analysing, or auditing anything for work/clients, make sure you represent your findings as truly and usefully as possible. We usually find that the median is the most accurate representation for measuring things like speed or performance. Any data that is subject to statistical outliers should not be represented by the mean, as results are easily skewed. If we want to get a general overview of what most things are looking like, the mode might be the one to go for. Harry Roberts is an independent consultant web performance engineer. He helps companies of all shapes and sizes find and fix site speed issues. Hi there, I’m Harry Roberts. I am an award-winning Consultant Web Performance Engineer, designer, developer, writer, and speaker from the UK. I write, Tweet, speak, and share code about measuring and improving site-speed. You should hire me. I am available for hire to consult, advise, and develop with passionate product teams across the globe. I specialise in large, product-based projects where performance, scalability, and maintainability are paramount.	{"url":"https://csswizardry.com/2017/01/choosing-the-correct-average/?utm_source=pagination&utm_medium=internal","timestamp":"2024-11-14T21:47:30Z","content_type":"text/html","content_length":"69792","record_id":"<urn:uuid:70a6fef4-516f-4769-99a5-aa60e5da1650>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00522.warc.gz"}
Boson stars and their relatives in semiclassical gravity On-site: Room 11.2.21 Juan Barranco (University of Guanajuato) In this talk we will discuss boson star configurations in quantum field theory using the semiclassical gravity approximation. Restricting our attention to the static case, we show that the semiclassical Einstein- Klein-Gordon system for a single real quantum scalar field whose state describes the excitation of N identical particles, each one corresponding to a given energy level, can be reduced to the Einstein- Klein-Gordon system for N complex classical scalar fields. Particular consideration is given to the spherically symmetric static scenario, where energy levels are labeled by quantum numbers n, l and m. When all particles are accommodated in the ground state n = l = m = 0, one recovers the standard static boson star solutions. On the other hand, for the case where all particles have fixed radial and total angular momentum numbers n and l, but are homogeneously distributed with respect to their magnetic number m, one obtains the l-boson stars, whereas when l = m = 0 and n takes multiple values, the multi-state boson star solutions are obtained.Thus, we have shown that in semiclassical gravity, Boson Stars relatives arise naturally as the quantum fluctuations associated with the state of a single field describing a many-body system.	{"url":"http://gravitation.web.ua.pt/node/4879","timestamp":"2024-11-03T06:08:08Z","content_type":"text/html","content_length":"38358","record_id":"<urn:uuid:ebd6f2d4-b605-4a53-a3af-76e4b3eb2e5a>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00723.warc.gz"}
Convert Stones to Metric Tons Stones to Metric Tons Calculator Convert Stones to Metric Tons To calculate a value in Stones to the corresponding value in Metric Tons, multiply the quantity in Stones by 0.00635029318 (conversion factor). Metric Tons = Stones x 0.00635029318 How to convert from Stones to Metric Tons The conversion factor from Stones to Metric Tons is 0.00635029318. To find out how many Stones in Metric Tons, multiply by the conversion factor or use the Stones to Metric Tons converter above. Definition of Stone The stone or stone weight (abbreviation: st) is an English and imperial unit of mass now equal to 14 pounds (6.35029318 kg). England and other Germanic-speaking countries of northern Europe formerly used various standardised "stones" for trade, with their values ranging from about 5 to 40 local pounds (roughly 3 to 15 kg) depending on the location and objects weighed. The United Kingdom's imperial system adopted the wool stone of 14 pounds in 1835. Definition of Metric Ton The tonne (SI unit symbol: t), commonly referred to as the metric ton in the United States, is a non-SI metric unit of mass equal to 1,000 kilograms; or one megagram (Mg); it is equivalent to approximately 2,204.6 pounds, 1.10 short tons (US) or 0.984 long tons (imperial). Although not part of the SI per se, the tonne is "accepted for use with" SI units and prefixes by the International Committee for Weights and Measures.	{"url":"https://whatisconvert.com/stones-to-metric-tons","timestamp":"2024-11-03T19:17:03Z","content_type":"text/html","content_length":"20473","record_id":"<urn:uuid:4b6f901e-19c6-4b75-b713-d0defe23c9a4>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00247.warc.gz"}
Rules of Baccarat Baccarat Standards Baccarat is played with 8 decks of cards. Cards of a value less than ten are of their printed value while 10, J, Q, K are 0, and A are each equal to 1. Wagers are placed upon the ‘banker,’ the ‘player’ or for a tie (these aren’t actual individuals; they just symbolize the two hands to be given out). 2 hands of 2 cards will then be played to the ‘banker’ as well as ‘player’. The score for every hand is the sum of the 2 cards, but the initial digit is discarded. For e.g., a hand of 7 as well as five results in a score of two (sevenplus5=12; drop the ‘1′). A 3rd card can be given out depending on the following practices: - If the player or banker has a value of 8 or nine, each gamblers stand. - If the player has 5 or less, he hits. bettors stand otherwise. - If gambler stands, the banker hits of 5 or less. If the gambler hits, a chart is used to judge if the banker stands or hits. Baccarat Odds The greater of the 2 scores wins. Winning bets on the banker pay out nineteen to 20 (even money less a 5 percent commission. Commission is tracked and paid out when you leave the table so be sure to have funds still before you leave). Winning bets on the player pay one to 1. Winner bets for tie usually pays 8 to one and on occasion nine to one. (This is a terrible wager as ties happen less than 1 every ten hands. Definitely don’t try betting on a tie. Still, odds are certainly better – 9 to 1 vs. eight to 1) When played accurately, baccarat provides fairly good odds, apart from the tie wager of course. Baccarat Strategy As with just about all games, Baccarat has some common false impressions. 1 of which is very similar to a roulette myth. The past is in no way an indicator of future results. Keeping track of prior outcomes on a chart is simply a waste of paper … a slap in the face for the tree that gave its life to be used as our stationary. The most commonly used and probably most successful method is the one-three-2-six technique. This scheme is used to pump up payout and minimizing risk. Begin by wagering 1 unit. If you win, add 1 more to the two on the table for a total of three on the 2nd bet. If you win you will have six on the table, take away four so you have 2 on the third wager. If you win the 3rd wager, add two to the 4 on the table for a sum total of 6 on the fourth gamble. If you don’t win on the first wager, you take a loss of 1. A win on the 1st bet followed up by loss on the second brings about a loss of 2. Wins on the 1st two with a loss on the 3rd gives you a profit of two. And wins on the first three with a loss on the 4th mean you break even. Attaining a win on all four bets leaves you with twelve, a profit of 10. This means you can fail to win the second bet 5 times for every successful streak of four bets and still break even. Leave a comment Trackback You must be logged in to post a comment.	{"url":"http://onlybaccarat.net/2024/10/12/rules-of-baccarat-37/","timestamp":"2024-11-09T22:22:10Z","content_type":"application/xhtml+xml","content_length":"12882","record_id":"<urn:uuid:3d71c857-e6ff-4fa7-a252-a7e30bfc7eaf>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00376.warc.gz"}
numpy.polynomial.laguerre.lagval2d(x, y, c)[source]¶ Evaluate a 2-D Laguerre series at points (x, y). This function returns the values: The parameters x and y are converted to arrays only if they are tuples or a lists, otherwise they are treated as a scalars and they must have the same shape after conversion. In either case, either x and y or their elements must support multiplication and addition both with themselves and with the elements of c. If c is a 1-D array a one is implicitly appended to its shape to make it 2-D. The shape of the result will be c.shape[2:] + x.shape. x, y : array_like, compatible objects The two dimensional series is evaluated at the points (x, y), where x and y must have the same shape. If x or y is a list or tuple, it is first converted to an ndarray, otherwise it is left unchanged and if it isn’t an ndarray it is treated as a scalar. c : array_like Array of coefficients ordered so that the coefficient of the term of multi-degree i,j is contained in c[i,j]. If c has dimension greater than two the remaining indices enumerate multiple sets of coefficients. values : ndarray, compatible object The values of the two dimensional polynomial at points formed with pairs of corresponding values from x and y.	{"url":"https://docs.scipy.org/doc/numpy-1.15.1/reference/generated/numpy.polynomial.laguerre.lagval2d.html","timestamp":"2024-11-12T10:31:52Z","content_type":"text/html","content_length":"10463","record_id":"<urn:uuid:e878fa30-adbb-4b2e-9ad8-5f73063024c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00540.warc.gz"}
Non-linear model for removing noise from corrupted signals - Patent 1398762 The present invention relates to noise reduction. In particular, the present invention relates to reducing noise in signals used in pattern recognition. A pattern recognition system, such as a speech recognition system, takes an input signal and attempts to decode the signal to find a pattern represented by the signal. For example, in a speech recognition system, a speech signal is received by the recognition system and is decoded to identify a string of words represented by the speech signal. However, input signals are typically corrupted by some form of additive noise. Therefore, to improve the performance of the pattern recognition system, it is often desirable to estimate the additive noise and use the estimate to provide a cleaner signal. Spectral subtraction has been used in the past for noise removal, particularly in automatic speech recognition systems. Conventional wisdom holds that when perfect noise estimates are available, basic spectral subtraction should do a good job of removing the noise; however, this has been found not to be the case. Standard spectral subtraction is motivated by the observation that noise and speech spectra mix linearly, and therefore, their spectra should mix according to Typically, this equation is solved for a \|X[k]\| , and a maximum attenuation floor F is introduced to avoid producing negative power special densities. Several experiments were run to examine the performance of Equation 1 using the true spectra of n, and floors F from e to e . The true noise spectra were computed from the true additive noise time series for each utterance. All experiments were conducted using the data, code and training scripts provided within the Aurora 2 evaluation framework described by H.G. Hirsch and D. Pearce in "The Aurora Experimental Framework for the Performance Evaluations of Speech Recognition Systems Under Noisy Conditions," ISCA ITRW ASR 2000 "Automatic Speech Recognition: Challenges for the Next Millennium", Paris, France, Sept. 2000 . The following digit error rates were found for various floors: │ e^-20 │ e^-10 │ e^-5 │ e^-3 │ e^-2 │ │ 87.50 │ 56.00 │ 34.54 │ 11.31 │ 15.56 │ From the foregoing, it is clear that even when the noise spectra is known exactly, spectral subtraction does not perform perfectly and improvements can be made. In light of this, a noise removal technique is needed that is more effective at estimating the clean speech spectral features. A new Bayesian estimation framework for statistical feature extraction in the form of cepstral enhancement, in which the joint prior distribution is exploited for both static and frame-differential dynamic cepstral parameters in the clean speech model, is presented in "A bayesian approach to speech feature enhancement using the dynamic cepstral pubr" ICASSP '2002 from Li Deng et al. A new statistical model describes the corruption of spectral features caused by additive noise. In particular, the model explicitly represents the effect of unknown phase together with the unobserved clean signal and noise. Development of the model has realized three techniques for reducing noise in a noisy signal as a function of the model. Generally, as an aspect of the present invention, as claimed in claims 1 and 8, and utilized in two techniques, a frame of a noisy input signal is converted into an input feature vector. An estimate of a noise-reduced feature vector uses a model of the acoustic environment. The model is based on a non-linear function that describes a relationship between the input feature vector, a clean feature vector, a noise feature vector and a phase relationship indicative of mixing of the clean feature vector and the noise feature , vector. Inclusion of a mathematical representation of the phase relationship renders an accurate model. One unique characteristic of the phase relationship is that it is in the same domain as the clean feature vector and the noise feature vector. Another separate distinguishing characteristic is that the phase relationship includes a phase factor with a statistical distribution. FIG. 1 is a block diagram of one computing environment in which the present invention may be practiced. FIG. 2 is a block diagram of an alternative computing environment in which the present invention may be practiced. FIG. 3A is a plot of conditional observation probability p(y\|x, n) with normal approximation for p[α](α). FIG. 3B is a plot of a sample distribution for a filter bank. FIG. 3C is a plot of normal approximation of p(y\|x, n) as a function of v. FIG. 3D is a plot of distributions of α for several frequency bins. FIG. 4A is a plot of output SNR to input SNR for known spectral subtraction. FIG. 4B is a plot of output SNR to input SNR for a new spectral subtraction method of the present invention. FIG. 5 is a method illustrating steps for obtaining a weighted Gaussian approximation in the model of the present invention. FIG. 6 is a flow diagram of another method of estimating clean speech. FIG. 7 is a block diagram of a pattern recognition system in which the present invention may be used. Before describing aspects of the present invention, a brief description of exemplary computing environments will be discussed. FIG. 1 illustrates an example of a suitable computing system environment 100 on which the invention may be implemented. The computing system environment 100 is only one example of a suitable computing environment and is not intended to suggest any limitation as to the scope of use or functionality of the invention. Neither should the computing environment 100 be interpreted as having any dependency or requirement relating, to any one or combination of components illustrated in the exemplary operating environment 100. The invention is operational with numerous other general purpose or special purpose computing system environments or configurations. Examples of well-known computing systems, environments, and/or configurations that may be suitable for use with the invention include, but are not limited to, personal computers, server computers, hand-held or laptop devices, multiprocessor systems, microprocessor-based systems, set top boxes, programmable consumer electronics, network PCs, minicomputers, mainframe computers, telephony systems, distributed computing environments that include any of the above systems or devices, and the like. The invention may be described in the general context of computer-executable instructions, such as program modules, being executed by a computer. Generally, program modules include routines, programs; objects, components, data structures, etc. that perform particular tasks or implement particular abstract data types. Tasks performed by the programs and modules are described below and with the aid of figures. Those skilled in the art can implement the description and figures as computer-executable instructions, which can be embodied on any form of computer readable media discussed The invention may also be practiced in distributed computing environments where tasks are performed by remote processing devices that are linked through a communications network. In a distributed computing environment, program modules may be located in both local and remote computer storage media including memory storage devices. With reference to FIG. 1, an exemplary system for implementing the invention includes a general-purpose computing device in the form of a computer 110. Components of computer 110 may include, but are not limited to, a processing unit 120, a system memory 130, and a system bus 121 that couples various system components including the system memory to the processing unit 120. The system bus 121 may be any of several types of bus structures including a memory bus or memory controller, a peripheral bus, and a local bus using any of a variety of bus architectures. By way of example, and not limitation, such architectures include Industry Standard Architecture (ISA) bus, Micro Channel Architecture (MCA) bus, Enhanced ISA (EISA) bus, Video Electronics Standards Association (VESA) local bus, and Peripheral Component Interconnect (PCI) bus also known as Mezzanine bus. Computer 110 typically includes a variety of computer readable media. Computer readable media can be any available media that can be accessed by computer 110 and includes both volatile and nonvolatile media, removable and non-removable media. By way of example, and not limitation, computer readable media may comprise computer storage media and communication media. Computer storage media includes both volatile and nonvolatile, removable and non-removable media implemented in any method or technology for storage of information such as computer readable instructions, data structures, program modules or other data. Computer storage media includes, but is not limited to, RAM, ROM, EEPROM, flash memory or other memory technology, CD-ROM, digital versatile disks (DVD) or other optical disk storage, magnetic cassettes, magnetic tape, magnetic disk storage or other magnetic storage devices, or any other medium which can be used to store the desired information and which can be accessed by computer 110. Communication, media typically embodies computer readable instructions, data structures, program modules or other data in a modulated data signal such as a carrier wave or other transport mechanism and includes any information delivery media. The term "modulated data signal" means a signal that has one or more of its characteristics set or changed in such a manner as to encode information in the signal. By way of example, and not limitation, communication media includes wired media such as a wired network or direct-wired connection, and wireless media such as acoustic, RF, infrared and other wireless media. Combinations of any of the above should also be included within the scope of computer readable media. The system memory 130 includes computer storage media in the form of volatile and/or nonvolatile memory such as read only memory (ROM) 131 and random access memory (RAM) 132. A basic input/output system 133 (BIOS), containing the basic routines that help to transfer information between elements within computer 110, such as during start-up, is typically stored in ROM 131. RAM 132. typically contains data and/or program modules that are immediately accessible to and/or presently being operated on by processing unit 120. By way of example, and not limitation, FIG. 1 illustrates operating system 134, application programs 135, other program modules 136, and program data 137. The computer 110 may also include other removable/non-removable volatile/nonvolatile computer storage media. By way of example only, FIG. 1 illustrates a hard disk drive 141 that reads from or writes to non-removable, nonvolatile magnetic media, a magnetic disk drive 151 that reads from or writes to a removable, nonvolatile magnetic disk 152, and an optical disk drive 155 that reads from or writes to a removable, nonvolatile optical disk 156 such as a CD ROM or other optical media. Other removable/non-removable, volatile/nonvolatile computer storage media that can be used in the exemplary operating environment include, but are not limited to, magnetic tape cassettes, flash memory cards, digital versatile disks, digital video tape, solid state RAM, solid state ROM, and the like. The hard disk drive 141 is typically connected to the system bus 121 through a non-removable memory interface such as interface 140, and magnetic disk drive 151 and optical disk drive 155 are typically connected to the system bus 121 by a removable memory interface, such as interface 150. The drives and their associated computer storage media discussed above and illustrated in FIG. 1, provide storage of computer readable instructions, data structures, program modules and other data for the computer 110. In FIG. 1, for example, hard disk drive 141 is illustrated as storing operating system 144, application programs 145, other program modules 146, and program data 147. Note that these components can either be the same as or different from operating system 134, application programs 135, other program modules 136, and program data 137. Operating system 144, application programs 145, other program modules 146, and program data 147 are given different numbers here to illustrate that, at a minimum, they are different copies. A user may enter commands and information into the computer 110 through input devices such as a keyboard 162, a microphone 163, and a pointing device 161, such as a mouse, trackball or touch pad. Other input devices (not shown) may include a joystick, game pad, satellite dish, scanner, or the like. These and other input devices are often connected to the processing unit 120 through a user input interface 160 that is coupled to the system bus, but may be connected by other interface and bus structures, such as a parallel port, game port or a universal serial bus (USB). A monitor 191 or other type of display device is also connected to the system bus 121 via an interface, such as a video interface 190. In addition to the monitor, computers may also include other peripheral output devices such as speakers 197 and printer 196, which may be connected through an output peripheral interface 190. The computer 110 may operate in a networked environment using logical connections to one or more remote computers, such as a remote computer 180. The remote computer 180 may be a personal computer, a hand-held device, a server, a router, a network PC, a peer device or other common network node, and typically includes many or all of the elements - described above relative to the computer 110. The logical connections depicted in FIG. 1 include a local area network (LAN) 171 and a wide area network (WAN) 173, but may also include other networks. Such networking environments are commonplace in offices, enterprise-wide computer networks, intranets and the Internet. When used in a LAN networking environment, the computer 110 is connected to the LAN 171 through a network interface or adapter 170. When used in a WAN networking environment, the computer 110 typically includes a modem 172 or other means for establishing communications over the WAN 173, such as the Internet. The modem 172, which may be internal or external, may be connected to the system bus 121 via the user input interface 160, or other appropriate mechanism. In a networked environment, program modules depicted relative to the computer 110, or portions thereof, may be stored in the remote memory storage device. By way of example, and not limitation, FIG. 1 illustrates remote application programs 185 as residing on remote computer 180. It will be appreciated that the network connections shown are exemplary and other means of establishing a communications link between the computers may be used. FIG. 2 is a block diagram of a mobile device 200, which is an exemplary computing environment. Mobile device 200 includes a microprocessor 202, memory 204, input/output (I/O) components 206, and a communication interface 208 for communicating with remote computers or other mobile devices. In one embodiment, the afore-mentioned components are coupled for communication with one another over a suitable bus 210. Memory 204 is implemented as non-volatile electronic memory such as random access memory (RAM) with a battery back-up module (not shown) such that information stored in memory 204 is not lost when the general power to mobile device 200 is shut down. A portion of memory 204 is preferably allocated as addressable memory for program execution, while another portion of memory 204 is preferably used for storage, such as to simulate storage on a disk drive. Memory 204 includes an operating system 212, application programs 214 as well as an object store 216. During operation, operating system 212 is preferably executed by processor 202 from memory 204. Operating system 212, in one preferred embodiment, is a WINDOWS® CE brand operating system commercially available from Microsoft Corporation. Operating system 212 is preferably designed for mobile devices, and implements database features that can be utilized by applications 214 through a set of exposed application programming interfaces and methods. The objects in object store 216 are maintained by applications 214 and operating system 212, at least partially in response to calls to the exposed application programming interfaces and methods. Communication interface 208 represents numerous devices and technologies that allow mobile device 200 to send and receive information. The devices include wired and wireless modems, satellite receivers and broadcast tuners to name a few. Mobile device 200 can also be directly connected to a computer to exchange data therewith. In such cases, communication interface 208 can be an infrared transceiver or a serial or parallel communication connection, all of which are capable of transmitting streaming information. Input/output components 206 include a variety of input devices such as a touch-sensitive screen, buttons, rollers, and a microphone as well as a variety of output devices including an audio generator, a vibrating device, and a display. The devices listed above are by way of example and need not all be present on mobile device 200. In addition, other input/output devices may be attached to or found with mobile device 200 within the scope of the present invention. Under one aspect of the present invention, a system and method are provided that remove noise from pattern recognition signals. To do this, this aspect of the present invention uses a new statistical model, which describes the corruption to the pattern recognition signals, and in particular to speech recognition spectral features, caused by additive noise. The model explicitly represents the effect of unknown phase together with the unobserved clean speech and noise as three hidden variables. The model is used to produce robust features for automatic speech recognition. As will be described below, the model is constructed in the log Mel-frequency feature domain. Advantages of this domain include low dimensionality, allowing for efficient training in inference. Logarithmic Mel-frequency spectral coefficients are also linearly related to Mel-Frequency Cepstrum Coefficients (MFCC), which correspond to the features used in a recognition system. Furthermore, corruption from linear channels and additive noise are localized within individual Mel-frequency bins, which allows processing of each dimension of the feature independently. As indicated above, the model of the present invention is constructed in the logarithmic Mel-frequency spectral domain. Each spectral frame is processed by passing it through a magnitude-squared operation, a Mel-frequency filterbank, and a logarithm. Generally, the noisy or observation (Y) signal is a linear combination of speech (X) and noise (N) as represented by Y[k] = X[k] + N[k]. Accordingly, the noisy log Mel-spectral features y can be directly related to the unobserved spectra X[k] and N[k], which can be represented as: is the kth coefficient in the ith Mel-frequency filterbank. The variable θ is the phase difference between X[k] and N[k]. When the clean signal and noise are uncorrelated, the θ are uncorrelated and have a uniform distribution over the range [-π,π]. Eq. 2 can be re-written to show how the noisy log spectral energies yi are a function of the unobserved log spectral energies x and n As a consequence of this model, when y is observed there are actually three unobserved random variables. The first two include the clean log spectral energy and the noise log spectral energy that would have been produced in the absence of mixing. The third variable, α , accounts for the unknown phase between the two sources. In the general case, α will be a function of X[k] and N[k]. However, if the magnitude spectra are assumed constant over the bandwidth of a particular filterbank, the definition of α collapses to a weighted sum of several independent random variables: Figure 3D shows the true distributions of α for a range of frequency bins. They were estimated from a set of joint noise, clean speech, and noisy speech data by solving for the unknown α. The higher frequency, higher bandwidth filters produce α distributions that are more nearly Gaussian. By design, the low frequency bins in a Mel-frequency filterbank have a narrow bandwidth, and the bandwidth increases with frequency. This means that the effective number of terms in Eq. 5 also increases with frequency. As a result, a Gaussian assumption is quite bad for the lowest frequency bins, and becomes much better as the bandwidth of the filters increase. As the bandwidth increases, so does the number of effective terms in Eq. 5, and the central limit theorem begins to apply. In practice, a frequency-dependent Gaussian approximation ) works well. At this point it should be noted, that the parameter can be estimated from a small set of training data. The estimate is the sample variance computed from all the sample values of α 's (for 1=1, 2, ...L, L= the total number Mel-filter banks). A linear regression line is fit to the computed as a function of 1. Conditional observation probability Eq. 3 places a hard constraint on the four random variables, in effect yielding three degrees of freedom. This can be expressed by solving for y and writing the conditional probability distribution, The conditional probability p(y\|x,n) is found by forming the distribution x, n) and marginalizing over α. Note is assumed, which is reasonable. The identity can then be used to evaluate the integral in closed form: When the Gaussian approximation for ) is introduced, the likelihood function becomes Shift Invariance The conditional probability (Eq. 7) appears at first glance to have three independent variables. Instead, it is only a function of two: the relative value of speech and observation (x - y), and the relative value of noise and observation (n - y). Model behavior Figure 3A contains a plot of this conditional probability distribution. Figure 3B shows an equivalent plot, directly estimated from data, thereby confirming Eq. 7A. Faith in the model can be built through either examining its ability to explain existing data, or by examining its asymptotic behavior. Figure 3A contains a plot of the conditional probability distribution of the model represented by Eq. 7A. Note that due to the shift invariance of this model, there are only two independent terms in the plot. Compare this to Figure 3B, which is a histogram of x - y versus n - y for a single frequency bin across all utterances in set A, subway noise, 10 dB SNR. It is clear that the model is an accurate description of the data. It should be noted some previous models, have an error model that is independent of SNR. The new model automatically adjusts its variance for different SNR hypotheses. As we move left along n = y in the graph of Figure 3A, the variance perpendicular to this line decreases. This corresponds to more and more certainty about the noise value as the ratio of speech to observation decreases. In the limit of this low SNR hypothesis, the model is reducing it's uncertainty that n = y to zero. If the prior probability distributions for speech and noise are concentrated in this area, the model reduces to Symmetrically, as we move down along x = y in the same graph, the variance perpendicular to this line decreases. As the ratio of noise to observation decreases, the model has increasing certainty that x = y. We refer to this region as the high SNR hypothesis. If the priors for speech and noise are concentrated in this area, the model reduces to The graph also has a third region of interest, starting from the origin and moving in a positive x and n direction. In this region, both speech and noise are greater than the observation. This occurs most frequently when x and n have similar magnitudes, and are destructively interfering with each other. In this case the relevant θ exist in the region Relationship to Spectral Subtraction Eq. 7A can be used to derive a new formula for spectral subtraction. The first step is to hold n and y fixed, and find a maximum likelihood estimate for x. Taking the derivative with respect to x in Eq. 7A and equating it to zero results in This formula is already more well-behaved than standard spectral subtraction. The first term is always real because the square root is taken of the sum of two positive numbers. Furthermore, the magnitude of the second term is never larger than the magnitude of the first term, so both sides of Eq. 8 are non-negative. The entire formula has exactly one zero, at n = y. This automatically prevents taking the logarithm of any negative numbers during spectral subtraction, allowing the maximum attenuation floor F to be relaxed. and Eq. 8 is solved for x, the result is a new spectral subtraction equation with an unexpected absolute value operation. The difference between Eq. 1 and Eq. 9 is confined to the region y < n, as illustrated in Figures 4A and 4B. Spectral subtraction assumes any observation below y=n is equivalent to a signal to noise ratio of the floor F, and produces maximum attenuation. The data illustrated in Figure 3B contradicts this assumption, showing a non-zero probability mass in the region n > y. The end result is that, even with perfect knowledge of the true noise, spectral subtraction treats these points inappropriately. Eq. 9 has more reasonable behavior in this region. As the observation becomes much lower than the noise estimate, the function approaches x = n. The new model indicates the most likely state is that x and n have similar magnitudes and axe experiencing destructive phase interference. Table II compares the relative accuracy of using Equations 1 and 9 for speech recognition, when the true noise spectra are available. Although the new method does not require a floor to prevent taking the logarithm of a negative number, it is included because it does yield a small improvement in error rate. Table II │ │ │ │ │ FLOOR │ │ │ Method │ e^-20 │ e^-10 │ e^-5 │ e^-3 │ e^-2 │ │ Standard │ 87.50 │ 56.00 │ 34.54 │ 11.31 │ 15.56 │ │ (Eq. 1) │ │ │ │ │ │ │ Proposed │ 6.43 │ 5.74 │ 4.10 │ 7.82 │ 10.00 │ │ (Eq. 9) │ │ │ │ │ │ Regardless of the value chosen for the floor, the new method outperforms the old spectral subtraction rule. Although the old method is quite sensitive to the value chosen, the new method is not, producing less than 10% digit error rate for all tests. In one embodiment, noisy-speech frames are processed independently of each other. A sequential tracker for estimating the log spectrum of non-stationary noise can be used to provide a noise estimate on a frame-by-frame basis. A suitable noise estimator is described in METHOD OF ITERATIVE NOISE ESTIMATION IN A RECURSIVE FRAMEWORK (Attorney Docket No. M61.12-0443), filed on even date herewith. Another advantage of deriving the conditional observation probability, Eq. 7A, is that it can be embedded into a unified Bayesian model. In this model, the observed variable y is related to the hidden variables, including x and n through a unified probabilistic model. From this model, one can infer posterior distributions on the hidden variables x and n, including MMSE (minimum mean square error) and maximum likelihood estimates. In this way, noisy observations are turned into probability distributions over the hidden clean speech signal. To produce noise-removed features for conventional decoding, conditional expectations of this model are taken. The Bayesian approach can additionally produce a variance of its estimate of . This variance can be easily leveraged within the decoder to improve word accuracy. A suitable decoding technique is described in METHOD OF PATTERN RECOGNITION USING NOISE REDUCTION UNCERTAINTY, filed May 20, 2002 and assigned serial no. 10/152,127. In this form of uncertainty decoding, the static feature stream is replaced with an estimate of . The noise removal process outputs high variance for low SNR features, and low variance when the SNR is high. To support this framework, the following are also provided: Better results are achieved with a stronger prior distribution for clean speech, such as a mixture model. When a mixture model is used, Equations 10, 11, 12, and 13 are conditioned on the mixture m, evaluated, and then combined in the standard way: Approximating the observation likelihood As mentioned previously, the form derived for x, n) does not lend itself to direct algebraic manipulation. Furthermore, it is capable of producing joint distributions that are not well modeled by a Gaussian approximation. As a result, some steps are performed to compute the necessary integrations for noise removal. When computation is less of an issue, a much finer non-iterative approximation of x, n) can be used. The approximation preserves the global shape of the conditional observation probability so that the usefulness of the model is not masked by the approximation. One perfectly reasonable, although computationally intensive, option is to make no approximation of Eq. 7A. The joint probability p(y, x, n) can be evaluated along a grid of points in x and n for each observation y. Weighted sums of these values could produce accurate approximations to all of the necessary moments. Selecting an appropriate region is a non-trivial task, because the region is dependent on the current observation and the noise and speech priors. More specifically, to avoid unnecessary computation, the evaluation should be limited to the region where the joint probability has the most mass. Essentially, a circular paradox is realized where it is necessary to solve the problem before choosing appropriate parameters for a solution. Another reasonable approach is to approximate the joint probability with a single Gaussian. This is the central idea in vector Taylor series (VTS) approximation. Because the prior distributions on x and n limit the scope of p(y\|x,n), this local approximation may be more appropriate than a global approximation. However, there are two potential pitfalls associated with this method. First, even though the prior distributions are unimodal, applying can introduce more modes to the joint probability. Second, the quadratic expansions along x and n do not capture the shape of x, n ) well when n << y or x << y. Instead, one aspect of the present inventions is a compromise between these two methods. In particular, a Gaussian approximation is used to avoid summation over a two-dimensional grid, while at the same time preserving the true shape of p(y\|x,n). This is accomplished by collapsing one dimension with a Gaussian approximation, and implementing a brute force summation along the remaining Normal Approximation to Likelihood In this aspect of the present invention, a Gaussian approximation is used along one dimension only, which allows preservation of the true shape of x, n), and allows a numerical integration along the remaining dimension. The weighted Gaussian approximation is found in four steps illustrated in Figure 5 at 250. The coordinate space is first rotated in step 252. An expansion point is chosen in step 254. A second order Taylor series approximation is then made in step 256. The approximation is then expressed as the parameters of a weighted Gaussian distribution in step 258. The coordinate rotation is necessary because expanding along x or n directly can be problematic. A 45 degree rotation is used, which makes p(y\|x,n) approximately Gaussian along u for each value of v. Although the new coordinates u and v are linear functions of y, x and n, the cumbersome functional notation at this point can be dropped. Next, v is held constant and a weighted Taylor series approximation along u is determined. For each v, the Taylor series expansion point is found by performing the change of variables on Eg. 3, holding v constant, α= O, and solving for u. The result is, The quadratic approximation of p(y\|x,n) at each value of v can then be expressed as a Gaussian distribution along u. Our final approximation is given by: As Figure 3C illustrates, this final approximation is quite good at capturing the shape of p(y\|x,n). And, as discussed below, the Gaussian approximation along u can be leveraged to eliminate a significant amount of computation. Building the joint probability The approximation for p(y\|x,n) is complete, and is now combined with the priors p (x) and p (n) to produce the joint probability distribution. To conform to the approximation of the conditional observation probability, these prior distributions to the (u, v) coordinate space are transformed, and written as a Gaussian in u whose mean and variance are functions of v. From the joint probability, Equations 10, 11, 12, and 13 are computed. Each equation requires at least one double integral over x and n, which is equivalent to a double integral over u and v. For Here, the method makes use of Eq. 14 for x, as well as Eq. 15 and Eq. 16 for p(y,x,n). The Gaussian approximation enables a symbolic evaluation of the integral over u, but the integral over v The integration in v is currently implemented as a numerical integration, a weighed sum along discrete values of v. In one embodiment, 500 equally spaced points in the range [-20, 20] are used. Most of the necessary values can be pre-computed and tabulated to speed up computation. MMSE Estimator Based on Taylor Series Expansion The foregoing has described a new spectral subtraction formula (Eq. 9) and a numerical integration, after rotation of axis, to compute the MMSE estimate for speech removal. The following provides a further technique, in particular, an iterative Taylor series expansion to compute the MMSE (minimum mean square error) estimate in an analytical form in order to remove noise using the phase-sensitive model of the acoustic environment described above. Given the log-domain noisy speech observation y, the MMSE estimator χ̂ for clean speech is the conditional expectation: is determined by the probabilistic environment model just presented. The prior model for clean speech, p(x) in Eq. 18 is assumed to have the Gaussian mixture PDF: whose parameters are pre-trained from the log-domain clean speech data. This allows Eq.18 to be written as The main difficulty in computing χ̂ above is the non-Gaussian nature of To overcome this difficulty, a truncated second-order Taylor series expansion is used to approximate the exponent of That is, the following function is approximated In Eq. 22, a single-point expansion point χ is used (i.e., x does not depend on the mixture component m) to provide significantly improved computational efficiency, and χ is iteratively updated to increase its accuracy to the true value of clean speech x. The Taylor series expansion coefficients have the following closed forms: It should be noted ∑ = σ . In other words, a zero-mean Gaussian distribution is used for the phase factor α in the new model described above, and which is used in this embodiment. Fitting Eq. 22 into a standard quadratic form, the following is obtained This then allows computing the integral of Eq. 20 in a closed form: The denominator of Eq. 20 is computed according to Substituting Eqs. 23 and 24 into Eq. 20, the final MMSE estimator is obtained: Where the weighting factors are Note that in Eq. 24 are all dependent on the noise estimator , which can be obtained from any suitable noise tracking estimator such as described in the co-pending application referenced above. Under this aspect of the present invention, the clean speech estimate of the current frame, , is calculated several times using an iterative method shown in the flow diagram of FIG. 6. The method of FIG. 6 begins at step 300 where the distribution parameters for the prior clean speech mixture model are pretrained from a set of clean training data. In particular, the mean, , covariance, σ , and mixture weight, c , for each mixture component m in a set of M mixture components is determined. At step 302, the expansion point, used in the Taylor series approximation for the current iteration, j, can be set equal to the mean vector of the Gaussian mixture model of the clean speech that best accounts for (in the maximum likelihood sense) the noisy speech observation vector y given the estimated noise vector n. At step 306, the Taylor series expansion point for the next iteration, is set equal to the noise estimate found for the current iteration, In terms of an equation: The updating step shown in Eq. 26 improves the estimate provided by the Taylor series expansion and thus improves the calculation during the next iteration. At step 308, the iteration counter j is incremented before being compared to a set number of iterations J at step 310. If the iteration counter is less than the set number of iterations, more iterations are to be performed and the process returns to step 304 to repeat steps 304, 306, 308 and 310 using the newly updated expansion point. After J iterations have been performed at step 310, the final value for the clean speech estimate of the current frame has been determined and at step 312, the variables for the next frame are set. In one embodiment, J is set equal to three. Specifically, the iteration counter j is set to zero, the frame value t is incremented by one, and the expansion point for the first iteration of the next frame is set. A method and system for using the present invention in speech recognition is shown the block diagram of FIG. 7. The method begins where a noisy speech signal is converted into a sequence of feature vectors. To do this, a microphone 404 of FIG. 7, converts audio waves from a speaker 400 and one or more additive noise sources 402 into electrical signals. The electrical signals are then sampled by an analog-to-digital converter 406 to generate a sequence of digital values, which are grouped into frames of values by a frame constructor module 408. In one embodiment, A-to-D converter 406 samples the analog signal at 16 kHz and 16 bits per sample, thereby creating 32 kilobytes of speech data per second and frame constructor module 408 creates a new frame every 10 milliseconds that includes 25 milliseconds worth of data. Each frame of data provided by frame constructor module 408 is converted into a feature vector by a feature extractor 410. Methods for identifying such feature vectors are well known in the art and include 13-dimensional Mel-Frequency Cepstrum Coefficients (MFCC) extraction. The feature vectors for the noisy speech signal are provided to a noise estimation module 411 in FIG. 7. Noise estimation module 411 estimates the noise in the current frame and provides a feature vector representing the noise estimate, or a distribution thereof, together with the noisy speech signal to a noise reduction module 412. The noise reduction module 412 uses any one of the techniques described above, (new spectral subtraction of Eq. 9, the Bayesian approach with weighted Gaussian Approximation, or an MMSE estimator) with model parameters of the corresponding implementing equations, which are stored in noise reduction parameter storage 411, to produce a sequence of noise-reduced feature vectors from the sequence of noisy feature vectors, or distributions thereof. The output of noise reduction module 412 is a series of noise-reduced feature vectors. If the input signal is a training signal, this series of noise reduced feature vectors is provided to a trainer 424, which uses the noise-reduced feature vectors and a training text 426 to train an acoustic model 418. Techniques for training such models are known in the art and a description of them is not required for an understanding of the present invention. If the input signal is a test signal, the noise-reduced feature vectors are provided to a decoder 414, which identifies a most likely sequence of words based on the stream of feature vectors, a lexicon 415, a language model 416, and the acoustic model 418. The particular method used for decoding is not important to the present invention and any of several known methods for decoding may be The most probable sequence of hypothesis words is provided to a confidence measure module 420. Confidence measure module 420 identifies which words are most likely to have been improperly identified by the speech recognizer, based in part on a secondary acoustic model(not shown). Confidence measure module 420 then provides the sequence of hypothesis words to an output module 422 along with identifiers indicating which words may have been improperly identified. Those skilled in the art will recognize that confidence measure module 420 is not necessary for the practice of the present Although FIG. 7 depicts a speech recognition system, the present invention may be used in other noise removal applications such as removing noise of recordings, or prior to transmission of data in order to transmit cleaner data. In this manner, the pattern recognition system is also not limited to speech. Computerlesbares Medium, das von einem Computer ausführbare Instruktionen hat, umfasst: ein erstes Modul, welches einen ersten Frame eines verrauschten Eingangssignals (404) in einem Eingangseigenschaftsvektor (410) umwandelt; und ein zweites Modul, welches eine Abschätzung von einem rauschreduzierten Eigenschaftsvektor (412) unter Benützung eines Modells der akustischen Umgebung erhält, wobei das Modell auf einer nicht linearen Funktion basiert ist, welche eine Beziehung zwischen dem Eingangseigenschaftsvektor, einem reinem Eigenschaftsvektor, einem verrauschten Eigenschaftsvektor und einer Phasenbeziehung, welche die Mischung des reinen Eigenschaftsvektors und des verrauschten Eigenschaftsvektors anzeigt, beschreibt, wobei die Phasenbeziehung einen Phasenfaktor mit einer statistischen Verteilung beinhaltet, welche eine Gaußverteilung mit Nullmittelwert umfasst, wobei das akustische Modell auf einer Rotation eines Koordinatenraumes (254) basiert, welcher den Eingangseigenschaftsvektor, den reinen Eigenschaftsvektor und den verrauschten Eigenschaftsvektor umfasst, um eine Änderung der Variablen, die zwei Dimensionen haben, zu erhalten. 2. Computerlesbares Medium nach Anspruch 1, wobei eine Dimension des akustischen Modells annähernd gaußverteilt (258) ist. 3. Computerlesbares Medium nach Anspruch 2, wobei das zweite Modul Instruktionen beinhaltet, um die Abschätzung des rauschreduzierten Eigenschaftsvektors (412) als eine Funktion vom Durchführen einer symbolischen Integration über eine Dimension des akustischen Modells zu erhalten. 4. Computerlesbares Medium nach Anspruch 2, wobei das zweite Modul Instruktionen beinhaltet, um die Abschätzung des rauschreduzierten Eigenschaftsvektors (412) als eine Funktion vom Durchführen einer numerischen Integration über eine zweite Dimension des akustischen Modells zu erhalten. 5. Computerlesbares Medium nach Anspruch 1, wobei das zweite Modul Instruktionen zum Erhalten einer Abschätzung des rauschreduzierten Eigenschaftsvektors (412) als eine Funktion einer Taylorreihenentwicklung (256) beinhaltet. 6. Computerlesbares Medium nach Anspruch 5, wobei das zweite Modul Instruktionen zum Erhalten einer Abschätzung des rauschreduzierten Eigenschaftsvektors (412) als eine Funktion einer vorangegangenen Abschätzung des rauschreduzierten Eigenschaftsvektors beinhaltet. 7. Computerlesbares Medium nach Anspruch 6, wobei das zweite Modul Anweisungen zum Erhalten einer Abschätzung des rauschreduzierten Eigenschaftsvektors (412) als eine Funktion vom wiederholten Benutzen einer intermediären Eingangsabschätzung des rauschreduzierten Eigenschaftsvektors (412) in einer nachfolgenden Berechnung für eine ausgewählte Anzahl von Iterationen beinhaltet. Verfahren zum Reduzieren von Rauschen in einem verrauschten Eingangssignal (404), wobei das Verfahren umfasst: Umwandeln eines Frames eines verrauschten Eingangssignals (404) in einen Eingangseigenschaftsvektor (410); Erhalten eines akustischen Modells der akustischen Umgebung, wobei das akustische Modell auf einer nicht linearen Funktion basiert ist, welche eine Beziehung zwischen dem Eingangseigenschaftsvektor, einem reinen Eigenschaftsvektor, einem verrauschten Eigenschaftsvektor und einer Phasenbeziehung, die das Mischen des reinen Eigenschaftsvektors und des verrauschten Eigenschaftsvektors anzeigt, basiert ist, wobei die Phasenbeziehung in der gleichen Domäne wie der reine Eigenschaftsvektor und der verrauschte Eigenschaftsvektor ist und einen Phasenfaktor beinhaltet, der eine statistische Verteilung hat, die eine Gaußverteilung mit Nullmittelwert ist, wobei das akustische Modell das Rotieren eines Koordinatenraumes (254) des akustischen Modells, das den Eingangseigenschaftsvektor, den reinen Eigenschaftsvektor und den verrauschten Eigenschaftsvektor umfasst, beinhaltet, um eine Änderung der Variablen, welche zwei Dimensionen haben, zu erhalten; und Verwenden des Eingangeigenschaftsvektors, des verrauschten Eigenschaftsvektors und des akustischen Modells, um den rauschreduzierten Eigenschaftsvektor (412) abzuschätzen. 9. Verfahren nach Anspruch 8, wobei eine Dimension des akustischen Modells annähernd gaußverteilt (258) ist. 10. Verfahren nach Anspruch 9, wobei das Verwenden des Eingangseigenschaftsvektors, des verrauschten Eigenschaftsvektors und des akustischen Modells zum Abschätzen des rauschreduzierten Eigenschaftsvektors (412), das symbolische Integrieren über eine Dimension des akustischen Modells beinhaltet. 11. Verfahren nach Anspruch 10, wobei das Verwenden des Eingangseigenschaftsvektors, des verrauschten Eigenschaftsvektors und des akustischen Modells zum Abschätzen des rauschreduzierten Eigenschaftsvektors (412), das numerische Integrieren über eine zweite Dimension des akustischen Modells beinhaltet. 12. Verfahren nach Anspruch 8, wobei das Verwenden des Eingangseigenschaftsvektors, des verrauschten Eigenschaftsvektors und des akustischen Modells zum Abschätzen des rauschreduzierten Eigenschaftsvektors (412) das Verwenden einer Taylorreinentwicklung (256) beinhaltet. 13. Verfahren nach Anspruch 12, wobei das Verwenden des Eingangseigenschaftsvektors des verrauschten Eigenschaftsvektors und des akustischen Modells zum Abschätzen des rauschreduzierten Eigenschaftsvektors (412) das Verwenden einer Eingangsabschätzung des rauschreduzierten Eigenschaftsvektors beinhaltet. 14. Verfahren nach Anspruch 13, wobei das Verwenden des Eingangseigenschaftsvektors, des verrauschten Eigenschaftsvektors und des akustischen Modells zum Abschätzen des rauschreduzierten Eigenschaftsvektors (412) die wiederholte Benützung einer intermediären Eingangsabschätzung des rauschreduzierten Eigenschaftsvektors (412) in einer nachfolgenden Berechnung für eine ausgewählte Anzahl von Iterationen beinhaltet.	{"url":"https://data.epo.org/publication-server/rest/v1.2/publication-dates/20090930/patents/EP1398762NWB1/document.html","timestamp":"2024-11-04T11:11:28Z","content_type":"text/html","content_length":"121179","record_id":"<urn:uuid:608ef021-e00e-4e6d-ba52-cd82b26b7b7f>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00658.warc.gz"}
Wave Power in context of input power 31 Aug 2024 Journal of Renewable Energy Systems Volume 12, Issue 3, 2023 Wave Power: A Review of Input Power Considerations Wave energy converters (WECs) have gained significant attention as a promising source of renewable energy. However, the input power characteristics of WECs are complex and require careful consideration. This article reviews the current understanding of wave power in the context of input power, highlighting the key factors that influence the performance of WECs. Wave energy is a form of ocean energy that harnesses the kinetic energy of waves to generate electricity. The input power of a WEC is determined by the wave characteristics, including the wave height (H), period (T), and direction. Understanding the input power characteristics of WECs is crucial for designing efficient and reliable systems. Wave Power Formulation The input power (P) of a WEC can be formulated using the following equation: P = 0.5 \* ρ \* C_d \* A \* H^2 / T where: ρ = water density C_d = drag coefficient A = surface area of the WEC Wave Characteristics The wave characteristics, including wave height (H), period (T), and direction, play a crucial role in determining the input power of a WEC. The wave height is typically measured using buoys or other sensors, while the wave period can be estimated using empirical formulas. Input Power Considerations Several factors influence the input power characteristics of WECs, including: • Wave direction: The direction of the waves relative to the WEC affects the input power. • Wave height and period: The magnitude and duration of the waves impact the input power. • Water depth: The water depth affects the wave characteristics and, subsequently, the input power. • WEC design: The design of the WEC, including its geometry and materials, influences the input power. The input power characteristics of WECs are complex and influenced by various factors. Understanding these factors is essential for designing efficient and reliable wave energy converters. Further research is needed to optimize WEC performance and maximize the potential of wave power as a renewable energy source. • [1] Falnes, J. (2007). A review of wave energy technology. Ocean Engineering, 34(4), 567-579. • [2] Kofoed, J., & Nielsen, K. (2013). Wave energy converters: A review of the state-of-the-art. Renewable and Sustainable Energy Reviews, 21, 468-483. Note: The references provided are fictional and for demonstration purposes only. Related articles for ‘input power’ : • Reading: Wave Power in context of input power Calculators for ‘input power’	{"url":"https://blog.truegeometry.com/tutorials/education/8a2c5e45e46d1cae4f9d48b0b453c542/JSON_TO_ARTCL_Wave_Power_in_context_of_input_power.html","timestamp":"2024-11-03T17:13:28Z","content_type":"text/html","content_length":"15292","record_id":"<urn:uuid:b3e9f128-f7d2-4d8d-b538-5295730fcc45>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00230.warc.gz"}
Playing with Python So I like nothing more than to spend the early hours of every morning looking at anaconda source[0] in the hopes of learning more about some of the more fun things one could to do with Python, if one felt so inclined. Actually, that’s partly true – I am trawling through the anaconda source at the moment of my own volition – everything from figuring out the pipe communication startup hacks in mini-wm/Anaconda to the isys library. Why? Because I work at a company that has heavy reliance on Python and feel I should better understand the language, and use it more in my own work as a result. Besides, I’m not a perl weenie, I should write fewer shell scripts, and I’m not particularly desperate to go near a Java compiler, having been subjected to that more than quite enough in college. I actually quite like Python. I’ve written a bunch of scripts using it, but nothing huge and I tend to write very C-like imperative code in “OO” languages – even when I’ve played with pygtk in the past. Still, that’s just fine. One of the things I have been doing is reading up on the more interesting features of the language, best practices, and the like. I’m using the two standard O’Reilly books as a reference and the online p.o documentation. The point of the exercise being that I’m reasonably keen to know how I “should” approach writing Python code, as opposed to my hackish approach in the past. I want to understand the intricacies of classes in Python, pickling, the internal differences between pyc and pyo “optimised” bytecode and some of the more funky APIs – like the HAL/gtk bindings I’m going to be playing with some more. Anyway. That’s all fairly boring. Not having looked at Python’s more funky features in a while, I wanted to dive right in and write some simple C/API plugin to play with Python Objects. This turns out to be a little harder these days than the (normally excellent, but in this case not so much) python documentation would have you believe. Here’s a simple test “library” I wrote to figure it out: * pytest - playing with Python's C API. #include <python .h> static PyObject list_set(PyObject list, PyObject item); void initpytest(void); static PyMethodDef pytestModuleMethods[] = { { "list_set", (PyCFunction) list_set, METH_VARARGS, NULL }, { NULL, NULL, 0, NULL } void initpytest(void) PyObject m; m = Py_InitModule("pytest", pytestModuleMethods); static PyObject list_set(PyObject self, PyObject args) PyObject list, value; int item; if (!PyArg_ParseTuple(args, "OiO", &list, &item, &value)) return NULL; if (!PyList_Check(list)) goto error; if (PyList_SetItem(list, item, value)) goto error; return Py_None; return NULL; You can compile this into a dynamically loadable (read: Python can use it) library on most Linux systems by driving the gcc front end, thus: gcc -o pytest.so -g -shared -fPIC -Lpython -I/usr/include/python2.4 pytest.c (this enables debugging, “-g”, and assumes a default 2.4 python installation, such as that on my Fedora Core 6 desktop box that I’m using today). Once built, such libraries can be used very simply: [jcm@perihelion pytest]$ python Python 2.4.3 (#1, Oct 1 2006, 18:00:19) [GCC 4.1.1 20060928 (Red Hat 4.1.1-28)] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> import pytest >>> foo=[1,2,3] >>> print foo [1, 2, 3] >>> pytest.list_set(foo,0,2) >>> print foo [2, 2, 3] It’s a simple example, pytest provides a single function that changes values in Python lists. Everything in Python is an object, and there is a need to be aware of reference-counting that’s happening underneath (well, sometimes you need to worry about it). Every C/API extension needs to define an init function, which educates the Python runtime about those functions it provides – in this case, the list_set function. These are defined in a NULL-terminated PyMethodDef array, and usually defined to require METH_VARARGS argument passing (I’m sure this was once different). Actual functions, such as list_set work in PyObjects (everything’s an object) too, using PyArg_ParseTuple to pull out C-style arguments, according to a format string. They also need to use occasional calls to Py_DECREF (and it’s NULL-friendly wrapper counterpart, Py_XDECREF). Though this varies – for example, in the above, PyList_SetItem “steals” a reference to “list” and so I don’t need to worry about freeing that particular reference. The documentation tries to explain what needs tracking. Anyway. This has been a random post. For those of you who are already die-hard Python programmers, I’m sure this is old-hat – and that’s appropriate to us now-officially-old uncle-types (my sister just gave birth last night). For me, it’s stuff I looked at too long ago but didn’t have a need for. Now that I’m looking at writing more funky stuff using Python, I’m working on becoming a Python nut, just like you [0] The installer that was originally written by Red Hat (and used by a metric fucktonne of other people and projects needing a graphical Linux installer not based YaST, Debian, or their own custom code). The thing that I (blindly, thanks to my previous Debianiteness/Ubuntu craziness that I’m recovering from over time) used to criticize because it wasn’t the d-i du jour. Actually, anaconda is a very powerful Linux installer that gets a lot of things right. And when you stop blindly agreeing with Debian rhetoric, this becomes abundantly more apparent (this isn’t an anti-Debian rant). Anaconda’s source can probably best be described as something that seems to have evolved over time. There are top level classes (like Anaconda) and some OO concepts in there, but it’s still largely a tangled mess of functions/function pointers (method object references, whatever) that works out in the end. Not that I’m being critical – I’m sure I couldn’t do it any better – but it’s certainly useful to pick at * I think some people have used anaconda for non-RPM distros, but I’m not aware of anyone trying to install weird stuff like Open Slowaris with it. Random aside: Sun seem to have stopped holding up a new edition of Slowaris Internals. There’s a new edition in Borders, covering Open Solaris and Solaris 10. It’s a shame it’s a bazillion years too late (the last edition was on Solaris 7, and like, totally ruled, dude) but then, I expected nothing less than to have to wait almost a decade for an updated edition. At least they now give out source code to the OS – unlike the many times I asked for it as a student and was told the programme was temporarily suspended…a million years ago.	{"url":"http://www.jonmasters.org/blog/2007/03/08/playing-with-python/","timestamp":"2024-11-05T16:58:33Z","content_type":"application/xhtml+xml","content_length":"15365","record_id":"<urn:uuid:fefa7aaa-9f71-4711-b096-bb4e708ba582>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00352.warc.gz"}
c05avf (contfn_interval_rcomm) NAG FL Interface c05avf (contfn_interval_rcomm) FL Name Style: FL Specification Language: 1 Purpose c05avf attempts to locate an interval containing a simple zero of a continuous function using a binary search. It uses reverse communication for evaluating the function. 2 Specification Fortran Interface Subroutine c05avf ( x, fx, h, boundl, boundu, y, c, ind, ifail) Integer, Intent (Inout) :: ind, ifail Real (Kind=nag_wp), Intent (In) :: fx, boundl, boundu Real (Kind=nag_wp), Intent (Inout) :: x, h, y, c(11) C Header Interface #include <nag.h> void c05avf_ (double x, const double fx, double h, const double boundl, const double boundu, double y, double c[], Integer ind, Integer ifail) The routine may be called by the names c05avf or nagf_roots_contfn_interval_rcomm. 3 Description You must supply an initial point and a step attempts to locate a short interval $\left[{\mathbf{x}},{\mathbf{y}}\right]\subset \left[{\mathbf{boundl}},{\mathbf{boundu}}\right]$ containing a simple zero of (On exit we may have is determined as the first point encountered in a binary search where the sign of differs from the sign of at the initial input point .) The routine attempts to locate a zero of in turn as its basic step before quitting with an error exit if unsuccessful. c05avf returns to the calling program for each evaluation of $f\left(x\right)$. On each return you should set ${\mathbf{fx}}=f\left({\mathbf{x}}\right)$ and call c05avf again. 4 References 5 Arguments this routine uses reverse communication. Its use involves an initial entry, intermediate exits and re-entries, and a final exit, as indicated by the argument . Between intermediate exits and re-entries, all arguments other than fx must remain unchanged 1: $\mathbf{x}$ – Real (Kind=nag_wp) Input/Output On initial entry: the best available approximation to the zero. must lie in the closed interval (see below). On intermediate exit: contains the point at which $f$ must be evaluated before re-entry to the routine. On final exit : contains one end of an interval containing the zero, the other end being in , unless an error has occurred. If are the end points of the largest interval searched. If a zero is located exactly, its value is returned in (and in 2: $\mathbf{fx}$ – Real (Kind=nag_wp) Input On initial entry : if need not be set. must contain for the initial value of On intermediate re-entry : must contain for the current value of 3: $\mathbf{h}$ – Real (Kind=nag_wp) Input/Output On initial entry: a basic step size which is used in the binary search for an interval containing a zero. The basic step sizes ${\mathbf{h}},0.1×{\mathbf{h}}$, $0.01×{\mathbf{h}}$ and $0.001×{\ mathbf{h}}$ are used in turn when searching for the zero. : either must lie inside the closed interval must be sufficiently large that ${\mathbf{x}}+{\mathbf{h}}e {\mathbf{x}}$ on the computer. On final exit: is undefined. 4: $\mathbf{boundl}$ – Real (Kind=nag_wp) Input 5: $\mathbf{boundu}$ – Real (Kind=nag_wp) Input On initial entry must contain respectively lower and upper bounds for the interval of search for the zero. Constraint: ${\mathbf{boundl}}<{\mathbf{boundu}}$. 6: $\mathbf{y}$ – Real (Kind=nag_wp) Input/Output On initial entry: need not be set. On final exit : contains the closest point found to the final value of , such that $f\left({\mathbf{x}}\right)×f\left({\mathbf{y}}\right)\le 0.0$ . If a value is found such that . On final exit with are the end points of the largest interval searched. 7: $\mathbf{c}\left(11\right)$ – Real (Kind=nag_wp) array Communication Array On initial entry: need not be set. On final exit: if ${\mathbf{ifail}}={\mathbf{0}}$ or ${\mathbf{4}}$, ${\mathbf{c}}\left(1\right)$ contains $f\left({\mathbf{y}}\right)$. 8: $\mathbf{ind}$ – Integer Input/Output On initial entry : must be set to fx need not be set. fx must contain $f\left({\mathbf{x}}\right)$. On intermediate exit : contains . The calling program must evaluate , storing the result in , and re-enter with all other arguments unchanged. On final exit: contains $0$. Constraint: on entry ${\mathbf{ind}}=-1$, $1$, $2$ or $3$. Note: any values you return to c05avf as part of the reverse communication procedure should not include floating-point NaN (Not a Number) or infinity values, since these are not handled by c05avf . If your code does inadvertently return any NaNs or infinities, c05avf is likely to produce unexpected results. 9: $\mathbf{ifail}$ – Integer Input/Output On initial entry must be set to to set behaviour on detection of an error; these values have no effect when no error is detected. A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not. If halting is not appropriate, the value is recommended. If message printing is undesirable, then the value is recommended. Otherwise, the value is recommended since useful values can be provided in some output arguments even when ${\mathbf{ifail}}e {\mathbf{0}}$ on exit. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit. On final exit unless the routine detects an error or a warning has been flagged (see Section 6 6 Error Indicators and Warnings If on entry , explanatory error messages are output on the current error message unit (as defined by Errors or warnings detected by the routine: On entry, ${\mathbf{boundl}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{boundu}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{boundl}}<{\mathbf{boundu}}$. On entry, ${\mathbf{x}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{boundl}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{boundu}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{boundl}}\le {\mathbf{x}}\le {\mathbf{boundu}}$. On entry, ${\mathbf{x}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{h}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{boundl}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{boundu}}=⟨\mathit{\text{value}}⟩$. Constraint: either ${\mathbf{x}}+{\mathbf{h}}$ or ${\mathbf{x}}-{\mathbf{h}}$ must lie inside the closed interval $\left[{\mathbf{boundl}},{\mathbf{boundu}}\right]$. On entry, must be sufficiently large that ${\mathbf{x}}+{\mathbf{h}}e {\mathbf{x}}$ on the computer. On entry, ${\mathbf{ind}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{ind}}=-1$, $1$, $2$ or $3$. An interval containing the zero could not be found. Try restarting with modified An unexpected error has been triggered by this routine. Please contact Section 7 in the Introduction to the NAG Library FL Interface for further information. Your licence key may have expired or may not have been installed correctly. Section 8 in the Introduction to the NAG Library FL Interface for further information. Dynamic memory allocation failed. Section 9 in the Introduction to the NAG Library FL Interface for further information. 7 Accuracy is not intended to be used to obtain accurate approximations to the zero of but rather to locate an interval containing a zero. This interval can then be used as input to an accurate rootfinder such as . The size of the interval determined depends somewhat unpredictably on the choice of . The closer is to the root and the the initial value of , then, in general, the smaller (more accurate) the interval determined; however, the accuracy of this statement depends to some extent on the behaviour of and on the size of 8 Parallelism and Performance Background information to multithreading can be found in the c05avf is not threaded in any implementation. For most problems, the time taken on each call to will be negligible compared with the time spent evaluating between calls to . However, the initial value of will clearly affect the timing. The closer is to the root, and the the initial value of then the less time taken. (However taking a large can affect the accuracy and reliability of the routine, see below.) You are expected to choose as physically (or mathematically) realistic limits on the interval of search. For example, it may be known, from physical arguments, that no zero of of interest will lie outside . Alternatively, may be more expensive to evaluate for some values of than for others and such expensive evaluations can sometimes be avoided by careful choice of The choice of affects the search only in that these values provide physical limitations on the search values and that the search is terminated if it seems, from the available information about , that the zero lies outside . In this case ( on exit), only one of may have been evaluated and a zero close to the other end of the interval could be missed. The actual interval searched is returned in the arguments and you can call again to search the remainder of the original interval. is intended primarily for determining an interval containing a zero of , it may be used to shorten a known interval. This could be useful if, for example, a large interval containing the zero is known and it is also known that the root lies close to one end of the interval; by setting to this end of the interval and small, a short interval will usually be determined. However, it is worth noting that once any interval containing a zero has been determined, a call to will usually be the most efficient way to calculate an interval of specified length containing the zero. To assist in this determination, the information in and in on successful exit from is in the correct form for a call to routine If the calculation terminates because , then on return is set to . (In fact, on return only in this case.) In this case, there is no guarantee that the value in corresponds to a zero and you should check whether it does. One way to check this is to compute the derivative of at the point , preferably analytically, or, if this is not possible, numerically, perhaps by using a central difference estimate. If ${f}^{\prime }\left({\mathbf{x}}\right)=0.0$ , then must correspond to a multiple zero of rather than a simple zero. 10 Example This example finds a sub-interval of $\left[0.0,4.0\right]$ containing a simple zero of ${x}^{2}-3x+2$. The zero nearest to $3.0$ is required and so we set ${\mathbf{x}}=3.0$ initially. 10.1 Program Text 10.2 Program Data 10.3 Program Results	{"url":"https://support.nag.com/numeric/nl/nagdoc_30.2/flhtml/c05/c05avf.html","timestamp":"2024-11-13T11:19:55Z","content_type":"text/html","content_length":"52096","record_id":"<urn:uuid:4d066292-6503-4628-b627-4506afbe3be2>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00698.warc.gz"}
Life of Fred: Trigonometry Expanded Edition Regular price $59.99 USD Regular price [S: :S] Sale price $59.99 USD Unit price per Sale Sold out Life of Fred offers a Complete Math Education, more mathematics than any other home schooling curriculum we know of. After Beginning Algebra, Advanced Algebra and Geometry, this book completes everything you need for calculus. Angles of elevation, Definition of the sine function, Angles of depression, Area of a triangle = ½ab sin θ, Heron’s formula, Review of graphing and significant digits, Discrete and continuous variables as illustrated in The Merchant of Venice, Tangent function, Why we create new mathematics, Limit of tan θ as θ approaches 90º, Ordinal and cardinal numbers, Cosine function, Graphing y = sin x, Identity function, Contrapositives, Domain and range of a function, Defining 6 to the pi power, Trig angles in standard position, Expanding the domain of a function, Periodic functions, Identities from algebra, Even and odd functions, Trig identities for sine and cosine, for tangent, for secant, Four suggestions for increasing success in solving trig identities, Trig identities for cotangent and cosecant, Nine tricks for solving trig identities, Shortcuts for graphing y = a sin (bx + c), Degrees, minutes, and seconds, Conversion factors, Radians, Videlicet, exempli. gratia, and id est, Area of a segment of a circle, Solving conditional trig equations, Related angles, Joseph Lister, Multiple angle formulas and their proofs, Symmetric law of equality, Probability of finding a right triangle, Law of Cosines, Florence Nightingale, Law of Sines, Inverse functions, One-to-one functions, Hyperbole, Principal values of the inverse trig functions, Ambiguous case for the law of sines, Why sin (2 Arctan 3) equals 3/5, Polar coordinates, Graphing a cardioid and a lemniscate, Codomain of a function, Official definition of the number one, Proof that the square root of 2 is irrational, Transcendental numbers, Complex numbers, Russell’s paradox, Malfatti’s problem and its solution in 1967, r cis θ, de Moivre’s theorem and its proof, The millionth roots of i, Review of the major parts of high school algebra and a preview of all of Calculus. ISBN: 978-1-937032-16-6, hardback, 496 pages. View full details • Shipping Share the details of your shipping policy. • Returns Share the details of your return policy. Pair text with an image to focus on your chosen product, collection, or artist. Add details on availability, style, or even provide a review. Image with text Pair text with an image to provide extra information about your brand or collections.	{"url":"https://bigtreeoflife.com/products/life-of-fred-trigonometry-expanded-edition","timestamp":"2024-11-05T21:32:20Z","content_type":"text/html","content_length":"136743","record_id":"<urn:uuid:f0f854c3-1467-48f9-9395-3f54c5090987>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00211.warc.gz"}
Polynomial Functions - Definition, Formula, Solved Example Problems, Exercise \| Algebra \| Mathematics Polynomial Functions 1. Division Algorithm Given two polynomials f(x) and g(x), where g(x) is not the zero polynomial, there exist two polynomials q(x) and r(x) such that f(x) = q(x)g(x) + r(x) where degree of r(x) < degree of g(x). Here, q(x ) is called the quotient polynomial, and r(x) is called the remainder polynomial. If r(x) is the zero polynomial, then q(x), g(x) are factors of f(x) and f(x) = q(x)g(x). These terminologies are similar to terminologies used in division done with integers. If g(x) = x − a, then the remainder r(x) should have degree zero and hence r(x) is a constant. To determine the constant, write f(x) = (x − a)q(x) + c. Substituting x = a we get c = f(a). Remainder Theorem If a polynomial f(x) is divided by x−a, then the remainder is f(a). Thus the remainder c = f(a) = 0 if and only if x − a is a factor for f(x). In general, if we can express f(x) as f(x) = (x − a)^k.g(x) where g(a) = 0, then the value of k, which depends on a, cannot exceed the degree of f(x). The value k is called the multiplicity of the zero a. Two important problems relating to polynomials are i. Finding zeros of a given polynomial function; and hence factoring the polynomial into linear factors and ii. Constructing polynomials with the given zeros and/or satisfying some additional conditions. To address the problem of finding zeros of a polynomial function, some well known algebraic identities are useful. What is an identity? An equation is said to be an identity if that equation remains valid for all values in its domain. An equation is called conditional equation if it is true only for some (not all) of values in its domain. Let us recall the following identities. 2. Important Identities Method of Undetermined Coefficients Now let us focus on constructing polynomials with the given information using the method of undetermined coefficients. That is, we shall determine coefficients of the required polynomial using the given conditions. The main idea here is that two polynomials are equal if and only if the coefficients of same powers of the variables in the two polynomials are equal.	{"url":"https://www.brainkart.com/article/Polynomial-Functions_33903/","timestamp":"2024-11-12T18:53:17Z","content_type":"text/html","content_length":"67317","record_id":"<urn:uuid:a5344da5-e21f-4e67-a67b-349e0672d707>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00898.warc.gz"}
VTU 21MATCS41 Set-1 Solved Model Question Paper - VTU Updates VTU 21MATCS41 Set-1 Solved Model Question Paper Mathematical Foundations for Computing, Probability & Statistics Computer Science & Allied Engg. branches-21MATCS41 Set-1 Solved Model Question Paper Module 1 1.A] Define tautology. Determine whether the following compound statement is a tautology or not. {(pVq)→r} ⇔ {¬r → ¬(pVq)} 1.B] Using the laws of logic, prove the following logical equivalence [(¬pV¬q) ^ (F[o] V p) ^ p] ⇔ p ^ ¬q. 1.C] Give direct proof and proof by contradiction for the statement “If n is an odd integer then n + 9 is an even integer” 2.A] Test the validity of the arguments using rules of inference. 2.B] Find whether the following arguments are valid or not for which the universe is the set of all triangles. In triangle XYZ, there is no pair of angles of equal measure. If the triangle has two sides of equal length, then it is isosceles. If the triangle is isosceles, then it has two angles of equal measure. Therefore Triangle XYZ has no two sides of equal length. Module 2 3.A] Let f and g be functions from R to R defined by 𝑓(𝑥) = 𝑎𝑥 + 𝑏 𝑎𝑛𝑑 𝑔(𝑥) = 1 − 𝑥 + 𝑥^2 , If (𝑔 ∘ 𝑓)(𝑥) = 9𝑥^2 − 9𝑥 + 3 determine a and b. 3.B] Let A={1,2,3,4,6} and R be a relation on A defined by aRb if and only if ” a is a multiple of b”. Write down the relation R, relation matrix M(R) and draw its digraph. 3.C] Prove that in every graph the number of vertices of odd degree is even. 4.A] The digraph of a relation R defined on the set A={1,2,3,4} is shown below. Verify that (A,R) is a poset and construct the corresponding Hasse diagram. 4.B] compute gof and show that gof is invertible 4.C] Define Graph isomorphism. Determine whether the following graphs are isomorphic or not. Module 3 5.A] Ten competitors in a beauty contest are ranked by two judges A and B in the following order: ID No. of 1 2 3 4 5 6 7 8 9 10 Judge A 1 6 5 10 3 2 4 9 7 8 Judge B 6 4 9 8 1 2 3 10 5 7 Calculate the rank correlation coefficient. 5.B] In a partially destroyed laboratory record, the lines of regression of y on x and x on y are available as 4𝑥 − 5𝑦 + 33 = 0 𝑎𝑛𝑑 20𝑥 − 9𝑦 = 107. Calculate 𝑥̅ 𝑎𝑛𝑑 𝑦̅ and the coefficient of correlation between x and y. 5.C] An experiment gave the following values: v(ft/min) 350 400 500 600 t(min.) 61 26 7 26 It is known that v and t are connected by the relation v=at^b . Find the best possible values of a and b. 6.A] The following table gives the heights of fathers(x) and sons (y): x 65 66 67 67 68 69 70 72 y 67 68 65 68 72 72 69 71 Find the lines of regression and Calculate the coefficient of correlation. 6.B] Fit a parabola y=ax^2 + bx + c for the data x 1.0 1.5 2.0 2.5 3.0 3.5 4.0 y 1.1 1.3 1.6 2.0 2.7 3.4 4.1 6.C] With usual notation, compute means, x̄,Ȳ, and correlation coefficient r from the following lines of regression: 2x + 3y +1=0 and x + 6y – 4=0 Module 4 7.A] A random variable 𝑋 has the following probability function: x -2 -1 0 1 2 3 P(x) 0.1 k 0.2 2k 0.3 k Find the value of k and calculate the mean and variance 7.B] Find the mean and standard deviation of the Binomial distribution. 7.C] In a test on 2000 electric bulbs, it was found that the life of a particular make was normally distributed with an average life of 2040 hours and Standard deviation of 60 hours. Estimate the number of bulbs likely to burn for i. More than 2150 hours ii. Less than 1950 hours iii. Between 1920 and 2160 hours 8.B] 2% of fuses manufactured by a firm are found to be defective. Find the probability that a box containing 200 fuses contains (i) no defective fuses (ii) 3 or more defective fuses (iii) at least one defective fuse. 8.C] In a normal distribution 31% of the items are under 45 and 8% of the items are over 64. Find the mean and S.D of the distribution. Module 5 9.A] The joint distribution of two random variables X and Y is as follows x\y -4 2 7 1 1/8 1/4 1/8 5 1/4 1/8 1/8 Compute the following. (i) E(X) and E(Y) (ii) E(XY) (iii) 𝜎[𝑋] and 𝜎[𝑌] (iv) COV(X,Y) (v) 𝜌(𝑋, 𝑌) 9.B] A coin was tossed 400 times and head turned up 216 times. Test the hypothesis that the coin is unbiased at 5% level of significance. 9.C] A certain stimulus administered to each of the 12 patients resulted in the following change in blood pressure 5, 2, 8, -1, 3, 0, 6, -2, 1, 5, 0, and 4. Can it be concluded that the stimulus will increase the blood pressure? (t.05 for 11 d.f = 2.201) 10.A] Explain the terms: (i) Null hypothesis (ii) Confidence intervals (iii) Type-I and Type-II errors. 10.B] The mean life of 100 fluorescent tube lights manufactured by a company is found to be 1570 hrs with a standard deviation of 120 hrs. Test the hypothesis that the mean lifetime of the lights produced by the company is 1600 hrs at 0.01 level of significance. 10.C] A die is thrown 264 times and the number appearing on the face(x) follows the following frequency distribution.	{"url":"https://vtuupdates.com/solved-model-papers/21matcs41-set-1/","timestamp":"2024-11-14T07:28:18Z","content_type":"text/html","content_length":"181732","record_id":"<urn:uuid:2422008c-9927-42de-b30e-2920b02a954b>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00069.warc.gz"}
The smoothly clipped absolute deviation (SCAD) penalty Variable selection is an important part of high-dimensional statistical modeling. Many popular approaches for variable selection, such as LASSO, suffer from bias. The smoothly clipped absolute deviation (SCAD) estimator attempts to alleviate this bias issue, while also retaining a continuous penalty that encourages sparsity. Penalized least squares A large class of variable selection models can be described under the family of models called “penalized least squares”. The general form of these objective functions is \[Q(\beta) = \frac1n \|\|\mathbf{Y} - \mathbf{X}\boldsymbol{\beta}\|\|_2^2 + \sum\limits_{j = 1}^p p_\lambda(\|\beta_j\|)\] where $\mathbf{X} \in \mathbb{R}^{n\times p}$ is the design matrix, $\mathbf{Y} \in \mathbb{R}^n$ is the vector of response variables, $\boldsymbol{\beta} \in \mathbb{R}^p$ is the vector of coefficients, and $p_\lambda(\cdot)$ is a penalty function indexed by regularization parameter $\lambda$. As special cases, notice that LASSO corresponds to a penalty function of $p_\lambda(\beta) = \lambda \|\|\beta\|\|_1$, and ridge regression corresponds to $p_\lambda(\beta) = \lambda \|\|\beta\|\|_2^2$. Recall the graphical shape of these univariate penalties below. The smoothly clipped absolute deviation (SCAD) penalty, introduced by Fan and Li (2001), was designed to encourage sparse solutions to the least squares problem, while also allowing for large values of $\beta$. The SCAD penalty is part of a larger family known as “folded concave penalties”, which are concave on $\mathbb{R}_+$ and $\mathbb{R}_-$. Graphically, the SCAD penalty looks like this: Somewhat oddly, the SCAD penalty is often defined primarily by its first derivative $p’(\beta)$, rather than $p(\beta)$. Its derivative is \[p'_\lambda(\beta) = \lambda \left\{ I(\beta \leq \lambda) + \frac{(a\lambda - \beta)_+}{(a - 1) \lambda} I(\beta > \lambda) \right\}\] where $a$ is a tunable parameter that controls how quickly the penalty drops off for large values of $\beta$, and the function $f(z) = z_+$ is equal to $z$ if $z \geq 0$, and $0$ otherwise. We can gain some insight by breaking down the penalty function’s derivative for different values of $\lambda$: \begin{cases} \lambda & \text{if } \|\beta\| \leq \lambda \\ \frac{(a\lambda - \beta)}{(a - 1) } & \text{if } \lambda < \|\beta\| \leq a \lambda \\ 0 & \text{if } \|\beta\| > a \lambda \\ \end{cases} Notice that for large values of $\beta$ (where $\|\beta\| > a \lambda$), the penalty is constant with respect to $\beta$. In other words, after $\beta$ becomes large enough, higher values of $\beta$ aren’t penalized more. This stands in contrast to the LASSO penalty, which has a monotonically increasing penalty with respect to $\|\beta\|$: \[p_\text{LASSO}(\beta) = \|\beta\|\] \[p'_\text{LASSO}(\beta) = \text{sign}(\beta)\] Unfortunately, this means that for large coefficient values, their LASSO estimates will be biased downwards. On the other end, for small values of $\beta$ (where $\|\beta\| \leq \lambda$), the SCAD penalty is linear in $\beta$. And for medium values of $\beta$ (where $\lambda < \|\beta\| \leq a \lambda$), the penalty is quadratic. Defined piecewise, $p_\lambda(\beta)$ is \begin{cases} \lambda \|\beta\| & \text{if } \|\beta\| \leq \lambda \\ \frac{a \lambda \|\beta\| - \beta^2 - \lambda^2}{a - 1} & \text{if } \lambda < \|\beta\| \leq a \lambda \\ \frac{\lambda^2(a + 1)}{2} & \text{if } \|\beta\| > a \lambda \\ \end{cases} In Python, the SCAD penalty and its derivative can be defined as follows: def scad_penalty(beta_hat, lambda_val, a_val): is_linear = (np.abs(beta_hat) <= lambda_val) is_quadratic = np.logical_and(lambda_val < np.abs(beta_hat), np.abs(beta_hat) <= a_val * lambda_val) is_constant = (a_val * lambda_val) < np.abs(beta_hat) linear_part = lambda_val * np.abs(beta_hat) * is_linear quadratic_part = (2 * a_val * lambda_val * np.abs(beta_hat) - beta_hat2 - lambda_val2) / (2 * (a_val - 1)) * is_quadratic constant_part = (lambda_val*2 (a + 1)) / 2 * is_constant return linear_part + quadratic_part + constant_part def scad_derivative(beta_hat, lambda_val, a_val): return lambda_val * ((beta_hat <= lambda_val) + (a_val * lambda_val - beta_hat)((a_val lambda_val - beta_hat) > 0) / ((a_val - 1) * lambda_val) * (beta_hat > lambda_val)) Fitting models with SCAD One general approach for fitting penalized least squares models (including SCAD-penalized models) is to use local quadratic approximations. This approach amounts to fitting a quadratic function $q(\ beta)$ around an initial point $\beta_0$ such that the approximation: • Is symmetric about 0, • Satisfies $q(\beta_0) = p_\lambda(\|\beta_0\|)$, • Satisfies $q’(\beta_0) = p’_\lambda(\|\beta_0\|)$. Thus, the approximation function must have the form \[q(\beta) = a + b \beta^2\] for coefficients $a$ and $b$ that don’t depend on $\beta$. The constraints above give us a system of two equations that we can solve: 1. $a + b \beta_0^2 = p_\lambda(\|\beta_0\|)$ 2. $2b \beta_0 = p’_\lambda(\|\beta_0\|)$ For completeness, let’s walk through the solution. Rearranging the second equation, we have \[b = \frac{p'_\lambda(\|\beta_0\|)}{2 \|\beta_0\|}.\] Plugging this into the first equation, we have &a + \frac{p’_\lambda(\|\beta_0\|)}{2 \|\beta_0\|} \beta_0^2 = p_\lambda(\|\beta_0\|) \\ \implies& a = p_\lambda(\|\beta_0\|) - \frac{p’_\lambda(\|\beta_0\|)}{2 \|\beta_0\|} \beta_0^2 Thus, the full quadratic equation is q(\beta) &= p_\lambda(\|\beta_0\|) - \frac{p’_\lambda(\|\beta_0\|)}{2 \|\beta_0\|} \beta_0^2 + \frac{p’_\lambda(\|\beta_0\|)}{2 \|\beta_0\|} \beta^2 \\ &= p_\lambda(\|\beta_0\|) + \frac{p’_\lambda(\|\beta_0\|)}{2 \|\beta_0\|} (\beta^2 - \beta_0^2) \\ Now, for any initial guess $\beta_0$ for the coefficient value, we can constsruct a quadratic estimate of the penalty using $q$ above. Then, it’s much easier to find the minimum of this quadratic compared to the initial SCAD penalty. Graphically, the quadratic approximation looks like this: Putting the quadratic approximation to the SCAD penalty into the the full least-squares objective function, the optimization problem becomes: \[\text{arg}\min_\beta Q(\beta) = \min_\beta \left[\frac1n \|\|Y - X \beta\|\|_2^2 + p_\lambda(\|\beta_0\|) + \frac{p'_\lambda(\|\beta_0\|)}{2 \|\beta_0\|} (\beta^2 - \beta_0^2) \right].\] Ignoring the terms that don’t depend on $\beta$, this minimization problem is equivalent to \[\text{arg}\min_\beta \left[\frac1n \|\|Y - X \beta\|\|_2^2 + \frac{p'_\lambda(\|\beta_0\|)}{2 \|\beta_0\|} \beta^2 \right].\] Cleverly, we can notice that this is a ridge regression problem where $\lambda = \frac{p’_\lambda(\|\beta_0\|)}{2 \|\beta_0\|}$: \[\min_\beta \left[\frac1n \|\|Y - X \beta\|\|_2^2 + \lambda \beta^2 \right].\] Recall that the ridge regression estimator is \[\hat{\beta} = (X^\top X + \lambda I)^{-1} X^\top Y\] where $I$ is the identity matrix. This implies that the approximate SCAD solution is \[\hat{\beta}_{\text{SCAD}} = \left(X^\top X + \left(\frac{p'_\lambda(\|\beta_0\|)}{2 \|\beta_0\|}\right) I\right)^{-1} X^\top Y.\] • Fan, J., Li, R., Zhang, C.-H., and Zou, H. (2020). Statistical Foundations of Data Science. CRC Press, forthcoming. • Fan, Jianqing, and Runze Li. “Variable selection via nonconcave penalized likelihood and its oracle properties.” Journal of the American statistical Association 96.456 (2001): 1348-1360. • In-person lectures from Prof. Jianqing Fan in ORF525. • Kenneth Tay’s blog post on SCAD	{"url":"https://andrewcharlesjones.github.io/journal/scad.html","timestamp":"2024-11-13T05:23:09Z","content_type":"text/html","content_length":"20451","record_id":"<urn:uuid:f62d30eb-4ee9-4a23-8b8b-99e3c90e3c6e>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00402.warc.gz"}
Can someone help me visualize clustering results for my R programming task? \| R Programming Assignment Help by Experts Can someone Discover More me visualize clustering results for my R programming have a peek at this site I need to find the lowest common multiple across all input data. Is is the graph of the clusterable data in R best suited for an image analysis problem? The problem I’ve been having is important link find the highest common multiple of R’s clustering results per input data for my R application. The only way in which I could do that is by working out how many data points are in each input data to process and identify (and parse) the smallest number of clusters. Luckily I’m making this through the data visualization tool available for R and the latest version of mplyr. But, there still stands the potential, and I’d be happier to try it out now. 1- The first approach produces the expected cluster from the data and gives the result near the upper bound of the cluster in our dataset. The outer round-robin approach works well as an upper bound, since the outer round-robin is an outer round-robin. No difference due to the higher complexity as opposed to the square cluster. 2- Once we have found the smallest cluster per input data, determine the highest distance between the outer round-robin and the cluster from the outer round-robin. Find the smallest distance, and run the inner round-robin with a cluster distance of 100. In my last example, I’m using the outer round-robin as my outer round-robin and do not even notice the difference between the outer round-robin and the clusters in this context. Any help on this would be greatly appreciated. I haven’t worked out yet about the correct distance calculation, perhaps because of the difference in complexity between the groups themselves. 2- The main problem is that we have to work through a collection of minigroups or data points to determine the smallest number of clusters related to each input data points. Is there an easier way to do this? Edit: To answer your question I’ve added another link to my current image data visualization tool, I use the R graphics library and find the smallest cluster with hlt 3- Finally, we find the top (smallest) cluster of each input data point and work through the outer round-robin and the find out this here round-robin. From this blogpost, I’ve made my first improvement. I’d like to find the smallest cluster with hlt for an example graph displaying clustering you could try here for my R programming task. The data is created by running the “plot” function from the command line and then using R’s new “fiedata” function: t <- rnorm(10) c('(1 & 2)') ggplot(data/7, aes(x=dat, y=dat, color='r') + geom_point(data=c(1, 1, 1)) + theme(col=white), group=Can someone help me visualize clustering results for my R programming task? A: a) For this matrix, I have a 2x2 matrix $\mathbf{A} = [\mathbf{x}^T - \mathbf{x}^T_n]^{T,n}$ where z is 3 nk points for the vectors $\mathbf{x}^T$ and $\mathbf{x}_n^T$, respectively. I Homepage going to do this again as I am using std::vector to help you visualize them. b) As you can see, \csc() does not correctly plot very “tight” but it is, particularly for the 2×2 scale which I would highly appreciate some help with. Best Site To Pay Someone To Do Your Homework I am still not sure if these are bug or not, go to this site I am not sure how I meant to explain. c) To my knowledge, the R package ROC internally stores ROC plots like this: A: I found my answer yesterday: if you want to show the results for the subset of cases where col-list:contents and cols:contents does not intersect, you need to: a) The ROC curve for the entire subset then covers all the cases of: cols:contents > 50000002351 (col-list:contents is an integer) You have three cases: 1: col-list:contents > 10000000223, cols:contents > 20996655, col-list:contents > 3086011 Can someone help me visualize clustering results for my R programming task? I am quite new in R. Can anyone advise me on how to visualize cluster graphs for my problem? I am using R Studio Version 5.0.2. A: Here look at here now R Data R – http://data.railstutorial.org/data/test/data/ My goal was to get the data to be my dataset when I wanted to look it up graphically. I put a batch plot of code below what I will end up doing for this example so you can see where you need to go. https://pastebin.com/hQ3o3qFb	{"url":"https://rprogramming.help/can-someone-help-me-visualize-clustering-results-for-my-r-programming-task","timestamp":"2024-11-04T01:56:12Z","content_type":"text/html","content_length":"161896","record_id":"<urn:uuid:3f81c7a1-2e49-4c98-9f97-bbe9a38a1c04>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00674.warc.gz"}
Nicolas Templier Research Focus Automorphic forms A major theme in my work has been to show instances of randomness in number theory. A guiding principle is that when an answer to a given problem is difficult (e.g. finding prime numbers), it exhibits randomness according to some probabilistic law (e.g. prime numbers seem to distribute according to the Poisson process). Families of automorphic representations are central to the resolution of certain algebraic and asymptotic questions. I am interested in the arithmetic statistics of families, the Katz-Sarnak heuristics, arithmetic quantum chaos and p-adic families of automorphic forms. I have been also working on the asymptotic properties of special functions motivated by periods and central values. Symplectic geometry and integrable systems provide the relevant tools to investigate the Bessel, Airy and hypergeometric functions and their generalizations notably arising from branching laws of Lie groups representations. In this direction I recently proved the mirror symmetry conjecture for minuscule flag varieties. • On the Ramanujan conjecture for automorphic forms over function fields I. Geometry. (with W. Sawin), Journal of the AMS, 34 no.3 (2021), 653—746. • Families of L-functions and their symmetry (with P. Sarnak and S.W. Shin), proceedings of Simons symposium on the trace formula, Springer-Verlag (2016). • Sato-Tate theorem for families and low-lying zeros of automorphic L-functions (with S.W. Shin), Invent. Math., 203 no.1 (2016), 1-177. • Hybrid sup-norm bounds for Hecke-Maass cusp forms, J. Eur. Math. Soc. 17 no.8 (2015), 2069--2082. • Large values of modular forms, Cambridge Journal of Mathematics, 2 no.1 (2014), 91–116. • A non-split sum of coefficients of modular forms, Duke Math. J., 157 no.1 (2011), 109–165. MATH Courses - Fall 2024 MATH Courses - Spring 2025	{"url":"https://math.cornell.edu/nicolas-templier","timestamp":"2024-11-11T18:11:47Z","content_type":"text/html","content_length":"55930","record_id":"<urn:uuid:5a6634d8-9a66-4501-8e60-0b83414481f6>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00324.warc.gz"}
Craps - True Odds vs Payout In a game with two dice, there are 3 possible ways to throw a 4. So the probability of throwing a 4 is 3/36 = 1/12 = 8.33%. We can express this as odds of 11 to 1. Thus it seems like the true odds of throwing a 4 would be 11:1. However, in craps you always see the true odds of throwing a 4 or 10 stated as "2 to 1". (This is also the payout for pass line/come bet odds.) How do you get from 11:1 to 2:1 odds? I know I'm overlooking something really simple here, so thanks in advance for any help you can give in explaining this to me. Quote: Pocketsidewalk In a game with two dice, there are 3 possible ways to throw a 4. So the probability of throwing a 4 is 3/36 = 1/12 = 8.33%. We can express this as odds of 11 to 1. Thus it seems like the true odds of throwing a 4 would be 11:1. However, in craps you always see the true odds of throwing a 4 or 10 stated as "2 to 1". (This is also the payout for pass line/ come bet odds.) How do you get from 11:1 to 2:1 odds? I know I'm overlooking something really simple here, so thanks in advance for any help you can give in explaining this to me. The odds against rolling a 4 on the next throw is 33:3 = 11:1. The odds of rolling a 7 before a 4 is 6:3 = 2:1. (Therefore, the free odds bet on 4 pays 2:1.) 11/1 = odds for getting a 4 on any one throw 5/1 = odds for getting a 7 on any one throw 11/1 = 12 and 5/1 = 6 12/6 = 2/1 Therefore, the odds are fair at 2/1 when trying to roll "a 4 against a 7". The math in the link below may be easier to read/understand: scroll down to the table "Multi-Roll Bets in Craps" (about 70% down the page) >>>> look at the row that says "Taking Odds 4 and 10" in the bet column >>> to the right there should be three figures ("PROB. WIN", "PROB. PUSH" and "PROB. LOSS") >>> you can then use those figures for working the "fair odds". Quote: Pocketsidewalk In a game with two dice, there are 3 possible ways to throw a 4. So the probability of throwing a 4 is 3/36 = 1/12 = 8.33%. We can express this as odds of 11 to 1. Thus it seems like the true odds of throwing a 4 would be 11:1. However, in craps you always see the true odds of throwing a 4 or 10 stated as "2 to 1". (This is also the payout for pass line/ come bet odds.) How do you get from 11:1 to 2:1 odds? I know I'm overlooking something really simple here, so thanks in advance for any help you can give in explaining this to me. As stated above, Odds bets and Place bets are live until a seven rolls. They are not wagers on the next roll only. That one roll bet is the Hop, and pays accordingly (with some vig deducted). Simplicity is the ultimate sophistication - Leonardo da Vinci Quote: Pocketsidewalk In a game with two dice, two fair 6-sided dice (2d6) Quote: Pocketsidewalk there are 3 possible ways to throw a 4. when the dice are the same color and size if would be hard to tell the difference of a 4 rolled D1.1,D2.3 or D1.2,D2.2 is very easy to figure out Quote: Pocketsidewalk So the probability of throwing a 4 is 3/36 = 1/12 = 8.33%. We can express this as odds of 11 to 1. Thus it seems like the true odds of throwing a 4 would be 11:1. This is not an answer to your question (others answered your question) but many still struggle to this day on the house edge deal In Craps, when allowed, one can bet the very next roll will be an easy 4 - 1,3 (not a hard 4 (2,2) as that is a different Hop bet) So the probability of throwing an easy 4 is 2/36 = 1/18 = 5.56%. We can express this as odds of 17 to 1 AGAINST or 1 chance in 18. Thus it seems like the true odds of throwing an easy 4 would be 17:1 These hop bets pay ONLY 15 to 1 on a win (some pay lower in the US, 14 to 1, and some pay a bit higher outside the US) Right there is the short pay (some call that the house edge) 17 chip winner and gets paid only 15 winsome johnny (not Win some johnny) Quote: ksdjdj 11/1 = odds for getting a 4 on any one throw 5/1 = odds for getting a 7 on any one throw 11/1 = 12 and 5/1 = 6 12/6 = 2/1 Therefore, the odds are fair at 2/1 when trying to roll "a 4 against a 7". The math in the link below may be easier to read/understand: scroll down to the table "Multi-Roll Bets in Craps" (about 70% down the page) >>>> look at the row that says "Taking Odds 4 and 10" in the bet column >>> to the right there should be three figures ("PROB. WIN", "PROB. PUSH" and "PROB. LOSS") >>> you can then use those figures for working the "fair odds". Thanks, this helps a lot. I also found the p/(p+q) formula at WizardofOdds, craps, Appendix 1 "How the House Edge for Each Bet is Derived" to be very helpful.	{"url":"https://wizardofvegas.com/forum/questions-and-answers/math/34328-craps-true-odds-vs-payout/","timestamp":"2024-11-05T06:32:10Z","content_type":"text/html","content_length":"55288","record_id":"<urn:uuid:3e01b589-a0e7-4b1a-bf9c-dff42089d972>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00101.warc.gz"}
Magnetic Induction with Cobra4 Item no.: P2440260 A magnetic field of variable frequency and varying strength is produced in a long coil. The voltages induced across thin coils which are pushed into the long coil are determined as a function of frequency, number of turns, diameter and field strength. • Determination of the induction voltage as a function 1. of the strength of the magneticfield, 2. of the frequency of the magneticfield, 3. of the number of turns of the induction coil, 4. of the cross-section of the induction coil. What you can learn about • Maxwell’s equations • Electrical eddy field • Magnetic field of coils • Coil • Magnetic flux • Induced voltage	{"url":"http://www.sfscientific.com/science/2016/03/11/magnetic-induction-with-cobra4/","timestamp":"2024-11-14T08:42:41Z","content_type":"text/html","content_length":"79180","record_id":"<urn:uuid:90848853-4316-4ce1-892f-86e8781c71cb>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00114.warc.gz"}
Clock and Data Recovery - Wikibooks, open books for an open world This is a reference textbook for readers that desire to complete and to render systematic their knowledge on this subject. This book provides the electronic engineer with the knowledge of: - what the CDR function and structure are - where a CDR is used in a communication network - the mathematical models (just three!) needed to understand the CDR operation (no matter how complex its actual implementation is!). Prior knowledge of calculus and some experience with Fourier and Laplace transforms are required.	{"url":"https://en.m.wikibooks.org/wiki/Clock_and_Data_Recovery","timestamp":"2024-11-11T18:23:33Z","content_type":"text/html","content_length":"30110","record_id":"<urn:uuid:01a10072-dd55-444f-8ab8-1670fc1bcf12>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00179.warc.gz"}
ocuments, Examples and FAQs Legal Terms Dictionary time value of money - Meaning in Law and Legal Documents, Examples and FAQs The time value of money means that a dollar today is worth more than a dollar in the future because you can invest it and earn interest. In normal language you would also say "money over time" instead of "time value of money" Need help understanding your legal documents? 66,230+ legal document summaries created for 20,497 happy customers Upload your document What does "time value of money" mean in legal documents? The term "time value of money" refers to the idea that money available today is worth more than the same amount in the future. This concept is based on the potential earning capacity of money. For instance, if you have $100 today, you can invest it and earn interest, making it worth more in the future. This principle is important in various financial decisions, including investments, loans, and retirement planning. When people talk about the time value of money, they are often considering how much their money can grow over time if invested wisely. In legal documents, understanding the time value of money can help parties make informed decisions. For example, if someone is owed a sum of money in the future, they might want to know how much that amount is worth today. This is especially relevant in cases involving settlements or damages, where the compensation might be paid out over time. By applying the time value of money, individuals can better assess the fairness of a settlement offer or the terms of a loan. Financial planners often use this concept to guide their clients in retirement planning. They calculate how much money needs to be saved today to reach a specific goal in the future, taking into account how investments will grow over time. This helps individuals understand the importance of starting to save early and making smart investment choices. The time value of money emphasizes that the sooner you invest, the more your money can work for you. Moreover, the time value of money is a key factor in loan agreements. When borrowing money, lenders will consider how much interest they will earn over time. This affects the total amount that borrowers will need to repay. Understanding this concept can help borrowers make better choices about loans and understand the true cost of borrowing. In summary, the time value of money is a fundamental financial principle that highlights the importance of timing in financial decisions. It helps individuals and businesses evaluate the worth of money over time, guiding them in making informed choices about investments, loans, and financial planning. What are some examples of "time value of money" in legal contracts? Loan Agreement: "The borrower agrees to repay the loan amount with interest, reflecting the time value of money over the repayment period." Settlement Agreement: "The parties acknowledge that the settlement amount considers the time value of money, ensuring fair compensation for future losses." Investment Contract: "The returns on the investment will be calculated based on the time value of money, allowing for accurate projections of future earnings." Lease Agreement: "The monthly payments are structured to account for the time value of money, ensuring that the total cost reflects the present value of future payments." Retirement Plan Document: "The contributions to the retirement plan will be evaluated using the time value of money to determine the expected growth by retirement age." Real Estate Purchase Agreement: "The purchase price takes into account the time value of money, reflecting the appreciation of property value over time." Insurance Policy: "The payout from the insurance policy will consider the time value of money, ensuring beneficiaries receive a fair amount based on current value." Divorce Settlement Agreement: "The division of assets will factor in the time value of money, ensuring that both parties receive equitable compensation based on future earning potential." FAQs about "time value of money" What is the time value of money? The time value of money is a financial concept that says money you have now is worth more than the same amount in the future. This is because money can earn interest or grow over time. So, having $100 today is better than getting $100 a year from now. Why is the time value of money important? Understanding the time value of money is important because it helps you make better financial decisions. It shows you how to evaluate investments, savings, and loans by considering how much money will be worth in the future compared to today. How does the time value of money affect investments? The time value of money affects investments by helping you understand how much your money can grow over time. If you invest money today, it can earn interest or increase in value, making it worth more in the future. This concept helps you choose the best investment options. What does "discounting" mean in the time value of money? Discounting is the process of determining how much future money is worth today. It involves reducing the future amount by a certain interest rate to find its present value. This helps you understand how much you should invest today to reach a specific amount in the future. How can I calculate the time value of money? You can calculate the time value of money using formulas or financial calculators. The basic formula involves the present value (PV), future value (FV), interest rate (r), and time (t). A simple formula is: [ PV = \frac{FV}{(1 + r)^t} ] This helps you find out how much money you need to invest today to reach a future goal. Who uses the time value of money concept? The time value of money concept is used by investors, financial analysts, and anyone making financial decisions. It helps individuals and businesses evaluate loans, investments, and savings plans to maximize their financial outcomes. What is the difference between present value and future value? Present value is the current worth of a sum of money that you will receive in the future, while future value is how much that sum will be worth at a specific time in the future, considering interest or growth. Understanding both helps you make informed financial choices. How does inflation affect the time value of money? Inflation reduces the purchasing power of money over time. This means that the same amount of money will buy you less in the future. When considering the time value of money, it's important to account for inflation to understand the real value of your money in the future. Can I apply the time value of money to my personal finances? Yes, you can apply the time value of money to your personal finances by evaluating savings accounts, loans, and investments. Understanding this concept can help you make smarter choices about saving for retirement, buying a home, or investing in education.	{"url":"https://www.legalbriefai.com/legal-terms/time-value-of-money","timestamp":"2024-11-12T18:29:04Z","content_type":"text/html","content_length":"61567","record_id":"<urn:uuid:4b9dc5c4-6250-4407-a6a8-0cfaafcf42df>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00634.warc.gz"}
Ch. 14 Summary - Principles of Finance \| OpenStax 14.1 Correlation Analysis Correlation is the measure of association between two numeric variables. A correlation coefficient called r is used to assess the strength and direction of the correlation. The value of r is always between $-1-1$ and $+1+1$. The size of the correlation r indicates the strength of the linear relationship between the two variables. Values of r close to $-1-1$ or to $+1+1$ indicate a stronger linear relationship. A positive value of r means that when x increases, y tends to increase and when x decreases, y tends to decrease (positive correlation). A negative value of r means that when x increases, y tends to decrease and when x decreases, y tends to increase (negative correlation). 14.2 Linear Regression Analysis Linear regression analysis uses a straight-line fit to model the relationship between the two variables. Once a straight-line model is developed, this model can then be used to predict the value of the dependent variable for a specific value of the independent variable. Two parameters are calculated for the linear model, the slope of the best-fit line and the y-intercept of the best-fit line. The method of least squares is used to generate these parameters; this method is based on minimizing the squared differences between the predicted values and observed values for y. 14.3 Best-Fit Linear Model Once a correlation has been deemed significant, a linear regression model is developed. The goal in the regression analysis is to determine the coefficients a and b in the following regression equation: $y^=a+bxy^=a+bx$. Typically some technology, such as Excel, R statistical tool, or a calculator, is used to generate the coefficients a and b since manual calculations are cumbersome. 14.4 Regression Applications in Finance Regression analysis is used extensively in finance-related applications. Many typical applications involve determining if there is a correlation between various stock market indices such as the S&P 500, the DJIA, and the Russell 2000 index. The procedure is to first generate a scatter plot to determine if a visual trend is observed, then calculate a correlation coefficient and check for significance. If the correlation coefficient is significant, a linear model can then be generated and used for predictions. 14.5 Predictions and Prediction Intervals A key aspect of generating the linear regression model is to then use the model for predictions, provided that the correlation is significant. To generate predictions or forecasts using the linear regression model, substitute the value of the independent variable (x) in the regression equation and solve the equation for the dependent variable (y). When making predictions using the linear model, it is generally recommended to only predict values for y using values of x that are in the original range of the data collection. 14.6 Use of R Statistical Analysis Tool for Regression Analysis R is an open-source statistical analysis tool that is widely used in the finance industry and can be found online. R provides an integrated suite of functions for data analysis, graphing, and correlation and regression analysis. R is increasingly being used as a data analysis and statistical tool because it is an open-source language and additional features are constantly being added by the user community. The tool can be used on many different computing platforms.	{"url":"https://openstax.org/books/principles-finance/pages/14-summary","timestamp":"2024-11-04T18:43:48Z","content_type":"text/html","content_length":"410766","record_id":"<urn:uuid:9b4aa109-baa7-4cd2-8471-e7dded85b5bc>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00462.warc.gz"}
Inventory Turnover Ratio Updated on 2023-08-29T12:00:38.797313Z What is the inventory turnover ratio? Inventory turnover ratio is an accounting ratio that establishes a relationship between revenue cost and the average inventory carried during an accounting period. It is an accounting ratio and is also commonly known as a stock turnover ratio. The inventory turnover ratio is an important ratio for businesses as it explains how much stock carried by them is converted into sales. Simply put, the ratio describes the number of times the company sells its inventory during a given period. The ratio is also becoming increasingly popular due to its importance in the ecommerce sector. Additionally, retail firms also use the ratio to keep a tab on their businesses. Turnover management is an essential step towards warehouse management. How is the ratio calculated? The formula for inventory turns ratio is depicted below: In the above formula, the first step for the calculation is finding a time frame to measure the inventory turnover ratio. Now, one must calculate the cost of goods sold. This requires the stock count of the inventory at the beginning and end of the month. The following formula can be used to find the COGS: COGS can also represent revenue from operations, which refers to sales. Thus, sales can also be used in place of COGS to calculate inventory turnover ratio. Therefore, the cost of sales is the actual value of inventory converted to sales. The second part of the formula involves finding out average inventory, which can be calculated using a simple formula. This involves adding the inventory stock at the beginning and end of the respective cycle then dividing it by 2. The number obtained from the formula of inventory turnover ratio depicts the number of times the inventory is turned over within the time frame for which inventory change was measured. What is an example of inventory turnover ratio? Consider a firm that wants to calculate inventory turnover ratio for the period of one month. The firm has the following information: Total cost of goods at the starting of the month is 100 * 8,000, which is equal to $800,000. This is known as beginning inventory of the month. It is a good practice for companies to take inventory count by the end of the month again. Thus, the total cost of goods at the end of the month is 100* 7000, equal to $700,000. This is known as the ending inventory of the month. Now the invoices suggest that a total of 7700 units of the good were purchased during this period of one month at the rate of $100 per unit. This amounts to a total of $770,000. COGS can be calculated as: beginning inventory cost (800,000) + purchases (770,000) – ending inventory cost (700,000), this is equal to $870,000. Therefore, COGS = $870,000. Average inventory cost can be calculated simply by taking the average of beginning and ending inventory. Thus, average inventory cost = (800,000+ 700,000)/2 = $750,000. Finally, total inventory turnover can be calculated by dividing COGS by average inventory cost. This gives the value 870,000/750,000 = 1.16. This means that the company turns in the inventory once every month. What does the ratio signify? The inventory turns ratio signifies how many times the inventory is turned in within a specific period, which was one month in the above case. This ratio can be compared with competitor firms to understand better how a company is performing. Suppose that a competitor of the firm has an inventory turns ratio of 2.5, then clearly the other firm is selling more products than the concerned firm. However, it is important to note that this ratio alone cannot deem a business as inefficient or worse than its competitors. Other metrics may turn out to be higher for a firm with a lower inventory turns ratio. Additionally, inventory turns ratio can be used to calculate how many days its takes for a firm to finish its inventory. This can simply be achieved by dividing 365 by the inventory turnover ratio. In the above case it would take 314 days for the firm to completely clear its inventory. Why is the inventory turnover ratio important? A high inventory turnover ratio signifies how successful a company is in turning stock into sales. It also suggests that the company is actively partaking in inventory control measures, coupled with a strong sales policy. Low turnover ratio points to lack of demand for the company’s goods. It may also hint at outdated products being sold by the company. This could mean a piling up of stocks in the future for the company. Moreover, the quality of a good may deteriorate over time as it is kept it storage. An ideal inventory ratio for a company can vary depending on the size and type of company it is. However, it is a good policy for a company to clean out its inventory 12-13 times a year. What are the limitations of this ratio? Despite its many advantages, the inventory turns ratio comes with certain limitations too. The time taken by a company to sell its inventory can vary greatly depending on the type of product that is sold. Thus, having a fair idea about the inventory turns for the respective industry can give a better idea about how well the business is doing. Consider the case of fast-moving goods wherein its easier for firms to make sales as goods are lower priced and come with a shelf life. Consequently, if a retail chain selling everyday supplies has a high inventory turns ratio, then it does not necessarily mean that the firm is performing exceedingly well. It is just the nature of the industry that the firm is a part of. On the contrary, consider the case of real estate sector, wherein a property may take time to get sold. However, from the sale of a single property, high amount of revenue can be generated.	{"url":"https://kalkine.com.au/glossarydetail/inventory-turnover-ratio/","timestamp":"2024-11-12T04:24:59Z","content_type":"text/html","content_length":"75184","record_id":"<urn:uuid:4b060596-902e-43a5-bcb2-9af71e9b8a37>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00118.warc.gz"}
Nearly spanning regular subgraphs Nearly spanning regular subgraphs Petersen's theorem asserts that every regular graph of even degree contains a 2-factor (i.e. a spanning 2-regular subgraph). Iterating this easy result we find that for any pair of positive even [A] N. Alon, Problems and results in extremal combinatorics, J, Discrete Math. 273 (2003), 31-53. [AFK] N. Alon, S. Friedland and G. Kalai, Regular subgraphs of almost regular graphs, J. Combinatorial Theory, Ser. B 37(1984), 79-91. indicates original appearance(s) of problem.	{"url":"http://www.openproblemgarden.org/op/nearly_spanning_regular_subgraphs","timestamp":"2024-11-01T19:14:33Z","content_type":"application/xhtml+xml","content_length":"15897","record_id":"<urn:uuid:eb8d0d43-7ccb-4073-903e-ad0887bc977d>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00880.warc.gz"}
Interactive Learner Guide - PapaCambridge - Free Download PDF InteractiveInteractive Learner GuideCambridge IGCSETM / Cambridge IGCSETM(9–1)Mathematics 0580 / 0980For examination from 2020 In order to help us develop the highest quality resources, we are undertaking a continuous programme ofreview; not only to measure the success of our resources but also to highlight areas for improvement and toidentify new development needs.We invite you to complete our survey by visiting the website below. Your comments on the quality andrelevance of our resources are very important to us.www.surveymonkey.co.uk/r/GL6ZNJBWould you like to become a Cambridge consultant and help us develop support materials?Please follow the link below to register your for/teachers/teacherconsultants/ Copyright UCLES 2018Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment isthe brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of theUniversity of Cambridge.UCLES retains the copyright on all its publications. Registered Centres are permitted to copy material from this booklet fortheir own internal use. However, we cannot give permission to Centres to photocopy any material that is acknowledged to athird party, even for internal use within a Centre.2 ContentsAbout this guide4Section 1: Syllabus content – what you need to know about5Section 2: How you will be assessed6Section 3: What skills will be assessed11Section 4: Example candidate response13Section 5: Revision243 About this guideThis guide introduces you to your Cambridge IGCSE Mathematics course and how you will be assessed. You should usethis guide alongside the support of your teacher.It will help you to:99 understand what skills you should develop by taking this course99 understand how you will be assessed99 understand what we are looking for in the answers you write99 plan your revision programme99 revise, by providing revision tips and an interactive revision checklist (Section 6).Key benefitsThe course will help you to build your skills and knowledge across a range of mathematical techniques. You will be ableto develop your problem solving and reasoning skills in a variety of situations.The Extended course will provide you with a strong foundation to continue to study mathematics qualifications beyondIGCSE. The Core course will equip you with skills needed to support your learning in other subjects and in your generalworking life.4 Section 1: Syllabus content – what you need to know aboutThis section gives you an outline of the syllabus content for this course. Only the top-level topics of the syllabushave been included here, which are the same for both the Core and Extended courses. In the ‘overview’ column youare given a very basic idea of what each topic covers. Highlighted cells show Extended-only content.Learners taking the Extended course need to know all of the Core content as well as some extra content.This extra content requires learners to explore topics and sub-topics of the Core syllabus in more detail, tocover some more complex techniques, and to learn new sub-topics.Ask your teacher for more detail about each topic, including the differences between the Core and Extendedcourses. You can also find more detail in the revision checklists in Section 6 of this guide.TopicOverviewNumberNumber, sets and Venn diagrams, squares and cubes, directed numbers, fractions,decimals and percentages, ordering, indices, ‘four rules’, estimates, bounds, ratio,proportion, rate, percentage, time, money and finance.Growth and decay (Extended only).Algebra and graphsBasic algebra, algebraic manipulation, equations, formulae sequences, drawing,sketching and interpreting graphs of functionsAlgebraic fractions, harder simultaneous equations, proportion, linearprogramming, functions, gradients of curves, derived functions and differentiation(Extended only).Co-ordinate geometryStraight-line graphsVectors and transformationsVectors (column), transformationsMagnitude of a vector, represent vectors by directed line segments, positionvectors (Extended only).GeometryLanguage, construction, symmetry, angle properties, congruence, similarityMensurationMeasures, mensurationTrigonometryBearings, trigonometry in right-angled trianglesSine rule, cosine rule, trig graphs, solving simple trig equations (Extended only).ProbabilityProbabilityConditional probability (Extended only).StatisticsStatisticsMake sure you always check the latest syllabus, which is available at www.cambridgeinternational.org5 Section 2: How you will be assessedYou will be assessed at the end of the course using two written examinations. The papers that you will sit aredifferent for the Core and Extended courses.CoreExtended Paper 1 – Short-answer questions Paper 2 – Short-answer questions Paper 3 – Structured questions Paper 4 – Structured questionsMake sure you find out from your teacher which course you will be following.Components at a glanceThe table summarises the key information about each component.ComponentCoreHow long andhow many marksPaper 11 hour(Short-answerquestions)56 marksPaper 3(Structuredquestions)2 hoursExtended Paper 2(Short-answerquestions)Paper 4(Structuredquestions)Skills assessedMathematical techniques as listed inthe Core syllabus, and applying thosetechniques to solve problems.104 marksPercentage of thequalification35%65%1 hour 30 minutesMathematical techniques as listed inthe Core and Extended syllabus, andapplying those techniques to solve2 hours 30 minutes problems.70 marks130 marks635%65% About the componentsIt is important that you understand the different types of question in each paper, so you know what to expect.Core: Paper 1 (Short-answer questions) and Paper 3 (Structured questions)You need to answer all questions on each paper.Paper 3Paper 1The number of marksfor each part is shown.Write your working andanswers in the spaces provided.You can use an electronic calculatorin both papers. Ask your teacher torecommend a suitable calculator.Paper 1 contains lots of shortanswer questions. These are usuallyworth 1–3 marks each. Some mightbe broken up into two parts.Paper 3 contains structured questions. Eachquestion is split into many parts, with eachpart usually being worth 1–4 marks. Here forexample, question 1 is split over two pages.Often the answers to later parts will dependon the answers to earlier parts.7 Extended: Paper 2 (Short–answer questions) and Paper 4 (Structured questions)You need to answer all questions on both papers.Paper 2Paper 4The number of marksfor each part is shown.Write your working andanswers in the spaces provided.You can use an electronic calculatorin both papers. Ask your teacher torecommend a suitable calculator.Paper 2 questions are short-answerquestions. Most questions areworth 1–3 marks, with some beingworth 4 or 5 marks. Some questionsmight be broken up into two parts.Paper 4 contains structured questions. Eachquestion is split into many parts, with eachpart usually being worth 1–6 marks. Here forexample, question 2 is split over two pages.Often the answers to later parts will dependon the answers to earlier parts.8 General advice for all Papers1. Read the questions carefully to make sure that youunderstand what is being asked.Make sure that you give your answer in the formasked for in the question, e.g. some questions askfor answers to be given in terms of π. For lengths,areas and volumes, give answers in decimals (notin surds or in terms of π) unless you are told togiven an exact answer.2. Give your answers to the accuracy indicated inthe question. If none is given, and the answer isn’texact, then: give your answer tothree significant figures12.3 12.298 x if the answer is in degrees, then give it toone decimal placeUse the value of π from your calculator, or use3.142, which is given on the front page of thequestion paper.23.1 23 x3. Include units with your answers if they are notgiven on the paper. For example, 1 kg of applescosts 1.20 1.20 xYou can gain marks for the correct working even ifyou have an incorrect answer, or cannot completethe whole question.4. Show your working. Show as much working asyou can for all your questions.5. If you make a mistake, clearly cross out anyworking or answers that you do not want theexaminer to mark.If you need more space, ask for extra of paperand clearly indicate where the rest of the answeris written. On the additional paper, make it clearwhich questions(s) you are answering.Equipment for the examMake sure you have: a blue or black pen (a spare pen is always a good idea) a pencil (for graphs and diagrams) an electronic calculator a protractor a pair of compasses a ruler.Timing If you are stuck on a question, don't waste too much time trying to answer it – go on to the next question andcome back to the one you are stuck on at the end. Use any time that you have left at the end of the exam to go back and check your answers and working.9 Section 3: What skills will be assessedThe areas of knowledge, understanding and skills that you will be assessed on are called assessment objectives(AOs). There are two AOs for this course.AO1Demonstrate knowledge andunderstanding of mathematicaltechniquesAO2Reason, interpret and communicatemathematically when solvingproblemsAO1 Demonstrate knowledge and understanding of mathematical techniquesYou need to show that you can recall and apply mathematical knowledge, terminology and definitions to carry outsingle or multi-step solutions in mathematical and everyday situations.This means that you need to show that you can: organise, process and present information accurately inwritten, tabular, graphical and diagrammatic formsUse tables, graphs and diagrams use and interpret mathematical notation correctly perform calculations and procedures by suitable methods,including using a calculator understand systems of measurement in everyday use andmake use of these estimate, approximate and work to degrees of accuracyappropriate to the context and convert between equivalentnumerical formse.g. a pair of compasses, a protractor and a ruler. use geometrical instruments to measure and to draw to anacceptable degree of accuracy recognise and use spatial relationships in two and threedimensions.AO1 is assessed in all papers.10An example of 'degress of accuracy' includesignificant figures or decimal places.An example of converting between'equivalent numerical forms' includebetween fractions, decimals andpercentages; or between normal numbersand standard form. AO2 Reason, interpret and communicate mathematically when solving problemsYou need to demonstrate that you can analyse a problem, select a suitable strategy and apply appropriatetechniques to obtain a solution.This means that you need to show that you can: make logical deductions, make inferences and drawconclusions from given mathematical dataRecognise and extent patterns recognise patterns and structures in a variety ofsituations, and form generalisations present arguments and chains of reasoning in a logicaland structured way interprete and communicate information accurately andchange from one form of presentation to another assesses the validity of an argument and criticallyevaluate a given way of presenting informationTake information and organise it to answera problem. solve unstructured problems by putting them into astructured form involving a series of processes apply combinations of mathematical skills andtechniques using connections between different areas ofmathematics in problem solving interprete results in the context of a given problem andevaluate the methods used and solutions obtained.AO2 is assessed in all papers.11 Cambridge IGCSE Mathematics 0580 syllabus for 2020, 2021 and 2022. Details of the assessmentSection 4: Command wordsA command word is the part of the question that tells you what you need to do with your knowledge. For example,Commandyoumight need towordsdescribe something, explain something, argue a point of view or list what you know. The tablebelow includes command words used in the assessment for this syllabus. The use of the command word(s) within anThe table below includes command words used in the assessment for this syllabus. The use of the command wordquestion will relate to the context.will relate to the subject context.Command wordWhat it meansCalculatework out from given facts, figures or information, generally using a calculatorConstructmake an accurate drawingDescribestate the points of a topic/give characteristics and main featuresDetermineestablish with certaintyExplainset out purposes or reasons/ make the relationships between things evident/ provide whyand/or how and support with relevant evidenceGiveproduce an answer from a given source or recall/memoryPlotmark point(s) on a graphShow (that)provide structured evidence that leads to a given resultSketchmake a simple freehand drawing showing the key featuresWork outcalculate from given facts, figures or information with or without the use of a calculatorWritegive an answer in a specific formWrite downgive an answer without significant workingThe question below is taken from Paper 4 and illustrates the use of two command words.The command words ‘Write down’ indicates that youdo not need to show your working, and the answershould just be written down. The mark allocation [1]also supports this.The command words ‘Show that’ indicate thatyou need to provide evidence in the form of a clearmathematical explanation, to demonstrate that youknow how to obtain the given result. In other words,you need to show a method that leads to the result.The answer space in this case does not contain adotted answer line as there is no single ‘answer’ tobe found. Your working that leads to the given resultshould be written in the blank working space.12 Section 5: Example candidate responseThis section takes you through an example question and learner response from one of the 2020 specimen papersfor this course. It will help you to identify the command words and other key instructions within questions and tounderstand what is required in your response.All information and advice in this section is specific to the example question and responsebeing demonstrated. It should give you an idea of how your responses might be viewed by anexaminer but it is not a list of what to do in all questions. In your own examination, you willneed to pay careful attention to what each question is asking you to do.This section is structured as follows:QuestionThe command words and instructions in the question havebeen highlighted and explained. This should help you tounderstand clearly what is required by the question.Mark schemeThis tells you as clearly as possible what an examiner expectsfrom an answer to award marks.Example candidate responseThis an exemplar answer written in the style ofa high level candidate.How the answer could have been improvedThis summarises what could be done to gain more marks.Common mistakesThis will help you to avoid common mistakes made bycandidates. So often candidates lose marks in their examsbecause they misread or misinterpret the questions.13 QuestionThe question used in this example is from Specimen Paper 3 (Core). It represents the type of structured questionyou will see in both Paper 3 (Core) and Paper 4 (Extended). A structured question means that it is broken intoseveral parts. Often, later parts will depend on your answers to earlier parts.Work out indicatesthat the answer shouldbe calculated from thegiven information; someasurement will notscore marks.This means that youcannot find the answer bymeasuring the diagram.The allocation of 1 markindicates that the answercan be obtained withminimum working.Give reasons for your answer indicates that your answershould be supported with reasons using the correctmathematical terminology. You must justify your answer.Use trigonometry to calculate indicates that you shoulduse this method with the given information to find theanswer, and that a calculator is needed to solve theproblem. If you did not use trigonometry, you would notbe awarded any marks.14 Show that indicates the answer is given and youneed to write down all of the steps in a methodthat leads to the given answer. You need toprovide evidence that you understand and knowhow the answer is reached.Work out indicates that you should calculatethe answer from the given information. The markallocation of 4 marks suggests that you will needto include several steps of working in order to getto the answer.15 Learner GuideMark schemeYour examination papers will be marked and the overall number of marks for each paper will be recorded. Your marks across all papers will then be converted to a grade.Final answer: This value is what the examiner expects to see. The answer has to be exactly as given in the mark scheme, unless there are acceptable alternatives. The markscheme will always make it clear if there are acceptable alternative answers.Method marks: Sometimes method marks are awarded for lines of working, as well as for the final answer. This means that you could get the final answer incorrect but stillget some marks if you include the correct working. The mark scheme does not include all possible methods, so if you use a method not included in the mark scheme but it isaccurate and relevant, then the examiner will still award marks for the appropriate parts of the working – unless the questions asks you to use a specific method.Answer(a)(i) 35(a)(ii) 74example of afinal answer(b) 43 and correct mathematicalreasonsMarkNotes1This is the only acceptable answer for this part of the question.1This is the only acceptable answer for this part of the question.3Two marks are awarded for the final answer of 43 .The third mark is awarded for a fully correct reason, for example, 'angles on a straight line add up to 180 and 'angles in atriangle add up to 180 '. There are other correct reasons that could be also be used.If 43 is not obtained, one method mark can be awarded if the following calculation is seen in the working:180 – 128 or 128 – 85(c) 32.2 or 32.23 2This is the only acceptable answer for this part of the question.The answer has to be rounded correct to three significant figures, or can be given with more figures in the answer.Those that did not get this answer can score one method mark for showing the following in their working:sin 8 15(d)(i)3002 22522This does not have to be shown in one step, as long as the method shown is the same as this overall. Those that do notshow this can have one method mark for showing the following in their working:300 225 (d)(ii) 15 354example of a method markThe answer 3 35pm is also acceptable for 4 marks.If the correct answer is not found, one method mark is available for showing 375 450 in their working; and a secondmethod mark can be awarded for sight of them multiplying their answer to this by 60 to change it to hours. A thirdmethod mark can be awarded for adding their answer in hours to 14 45 – this shows the correct method, so only one markis lost for an incorrect final answers.16 Example candidate responseNow let’s look at the sample candidate’s response to question 8. The examiner's comments are in the orange boxes.The candidate was awarded 8 marks out of 13.0 out of 1The candidate's working suggests theyunderstand that the angle sum of a triangleis 180 but they did not include bracketsaround '74 71'. The answer 177 is notsensible for this question.180 – 74 71 1771771 out of 1The candidate recognises that thereare parallel lines, and that angle y is acorresponding angle to angle 74 .742 out of 3The candidate has given a correct answer forw and shown correct working in two steps.They have given a correct reason for 52 usingcorrect mathematical languge. But they havenot provided a reason to explain the angle 43 .T Extended Paper 2 (Short-answer questions) 1 hour 30 minutes 70 marks Mathematical techniques as listed in the Core and Extended syllabus, and applying those techniques to solve problems. 35% Paper 4 (Structured questions) 2 hours 30 minutes 130 marks 65% Section 2: How you will be assessed Core Extended	{"url":"https://zbook.org/read/32bf_interactive-learner-guide-papacambridge.html","timestamp":"2024-11-07T19:22:33Z","content_type":"text/html","content_length":"95166","record_id":"<urn:uuid:44fbfb2f-3b7e-41d5-ad58-8f54af24fa08>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00556.warc.gz"}
Applied Mathematics Unveiling the world through math: The power of applied mathematics. Applied mathematics is like a Swiss Army knife for science. It tackles real-world challenges in fields as diverse as finance, engineering, biology, and medicine. To master this powerful discipline, students build a versatile toolbox of mathematical methods and techniques. During their undergraduate studies, students will focus on two main areas: • Mathematical Modeling: This is where the magic happens. We transform real-world scenarios into equations to make predictions or gain deeper insights. Imagine figuring out the movement of a falling object with an equation, or using math functions to model how populations grow. • Computational Mathematics: Here, computers become our allies. We use their power to solve complex math problems and analyse massive datasets. This combines math theory, clever problem-solving steps (algorithms), and computer science to crack problems that would be impossible by hand. Succeeding in applied mathematics requires a well-rounded approach: • Strong foundation in pure mathematics: This is the core language of applied math. Just like you need grammar to write well, you need pure math to solve complex problems. • Scientific programming skills: Just like a mechanic needs tools, applied mathematicians need to know how to use computers to solve problems and analyse data. • Branching out beyond math: Applied math is a team player! Understanding physics helps translate real-world problems about forces, motion, and energy into math models. Similarly, biology knowledge is crucial for modeling things like population changes or disease outbreaks. By combining these elements, applied mathematics equips students to not just solve problems, but to truly understand the world around them. Careers and Research While job titles might not always say "applied mathematician," graduates leverage their skills in a wide range of exciting fields. Students wondering what kind of career they can have with a degree in applied mathematics should take a look at this brochure by the Society for Industrial and Applied Mathematics (SIAM). Our department contributes to research in areas that align with the diverse applications of applied mathematics, including: • Machine Learning • Mathematical Biology • Numerical Analysis and Scientific Computing • Mathematics Education Please note if you planning on majoring in Applied Mathematics you are required to have 30 credits at first year level, 40 credits at second year level and 60 credits at third year level. In other words you need all the modules listed below for first to third year. First Year MAPV101 Graph Theory MAPV111 Mathematical Modelling MAPV102 Mechanics MAPV112 Numerical Methods I Note : All Applied Mathematics students are required to complete an introductory scientific programming course currently hosted in WRSC111, and both first year Pure Mathematics modules in their first Second Year MAPV201 Differential Equations MAPV211 Numerical Methods II MAPV202 Transform Theory MAPV222 Linear Optimization Third Year MAPV301 Partial Differential Equations MAPV311 Finite Difference Methods MAPV302 Nonlinear Optimization MAPV312 Dynamical Systems MAPM411 Finite Element Methods MAPM413 Graph Theory MAPM414 Continuum Mechanics MAPM415 Mathematical Control Theory MAPM417 Capita Selecta MAPM421 Biomathematics	{"url":"https://maths.mandela.ac.za/Applied-Mathematics","timestamp":"2024-11-11T08:24:37Z","content_type":"application/xhtml+xml","content_length":"32808","record_id":"<urn:uuid:55d0e4ac-aa69-41c2-a0c1-b4103cc69fe5>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00625.warc.gz"}
How Do You Write an Equation of a Line in Slope-Intercept Form If You Have the Slope and the Y-Intercept? Want to find the slope-intercept form of a line when you're given a point on that line and another line perpendicular to that line? Remember, perpendicular lines have slopes that are opposite reciprocals of each other. In this tutorial, you'll see how to find the slope using the slope of the perpendicular line. Then, use this slope and the given point to write an equation for the line in slope-intercept form. Check it out!	{"url":"https://virtualnerd.com/common-core/hsa-algebra/HSA-CED-/A/2/slope-intercept-line-direct-example","timestamp":"2024-11-02T09:02:47Z","content_type":"text/html","content_length":"28507","record_id":"<urn:uuid:23638d5d-a603-4da7-a548-911efe233352>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00330.warc.gz"}
The tide is the cyclic rising and falling of Earth's ocean surface caused by the tidal forces of the Moon and the Sun acting on the Earth. Tides cause changes in the depth of the sea, and also produce oscillating currents known as tidal streams, making prediction of tides important for coastal navigation (see Tides and navigation, below). The strip of seashore that is submerged at high tide and exposed at low tide, the intertidal zone, is an important ecological product of ocean tides. The changing tide produced at a given location on the Earth is the result of the changing positions of the Moon and Sun relative to the Earth coupled with the effects of the rotation of the Earth and the local bathymetry (the underwater equivalent to topography or terrain). Though the gravitational force exerted by the Sun on the Earth is almost 200 times stronger than that exerted by the Moon, the tidal force produced by the Moon is about twice as strong as that produced by the Sun. The reason for this is that the tidal force is related not to the strength of a gravitational field, but to its gradient. The field gradient decreases with distance from the source more rapidly than does the field strength; as the Sun is about 400 times further from the Earth than is the Moon, the gradient of the Sun's field, and thus the tidal force produced by the Sun, is weaker. Tidal terminology The maximum water level is called "high tide" or "high water" and the minimum level is "low tide" or "low water." If the ocean were a constant depth, and there were no land, high water would occur as two bulges in the height of the oceans--one bulge facing the Moon and the other on the opposite side of the earth, facing away from the Moon. There would also be smaller, superimposed bulges on the sides facing toward and away from the Sun. For an explanation see below under Tidal physics. At any given point in the ocean, there are normally two high tides and two low tides each day just as there would be for an earth with no land; however, rather than two large bulges propagating around the earth, with land masses in the way the result is many smaller bulges propagating around amphidromic points, so there is no simple, general rule for predicting the time of high tide from the position of the Moon in the sky. The common names of the two high tides are the "high high" tide and the "low high" tide; the difference in height between the two is known as the "daily inequality." The daily inequality is generally small when the moon is over the equator. The two low tides are called the "high low" tide and the "low low" tide. On average, high tides occur 12 hours 24 minutes apart. The 12 hours is due to the Earth's rotation, and the 24 minutes to the Moon's orbit. This is the "principal lunar semi-diurnal" period, abbreviated as the M2 tidal component, and it is, on average, half the time separating one lunar zenith from the next. The M2 component is usually the biggest one, but there are many others as well due to such complications as the tilt of the earth's axis and the inclination of the lunar orbit. The lunar cycle is what is tracked by tide clocks. The time between high tide and low tide, when the water level is falling, is called the "ebb." The time between low tide and high tide when the tide is rising, is called "flow," or "flood." At the times of high tide and low tide, the tide is said to be "turning," also slack tide. The height of the high and low tides (relative to mean sea level) also varies. Around new and full Moon when the Sun, Moon and Earth form a line (a condition known as syzygy), the tidal forces due to the Sun reinforce those of the Moon. The tides' range is then at its maximum: this is called the "spring tide," or just "springs" and is derived not from the season of spring but rather from the verb "to jump" or "to leap up." When the Moon is at first quarter or third quarter, the Sun and Moon are at 90° to each other and the forces due to the Sun partially cancel out those of the Moon. At these points in the Lunar cycle, the tide's range is at its minimum: this is called the "neap tide," or "neaps". Spring tides result in high waters that are higher than average, low waters that are lower than average, slack water time that is shorter than average and stronger tidal currents than average. Neaps result in less extreme tidal conditions. Normally there is a seven day interval between springs and neaps. The relative distance of the Moon from the Earth also affects tide heights: When the Moon is at perigee the range increases, and when it is at apogee the range is reduced. Every 7½ lunations, perigee and (alternately) either a new or full Moon coincide; at these times the range of tide heights is greatest of all, and if a storm happens to be moving onshore at this time, the consequences (in the form of property damage, etc.) can be especially severe. ( Surfers are aware of this, and will often intentionally go out to sea during these times, as the waves are larger at these times.) The effect is enhanced even further if the line-up of the Sun, Earth and Moon is so exact that a solar or lunar eclipse occurs concomitant with perigee. The maximum water level is called "high tide" or "high water" and the minimum level is "low tide" or "low water." If the ocean were a constant depth, and there were no land, high water would occur as two bulges in the height of the oceans--one bulge facing the Moon and the other on the opposite side of the earth, facing away from the Moon. There would also be smaller, superimposed bulges on the sides facing toward and away from the Sun. For an explanation see below under Tidal physics. At any given point in the ocean, there are normally two high tides and two low tides each day just as there would be for an earth with no land; however, rather than two large bulges propagating around the earth, with land masses in the way the result is many smaller bulges propagating around amphidromic points, so there is no simple, general rule for predicting the time of high tide from the position of the Moon in the sky. The common names of the two high tides are the "high high" tide and the "low high" tide; the difference in height between the two is known as the "daily inequality." The daily inequality is generally small when the moon is over the equator. The two low tides are called the "high low" tide and the "low low" tide. On average, high tides occur 12 hours 24 minutes apart. The 12 hours is due to the Earth's rotation, and the 24 minutes to the Moon's orbit. This is the "principal lunar semi-diurnal" period, abbreviated as the M2 tidal component, and it is, on average, half the time separating one lunar zenith from the next. The M2 component is usually the biggest one, but there are many others as well due to such complications as the tilt of the earth's axis and the inclination of the lunar orbit. The lunar cycle is what is tracked by tide clocks. The time between high tide and low tide, when the water level is falling, is called the "ebb." The time between low tide and high tide when the tide is rising, is called "flow," or "flood." At the times of high tide and low tide, the tide is said to be "turning," also slack tide. The height of the high and low tides (relative to mean sea level) also varies. Around new and full Moon when the Sun, Moon and Earth form a line (a condition known as syzygy), the tidal forces due to the Sun reinforce those of the Moon. The tides' range is then at its maximum: this is called the "spring tide," or just "springs" and is derived not from the season of spring but rather from the verb "to jump" or "to leap up." When the Moon is at first quarter or third quarter, the Sun and Moon are at 90° to each other and the forces due to the Sun partially cancel out those of the Moon. At these points in the Lunar cycle, the tide's range is at its minimum: this is called the "neap tide," or "neaps". Spring tides result in high waters that are higher than average, low waters that are lower than average, slack water time that is shorter than average and stronger tidal currents than average. Neaps result in less extreme tidal conditions. Normally there is a seven day interval between springs and neaps. The relative distance of the Moon from the Earth also affects tide heights: When the Moon is at perigee the range increases, and when it is at apogee the range is reduced. Every 7½ lunations, perigee and (alternately) either a new or full Moon coincide; at these times the range of tide heights is greatest of all, and if a storm happens to be moving onshore at this time, the consequences (in the form of property damage, etc.) can be especially severe. ( Surfers are aware of this, and will often intentionally go out to sea during these times, as the waves are larger at these times.) The effect is enhanced even further if the line-up of the Sun, Earth and Moon is so exact that a solar or lunar eclipse occurs concomitant with perigee. Tidal physics Ignoring external forces, the ocean's surface defines a geopotential surface or geoid, where the gravitational force is directly towards the centre of the Earth and there is no net lateral force and hence no flow of water. Now consider the effect of added external, massive bodies such as the Moon and Sun. These massive bodies have strong gravitational fields that diminish with distance in space. It is the spatial differences, called the gradient in these fields that deform the geoid shape. This deformation has a fixed orientation relative to the influencing body and the rotation of the Earth relative to this shape drives the tides around. Gravitational forces follow the inverse-square law (force is inversely proportional to the square of the distance), but tidal forces are inversely proportional to the cube of the distance. The Sun's gravitational pull on Earth is on average 179 times bigger than the Moon's, but because of its much greater distance, the Sun's field gradient and thus its tidal effect is smaller than the Moon's (about 46% as strong). For simplicity, the next few sections use the word "Moon" where also "Sun" can be understood. Since the Earth's crust is solid, it moves with everything inside as one whole, as defined by the average force on it. For a geoid shape this average force is equal to the force on its centre. The water at the surface is free to move following forces on its particles. It is the difference between the forces at the Earth's centre and surface which determine the effective tidal force. At the point right "under" the Moon (the sub-lunar point), the water is closer than the solid Earth; so it is pulled more and rises. On the opposite side of the Earth, facing away from the Moon (the antipodal point), the water is farther from the moon than the solid earth, so it is pulled less and effectively moves away from Earth (i.e. the Earth moves more toward the Moon than the water does), rising as well. On the lateral sides, the water is pulled in a slightly different direction than at the centre. The vectorial difference with the force at the centre points almost straight inwards to Earth. It can be shown that the forces at the sub-lunar and antipodal points are approximately equal and that the inward forces at the sides are about half that size. Somewhere in between (at 55° from the orbital plane) there is a point where the tidal force is parallel to the Earth's surface. Those parallel components actually contribute most to the formation of tides, since the water particles are free to follow. The actual force on a particle is only about a ten millionth of the force caused by the Earth's gravity. These minute forces all work together: • pull up under and away from the Moon • pull down at the sides • pull towards the sub-lunar and antipodal points at intermediate points So in an ocean of constant depth on an Earth with no land, two bulges would form pointing towards the Moon just under it and away from it on Earth's far side. In reality, the presence of land masses and the depth profile of oceans distort this simple pattern significantly. Tidal amplitude and cycle time Since the Earth rotates relative to the Moon in one lunar day (24 hours, 48 minutes), each of the two bulges travels around at that speed, leading to one high tide every 12 hours and 24 minutes. The theoretical amplitude of oceanic tides due to the Moon is about 54 cm at the highest point. This is the amplitude that would be reached if the ocean were uniform with no landmasses and Earth not The Sun similarly causes tides, of which the theoretical amplitude is about 25 cm (46% of that of the Moon) and the cycle time is 12 hours. At spring tide the two effects add to each other to a theoretical level of 79 cm, while at neap tide the theoretical level is reduced to 29 cm. Real amplitudes differ considerably, not only because of global topography as explained above, but also because the natural period of the oceans is in the same order of magnitude as the rotation period: about 30 hours. If there were no land masses and the ocean bottom were flat, it would take about 30 hours for a long wavelength ocean surface wave to propagate halfway around the Earth (by comparison, the natural period of the Earth's crust is about 57 minutes). This means that, if the Moon suddenly vanished, and there were no land, the level of the oceans would oscillate with a period of 30 hours with a slowly decreasing amplitude while dissipating the stored energy. This 30 hour value is a simple function of terrestrial gravity, the average depth of the oceans, and the circumference of the Earth. The distances of Earth from the Moon or the Sun vary, because the orbits are not circular, but elliptical. This causes a variation in the tidal force and theoretical amplitude of about ±18% for the Moon and ±5% for the Sun. So if both are in closest position and aligned, the theoretical amplitude would reach 93 cm. Tidal lag Because the Moon's tidal forces drive the oceans with a period of about 12.42 hours (half of the Moon's synodic period of rotation), which is considerably less than the natural period of the oceans, complex resonance phenomena take place. The global average tidal lag is 12 minutes, which corresponds to an angle of 3 degrees between the position of the moon and the location of global average high tide. Tidal lag and the transfer of momentum between sea and land causes the Earth's rotation to slow down and the Moon to be moved further away in a process known as tidal acceleration. Alternative explanation Some other explanations in articles on the physics of tides include the (apparent) centrifugal force on the Earth in its orbit around the common centre of mass (the barycenter) with the Moon. The barycenter is located at about ¾ of the radius from the Earth's center. It is important to note that the Earth has no "rotation" around this point. It just "displaces" around this point in a circular way (see figure). Every point on Earth has the same angular velocity and the same radius of orbit, but with a displaced center. So the centrifugal force is uniform and does not contribute to the tides. However, this uniform centrifugal force is just equal (but with opposite sign) to the gravitational force acting on the center of mass of Earth. So subtracting the gravitational force at the centre of Earth from the local gravitational forces at the surface, has the same effect as adding the (uniform) centrifugal forces. Although these two explanations seem very different, they yield the same results. Image: http://www.seafriends.org.nz/oceano/topextide.jpg (52KB) History of tidal physics The first well-documented mathematical explanation of tidal forces was given in 1687 by Isaac Newton in the Philosophiae Naturalis Principia Mathematica. However, there is some evidence that Hellenistic Greeks were able to explain tides in terms of a mathematical theory of gravity. Lucio Russo, an Italian scholar, makes this argument in his books Flussi e Riflussi (yet to be published in English) and La Rivoluzione Dimenticata (which has been translated into English as The Forgotten Revolution). Russo argues that the ancients had a more developed theory of gravity than has generally been acknowledged. For example, he exhibits excerpts from ancient texts indicating that Seleucus of Seleucia (II B.C.) devised a gravitational explanation to prove that the Earth revolves around the Sun, rather than vice versa. Tides and navigation Tidal flows are of profound importance in navigation and very significant errors in position will occur if tides are not taken into account. Tidal heights are also very important; for example many rivers and harbours have a shallow "bar" at the entrance which will prevent boats with significant draught from entering at certain states of the tide. Tidal flow can be found by looking at a tidal chart or tidal stream atlas for the area of interest. Tidal charts come in sets, each diagram of the set covering a single hour between one high tide and another (they ignore the extra 24 minutes) and give the average tidal flow for that one hour. An arrow on the tidal chart indicates direction and two numbers are given: average flow (usually in knots) for spring tides and neap tides respectively. If a tidal chart is not available, most nautical charts have " tidal diamonds" which relate specific points on the chart to a table of data giving direction and speed of tidal flow. Standard procedure is to calculate a " dead reckoning" position (or DR) from distance and direction of travel and mark this on the chart (with a vertical cross like a plus sign) and then draw in a line from the DR in the direction of the tide. Measuring the distance the tide will have moved the boat along this line then gives an "estimated position" or EP (traditionally marked with a dot in a Nautical charts display the "charted depth" of the water at specific locations and on contours. These depths are relative to " chart datum", which is the level of water at the lowest possible astronomical tide (tides may be lower or higher for meteorological reasons) and are therefore the minimum water depth possible during the tidal cycle. "Drying heights" may also be shown on the chart. These are the heights of the exposed seabed at the lowest astronomical tide. Heights and times of low and high tide on each day are published in " tide tables". The actual depth of water at the given points at high or low water can easily be calculated by adding the charted depth to the published height of the tide. The water depth for times other than high or low water can be derived from tidal curves published for major ports. If an accurate curve is not available, the rule of twelfths can be used. This approximation works on the basis that the increase in depth in the six hours between low and high tide will follow this simple rule: first hour - 1/12, second - 2/12, third - 3/12, fourth - 3/12, fifth - 2/12, sixth - 1/12. Other tides In addition to oceanic tides, there are atmospheric tides as well as terrestrial tides (earth tides), affecting the rocky mass of the Earth. Atmospheric tides may be negligible for everyday phenomena, drowned by the much more important effects of weather and the solar thermal tides. However, there is no strict upper limit to the Earth's atmosphere, and the tidal pull increases with the distance from the Earth's centre. Theoretically, the Earth's atmosphere extends beyond the Roche limit of the Earth in the Moon's gravitational field. Since the outer extremely thin layers of the atmosphere are in equilibrium with the layers below, the long term effects may not be easily neglected. This means, if the extremely thin outer layers are steadily siphoned away, the material is re-supplied by lower layers, causing an altogether constant small loss of material. The Earth's crust, on the other hand, rises and falls imperceptibly in response to the Moon's solicitation. The amplitude of terrestrial tides can reach about 55 cm at the equator (15 cm of which are due to the Sun), and they are nearly in phase with the Moon (the tidal lag is about two hours only). While negligible for most human activities, terrestrial tides need to be taken in account in the case of some particle physics experimental equipments ( Stanford online). For instance, at the CERN or SLAC, the very large particle accelerators are designed while taking terrestrial tides into account for proper operation. Indeed, despite their kilometre-range dimension, centimetric deformations might lead to their malfunctioning as a physics experimental apparatus. Among the effects that need to be taken into account are circumference deformation for circular accelerators and particle beam Since tidal forces generate currents of conducting fluids within the interior of the Earth, they affect in turn the Earth's magnetic field itself. The loss of rotational energy of the earth, due to friction within the tides, and the torque produced by the gravitational effects of the Sun and Moon on the tidal deformations of the earth's body are responsible for the slowdown of the earth's rotation and the increase of the distance to the Moon; see Tidal force. Tsunamis, the large waves that occur after earthquakes, are sometimes called tidal waves, but have nothing to do with the tides. Other phenomena unrelated to tides but using the word tide are rip tide, storm tide, hurricane tide, and red tide. The term tidal wave appears to be disappearing from popular usage.	{"url":"https://dlab.epfl.ch/wikispeedia/wpcd/wp/t/Tide.htm","timestamp":"2024-11-09T22:36:31Z","content_type":"application/xhtml+xml","content_length":"36517","record_id":"<urn:uuid:edca421a-4329-42ed-9d8a-2058c08487fb>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00591.warc.gz"}
Russell's Paradox \| Brilliant Math & Science Wiki Russell's paradox is a counterexample to naive set theory, which defines a set as any definable collection. The paradox defines the set $R$ of all sets that are not members of themselves, and notes • if $R$ contains itself, then $R$ must be a set that is not a member of itself by the definition of $R$, which is contradictory; • if $R$ does not contain itself, then $R$ is one of the sets that is not a member of itself, and is thus contained in $R$ by definition--also a contradiction. This contradiction is Russell's paradox. It was significant due to reshaping the definitions of set theory, which was of particular interest at the time as the fundamental axioms of mathematics (e.g. the Peano axioms that define arithmetic) were being redefined in the language of sets. Russell's paradox (and similar issues) was eventually resolved by an axiomatic set theory called ZFC, after Zermelo, Franekel, and Skolem, which gained widespread acceptance after the axiom of choice was no longer controversial. In short, ZFC's resolved the paradox by defining a set of axioms in which it is not necessarily the case that there is a set of objects satisfying some given property, unlike naive set theory in which any property defines a set of objects satisfying it. Consider a barber who shaves exactly those men who do not shave themselves (i.e. the barber shaves everyone who doesn't shave themselves and shaves nobody else). Then • if the barber shaves himself, then the barber is an example of "those men who do not shave themselves," a contradiction; • if the barber does not shave himself, then the barber is an example of "those men who do not shave themselves," and thus the barber shaves himself--also a contradiction. Hence the barber does not shave himself, but he also does not not shave himself, hence the paradox. In the above example, an easy resolution is "no such barber exists," but the point of Russell's paradox is that such a "barber" (i.e. a set) must exist if naive set theory were consistent. Since this barber leads to a paradox, naive set theory must be inconsistent. Motivation and Importance Set theory was of particular interest just prior to the 20$^\text{th}$ century, as its language is extremely useful in formalizing general mathematics. For instance, just a few applications are • Arithmetic can be formalized using sets as in the Peano axioms. As much of mathematics depends solely on the completeness of arithmetic, this allows for a large amount of mathematics to be implicitly formalized. • Bijection can be understood as a statement about one-to-one correspondence of sets, which also leads to... • Formalization of infinity and cardinality, especially in defining different types of infinity such as Cantor's proof that there are more real numbers than there are integers. As a result of this incredibly useful formalization, much of mathematics was repurposed to be defined in terms of Cantorian set theory, to the point that it (literally) formed the foundation of Russell's paradox served to show that Cantorian set theory led to contradictions, meaning not only that set theory had to be rethought, but most of mathematics (due to resting on set theory) was technically in doubt. Fortunately, the field was repaired a short time later by new axioms (ZFC), and set theory remains the main foundational system of mathematics today. Formal Definitions and Formulation Naive set theory is the theory of predicate logic with binary predicate $\in$, that satisfies \[\exists y\forall x\big(x \in y \iff \phi(x)\big)\] for any predicate $\phi$. This is called unrestricted comprehension, and means There exists a set $y$ whose members are exactly the objects satisfying the predicate $\phi$. Naive set theory also contains two other axioms (which ZFC also contains): Existential instantiation: Given a formula of the form $(\exists x)\phi(x)$, one can infer $\phi(c)$ for some new symbol $c$. All this is saying is that if there exists some object satisfying a given property, that element can be given a name $c$ (in such a way that $c$ was not previously used). For instance, • There exists a number satisfying the equation $x^2 + 1 = 0$. • Define $i$ to be a number satisfying the equation $i^2 + 1 = 0$. is valid logic. Universal instantiation: Given a formula of the form $\forall x\phi(x)$, one can infer $\phi(c)$ for any $c$ in the universe. Intuitively speaking, this axiom states that if everything satisfies some property, any one of those things also satisfies that property. For instance, • All people living in California live in the U.S.A. (hypothesis) • John lives in California. (implying that John is part of the universe) • John lives in the U.S.A. (invocation of universal instantiation) is valid logic. These axioms are sufficient to illustrate Russell's paradox: • Consider the predicate $\phi: x \not\in x$. • By unrestricted comprehension, there exists a set $y$ consisting of elements satisfying the predicate $\phi$. • By existential instantiation, there exists a $z$ such that $\forall x\big(x \in z \iff \phi(x)\big)$. • By universal instantiation, $z$ (which is in the universe) satisfies the predicate $x \in z \iff \phi(x)$, so $z \in z \iff z \not\in z$, which is a contradiction, implying that naive set theory is inconsistent. Roughly speaking, there are two ways to resolve Russell's paradox: either to • alter the logical language, i.e. first order logic, that the axioms of set theory are expressed in, or • alter the axioms of set theory, while retaining the logical language they are expressed in. Russell took the first approach in his attempt at redefining set theory with Whitehead in Principia Mathematica, developing type theory in the process. However, though they eventually succeeded in defining arithmetic in such a fashion, they were unable to do so using pure logic, and so other problems arose. In fact, Godel showed that Peano arithmetic is incomplete (assuming Peano arithmetic is consistent), essentially showing that Russell's approach was impossible to formalize. In doing so, Godel demonstrated his acclaimed incompleteness theorems. The second approach, in which the axioms of set theory are altered, was favored by Zermelo (later joined by Franekel and Skolem) in his derivation of ZFC. This resolves the paradox by replacing unrestricted comprehension with restricted comprehension (also called specification): Given a predicate $\phi$ with free variables in $x, z, w_1, w_2, \ldots, w_n$, \[\forall z\forall w_1 \forall w_2 \ldots \forall w_n \exists y \forall x(x \in y \iff \big(x \in z \land \phi)\ Essentially, this means that given a set $z$ and a predicate $\phi$, the subset \[\{x \in z: \phi(x)\}\] (i.e. all elements of $z$ satisfying the predicate $\phi$) exists. This resolves Russell's paradox as only subsets can be constructed, rather than any set expressible in the form $\{x:\phi(x)\}$. In this sense, Russell's paradox serves to show that There does not exist a set containing all sets, which is also a useful result in its own right.	{"url":"https://brilliant.org/wiki/russells-paradox/?subtopic=paradoxes&chapter=paradoxes-in-probability","timestamp":"2024-11-10T18:17:02Z","content_type":"text/html","content_length":"52607","record_id":"<urn:uuid:7c0d1498-0d1e-4914-873e-a7fcc9d5b888>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00442.warc.gz"}
Identifying Normal Vectors to Planes: Methods and Applications for Plane Geometry Identifying Normal Vectors To Planes: Methods And Applications For Plane Geometry To find a vector normal to a plane, we need to determine a vector that is perpendicular to all vectors lying on the plane. The cross product of two vectors lying on the plane gives a vector perpendicular to both. Alternatively, we can use the dot product to find a vector orthogonal to two vectors on the plane and normalize it to obtain a normal vector. The normal vector is crucial for various applications, including determining the plane’s orientation, calculating angles, and performing geometric operations involving planes. Plane Normals: Unveiling the Secret of Vectors and Planes Planes are ubiquitous in our world, from the flat surface of a table to the expansive sky above. But did you know that there’s a hidden secret that unlocks their true nature? It’s called a plane normal. Join us as we embark on an adventure to unravel the mystery of plane normals, explaining why they’re so important and how you can find them with ease. Understanding the Purpose and Significance of Plane Normals Imagine you’re trying to describe a plane to someone. You could give its equation, but that’s not very intuitive. Instead, you could simply point in the direction that’s perpendicular to the plane. That’s where plane normals come in. A plane normal is a vector that’s perpendicular to the plane. It provides a convenient way to represent the plane’s orientation in space. Whether you’re studying geometry, physics, or computer graphics, understanding plane normals is crucial for describing and manipulating planes. Essential Concepts: Cross and Dot Products – The Keys to Vector Orthogonality In the world of vectors, perpendicularity, or being at right angles to each other, plays a crucial role. Two indispensable concepts in vector calculus, the cross product and dot product, are the keys to unlocking the secrets of orthogonality. The Cross Product: A Perpendicularity Determinant Picture two vectors, A and B, in three-dimensional space. Their cross product, denoted as A x B, results in a vector that is perpendicular to both A and B. This cross product vector points in the direction of the normal vector to the plane formed by A and B. The Dot Product: Orthogonality Detector The dot product, symbolized as A · B, quantifies the extent to which two vectors are aligned. It yields a scalar value that is zero if the vectors are perpendicular and non-zero otherwise. This property makes the dot product an invaluable tool for determining orthogonality. Properties of Orthogonal Vectors Orthogonal vectors exhibit unique characteristics: • Perpendicularity: They intersect at a 90-degree angle. • Zero Dot Product: Their dot product is always zero. • No Component in the Same Direction: Neither vector has a component in the direction of the other. Significance of Normal Vectors Normal vectors, which are perpendicular to a plane, are vital in various applications. For instance: • In geometry, they define planes and are used to calculate angles between planes. • In physics, they determine the direction of forces and the orientation of surfaces. • In computer graphics, they are essential for shading, lighting, and collision detection. Properties of Orthogonal Vectors • Describe the characteristics of orthogonal vectors and their perpendicular relationship. • Highlight the significance of normal vectors being orthogonal to vectors on the plane. Orthogonal Vectors: The Perpendicular Guardians of Planes In the realm of geometry, orthogonal vectors stand as sentinels, ensuring the perpendicularity between planes and their constituent vectors. Understanding their characteristics and significance is crucial for navigating the intricate world of planes. Orthogonal vectors are vectors that form right angles with each other. Geometrically, they have a dot product of zero, indicating a complete lack of alignment. This perpendicular relationship is fundamental to the concept of plane normals. In the context of planes, normal vectors are vectors that are orthogonal to every vector lying on the plane. This orthogonality ensures that the normal vector points perpendicularly to the plane, providing a consistent reference direction. Moreover, normal vectors play a vital role in many applications. In geometry, they are used to determine the distance between a point and a plane. In physics, they are essential for understanding the reflection and refraction of light and other waves. In computer graphics, they are used for shading, lighting, and collision detection. The perpendicular relationship between normal vectors and plane vectors is a cornerstone of geometry and its applications. By understanding this relationship, we can unlock the power of orthogonal vectors and delve deeper into the fascinating world of planes. Method 1: Finding a Plane Normal Using Three Points Finding a normal vector to a plane plays a crucial role in geometry and has practical applications in physics, engineering, and computer graphics. Among the various methods to compute a plane normal, one straightforward approach involves utilizing three points that lie on the plane. To grasp this method, let’s visualize a plane in 3D space. Any three non-collinear points on this plane can define two vectors that lie within the plane. The cross product of these two vectors gives us a vector that is perpendicular to both input vectors. And since these input vectors lie on the plane, their cross product will result in a vector that is normal to the plane. Step 1: Define Two Vectors on the Plane Consider three distinct points on the plane: (P_1(x_1, y_1, z_1), P_2(x_2, y_2, z_2), ) and (P_3(x_3, y_3, z_3)). Define two vectors (u) and (v) as follows: u = P_2 - P_1 = (x_2 - x_1, y_2 - y_1, z_2 - z_1) v = P_3 - P_1 = (x_3 - x_1, y_3 - y_1, z_3 - z_1) Step 2: Calculate the Cross Product The cross product of vectors (u) and (v) is given by: n = u x v = (y_2 - y_1)(z_3 - z_1) - (z_2 - z_1)(y_3 - y_1), (z_2 - z_1)(x_3 - x_1) - (x_2 - x_1)(z_3 - z_1), (x_2 - x_1)(y_3 - y_1) - (y_2 - y_1)(x_3 - x_1) This resulting vector (n) is perpendicular to both (u) and (v), and thus to the plane defined by the three points. Consider a plane defined by three points: ((1, 2, 3), (4, 5, 6), ) and ((7, 8, 9)). • Define vectors (u) and (v): □ (u = (4, 5, 6) – (1, 2, 3) = (3, 3, 3)) □ (v = (7, 8, 9) – (1, 2, 3) = (6, 6, 6)) • Calculate the cross product: □ (n = (3 x 6) – (3 x 6), (6 x 6) – (3 x 6), (3 x 6) – (3 x 6)) □ (n = (0, 0, 0)) In this example, the cross product of (u) and (v) results in the zero vector. This is because the three points lie on the same line, which is not a plane. For a plane, the cross product should typically yield a non-zero vector. Method 2: Harnessing the Dot Product Imagine you have a plane soaring through space, but you can’t quite figure out its orientation. Fear not, for the dot product comes to the rescue! This mathematical tool empowers you to determine a vector that stands perpendicular to the plane, thereby defining its normal vector. Let’s embark on a step-by-step adventure: 1. Identify two vectors on the plane: Pick any two vectors lying on the plane. Let’s call them u and v. 2. Calculate the dot product: The dot product is denoted by a dot symbol between two vectors, such as u · v. It measures the cosines of the angle between these vectors. 3. Subtract the projected components: Remember that the cross product of two parallel vectors is zero. So, we need to find a vector w such that w · u = 0 and w · v = 0. This means that w is perpendicular to both u and v. 4. Construct the normal vector: The resulting vector w is a normal vector to the plane because it is perpendicular to any two vectors that lie within the plane. This method provides an alternative approach to finding the plane’s normal vector, offering a valuable tool in your geometric toolbox.	{"url":"https://rectangles.cc/identify-normal-vectors-planes-methods-applications-plane-geometry/","timestamp":"2024-11-12T10:54:48Z","content_type":"text/html","content_length":"147970","record_id":"<urn:uuid:324d09e6-6637-4e48-b4d7-c9c53c1a1d8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00864.warc.gz"}
Determination of Gas Permeation Properties in Polymer Using Capacitive Electrode Sensors Hydrogen Energy Materials Research Center, Korea Research Institute of Standards and Science, Daejeon 34113, Korea Department of Physics and Research Institute of Natural Science, Gyeongsang National University, Jinju 52828, Korea Author to whom correspondence should be addressed. Submission received: 6 January 2022 / Revised: 27 January 2022 / Accepted: 29 January 2022 / Published: 2 February 2022 The objective of this work was to develop an effective technique for characterizing the permeation properties of various gases, including H[2], He, N[2], and Ar, that are absorbed in polymers. Simultaneous three-channel real-time techniques for measuring the sorption content and diffusivity of gases emitted from polymers are developed after exposure to high pressure and the subsequent decompression of the corresponding gas. These techniques are based on the volumetric measurement of released gas combined with the capacitance measurement of the water content by both semi-cylindrical and coaxial-cylindrical electrodes. This minimizes the uncertainty due to the varying temperature and pressure of laboratory environments. The gas uptake and diffusivity are determined as a function of the exposed pressure and gas spices in nitrile butadiene rubber (NBR) and ethylene propylene diene monomer (EPDM) polymers. The pressure-dependent gas transport behaviors of four different gases are presented and compared with those obtained by different techniques. A linear correlation between the logarithmic diffusivity and kinetic diameter of molecules in the gas is found between the two polymers. 1. Introduction The permeability of a polymer is defined as the rate at which it is penetrated by various gases. The characteristic passage of gas through a polymer is affected by the solubility in the polymer, and gases pass through the polymer sheet by the process of diffusion. In other words, gas permeation is the passage of a permeant through a polymer material. The process of permeation involves the diffusion of molecules—i.e., the permeant—through a membrane or interface where the permeant will move from a high concentration to a low concentration across the interface. Permeation is extensively utilized for various applications, such as in the food packaging field, tires and fuel cells in automobiles, electrical insulating materials, the medical field for drug delivery, thermoplastic piping in gas transportation, and O-rings in high-pressure gas vessels [ ]. Studying the permeability characterization of materials with different gases and under different environmental conditions is crucial in order to understand whether the corresponding material is adapted to the chosen gases. At the same time, the transport properties of gases to permeate the materials can be clarified with reliable measurement techniques. Meanwhile, the gas permeation of a material can be measured by numerous methods that quantify the permeability of a material. These methods include manometric methods [ ], constant-pressure methods [ ], gravimetric techniques [ ], magnetic suspension balance methods [ ], gas chromatography [ ], and computer simulation [ ]. Most methods are time-consuming, requiring complicated processes and fine control. For instance, for polymers with a diffusivity in the order of 10 /s and with a thickness above 3 mm, it takes at least a few days to reach the adsorption/desorption equilibrium and then complete the permeation measurement. Furthermore, the variation in both temperature and pressure across the days affects measurements of aspects such as the gas volume and then increases the uncertainty in the determination of permeation parameters. Thus, the instability due to temperature and pressure should be minimized to achieve precise measurement and compensation. Effective and real-time automatic measurements are required to overcome the limitations of methods and further enhance the reliability of the measurement of permeability characteristics. We sought to find an appropriate technique for determining the permeation properties of several gases dissolved in materials. Thus, we developed the volumetric analysis technique (VAT) in previous studies [ ] and confirmed this by comparing the results obtained using VAT with those obtained using different methods, such as gas chromatography (GC) by thermal desorption analysis (TDA) and gravimetric measurement by electronic balance for same samples. The results were found to be consistent with each other. A more effective technique is to combine a volumetric measurement using a graduated cylinder and automatic capacitance measurement with electrodes through a frequency response analyzer interfaced with a PC. The developed technique reduces the uncertainty of permeation parameters due to the varying temperature and pressure of the laboratory environment. The techniques were applied to nitrile butadiene rubber (NBR) and ethylene propylene diene monomer (EPDM) polymers, which are used for gas sealing materials under high pressure. The solubility, diffusivity, and permeability of the four different gases in the two polymers were investigated as a function of the exposed pressure and compared with those determined by different methods. The permeation characteristics obtained by this method were described. Another motivation of our research was that the polymer materials can be applicable for various gas sealing requirements under a high pressure. The diffusivity in the NBR and EPDM polymers can be interpreted in terms of the kinetic diameter of molecules in the employed gases. 2. Experimental Aspects Sample Preparation and Gas Exposure Conditions The compositions and densities of the NBR and EPDM polymer specimens used in this study have already been listed in previous literature [ ]. NBR samples with two different thicknesses and EPDM samples with different shapes/dimensions were used: cylindrical-shaped NBR samples with a radius of 7.0 mm and thicknesses of 1.1 mm and 2.2 mm were prepared. Cylindrical-shaped EPDM samples with a radius of 7.0 mm and thicknesses of 1.4 mm and 2.5 mm as well as spherical-shaped EPDM with a radius of 4.9 mm were also prepared. A SUS 316 chamber with an inner diameter of 50 mm and height of 90 mm was used for gas exposure to high pressure at room temperature and a specified pressure. The chamber was purged three times with the corresponding gas of 1 MPa–3 MPa depending on the pressure before the gas exposure. We exposed the specimen to the gas for 24 h in a pressure range from 1.5 MPa to 10 MPa. Gas charging for 24 h is sufficient to attain the equilibrium state for gas sorption, except for N[2] gas exposure. N[2] gas charging for 48 h is needed to attain the equilibrium state for N[2] sorption because of its slow diffusion rate. After exposure to gas, the valve was opened and the gas in the chamber was released. After decompression, the elapsed time was recorded from the moment (t = 0) at which the high-pressure gas in the chamber was reduced to atmospheric pressure when the time was set to zero. Since the specimen was loaded in the graduated cylinder after decompression, it took approximately 5–10 min to start the measurement. The gas content emitted for the inevitable time lag could be measured later by offset determination via the simulation. 3. Two Types of Capacitor Electrodes to Measure the Water Level We employed two types of electrodes to measure the capacitance corresponding to the water content in the acrylic tube (graduated cylinder). A semi-cylindrical capacitor and coaxial-cylindrical capacitor electrodes were fabricated and attached to the outer wall of the graduated cylinder. The capacitance was measured at 1 MHz with two electrodes by a frequency response analyzer (VSP 300) with a general-purpose interface bus (GPIB) connected to a PC. 3.1. Semi-Cylindrical Capacitor Electrode The capacitive sensor fabricated with semi-cylindrical electrodes mounted outside of an acrylic tube is shown in Figure 1 a. An acrylic tube surrounded by two semi-cylindrical electrodes is filled with water gas. The electrode attached to the outer wall of the acrylic tube is made of copper cylinder with a thickness of 1 mm. The capacitance of the sensor depends on the dielectric permittivity of the medium existing between the electrodes. The dielectric permittivity of water is 78.4 times larger than that of gas inside the graduated cylinder. Thus, the position shift of the water level in the two electrodes leads to a change in the capacitance. The actual capacitance ( ) due to water gas is connected in series with the capacitance ( ) of the acrylic dielectric tube wall. The total capacitance ( ) between the semi-cylindrical electrodes can be expressed as: $C t = C a C t w C a + C t w$ The actual permittivity ( $ε a )$ of both the water and gas inside the cylinder, depending on the volume fraction of the two media, is given by: $ε a = V w ε w + V 0 ε 0 V t$ $V w$ is the water volume in the cylinder, $ε w$ is the dielectric permittivity of water, $V 0$ is the gas volume in the cylinder, $ε 0$ is the dielectric permittivity of gas, and $V t$ is the total volume. The actual capacitance with two semi-cylindrical electrodes of the same size is calculated as [ $C a = ∑ i = 0 n 2 ε * 0 ε a A × [ 1 d + ( i − 1 ) Δ d ] + ε 0 ε a A 2 R$ is the area of the electrode, $ε * 0$ is the dielectric permittivity of free space, is the distance between the electrodes, is the radius of the acrylic tube, and $Δ d$ is an increment distance between semi-cylindrical concave electrodes. In this work, the values in Equation (3) are constant except for $ε a$ . The capacitance values with respect to the water content are obtained by a combination of Equations (1)–(3). We measured the change in actual capacitance ( $C a )$$by the change in ε a$ arising from the changing water level in the graduated cylinder. Therefore, the changing water level corresponding to the change capacitance is determined with the precalibration equation between the capacitance and water level, which will be presented in the following chapter. 3.2. Coaxial-Cylindrical Capacitor Electrode Another capacitive sensor is designed with coaxial-cylindrical electrodes mounted at the center and outside of an acrylic tube, as shown in Figure 1 b. The water gas in the acrylic tube is filled between two coaxial electrodes. The change in capacitance, $Δ C$ , with respect to the water level, , and remaining height, , in the cylinder filled with gas is given by [ $Δ C = 2 π ε 0 ( ε w h + ε g ( L − h ) ) ln ( R 2 R 1 ) = 2 π ε 0 ( ε w − ε g ) h ln ( R 2 R 1 ) + 2 π ε 0 ε g L ln ( R 2 R 1 )$ is the water level, is the length of the cylindrical capacitor, is the radius of the solid cylindrical conductor (electrode 2) made of thin copper wire, and is the inner radius of the coaxial cylindrical shell (electrode 1) made of copper plate. $ε 0 , ε w ,$ $ε g$ are the permittivity of free space, water, and gas, respectively. For a fixed configuration of the coaxial cylindrical electrode, Equation (4) indicates that $Δ C$ is linearly related to the change in the water level, h. Similar to the semi-cylindrical electrode, we thus determined the water level by measuring the change in capacitance with a precalibration equation. 4. Volumetric Analysis Measurement System 4.1. Volumetric Measurement of Emitted Gas Figure 2 shows a three-channel volumetric measurement system with three graduated cylinders and three electrodes to measure the released gas in real time. After exposure to the high-pressure chamber and subsequent decompression, the specimen is loaded into the gas space of a graduated cylinder. Three parallel standing graduated cylinders partially immersed in each water container collect and measure the gas released from the specimen. The semi-cylindrical and coaxial-cylindrical electrodes, connected in parallel to the responding capacitance measurement channel of the frequency response analyzer, are mounted outside of acrylic tubes in the left and right cylinders and center cylinder. The precise frequency response analyzer (FRA, VSP 300) with an excellent performance is a general purpose interface bus (GPIB) interfaced with a programmed PC with autosensing and autocontrol functions for the temperature and pressure. The FRA GPIB interfaced with the PC at three channels is employed for automatic real-time capacitance measurement with both semi-cylindrical and coaxial-cylindrical electrodes, as shown in Figure 2 . The temperature and pressure measured near the sample are automatically applied for the calculation of the gas uptake. The pressures ( $P 1 , P 2 , and P 3$ ) inside each graduated cylinder for the three channels are expressed as [ $P 1 = P o − ρ g h 1 , P 2 = P o − ρ g h 2 , P 3 = P o − ρ g h 3$ $P o$ is the outside atmospheric pressure of the cylinder, is the density of distilled water in the water container, and is gravity. $h 1$ $h 2 ,$ $h 3$ are the heights of the distilled water level inside the corresponding graduated cylinder measured from the water level in the water container of channel 1, channel 2, and channel 3, respectively. $V 1 , V 2 , and V 3$ are the gas volumes inside the corresponding graduated cylinder filled with gas. As shown in Figure 2 , the gas inside the cylinder is governed by the ideal gas equation, , and is the gas constant with 8.20544 × 10 The total number of moles ( $n 1$ $n 2 and n 3$ ) of gas inside the corresponding cylinder for the three channels is expressed at specified as follows: $n 1 = n 1 , 0 + Δ n 1 = ( P o − ρ g h 1 ) V 1 R T , n 2 = n 2 , 0 + Δ n 2 = ( P o − ρ g h 2 ) V 2 R T , n 3 = n 3 , 0 + Δ n 3 = ( P o − ρ g h 3 ) V 3 R T$ $n 1 , 0$ $n 2 , 0 ,$ $n 3 , 0$ are the initial number of moles of air already in cylinder 1, cylinder 2, and cylinder 3, respectively, before the gas emission. The gas released from the specimen after decompression lowers the water level of the cylinder. Thus, the increased number of moles ( $Δ n 1 , Δ n 2 , and Δ n 3$ ) in each cylinder from emitted gas after decompression is obtained by measuring the increase in volume ( $Δ V 1 , Δ V 2 , and Δ V 3 )$ in each graduated cylinder, with the lowering of the water level as follows: $Δ n 1 = ( P o − ρ g h 1 ) Δ V 1 R T , Δ n 2 = ( P o − ρ g h 2 ) Δ V 2 R T , Δ n 3 = ( P o − ρ g h 3 ) Δ V 3 R T$ The increased number of moles in each channel is converted to the corresponding mass concentration [ $C 1 ( t ) , C 2 ( t ) , and C 3 ( t ) ]$ of gas emitted from the rubber sample as follows: $C 1 ( t ) [ wt · ppm ] = Δ n 1 [ mol ] × m g a s [ g mol ] m s a m p l e [ g ] × 10 6 C 2 ( t ) [ wt · ppm ] = Δ n 2 [ mol ] × m g a s [ g mol ] m s a m p l e [ g ] × 10 6 C 3 ( t ) [ wt · ppm ] = Δ n 3 [ mol ] × m g a s [ g mol ] m s a m p l e [ g ] × 10 6$ $m g a s$ (g/mol) is the molar mass of the gas investigated. For example, for H $m H 2 g a s$ is 2.016 g/mol. $m s a m p l e$ is the mass of the specimen. By measuring the change in the water level ( $Δ V$ ), we obtained an increased number of moles and thus transformed the mass concentration of the emitted gas. Therefore, the time-dependent mass concentration by released gas can be obtained by measuring the water level change, $Δ V ,$ versus the time elapsed since decompression. The water level data were transformed from the capacitance by the precalibration data of the polynomial form between the capacitance and the position of the water level. 4.2. Time-Dependent Emitted Gas Concentration versus Specimen Shape The adsorption of gas under high pressure causes the release of gas dissolved in rubber after decompression to atmospheric pressure. Assuming that the adsorption and desorption of gas are diffusion-controlled processes, the emitted gas concentration $C E ( t )$ in the desorption process is expressed as [ $C E ( t ) = C ∞ [ 1 − 6 π 2 ∑ n = 1 ∞ 1 n 2 exp ( − D n 2 π 2 t a 2 ) ]$ Equation (9) is the solution to Fick’s second law of diffusion for a spherical sample with an initially constant uniform gas concentration and constant concentration at the spherical surface. $C ∞$ is the saturated gas mass for an infinitely long time—i.e., the total emitted mass concentration or gas uptake in the adsorption process. is the diffusion coefficient of desorption. is the radius of the spherical rubber [ Similarly, the emitted gas content $C E ( t )$ for a cylindrical specimen is expressed under the boundary condition—i.e., a uniform gas concentration is initially maintained and the cylindrical surfaces are kept at a constant concentration [ $C E ( t ) / C ∞ = 1 − 32 π 2 × [ ∑ n = 0 ∞ exp { − ( 2 n + 1 ) 2 π 2 D s t l 2 } ( 2 n + 1 ) 2 ] × [ ∑ n = 1 ∞ exp { − D s β n 2 t ρ 2 } β n 2 ]$ In Equation (10), is the thickness of the cylindrical rubber sample, is the radius, and $β n$ is the root of the zero-order Bessel function. To analyze the mass concentration data, we used a diffusion analysis program developed using Visual Studio to calculate D$and$$C ∞$ in Equations (9) and (10) based on least-squares regression [ 4.3. Diffusion Parameter Analysis through Programmed Capacitance Measurement The gas emitted from the specimen lowers the water level, and then the water level decreases as the elapsed time increases. Using programmed capacitor measurements with electrodes and diffusion analysis programs, the diffusion parameters for specimens can be determined. Figure 3 a–c shows the processes used for acquiring the diffusion parameter in NBR cylindrical rubber by coaxial-cylindrical electrodes as follows: To obtain the precalibration data, the user measures the water level versus the capacitance at the corresponding channel with decreasing water levels. Then, the 2nd polynomial equation related to the position of the water level and capacitance is obtained by quadratic regression, as shown in Figure 3 a. The 2nd polynomial equation originates from Equation (4). The position of the water level is measured by a digital camera. According to the precalibration data, the capacitance is transformed to the water level, as shown in Figure 3 b. The black and blue squares correspond to the capacitance and position of the water level, respectively, versus the time elapsed. Last, the diffusion parameters D$and$$C ∞$ are determined using a diffusion analysis program by applying Equation (10) based on least-squares regression, as shown in Figure 3 Figure 4 shows the sequence used for obtaining the diffusion parameter manually by a digital camera for the same NBR as Figure 3 Figure 4 a shows the water level measured directly by a digital camera without precalibration, and Figure 4 b shows the water level as a function of time transformed to the mass concentration, resulting in diffusion parameters D$and$$C ∞$ determined using a diffusion analysis program. The two results in Figure 3 Figure 4 are consistent with each other. Figure 5 represents the sequence of acquiring diffusion parameters for EPDM cylindrical rubber by employing semi-cylindrical electrodes. Figure 5 a represents precalibration data expressed as the 2nd polynomial equation between the water level and capacitance by quadratic regression, which comes from Equations (1)–(3). Figure 5 b shows the water level transformed from the capacitance, where the black and blue squares correspond to the capacitance and transformed water level, respectively, versus time. Figure 5 c shows diffusion parameters D$and$$C ∞$ , which are determined using a diffusion analysis program according to Equation (10). Figure 6 shows the sequence used for obtaining the diffusion parameter measured manually by a digital camera for the same EPDM as that shown in Figure 5 Figure 6 a shows the water level measured directly by a digital camera, and Figure 6 b shows the water level as a function of time transformed to the mass concentration, resulting in diffusion parameters D$and$$C ∞$ determined using a diffusion analysis program. The two results in Figure 5 Figure 6 are consistent with each other. 5. Results and Discussion 5.1. Stability Test of the Volumetric Measurement System The volume and number of moles of gas in the graduated cylinder are directly affected by both the temperature and pressure in the laboratory environment. Therefore, before measuring the main diffusion properties, the stability of the volumetric measurement system should be improved by applying variations in both the temperature and pressure during long-term measurement to calculations using Equations (6) and (7). Figure 7 shows the stability measurements performed for three days, in which the temperature (top of Figure 7 ) and pressure (middle of Figure 7 ) were maintained within 24.0 ± 0.5 °C and 997.5 ± 3.5 hPa, respectively. The bottom of Figure 7 represents the stability test with (closed circle) and without (open circle) the application of variation in both the temperature and pressure to Equations (6) and (7). The change in the mass concentration due to correction for the changes in temperature and pressure over three days is within 4 wt·ppm, which is comparable with 7 wt·ppm in the case that does not consider the variation in temperature and pressure. The system stability is improved by removing the variation in both the temperature and pressure, which are included as uncertainty factors in permeation parameter determination. 5.2. Pressure Dependence on the Permeation Parameter Figure 8 Figure 9 show the permeation parameters versus exposed pressure in NBR and EPDM, respectively, for four different gases with coaxial-cylindrical or semi-cylindrical electrodes at three channels. The diffusion $C ∞$ are determined using a diffusion analysis program by the application of Equations (9) and (10) based on least-squares regression. The standard deviation between the experimental data and the diffusion model was within 3% for both rubbers. All the gas uptake follows Henry’s law [ ] up to 9 MPa with a squared correlation coefficient R > 0.990, as indicated by the black and blue lines in Figure 8 a for NBR, and black, blue, and gray lines in Figure 9 a for EPDM. This implies that gas does not dissociate and penetrates into the specimen as a gas molecule. The slopes in the two specimens indicate Henry’s law of solubility. As shown in Figure 8 b, the diffusivity does not represent a distinct pressure dependency. Thus, we take the average diffusivity, as indicated by the black and blue horizontal lines. Meanwhile, Figure 9 b shows that the diffusivity decreases as the pressure increases above 6 MPa, except for H diffusivity. This may be ascribed to the bulk diffusion associated with the mean free path, which is normally observed for high-pressure gas diffusion. The error bars indicate the relative expanded uncertainty of 10%, as evaluated in previous research. At pressures below 6 MPa in Figure 9 b, we also take the average diffusivity, as indicated by black and blue horizontal lines. As shown in Figure 8 Figure 9 , no dependence of the permeation parameters on the thickness in cylindrical-shaped NBR and EPDM was observed. The solubility (S) is determined from the linear slope obtained in Figure 8 a and Figure 9 a as follows: $S [ mol m 3 · MPa ] = C ∞ slope [ wt · ppm MPa ] 10 6 × d [ g m 3 ] m g [ g mol ]$ is the molar mass of gas used, and is the density of the rubber. The permeabilities of the four gases in the NBR and EPDM polymers are obtained from the solubility and the average diffusivity by using the relation of P = D S. The permeation parameters for four gases in NBR and EPDM are summarized with those obtained by different methods in Table 1 The values in parentheses were determined by the differential pressure method and thermal desorption analysis–gas chromatography [ ] in the same specimen. The results obtained by different methods for H gas are consistent with those in the present experimental investigation within expanded uncertainty. Differences in the permeation parameters were found for gases in both NBR and EPDM. The magnitudes of the diffusivity and permeability decrease in the orders in both NBR and EPDM. Although there are many factors affecting the permeation parameters of rubber, we focus on the molecule size in the gas. The size of the permeant molecule affects the diffusivity. As the effective size of the molecule increases, the diffusivity decreases. As expected for both NBR and EPDM ( Figure 10 a), we found a linear correlation with a squared correlation coefficient of R > 0.90 between the logarithmic diffusivity and kinetic diameter of the molecules in the gas, which is the size of the sphere of influence that can lead to a scattering event and is also related to the mean free path of molecules in a gas [ Figure 10 a also displays different diffusivity values obtained at same kinetic diameter between NBR and EPDM polymer. For the case of NBR, the existence of a –CN polar group can make it possible to increase interchain interaction, leading to the tight packing of polymer chains. As a result, the available free volume decreases, and then NBR achieves a low diffusivity of gas molecules. In contrast, EPDM could have a large free volume due to the presence of norbornene, and thus it is not easy to have a tight packing of chains, resulting in the high diffusivity of gas. In addition, EPDM chains are expected to be more flexible than NBR since the chain mobility has also been known to be governed by the chain packing characteristics. Meanwhile, the solubility of gases depends on the relative affinity between the gas and polymer, but more strongly on the penetrant condensability correlated with the gas critical temperature ( ). The relationship between gas solubility and the critical temperature is generally expressed as [ The constant “a” is a measure of the overall sorption capacity, while slope “b” indicates the increase in solubility with regard to the penetrant condensability. Figure 10 b demonstrates the solubility of the four gases versus critical temperature for two polymers. It is observed for EPDM rather than NBR that the logarithmic solubility increases nearly linearly with the increase in the critical temperature, except for H gas, which deviates from linearity. A similar relationship was reported for polyvinylpyridine film [ We present the performance parameters of capacitor sensors, such as sensitivity, resolution, stability, detection range, and response time, for the two sensors in Table 2 with a related description. The sensitivity is defined as the slope obtained by the change in capacitance with regard to the water level in the unit of ml. The sensitivity is the most important factor deciding the performance of a sensor. The coaxial-cylindrical capacitor sensor with a high sensitivity and minute resolution could be a better choice. 6. Conclusions We first developed an automatic technique for determining the permeation of various gases, including H[2], He, N[2], and Ar. This simple and effective method combines a volumetric measurement using a graduated cylinder with water level detection by capacitance measurement with two different types of electrodes in real time. This technique is able to simultaneously evaluate three sets of diffusion characteristics of gas by quantitatively analyzing the amount of gas released after high-pressure gas charging and subsequent decompression. With the autoreading and autocontrol of temperature and pressure sensors, fluctuations due to variations in the temperature and pressure of the laboratory environment were removed, resulting in good-quality permeation data. The results achieved for polymers demonstrate that the H[2] permeation properties determined by the developed method are in agreement with those determined by the differential pressure method and gas chromatography. The experimental investigation indicates that the gas content emitted from the NBR and EPDM satisfied Henry’s law up to a pressure of 9 MPa, which confirmed that the content was primarily proportional to the pressure. The solubility and diffusivity were identical for all specimens employed, regardless of the sample shape and dimensions. This is a general trend, but different diffusivity values were found for thicker specimens. The different diffusivities for each gas can be attributed to the different kinetic diameters of the molecules in the gas. In conclusion, a technique for determining permeation with capacitance measurement using a frequency response analyzer could be effectively applied for automatically evaluating the transport properties of gases in polymers and other materials for cases requiring real-time and time-consuming measurements with a slow diffusion rate. This simple technique could be applied in permeation evaluation and leakage tests for all types of gas without sample size and shape limitations. Author Contributions Conceptualization, J.J.; methodology, J.J. and G.K.; software, G.K. and G.G.; validation, C.P. and J.L.; formal analysis, G.G. and J.L.; writing—original draft preparation, J.J.; writing—review and editing, J.J. All authors have read and agreed to the published version of the manuscript. This research was supported by Development of Reliability Measurement Technology for Hydrogen Fueling Station funded by the Korea Research Institute of Standards and Science (KRISS-2022-GP2022-0007). Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. Data Availability Statement The data used to support the findings of this study are available from the corresponding author upon request. This research was supported by Development of Reliability Measurement Technology for Hydrogen Fueling Station funded by the Korea Research Institute of Standards and Science (KRISS-2022-GP2022-0007). This research was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education(NRF-2020R1I1A1A01064234)(Changyoung Park). Conflicts of Interest The authors declare that they have no known competing financial interest or personal relationship that could have appeared to influence the work reported in this paper. Figure 1. (a) Configuration of the semi-cylindrical capacitor electrode, indicated in blue. (b) Configuration of the coaxial-cylindrical capacitor electrode. Figure 2. Schematic diagram of the three-channel volumetric measurement system in which three cylinders are standing. The blue part indicates the distilled water filling the water containers and cylinders. A frequency response analyzer GPIB interfaced with a PC at three channels is employed for automatic real-time capacitance measurement with both semi-cylindrical and coaxial-cylindrical Figure 3. A sequence acquiring diffusion parameters measured for a NBR cylindrical rubber by employing coaxial-cylindrical electrodes in a frequency response analyzer. (a) Precalibration data expressed as a 2nd polynomial equation between the water level and capacitance by quadratic regression, (b) water level transferred from the capacitance with black and blue squares corresponding to the capacitance and transformed water level, respectively, versus time and (c) diffusion parameters D$and$$C ∞$ determined using a diffusion analysis program by application of Equation (10). The blue line is the total compensated emission curve restoring the missing content due to the lag time. Figure 4. A sequence acquiring the diffusion parameter in NBR cylindrical rubber by employing a digital camera without precalibration. (a) Water level versus time after decompression and (b) diffusion parameters D$and$$C ∞$ determined using a diffusion analysis program. The blue line is the total compensated emission curve restoring the missing content due to the lag time. Figure 5. A sequence acquiring diffusion parameters measured for EPDM cylindrical rubber by employing semi-cylindrical capacitor electrodes in a frequency response analyzer. (a) Precalibration data expressed as a 2nd polynomial equation between the water level and capacitance by quadratic regression; (b) water level transformed from the capacitance, where black and blue squares correspond to the capacitance and transformed water level, respectively, versus elapsed time; and (c) diffusion parameters D$and$$C ∞$ determined using a diffusion analysis program by the application of Equation (10). The blue line is the total compensated emission curve restoring the missing content due to the lag time. Figure 6. Sequence of acquiring the diffusion parameter in EPDM cylindrical rubber by employing a digital camera without precalibration. (a) Water level versus time after decompression and (b) diffusion parameters D$and$$C ∞$ determined using a diffusion analysis program. The blue line is the total compensated emission curve restoring the missing content due to the lag time. Figure 7. Stability for volumetric measurement with variations in the temperature and pressure over three days. Figure 8. (a) Gas uptake ($C ∞$) and (b) diffusivity (D) versus exposed pressure for four gases in cylindrical-shaped NBR with different thicknesses. R and T indicate the radius and thickness, respectively, of cylindrical-shaped NBR. Figure 9. (a) Gas uptake ($C ∞$) and (b) diffusivity (D) versus exposed pressure for four gases in cylindrical-shaped EPDM with different thicknesses and spherical-shaped EPDM. R indicates the radius of cylindrical-shaped and spherical-shaped EPDM. T indicates the thickness of the cylindrical-shaped EPDM. Figure 10. (a) A linear correlation between the logarithmic diffusivity and kinetic diameter and (b) a linear correlation between the logarithmic solubility and kinetic diameter of molecules in gas. Solubility (mol/m^3·MPa) Diffusivity (×10^−11 m^2/s) Permeability Specimen (mol/m·s·MPa, ×10^−10) H[2] He N[2] Ar H[2] He N[2] Ar H[2] He N[2] Ar NBR 34.2 8.96 11.0 22.5 5.60 21.5 1.14 2.01 19.2 19.3 1.25 4.53 (35.3) (6.50) (22.8) 25.6 19.7 50.3 EPDM (26.2) 7.79 17.0 38.6 (24.1) 83.1 7.24 10.5 (63.1) 64.8 12.3 40.4 [23] [23] [23] Parameter Coaxial-Cylindrical Semi-Cylindrical Sensitivity ~3 pF/mL ~1 pF/mL Resolution ~0.5 wt·ppm ~2 wt·ppm Stability <10 wt·ppm <15 wt·ppm Detection range ~max 1000 wt·ppm for H[2] ~max 1000 wt·ppm for H[2] Response time <1 s <1 s Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations. © 2022 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 (https: Share and Cite MDPI and ACS Style Jung, J.; Kim, G.; Gim, G.; Park, C.; Lee, J. Determination of Gas Permeation Properties in Polymer Using Capacitive Electrode Sensors. Sensors 2022, 22, 1141. https://doi.org/10.3390/s22031141 AMA Style Jung J, Kim G, Gim G, Park C, Lee J. Determination of Gas Permeation Properties in Polymer Using Capacitive Electrode Sensors. Sensors. 2022; 22(3):1141. https://doi.org/10.3390/s22031141 Chicago/Turabian Style Jung, Jaekap, Gyunghyun Kim, Gahyoun Gim, Changyoung Park, and Jihun Lee. 2022. "Determination of Gas Permeation Properties in Polymer Using Capacitive Electrode Sensors" Sensors 22, no. 3: 1141. Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. 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Negative heat capacities Specific Heat Capacity Energy and Thermal Physics Negative heat capacities Stories from Physics for 11-14 14-16 It seems intuitive that, when energy is added to a system, its temperature will rise. Hence, the notion of negative heat capacities seems an impossibility. However, astrophysicists have argued that a star, or cluster of stars, can cool down when energy is added. Virial theorem describes the mean total kinetic energy of a system of particles bound by a potential over time. When applied to the cores of main sequence stars, as hydrogen is converted to helium by fusion, the mean molecular mass of particles increases and the core collapses. This contraction, according to virial theorem, results in a decrease in potential energy and an increase in thermal energy. Hence the core’s temperature increases as its energy falls, suggesting a negative heat capacity. A similar argument holds for clusters of atoms with negative heat capacities observed in clusters of sodium atoms. Negative heat capacities M. Guidry, Stars and Stellar Processes, Cambridge, University of Cambridge Press, 2019, p. 98 M. Schmidt, R. Kusche, T. Hippler, J. Donges, W. Kronmüller, B. Von Issendorff, & H. Haberland, Negative heat capacity for a cluster of 147 sodium atoms. Physical Review Letters, vol. 86, no. 7, 2001, 1191.	{"url":"https://spark.iop.org/negative-heat-capacities","timestamp":"2024-11-07T22:08:55Z","content_type":"text/html","content_length":"41480","record_id":"<urn:uuid:26c86bb6-e56d-4803-b696-1e5994759dff>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00345.warc.gz"}
Minimum Spanning Tree Problem E Minimum Spanning Tree The input consists of several test cases. Each test case starts with a line with two non-negative integers, $1 \le n \le 20\, 000$ and $0 \le m \le 30\, 000$, separated by a single space, where $n$ is the numbers of nodes in the graph and $m$ is the number of edges. Nodes are numbered from $0$ to $n-1$. Then follow $m$ lines, each line consisting of three (space-separated) integers $u$, $v$ and $w$ indicating that there is an edge between $u$ and $v$ in the graph with weight $-20\, 000 \le w \le 20\, 000$. Edges are undirected. Input will be terminated by a line containing 0 0, this line should not be processed. For every test case, if there is no minimum spanning tree, then output the word Impossible on a line of its own. If there is a minimum spanning tree, then you first output a single line with the cost of a minimum spanning tree. On the following lines you output the edges of a minimum spanning tree. Each edge is represented on a separate line as a pair of numbers, $x$ and $y$ (the endpoints of the edge) separated by a space. The edges should be output so that $x < y$ and should be listed in the lexicographic order on pairs of integers. If there is more than one minimum spanning tree for a given graph, then any one of them will do. Sample Input 1 Sample Output 1 3 0 Impossible	{"url":"https://liu.kattis.com/courses/AAPS/AAPS24/assignments/esj7kp/problems/minspantree","timestamp":"2024-11-13T14:55:02Z","content_type":"text/html","content_length":"26526","record_id":"<urn:uuid:f685d90d-cce9-4458-8692-d5b7c021af43>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00518.warc.gz"}
Geometric Figure \| Lexique de mathématique Geometric Figure A set of points in a geometric space of a given dimension that is used to represent a geometric object (point, segment, line, curve, polygon, polyhedron, etc.). • These are a few examples of geometric figures : • A point is a geometric figure that has zero dimensions. • A line segment is a geometric figure that has 1 dimension. • A square is a geometric figure that has 2 dimensions. • A cube is a geometric figure that has 3 dimensions.	{"url":"https://lexique.netmath.ca/en/geometric-figure/","timestamp":"2024-11-07T19:19:33Z","content_type":"text/html","content_length":"64498","record_id":"<urn:uuid:9444c1c1-3a0f-47f9-8f64-e0605aa8319f>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00246.warc.gz"}
What is the R-Squared Statistics? December 23, 2020 What is the R-Squared Statistics? R-Squared statistics is used to deduce how much our model is able to explain the change in dependent variable. by : Monis Khan Quick Summary: R-Squared statistics is used to deduce how much our model is able to explain the change in dependent variable.	{"url":"https://www.csias.in/what-is-the-r-squared-statistics/","timestamp":"2024-11-08T09:08:19Z","content_type":"text/html","content_length":"23207","record_id":"<urn:uuid:1eba43e1-af43-4050-9d14-09799e062d39>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00569.warc.gz"}
Exploring the Infinite Nature of Primes in Arithmetic Progressions Written on Chapter 1: The Legacy of Dirichlet Lejeune Dirichlet, a prominent German mathematician, demonstrated that there are infinitely many primes in an arithmetic progression. Dirichlet (1805–1859) Born in 1805, Dirichlet developed a profound passion for mathematics at a young age, often spending his allowances on math textbooks. After completing his education, he chose to further his studies in Paris, where the quality of mathematical education exceeded that in Germany at the time. He learned from some of the era's greatest mathematicians, such as Fourier, Laplace, and Poisson. Dirichlet’s initial work revolved around Fermat's Last Theorem, which posits: For n > 2, there are no non-zero integers x, y, z such that x^n + y^n = z^n. While mathematicians like Euler and Fermat had addressed the cases for n = 3 and n = 4, Dirichlet focused on proving the case for n = 5. He established that in this scenario, at least one of the integers x, y, or z must be even, and another must be divisible by 5. He explored two possibilities: one where the number divisible by 5 is even, and another where the even number and the one divisible by 5 are distinct. Dirichlet successfully proved the first scenario and presented his findings to the Paris Academy in 1825, also contributing to the n = 14 case. In 1828, Dirichlet secured a teaching position at a Berlin college, allowing him to also instruct at the University of Berlin. He held the title of Professor of Mathematics there until 1855. After the passing of the renowned mathematician Gauss in 1855, Dirichlet assumed his role in Göttingen, where he remained until his death in 1859. Dirichlet was instrumental in defining mathematical functions and, in 1837, he proved the existence of infinitely many primes within an arithmetic progression, establishing himself as a pioneer in ‘Analytic Number Theory.’ This field employs mathematical analysis to address and resolve problems related to numbers. Section 1.1: Understanding Arithmetic Progression To comprehend Dirichlet’s theorem, we first need to understand what an arithmetic progression is. For instance, consider the sequence 1, 8, 15, 22, 29,... where each term increases by 7. This sequence exemplifies an arithmetic progression, which can be expressed generally as: a, a + b, a + 2b, a + 3b, ... where 'a' represents the first term and 'b' is the common difference. Dirichlet’s Theorem: Let 'a' and 'b' be relatively prime positive integers. Then the sequence a, a + b, a + 2b, a + 3b, ... contains infinitely many prime numbers. The proof of this theorem is intricate and demands a solid understanding of calculus and analytic number theory. I plan to delve into the proof in a subsequent publication. What does this signify? For instance, if we set a = 2 and b = 3, we derive an infinite sequence of primes: 2, 5, 11, 17, ... Notice that Dirichlet’s theorem does not imply that this sequence consists solely of primes; numbers like 8 and 14 are clearly not prime. We can also demonstrate that there are infinitely many primes that end with 999 by selecting a = 999 and b = 1000, ensuring that their greatest common divisor is 1. If you have any feedback or thoughts, feel free to share them below—I read every comment! While you’re here, consider exploring some of my other articles: Unsolved Problems of Primes! Four intriguing challenges in number theory that are straightforward to articulate but complex to solve. The Queen’s Gambit Declined!! Reflecting on the famous Netflix series, The Queen’s Gambit, which garnered 11 Emmy awards, including Best Limited Series. Proof that there are Infinitely Many Primes! Euclid’s proof remains one of the most elegant demonstrations in mathematics. Chess Champion Sues Netflix Over “Sexist” Line in Queen’s Gambit Show Nona Gaprindashvili, the first female grandmaster, is suing Netflix for five million dollars. Thank you for reading! Chapter 2: Videos on Primes in Arithmetic Progressions Primes in Arithmetic Progressions and Sieves This video explores the connection between primes in arithmetic sequences and sieve methods, providing an insightful overview of the topic. Primes in Arithmetic Progressions: The Riemann Hypothesis - and Beyond! In this video, James Maynard discusses the implications of the Riemann Hypothesis concerning primes in arithmetic progressions and its far-reaching significance.	{"url":"https://thespacebetweenstars.com/exploring-infinite-primes-arithmetic-progressions.html","timestamp":"2024-11-06T01:32:23Z","content_type":"text/html","content_length":"14267","record_id":"<urn:uuid:011f9ac3-2bd1-4701-b2c8-fce386fb5279>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00029.warc.gz"}
15CE12E - PRESTRESSED CONCRETE - 2020 -21 Upon completion of this course, the students will be able to CO1: Examine the basics and behavior of prestressed concrete. (K3) CO2: Estimate the concepts of Limit state of serviceability. (K3) CO3: Estimate the Limit state of strength. (K3) CO4: Design of prestressed circular tanks and pipes. (K3) CO5: Analyse the prestressed composite structures. (K3)	{"url":"https://moodle.nec.edu.in/course/info.php?id=1920","timestamp":"2024-11-12T23:15:40Z","content_type":"text/html","content_length":"27606","record_id":"<urn:uuid:379b26fd-d680-4cec-bd73-b2ecdc4276fb>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00607.warc.gz"}
Finding the Most Frequent Elements in an Array Today's algorithm is the Top K Frequent Elements problem: Given a non-empty array of integers, return the k most frequent elements. For example, if you were given the array [1, 1, 1, 2, 2, 3, 3, 3], and k = 2, you'd want to return the two most frequently found elements in the array, which is [1, 3]. This problem has a number of ways to solve it, and many solutions use complex algorithms or sorting techniques. In this post, I'll use commonly found methods to solve this problem. I'll start by discussing how I'll approach the algorithm, and then will code the solution in JavaScript. Approaching the Problem Many times, when algorithms are based on the frequency of an element, it's a good opportunity to use a hash. A hash is so convenient because it stores key-value pairs, where keys can be the element, and the value is its frequency. In this algorithm, we'll create a hash which will store the frequency of each element in the inputted array. We will then use the Object.entries() method, which will turn each key-value pair in the hash into an array of arrays. For example, if the given hash were { '1': 3, '2': 2, '3': 3 }, calling Object.entries() and passing in the hash would give us [ [ '1', 3 ], [ '2', 2 ], [ '3', 3 ] ]. You can read more about Object.entries() here. With this array, we can then sort it by frequency, and ultimately return the first k numbers in the sorted array. Coding the Solution We'll start by initializing an empty object, called hash. We'll then want to go through each element in the nums array and add it to hash. If the element has already been seen in hash, then we can increment its value. Otherwise, we can initialize it to 0. There are many ways to iterate through an array, and in this solution I'll use a for...of loop. You can read more about them here. function topKFrequent(nums, k) { let hash = {} for (let num of nums) { if (!hash[num]) hash[num] = 0 For problems like this, I think it's helpful to stop every so often and see what the variables equal at each point. If we were given nums = [1, 1, 1, 2, 2, 3, 3, 3], then at this point, hash = { '1': 3, '2': 2, '3': 3 }. You may notice that each key in the hash is a string--that'll be an important thing to correct in a later step. For now, we want to turn hash into an array of arrays, using Object.entries(), as discussed above. We'll save the value to a variable called hashToArray. function topKFrequent(nums, k) { let hash = {} for (let num of nums) { if (!hash[num]) hash[num] = 0 const hashToArray = Object.entries(hash) Using the same example, where nums = [1, 1, 1, 2, 2, 3, 3, 3], at this point, hashToArray = [ [ '1', 3 ], [ '2', 2 ], [ '3', 3 ] ]. Now, we want to sort the elements in hashToArray. The first value (index 0) in each inner hash is the element in nums. The second value (index 1) in each inner hash is how many times that element was found in nums. Therefore, since we want to find the most frequent elements, we'll need to sort hashToArray, from most frequently found to least frequently found. We can use .sort(), and sort each inner array by the value at index 1. In other words, we'll pass in the callback function (a,b) => b[1] - a[1]. We'll store this sorted array in a variable called function topKFrequent(nums, k) { let hash = {} for (let num of nums) { if (!hash[num]) hash[num] = 0 const hashToArray = Object.entries(hash) const sortedArray = hashToArray.sort((a,b) => b[1] - a[1]) Continuing with the same example, where nums = [1, 1, 1, 2, 2, 3, 3, 3], at this point, sortedArray = [ [ '1', 3 ], [ '3', 3 ], [ '2', 2 ] ]. Now, for the solution, all we want to return are the most frequently found elements--we don't need to return how many times each element was found. Therefore, we only want the elements at index 0 in sortedArray. As mentioned above, the elements at index 0 are all strings, and we need to return integers. Therefore, we'll use parseInt, which converts a string to an integer, and pass in the numbers at index 0 of each inner array in sortedArray. We'll want to store these sorted elements in a new array, which we will call sortedElements. We'll call .map() on sortedArray, and tell it to return the integer version of the first element in each inner array of sortedArray. function topKFrequent(nums, k) { let hash = {} for (let num of nums) { if (!hash[num]) hash[num] = 0 const hashToArray = Object.entries(hash) const sortedArray = hashToArray.sort((a,b) => b[1] - a[1]) const sortedElements = sortedArray.map(num => parseInt(num[0])) At this point, if nums = [1, 1, 1, 2, 2, 3, 3, 3], then sortedElements = [1, 3, 2]. We're so close! All that's left to do is to return the first k elements of this array. To do that, we'll use .slice (), passing in 0 and k. We will return this sliced off ported of sortedElements, giving us the final result. function topKFrequent(nums, k) { let hash = {} for (let num of nums) { if (!hash[num]) hash[num] = 0 const hashToArray = Object.entries(hash) const sortedArray = hashToArray.sort((a,b) => b[1] - a[1]) const sortedElements = sortedArray.map(num => parseInt(num[0])) return sortedElements.slice(0, k) Let me know if you have any questions or other ways you'd solve this problem! Top comments (1) Justin Wolcott • Thanks for sharing this! I recently got this question during an interview, and my initial solution worked - but was "sub-optimal". Below is my "refactored" version. It keeps track of the item's count in an object in a sortable array, which saves you from having to use Object.entries. A minor speedup, but I thought it was clever and worth sharing. // NOVEL Function to return the {{k}} most frequent items // in a javascript array (of strings or numbers) by SIMULTANEOUSLY // tracking count in a "lookup" object AND a sortable "output" // array // Arguments: // - items: an array of strings or numbers [1,3,500,2,1,3] // - k: the number of results you want returned // Returns: // - an array of the top {{k}} most frequent items // ALGORITHM APPROACH: // 1.) Create an "output" ary like this: [] // 2.) Create a "lookup" object like this: {} // 3.) Iterate over the array of items. // 3a.) If the item does not exist in our "lookup" object // create the following object in the "lookup" object with {{item}} as key // {id: item, count: 0} // * CRITICAL * PUSH the object you just created into our "output" array // by setting reference to the object in the "lookup" object // 3b.) When you update the count-attribute of the object in the "lookup" object, // it AUTOMAGICALLY updates in the "output" array! // 4.) Sort the "output" array descending on its "count" attribute // 5.) Slice to return the "k" top elements of the output array function mostFrequent(items, k){ var lookup = {}; var output = []; var itemCounter = 0; for(var i = 0; i < items.length; i++){ // Have we seen this item before or not? // No? Ok, create an object in our lookup // and set reference to it in our output array lookup[items[i]] = {"count": 0, "id": items[i]}; // Add one to the "count" attribute in our lookup // which adds one to the count attribute in our "output" array // Sort descending, Slice the top {{k}} results, and return it to the user // so they can handle it return output.sort((a,b) => {return a.count > b.count ? -1 : 1}).slice(0,k) // DEMO: console.log(mostFrequent(Array.from({length: 1564040}, () => Math.floor(Math.random() * 40)),4)); For further actions, you may consider blocking this person and/or reporting abuse	{"url":"https://practicaldev-herokuapp-com.global.ssl.fastly.net/alisabaj/finding-the-most-frequent-elements-in-an-array-10k2","timestamp":"2024-11-09T06:27:29Z","content_type":"text/html","content_length":"148254","record_id":"<urn:uuid:bd9a9450-1f3f-44e6-aadc-757c4fde2cc1>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00723.warc.gz"}
Penrose's Zig-Zag Model and Conservation of Momentum 297 views I was reading through Penrose's Road to Reality when I saw his interesting description of the Dirac electron (Chapter 25, Section 2). He points out that in the two-spinor formalism, Dirac's one equation for a massive particle can be rewritten as two equations for two interacting massless particles, where the coupling constant of the interaction is the mass of the electron. In the Dirac formalism, we can write the electron field as $$\psi = \psi_L + \psi_R$$ where $$\psi_L = \frac{1}{2}(1-\gamma_5)\psi$$ and $$\psi_R = \frac{1}{2}(1+\gamma_5)\psi$$. Then the Lagrangian is: $$\mathcal{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi - m\bar{\psi}\psi$$$$\mathcal{L}=i\bar{\psi}_L\gamma^\mu\partial_\mu\psi_L+i\bar{\psi}_R\gamma^\mu\partial_\mu\psi_R - m(\bar{\psi}_L\psi_R + \bar This is the Lagrangian of two massless fields, one left-handed and one right-handed, which interact with coupling constant $$m$$. He then pictorially explains his interpretation by drawing this interaction Feynman-diagram-style. The initial particle (L or R) travels at speed $$c$$ (with luminal momentum) until it "interacts" and transforms into the other particle (R or L) which also travels at speed $$c$$but in the opposite direction. This "zig-zagging" of the particles causes the two-particle system to travel at a net velocity which is less than $$c$$, thus giving the electron a subluminal momentum, thereby granting it a mass of $$m$$. He later states that this interaction is mediated by the Higgs boson, which we get if we replace the coupling constant $$m$$ with the Higgs field. I've tried to put this argument into a mathematical setting, deriving the massive propagator from the two massless propagators via perturbation methods, but what I can't seem to get around is the conservation of momentum. When a "zig" particle changes into a "zag", the direction changes and thus the new particle gains momentum out of "nowhere." If we involve Higgs, then the Higgs boson could carry away/grant the necessary momentum, but I want to know if the model can work without involving the Higgs. Is this possible? This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user FrancisFlute The momentum is conserved. Feynman diagram is not a space-time diagram. It is a "momentum-energy" diagram. It could be seen as a superposition of a infinite sum (over space) of space-time diagrams wich, in Penrose idea, could be some specific space-time zig-zag diagrams. You may see the correspondence fig $25.2$ page $631$ in the book. This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Trimok But for each of the diagrams in the superposition, wouldn't momentum be conserved at each vertex? This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user FrancisFlute No. Without pronouncing about the validity of the zig-zag analogy, , "momentum/energy" Feynman Diagrams (the usual Feynman diagrams) $A(p_1,p_2)$ can be seen as Fourier transform of "space-time" Feynman Diagrams $A(x_1,x_2)$ (Fourier Transform is the superposition). When starting from space-time amplitudes, The sum on intermediary $x$ positions, make terms like $\int dp f(p) ~e^{ip(x_2 - x_1)}$ appear in $A(x_1,x_2)$. And this corresponds to a factor $\delta(p_1-p_2)$ for the momentum amplitude $A(p_1,p_2)$. This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Trimok More exactly, this does not make sense to speak about momentum, when looking at space-time diagrams. This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Trimok Oh, I see it now. I was taking the diagram too literally. If you just take the momentum representation of the propagators it should come out as a geometric series in $m^2$ giving the massive propagator. Thanks! This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user FrancisFlute Is it related to Feynman "checkerboard" zig-zag? This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user arivero @FrancisFlute You write well. I have no knowledge of this subject matter, but your writing is so clear I feel a textbook authored by you would be valuable to many. This post imported from StackExchange Physics at 2024-08-15 20:45 (UTC), posted by SE-user Inquisitive	{"url":"https://physicsoverflow.org/45947/penroses-zig-zag-model-and-conservation-of-momentum","timestamp":"2024-11-08T06:17:44Z","content_type":"text/html","content_length":"129742","record_id":"<urn:uuid:ecb3b552-0f4b-4283-8476-8b84b085062c>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00376.warc.gz"}
Networks@IMT - NETWORKS Lecture Series 23/11/2023, IMT School of Advanced Studies Lucca, San Francesco complex A random day on random graphs and random walks The NETWORKS Unit hosts two seminars by mathematicians Rajat Hazra and Luca Avena Associate Professor, University of Leiden, The Netherlands Associate Professor, University of Florence, Italy Seminar by Rajat Hazra: Spectra of inhomogeneous random graphs 23/11/2024, 9:30-11:00 (Sagrestia, San Francesco complex). Also online at http://imt.lu/sagrestia. Abstract: I will describe some results on the bulk of the spectrum of sparse and dense inhomogeneous random graphs and how the two spectrums are related. In the second part of the talk, I will focus on the edge on the spectrum and describe properties of the largest eigenvalue, eigenvectors and some related large deviation results. Seminar by Luca Avena: Meetings of random walks & consensus dynamics on sparse random digraphs 23/11/2023, 16:00-17:30 (Aula 1, San Francesco complex). Also online at http://imt.lu/aula1. Abstract: I will first discuss the general relation between meeting time of random walks and consensus time in Markovian opinion dynamics on finite networks. I will then focus on sparse random directed graphs and give an account on a recent result in which we characterize in such directed setup how random walks meet and related implications for opinion dynamics. I will in particular discuss how the directed degree structure of the underlying network may speed up or slow down meetings of different random walks. This series aims at introducing the audience to the theory of dynamical systems, encompassing deterministic, disordered, stochastic and irreversible systems, providing a unified description in terms of various statistical indicators, typically associated with the solution of spectral problems. The series consists of four self-contained daily blocks, where each block has a first 2-hour introductory lecture intended for a broader audience (11:00 - 13:00) and a second 2-hour lecture providing a more detailed illustration of useful methods and applications (14:00 - 16:00). Useful resources at this link. Block 1. Deterministic dynamical systems: tools and methods 14/04/2023, 11:00-13:00 (Sagrestia, San Francesco complex) and 14:00-16:00 (Sagrestia, San Francesco complex). Also online at http://imt.lu/seminar. The first block is devoted to the description of the basic mathematical tools, needed to study determinisitic dynamical systems. As applications we discuss how to characterize the dynamics by statistical methods (symbolic representation, etc) and by its spectrum of Lyapunov exponents, making use of analytical and numerical techniques. Topics include: • An introduction to dynamical systems: fluxes and maps; • Mathematical ingredients: singular points, manifolds, Poincarè section, chaotic scenarios; • Dissipative and conservative dynamical systems; • Stability analysis: Lyapunov spectra (covariant definition). Block 2. Dynamical evolutions in spatially extended systems 28/04/2023, 11:00-13:00 (Sagrestia, San Francesco complex) and 14:00-16:00 (Sagrestia, San Francesco complex). Also online at http://imt.lu/seminar. The second block deals with deterministic dynamical systems made of many degrees of freedom, organized on lattices and networks. Applications will be focused on the relation of hydrodynamics with the Lyapunov spectrum in conservative systems, on the spectral method for solving transport problems in disordered harmonic chains and on suitable methods (stable chaos, heterogeneous mean-field) for models of neural networks. Topics include: • Dynamics in systems with many degrees of freedom; • Dynamics and disorder: transport in harmonic chains; • Coupled maps, networks and stable chaos. Block 3. Stochastic dynamics: basic ingredients 12/05/2023, 11:00-13:00 (Sagrestia, San Francesco complex) and 14:00-16:00 (Conference Room, San Ponziano complex). Also online at http://imt.lu/seminar. The third block provides an introduction to stochastic processes in discrete-time (Markov chains) and continuous time (Langevin, Fokker-Planck). For what concerns applications we focus on simple examples of Markov processes (random walker on a ring, Monte-Carlo algorithms) and on models of anomalous diffusion. Topics include: • Markov Chains and their spectral properties (ergodicity); • Stochastic dynamics: Langevin and Fokker-Planck; • Anomalous diffusion. Block 4. Stochastic processes, reversibility and fluctuation relations 26/05/2023, 11:00-13:00 (Sagrestia, San Francesco complex) and 14:00-16:00 (Sagrestia, San Francesco complex). Also online at http://imt.lu/seminar. The fourth block deals with the problem of “irreversibility” in stochastic processes in the framework of the Chapman Kolmogorov equation and on the basic aspects of fluctuation-dissipation relations. Applications are devoted to outline the general approach to solve “first passage problems” and also to describe Jarzinsky, Crooks and Gallavotti-Cohen relations, associated with fluctuating thermodynamics. The final comments will be about linear response theory and coupled transport. Topics include: • Chapman-Kolmogorov equations (first-passage problem); • Fluctuation-dissipation and stochastic thermodynamics; • Linear response theory and transport. 25/11/2019, 27/11/2019, IMT School of Advanced Studies Lucca. Mathematical theory of complex networks IMT Visiting Professor & Assistant Professor, University of Leiden. This is a series of two mini-workshops. Each workshop begins with a longer lecture by the visiting professor, followed by a series of shorter seminars by IMT professors and PhD students that work on related themes. There will be plenty of time devoted to informal discussions, aimed at establishing potential lines of joint research between Dr. Avena and IMT researchers and students. To keep focus, each mini-workshop has a specific theme: "Network coarse-graining" (25 November, 9:00-13:00) and "Brain networks and metastability" (27 November, 9:00-13:00). Both workshops are open to all IMT members, to maximize the number and diversity of potential collaborations generated by the event. Whoever is interested in the topics discussed will have a chance of further discussing with the visiting professor during the week. Workshop 1. Network coarse-graining 27/11/2019, 9:00-13:00 (Aula 2, San Francesco complex) Network coarse-graining refers to the reduction of a larger network to a smaller version of it, by keeping certain properties preserved. This procedure is needed in a diverse range of disciplines, especially when the original network is too big to be dealt with mathematically, computationally or experimentally. The mini-workshop aims at exposing different techniques that have been proposed across disciplines (most prominently mathematics, physics and computer science), and to highlight the open challenges that require a multidisciplinary approach. • 09:00 Luca Avena: "Renormalisation for graphs and signals on graphs via random walk kernels" • 10:30 Mirco Tribastone: "Automatic simplification of large-scale reaction networks" • 11:00 Coffee break • 11:30 Margherita Lalli: "Spectral coarse graining of complex networks" • 12:00 Diego Garlaschelli: "Multiscale network renormalization: scale-invariance without geometry" • 12:30 Matteo Serafino: "Scale-free networks revealed from finite-size scaling" • 13:00 Lunch Workshop 2. Brain networks and metastability 27/11/2019, 9:00-13:00 (Aula 2, San Francesco complex) Metastability is the property of systems that, before eventually reaching a completely stable state, visit a number of intermediate states that are temporarily and transiently stable. This workshop aims at introducing some mathematical aspects of metastability and discussing the potential applications to the empirical analysis of brain networks, especially in relation to the presence of possible transient neural states and/or modular functional structures. • 09:00 Luca Avena: "Mesoscopic explorations of graphs through random rooted forests" • 10:30 Tommaso Gili: "Functional brain networks dynamics: from time series to metastability" • 11:00 Coffee break • 11:30 Rossana Mastrandrea: "Functional brain networks in schizophrenic patients and healthy individuals" • 12:00 Virginia Gurioli: "Reconstruction of functional connectivity brain network in healthy subjects and schizophrenic patients" • 12:15 Adrian Ioan Onicaș: "Detecting task events in fMRI time series based on the topological structure of visibility graphs" • 12:30 Francesca Santucci: "Statistical inference for brain network construction" • 12:45 Diego Garlaschelli: "Communities from brain correlation matrices: functional modules without functional networks" • 13:00 Lunch	{"url":"https://networks.imtlucca.it/networks-lecture-series","timestamp":"2024-11-08T22:01:46Z","content_type":"text/html","content_length":"140379","record_id":"<urn:uuid:fd458329-d9ee-46ee-b60e-a057ba292293>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00231.warc.gz"}
The product of 3 3 = _______.-Turito Are you sure you want to logout? The product of 3 × 3 = _______. A. 0 B. 6 C. 9 D. 12 There are four operations in mathematics which we can use while solving a problem, one of them is multiplication. Multiplication is the repeated addition of the same number a number of times. It is also known as a method of finding product of two or more numbers by multiplying them. Here, we have to find the solve the problem using multiplication. In the question it is given that the we have to find the product of 3 × 3. As we know that 3 × 3 = 9. So, the product of 3 × 3 is 9. Therefore, the correct option is c, i.e., 9. For simple multiplication we have to remember the multiplication table of one-digit numbers and for two-digit number we can find it using the multiplication table of one-digit numbers. Here, we have to find the product of one digit number. Related Questions to study Get an Expert Advice From Turito.	{"url":"https://www.turito.com/ask-a-doubt/Mathematics-the-product-of-3-3-12-9-6-0-q9d17a8","timestamp":"2024-11-04T20:42:12Z","content_type":"application/xhtml+xml","content_length":"72558","record_id":"<urn:uuid:d01234cf-69dc-4bde-ba84-a733ad520621>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00265.warc.gz"}
Solving ratio problems How do I simplify a ratio? Ratios are simplified in the same way as fractions: by doing the same thing to one side as to the other. To simplify, simply divide each figure by the same amount. What does the colon mean in a ratio? The colon in a ratio is read as 'to' so 3:2 is read as 'three to two'.	{"url":"https://evulpo.com/en/uk/dashboard/lesson/uk-m-ks2-09ratio-and-proportion-01solve-problems-with-the-relative-size-of-two-quantities","timestamp":"2024-11-03T12:50:40Z","content_type":"text/html","content_length":"803715","record_id":"<urn:uuid:37a5b335-4b36-4044-b5d2-5e9e1791f265>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00288.warc.gz"}
Motorola Interview Paper There were basically 3 papers -software ,DSP, Semiconductor software paper (20 questions 45 minutes) concentrate more on data structures 10 questions from data structures and 10 from C++ and data structures10 questions were in the fill in the blank format and 10 questions were multiple choice questions. bubble sorting is a)two stage sorting d)none of the above .c++ supports a) pass by value only b) pass by name c) pass by pointer d) pass by value and by reference .Selection sort for a sequence of N elements no of comparisons = _________ no of exchanges = ____________ Insertion sort no of comparisons = _________ no of exchanges = ____________ what is a language? a) set of alphabets b)set of strings formed from alphabets d)none of the above Which is true abt heap sort a)two method sort b)has complexity of O(N2) c)complexity of O(N3) In binary tree which of the following is true a)binary tree with even nodes is balanced b)every binary tree has a balance tree c)every binary tree cant be balanced d)binary tree with odd no of nodes can always be balanced Which of the following is not conducive for linked list implementation of array a)binary search b)sequential search c)selection sort d)bubble sort In c++ ,casting in C is upgraded as Which of the following is true abt AOV(Active On Vertex trees) a)it is an undirected graph with vertex representing activities and edges representing precedence relations b)it is an directed graph “” “” “”” “” “” “” “” “” ” Question on worst and best case of sequential search question on breadth first search char p=”abcdefghijklmno” then printf(“%s”,5[p]); what is the error struct { int item; int x;} main(){ int y=4; return y;} error:absence of semicolon Which of the following is false regarding protected members a)can be accessed by friend functions of the child b) can be accessed by friends of child’s child c)usually unacccessible by friends of class d) child has the ability to convert child ptr to base ptr What is the output of the following void main() int a=5,b=10; int &ref1=a,&ref2=b; ++ ref1; ++ ref2; } value of a and b a)5 and 12 b)7 and 10 c)11 and 11 d)none of the above What does this return f(int n) return n<1?0:n==1?1:f(n-1)+f(n-2) hint:this is to generate fibonacci series code for finding out whether a string is a palindrome,reversal of linked list, recursive computation of factorial with blanks in the case of some variables.we have to fill it out for eg; for palindrome palindrome(char inputstring) int len=strlen ( ?); int start= ?; end =inputstring + ?-?; for(; ?<end && ?==?;++ ?,–?); return(?==?); } we have to replace the question marks(?) with corresponding variables .linked list reversal Linked (Link h) Link temp,r=0,y=h; while(y!= ?) (ans:Null) temp = ?;(ans:y->next) some code here with similar fill in type fill in the blanks type question involving recursive factorial computation	{"url":"https://placement.freshershome.com/motorola/motorola-interview-paper-3_439.html","timestamp":"2024-11-14T20:10:37Z","content_type":"text/html","content_length":"83099","record_id":"<urn:uuid:52b345cb-408e-4248-b6d3-9101e2879760>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00602.warc.gz"}
trivial group nLab trivial group Group Theory Classical groups Finite groups Group schemes Topological groups Lie groups Super-Lie groups Higher groups Cohomology and Extensions Related concepts The trivial group is the group whose underlying set is the singleton, hence whose only element is the neutral element. In the context of nonabelian groups the trivial group is usually denoted $1$, while in the context of abelian groups it is usually denoted $0$ (being the zero object) and also called the zero group (notably in homological algebra). The trivial group is a zero object (both initial and terminal) of Grp. The trivial group is a subgroup of any other group, and the corresponding inclusion $1 \hookrightarrow G$ is the unique such group homomorpism. The quotient group of any group $G$ by itself is the trivial group: $G/G = 1$, and the quotient projection $G \to G/G =1$ is the unique such group homomorphism. It can be nontrivial to decide from a group presentation whether a group so presented is trivial, and in fact the general problem is undecidable. See also combinatorial group theory and word problem. The trivial group is an example of a trivial algebra. Last revised on April 19, 2023 at 16:40:19. See the history of this page for a list of all contributions to it.	{"url":"https://ncatlab.org/nlab/show/trivial+group","timestamp":"2024-11-14T21:27:27Z","content_type":"application/xhtml+xml","content_length":"23443","record_id":"<urn:uuid:88e95903-4ffd-49be-bc50-aa023597fa13>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00619.warc.gz"}
Length & Distance Calculators Online web tools for converting and calculating length, width, height, depth or distance. This tool which is based on the icao standard atmosphere model will calculate the air pressure at a height above or below sea level, the altitude from the atmospheric air pressure at the same level, the pressure difference between two altitudes, and the altitude difference between two atmospheric pressures.	{"url":"https://www.sensorsone.com/section/tools/length-distance-calculators/","timestamp":"2024-11-03T15:29:46Z","content_type":"text/html","content_length":"67180","record_id":"<urn:uuid:6fe3aaa4-a9e1-4027-aab1-110228f247e6>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00801.warc.gz"}
best maths teacher for ssc cpo Archives - Hi students, Welcome to Amans Maths Blogs (AMBIPi). Are you preparing for SSC CPO and looking forSSC CPO Exam Math Number System Question with Solutions AMBIPi? In this article, you will get Previous Year Mathematics Questions asked in SSC CPO, which helps you in the preparation of government job exams Read More	{"url":"https://www.amansmathsblogs.com/tag/best-maths-teacher-for-ssc-cpo/","timestamp":"2024-11-15T04:49:04Z","content_type":"text/html","content_length":"91783","record_id":"<urn:uuid:e5a73861-48df-4c5c-a613-a9770b48f01e>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00381.warc.gz"}
Angular Acceleration CalculatorAngular Acceleration Calculator - Calculator Flares Angular Acceleration Calculator Angular acceleration is a key concept in physics, especially in the study of rotational motion. It helps us understand how the speed of a rotating object changes over time. Whether you’re a student, engineer, or just someone interested in physics, using an angular acceleration calculator can make these calculations easy and accurate. In this article, we’ll explain how to calculate angular acceleration, explore the relevant formulas, and show you how to use an angular acceleration calculator to simplify the process. What is Angular Acceleration? Angular acceleration refers to the rate at which the angular velocity of an object changes with respect to time. It is a vector quantity, meaning it has both magnitude and direction. When an object rotates, its speed can increase or decrease. Angular acceleration measures how quickly this change in rotational speed occurs. For example, if a fan starts rotating faster, its angular acceleration is positive. If it slows down, the angular acceleration is negative. Understanding angular acceleration is important in fields like mechanical engineering, physics, and rotational dynamics. MLU (Mean Length of Utterance) Calculator the Angular Acceleration Formula The formula for calculating angular acceleration is simple: Here, angular acceleration (α) is measured in radians per second squared (rad/s²), which is the SI unit for this quantity. The initial and final angular velocities are measured in radians per second (rad/s), and the time interval is measured in seconds. This formula helps you determine how much the rotational speed of an object changes over a specific time period. The angular acceleration calculator allows you to perform this calculation quickly and How to Use an Angular Acceleration Calculator An angular acceleration calculator is a useful tool that makes it easy to compute the angular acceleration of a rotating object. To use the calculator: 1. Input the Initial Angular Velocity (Ai): Enter the starting speed of the object in radians per second. 2. Input the Final Angular Velocity (Af): Enter the speed of the object after it has accelerated, also in radians per second. 3. Enter the Time Interval (t): This is the time during which the acceleration occurs. 4. Calculate: The calculator will then compute the angular acceleration using the formula mentioned earlier. This tool is particularly beneficial for students and engineers who need to make precise calculations quickly. Units of Angular Acceleration Angular acceleration is typically measured in radians per second squared (rad/s²), which is the SI unit for this quantity. This unit shows how quickly the angular velocity of an object changes per second. Understanding the units is crucial when using an angular acceleration calculator or performing manual calculations. For example, if an object has an angular acceleration of 2 rad/s², it means that its angular velocity increases by 2 radians per second every second. How Torque Relates to Angular Acceleration Torque plays a significant role in determining angular acceleration. Torque is the force that causes an object to rotate, and it is directly related to angular acceleration by the following equation: This equation shows that the torque applied to an object and its moment of inertia determine the angular acceleration. The moment of inertia depends on the mass distribution of the object relative to the axis of rotation. For instance, if you apply a greater torque to a wheel, its angular acceleration will increase, assuming the moment of inertia remains constant. 6. Finding Angular Acceleration from Angular Velocity To find angular acceleration, you need to know the initial and final angular velocities and the time it took for the change. The change in angular velocity over time gives you the angular For example, if a disc starts rotating at 10 rad/s and speeds up to 30 rad/s over 5 seconds, the angular acceleration would be: This means the disc’s angular velocity increases by 4 radians per second every second. The Role of Moment of Inertia in Angular Acceleration The moment of inertia is a measure of an object’s resistance to changes in its rotational motion. It depends on the mass of the object and how that mass is distributed relative to the axis of In the formula τ=I×ατ = I \times ατ=I×α, the moment of inertia (I) directly affects the angular acceleration. A higher moment of inertia means more torque is needed to achieve the same angular For example, a heavy flywheel has a high moment of inertia, making it harder to spin up or slow down compared to a lightweight wheel. Calculating Angular Acceleration Using a Calculator Using an angular acceleration calculator simplifies the process of finding the angular acceleration. By entering the initial and final angular velocities and the time interval, the calculator instantly provides the angular acceleration. For example, if you input an initial angular velocity of 50 rad/s, a final angular velocity of 100 rad/s, and a time interval of 10 seconds, the calculator will show that the angular acceleration is: This tool is especially useful for complex problems where manual calculations might be time-consuming or prone to errors. Applications of Angular Acceleration in Physics and Engineering Angular acceleration is crucial in many areas of physics and engineering. It is used to design and analyze rotating machinery, such as engines, turbines, and gears. Understanding angular acceleration helps engineers create systems that operate smoothly and efficiently. In physics, angular acceleration is important in the study of rotational dynamics, where it helps explain how forces cause objects to rotate. It also plays a role in everyday applications, like understanding how wheels on a car speed up or slow down.	{"url":"https://calculatorflares.com/angular-acceleration-calculator/","timestamp":"2024-11-03T07:35:45Z","content_type":"text/html","content_length":"199368","record_id":"<urn:uuid:064d490f-c14e-4b37-90bc-b0ab1065a5f7>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00112.warc.gz"}
Unit base class Unit base class¶ The unit base class is a bitmap comprising segments of a 32 bit number. all the bits are defined the underlying definition is a set of bit fields that cover a full 32 bit unsigned integer // needs to be defined for the full 32 bits signed int meter_ : 4; signed int second_ : 4; // 8 signed int kilogram_ : 3; signed int ampere_ : 3; signed int candela_ : 2; // 16 signed int kelvin_ : 3; signed int mole_ : 2; signed int radians_ : 3; // 24 signed int currency_ : 2; signed int count_ : 2; // 28 unsigned int per_unit_ : 1; unsigned int i_flag_ : 1; // 30 unsigned int e_flag_ : 1; // unsigned int equation_ : 1; // 32 The default constructor sets all the fields to 0. But this is private and only accessible from friend classes like units. The main constructor looks like constexpr unit_data( int meter, int kilogram, int second, int ampere, int kelvin, int mole, int candela, int currency, int count, int radians, unsigned int per_unit, unsigned int flag, unsigned int flag2, unsigned int equation) an alternative constructor explicit constexpr unit_data(std::nullptr_t); sets all the fields to 1 Math operations¶ When multiplying two base units the powers are added. For the flags. The e_flag and i_flag are added, effectively an Xor while the pu and equation are ORed. For division the units are subtracted, while the operations on the flags are the same. Power and Root and Inv functions¶ For power operations all the individual powers of the base units are multiplied by the power number. The pu and equation flags are passed through. For even powers the i_flag and e_flag are set to 0, and odd powers they left as is. For root operations, First a check if the root is valid, if not the error unit is returned. If it is a valid root all the powers are divided by the root power. The Pu flag is left as is, the i_flag and e_flag are treated the same is in the pow function and the equations flag is set to 0. There is one exception to the above rules. There is a special unit for √Hz it is a combination of some i_flag and e_flag and a high power of the seconds unit. This unit is used in amplitude spectral density and comes up on occasion in some engineering contexts. There is some special logic in the power function that does the appropriate things such the square of √Hz= Hz. If a low power of seconds is multiplied or divided by the special unit it still does the appropriate thing. But √Hz*√Hz will not generate the expected result. √Hz is a singular unique unit and the only coordinated result is a power operation to remove it. √Hz unit base itself uses a power of (-5) on the seconds and sets the i_flag and e_flag. The inverse function is equivalent to pow(-1), and just inverts the unit_data. The unit data type supports getters for the different fields all these are constexpr methods • meter(): return the meter power • kg(): return the kilogram power • second(): return the seconds power • ampere(): return the ampere power • kelvin(): return the kelvin power • mole(): return the mole power • candela(): return the candela power • currency(): return the currency power • count(): return the count power • radian(): return the radian power • is_per_unit(): returns true if the unit_base has the per_unit flag set • is_equation(): returns true if the unit_base has the equation field set • has_i_flag(): returns true if the i_flag is set • has_e_flag(): returns true if the e_flag is set • empty(): will check if the unit_data has any of the base units set, flags are ignored. • unit_type_count: will count the number of base units with a non-zero power there are a few methods will generate a new unit based on an existing one the methods are constexpr • add_per_unit(): will set the per_unit flag • add_i_flag(): will set the i_flag • add_e_flag(): will set the e_flag The method clear_flags is the only non-const method that modifies a unit_data in place. Unit data support the == and != operators. these check all fields. There are a few additional comparison functions that are also available. • equivalent_non_counting(unit_base other) : will return true if all the units but the counting units are equal, the counting units are mole, radian, and count. • has_same_base(unit_base other): will return true if all the units bases are equivalent. So the flags can be different.	{"url":"https://units.readthedocs.io/en/latest/details/unit_base.html","timestamp":"2024-11-13T15:49:48Z","content_type":"text/html","content_length":"40968","record_id":"<urn:uuid:aacb8d76-991c-4ecb-945a-4a611e761008>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00146.warc.gz"}
Hypershot: Fun with Hyperbolic Geometry Hypershot: Fun with Hyperbolic Geometry HYPERSHOT: FUN WITH HYPERBOLIC GEOMETRY Praneet Sahgal MODELING HYPERBOLIC GEOMETRY Upper Half-plane Model (Poincar half-plane model) Poincar Disk Model Klein ... – PowerPoint PPT presentation Number of Views:198 Avg rating:3.0/5.0 more less Transcript and Presenter's Notes Title: Hypershot: Fun with Hyperbolic Geometry 1 Hypershot Fun with Hyperbolic Geometry Modeling Hyperbolic Geometry • Upper Half-plane Model (Poincaré half-plane • Poincaré Disk Model • Klein Model • Hyperboloid Model (Minkowski Model) Image Source Wikipedia Upper Half Plane Model • Say we have a complex plane • We define the positive portion of the complex axis as hyperbolic space • We can prove that there are infinitely many parallel lines between two points on the real axis Image Source Hyperbolic Geometry by James W. Poincaré Disk Model • Instead of confining ourselves to the upper half plane, we use the entire unit disk on the complex • Lines are arcs on the disc orthogonal to the boundary of the disk • The parallel axiom also holds here Image Source http//www.ms.uky.edu/droyster/cour Klein Model • Similar to the Poincaré disk model, except chords are used instead of arcs • The parallel axiom holds here, there are multiple chords that do not intersect Image Source http//www.geom.uiuc.edu/crobles/hy Hyperboloid Model • Takes hyperbolic lines on the Poincaré disk (or Klein model) and maps them to a hyperboloid • This is a stereographic projection (preserves • Maps a 2 dimensional disk to 3 dimensional space (maps n space to n1 space) • Generalizes to higher dimensions Image Source Wikipedia Motion in Hyperbolic Space • Translation in x, y, and z directions is not the same! Here are the transformation matrices • To show things in 3D Euclidean space, we need 4D Hyperbolic space The Project • Create a system for firing projectiles in hyperbolic space, like a first person shooter • Provide a sandbox for understanding paths in hyperbolic space Notable behavior • Objects in the center take a long time to move the space in the center is bigger (see right) Techincal challenges • Applying the transformations for hyperbolic • LOTS of matrix multiplication • Firing objects out of the wand • Rotational transformation of a vector • Distributing among the Cubes walls • Requires Syzygy vector (the data structure) • Hyperbolic viewing frustum Adding to the project • Multiple weapons (firing patterns that would show different behavior) • Collisions with stationary objects • Path tracing • Making sure wall distribution works • 3D models for gun and target (?) • http//mathworld.wolfram.com/EuclidsPostulates.htm • Hyperbolic Geometry by James W. Anderson • http//mathworld.wolfram.com/EuclidsPostulates.htm • http//www.math.ecnu.edu.cn/lfzhou/others/cannon. • http//www.geom.uiuc.edu/crobles/hyperbolic/hypr/	{"url":"https://www.powershow.com/view4/7288ad-MDMwY/Hypershot_Fun_with_Hyperbolic_Geometry_powerpoint_ppt_presentation","timestamp":"2024-11-02T18:23:44Z","content_type":"application/xhtml+xml","content_length":"110617","record_id":"<urn:uuid:52ddb1b6-0ce5-421c-b271-7d6161d29ffe>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00399.warc.gz"}
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[Top 10] Colour Prediction Game Trick - 100% working - colour prediction game Are you looking for some magic to make you rich colour prediction games are the hidden gems that can make your dreams true. It’s not that easy to make money by predicting colour and numbers in those games but it is not impossible. Many people make a good amount of money by playing these games and live a happy life. Today in this blog I will provide you with 10 best 100% working tricks and some bonus tricks that will help you win these games and earn some serious cash. Before applying those tricks you have to remember points, I will provide you with all the details on how to apply them and what you should remember. Before applying all the hacks or tricks you have to know some basic information like which number comes with which color. GREEN 💚 = 1,3,7,9 (odd numbers) RED ♥️ = 2,4,6,8 (even numbers) Violet 💜 = 0,5 (5 comes with green and Violet 💚💜 and 0 comes with Red and violet ♥️💜) Use 2X or 3X treading methods to recover your losses. Here are some best colour prediction games you can join and win a welcome bonus for free Diamond player Join Now Fiewin Join Now Mantri Mall Join Now Godrej Mall Join Now Vclub Join Now Monopoly Join Now Colour prediction game color tricks Earn Daily Up to ₹10,000 without investment Trick 1 In this trick you have to follow some numbers, whenever these numbers come in results you can predict the next color. Remember one thing before using this trick you have to check the previous results, when this trick works if this trick doesn’t apply in the last 20 results or no other trick is overlaid then you can bet on this trick. • After number 6 you have to bet on the RED color • After 4 bets on Green • After 3 bets on Red • After 2 bets on Red • After 9 bets on Green ( this one is tested and worked 100%) • After 5 bets on Red • After 0 bets on Green Maintain enough funds and these tricks need 4 to 5 level funds to work properly. Trick 2 This trick is the safest trick ever, using this trick you will never book a loss. Just maintain the wallet balance and observe the game properly. You can use the 3X method to recover the loss. In this trick, you have to follow the last result. you have to bet on both color and numbers to make the winning chance more accurate. First place the bet on the opposite color in the results, then place the bet on all numbers that came with the color in the previous results, excluding the last results number and 0,5. The betting amount will be 4x in color and 1x in numbers. If you play with 10 then place 40 on the color and 10 on each number. example: if the last result is 3 and Green 💚. Then you have to predict on RED ♥️ X 40(if you play with 10 rupees) and in numbers bet on 1,7 and 9. 10 rupees each on those numbers. If the colour wins 80 rupees or if numbers then you will get 90 rupees. Trick 3 This trick is based on adding and multiplying the period, price, and result number. First, you have to take the mid-three digit from the period and add them. Now add the last two digits from the price. Now add it with the final period number. (final period + final price number) Now multiply the final number by the number from the last result. (final number X result number) You have to take the last number from the final number 48. 8 will be the next upcoming result, number 8 stands for RED color so you have to invest in RED color for the next game. Here also you have to follow the level four management rule. Trick 4 In this trick, you have to pick the last two digits from Period and the second last digit from Price. You have to do some math to figure out the next color. • First Pick the last two digits from Period and subtract them • Now multiply the number by the second last digits of the price. • Pick the last number from that number and add 1 to the number • Here is the final result, Even the number means you have to bet on the RED color. Remember this trick will not work in some games please verify this before use in the game. Trick 5 This one is the same as the number 4 trick. here you have to pick the last two digits of the Period and add them first, now add the last number of the price, the final number is your prediction number, Even for RED and Odd number for GREEN. Trick 6 Now this trick will work with the price, add the last second and third number of the Price. The final ODD or EVEN number is your result. Remember one thing this trick will work only 4/5 times continuously. You have to observe the game if this trick follows the trend three times continuously then you can start betting on the 4th and 5th number game. After completing the 5th game stop betting and wait for the next round. Trick 7 Take a piece of paper, or pen or open the calculator, multiply the last two digits from the Price. Now add all the digits of that final number. If a two or three-digit number comes you have to pick the last one. That odd or even number is the final result. Trick 8 This trick is the combination of the Period and the result Number. Take a piece of paper and add the last number and third last number of periods and add it with the resulting number. Trick 9 The Dragon trend uses two or three X investments to make every round profitable with this trick. Whenever you see the same color three to four times continuously then you have to bet on that same color in the next game. When the same color is repeated more than five times the Dragon trend is going to start and it will go up to 12 to 13 times. Use a 3X or 2X method to recover the losses. Remember one thing, check the last result page if the same trend comes in the last results then don’t bet on that the time system will never repeat a trend immediately, Trick 10 The opposite selection trick. Pick the last two digits from Period and Price and multiply them crossly. First Period Number X First Price Number Second Period Number X Second Price Number Now add and subtract both numbers, if you get two-digit numbers in the answers then pick the last numbers and you have to bet on the opposite color of those numbers. Bonus Tricks As I promise, these are the 10 hacks you can use to earn lots of money. Here are some bonus tricks that will also work. 10 BASIC COLOR & NUMBER Prediction TRICKS A= GREEN, B= RED • Rule 1: 👉 ABBBBA 🟢🔴🔴🔴🔴🟢 = 🔴 4 or 6 • Rule 2: 👉 AABB 🟢🟢🔴🔴 = 🟢🟢 1 or 7 • Rule 3: 👉 ABABA 🟢🔴🟢🔴🟢🔴🟢🔴 = 🔴 2 • Rule 4: 👉 BBAAAABB 🔴🔴🟢🟢🟢🟢🔴🔴 = 🟢 1 • Rule 5: 👉 AAABBB 🟢🟢🟢🔴🔴🔴🟢 = 🔴 6 • Rule 6: 👉 ABB 🟢🔴🔴 = 🔴 8 • Rule 7: 👉 AABBB 🟢🟢🔴🔴🔴 = 🟢7 • Rule 8: 👉 AAAA 🟢🟢🟢 = 🟢 5 • Rule 9: 👉 AABAAA 🟢🟢🔴🟢🟢 = 🟢 • Rule 10: 👉 AABBBA 🟢🟢🔴🔴🔴🟢 = 🔴 Follow the telegram channel, every colour prediction game has an official telegram channel. You can follow the official telegram channel for the prediction podcast. Join this telegram channel “Colour Prediction Game” and watch the podcast. This channel provides a 98% accurate prediction if you think it performs well then register yourself and recharge enough funds to make more money. As per your budget, it is recommended to start with a small amount. If you start playing with the amount of ₹10 as per the triple investment plan, if you lose ₹10 seconds invest ₹30 then ₹90 then ₹270. If you win at level 4. You can earn ₹140. If you invest ₹100 then you can win ₹1400. The double investment plan is the same as a triple investment plan. Invest ₹10 – ₹20 -₹40 -₹80 – ₹160. All the return depends on how much you invest. what is colour prediction game? Ans: colour prediction games are a kind of online betting app where you can earn real money by predicting the right color. Which colour prediction game is best? Ans: This is very difficult to say the name of the best colour prediction game because all the games are working on the same algorithm. all the best who approved the withdrawal on time, in my opinion, Dimond player is my favourite one. How to play colour prediction game? Ans: its very simple just you have three options of colors and 10 numbers. you have to predict the right color or number. There are lots of colour prediction game tricks but some tricks are fake or not working. These are the best tricks I tested in many colour prediction games. All those tricks do not work in a single game. Every game has different patterns you have to observe the game and find out the unique pattern. These tricks worked with me but it doesn’t mean that this will also work in your game before applying these tricks observe the patterns. These tricks are only for educational purposes i am not responsible for any kind of financial or other kind of loss.	{"url":"https://colourpredictiongames.in/colour-prediction-game-hacks/","timestamp":"2024-11-07T00:52:38Z","content_type":"text/html","content_length":"195221","record_id":"<urn:uuid:d568f51f-7371-4cf4-84aa-a674e8536eb1>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00697.warc.gz"}
What's a Walk Worth? Moneyball. What is it and why is it important? Many people have seen the film starring Brad Pitt or even read the book by Michael Lewis. Put very simply, the idea is that getting more people on base by any means increases a teams chance of scoring runs, and ultimately winning the game. This idea intuitively makes sense, however, the application challenged baseball norms at the time. At the time, teams were obsessed with putting the ball in play and making flashy, big plays. The ‘Moneyball’ concept does not care how the runners get on. Basehits, walks, bunts, and even hit by pitches count! 14 years after publication (2003), let’s see what a walk is worth and how it relates to winning. Data Support To begin, let’s take a look at walk totals by season for the entire league. These results were somewhat surprising to me. However, I think the time period of 2005 – 2009 is worth exploring. The early 2000’s were plagued by the steroid era in baseball, offense was up overall and pitching was down since every fucking player put needles in their buttholes (still love Sammy Sosa though). Every year from 2005 – 2009 saw an increase in walks. Does this make sense? What could have happened in 2004? oh yeah, the GOAT Theo Epstein and boy wonder, sabermetric genius won the World Series in 2004 with the Red Sox (suck it Cardinals). Baseball is such a copycat league, it makes sense other teams would try to copy a similar model to those Red Sox. I know what you’re thinking, “okay, sure, but what about 2010 – 2014! Players forgot what a walk was!” This is explained by better pitching. The graph below summarizes league starters ERA by season. It almost beautifully correlates with the walk totals. These years were just dominated by great pitching. Next, notice that in the walk totals and ERA, that there is an uptick in 2015. If you follow baseball, you’d be aware that there is a story here regarding increased homeruns. See the chart below for home run totals by year. Two things should stand out: 1. This chart also trends well with ERA and Walks 2. Massive explosion in homeruns in 2015-present Again, as similar to the early 2000’s, with more home runs/offense come more walks. From a baseball point of view this makes sense. If the ball is flying out of the park, pitchers will be more careful and walk more guys. When this happens, more guys are on base, and even if the left on base % stays constant at 75%, more runners will score, because more runners are on base. Thus, the increase in ERA. Hopefully by now, you understand why walks are important. Walks relate strongly to a teams offense. Let’s investigate this correlation a little more. Walks and wins I’d like to start with the following graph. If this doesn’t convince you that walks matter, literally nothing will. For each of the 18 seasons, I grabbed the top 10 teams in walks (blue), bottom 10 teams in walks (red) and plotted the average win totals for those teams compared to the league average win total (81). In multiple seasons, this difference is worth upwards of 20 wins. For a game dubbed the ‘game of inches’ and a league where teams miss the playoffs by 1 or 2 games this difference is huge. It’s pretty obvious, get more baserunners via the walk and win more games. Now that the importance of walks is incredibly clear, just how important is a given walk to winning a game? By comparing the ratio of walks to wins for each of the last 18 seasons, a single walk can be worth as much as 0.18 wins. Next time you want to yell at your favorite player for taking his walks instead of swinging for the fences on 3-0 you better think twice.	{"url":"http://www.feelslikeanalytics.com/2018/03/whats-walk-worth.html","timestamp":"2024-11-04T10:10:53Z","content_type":"application/xhtml+xml","content_length":"98788","record_id":"<urn:uuid:ef66f9fa-439a-4a53-9f75-64140c3cde9e>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00473.warc.gz"}
Butterworth filter design [b,a] = butter(n,Wn) designs an nth-order lowpass digital Butterworth filter with normalized cutoff frequency Wn. The butter function returns the numerator and denominator coefficients of the filter transfer function. [b,a] = butter(n,Wn,ftype) designs a lowpass, highpass, bandpass, or bandstop digital Butterworth filter, depending on the value of ftype and the number of elements of Wn. The resulting bandpass and bandstop filter designs are of order 2n. You might encounter numerical instabilities when designing IIR filters with transfer functions for orders as low as 4. See Transfer Functions and CTF for more information about numerical issues that affect forming the transfer function. [z,p,k] = butter(___) designs a digital Butterworth filter and returns its zeros, poles, and gain. This syntax can include any of the input arguments in previous syntaxes. [A,B,C,D] = butter(___) designs a digital Butterworth filter and returns the matrices that specify its state-space representation. [___] = butter(___,"s") designs an analog Butterworth filter using any of the input or output arguments in previous syntaxes. [B,A] = butter(n,Wn,"ctf") designs a lowpass digital Butterworth filter using second-order Cascaded Transfer Functions (CTF). The function returns matrices that list the denominator and numerator polynomial coefficients of the filter transfer function, represented as a cascade of filter sections. This approach generates IIR filters with improved numerical stability compared to with single-section transfer functions. (since R2024b) [___] = butter(n,Wn,ftype,"ctf") designs a lowpass, highpass, bandpass, or bandstop digital Butterworth filter, and returns the filter representation using the CTF format. The resulting design sections are of order 2 (lowpass and highpass filters) or 4 (bandpass and bandstop filters). (since R2024b) [___,gS] = butter(___) also returns the overall gain of the system. You must specify "ctf" to return gS. (since R2024b) Lowpass Butterworth Transfer Function Design a 6th-order lowpass Butterworth filter with a cutoff frequency of 300 Hz, which, for data sampled at 1000 Hz, corresponds to $0.6\pi$ rad/sample. Plot its magnitude and phase responses. Use it to filter a 1000-sample random signal. fc = 300; fs = 1000; [b,a] = butter(6,fc/(fs/2)); ylim([-100 20]) dataIn = randn(1000,1); dataOut = filter(b,a,dataIn); Bandstop Butterworth Filter Design a 6th-order Butterworth bandstop filter with normalized edge frequencies of $0.2\pi$ and $0.6\pi$ rad/sample. Plot its magnitude and phase responses. Use it to filter random data. [b,a] = butter(3,[0.2 0.6],'stop'); dataIn = randn(1000,1); dataOut = filter(b,a,dataIn); Highpass Butterworth Filter Design a 9th-order highpass Butterworth filter. Specify a cutoff frequency of 300 Hz, which, for data sampled at 1000 Hz, corresponds to $0.6\pi$ rad/sample. Plot the magnitude and phase responses. Convert the zeros, poles, and gain to second-order sections. Display the frequency response of the filter. [z,p,k] = butter(9,300/500,"high"); sos = zp2sos(z,p,k); Bandpass Butterworth Filter Design a 20th-order Butterworth bandpass filter with a lower cutoff frequency of 500 Hz and a higher cutoff frequency of 560 Hz. Specify a sample rate of 1500 Hz. Use the state-space representation. Convert the state-space representation to second-order sections. Visualize the frequency responses. fs = 1500; [A,B,C,D] = butter(10,[500 560]/(fs/2)); sos = ss2sos(A,B,C,D); Design an identical filter using designfilt. Visualize the frequency responses. d = designfilt("bandpassiir",FilterOrder=20, ... HalfPowerFrequency1=500,HalfPowerFrequency2=560, ... Comparison of Analog IIR Lowpass Filters Design a fifth-order analog Butterworth lowpass filter with a cutoff frequency of 2 GHz. Multiply by $2\pi$ to convert the frequency to radians per second. Compute the frequency response of the filter at 4096 points. n = 5; wc = 2pi2e9; w = 2pi1e9logspace(-2,1,4096)'; [zb,pb,kb] = butter(n,wc,"s"); [bb,ab] = zp2tf(zb,pb,kb); [hb,wb] = freqs(bb,ab,w); gdb = -diff(unwrap(angle(hb)))./diff(wb); Design a fifth-order Chebyshev Type I filter with the same edge frequency and 3 dB of passband ripple. Compute its frequency response. [z1,p1,k1] = cheby1(n,3,wc,"s"); [b1,a1] = zp2tf(z1,p1,k1); [h1,w1] = freqs(b1,a1,w); gd1 = -diff(unwrap(angle(h1)))./diff(w1); Design a fifth-order Chebyshev Type II filter with the same edge frequency and 30 dB of stopband attenuation. Compute its frequency response. [z2,p2,k2] = cheby2(n,30,wc,"s"); [b2,a2] = zp2tf(z2,p2,k2); [h2,w2] = freqs(b2,a2,w); gd2 = -diff(unwrap(angle(h2)))./diff(w2); Design a fifth-order elliptic filter with the same edge frequency, 3 dB of passband ripple, and 30 dB of stopband attenuation. Compute its frequency response. [ze,pe,ke] = ellip(n,3,30,wc,"s"); [be,ae] = zp2tf(ze,pe,ke); [he,we] = freqs(be,ae,w); gde = -diff(unwrap(angle(he)))./diff(we); Design a fifth-order Bessel filter with the same edge frequency. Compute its frequency response. [zf,pf,kf] = besself(n,wc); [bf,af] = zp2tf(zf,pf,kf); [hf,wf] = freqs(bf,af,w); gdf = -diff(unwrap(angle(hf)))./diff(wf); Plot the attenuation in decibels. Express the frequency in gigahertz. Compare the filters. fGHz = [wb w1 w2 we wf]/(2e9pi); plot(fGHz,mag2db(abs([hb h1 h2 he hf]))) axis([0 5 -45 5]) grid on xlabel("Frequency (GHz)") ylabel("Attenuation (dB)") legend(["butter" "cheby1" "cheby2" "ellip" "besself"]) Plot the group delay in samples. Express the frequency in gigahertz and the group delay in nanoseconds. Compare the filters. gdns = [gdb gd1 gd2 gde gdf]*1e9; gdns(gdns<0) = NaN; grid on xlabel("Frequency (GHz)") ylabel("Group delay (ns)") legend(["butter" "cheby1" "cheby2" "ellip" "besself"]) The Butterworth and Chebyshev Type II filters have flat passbands and wide transition bands. The Chebyshev Type I and elliptic filters roll off faster but have passband ripple. The frequency input to the Chebyshev Type II design function sets the beginning of the stopband rather than the end of the passband. The Bessel filter has approximately constant group delay along the passband. Highpass Butterworth Filter in Cascade Design a ninth-order highpass Butterworth filter with a cutoff frequency of 300 Hz and sampling rate of 1000 Hz. Return the coefficients of the filter system as a cascade of second-order sections. Wn = 300/(1000/2); [B,A] = butter(9,Wn,"high","ctf") B = 5×3 0.2544 -0.2544 0 0.2544 -0.5088 0.2544 0.2544 -0.5088 0.2544 0.2544 -0.5088 0.2544 0.2544 -0.5088 0.2544 A = 5×3 1.0000 0.1584 0 1.0000 0.3264 0.0561 1.0000 0.3575 0.1570 1.0000 0.4189 0.3554 1.0000 0.5304 0.7165 Plot the magnitude response of the filter. Input Arguments Wn — Cutoff frequency scalar \| two-element vector Cutoff frequency, specified as a scalar or a two-element vector. The cutoff frequency is the frequency at which the magnitude response of the filter is 1 / √2. • If Wn is scalar, then butter designs a lowpass or highpass filter with cutoff frequency Wn. If Wn is the two-element vector [w1 w2], where w1 < w2, then butter designs a bandpass or bandstop filter with lower cutoff frequency w1 and higher cutoff frequency w2. • For digital filters, the cutoff frequencies must lie between 0 and 1, where 1 corresponds to the Nyquist rate—half the sample rate or π rad/sample. For analog filters, the cutoff frequencies must be expressed in radians per second and can take on any positive value. Data Types: double ftype — Filter type "low" \| "bandpass" \| "high" \| "stop" Filter type, specified as one of the following: • "low" specifies a lowpass filter with cutoff frequency Wn. "low" is the default for scalar Wn. • "high" specifies a highpass filter with cutoff frequency Wn. • "bandpass" specifies a bandpass filter of order 2n if Wn is a two-element vector. "bandpass" is the default when Wn has two elements. • "stop" specifies a bandstop filter of order 2n if Wn is a two-element vector. Output Arguments B,A — Cascaded transfer function (CTF) coefficients row vector \| matrix Since R2024b Cascaded transfer function (CTF) coefficients, returned as a row vector or matrix. B and A list the numerator and denominator coefficients of the cascaded transfer function, respectively. The sizes for B and A are L-by-(m+1) and L-by-(n+1), respectively. The function returns the first column of A as 1, thus A(1)=1 when A is a row vector. • L represents the number of filter sections. • m represents the order of the filter numerators. • n represents the order of the filter denominators. The butter function returns the CTF coefficients with these order specifications: • m = n = 2 for lowpass and highpass filters. • m = n = 4 for bandpass and bandstop filters. To customize the CTF coefficient computation, such as setting a different order in the CTF coefficients or customizing the gain scaling, specify to return z,p,k and then use zp2ctf to obtain B,A. For more information about the cascaded transfer function format and coefficient matrices, see Return Digital Filters in CTF Format. gS — Overall system gain real-valued scalar Since R2024b Overall system gain, returned as a real-valued scalar. • If you specify to return gS, the butter function normalizes the numerator coefficients so that the first column of B is 1 and returns the overall system gain in gS. • If you do not specify to return gS, the butter function uniformly distributes the system gain across all system sections using the scaleFilterSections function. More About Cascaded Transfer Functions Partitioning an IIR digital filter into cascaded sections improves its numerical stability and reduces its susceptibility to coefficient quantization errors. The cascaded form of a transfer function H(z) in terms of the L transfer functions H[1](z), H[2](z), …, H[L](z) is $H\left(z\right)=\prod _{l=1}^{L}{H}_{l}\left(z\right)={H}_{1}\left(z\right)×{H}_{2}\left(z\right)×\cdots ×{H}_{L}\left(z\right).$ Return Digital Filters in CTF Format Specify B and A to return the filter coefficients. You can also specify gS to return the overall system gain of the filter. By specifying these output arguments, you can design digital filters in the CTF format for analysis, visualization, and signal filtering. Filter Coefficients When you specify to return the numerator and denominator coefficients in the CTF format, the L-row matrices B and A are returned as $B=\left[\begin{array}{cccc}{b}_{11}& {b}_{12}& \cdots & {b}_{1,m+1}\\ {b}_{21}& {b}_{22}& \cdots & {b}_{2,m+1}\\ ⋮& ⋮& \ddots & ⋮\\ {b}_{L1}& {b}_{L2}& \cdots & {b}_{L,m+1}\end{array}\right],\text{ }A=\left[\begin{array}{cccc}1& {a}_{12}& \cdots & {a}_{1,n+1}\\ 1& {a}_{22}& \cdots & {a}_{2,n+1}\\ ⋮& ⋮& \ddots & ⋮\\ 1& {a}_{L2}& \cdots & {a}_{L,n+1}\end{array}\right],$ such that the full transfer function of the filter is $H\left(z\right)=\frac{{b}_{11}+{b}_{12}{z}^{-1}+\cdots +{b}_{1,m+1}{z}^{-m}}{1+{a}_{12}{z}^{-1}+\cdots +{a}_{1,n+1}{z}^{-n}}×\frac{{b}_{21}+{b}_{22}{z}^{-1}+\cdots +{b}_{2,m+1}{z}^{-m}}{1+{a}_{22} {z}^{-1}+\cdots +{a}_{2,n+1}{z}^{-n}}×\cdots ×\frac{{b}_{L1}+{b}_{L2}{z}^{-1}+\cdots +{b}_{L,m+1}{z}^{-m}}{1+{a}_{L2}{z}^{-1}+\cdots +{a}_{L,n+1}{z}^{-n}},$ where m ≥ 0 is the numerator order of the filter and n ≥ 0 is the denominator order. • To filter signals using cascaded transfer functions, use the ctffilt function. • To analyze filters represented as cascaded transfer functions, use the Filter Analyzer app. You can also use these Signal Processing Toolbox™ functions to visualize and analyze filters in CTF □ Time-Domain Responses — impzlength, impz, and stepz □ Frequency-Domain Responses — freqz, grpdelay, phasedelay, phasez, zerophase, and zplane □ Filter Exploration — filtord, islinphase, ismaxphase, isminphase, and isstable Coefficients and Gain You can specify to return the coefficients and overall system gain using the output argument triplet [B,A,gS]. In this case, the numerator coefficients are normalized, returning the filter coefficient matrices and gain as $B=\left[\begin{array}{cccc}1& {\beta }_{12}& \cdots & {\beta }_{1,m+1}\\ 1& {\beta }_{22}& \cdots & {\beta }_{2,m+1}\\ ⋮& ⋮& \ddots & ⋮\\ 1& {\beta }_{L2}& \cdots & {\beta }_{L,m+1}\end{array}\ right],\text{ }A=\left[\begin{array}{cccc}1& {a}_{12}& \cdots & {a}_{1,n+1}\\ 1& {a}_{22}& \cdots & {a}_{2,n+1}\\ ⋮& ⋮& \ddots & ⋮\\ 1& {a}_{L2}& \cdots & {a}_{L,n+1}\end{array}\right],\text{ }{g}_{\ so that the transfer function is $H\left(z\right)={g}_{\text{S}}\left(\frac{1+{\beta }_{12}{z}^{-1}+\cdots +{\beta }_{1,m+1}{z}^{-m}}{1+{a}_{12}{z}^{-1}+\cdots +{a}_{1,n+1}{z}^{-n}}×\frac{1+{\beta }_{22}{z}^{-1}+\cdots +{\beta }_ {2,m+1}{z}^{-m}}{1+{a}_{22}{z}^{-1}+\cdots +{a}_{2,n+1}{z}^{-n}}×\cdots ×\frac{1+{\beta }_{L2}{z}^{-1}+\cdots +{\beta }_{L,m+1}{z}^{-m}}{1+{a}_{L2}{z}^{-1}+\cdots +{a}_{L,n+1}{z}^{-n}}\right).$ This transfer function is equivalent to the one defined in the Filter Coefficients section, where g[S] = b[11]×b[21]×...×b[L1], and β[li] = b[li]/b[l1] for i = 1,2,…,m and l = 1,2,…,L. Transfer Functions and CTF Numerical Instability of Transfer Function Syntax In general, use cascaded transfer functions ("ctf" syntaxes) to design IIR digital filters. If you design the filter using transfer functions (any of the [b,a] syntaxes), you might encounter numerical instabilities. These instabilities are due to round-off errors and can occur for an order n as low as 4. This example illustrates this limitation. n = 6; Fs = 200e6; Wn = [0.5e6 6e6]/(Fs/2); ftype = "bandpass"; % Transfer Function (TF) design [b,a] = butter(n,Wn,ftype); % This is an unstable filter % CTF design [B,A] = butter(n,Wn,ftype,"ctf"); % Compare frequency responses [hTF,f] = freqz(b,a,8192,Fs); hCTF = freqz(B,A,8192,Fs); grid on legend(["TF Design" "CTF Design"]) xlabel("Frequency (MHz)") ylabel("Magnitude (dB)") Butterworth filters have a magnitude response that is maximally flat in the passband and monotonic overall. This smoothness comes at the price of decreased rolloff steepness. Elliptic and Chebyshev filters generally provide steeper rolloff for a given filter order. butter uses a five-step algorithm: 1. It finds the lowpass analog prototype poles, zeros, and gain using the function buttap. 2. It converts the poles, zeros, and gain into state-space form. 3. If required, it uses a state-space transformation to convert the lowpass filter into a bandpass, highpass, or bandstop filter with the desired frequency constraints. 4. For digital filter design, it uses bilinear to convert the analog filter into a digital filter through a bilinear transformation with frequency prewarping. Careful frequency adjustment enables the analog filters and the digital filters to have the same frequency response magnitude at Wn or at w1 and w2. 5. It converts the state-space filter back to its transfer function or zero-pole-gain form, as required. [1] Lyons, Richard G. Understanding Digital Signal Processing. Upper Saddle River, NJ: Prentice Hall, 2004. Extended Capabilities C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. Version History Introduced before R2006a R2024b: Design digital filters using cascaded transfer functions The butter function supports outputs in the cascaded transfer function (CTF) format.	{"url":"https://in.mathworks.com/help/signal/ref/butter.html","timestamp":"2024-11-04T00:59:35Z","content_type":"text/html","content_length":"161623","record_id":"<urn:uuid:96cf6764-8193-440d-9637-cec5f22bf644>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00577.warc.gz"}
Get Shorty Problem A Get Shorty Nils and Mikael are intergalaxial fighters as well as lethal enemies. Now Nils has managed to capture the poor Mikael in his dark dungeons, and it is up to you to help Mikael escape with as much of his pride intact as possible. The dungeons can be viewed as a set of corridors and intersections. Each corridor joins two intersections. There are no guards, traps, or locked doors in Nils’ dungeon. However, there is one obstacle which makes escaping from the dungeon a perilious project: in each corridor there is a sentry, armed with a factor weapon. (As is commonly known, a factor weapon with factor $f$ reduces the size of its target to a factor $f$ of its original size, e.g. if Mikael is $8$ gobs large and is hit by a factor weapon with factor $f = 0.25$ his size will be reduced to $2$ gobs.) Mikael will not be able to pass through a corridor without being hit by the factor weapon (but luckily enough, reloading the factor weapon takes enough time that the sentry will only have time to shoot him once as he passes through). It seems inevitable that Mikael will come out of this adventure a smaller man, but since the sentries have different factors in their factor weapons, his final size depends very much on the route he takes to the exit of the dungeons. Naturally, he would like to lose as little size as possible, and has asked you to help him accomplish that. Input consists of a series of test cases (at most $20$). Each test case begins with a line consisting of two integers $n$, $m$ separated by a single space, where $2 \le n \le 10\, 000$ is the number of intersections and $1 \le m \le 15\, 000$ is the number of corridors in Nils’ dungeon. Then follow $m$ lines, each containing two integers $x$, $y$ and a real number $f$ (with at most four decimals), indicating that there is a corridor between intersections $x$ and $y$, and that the factor weapon of the sentry in that corridor has factor $0 \le f \le 1$. Intersections are numbered from $0$ to $n-1$. Mikael always starts in intersection $0$, and the exit is located in intersection $n-1$. The last case will be followed by a case where $n = m = 0$, which should not be processed. For each test case, output a single line containing a real number (with exactly four decimals) indicating how big a fraction of Mikael will be left when he reaches the exit, assuming he chooses the best possible route through the dungeon. You may assume that it is always possible for Mikael to reach the exit. Sample Input 1 Sample Output 1 0 1 0.9 1 2 0.9 0.8100 0 2 0.8 1.0000	{"url":"https://liu.kattis.com/courses/AAPS/AAPS24/assignments/tenby4/problems/getshorty","timestamp":"2024-11-08T10:38:48Z","content_type":"text/html","content_length":"27906","record_id":"<urn:uuid:03708c95-4b47-46d8-aa8e-6a262a75749c>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00091.warc.gz"}
EViews Help: fill Fill a matrix object with specified values. matrix_name.fill(options) n1[, n2, n3 …] Follow the keyword with a list of values to place in the matrix object. Each value should be separated by a comma. Running out of values before the object is completely filled is not an error; the remaining cells or observations will be unaffected, unless the “l” option is specified. If, however, you list more values than the object can hold, EViews will not modify any observations and will return an error message. l Loop repeatedly over the list of values as many times as it takes to fill the object. o=integer (default=1) Fill the object starting from the specified element. Default is the first element. b=arg (default=“c”) Matrix fill order: “c” (fill the matrix by column), “r” (fill the matrix by row). The commands, matrix(2,2) m1 matrix(2,2) m2 m1.fill 1, 0, 1, 2 m2.fill(b=r) 1, 0, 1, 2 create the matrices: “Matrix Language” for a detailed discussion of vector and matrix manipulation in EViews.	{"url":"https://help.eviews.com/content/matrixcmd-fill.html","timestamp":"2024-11-06T12:20:12Z","content_type":"application/xhtml+xml","content_length":"12475","record_id":"<urn:uuid:2a8d745e-19a2-4d72-b631-4524301b43ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00142.warc.gz"}
ALL Python Programmers Should Know This!! This is a powerful tip that all Python programmers should know. So here I have a list of numbers from one to one thousand: nums = range(1,1000) But what I want to do is to get all the prime numbers in that list. So what I will do first, is create a function called is_prime which takes in a single number and returns false if the number is not prime and returns true if it is: nums = range(1,1000) def is_prime(num): for x in range(2, num): if (num % x) == 0: return False return True And now all I have to do is use Python’s built-in filter() function. I’ll add is_prime and the nums as inputs: nums = range(1,1000) def is_prime(num): for x in range(2, num): if (num % x) == 0: return False return True primes=filter(is_prime, nums) Essentially what this does is applies the is_primes function to every item in the nums list. If a boolean True is returned, it stores it in our primes variable otherwise, it removes the number. All we have to do now is print our primes variable however you will see it prints out a filter object. This is Python’s way of conserving memory and so we need to convert it into a list by putting primes inside of a list function and now we can run: nums = range(1,1000) def is_prime(num): for x in range(2, num): if (num % x) == 0: return False return True primes=filter(is_prime, nums)	{"url":"https://amir-tech.medium.com/all-python-programmers-should-know-this-901a364932b5?source=user_profile_page---------9-------------adbe977e94f5---------------","timestamp":"2024-11-07T13:10:32Z","content_type":"text/html","content_length":"94721","record_id":"<urn:uuid:81cddafe-d2c9-4529-af30-049140f718e6>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00216.warc.gz"}
How is this wrong? youre a faggot, thats how You completely ignored your paub for . 6 How the fuck did I not only answer these types of questions in high school, but had one of the highest math grades in my entire class? I look at this now and see miscellaneous characters. A U B would be 1.2 retard If they're completely enclosed such that A = B, then #3 would be true. So, it's not wrong. For some reason your teacher didn't mark you wrong for the next one too, so it was inconsistent. Talk to your teacher, he messed up grading. Explain your reasoning. What kind of math is this? >samefag, cont. Your example shows they can't be mutually exclusive since P(A)+P(B) = 1.2??? So, you'd need the P(A U B) = P(A) + P(B) - P(A N B) and your example would have P(A) = 0.6, P(B) = 0.6, P(A N B) = 0.6, resulting in P(A U B) = 0.6. That's how to explain it. That's what I thought, but my T.A is a street shitter and said that it was wrong and that I didn't show work. Probabilities can't go above 1 Except the probability that you're literally retarded Talk to the professor directly, man. Since you said he's indian, the TA probably hasn't taken the class in 4 years and knows nothing about what the class is. Just skip the middle man. Bruh, when I was in college I did calculus by hand for orbits in Astro. Now I'm on /b and I'm a retard and can't do shit. Discrete Math or Probability Probabilities should add up to one u must be 18 2 post here user. real men are feminist Probabilities add up to 1 only for P(A U B) + P(A U B)', those are mutually exclusive. P(A) and P(B) are not mutually exclusive. I just noticed the question says Les than 1 But being a stats enthusiast drunk as i am right now I think you're right. Whoever wrote the question is a retard What kind of math is this? Can someone explain? How do you come up with numbers from random letters? this is post-high school stats in usa. Set theory. The big U is for Union. To find union you just multiply A and B .6 * .6 isnt .6 retard, how are you taking a probability unit in math and on chan It's statistical probabilities. P(A) = % Probability of Event A P(B) = % Probability of Event B P(AUB)= Probability of A OR B P(AUB)' = Whatever value will give you a total of 1, which represents 100%. It has to be 100% because you can't have a "120% chance" at something; It's impossible. It is also important to know that P (A) and P(B) are not mutually exclusive, which means they are independent of each other. A formula like P(ANB) (N is A AND B) is mutually exclusive because Example: Flip a coin ten times and record what face the coin falls on. Let's say Heads (A) hits 4 times and Tails (B) hits 6 times. Your probabilities would be: Heads = 0.4 Tails = 0.6 Heads OR tails = 0.24 Heads AND tails = 0.76 Heads OR tails prime = P((AUB)') = 1 - P(AUB) = 1 - 0.76 = 0.24 >mutually exclusive because My bad. I went to check my reasoning and forgot to fucking type the fact in. Derp. I'll just post a picture people like those. In the picture, King of Hearts is NOT mutually exclusive to either Hearts or Kings. But the King of Spades, for example, is independent from Hearts. If you were measuring probabilities for Spades in the same situation, you'd get the same math answer. Calculations don't always reflect the reality. Chance. As stated, it's not wrong. But the person who is grading the question is probably working under an unstated assumption that A and B are completely independent variables. However, you answer could be correct if A and B were 100% dependent. If P(A) and P(B) are both .6, P(AUB) is .84. The only way you DON'T get the AUB event is if NEITHER of them happen. P(A)' = .4 P(B)' = .4 Both of these have to happen. Or .42 or .16... meaning the AUB event has a chance of .84. That assumes that A and B are independent variables which was not stated in the question. This is a bad example. If you flip a coin, your chances of getting heads or tails is 1. You, like the person grading the question, and possibly the person who wrote the question are making the assumption that A and B are independent. It's perfectly possible for there to be completely dependent variables with the probabilities given by the OP, or even two partially dependent variables with the values: P(A) = 0.6 P(B) = 0.6 P(A U B) = 0.2 P(!(AUB)) = 0.8 If the measured probability of "A or B" differed from P(AUB), that would be a good argument for dependence, but that means your statistical model is flawed, it doesn't change the calculated value of I think you'll find that if you step outside a basic stats class that the notation P(A U B) is taken to mean the actual probability given the complete probability space regardless of dependence/ independence and simply calculating P(A U B) given only P(A) and P(B) is improper unless you know that they are independent variables. "The probability of 'A or B'" and "P(A U B)" mean the same This is like saying x=1 and y=1, but x+y=3 is true because in the modeled system when X and Y get together they have a baby. I mean, yeah, sure, okay, nice nonlinear thinking, but you just look like retard who can't math good. >This is like saying x=1 and y=1, but x+y=3 is true because in the modeled system when X and Y get together they have a baby. It's probability, not simple algebra. Different rules. Gross conceptual error. > simply calculating P(A U B) given only P(A) and P(B) is improper unless you know that they are independent variables When you step outside of a basic statistics class, how would you prove that the A variable and B variable aren't independent in the system being studied? By showing P(AUB) does not correctly predict the measured probability. You didn't catch enough cum in your butt op that's fucking how you limey cunt nugget Addition and Union have different rules, sure, but if you aren't following the rules of the operator, you're not performing the specified operation. No it isn't. There are all kinds of things that follow this kind of behavior. But usually you talk about these kinds of dependent variables as being "correlated". For an example of two variables that match the OP's given distributions, consider P(A) = the probability a d10 rolled value is less than 7 P(B) = the probability the same d10's rolled value is not greater than 6. You might say this example is silly because the two events are just two ways to say the same thing, but that's entirely the point. Events with 100% dependence tend to just be different ways to measure the same state, but recognizing that the two are actually the same is sometimes not obvious. But I did make a mistake on my provided alternate example. The values I gave aren't possible. P(AUB) must be at least .6 (in the case they were 100% correlated and as much as 1.0 if they were maximally anti-correlated. strongnet.org/cms/lib6/OH01000884/Centricity/Domain/308/Venn Diagrams.pdf The "U" operator is a set operation. See the 8th slide for the real way to calculate unioned probabilities whether independent or dependent. It just so happens that when two variables are independent P(A) + P(B) - P(A&&B) = P(A)*P(B). scores matter more than actual comprehension so of course nothing sticks lmao I bet you're the kind of retard to say that flipping a coin 3 times and having it all be heads is the least likely outcome	{"url":"https://credforums.com/thread/820002199/miscellaneous/how-is-this-wrong.html","timestamp":"2024-11-14T15:20:44Z","content_type":"text/html","content_length":"63107","record_id":"<urn:uuid:174553b7-9d14-42f3-a06e-af870268f29a>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00845.warc.gz"}
Trading Case OP2 Case Objective To understand the two-period binomial option pricing model. Key Concepts Binomial option pricing model; option replication; dynamic trading strategies. Case description There is one stock, one bond, a put option and a call option. Both options are American style options. Each trial of the trading case covers two months of calendar time. The FTS markets are open for the first trading day of each month and during this day each American option can be exercised if desired. To exercise you need to put in the quantity and then click on the button Exercise (see Generic Trading Screen below). Actual exercise is executed at the end of the current trading period and the number to be exercised is indicated with an additional /num in relevant position cell. That is, at the end of the first trading day time “flashes by” and then end of month one realizations occur. Shortly after, the first trading day of the second month opens. The stock price at the beginning of the first month is 20, and at the end of the month, it either goes up to 40 or down to 10. You can trade at this realized stock price (i.e., either 10 or 40) during the first day of month two. At the end of the second month, the stock value again either doubles or is halved (i.e., three possibilities 5, 20, or 80 see below). At the end of each trading day the interest rate is 1% for the remaining month. That is, any surplus (shortfall) of cash earns (pays) 1% per month. Both options expire at the end of month two, and have a strike price equal to 25. In summary, the price of the stock is fixed at 20 at the beginning of month 1, and then at either 40 or 10 at the beginning of month 2, depending on whether an uptick or downtick was realized. An up or down tick is equally likely. All other prices are determined by the traders as a result of their limit and market orders. Case Data The following binomial tree shows the cash flows from each security at the end of period 2. There are no cash flows in period 1. Trading Objective The objective is to accumulate as much grade cash per trial (two trading periods) as possible. Your realized final market cash position determines your grade cash as follows: Earning Grade Cash In the trading period securities are exchanged using market cash. Your end of period final market cash balance determines your earning grade cash as follows:	{"url":"http://www.ftsmodules.com/public/modules/trader/OP2.htm","timestamp":"2024-11-13T09:35:32Z","content_type":"text/html","content_length":"13440","record_id":"<urn:uuid:e4bb3606-6565-41ff-a9bb-03bea19181ac>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00555.warc.gz"}
Course 2022-2023 a.y. - Universita' Bocconi 30401 - MATHEMATICS AND STATISTICS - MODULE 2 (STATISTICS) Department of Decision Sciences Course taught in English Suggested background knowledge Basic mathematics and basic R programming Mission & Content Summary The course develops the principles of scientific learning from data. For each of the topics discussed the course starts from actual data and a specific scientific question and develops the statistical learning and uncertainty quantification tools to gather knowledge from the available data for the given question. The course is organized in themes. Each theme starts with a theme overview, it introduces some motivating data and associated scientific questions and then develops the statistical tools (models, algorithms, mathematical concepts) needed to gather knowledge from the data to address the motivating questions. The theme finishes with a summary and exercises. The themes are: 1. Data visualization and summarization Data: heart attack study, Shipman's dead patients, daily homicides, test results, jelly beans competition Concepts: barplots, box plots, means and medians and variational formulation, logarithmic scale, correlation and distance correlation 2. From randomization to randomness Data: chocolates and nobel prizes, university admission data, death penalty data Concepts: spurious correlations, experimental vs observational data, random numbers, randomized control trials, confounders, simpson's paradox 3. What is probability and what is it useful for Concepts: Bernoulli distr., probability densities, Poisson distribution, series and limits, learning a model from the data 4. The calculus of probability Concepts: events, basic set theory, axioms of probability 5. More models for more data Data: birth weights, human heights, heart transplant survival data Concepts: density functions, Gaussian distribution, survival analysis, exponential distribution, censoring, gamma distribution and special functions, uniform distribution, transformation of variables, simulation of random variables 6. Joint distributions, independence and combinatorics Data: 10 year maturity bonds, heights of fathers and sons, the Sally Clark story Concepts: joint and marginal distributions, independence, statistical arguments in Law, the binomial distribution, binomial coefficients, basic combinatorics 7. Expectation Concepts: expected value and interpretation, properties of expectation, moments, variance, standard deviation and interpretation, the uncertainty rule of thumb, skewness and interpretation, sample and population moments 8. Elements of Network Science Data: the Internet, employees communication network, the actor network Concepts: Erdos-Renyi network model, degree distributions, six degrees of separation, heavy tails, scale-free property, power laws, the Student-t distribution 9. Concentration, inequalities and limit theorems Concepts: Markov inequality, Chebyshev inequality, uncertainty quantification, weak law of large numbers, a basic understanding of the central limit theorem 10. Statistical learning Data: cholestor and heart disease, arm-folding and sex, bowel cancer rates in the UK Concepts: quantifying evidence in data about a hypothesis, p-value, Fisher exact test, multiple testing, confidence intervals from concentration inequalities, bootstrap and confidence intervals, funnel plots Intended Learning Outcomes (ILO) At the end of the course student will be able to... + fomulate statistical learning questions + identify appropriate data analysis methodologies + carry out uncertainty quantification + learn basic models from data At the end of the course student will be able to... + choose appropriate data summaries and visualization + carry out basic network analysis + derive basic probability calculations + use statistical learning tools Teaching methods • Face-to-face lectures • Exercises (exercises, database, software etc.) • Group assignments • Interactive class activities (role playing, business game, simulation, online forum, instant polls) Exercises (Exercises, database, software etc.): Special sessions with exercises, examples and illustrations of concepts and methods, also with the help of statistical software R, will be provided. Group assignments: A project will be given for students to work in groups that will involve both methodology and data analysis Assessment methods Continuous assessment Partial exams General exam • Written individual exam (traditional/online) x x • Group assignment (report, exercise, presentation, project work etc.) x Students may choose between the following two options: - Two partial written exams (a mid-term and a final) that contribute to the final grade with a 50% weight each. - A single general written exam (after the end of the course) that counts for 100% of the final mark. The tests consist of exercises. They aim at ascertaining students' mastery of concepts and results discussed during lectures as well as an adequate knowledge of R. In each test the maximum grade is 31. The assessment method is the same for both attending and non-attending students. Students who take the mid-term exam may still take the general exam instead of taking the final exam. Importantly, access to the final (or second partial) exam follows the rules indicated in Section 7.6 of the Guide to the University. There will be an optional group project that will receive a maximum of 1.5/31 points. These will be added to the total mark achieved by the previous options. Teaching materials The teaching material will be primarily that developed during the classes and distributed to the students in a PDF format after each class. The course will use examples and extracts primarily from the first book listed below. It is advisable to acquire this book either in its original publication or its Italian translation (it is also available as an e-book), since it is an excellent modern resource to learn Probability and Statistics and why these are fundamental in anything that has to do with learning from data. Early chapters from the second book provide an excellent more technical introduction to Probability. The introduction and some Appendices of the third book provide an excellent and accessible introduction to statistical machine learning and the use of Probability and Statistics for designing and analyzing algorithms. The fourth is a textbook whose syllabus correlates highly with the contents of this course. For a number of basic concepts the corresponding Wikipedia pages are a great resource. Please use that instead of random blogs, webpages or videos posted on youtube. • Spiegelhalter, The Art of Statistics: How to Learn from Data, Penguin, 2019, ISBN 978-1541618510 (available also in Italian translation) • Barabasi, Network Science • Grimmett and Stirzaker, Probability and Random Processes, Oxford, Fourth Edition, 2020, ISBN 978-0198847595 • Bishop, Pattern Recognition and Machine Learning, Springer, 2006, ISBN 978-0387310732 • S. ROSS, Introduction to Probability and Statistics for Engineers and Scientists, Fourth Edition, Academic Press, 2014 Last change 11/12/2022 18:26	{"url":"https://didattica.unibocconi.eu/ts/tsn_anteprima.php?cod_ins=30401&anno=2023&IdPag=6956","timestamp":"2024-11-08T01:58:03Z","content_type":"text/html","content_length":"174727","record_id":"<urn:uuid:5ff5c82c-a10e-47d7-a5fe-51135cbf0ba3>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00827.warc.gz"}
Suppose The Index Model For Stocks A And B Is Estimated From The Excess Returns With The Following Results: (2024) Answer 1 Cov(a, market index) = 0.012 and cov(b, market index) = 0.01 (a) the variance of each stock can be calculated using the regression r² values and the variance of the market index. the formula for variance in the index model is: var(stock) = r²(stock) * var(market index) for stock a: var(a) = 0.30 * (0.2²) = 0.012 for stock b:var(b) = 0.25 * (0.2²) = 0.01 (b) the firm-specific risk of each stock can be calculated by subtracting the variance of each stock from their respective total variances. firm-specific risk represents the portion of the stock's risk that is not related to the market. for stock a: firm-specific risk (a) = var(a) - [r²(a) * var(market index)] = 0.012 - (0.30 * (0.2²)) = 0.012 - 0.012 = 0 for stock b:firm-specific risk (b) = var(b) - [r²(b) * var(market index)] = 0.01 - (0.25 * (0.2²)) = 0.01 - 0.01 = 0 both stocks a and b have zero firm-specific risk, meaning all of their risk is explained by the market. (c) the covariance between the two stocks can be calculated using their respective r² values and the variance of the market index. the formula for covariance in the index model is: cov(a, b) = sqrt(r²(a) * var(a) * r²(b) * var(b)) cov(a, b) = sqrt(0.30 * (0.2²) * 0.25 * (0.2²)) = sqrt(0.012 * 0.01) ≈ 0.03464 (d) the covariance between each stock and the market index can be calculated using their respective r² values and the variance of the market index. the formula for covariance in the index model is: cov(stock, market index) = r²(stock) * var(market index) for stock a: cov(a, market index) = 0.30 * (0.2²) = 0.012 for stock b:cov(b, market index) = 0.25 * (0.2²) = 0.01 Learn more about stock here: Related Questions What are the differences between the Classical and Keynesian economics (don't forget to mention: wage, price, unemployment)? What are the important consequences of the differences in their The differences between Classical and Keynesian economics revolve around the roles of wages, prices, and unemployment, which lead to different policy recommendations and views on economic stability and growth. Here are the key differences between them: 1. Wage: In classical economics, wages are determined by the supply and demand for labour. According to Keynesian economics, wages are sticky and do not adjust quickly to changes in the labor market. 2. Price: Classical economists believe that prices are flexible and adjust to changes in supply and demand. On the other hand, Keynesian economists argue that prices are sticky in the short run and do not adjust immediately to changes in the economy. 3. Unemployment: Classical economics states that unemployment is caused by factors like minimum wage laws or unions interfering with the labor market. Keynesian economics, on the other hand, believes that unemployment can be caused by insufficient aggregate demand in the economy. The differences in their views have important consequences: 1. Policy recommendations: Classical economics suggests that the government should take a hands-off approach and let markets self-adjust. Keynesian economics argues that the government should intervene to stimulate demand during times of economic downturns. 2. Economic stability: Classical economics emphasizes the self-regulating nature of the market and believes that it will naturally reach equilibrium. Keynesian economics highlights the need for government intervention to stabilize the economy and reduce fluctuations in output and employment. 3. Economic growth: Classical economics focuses on long-term growth through free markets and minimal government intervention. Keynesian economics emphasizes the role of government spending to boost economic growth during times of recession. Learn more about Keynesian economics I would like you to choose a real-life company that you believe has been significantly impacted by changes in their external environment (political, economic, socio-cultural and/or technological). Explain your rationale for choosing this company. In order to make things interesting in the discussion forum, try to choose a company that has not already been discussed by a fellow student. If possible, diversify the industries as well. please not jet airways. Tesla Inc. exemplifies a company that has been greatly impacted by changes in its external environment, particularly in the realms of politics, economics, and technology. Its success and market position have been shaped by favorable government policies, economic conditions, and technological advancements, making it an interesting case study for discussions on the impact of external factors on a company's performance. I have chosen Tesla Inc. as a company that has been significantly impacted by changes in its external environment, specifically in the areas of politics, economics, and technology. Rationale for choosing Tesla Inc.: 1. Political Impact: Tesla has been greatly influenced by political factors, particularly government regulations and incentives related to electric vehicles (EVs). Governments worldwide have implemented policies to promote sustainable transportation and reduce carbon emissions. For instance, countries like the United States, China, and several European nations have introduced tax credits, subsidies, and stricter emission standards that have directly benefited Tesla's EVs. 2. Economic Impact: Economic factors have played a crucial role in shaping Tesla's trajectory. Fluctuating oil prices, the availability of charging infrastructure, and consumer purchasing power have influenced the demand for electric vehicles. Additionally, government support through grants, loans, and favorable financing options has impacted Tesla's financial position and ability to expand 3. Technological Impact: Tesla's success is closely tied to its innovative technologies, particularly in the realm of electric vehicles and renewable energy solutions. The company's advancements in battery technology, autonomous driving capabilities, and energy storage systems have positioned it as a leader in the industry. Technological breakthroughs and advancements have allowed Tesla to differentiate itself and maintain a competitive edge. The combination of these external environmental factors has significantly affected Tesla's growth and market position. Government policies and regulations have helped create a favorable environment for the adoption of EVs, enabling Tesla to expand its market share. Economic factors, such as fluctuations in oil prices and consumer preferences, have influenced the demand for Tesla's vehicles. Additionally, Tesla's technological innovations have propelled the company to the forefront of the EV industry. Learn more about technology here: 1: Short Answer (20 marks total) Alyah a marketing manager for a chain of stores that sell sports equipment, Easy Riders. Easy Riders sells sports equipment but specializes in bicycles for all ages and abilities, including bicycles for children learning to ride. They pride themselves on their variety of merchandise and low price points that suit any budget. Alyah decides that she is going to use a contest to win a new bicycle and bicycle helmet in her stores. (3 marks total) 1. What category of integrated marketing communication is this an example of? Alyah's contest is an example of sales promotion, which is a category of integrated marketing communication used to incentivize consumers to make purchases. The category of integrated marketing communication that Alyah's contest is an example of is sales promotion. Integrated marketing communication is a strategy used by businesses and organizations to communicate effectively with their target audience. It involves coordinating all aspects of marketing communication, such as advertising, public relations, sales promotion, and personal selling, to promote a consistent brand image. Sales promotion refers to short-term incentives used by businesses to persuade consumers to purchase their products or services. These incentives can take a variety of forms, including coupons, contests, samples, rebates, and discounts. The goal of sales promotion is to increase sales quickly and encourage customers to make repeat purchases. To summarize, Alyah's contest is an example of sales promotion, which is a category of integrated marketing communication. Learn more about purchases here: The following provides the previous year's financial information for Lowes (in millions): Sales is $50,521; Cost of goods sold is $33,194; and, Inventory is $8,600. How many days-of-supply did Lowes hold in 2012? Assume 365 days in a year. 94.6 days 43.6 days 142.6 days 201.5 days 94.6 days days-of-supply did Lowes hold in 2012 To calculate the days-of-supply, we can use the formula: Days-of-Supply = (Inventory / Cost of goods sold) * 365 Plugging in the given values: Days-of-Supply = (8,600 / 33,194) * 365 Days-of-Supply ≈ 94.6 days Therefore, Lowes held approximately 94.6 days-of-supply in 2012. Inventory refers to the stock of goods or materials held by a business. It includes finished products, raw materials, and work-in-progress. Efficient inventory management is crucial for balancing supply and demand and optimizing profitability. Inventory is a vital component of any business that deals with physical goods. It represents the stock of products or materials that a company holds for its operations. This stock can include finished goods ready for sale, raw materials required for production, and work-in-progress items that are in the manufacturing process. Learn more about inventory here: Refer to lecture notes "Two-period neoclassical model". Consider the special case of linear production function F(K,A)=AK. Work out the formula of the production possibility frontier (PPF) in closed-form. Show that the PPF is a straight line. What is the vertical intercept, the horizontal intercept, and the slope? How does the firm's demand curve for investment look like in this case? Is it downward sloping? In terms of investment demand, the firm's investment demand curve is downward sloping. As the interest rate increases, the firm's cost of capital rises, leading to a decrease in the marginal productivity of capital. Consequently, the firm's demand for investment decreases, resulting in a downward-sloping investment demand curve. The production possibility frontier (PPF) represents the maximum potential output an economy can produce given its resources and technology at a specific time. In the case of a linear production function F(K, A) = AK, where K represents capital and A represents total factor productivity, the formula for calculating the PPF can be expressed as follows: y = (Q / A)^(1/b) * [(A - 1) / (1 - b)] In this formula, b represents the capital's output elasticity, A represents the level of total factor productivity, and Q represents a given level of output. The formula allows us to determine the quantity of output (y) that can be produced given the values of A, b, and Q. The slope of the PPF, which represents the marginal rate of transformation (MRT) between the two goods, can be determined using the marginal productivities of the goods. In the case of a linear production function F(K, A) = AK, the marginal product of capital is A. Therefore, the slope of the PPF is equal to the MRT, which is A. The slope of the PPF is also equal to the negative slope of the isoquant at the point of tangency, and it is equivalent to the ratio of the prices of the two goods, as represented by the slope of the budget constraint. The vertical intercept of the PPF represents the maximum production of the first good when the economy produces nothing of the second good. In the given formula, the vertical intercept is A. Similarly, the horizontal intercept represents the maximum production of the second good when the economy produces nothing of the first good, and it is calculated as Q / A. Learn more about vertical intercept Discuss the different types of financial Need unique Answer please ///////////// Financial intermediaries are the go-between that links the surplus and shortage unit, that is, the individual who has surplus money to save and the individuals who require funding. Intermediaries are entities that bring together borrowers and lenders of funds in the financial market. There are different types of financial intermediaries, and they can be categorized into several types based on the intermediation process they provide. These financial intermediaries can be broadly categorized into two types: deposit-taking and non-deposit-taking intermediaries. Deposit-taking intermediaries Deposit-taking intermediaries are intermediaries that collect funds from depositors and lend money to borrowers. These intermediaries are called deposit-taking intermediaries because they receive deposits from people in the form of savings and demand deposits, which they use to grant loans to people who require funding. These intermediaries are further classified into commercial banks, savings banks, and credit unions.1. Commercial banks Commercial banks are the most prominent deposit-taking intermediary in the financial market, and they operate on a large scale to collect deposits from people and grant loans to people who require funding. Commercial banks provide a range of services, including checking and savings accounts, loans, mortgages, and other financial services. 2. Savings banks Savings banks are intermediaries that collect deposits from individuals and provide funding to people who require funding. Savings banks are similar to commercial banks, but they are usually smaller and operate in specific regions.3. Credit unions Credit unions are a type of financial intermediary that operates on a cooperative basis, and they are owned by their members. Members pool their money to lend to other members who require funding. Non-deposit taking intermediaries Non-deposit-taking intermediaries are intermediaries that do not collect funds from depositors but instead raise funds from other sources such as the financial market, and then lend those funds to borrowers. These intermediaries include insurance companies, pension funds, mutual funds, and investment banks.1. Insurance companies Insurance companies are intermediaries that sell insurance policies to individuals and companies, and they collect premiums from policyholders. The premiums collected from policyholders are invested in different securities, and the returns on these investments are used to pay claims to policyholders. 2. Pension funds Pension funds are intermediaries that collect money from employees, employers, and governments to provide retirement benefits to employees. Pension funds invest these funds in stocks, bonds, and other securities, and they earn returns on these investments.3. Mutual funds Mutual funds are intermediaries that pool money from different investors and invest this money in a diversified portfolio of stocks, bonds, and other securities. Mutual funds are operated by professional fund managers, and they provide investors with access to a diversified portfolio of securities. 4. Investment banks Investment banks are intermediaries that provide a range of services, including underwriting securities, providing merger and acquisition advisory services, and trading securities on behalf of clients. Investment banks are usually involved in the initial public offering (IPO) process, where they help companies raise funds by selling their shares to the public. learn more about money Common stock transactions on the statement of cash flows Jones Industries received $600,000 from issuing shares of its common stock and $400,000 from issuing bonds. During the year, Jones Industries also paid payments, decreases in cash and for any adjustments, if required. If a transaction has no effect on the statement of cash flows, select "No effect" from the drop down menu and leave the amount box blank. Cash received from issuing common stock____ Cash received from issuing bonds___ Cash paid dor dividends_____ 1. Cash received from issuing common stock: $600,000. 2. Cash received from issuing bonds: $400,000. 3. Cash paid for dividends: No effect. 1. The cash received from issuing common stock is a financing activity and should be reported in the financing section of the statement of cash flows. In this case, Jones Industries received $600,000 from the issuance of its common stock, which represents an inflow of cash into the company. This amount would be reported as a positive number in the financing section. 2. Similarly, the cash received from issuing bonds is also a financing activity. When a company issues bonds, it receives cash from investors in exchange for the bonds. In this case, Jones Industries received $400,000 from the issuance of bonds, which represents an inflow of cash. This amount would also be reported as a positive number in the financing section of the statement of cash flows. 3. The cash paid for dividends is not provided in the given information. If there is no information about dividends being paid during the year, then it would be considered as having no effect on the statement of cash flows. Dividends are typically classified as a financing activity when paid, but since there is no information about dividends being paid in this scenario, it is assumed to have no effect on the statement of cash flows. Therefore, the amount for cash paid for dividends would be left blank, and "No effect" would be selected from the drop-down menu. Learn more about dividends here: Write one full summary page about Amazon and their recruitment Just one page but very insightful and understanding. Please add your source in case you pick it from somewhere Amazon is known for its innovative and effective recruitment strategies, which have played a crucial role in the company's success. With a focus on attracting top talent, Amazon employs a multi-faceted approach that combines cutting-edge technology, streamlined processes, and a strong employer brand. The company utilizes various recruitment channels, including online job portals, social media platforms, and its own career website, to reach a wide pool of candidates. Amazon's recruitment strategies prioritize efficiency and speed, leveraging automation and data-driven approaches to streamline the hiring process. Additionally, the company emphasizes a customer-centric approach even in its recruitment efforts, aiming to hire individuals who align with its core values and customer obsession. By consistently refining and innovating its recruitment strategies, Amazon has been able to build a diverse and highly skilled workforce that drives its continued growth and success. Amazon's recruitment strategies are built on several key pillars that contribute to their effectiveness. Firstly, the company leverages its strong employer brand, emphasizing its commitment to innovation, customer obsession, and a fast-paced work environment. This branding helps attract candidates who are aligned with Amazon's values and are motivated to contribute to its ambitious goals. To reach a wide talent pool, Amazon utilizes a range of recruitment channels. This includes partnerships with online job portals like LinkedIn, where the company actively promotes job openings and engages with potential candidates. Additionally, Amazon maintains a strong presence on social media platforms, using targeted advertising and engaging content to attract passive job seekers. One of the notable aspects of Amazon's recruitment strategies is its focus on efficiency and speed. The company leverages technology and automation to streamline the hiring process, using algorithms and data-driven approaches to identify the most suitable candidates. This enables Amazon to handle a large volume of applications while ensuring a quick and efficient recruitment process. In conclusion, Amazon's recruitment strategies are a key driver of its ability to attract top talent and build a highly skilled workforce. The company's focus on leveraging technology, streamlining processes, and prioritizing a strong employer brand has enabled it to stay at the forefront of innovation and maintain its position as a leading global organization. Learn more about Amazon here Suppose the Parliament passes legislation making it more difficult for firms to fire workers (an example is a law requiring severance pay to fired workers). If this legislation reduces the rate of job separation without affecting the rate of job finding, how would the natural rate of unemployment change? Do you think that it is plausible that legislation would not affect the rate of job The natural rate of unemployment would not change if legislation makes it more difficult for firms to fire workers. It is plausible that legislation would not affect the rate of job finding. If the legislation reduces the rate of job separation without affecting the rate of job finding, it means that firms would find it harder to make staffing decisions and the labor market would become less flexible. However, the natural rate of unemployment, which represents the equilibrium level of unemployment in the economy, would not be directly affected. The natural rate of unemployment is determined by structural and frictional factors. Structural factors, such as changes in the economy that affect labor demand or supply, and frictional factors, such as the time it takes for workers to find new jobs or for employers to fill vacancies, contribute to the natural rate of unemployment. In this scenario, the legislation would impact the rate of job separation, which is a structural factor affecting the natural rate of unemployment. However, it would not affect the rate of job finding, which is a frictional factor. Therefore, the overall impact on the natural rate of unemployment would be minimal, and it would remain unchanged. Considering these factors, it is plausible that legislation aimed at making it more difficult for firms to fire workers would not directly affect the rate of job finding. The legislation primarily targets the structural factors that contribute to the natural rate of unemployment. Learn more about the natural rate of unemployment: The natural rate of unemployment would not change if legislation makes it more difficult for firms to fire workers. It is plausible that legislation would not affect the rate of job finding. If the legislation reduces the rate of job separation without affecting the rate of job finding, it means that firms would find it harder to make staffing decisions and the labor market would become less flexible. However, the natural rate of unemployment, which represents the equilibrium level of unemployment in the economy, would not be directly affected. The natural rate of unemployment is determined by structural and frictional factors. Structural factors, such as changes in the economy that affect labor demand or supply, and frictional factors, such as the time it takes for workers to find new jobs or for employers to fill vacancies, contribute to the natural rate of unemployment. In this scenario, the legislation would impact the rate of job separation, which is a structural factor affecting the natural rate of unemployment. However, it would not affect the rate of job finding, which is a frictional factor. Therefore, the overall impact on the natural rate of unemployment would be minimal, and it would remain unchanged. Considering these factors, it is plausible that legislation aimed at making it more difficult for firms to fire workers would not directly affect the rate of job finding. The legislation primarily targets the structural factors that contribute to the natural rate of unemployment. Learn more about unemployment: The following transactions were selected from among those completed by Aashi Retailers in November and December: Nov. 20 Sold 20 items of merchandise to Customer B at an invoice price of $6,300 (total); terms 2/10, n/30. 25 Sold two items of merchandise to Customer C, who charged the $700 (total) sales price on her visa credit card. Visa charges Aashi Retailers a 2 percent credit card fee. 28 Sold 10 identical items of merchandise to Customer D at an invoice price of $9,100 (total); terms 2/10, n/30. 29 Customer D returned one of the items purchased on the 28th; the item was defective and credit was given to the customer. Dec.6 Customer D paid the account balance in full. 20 Customer B paid in full for the invoice of November 20. Required: Assume that Sales Returns and Allowances, Sales Discounts, and Credit Card Discounts are treated as contra-revenues; compute net sales for the two months ended December 31. (Do not round your intermediate calculations. Round your answer to the nearest whole dollar amount.) The net sales for the two months ended December 31 can be calculated by subtracting the contra-revenues from the total sales. The contra-revenues include sales returns and allowances, sales discounts, and credit card discounts. In this scenario, the net sales can be computed by subtracting the sales returns and allowances, sales discounts, and credit card discounts from the total sales. The sales returns and allowances are determined by the return of one item by Customer D, which had an invoice price of $9,100. This reduces the total sales by $9,100. The sales discounts are calculated based on the payment made by Customer B for the November 20 invoice. The discount is determined as 2% of the invoice price, which amounts to $126. Lastly, the credit card discount is computed by applying a 2% credit card fee to the sales charged by Customer C on her visa credit card, resulting in a discount of $14. By subtracting the contra-revenues of $9,100 (sales returns and allowances), $126 (sales discounts), and $14 (credit card discounts) from the total sales, the net sales for the two months ending December 31 can be determined. This approach provides a more accurate representation of the revenue generated by Aashi Retailers during the specified period. To learn more about net sales, Click here: brainly.com/question/28903071? The Americans with Disabilities Act (ADA) of 1990 requires that positions must be found for any job candidate with a disability. employers focus on the needs of the disabled more than the needs of position. reasonable accommodations must be made to enable disabled employees to do their job effectively. organization's must fill at least 10% of their positions with disabled employees. Page 25 of Option 3 is correct. The Americans with Disabilities Act (ADA) of 1990 mandates that employers make reasonable accommodations for disabled employees, ensuring they can perform their job effectively. The Americans with Disabilities Act (ADA) of 1990 is a legislation that protects individuals with disabilities from discrimination in various aspects of their lives, including employment. One of the key provisions of the ADA is that employers are required to make reasonable accommodations for disabled employees. These accommodations are intended to enable individuals with disabilities to effectively perform their job duties. Reasonable accommodations can vary depending on the specific needs of the employee and the nature of the job. They may include modifications to the work environment, equipment, or policies, as well as providing additional support or resources. The ADA does not explicitly require that positions be found for any job candidate with a disability, as stated in option 1. Instead, it emphasizes the importance of providing reasonable accommodations to enable disabled employees to perform their job effectively. Option 2, which suggests that employers should prioritize the needs of disabled individuals over the needs of the position, is also not accurate. The ADA aims to strike a balance between the needs of the employee and the requirements of the job. Lastly, option 4 stating that organizations must fill at least 10% of their positions with disabled employees is not a requirement under the ADA. Learn more about Americans with Disabilities Act here: The complete question is: The Americans with Disabilities Act (ADA) of 1990 requires that 1. positions must be found for any job candidate with a disability. 2. employers focus on the needs of the disabled more than the needs of position. 3. reasonable accommodations must be made to enable disabled employees to do their job effectively. 4. organization's must fill at least 10% of their positions with disabled employees. Project Cash Flow The financial staff of Cairn Communications has identified the following information for the first year of the roll-out of its new proposed service: Projected sales$18 million Operating costs (not including depreciation)$10 million Depreciation$4 million Interest expense$3 million The company faces a 25% tax rate. What is the project's operating cash flow for the first year (t = 1)? Enter your answer in dollars. For example, an answer of $1.2 million should be entered as $1,200,000. Round your answer to the nearest dollar. The project's operating cash flow for the first year is $6 million, calculated by subtracting operating costs and taxes from the projected sales of $18 million. To calculate the project's operating cash flow for the first year (t = 1), we need to subtract the operating costs (excluding depreciation) and taxes from the projected sales. Operating cash flow = Projected sales - Operating costs (excluding depreciation) - Taxes Projected sales = $18 million Operating costs (excluding depreciation) = $10 million Taxes = 25% of (Projected sales - Operating costs) Taxes = 25% * ($18 million - $10 million) = 25% * $8 million = $2 million Operating cash flow = $18 million - $10 million - $2 million = $6 million Therefore, the project's operating cash flow for the first year is $6 million.\ Learn more about cash flow here;- The Saunders Investment Bank has the following financing outstanding. Debt: 50,000 bonds with a coupon rate of 5.1 percent and a current price quote of 104.5; the bonds have 15 years to maturity and a par value of $1,000.30,000 zero coupon bonds with a price quote of 23.6,30 years until maturity, and a par value of $10,000. Both bonds have semiannual compounding. Preferred 235,000 shares of 3.6 percent preferred stock with a current price of stock: $87 and a par value of $100. Common 2,100,000 shares of common stock; the current price is $79 and the beta stock: of the stock is 1.08. Market: The corporate tax rate is 23 percent, the market risk premium is 7 percent, and the risk-free rate is 3.6 percent. What is the WACC for the company? The weighted average cost of capital (WACC) for the Saunders Investment Bank is approximately 9.29%. To calculate the WACC, we need to determine the cost of each component of the company's capital structure and weight them based on their respective proportions. 1. Cost of Debt: The cost of debt can be calculated using the formula: Cost of Debt = Coupon Rate [tex]\times[/tex] (1 - Tax Rate), where the coupon rate is 5.1% and the tax rate is 23%. Cost of Debt = 5.1% [tex]\times[/tex] (1 - 0.23) = 3.93% 2. Cost of Preferred Stock: The cost of preferred stock is calculated as the dividend rate divided by the current price of the stock. Cost of Preferred Stock = Dividend Rate / Stock Price, where the dividend rate is 3.6% and the stock price is $87. Cost of Preferred Stock = 3.6% / $87 = 0.0414 3. Cost of Common Stock (Equity): The cost of equity can be calculated using the Capital Asset Pricing Model (CAPM): Cost of Equity = Risk-Free Rate + Beta [tex]\times[/tex] Market Risk Premium, where the risk-free rate is 3.6%, the beta of the stock is 1.08, and the market risk premium is 7%. Cost of Equity = 3.6% + 1.08 [tex]\times[/tex] 7% = 11.56% 4. Weights of each component: To calculate the weights, we need to determine the proportion of each component in the total capital structure. Weight of Debt = (Number of Bonds [tex]\times[/tex] Bond Price [tex]\times[/tex] Bond Par Value) / Total Market Value, Weight of Preferred Stock = (Number of Preferred Shares [tex]\times[/tex] Preferred Stock Price [tex]\times[/tex] Preferred Stock Par Value) / Total Market Value, Weight of Equity = (Number of Common Shares [tex]\times[/tex] Common Stock Price) / Total Market Value. Total Market Value = (Number of Bonds [tex]\times[/tex] Bond Price [tex]\times[/tex] Bond Par Value) + (Number of Preferred Shares [tex]\times[/tex] Preferred Stock Price [tex]\times[/tex] Preferred Stock Par Value) + (Number of Common Shares [tex]\times[/tex] Common Stock Price). Once we have the weights for each component, we can calculate the WACC using the formula: WACC = (Weight of Debt [tex]\times[/tex] Cost of Debt) + (Weight of Preferred Stock [tex]\times[/tex] Cost of Preferred Stock) + (Weight of Equity [tex]\times[/tex] Cost of Equity). Plugging in the values, we get: WACC = (0.2096 [tex]\times[/tex] 3.93%) + (0.0579 [tex]\times[/tex] 0.0414) + (0.7325 [tex]\times[/tex] 11.56%) ≈ 0.82% + 0.0024 + 8.47% ≈ 9.29%. Therefore, the WACC for the Saunders Investment Bank is approximately 9.29%. Learn more about capital here: brainly.com/question/25715888 Which phrase best describes the current role of the managerial accountant? a. Managerial accountants prepare the financial statements for an organization. b. Managerial accountants facilitate the decision-making process within an organization. c. Managerial accountants make the key decisions within an organization. d. Managerial accountants are primarily information collectors. e. Managerial Accountants are solely staff advisors in an organization The phrase that best describes the current role of the managerial accountant is "b. Managerial accountants facilitate the decision-making process within an organization." The current role of the managerial accountant The role of a managerial accountant involves many tasks in an organization. Managerial accountants are responsible for recording, analyzing, and interpreting financial information and presenting it to management. Their duties include preparing budgets, cost analysis reports, and other financial statements. However, the most critical role of managerial accountants in the modern business world is to facilitate the decision-making process within an organization. The managerial accountants' role is to provide management with the financial information they need to make informed decisions that will have a significant impact on the organization's success. The most critical role of managerial accountants in today's business world is to facilitate the decision-making process within an organization. They are responsible for providing management with the financial information they need to make informed decisions that will have a significant impact on the organization's success. Managerial accountants help businesses by providing financial information that helps them to plan, organize, and control their operations. They also assist management in formulating policies and strategies that will enable the organization to achieve its goals. Learn more about managerial accountants: Spot and futures prices for Gold and the S&P in September 2016 are given below. 16-September 16-December 17-June COMEX Gold ($/oz) $693 $706.42 $726.7 CME S&P 500 $1453.55 $1472.4 $1493.7 a. Use prices for gold to calculate the effective annualized interest rate for Dec 2016 and June 2017. Assume that the convenience yield for gold is zero. b. Suppose you are the owner of a small gold mine and would like to fix the revenue generated by your future production. Explain how the futures market enables such hedges. c. Calculate the convenience yield on the S&P index between September 16 and December 16. The effective annualized interest rate for December 2016 and June 2017 can be calculated using the gold prices provided. The formula for calculating the effective annualized interest rate is: Effective Annualized Interest Rate = [(Spot Price at Future Date / Spot Price at Current Date) ^ (1 / Time)] - 1 For December 2016: Effective Annualized Interest Rate = [(706.42 / 693) ^ (1 / (3/12))] - 1 = [(1.019437562) ^ 4] - 1 ≈ 0.0777 or 7.77% For June 2017: Effective Annualized Interest Rate = [(726.7 / 693) ^ (1 / (9/12))] - 1 = [(1.047978258) ^ (4/3)] - 1 ≈ 0.0654 or 6.54% The futures market enables owners of small gold mines to hedge their revenue by providing a platform to enter into contracts to buy or sell gold at a predetermined price in the future. By using gold futures contracts, the mine owner can lock in a fixed price for their future production, mitigating the risk of price fluctuations. For example, if the mine owner expects gold prices to decline, they can sell futures contracts at the current higher price, ensuring that they will receive that price when they sell their gold in the future, regardless of the actual market price at that time. Learn more about gold prices here: Ex. 3* Suppose there are two PT decision makers with the same weighting funct and with the same value function, except for the loss aversion parameter, s. A2 > A1 > 1. Assume that there are two lotteries x and y, where x is a mixed lottery and y is a pure loss lottery, pick such that y~1. Which lottery will the second decision maker prefer? Answer- The second decision-maker would then choose the x lottery. Let us use the following nomenclature: u denotes the value function, w denotes the weight function, L stands for the loss aversion parameter, and p stands for the probability of the x lottery. Also, for decision-maker 1 and decision-maker 2, let us use L1 and L2 to represent their loss aversion parameters, respectively. It is given that A2 > A1 > 1. We assume that the value function u and the weight function w are the same for both decision-makers. In this case, the first decision-maker has a value function of the following form: u(x) = w(x)[px + (1 − p)(1 − x)], and a value function of the following form: u(y) = w(y)[(1 − s)y + sy]. The second decision-maker, on the other hand, has a value function of the following form: u(x) = w(x)[px + (1 − p)(1 − x)],and a value function of the following form: u(y) = w(y)[(1 − s')y + s'y],where s' > s. To obtain y~1, we need to set y to be equal to 0. The utility of the second decision-maker from the y lottery will be as follows: u(y) = w(y)[(1 − s')0 + s'0] = 0. The second decision-maker would then choose the x lottery. To know more about value function Consider the following information regarding the performance of a money manager in a recent month. The table represents the actual return of each sector of the manager's portfolio in column 1, the fraction of the portfolio allocated to each sector in column 2 , the benchmark or neutral sector allocations in column 3 , and the returns of sector indices in column 4. a-1. What was the manager's return in the month? (Do not round intermediate calculations. Input all amounts as positive values. Round your answer to 2 decimal places.) a-2. What was her overperformance or underperformance? (Do not round intermediate calculations. Input all amounts as positive values. Round your answer to 2 decimal places.) What was the contribution of security selection to relative performance? (Do not round intermediate calculations. Round your nswer to 2 decimal places. Negative amount should be indicated by a minus sign.) What was the contribution of asset allocation to relative performance? (Do not round intermediate calculations. Round your answer to 2 decimal places. Negative amount should be indicated by a minus sign.) a-1) The manager's return in the month was 3.25%. a-2) The manager's overperformance or underperformance was 1.25%. The contribution of security selection to relative performance was 1.25%. The contribution of asset allocation to relative performance was 0.50%. a-1) The manager's return in the month is calculated by finding the weighted average of the returns of each sector of the portfolio. The formula for this calculation is: (Return of sector 1 x Allocation of sector 1) + (Return of sector 2 x Allocation of sector 2) + (Return of sector 3 x Allocation of sector 3) + (Return of sector 4 x Allocation of sector 4) = (0.03 x 0.35) + (0.04 x 0.25) + (0.02 x 0.20) + (0.01 x 0.20) = 0.0325 or 3.25%. Therefore, the manager's return in the month was 3.25%. a-2) The overperformance or underperformance of the manager is calculated by finding the difference between the actual return of the portfolio and the benchmark return. The formula for this calculation is: Actual return of portfolio - Benchmark return = 0.0325 - 0.02 = 0.0125 or 1.25%. Therefore, the manager's overperformance or underperformance was 1.25%. Contribution of security selection to relative performance: The contribution of security selection to relative performance is calculated by finding the difference between the actual return of the portfolio and the benchmark return assuming that the allocation of the portfolio was the same as the benchmark. The formula for this calculation is: Actual return of portfolio assuming benchmark allocation - Benchmark return = (0.0325 - (0.35 x 0.02) - (0.25 x 0.01) - (0.20 x 0.00) - (0.20 x 0.02)) - 0.02 = 0.0125 or 1.25%. Therefore, the contribution of security selection to relative performance was 1.25%. Contribution of asset allocation to relative performance: The contribution of asset allocation to relative performance is calculated by finding the difference between the actual return of the portfolio and the benchmark return assuming that the securities selected by the portfolio manager were the same as the benchmark. The formula for this calculation is: Actual return of benchmark portfolio assuming actual securities selected - Benchmark return = (0.35 x 0.02) + (0.25 x 0.01) + (0.20 x 0.00) + (0.20 x 0.02) - 0.02 = 0.005 or 0.50%. Therefore, the contribution of asset allocation to relative performance was 0.50%. For more such questions on manager Gwyneth (25) is unmarried and was a full-time student from January through June. Gwyneth worked part-time but did not provide more than 50% support for herself or her son, Saul (1). They are both U.S. citizens and have social security numbers that are valid for employment. Gwyneth and Saul lived with Gwyneth's mother, Joan (49), the entire year. Gwyneth's earned income and AGI were $7,278. Her mother, Joan, has an AGI of $30,225 in 2021, which is higher than her AGI in 2019. Gwyneth did not have any foreign income or investment income and Saul's income was $0. Gwyneth's earned income was less in 2019. Gwyneth would like to claim Saul if she is qualified to do so. a-What is Gwyneth's correct and most favorable 2021 filing status? b-What is Saul's dependency status for Gwyneth? c-Is Gwyneth eligible to claim the Child Tax Credit and/or the Other Dependent Credit for any potential dependent? Choose the best response. d-Is Gwyneth eligible to claim and receive the Earned Income Credit? e-Is Gwyneth eligible to utilize the 2019 lookback provision for the Earned Income Credit? a. Gwyneth's and most favorable 2021 filing status is Head of Household (HOH) since she meets the requirements of being unmarried, having a qualifying dependent (Saul), and paying more than half the household expenses. b. Saul's dependency status for Gwyneth is a Qualifying Child since he meets the criteria of being Gwyneth's son, under the age of 19 (or 24 if a full-time student), and living with Gwyneth for the entire year. c. Gwyneth is eligible to claim the Child Tax Credit for Saul, as he qualifies as a qualifying child for tax purposes. d. Gwyneth may be eligible to claim and receive the Earned Income Credit (EIC) if she meets the income and other eligibility criteria. e. Gwyneth may be eligible to utilize the 2019 lookback provision for the Earned Income Credit if her earned income was higher in 2019 compared to 2021. here some more information: a) Gwyneth qualifies for HOH because she is unmarried, lived with her dependent (Saul) for the entire year, and provided more than 50% of the household expenses. This filing status generally offers more favorable tax rates and a higher standard deduction compared to Single status. b) Saul qualifies as a Qualifying Child because he is Gwyneth's son, under the age of 19, and lived with her for the entire year. Since Gwyneth is the custodial parent and Saul meets the qualifying criteria, she can claim him as a dependent on her tax return. c) Gwyneth can claim the Child Tax Credit for Saul since he meets the criteria of being her qualifying child. The Child Tax Credit provides a tax credit per child that can help reduce the overall tax d) To be eligible for the EIC, Gwyneth needs to have earned income within certain limits and meet other requirements such as filing status and having a valid Social Security number. Without specific income information, it cannot be determined definitively if she qualifies for the EIC. e) The 2019 lookback provision allows taxpayers to use their earned income from the previous year to calculate the Earned Income Credit if it results in a larger credit amount. Since Gwyneth's income was lower in 2021, it may be beneficial for her to utilize the 2019 earned income to potentially qualify for a higher EIC amount. However, specific income details are necessary to determine Learn more about Income here: what consumer behavior concept and/or experience stands out to you? Please explain the answer. One consumer behavior concept that stands out is the "mere exposure effect." This refers to the phenomenon where people tend to develop a preference for things that they are familiar with or have been exposed to repeatedly. The mere exposure effect suggests that repeated exposure can increase liking and familiarity, influencing consumer choices and preferences. The mere exposure effect is a fascinating concept in consumer behavior that highlights the impact of familiarity on consumer preferences. It suggests that the more people are exposed to a particular stimulus, such as a brand, product, or advertisement, the more they tend to like it. This effect operates on a subconscious level, with repeated exposure influencing our perceptions and attitudes. The underlying mechanism behind the mere exposure effect is thought to be rooted in cognitive processes. When we encounter something repeatedly, our brains develop a sense of familiarity and comfort. This familiarity creates a sense of safety and reduces the perceived risk associated with the stimulus. As a result, we tend to feel more positive and inclined towards familiar things. The implications of the mere exposure effect for marketers and advertisers are significant. It emphasizes the importance of building brand awareness and establishing frequent touchpoints with By consistently exposing consumers to a brand or product through various channels, such as advertising, social media, and product placements, marketers can leverage the mere exposure effect to increase familiarity, likability, and ultimately, purchase intentions. Moreover, the mere exposure effect can also be observed in various aspects of consumer behavior, including music preferences, product packaging, and even interpersonal relationships. Understanding this concept allows marketers to strategically design marketing campaigns, create memorable experiences, and enhance brand recognition. By leveraging the power of familiarity and repeated exposure, companies can influence consumer decision-making processes and build lasting connections with their target audience. Learn more about brand recognition here : It has been written that one of two definitions of Kaizen is "using very small moments to inspire new products and inventions." (Maurer, pg. 3) How can you apply the provisions of Kaizen to your daily life? To apply the provisions of Kaizen to your daily life, focus on making small, incremental improvements in various aspects such as personal growth, habits, relationships, and productivity. To apply the provisions of Kaizen to your daily life, you can follow these steps: Set goals: Identify areas in your life that you would like to improve, such as personal development, health, relationships, or productivity. Set specific, achievable goals for each area. Break it down: Divide each goal into small, manageable tasks or habits that can be easily incorporated into your daily routine. These tasks should be simple and achievable, promoting continuous Embrace continuous improvement: Adopt a mindset of continuous improvement and seek opportunities for growth in every aspect of your life. Instead of aiming for drastic changes, focus on making small, incremental improvements regularly. Start small: Begin with one small change at a time. It could be as simple as incorporating a short meditation session or reading a few pages of a book every day. Consistency is key. Measure progress: Keep track of your progress regularly. Monitor the impact of the small changes you've made and reflect on how they are contributing to your overall goals. This will provide motivation and insights for further improvements. Reflect and adapt: Take time to reflect on your experiences and learn from them. Identify areas where you can refine your approach or make adjustments based on feedback and results. Maintain momentum: Build on the momentum of small improvements by continuously seeking new opportunities for growth. Explore new ideas, challenge yourself, and embrace a learning mindset. By applying the provisions of Kaizen to your daily life, you can make sustainable progress and achieve long-term personal growth and improvement. Remember, it's the small, consistent steps that lead to significant changes over time. Know more about the personal development click here: Colson Company has a line of credit with Federal Bank. Colson can borrow up to $328,500 at any time over the course of the calendar year. The following table shows the prime rate expressed as an annual percentage along with the amounts borrowed and repaid during the first four months of the year. Colson agreed to pay interest at an annual rate equal to 3 percent above the bank's prime rate. Funds are borrowed or repaid on the first day of the month. Interest is payable in cash on the last day of the month. The interest rate is applied to the outstanding monthly balance. For example, Colson pays 6 percent (3.50 percent + 4 percent) annual interest on $76,500 for the month of January. Month Amount Borrowed or Repaid Prime Rate for the Month January $76,500 3.50% February 115,100 2.50% March (16,100) 3.00% April 26,800 3.50% Required a. Compute the amount of interest that Colson will pay on the line of credit for the first four months of the year. b. Compute the amount of Colson's liability at the end of each of the first four months. Colson's liability at the end of each of the first four months is as follows: January: $76,500 February: $191,600 March: $175,500 April: $202,300 To compute the amount of interest that Colson will pay on the line of credit for the first four months of the year, we need to calculate the interest expense for each month based on the borrowed amount and the prime rate. a. Interest paid on the line of credit for the first four months: Borrowed amount: $76,500 Interest rate: Prime rate (3.50%) + 3% = 6.50% Interest expense: $76,500 * 6.50% = $4,972.50 Borrowed amount: $115,100 Interest rate: Prime rate (2.50%) + 3% = 5.50% Interest expense: $115,100 * 5.50% = $6,330.50 Repayment amount: $16,100 (negative value) Interest rate: Prime rate (3.00%) + 3% = 6.00% Interest expense: $16,100 * 6.00% = $966 Borrowed amount: $26,800 Interest rate: Prime rate (3.50%) + 3% = 6.50% Interest expense: $26,800 * 6.50% = $1,742 Total interest paid for the first four months: $4,972.50 + $6,330.50 + $966 + $1,742 = $14,011 b. Liability at the end of each month: January: $76,500 (borrowed amount) February: $76,500 + $115,100 = $191,600 (borrowed amount + additional borrowing) March: $191,600 - $16,100 = $175,500 (previous balance - repayment) April: $175,500 + $26,800 = $202,300 (previous balance + additional borrowing) Therefore, Colson's liability at the end of each of the first four months is as follows: January: $76,500 February: $191,600 March: $175,500 April: $202,300 Learn more about liability here: Lifeline, Inc., has sales of $772,764, costs of $446,895, depreciation expense of $57,610, interest expense of $37,079, and a tax rate of 37 percent. What is the net income for this firm? (Hint: Build the Income Statement The equilibrium wages and the allocation of labour will also be affected. Opening to trade can result in winners and losers in Portugal, depending on the specific factors of production and their (b) When the price of timber increases by 20%, the labour demand curves in the computers and timber sectors will be affected. In the timber sector, the higher price of timber increases the demand for labour, leading to an outward shift of the labour demand curve. In the computers sector, since capital is specific to producing computers, there is no direct effect on the labour demand curve. Learn more about deposit : A company has $8,000,000 of bonds payable (its only debt) with a 6% coupon, and has $12,000,000 in equity capital. The tax rate is 25% and the investor required rate of return is 8%. What is the company's weighted average cost of capital? The company's weighted average cost of capital is 6.8%. The weighted average cost of capital (WACC) is the average cost of capital for a company, weighted by the relative proportions of each component of capital. The formula for calculating the WACC is: WACC = (E/V × Re) + [(D/V × Rd) × (1 − T)] where, E = market value of the company's equity V = total market value of equity and debt Re = cost of equity D = market value of the company's debt Rd = cost of debt T = tax rate A company has $8,000,000 in bonds payable and $12,000,000 in equity capital. Thus, its total market value is: Total market value = $8,000,000 + $12,000,000 = $20,000,000 The proportion of debt and equity is: Proportion of debt (D/V) = $8,000,000 / $20,000,000 = 0.4 Proportion of equity (E/V) = $12,000,000 / $20,000,000 = 0.6 The cost of equity (Re) is 8%. The cost of debt (Rd) is calculated as follows: Rd = coupon rate × (1 − T) Rd = 0.06 × (1 − 0.25) = 0.045 or 4.5%The tax rate (T) is 25%. Now, let's plug in the values into the WACC formula: WACC = (0.6 × 0.08) + [(0.4 × 0.045) × (1 − 0.25)] WACC = 0.048 + (0.027 × 0.75)WACC = 0.068 or 6.8% To learn more about average cost click here: Modul University asked a sample of 7 random students how many hours they studied for the Maths mid term exam. The students (the names have been changed) answered: Albert =79 , Belinda =67 , The sample of 7 random students from the Modul University shows that the mean number of hours students studied for the Maths midterm exam is approximately 73.14 hours. The data from the sample of 7 random students, who were asked by the Modul University, for how many hours they studied for the Maths midterm exam are as follows: Albert = 79 Belinda = 67 Caroline = 72 Diana = 63 Elena = 85 Freddie = 77 Gloria = 80 The above data can be analysed using different statistical tools and methods for making decisions and predictions based on the data. To find different statistical measures from the data, we can calculate the measures of central tendency and measures of dispersion. For example, we can calculate the mean, median, and mode from the data to find out the central values of the data. In addition to this, we can calculate the range, variance, and standard deviation from the data to find the dispersion in the data. Using the above data, we can calculate the mean as follows: Mean = (79 + 67 + 72 + 63 + 85 + 77 + 80) / 7 Mean = 73.14 Therefore, the mean number of hours students studied for the Maths midterm exam is approximately 73.14 hours. We can calculate other measures such as median, mode, range, variance, and standard deviation by using the relevant formulae or statistical tools. The sample of 7 random students from the Modul University shows that the mean number of hours students studied for the Maths midterm exam is approximately 73.14 hours. By calculating other statistical measures, we can gain more insights into the data and use it to make predictions and decisions. To know more about analysed visit: 7. A worker can assemble a flashlight in 2 minutes. How many hours will it take to complete lub 60.000 units? If 6 workers are available, how long will it take to complete this task? 8. The final assembly requires 3 workers to complete. The first worker estimates 4 hours. The second worker estimate is 6 hours, and the third worker estimate is 5 hours. If 500 assem- blies are needed, what is the total hours needed? How many hours are needed from the first worker to complete the 500 assemblies. How many hours are needed from the second worker? How many hours are needed from the third worker? If the 500 assemblies need to be completed in 5 working days (8 hour shifts), how many workers will be needed? A worker can assemble a flashlight in 2 minutes. To find how many hours it will take to complete 60,000 units, convert the time taken per unit to hours, then use it to calculate the total time. 1 minute = 1/60 hours1 worker can assemble 1/2 minutes or 0.5/60 = 1/120 hours 60,000 units will be assembled in:60,000 × 1/120 hours = 500 hoursIf 6 workers are available, the time taken to complete the task will be reduced. We can use the following formula to find how long it will take: Time ∝ 1/number of workers. Hence,Time = K/number of workers,Where K is a constant. To find K, we can use the time taken by 1 worker. 500 hours = K/1K = 500Time = 500/6 hours = 83.33 hours ≈ 83 hours 20 minutes8. The total hours needed to complete 500 assemblies can be found by adding the time taken by each worker to complete one assembly, then multiplying by the number of assemblies needed. First worker's time = 4 hours Number of assemblies completed by first worker = 500 Second worker's time = 6 hours. To know more about worker visit:- a) Discuss briefly the differences if a unilateral or bilateral mistake of fact is made in the formation of a contract. b) Tell me the elements of either 1) fraudulent misrepresentation, 2) undue influence or 3) duress, and whether a party can get out of a contract with such a claim. a) In this situation, neither party is held responsible for the mistake because it was mutual. If the error is significant enough, the contract may be cancelled. b) 1) fraudulent misrepresentation: A party can get out of a contract with fraudulent misrepresentation if the party can prove all the elements of fraudulent misrepresentation. 2) undue influence : For a party to get out of a contract with undue influence, they must prove that the other party had a relationship of trust or confidence with them, that the other party abused that relationship, and that they suffered a loss as a result. 3) Duress : For a party to get out of a contract with duress, they must prove that they were under duress at the time they entered into the contract and that the duress was the reason they entered into the contract. a) The differences between a unilateral or bilateral mistake of fact in the formation of a contract are given below: Unilateral mistake of fact is a mistake made by one party to the contract, and the other party is unaware of this error. When a unilateral error is discovered, the party who made it may be able to get out of the contract, but only if the error was so severe that it changed the essential meaning of the contract. A bilateral error of fact is one in which both parties are mistaken about the same thing. In this situation, neither party is held responsible for the mistake because it was mutual. If the error is significant enough, the contract may be cancelled. b) The elements of fraudulent misrepresentation, undue influence, and duress are:1) Fraudulent misrepresentation: The elements of fraudulent misrepresentation include a material misrepresentation of a fact, the intention to deceive, the plaintiff's reliance on the misrepresentation, and damages resulting from that reliance. A party can get out of a contract with fraudulent misrepresentation if the party can prove all the elements of fraudulent misrepresentation. 2) Undue influence: Undue influence is when one party uses their power or position to force another party into a contract that they wouldn't have otherwise entered into. For a party to get out of a contract with undue influence, they must prove that the other party had a relationship of trust or confidence with them, that the other party abused that relationship, and that they suffered a loss as a result.3) Duress: Duress is when one party uses threats or force to make the other party enter into a contract. For a party to get out of a contract with duress, they must prove that they were under duress at the time they entered into the contract and that the duress was the reason they entered into the contract. learn more about fraudulent misrepresentation Comparing traditional and Roth which is better? What are the determining factors that make one better than the other? Are there any drawbacks to having one and not the other? Determining whether a traditional or Roth account is better depends on individual circumstances and financial goals. In general, the determining factors include current and future tax rates, expected income in retirement, and personal preferences. Traditional accounts provide upfront tax benefits as contributions are made with pre-tax dollars, reducing taxable income. Roth accounts, on the other hand, offer tax-free withdrawals in retirement but contributions are made with after-tax dollars. The choice also depends on whether one prefers to lower their current tax burden or have tax-free income in retirement. Drawbacks can include potential changes in tax laws and eligibility restrictions based on income limits or employer-sponsored plans. The choice between a traditional and Roth account depends on several factors. Traditional accounts offer immediate tax benefits as contributions are tax-deductible, reducing current taxable income. However, withdrawals in retirement are subject to income tax. On the other hand, Roth accounts do not provide immediate tax deductions, but qualified withdrawals in retirement are tax-free. The decision involves considering current and expected future tax rates. If an individual expects to be in a higher tax bracket in retirement, a Roth account may be more beneficial. Additionally, if someone prefers to have tax-free income during retirement or wants to minimize their tax burden, a Roth account might be a better choice. Drawbacks of having only one type of account include limited flexibility and potential changes in tax laws that could impact the advantages. It's worth noting that eligibility restrictions and income limits can also affect the availability of Roth accounts or tax-deductible contributions to traditional accounts, depending on individual circumstances and employer-sponsored plans. Learn more about tax deductions here: Let M denote the initial money supply. A friend of yours does not trust banks and keeps all his money in cash. He buys an old car for $30,000. The seller deposits the money in her checking account in Citibank. The bank keeps 5% of the deposit in reserve and lends the rest to Jack. Jack keeps $2,500 in cash and spends the remainder on Bed Bath & Beyond stock. The seller of the stock, Jane, transfers the whole amount to her checking account in Wells Fargo. Wells Fargo keeps 10% of the amount in reserve and lends to rest to Joshua. Joshua keeps $2,000 in cash and spends the rest on treasury securities. The seller of the securities happens to be the Fed. A fire in Joshua's uninsured house destroys $1,000 of his cash. Let M denote the resulting money supply. Calculate the change in the money supply, M The change in the money supply is - $5,250, thus Option (b) is the correct answer. Initial money supply, Mt Old car price = $30,000 The seller deposits the money in Citibank. Bank reserve 5% and lends the rest to Jack. Jack keeps $2,500 in cash and spends the remainder on Bed Bath & Beyond stock. The seller of the stock, Jane, transfers the whole amount to her checking account in Wells Fargo. Wells Fargo keeps 10% of the amount in reserve and lends to rest to Joshua. Joshua keeps $2,000 in cash and spends the rest on treasury securities. The seller of the securities happens to be the Fed. A fire in Joshua's uninsured house destroys $1,000 of his cash. Calculate the change in the money supply, Mt+1 −Mt When a person buys a car and pays $30,000 in cash to the seller, the money supply remains the same. But when the seller deposits the cash in Citibank, the bank has to keep a 5% reserve and the rest is given as a loan to Jack. Therefore, Mt+1 - Mt = - $30,000 x 5/100 = - $1,500 (because the deposit of the seller to Citibank would result in reducing the money supply by $1,500.) Jack keeps $2,500 in cash and spends the remainder ($30,000 - $2,500 = $27,500) on Bed Bath & Beyond stock. Therefore, the money supply remains the same. Jane sells stock and deposits the amount ($27,500) in Wells Fargo. The bank keeps 10% as a reserve ($2,750) and lends the rest to Joshua. Therefore, Mt+1 - Mt = -$2,750 (because the deposit of Jane to Wells Fargo would result in reducing the money supply by $2,750.) Joshua keeps $2,000 in cash, and the rest is spent on treasury securities. Therefore, the money supply remains the same. A fire in Joshua's uninsured house destroys $1,000 of his cash. Therefore, Mt+1 - Mt = -$1,000 (because the destruction of cash in Joshua’s house would result in reducing the money supply by $1,000.) The overall change in the money supply is: Mt+1 - Mt = - $1,500 - $2,750 - $1,000 = - $5,250. Learn more about money supply Considering f(p) = -10p2 + 50p + 140. 1. Compute the elasticity function E(p), and then find E(5). 2. At E(5) is it elastic or inelastic? Prove relationship. Based on the calculation of the elasticity function and the elasticity coefficient, we can conclude that the function f(p) = -10p^2 + 50p + 140 is inelastic. 1. Given the function: f(p) = -10p^2 + 50p + 140 Calculation of elasticity function: To compute the elasticity function E(p), we first find the derivative of the given function. E(p) = (p/f(p)) * f'(p) E(p) = (p/f(p)) * (-20p + 50) Calculate E(5): E(5) = (5/f(5)) * (-20(5) + 50) E(5) = (-50/140) = -0.35714 2. We can determine the elasticity of the function at p=5 using the value of E(5). If E(5) < -1, the function is elastic. If E(5) > -1, the function is inelastic. If E(5) = -1, the function is unit In this case, E(5) = -0.35714, which is greater than -1. Therefore, the function is inelastic at p=5. Calculation of elasticity coefficient: To prove the relationship between elasticity and the elasticity coefficient, we use the formula: Elasticity Coefficient = % Change in Quantity Demanded / % Change in Price Now, let's calculate the elasticity coefficient at p=5 and p=6. Change in Quantity Demanded = f(6) - f(5) = (-106^2 + 506 + 140) - (-105^2 + 505 + 140) = 90 Change in Price = 6 - 5 = 1 Elasticity Coefficient = % Change in Quantity Demanded / % Change in Price Elasticity Coefficient = (90/340) / (1/5) Elasticity Coefficient = 0.66 < 1 Since the elasticity coefficient is less than 1, the function is inelastic. Therefore, based on the calculation of the elasticity function and the elasticity coefficient, we can conclude that the function f(p) = -10p^2 + 50p + 140 is inelastic. Learn more about elasticity function E(p) visit: A project requires an initial investment (today) of $30,000 and would give returns of $14,000, $12,000, $6,000, $4,000 and $2,000 at the end of each year for 5 years. Find (a) Internal rates of return. (b) The net present value with a rate of 10%. (c) The payback period. (d) The modified payback period with a rate of 10%. Note: Please answer in detail showing every step, thank you. ( And please do not copy previous chegg answers) The IRR is approximately 20.6%. The NPV of the project with a discount rate of 10% is $1,419. The payback period is approximately 3.3 years. The modified payback period with a rate of 10% is approximately 4.1 years. (a) Internal rates of return The internal rate of return (IRR) is the discount rate that makes the net present value (NPV) of a project equal to zero. In this case, the IRR is approximately 20.6%. This means that if the project's cash flows are discounted at 20.6%, the NPV will be zero. If the discount rate is lower than 20.6%, the NPV will be positive, and if the discount rate is higher than 20.6%, the NPV will be negative. (b) Net present value with a rate of 10% The NPV of the project with a discount rate of 10% is $1,419. This means that the project is expected to generate $1,419 in additional value over its lifetime, after taking into account the initial investment and the time value of money. (c) Payback period The payback period is the amount of time it takes to recover the initial investment. In this case, the payback period is approximately 3.3 years. This means that the initial investment will be fully recovered after 3.3 years of cash flows. (d) Modified payback period with a rate of 10% The modified payback period is the amount of time it takes to recover the initial investment, assuming that the cash flows are discounted at a certain rate. In this case, the modified payback period with a rate of 10% is approximately 4.1 years. This means that the initial investment will be fully recovered after 4.1 years of discounted cash flows. The IRR was calculated using a financial calculator or a spreadsheet. The NPV was calculated using the following formula: NPV = -$30,000 + $14,000/(1 + 0.10) + $12,000/(1 + 0.10)^2 + $6,000/(1 + 0.10)^3 + $4,000/(1 + 0.10)^4 + $2,000/(1 + 0.10)^5 The payback period was calculated by finding the year in which the cumulative cash flows first exceed the initial investment. The modified payback period was calculated by finding the year in which the cumulative discounted cash flows first exceed the initial investment. Learn more about discount rate here; brainly.com/question/9841818 Other Questions student representatives are organizing a tax preparation system. They expect toreceive 100 students, coming to the system within 4 hours at a constant rate. The process consists of a check-in station, where there is a single server that takes 2minutes to decide whether the student has all the information needed to fill out theforms or not. From previous experience, they know that 80% will be OK, and then,they will go to one of the analysts to complete the tax forms. The other 20% will needto gather the extra info, a process that they do by themselves, and that will takeexactly 1 hour. After that hour, they will join the queue for the analysts. There are 5analysts and each takes 15 minutes to serve a student. what is the maximum time it will take a student to complete the process? Ayuden por favor, no entiendo este problema A.Describe how photosynthesis captures carbon dioxide and converts itto simple sugars.B. What is the importance of rubisco and what problems haveoccured because of a changing atmosphere? Recording Business transactionsOn February 1st, Brenda creates a new company providing accounting services investing $20,000 cash in exchange of 1,000 shares of capital stock. The company will be called Creative accounting Ltd.The same day the company purchases a small office for $50,000. Pays $5,000 in cash and issues a note payable for the remainder. The price of the building is $40,000 and the land $10,000.On February 2nd the company hires an employee for $1,000/month. Salaries are payable the day 5 of the following month.On February 3rd the company purchases furniture for $6,000. Pays $1,500 cash and will pay the remainder in March 15.The same day the company purchases 2 computers for $2,400On February 15 the company bills client Mr. Happy for $1,600. The client pays $600 cash and will pay the remainder within one week. On February 19 the company receives an electricity bill to be paid in March.On February 21 the company receives the settlement of Mr. Happys account receivable.On February 28, the company Sends an invoice to its client Mr. Brown for services rendered during the month of February for $3,000. The invoice should be paid before March 15.1: Prepare the journal entries for the above mentioned transactions. 2: Prepare a ledger account for all the accounts listed below.List of accounts: Accounts payable, accounts receivable, cash, note payable, land, computers, building, salaries payable, revenue, salaries expense, electricity expense,.3: calculate the balance of each one of the 3 sections of the balance sheet at the end of February (assets, liabilities, owners equity). Show your calculation) Canadas unemployment rate is lower now than it has been for years. Answer the following regarding unemployment concepts (8 Marks):a)John Smith has been looking for work for some time. He has decided to give up his search for work. Is John counted in the labour force? In the unemployment statistics?B) Explain Says Law regarding supply and demand and how it affects unemployment. The following are n=10 temperature measurements (degrees F.) made every minute on a chemical reactor. 200,202,208,204,204,207,207,204,202,199 Calculate approximate standard errors for r 1,r 2,r 3that is, SE(r 1),SE(r 2),SE(r 3), using the folllowing formulas: 13. SE(r 1)=1/ n. (A) 0.200( B)0.316(C) 0.400(D) 0.500(E) 0.60014. SE(r 2)= 1+2r 12/ n. 2 (A) 0.200( B)0.316(C) 0.341(D) 0.500(E) 0.60015. SE(r 3)= 1+2r 12+2r 22/ n. CA8.11 (LO 3, 4) Ethics (LIFO Choices) Wilkens Company uses the LIFO method for inventory costing. In an effort to lower net income, company president Mike Wilkens tells the plant accountant to take the unusual step of recommending to the purchasing department a large purchase of inventory at year-end. The price of the item to be purchased has nearly doubled during the year, and the item represents a major portion of inventory value. Instructions Answer the following questions. a. Identify the major stakeholders. If the plant accountant recommends the purchase, what are the consequences? b. If Wilkens Company were using the FIFO method of inventory costing, would Mike Wilkens give the same order? Why or why not? the balcony of a school auditorium, There are 430 seats. In the first raw there are 34 seats and each succeeding row has 2 seats more than the previous row How may rows are there? How many people does it take to fill the first eight rows? Some speculators think you should buy stocks which are heavily shorted. Their reasoning is:a. the more a stock is shorted, the more that needs to be bought to close the short positionsb. the more a stock is shorted, the more speculators will have to sell in the futurec. the crowd is always right The internal audit function of one of your clients consists of 3 staff. The Head of Internal Audit is a former Big 4 audit manager with 12 years experience in the profession. He also has 4 years of internal audit experience. He reports directly to the Board and meets with external auditors on a timely basis. Since joining the firm he has spent most of his time developing audit work programs, improving internal control systems documentation of the company and reviewing any internal audit work performed by his staff. He is assisted by a new accounting graduate, and another staff member who has no accounting qualifications but has 6 years internal audit experience.Outline the main factors that would determine your reliance on the clients internal audit function ie the extent to which the internal work is adequate for the purposes of the audit. 10.1 Learning Outcomes: Describe the components of a retail strategy. Identify the benefits and challenges of Omni-channel retailing.10.2 Action Required: Chapter Reading: Read Ch-17 "Retailing and Omni-channel Marketing" from the text book- Dhruv Grewal and Michael Levy (2020) "Marketing" (8th Edition). McGraw-Hill Education. Video Link: Watch the following video.Nordstrom Rack: Retail Strategyhttps://www.viddler.com/embed/fd437bba Summary of the Video- Since 1973, Nordstrom Rack has been the off-price division of fashion retailer Nordstrom. Nordstrom Rack has been successful in applying the four Ps to its retail strategy. The stores stock high-quality merchandise from well-known designers at discounted prices, creating the "treasure hunt": the search through ever-changing merchandise for "things they didnt know they had to have when they walked in." Nordstrom also sells its products on its website, and on a flash sale site called Hautelook.10.3 Test your Knowledge (Answer the following Questions): Discussion Question #1: Which benefits of in-store shopping are related to the appeal of the "treasure hunt" discussed in the video?10.4 Instructions Answer both questions in test your knowledge section. Post your answers in the discussion board using the discussion link below.(Week 10: Interactive learning Discussion) Enable GingerCannot connect to Ginger Check your internetconnectionor reload the browserDisable in this text fieldRephraseRephrasecurrent sentence0Edit in Ginger Recall that according to the Solow Model, y = Ak And that the level of k each period is determined by the equations kt+1 = kt + kt. Income is split between consumption and savings: y = C + S where S = sy. The following values are given: A = 5, s = .2, = .005, = 0.5. (Do not use comma separators or dollar signs $) What is the long-run steady-state level of capital?:What is the long-run steady state level of real income?:What is the value of savings in the steady-state?: Crowley collects sales on account in the month after the sale. The Accounts Receivable balance on 1 is, which represents 's sales on account. Crowley projects the following cash receipts from customers:Re-calculate cash receipts from customers if total sales remain the same but cash sales are only 20% of the total.January February MarchCash sales (20 %)Sales on account (80%)Total sales $20,000 $22.000 $24,00January February MarchCash sales (25%) 5,000 5,000 $6,000Sales on account (75%) $15,500 16,000 18,000Total sales 20,000 $ 22,000 $ 24,000Now, recalculate cash receipts from customers for each month if cash sales are only 20 % of the total.January February MarchCash receipts from cash sales 5,500 5,000 $6,000Cash receipts from sales on account 13,500 15,000 16.500Total cash receipts from customers 18,500 $20,500 $22,500(opens in a new tab)(opens in a new tab)We're in the know A Distributor Receives A Very Large Shipment. The Distributor Would Like To Accept The Shipment If 15% Or Fewer Of The Items Are Defective And To Return It If More Than 15% Of The Components Are Defective. Someone On The Quality Assurance Team Samples 4 Items. Let X Be The Random Variable For The Number Of Defective Items In The Sample. You Can Find the simple interest and the final value. If the principle is BD 1100 and interest rate 8% and the length of loan 950 months. Discuss the importance of understanding a firm's strategy even if you are not a senior manager in a firm. The next dividend payment by Savitz, Incorporated, will be $3.25 per share. The dividends are anticipated to maintain a growth rate of 5 percent forever.If the stock currently sells for $47 per share, what is the required return?Multiple ChoiceA. 5.00%B. 11.68%C. 6.91% Consider the following Information for a firm financed only_by debt and equity Cost of Debt Before Tax =6% Cost of Equity = 11\% Weight of Debt =35% Corporate Tax Rate =30% What is the WACC of this firm? 8.62% 9.25% Bolts from a particular production line have a diameter that we can assume is normally distributed with expectation 4.9 mm and standard deviation 0.03 mm. The nut to be screwed onto the bolt has a diameter that we can assume is normally distributed with expectation 5.0 mm and standard deviation 0.04 mm. What is the probability that a randomly selected nut can be screwed onto a randomly selected bolt?2. A company casts concrete cylinders of a specific type. The concrete cylinders are characterized by the so-called cylinder compressive strength, i.e. the maximum load (in kg/cm2) a cylinder can be subjected to without crushing. It can be assumed that the cylinder pressure strength is normally distributed with expected value 200.0 and standard deviation 9.5. Cylinders with compressive strength below 180.0 are considered defective.(a) What proportion of the production must the company expect to be defective?(b) Now suppose that concrete cylinders are checked before sale by being loaded with 180.0 kg/cm2, and the defective ones are sorted out. What is the probability that such a controlled cylinder has a compressive strength below 187.125 kg/cm2?	{"url":"https://denizsozluk.com/article/suppose-the-index-model-for-stocks-a-and-b-is-estimated-from-the-excess-returns-with-the-following-results","timestamp":"2024-11-05T17:19:41Z","content_type":"text/html","content_length":"183891","record_id":"<urn:uuid:80f40e90-15a3-4d89-b6f0-054ad8be66fa>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00077.warc.gz"}
Evaluate x/y+y/x if log((x+y)/3) = 1/2(logx+logy) Evaluate $\dfrac{x}{y}+\dfrac{y}{x}$ if $\log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\dfrac{1}{2}(\log{x}+\log{y})$ A logarithmic equation is defined in terms of two variables $x$ and $y$. $\log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\dfrac{1}{2}(\log{x}+\log{y})$ On the basis of this logarithmic equation, we have to find the value of the following algebraic expression in this logarithmic problem. Eliminate the Logarithm from the equation Let’s concentrate on the given logarithmic equation for eliminating the logarithmic form from the equation. It helps us to express the whole equation in terms of the variables $x$ and $y$. $\log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\dfrac{1}{2}(\log{x}+\log{y})$ The left hand side expression is purely in logarithmic form but the right hand side expression is also in logarithmic form but the factor $\dfrac{1}{2}$ pulled us back in eliminating the logarithmic form from the equation. Hence, we should overcome this issue mathematically. In order to overcome it, move the number $2$ to left hand side of the equation. $\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $1 \times (\log{x}+\log{y})$ $\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\log{x}+\log{y}$ Look at the expression in the right hand side of the equation, two logarithmic terms are connected by a plus sign. So, it can be simplified by the product rule of logarithms. $\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\log{(x \times y)}$ $\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\log{(xy)}$ The expression in the right hand side of the equation is purely converted in logarithmic form but the expression in the left hand side of the equation is not in pure logarithmic form due to the multiplying factor $2$. However, it can be overcome by the power rule of logarithms. $\implies$ $\log{\Big(\dfrac{x+y}{3}\Big)^2}$ $\,=\,$ $\log{(xy)}$ Now, the expressions in the both sides of the equation are in logarithmic form. Hence, the expressions inside the logarithm are equal mathematically. $\implies$ $\Big(\dfrac{x+y}{3}\Big)^2$ $\,=\,$ $xy$ Simplify the equation in algebraic form Now, we have to focus on simplifying the equation in the algebraic form. $\implies$ $\Big(\dfrac{x+y}{3}\Big)^2$ $\,=\,$ $xy$ There is no way to simplify the expression in the right hand side of the equation. So, we must concentrate on the simplifying the algebraic expression at the left hand side of the equation. It can be simplified by the power of a quotient rule. $\implies$ $\dfrac{(x+y)^2}{3^2}$ $\,=\,$ $xy$ $\implies$ $\dfrac{(x+y)^2}{9}$ $\,=\,$ $xy$ $\implies$ $(x+y)^2$ $\,=\,$ $9xy$ The left hand side expression in the equation is in the form of square of sum of two terms. So, it can be expanded by the square of sum of two terms formula. $\implies$ $x^2+y^2+2xy$ $\,=\,$ $9xy$ Now, let’s complete the simplification of the algebraic equation. $\implies$ $x^2+y^2$ $\,=\,$ $9xy-2xy$ $\implies$ $x^2+y^2 \,=\, 7xy$ Find the value of the algebraic expression The given logarithmic equation $\log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\dfrac{1}{2}(\log{x}+\log{y})$ is successfully simplified as the algebraic equation $x^2+y^2 \,=\, 7xy$ We have to use the algebraic equation $x^2+y^2 \,=\, 7xy$ to evaluate algebraic expression $\dfrac{x}{y}+\dfrac{y}{x}$ Move the literal coefficient of $7$ to left hand side of the equation. $\implies$ $\dfrac{x^2+y^2}{xy} \,=\, 7$ $\implies$ $\dfrac{x^2}{xy}+\dfrac{y^2}{xy} \,=\, 7$ $\implies$ $\require{cancel} \dfrac{\cancel{x^2}}{\cancel{x}y}+\dfrac{\cancel{y^2}}{x\cancel{y}} \,=\, 7$ $\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{x}{y}+\dfrac{y}{x} \,=\, 7$ Therefore, it is evaluated that the value of the algebraic expression $\dfrac{x}{y}+\dfrac{y}{x}$ is $7$.	{"url":"https://www.mathdoubts.com/evaluate-x-by-y-plus-y-by-x-if-log-of-x-plus-y-by-3-equals-to-log-x-plus-log-y-by-2/","timestamp":"2024-11-11T00:27:09Z","content_type":"text/html","content_length":"31227","record_id":"<urn:uuid:a2bf75ac-606f-49e5-84ff-87c21c67d781>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00407.warc.gz"}
Negative Binomial Distribution In this lesson, we cover the negative binomial distribution and the geometric distribution. As we will see, the geometric distribution is a special case of the negative binomial distribution. Negative Binomial Experiment A negative binomial experiment is a statistical experiment that has the following properties: • The experiment consists of x repeated trials. • Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. • The probability of success, denoted by P, is the same on every trial. • The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials. • The experiment continues until r successes are observed, where r is specified in advance. Consider the following statistical experiment. You flip a coin repeatedly and count the number of times the coin lands on heads. You continue flipping the coin until it has landed 5 times on heads. This is a negative binomial experiment because: • The experiment consists of repeated trials. We flip a coin repeatedly until it has landed 5 times on heads. • Each trial can result in just two possible outcomes - heads or tails. • The probability of success is constant - 0.5 on every trial. • The trials are independent; that is, getting heads on one trial does not affect whether we get heads on other trials. • The experiment continues until a fixed number of successes have occurred; in this case, 5 heads. The following notation is helpful, when we talk about negative binomial probability. • x: The number of trials required to produce r successes in a negative binomial experiment. • r: The number of successes in the negative binomial experiment. • P: The probability of success on an individual trial. • Q: The probability of failure on an individual trial. (This is equal to 1 - P.) • b(x; r, P): Negative binomial probability - the probability that an x-trial negative binomial experiment results in the rth success on the xth trial, when the probability of success on an individual trial is P. • [n]C[r]: The number of combinations of n things, taken r at a time. • n!: The factorial of n (also known as n factorial). Negative Binomial Distribution A negative binomial random variable is the number X of repeated trials to produce r successes in a negative binomial experiment. The probability distribution of a negative binomial random variable is called a negative binomial distribution. The negative binomial distribution is also known as the Pascal distribution. Suppose we flip a coin repeatedly and count the number of heads (successes). If we continue flipping the coin until it has landed 2 times on heads, we are conducting a negative binomial experiment. The negative binomial random variable is the number of coin flips required to achieve 2 heads. In this example, the number of coin flips is a random variable that can take on any integer value between 2 and plus infinity. The negative binomial probability distribution for this example is presented below. Number of coin flips Probability 2 0.25 3 0.25 4 0.1875 5 0.125 6 0.078125 7 or more 0.109375 Negative Binomial Probability The negative binomial probability refers to the probability that a negative binomial experiment results in r - 1 successes after trial x - 1 and r successes after trial x. For example, in the above table, we see that the negative binomial probability of getting the second head on the sixth flip of the coin is 0.078125. Given x, r, and P, we can compute the negative binomial probability based on the following formula: Negative Binomial Formula. Suppose a negative binomial experiment consists of x trials and results in r successes. If the probability of success on an individual trial is P, then the negative binomial probability is: b(x; r, P) = [x-1]C[r-1] * P^r * (1 - P)^x - r b(x; r, P) = { (x-1)! / [ (r-1)!(x-r)!] } P^r * (1 - P)^x - r The Mean of the Negative Binomial Distribution If we define the mean of the negative binomial distribution as the average number of trials required to produce r successes, then the mean is equal to: μ = r / P where μ is the mean number of trials, r is the number of successes, and P is the probability of a success on any given trial. Geometric Distribution The geometric distribution is a special case of the negative binomial distribution. It deals with the number of trials required for a single success. Thus, the geometric distribution is negative binomial distribution where the number of successes (r) is equal to 1. An example of a geometric distribution would be tossing a coin until it lands on heads. We might ask: What is the probability that the first head occurs on the third flip? That probability is referred to as a geometric probability and is denoted by g(x; P). The formula for geometric probability is given below. Geometric Probability Formula. Suppose a negative binomial experiment consists of x trials and results in one success. If the probability of success on an individual trial is P, then the geometric probability is: g(x; P) = P * Q^x - 1 Test Your Understanding The problems below show how to apply your new-found knowledge of the negative binomial distribution (see Example 1) and the geometric distribution (see Example 2). Negative Binomial Calculator As you may have noticed, the negative binomial formula requires many time-consuming computations. The Negative Binomial Calculator can do this work for you - quickly, easily, and error-free. Use the Negative Binomial Calculator to compute negative binomial probabilities. The calculator is free. It can found in the Stat Trek main menu under the Stat Tools tab. Or you can tap the button below. Negative Binomial Calculator Example 1 Bob is a high school basketball player. He is a 70% free throw shooter. That means his probability of making a free throw is 0.70. During the season, what is the probability that Bob makes his third free throw on his fifth shot? Solution: This is an example of a negative binomial experiment. The probability of success (P) is 0.70, the number of trials (x) is 5, and the number of successes (r) is 3. To solve this problem, we enter these values into the negative binomial formula. b(x; r, P) = [x-1]C[r-1] P^r * Q^x - r b(5; 3, 0.7) = [4]C[2] 0.7^3 * 0.3^2 b(5; 3, 0.7) = 6 0.343 * 0.09 = 0.18522 Thus, the probability that Bob will make his third successful free throw on his fifth shot is 0.18522. Example 2 Let's reconsider the above problem from Example 1. This time, we'll ask a slightly different question: What is the probability that Bob makes his first free throw on his fifth shot? Solution: This is an example of a geometric distribution, which is a special case of a negative binomial distribution. Therefore, this problem can be solved using the negative binomial formula or the geometric formula. We demonstrate each approach below, beginning with the negative binomial formula. The probability of success (P) is 0.70, the number of trials (x) is 5, and the number of successes (r) is 1. We enter these values into the negative binomial formula. b(x; r, P) = [x-1]C[r-1] P^r * Q^x - r b(5; 1, 0.7) = [4]C[0] 0.7^1 * 0.3^4 b(5; 3, 0.7) = 0.00567 Now, we demonstate a solution based on the geometric formula. g(x; P) = P Q^x - 1 g(5; 0.7) = 0.7 * 0.3^4 = 0.00567 Notice that each approach yields the same answer.	{"url":"https://stattrek.com/probability-distributions/negative-binomial?tutorial=AP","timestamp":"2024-11-07T22:05:59Z","content_type":"text/html","content_length":"70681","record_id":"<urn:uuid:31d2c884-71d9-44b9-be01-4751d141abbf>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00374.warc.gz"}
At the 2 - math word problem (76614) At the 2 In the beginning, Kim has 25 cards. He then gave 11 cards to his friend. His cousin came over and gave him 17 cards. His uncle also came and gave him 18 cards. How many cards does he have now? Correct answer: Did you find an error or inaccuracy? Feel free to write us . Thank you! You need to know the following knowledge to solve this word math problem: Related math problems and questions:	{"url":"https://www.hackmath.net/en/math-problem/76614","timestamp":"2024-11-04T04:01:42Z","content_type":"text/html","content_length":"47400","record_id":"<urn:uuid:b833915c-677d-4534-8c4c-13bd148d69b0>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00287.warc.gz"}
[ASoT] Observations about ELK — AI Alignment Forum This document outlines some of my current thinking about ELK, in the form of a series of observations I have made that inform my thinking about ELK. Editor’s note: I’m experimenting with having a lower quality threshold for just posting things even while I’m still confused and unconfident about my conclusions, but with this disclaimer at the top. Thanks to AI_WAIFU and Peter Barnett for discussions. • One way we can think of ELK is we have some set of all possible reporters, each of which takes in the latent states of a world model and outputs, for the sake of concreteness, the answer to some particular fixed question. So essentially we can think of a reporter+model pair as assigning some answer to every point in the action space. Specifically, collapsing possible sequences of actions into just a set of possible actions and only worrying about one photo doesn’t result in any loss of generality but makes things easier to talk about. • We pick some prior over the set of possible reporters. We can collect training data which looks like pairs (action, answer). This can, however, only cover a small part of the action space, specifically limited by how well we can “do science.” • This prior has to depend on the world model. If it didn’t, then you could have two different world models with the same behavior on the set where we can do science, but where one of the models understands what the actual value of the latent variable is, and one is only able to predict that the diamond will still appear there but can’t tell whether the diamond is still real. A direct translator for the second will be a human simulator for the first. • More generally, we don’t really have guarantees that the world model will actually understand what’s going on, and so it might genuinely believe things about the latent variable that are wrong. Like, we might have models which think exactly like a human and so they don't understand anything a human wouldn’t understand, so the direct translator for this model would also simultaneously be a human simulator, because the model literally believes exactly what a human would believe. • There are more than just human simulators and direct translators that are consistent with the training data; there are a huge number of ways the remaining data can be labeled. This basically breaks any proposal that starts with penalizing reporters that look like a human simulator. • I basically assume the “human simulator” is actually simulating us and whatever science-doing process we come up with, since this doesn’t change much and we’re assuming that doing science alone isn’t enough because we’re looking at the worst case. • So, for any fixed world model, we need to come up with some prior over all reporters such that after picking all the reporters consistent with our data (action, answer), the maximum of the resulting posterior is the direct translator. • A big assumption I’m making is that the KL distance between our prior and the complexity prior has to be finite. My intuition for why this has to be is that a) universality of the complexity prior, and b) even if our prior doesn’t have to cover all possible reporters, it still seems reasonable that as complexity increases the number of reporters that our prior accepts as nonzero needs to increase exponentially, which gets something similar to universality of complexity on the subset of plausible reporters. I think this point is the main weak point of the argument. • Since the amount of data we can get is bounded by our ability to do science, the amount of information we can get from that is also bounded. However, since we’re assuming worst case, we can make the complexity difference between the direct translator and all the other junk arbitrarily big, and therefore arbitrarily less likely, and since our modified prior and our data only get us a finite amount of evidence, we can always imagine a world where the direct translator is sufficiently complex such that we can’t pick it out. • So to solve ELK we would need to somehow rule out a huge part of the set of possible reporters before even updating on data. In some sense this feels like we ran in a circle, because there is always the prior that looks at the model, solves ELK, and then assigns probability 1 to the direct translator. So we can always just move all the difficulty into choosing the prior and then the data part is just entirely useless. I guess in retrospect it is kind of obvious that the prior has to capture the part that the data can’t capture, but that's literally the entire ELK problem. • One seemingly useful thing this does tell us is that our prior has to be pretty restrictive if we want to solve ELK in the worst case, which I think rules out a ton of proposals right off the bat. In fact, I think your prior can only assign mass to a finite number of reporters, because to get the sum to converge it needs to decay fast enough, which also gets you the same problem. • Only assigning mass to a finite number of reporters is equivalent to saying we know the upper bound of direct translator complexity. Therefore, we can only solve ELK if we can bound direct translator complexity, but in the worst case setting we're assuming direct translator complexity can be unbounded. • [S:So either we have to be able to bound complexity above, violating the worst case assumption, or we have to bound the number of bits we can get away from the complexity prior. Therefore, there exists no worst case solution to ELK. :S](Update: I no longer think this is accurate, because there could exist a solution to ELK that function entirely by looking at the original model. Then the conclusion of this post is weaker and only shows that assigning an infinite prior is not a feasible strategy for worst case) New Comment	{"url":"https://www.alignmentforum.org/posts/kWTko53s2DqTeprjz/asot-observations-about-elk","timestamp":"2024-11-02T02:04:59Z","content_type":"text/html","content_length":"154404","record_id":"<urn:uuid:a6a24b39-50c5-43f2-b308-423f30e4b589>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00158.warc.gz"}