Subset (6)

UltraData-Math-L1 · 86M rows

Split (1)

train · 86M rows

content stringlengths 86 994k	meta stringlengths 288 619
Sampling from High-Dimensional Gaussian Distributions without the Full Covariance Matrix – The Dan MacKinlay stable of variably-well-consider’d enterprises Sampling from High-Dimensional Gaussian Distributions without the Full Covariance Matrix October 30, 2024 — October 31, 2024 Hilbert space kernel tricks Lévy processes stochastic processes time series Assumed audience: ML people When dealing with high-dimensional Gaussian distributions, sampling can become computationally expensive, especially when the covariance matrix is large and dense. Traditional methods like the Cholesky decomposition become impractical. However, if we can efficiently compute the product of the covariance matrix with arbitrary vectors, we can leverage Langevin dynamics to sample from the distribution without forming the full covariance matrix. I have been doing this recently in the setting where $\Sigma$ is outrageously large, but I can nonetheless calculate it for arbitrary vectors $\Sigma \mathbf{v}$; This arises, for example, when I have a kernel which I can evaluate and I need to use it to generate some samples from my random field, especially where the kernel arises as linear product under some feature map. TODO: evaluate actual computational complexity of this method. Note this is really just some notes I have made to myself. I need to sanity check the procedure on a real problem. 1 Problem Setting We aim to sample from a multivariate Gaussian distribution: \[ \mathbf{x} \sim \mathcal{N}(\boldsymbol{\mu}, \Sigma) \] • $\boldsymbol{\mu} \in \mathbb{R}^D$ is the known mean vector. • $\Sigma \in \mathbb{R}^{D \times D}$ is the notional known covariance matrix, which might be too large to actually compute, let alone factorise for sampling in the usual way. 2 Langevin Dynamics for Sampling Langevin dynamics provide a way to sample from a target distribution by simulating a stochastic differential equation (SDE) whose stationary distribution is the desired distribution. For a Gaussian distribution, the SDE simplifies due to the properties of the normal distribution (i.e. Gaussians all the way down). 2.1 The Langevin Equation The continuous-time Langevin equation is \[ d\mathbf{x}_t = -\nabla U(\mathbf{x}_t) \, dt + \sqrt{2} \, d\mathbf{W}_t \] • $U(\mathbf{x})$ is the potential function related to the target distribution $p(\mathbf{x})$ via $p(\mathbf{x}) \propto e^{-U(\mathbf{x})}$. • $d\mathbf{W}_t$ represents the increment of a Wiener process (standard Brownian motion). For our Gaussian distribution, the potential function is: \[ U(\mathbf{x}) = \frac{1}{2} (\mathbf{x} - \boldsymbol{\mu})^\top \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \] We discretize the Langevin equation using the Euler-Maruyama method with time step $\epsilon$: \[ \mathbf{x}_{k+1} = \mathbf{x}_k - \epsilon \nabla U(\mathbf{x}_k) + \sqrt{2\epsilon} \, \boldsymbol{\eta}_k \] where $\boldsymbol{\eta}_k \sim \mathcal{N}(\mathbf{0}, \mathbf{I}_D)$. Next, the gradient of the potential function is: \[ \nabla U(\mathbf{x}) = \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \] Instead of computing $\Sigma^{-1}$ directly, we can solve the linear system: \[ \Sigma \mathbf{v} = \mathbf{x} - \boldsymbol{\mu} \] for $\mathbf{v}$, which gives $\mathbf{v} = \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu})$. 3 Now, to solve that linear equation To solve $\Sigma \mathbf{v} = \mathbf{r}$ efficiently without forming $\Sigma$, we use the Conjugate Gradient (CG) method. The CG method is suitable for large, sparse, and positive-definite matrices and relies only on matrix-vector products $\Sigma \mathbf{v}$. Given $\Sigma \mathbf{v} = \mathbf{r}$: 1. Initialize $\mathbf{v}_0 = \mathbf{0}$, $\mathbf{r}_0 = \mathbf{r} - \Sigma \mathbf{v}_0$, $\mathbf{p}_0 = \mathbf{r}_0$. 2. For $k = 0, 1, \ldots$: □ $\alpha_k = \frac{\mathbf{r}_k^\top \mathbf{r}_k}{\mathbf{p}_k^\top \Sigma \mathbf{p}_k}$ □ $\mathbf{v}_{k+1} = \mathbf{v}_k + \alpha_k \mathbf{p}_k$ □ $\mathbf{r}_{k+1} = \mathbf{r}_k - \alpha_k \Sigma \mathbf{p}_k$ □ If $\\|\mathbf{r}_{k+1}\\| < \text{tolerance}$, stop. □ $\beta_k = \frac{\mathbf{r}_{k+1}^\top \mathbf{r}_{k+1}}{\mathbf{r}_k^\top \mathbf{r}_k}$ □ $\mathbf{p}_{k+1} = \mathbf{r}_{k+1} + \beta_k \mathbf{p}_k$ 4 Plug the bits together We have the following algorithm: 1. Initialization: □ Start with $\mathbf{x}_0 = \boldsymbol{\mu}$ or any arbitrary vector. 2. For $k = 0, 1, \ldots, N$: □ Compute $\mathbf{r}_k = \mathbf{x}_k - \boldsymbol{\mu}$. □ Solve $\Sigma \mathbf{v}_k = \mathbf{r}_k$ using CG to get $\mathbf{v}_k = \Sigma^{-1} (\mathbf{x}_k - \boldsymbol{\mu})$. □ Update$ _{k+1} = _k - _k + , _k$ where $\boldsymbol{\eta}_k \sim \mathcal{N}(\mathbf{0}, \mathbf{I}_D)$. 5 PyTorch Implementation For my sins, I am cursed to never escape PyTorch. Here is an implementation in that language that I got an LLM to construct for me from the above algorithm. 5.1 Define the Matrix-Vector Product First, we need a function to compute $\Sigma \mathbf{v}$ efficiently. 5.2 Conjugate Gradient Solver Oh dang, the LLM did a really good job on this. def cg_solver(b, tol=1e-5, max_iter=100): x = torch.zeros_like(b) r = b.clone() p = r.clone() rs_old = torch.dot(r, r) for _ in range(max_iter): Ap = sigma_mv_prod(p) alpha = rs_old / torch.dot(p, Ap) x += alpha * p r -= alpha * Ap rs_new = torch.dot(r, r) if torch.sqrt(rs_new) < tol: p = r + (rs_new / rs_old) * p rs_old = rs_new return x 5.3 Langevin Dynamics Sampler def sample_mvn_langevin(mu, num_samples=1000, epsilon=1e-3, burn_in=100): Samples from N(mu, Σ) using Langevin dynamics. - mu: Mean vector (torch.Tensor of shape [D]) - num_samples: Number of samples to collect after burn-in - epsilon: Time step size - burn_in: Number of initial iterations to discard D = mu.shape[0] x = mu.clone().detach() samples = [] total_steps = num_samples + burn_in for n in range(total_steps): # Compute gradient: v = Σ^{-1} (x - μ) r = x - mu v = cg_solver(r, tol=1e-5, max_iter=100) # Langevin update noise = torch.randn(D) x = x - epsilon * v + torch.sqrt(torch.tensor(2 * epsilon)) * noise if n >= burn_in: return torch.stack(samples) 6 Validation After sampling, it’s wise to verify that the samples approximate the target distribution. import matplotlib.pyplot as plt empirical_mean = samples.mean(dim=0) empirical_cov = torch.from_numpy(np.cov(samples.numpy(), rowvar=False)) print("Empirical Mean:\n", empirical_mean) print("Empirical Covariance Matrix:\n", empirical_cov) # Plot histogram for the first dimension plt.hist(samples[:, 0].numpy(), bins=30, density=True) plt.title("Histogram of First Dimension")	{"url":"https://danmackinlay.name/notebook/gp_simulation_langevin.html","timestamp":"2024-11-04T23:31:57Z","content_type":"application/xhtml+xml","content_length":"53931","record_id":"<urn:uuid:55387a9c-cd48-4700-91d9-a2c422721d70>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00775.warc.gz"}
Infinitesimal (idea) Most mathematicians throw a conniption fit if you happen to mention the unfortunate term "infinitesimal". This actually happened to me in my senior year of high school, when the head of the math department was a substitute teacher in Calc class one day. The merest mention of the word "infinitesimal" and we were subjected to another month of limit theory. The inventors of calculus imagined differentials as infinitesimals, and used them in their work. Unforunately, the notion of a quantity that is infinitely small causes some pretty severe paradoxes in arithmetic. Because of the problems associated with infinitesimals, calculus was reformulated by 18th century mathematicians to be based upon the theory of limits. The father of modern set theory, Georg Cantor, called infinitesimals "the Cholera-bacillus of mathematics." This may have due to the fact that the existence of infinitesimals would render his Continuum Hypothesis false.	{"url":"https://everything2.com/user/Gorgonzola/writeups/Infinitesimal","timestamp":"2024-11-08T18:00:10Z","content_type":"text/html","content_length":"27065","record_id":"<urn:uuid:a9fede3c-f615-4554-9dcd-7d6f1f92c4d6>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00370.warc.gz"}
Math Problem Statement A tank with some water in it begins to drain. The function v ( t ) 40 − 3.5 t determines the volume of the water in the tank (in gallons) given a number of minutes t since the water began draining. What is the vertical intercept of v ? What does the v -coordinate of your answer to part (a) represent? Select all that apply. The weight of the tank when it is empty The number of gallons of water in the tank when it starts draining How many minutes it takes for all of the water to drain from the tank List all horizontal intercepts of v . What does the t -coordinate of your answer to part (c) represent? Select all that apply. How many minutes it takes for all of the water to drain from the tank The weight of the tank when it is empty The number of gallons of water in the tank when it starts draining Ask a new question for Free By Image Drop file here or Click Here to upload Math Problem Analysis Mathematical Concepts Linear Functions Linear equation: v(t) = 40 − 3.5t Intercepts of a linear function Suitable Grade Level Grades 7-9	{"url":"https://math.bot/q/linear-function-intercepts-vt-40-3-5t-FMOFQuTn","timestamp":"2024-11-06T01:50:45Z","content_type":"text/html","content_length":"87877","record_id":"<urn:uuid:2d30c229-0f03-4508-a470-2aaa7d2487b5>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00851.warc.gz"}
using set theory show that (A n S) n (ø A ) = ø, where ø is a null set ,s is universal set and A is finite set Hope this is useful to you Can u tell me Who are u I didn't remember u Itz Harsh	{"url":"https://alumniagri.in/task/using-set-theory-show-that-a-n-s-n-oe-a-oe-where-oe-is-a-42345481","timestamp":"2024-11-05T16:49:16Z","content_type":"text/html","content_length":"24789","record_id":"<urn:uuid:9f87a081-ffd4-4698-9055-c95d229699ee>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00204.warc.gz"}
Weighted Secret Sharing from Wiretap Channels Paper 2022/1578 Weighted Secret Sharing from Wiretap Channels Secret-sharing allows splitting a piece of secret information among a group of shareholders, so that it takes a large enough subset of them to recover it. In \emph{weighted} secret-sharing, each shareholder has an integer weight, and it takes a subset of large-enough weight to recover the secret. Schemes in the literature for weighted threshold secret sharing either have share sizes that grow linearly with the total weight, or ones that depend on huge public information (essentially a garbled circuit) of size (quasi)polynomial in the number of parties. To do better, we investigate a relaxation, $(\alpha, \beta)$-ramp weighted secret sharing, where subsets of weight $\beta W$ can recover the secret (with $W$ the total weight), but subsets of weight $\alpha W$ or less cannot learn anything about it. These can be constructed from standard secret-sharing schemes, but known constructions require long shares even for short secrets, achieving share sizes of $\max\big(W,\frac{\|\ mathrm{secret}\|}{\epsilon}\big)$, where $\epsilon=\beta-\alpha$. In this note we first observe that simple rounding let us replace the total weight $W$ by $N/\epsilon$, where $N$ is the number of parties. Combined with known constructions, this yields share sizes of $O\big(\max(N,\|\mathrm{secret}\|)/{\epsilon}\big)$. Our main contribution is a novel connection between weighted secret sharing and wiretap channels, that improves or even eliminates the dependence on~$N$, at a price of increased dependence on $1/\epsilon$. We observe that for certain additive-noise $(R,A)$ wiretap channels, any semantically secure scheme can be naturally transformed into an $(\alpha,\beta)$-ramp weighted secret-sharing, where $\alpha,\beta$ are essentially the respective capacities of the channels $A,R$. We present two instantiations of this type of construction, one using Binary Symmetric wiretap Channels, and the other using additive Gaussian Wiretap Channels. Depending on the parameters of the underlying wiretap channels, this gives rise to $(\alpha, \beta)$-ramp schemes with share sizes $\|\mathrm{secret}\|/\mathrm{poly}(\epsilon\log N)$ or even just $\|\mathrm{secret}\|/\mathrm{poly}(\ Available format(s) Publication info Contact author(s) fbenhamo102 @ gmail com shai halevi @ gmail com lstamble @ andrew cmu edu 2023-02-10: revised 2022-11-14: received Short URL author = {Fabrice Benhamouda and Shai Halevi and Lev Stambler}, title = {Weighted Secret Sharing from Wiretap Channels}, howpublished = {Cryptology {ePrint} Archive, Paper 2022/1578}, year = {2022}, url = {https://eprint.iacr.org/2022/1578}	{"url":"https://eprint.iacr.org/2022/1578","timestamp":"2024-11-06T20:32:59Z","content_type":"text/html","content_length":"17715","record_id":"<urn:uuid:f7565872-1366-4cc9-a541-80ba5e512532>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00834.warc.gz"}
Self-limiting behavior of scanning-electron-beam-induced local oxidation of hydrogen-passivated silicon surfaces The mechanism and the kinetics of electron-beam-induced local oxidation of an H-passivated Si surface in the electron energy range from 10 to 40 keV was investigated using scanning-electron-beam lithography. The volume expansion of Si upon oxidation produces a negative image surface pattern that can be imaged by atomic force microscopy. This latent pattern was used to study the dependence of the height and width of dot and line patterns as a function of the electron-beam exposure parameters. Patterns with minimum linewidth below 50 nm have been obtained. Similarly to atomic-force-microscope-induced local oxidation of Si, the height and linewidth saturate with electron dose for a given accelerating voltage. The saturation height roughly scales with the accelerating voltage, and depends more strongly on the accelerating voltage than the linewidth. The experimental results are interpreted by a mechanism that is based on charge generation and transport through the evolving insulating SiO[2] layer. Dive into the research topics of 'Self-limiting behavior of scanning-electron-beam-induced local oxidation of hydrogen-passivated silicon surfaces'. Together they form a unique fingerprint.	{"url":"https://impact.ornl.gov/en/publications/self-limiting-behavior-of-scanning-electron-beam-induced-local-ox","timestamp":"2024-11-05T00:21:26Z","content_type":"text/html","content_length":"49105","record_id":"<urn:uuid:f1c67c4d-c85a-443c-8e60-603a09749f9b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00859.warc.gz"}
Natural Homogeneous Coordinates Edward J Total Page：16 File Type：pdf, Size：1020Kb [email protected] Center for Computational Data Sciences, George Mason University, 3a. Not all lines pass through the same point. Fairfax, VA, USA 4a. Two distinct lines meet at one and only one DOI: 10.1002/wics.122 point (cf. 5 above). 678 2010 John Wiley & Sons, Inc. Volume 2, November/December 2010 WIREs Computational Statistics Natural homogeneous coordinates 5a. Two distinct points lie on one and only one Line parallel to plane line (cf. 4 above). Ideal point 6a. There is a one-to-one correspondence between the real numbers and all but one line through a point. (0,0) MODELS FOR A PROJECTIVE PLANE Projective plane An alternate way of imagining the projective plane is to visualize a hemisphere with its South Pole sitting FIGURE 1\| Representation of the projective plane by a hemisphere at the origin of a Euclidean plane. Any point on the which can be deformed into a crosscap. ordinary Euclidean plane can be represented on the hemisphere by connecting that point to the center of the hemisphere with a straight line. The point where the line meets the hemisphere is the mapping of the point in the Euclidean plane into the hemisphere. Lines from the center of the hemisphere through the equator represent ideal points because they are parallel to the Euclidean plane. As described in Figure 1, antipodal points on the equator of the hemisphere represent the same ideal point and, hence, these points are identiﬁed in a topological sense. One can imagine deforming the equator in such a way that antipodal points are joined. See Figure 2 to illustrate this deformation partially completed. In Figure 3, we represent the completion of this FIGURE 2\| Partially deformed hemisphere so that antipodal points deformation. This structure is called a crosscap and along the equator are approaching each other. represents a topological model of the projective plane. Figure 4 is a color and shaded rendering of a crosscap. Solving simultaneously, we have (C − C)z = 0. Since we know C − C = 0, ⇒ z = 0. Therefore, NATURAL HOMOGENEOUS (x, y, 0) represents an ideal point. COORDINATES The two-dimensional natural homogeneous coordinate system will be a triple (x, y, z). If z = 0, In ordinary Euclidean space, we have the Cartesian then we have an ideal point. Notice that if the origin, Coordinate system. We wish to develop an analog given by (0, 0) in ordinary Cartesian coordinates, is to Cartesian Coordinates for the projective plane. joined to a point (x, y), again in Cartesian coordinates, Consider the following equations: they determine a line and the ideal point (x, y,0)ison that line. The line has slope y/x. Hence, (x, y,0)isthe + + = Ax By C 0, ideal point corresponding to all lines with slope y/x. = Ax + By + C = 0. For ordinary points, we want z 1sothat the ordinary equation Ax + By + C = 0 holds. Thus These are the ordinary linear equations for two the Cartesian point (x, y) is represented in natural homogeneous coordinates as (x, y, 1). However, if straight lines which are parallel. If we try to solve + + = + + = these equations simultaneously we obtain no solution. Ax By C 0, then also Apx Bpy Cp 0so However in the projective plane we know that the that (px, py, p) is also a valid representation of solution is the ideal point. We rewrite these equations (x, y). We may always rescale so that if we have = x y (x, y, z), z 0, then this is equivalent to z , z ,1 or as x y the Cartesian point z , z . Although this multiple representation for Cartesian points at ﬁrst appears to Ax + By + Cz = 0, be a handicap, it is indeed a useful representation as Ax + By + C z = 0. we shall see in our later examples. Volume 2, November/December 2010 2010 John Wiley & Sons, Inc. 679 Advanced Review wires.wiley.com/compstats This deﬁnition can be extended in the obvious way for higher dimensional projective planes. An equation in xi, u1x1 + u2x2 + u3x3 = 0, is the equation of a line in the projective plane. Notice that the triple, (x1, x2, x3), represents a point. However, the values in the triple, [u1, u2, u3], are the coefﬁcients of the line and hence represent the line. The natural homogeneous coordinates mirror the projective duality between points and lines. Notice that if u1 = u2 = 0, then the equation is satisﬁed by ideal points. In other words, u3x3 = 0 is the equation of the ideal line, u3 = 0. Deﬁnition: The set of all real number triples [u1, u2, u3], ui not all 0, are the natural homogeneous line coordinates in the real projective plane. FIGURE 3\| The completely deformed hemisphere with antipodal An equation in ui, u1x1 + u2x2 + u3x3 = 0, is the points identiﬁed. In this rendition, 2D view of a 3D structure, the equation of a point in the real projective plane. surfaces penetrate each other. However, embedded in a higher dimensional space these surfaces do not intersect. APPLICATIONS OF NATURAL HOMOGENEOUS COORDINATES Computer Graphics Application I—Representing Translations in Matrix Form In computer graphics applications,3 we are interested in representing translations, rotations, and scalings. Consider the situation with translations. If p = (x, y)T is a point in Cartesian coordinates and it is translated by an amount tx in the x direction and an amount ty T in the y direction, then p = T(p) = (x + tx, y + ty) . T(p) is of course the sum of two vectors, but this oper- ation is not directly translatable into matrix notation. For rotations, consider a point p = (x, y)T which is given in polar coordinates by (r, γ )T. Thus x = r cos(γ )andy = r sin(γ ). If the point is rotated through an angle θ,thenx = r cos(γ + θ)and y = r sin(γ + θ); see Figure 5. But FIGURE 4\| Crosscap rendered as a color shaded ﬁgure. (Reprinted with permission from Professor Paul Bourke, University of Western cos(γ + θ) = cos(γ )cos(θ) − sin(γ )sin(θ), Australia.	{"url":"https://docslib.org/doc/527273/natural-homogeneous-coordinates-edward-j","timestamp":"2024-11-02T08:14:32Z","content_type":"text/html","content_length":"62596","record_id":"<urn:uuid:987f382f-19ff-4824-85a2-ffb09ca0c6de>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00781.warc.gz"}
Finding the Greatest Common Factor \| sofatutor.com Finding the Greatest Common Factor Finding the Greatest Common Factor Basics on the topic Finding the Greatest Common Factor Finding the Greatest Common Factor – Introduction In mathematics, the concept of finding the greatest common factor (GCF) is incredibly useful for solving a variety of problems. Whether simplifying fractions, solving equations, or analyzing ratios, understanding how to determine the GCF can simplify computations and lead to more elegant solutions. Let's dive into what the greatest common factor is and how to find it. Understanding the Greatest Common Factor – Definition The greatest common factor (GCF), sometimes known as the greatest common divisor (GCD) or highest common factor (HCF), is the largest whole number that is a divisor of two or more given numbers. The GCF is instrumental in simplifying fractions, solving problems with ratios, and even in everyday situations like dividing a set of items into equal groups without leftovers. How to Find the GCF – Method There are several methods to find the GCF, including prime factorization, Euclidean algorithm, and inspection. The most intuitive of these, especially for smaller numbers, is via prime factorization. Steps to Find the GCF: 1. Decompose each number into prime factors. 2. Identify the common prime factors. 3. Multiply these common prime factors together to get the GCF. For instance, to find the GCF of 18 and 24: • Prime factors of 18: 2 × 3 × 3 • Prime factors of 24: 2 × 2 × 2 × 3 • Common factors are 2 and 3 • Thus, GCF(18, 24) = 2 × 3 = 6 Check your understanding so far. Finding the Greatest Common Factor – Guided Practice Suppose we want to find the GCF of 45 and 75. Follow the steps of prime factorization: 1. List the prime factors for both numbers. 2. Circle the common prime factors. 3. Multiply these to calculate the GCF. Let’s find the GCF of 45 and 75. Try practicing finding the GCF on your own! Finding the Greatest Common Factor – Summary Key Learnings from this Text: • The greatest common factor is the highest number that divides two or more numbers without leaving a remainder. • Prime factorization is a reliable method to find the GCF. • The GCF is useful in simplifying fractions, distributing goods evenly, and more. Explore other content on our website platform for interactive practice problems, videos, and printable worksheets that support your educational journey in understanding factors and multiples. Finding the Greatest Common Factor – Frequently Asked Questions Transcript Finding the Greatest Common Factor Luis and June seem to be struggling to find some common factors between them. Speaking of common factors, let's learn about finding the greatest common factor. The Greatest Common Factor, or GCF, is the largest number that divides equally into two or more numbers. We often use the GCF to solve problems involving equal sharing, like dividing a cake or distributing supplies between your friends. There are two methods of finding the GCF; factor pairs and factor trees. First, let's explore factor pairs. This method lists out factor pairs and is most useful for smaller numbers. For this strategy, we will use twelve and fifteen. Factor pairs for twelve are one and twelve, two and six, and three and four. Factor pairs for fifteen are one and fifteen, and three and five. Now find the largest factor that both numbers have in common. What is the highest factor twelve and fifteen have in common? Three, so the GCF of twelve and fifteen is three. Now let's explore the second method, factor trees which uses prime factorization. This method is most useful when dealing with larger numbers. First, start with the numbers at the top; here we have twenty-four and thirty-six. Now find the prime factorization for each number. For twenty-four, we have two and twelve, two and six, and two and three. For thirty-six, we have two and eighteen, two and nine, and three and three. Next, identify the shared prime factors. Both twenty-four and thirty-six share a two, another two, and a three. Finally, multiply these together to find the GCF. We need to solve two times two times three. Two times two equals four, and four times three equals twelve. The GCF of twenty-four and thirty-six is twelve. Now it's your turn! Find the GCF of thirty-six and fifty-four. Pause the video to work on the problem, and press play when you are ready for the solution. The factor tree of thirty-six is four and nine, two and two, and three and three. The factor tree of fifty-four is six and nine, two and three, and three and three. The shared factors are a two, a three, and another three. Multiplied together you get a GCF of eighteen. Let's summarize. There are two useful methods for finding the greatest common factor, or GCF. The first method is factor pairs. This is useful for finding the GCF with smaller numbers. The second method is factor trees. This is useful for finding the GCF with larger numbers. Finding the GCF can help in making fair and efficient decisions when it comes to sharing. Ah, it looks like Luis and June have finally found their greatest common factor, their love for math and factor trees! Finding the Greatest Common Factor exercise Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Finding the Greatest Common Factor. • What is GCF? When finding the GCF, we are looking for something that two or more numbers share or have in common. The GCF of 12 and 15 is 3. This is the biggest number that fits equally into 12 and 15. GCF stands for Greatest Common Factor. It is the largest number that divides equally into two or more numbers. • Find the GCF of 12 and 24. All of the factors of 12 are also factors of 24, which is the greatest? Greatest means the largest common factor. The Greatest Common Factor (GCF) of 12 and 24 is 12. This is the largest number that equally divides into both numbers. • Find the GCF of 54 and 72. Fill in the gaps in each factor tree first. How many times does 27 go into 54? This is a factor pair. Once you have completed each factor tree, look for the shared factors. You should find the shared factors of 2, 3 and 3. Multiply these together to find the GCF. The greatest common factor of 54 and 72 is 18. When we complete the factor trees for 54 and 72, we can find the shared factors of 2, 3 and 3. When we multiply 2 $\times$ 3 $\times$ 3 we get 18. • Can you find the GCF of these pairs of numbers? Draw out factor pairs for smaller numbers or factor trees for larger numbers. For 15 and 30, draw factor pairs. Here is one for 15, can you draw one for 30 and then find the GCF? For 126 and 154, factor trees would be helpful. Once you have drawn your factor trees, look for all of the shared factors and multiply them to find the GCF. □ The GCF of 15 and 30 is 15. □ The GCF of 84 and 144 is 12. □ The GCF of 126 and 154 is 14. □ The GCF of 21 and 35 is 7. • Find the GCF of 12 and 18. First look for common factors. Which numbers are factors of both 12 and 18? Out of these common factors, which one is the greatest or largest number? The correct answer is 6. The common factors are: 1, 2, 3 and 6. Out of these common factors, 6 is the greatest. • Can you solve the problem? You need to find the greatest common factor of 72, 96 and 104. Draw a factor tree for each of these numbers and look for the shared factors. You should find the shared factors of 2, 2 and 2. Multiply these together to find the answer. The greatest number of bunches is 8. The greatest common factor of 72, 96 and 108 is 8, therefore, this is the greatest number of bunches Cara can make with the same number of of each colour of balloon.	{"url":"https://us.sofatutor.com/math/videos/finding-the-greatest-common-factor","timestamp":"2024-11-05T10:12:28Z","content_type":"text/html","content_length":"187121","record_id":"<urn:uuid:cd47edb7-b484-429b-99a2-a7d4e81d44c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00114.warc.gz"}
Variation of singular Kähler-Einstein metrics: Kodaira dimension zero We study several questions involving relative Ricci-flat Kähler metrics for families of log Calabi-Yau manifolds. Our main result states that if p : W (X; B) → Y is a Kähler fiber space such that (X [y] B/[Xy]) is generically klt, K[X=Y] + B is relatively trivial and p[* ](m(K[X=Y] + B)) is Hermitian flat for some suitable integer m, then p is locally trivial. Motivated by questions in birational geometry, we investigate the regularity of the relative singular Ricci-flat Kähler metric corresponding to a family p : (X; B) → Y of klt pairs (X[y]; B[y]) such that k(K[Xy] + B[y]) = 0. Finally, we disprove a folkore conjecture by exhibiting a one-dimensional family of elliptic curves whose relative (Ricci-)flat metric is not semipositive. • Kähler fiber space • conic Kähler metrics • direct image of log pluricanonical bundles • log Calabi-Yau manifolds ASJC Scopus subject areas • General Mathematics • Applied Mathematics Dive into the research topics of 'Variation of singular Kähler-Einstein metrics: Kodaira dimension zero'. Together they form a unique fingerprint.	{"url":"https://nyuscholars.nyu.edu/en/publications/variation-of-singular-k%C3%A4hler-einstein-metrics-kodaira-dimension-z","timestamp":"2024-11-11T16:37:26Z","content_type":"text/html","content_length":"51869","record_id":"<urn:uuid:c977771c-0681-4382-823a-a74e161015c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00131.warc.gz"}
Convex Maximization via Adjustable Robust Optimization Maximizing a convex function over convex constraints is an NP-hard problem in general. We prove that such a problem can be reformulated as an adjustable robust optimization (ARO) problem where each adjustable variable corresponds to a unique constraint of the original problem. We use ARO techniques to obtain approximate solutions to the convex maximization problem. In order to demonstrate the complete approximation scheme, we distinguish the case where we have just one nonlinear constraint and the case where we have multiple linear constraints. Concerning the first case, we give three examples where one can analytically eliminate the adjustable variable and approximately solve the resulting static robust optimization problem efficiently. More specifically, we show that the norm constrained log-sum-exp (geometric) maximization problem can be approximated by (convex) exponential cone optimization techniques. Concerning the second case of multiple linear constraints, the equivalent ARO problem can be represented as an adjustable robust linear optimization (ARLO) problem. Using linear decision rules then returns a safe approximation of the constraints. The resulting problem is a convex optimization problem, and solving this problem gives an upper bound on the global optimum value of the original problem. By using the optimal linear decision rule, we obtain a lower bound solution as well. We derive the approximation problems explicitly for quadratic maximization, geometric maximization, and sum-of-max-linear-terms maximization problems with multiple linear constraints. Numerical experiments show that, contrary to the state-of-the-art solvers, we can approximate large-scale problems swiftly with tight bounds. In several cases, we have equal upper and lower bounds, which concludes that we have global optimality guarantees in these cases. Selvi A, Ben-Tal A, Brekelmans R, Den Hertog D (July 2020) Convex maximization via adjustable robust optimization. Corresponding author: a.selvi19@imperial.ac.uk	{"url":"https://optimization-online.org/2020/07/7881/","timestamp":"2024-11-04T21:03:24Z","content_type":"text/html","content_length":"85882","record_id":"<urn:uuid:28f990a4-c3f0-4b63-89c8-943c977ff4bc>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00858.warc.gz"}
郝成春研究员(Prof. Chengchun Hao) - 2019S: 011D9102Z* 调和分析I, II，数学所研究生核心基础课 Course No.： Course Hours：80 Course Points：4 Course Title：Harmonic Analysis I, II (调和分析I, II，数学所研究生核心基础课) Time： Monday, 13:30-15:10; Wednesday, 13:30-15:10 & 15:20-16:10 （2019 Spring, Feb.25-Jun.19） Place：N401, Teaching Building, Zhongguancun Campus Textbook: [Book I] Loukas Grafakos，Classical Fourier Analysis，GTM 249, 3rd Edition （2014）Springer. [Book II] Loukas Grafakos，Modern Fourier Analysis，GTM 250, 3rd Edition （2014）Springer. Grading: Grading for this course has two components: homework sets, whose average counts 40% for the final grade, and a final exam that counts 60%. The homework sets will be posted in this course's website in SEP system every two weeks. The students are supposed to solve the problems and write down the solutions individually and turn them in, by the indicated deadline, online in SEP system. Late homework assignments will not be accepted . You will be evaluated both on the mathematical rigor of your solutions as well as clarity of exposition, so please pay attention to details when preparing your answers. Of course, discussions about the problems, among the students, are a healthy and recommended practice, but each student should write down the solutions in his own words, showing a clear grasp of the material being used. Besides, the deep understanding of the subject is crucial as a form of continuous buildup of knowledge for the final exam. Final exam would be held on Jun.19. Book I: Classical Fourier Analysis 1 $L^p$ Spaces and Interpolation 1.1 $L^p$ and Weak $L^p$ 1.2 Convolution and Approximate Identities 1.3 Interpolation 2 Maximal Functions, Fourier Transform, and Distributions 2.1 Maximal Functions 2.2 The Schwartz Class and the Fourier Transform 2.3 The Class of Tempered Distributions 2.4 More about Distributions and the Fourier Transform 5 Singular Integrals of Convolution Type 5.1 The Hilbert Transform and the Riesz Transforms 5.2 Homogeneous Singular Integrals and the Method of Rotations 5.3 The Calderon–Zygmund Decomposition and Singular Integrals 5.5 Vector-Valued Inequalities 5.6 Vector-Valued Singular Integrals 2.5.5 The space of Fourier Multipliers $M_p(\Bbb{R}^n)$ 6 Littlewood-Paley Theory and Multipliers 6.1 Littlewood-Paley Theory 6.2 Two Multiplier Theorems Book II: Modern Fourier Analysis 1 Smoothness and Function Spaces 1.1 Smooth Functions and Tempered Distributions 1.2 Laplacian, Riesz Potentials, and Bessel Potentials 1.3 Sobolev Spaces 2.1 Hardy Spaces 2.1.1 Definition of Hardy Spaces 2.1.2 Quasi-norm Equivalence of Several Maximal Functions 3 BMO Spaces 3.1 Functions of Bounded Mean Oscillation 3.2 Duality between $H^1$ and BMO Typos corrections：http://faculty.missouri.edu/~grafakosl/FourierAnalysis.html Some other corrections for Book I: P. 34, in the 3rd line, $L^{p_j}(X_j)$ should be $L^{p_j}(X)$. P. 115, in the 5th line, $\|\xi_{j_0}\|>\|\xi\|/\sqrt{n}$ should be $\|\xi_{j_0}\| \geqslant \|\xi\|/\sqrt{n}$, since we can not exclude strictly the case of $=$, e.g., the cube. P. 128, in the 5th line from below, "$M>2\|\alpha\|$" should be "$M>2\max(\|\alpha\|,n)$", since it is necessray to prove the convergence of the integral over the complement of the cube. P. 129, in 2nd line, it is enough to replace "$(1+\|x-y\|)^M$" by "$(1+\|x-y\|)^{M/2}$". P. 130, in 8th line, the "$+$" symbol between two integrals should be "$-$". P. 319, 6th line from below, I think that "Theorem 1.4.19" might be replaced by "Theorem 1.3.2 with $p_0=1$ and $p_1=2$, and Remark 1.3.3 since $H$ is linear". This is only a suggestion, since Thm 1.4.19 was not taught as the suggestion in preface (1.1,1.2,1.3,2.1,...) which is in Section 1.4. P. 321, 6th line from below, "$\\|H(f)\\|_{L^{2p}}<\infty$" should be "$\\|f\\|_{L^{2p}}<\infty$". By the way, in the inequality just above this line, it is maybe better and more readable to take square for each side. P. 322, 8th line, "$0<x<\pi/2$" should include $\pi/2$, i.e., "$0<x\leqslant\pi/2$", since the case of $x=\pi/2$ is used later. P. 327, in the 2nd line, it is better to omit "$\|\xi\|$" in the denominator because it has been assumed to be $1$. P. 344, in the 10th line, $\frac{dr}{r}$ should be $dr$. P. 346, in the 10th line, $\Omega\in L^1$ should be $\Omega_j\in L^1$. P. 347, Theorem 5.2.11 should be added the condition "$n\geqslant 2$" because some estimates are not valid for $n=1$ in the proof. P. 349, in the 6th line from below, $\Omega()$ should be its absolute value $\|\Omega()\|$. P. 350, in (5.2.39), $\max(p,(p-1)^{-1})$ should be $\max(p^2,(p-1)^{-2})$. P. 352, in the middle long inequalities, $F_j(z)$ should be $G_j(z)$ or $F_j(z/\varepsilon)$; similarly, in next line $F_j(r\theta)$ should be $G_j(r\theta)$ or $F_j(r\theta/\varepsilon)$. P. 352, in (5.2.45), $\max(p,...)$ should be $\max(p^2,...)$. P. 352, Corollary 5.2.12 should be added the condition "$n\geqslant 2$".	{"url":"http://www.math.ac.cn/kyry/hcc/teach/201812/t20181222_470499.html","timestamp":"2024-11-09T00:00:40Z","content_type":"application/xhtml+xml","content_length":"18832","record_id":"<urn:uuid:3d941131-5d9e-4dbd-b65e-e0ef2171a2d0>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00377.warc.gz"}
fixadd: Safe function to add fixed point numbers clamping overflow. Allegro game programming library. - Linux Manuals (3) fixadd (3) - Linux Manuals fixadd: Safe function to add fixed point numbers clamping overflow. Allegro game programming library. fixadd - Safe function to add fixed point numbers clamping overflow. Allegro game programming library. #include <allegro.h> fixed fixadd(fixed x, fixed y); Although fixed point numbers can be added with the normal '+' integer operator, that doesn't provide any protection against overflow. If overflow is a problem, you should use this function instead. It is slower than using integer operators, but if an overflow occurs it will set `errno' and clamp the result, rather than just letting it wrap. Example: fixed result; /* This will put 5035 into `result'. / result = fixadd(itofix(5000), itofix(35)); / Sets `errno' and puts -32768 into `result'. / result = fixadd(itofix(-31000), itofix(-3000)); ASSERT(!errno); / This will fail. */ Returns the clamped result of adding `x' to `y', setting `errno' to ERANGE if there was an overflow.	{"url":"https://www.systutorials.com/docs/linux/man/3-fixadd/","timestamp":"2024-11-04T00:58:11Z","content_type":"text/html","content_length":"8471","record_id":"<urn:uuid:76c85af7-5fa6-4bb8-b427-b4883e0d10da>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00458.warc.gz"}
Conservative vector fields II \| JustToThePointConservative vector fields II Enjoy the little things, for one day you may look back and realize they were the big things, Robert Brault. A vector field is an assignment of a vector $\vec{F}$ to each point (x, y) in a space, i.e., $\vec{F} = M\vec{i}+N\vec{j}$ where M and N are functions of x and y. A vector field on a plane can be visualized as a collection of arrows, each attached to a point on the plane. These arrows represent vectors with specific magnitudes and directions. Work is defined as the energy transferred when a force acts on an object and displaces it along a path. In the context of vector fields, we calculate the work done by a force field along a curve or trajectory C using a line integral. The work done by a force field $\vec{F}$ along a curve C is: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy = \int_{C} \vec{F}·\hat{\mathbf{T}}ds$, where $\ hat{\mathbf{T}}$ is the unit tangent vector. Path Independence and Conservative Vector Fields Theorem. Path Independence and Conservative Vector Fields. Let $\vec{F}$ be a continuous vector field defined on an open, connected region D. If the line integral $\int_{C} \vec{F}\vec{dr}$ is independent of the path in D, then $\vec{F}$ is conservative. This means there exists a function f such that ∇f = $\vec{F}$, where ∇f represents the gradient of f. Suppose the line integral $\int_{C} \vec{F}\vec{dr}$ is independent of the path in D. This means that for any two arbitrary points A and B in D, the work done by $\vec{F}$ from A to B is the same no matter which path you take. $\vec{F}$ must be the gradient of some function, ∇f = $\vec{F}$. Let (a, b) be a fixed point in D. We define the function f(x, y) for any point (x, y) in D as the value of the line integral from (a, b) to (x, y): f(x, y) = $\int_{(a, b)}^{(x, y)} \vec{F}\vec{dr}$. Claim: This function f(x, y) will be our potential function, and our goal is to show that the gradient of f is $\vec{F}$, i.e., ∇f = $\vec{F}$. Consider a path from (a, b) to (x, y). We can break down this path into two parts: • C[1]: From (a, b) to (c, y). • C[2]: From (c, y) to (x, y) (Refer to Figure F for a visual representation and aid in understanding it) and recall that we can do so because we are working within an open, connected region. The function f(x, y) can then be written as: f(x, y) = $\int_{(a, b)}^{(c, y)} \vec{F}\vec{dr} + \int_{(c, y)}^{(x, y)} \vec{F}\vec{dr}$. This means that f(x, y) is the sum of the work done by $\vec {F}$ along each sub-path. Compute the partial derivative $\frac{∂f}{∂x}$ To find $f_x = \frac{∂f}{∂x}$, we focus on the path C[2] because C[1] does not involve x, so its derivative with respect to x is zero. On the path C[2]: $\frac{∂f}{∂x} = \frac{∂}{∂x}\int_{(c, y)}^{(x, y)} \vec{F}\vec{dr} =[\vec{F} = ⟨P, Q⟩] \frac{∂}{∂x}\int_{(c, y)}^{(x, y)} Pdx + Qdy$ = [On the path C[2], y is constant, hence dy = 0, and the line integral reduces to] $\frac{∂}{∂x}\int_{(c, y)}^{(x, y)} Pdx = \frac{∂}{∂x}[\hat{\mathbf{P}}(x, y)\bigg\|_{(c, y)}^{(x, y)}] = $ where $\hat{\mathbf{P}}$ is the anti-derivative of P with respect to x = $\frac{∂}{∂x}[\hat{\mathbf{P}}(x, y)-\hat{\mathbf{P}}(c, y)] =$[The derivative of an anti-derivate will give us back P, and the second term is x-independent (constant), so the derivative is zero] $P(x, y) +0$ Therefore, $f_x = \frac{∂f}{∂x} = P$. Our claim ∇f = ⟨f[x], f[y]⟩ = $\vec{F}$ = ⟨P, Q⟩. On the second hand of the proof everything is quite similar. The difference is that we are going to consider another path C composed of two parts, namely C[1] from (a, b) to (x, d) and C[2] from (x, d) to (x, y) (Refer to Figure G for a visual representation and aid in understanding it) and recall that we can do so because we are working within an open, connected region. Using a similar argument f[y] = Q ∎ Since f[x] = P (x, y) and f[y] = Q(x,y), we have ∇f= ⟨f[x], f[y]⟩ = ⟨P, Q⟩ = $\vec{F}$. Therefore, $\vec{F}$ is the gradient of f, meaning that $\vec{F}$ is conservative, which completes the proof. Testing whether $\vec{F}$ is a gradient field To determine if a vector field $\vec{F} = ⟨M, N⟩$ is a gradient field, we need to check whether there exists a scalar potential function f such that $\vec{F} = ∇f$. This means that the components of $\vec{F}$ can be expressed as the partial derivatives of f: $M = \frac{∂f}{∂x}$ and N = $\frac{∂f}{∂y}$. For $\vec{F}$ to be a gradient field, f must satisfy the condition that the mixed partial derivatives are equal, i.e., $\frac{∂^2f}{∂x∂y}=\frac{∂^2f}{∂y∂x}$. The condition that the mixed partial derivatives must be equal arises from a fundamental result in calculus known as Clairaut’s theorem (or Schwarz’s theorem). It states that if a function f(x,y) has continuous second-order partial derivatives, then the order in which you take the derivatives does not matter. This leads to the following criterion to check that $\vec{F}$ is conservative: $\frac{∂M}{∂y}=\frac{∂N}{∂x}$ Criterion for a Conservative Vector Field The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, and if the functions M and N have continuous first partial derivatives across the entire domain D, and the domain D is open and simply connected, then $\vec{F}$ is a gradient field, meaning it is conservative. Breaking Down the Criteria: 1. Equality of Mixed Partial Derivatives. The condition M[y] = N[x] ↭ $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, ensures that the curl of the vector field $\vec{F}$ is zero. This is a necessary condition for $\vec{F}$ to be a gradient field, which means that can be expressed as the gradient of some potential function f(x, y). 2. Continuous First Partial Derivatives. The functions M and N must have continuous first partial derivatives across the domain D. This ensures that the vector field $\vec{F}$ behaves smoothly, without any abrupt changes in direction or magnitude. 3. Open and Simply Connected Domain. The domain D must be open (meaning it does not include its boundary points) and simply connected (meaning it has no holes). This condition prevents the presence of obstacles or gaps in the domain that could disrupt the path independence of the line integral. In a simply connected domain, any closed loop can be continuously shrunk to a point without leaving the domain, ensuring that the vector field $\vec{F}$ is conservative. Conversely, if $\vec{F}$ = ⟨M, N⟩ is conservative, then it must satisfy M[y] = N[x] ↭ $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, assuming that $\vec{F}$ is defined and differentiable everywhere. • Example 0: Conservative Vector Field. Consider the vector field $\vec{F} = (x^2+y)\vec{i}+(y^2+x)\vec{j}$. We want to determine if this vector field is conservative. If it is, we will find its potential function. Check if the Vector Field is Conservative. It is conservative because: 1. Equality of Mixed Partial Derivatives: $\frac{∂M}{∂y} = \frac{∂}{∂y}(x^2+y) = 1 = \frac{∂N}{∂x} = \frac{∂}{∂x}(y^2+x)$. 2. Continuous First Partial Derivatives. The functions M(x, y) and N(x, y) are both polynomials, which means their partial derivatives are continuous everywhere in ℝ^2. 3. Open and Simply Connected Domain. The domain D = ℝ^2 is the entire plane, which is open (it does not include its boundary) and simply connected (it has no holes). This criterion is also Since all three conditions are met, the vector field $\vec{F}$ is indeed conservative. In the context of determining whether a vector field is conservative, conditions (2) (continuous first partial derivatives) and (3) (the domain being open and simply connected) are indeed fundamental. While it’s true that these conditions are often not explicitly mentioned in literature or problem-solving scenarios, they are nonetheless essential. Find its potential function To find the potential function f(x, y) such that $\vec{F} = ∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨M(x, y), N(x, y)⟩$, we can use the following methods: We start by integrating M(x,y) with respect to x: f[x] = x^2+y ⇒$f = \int (x^2+y)dx = \frac{1}{3}x^3+xy+g(y)$. Here, g(y) is a function of y alone. Next, we integrate M(x,y) with respect to y: f[y] = y^2+x ⇒$f = \int (y^2+x)dy = \frac{1}{3}y^3+xy+h(x)$. Here, h(x) is a function of x alone. Combining both results, $h(x) = \frac{1}{3}x^3, g(y) = \frac{1}{3}y^3, f(x, y) = \frac{1}{3}x^3+xy+\frac{1}{3}y^3.$ We start by integrating M(x,y) with respect to x: f[x] = x^2+y ⇒$f = \int (x^2+y)dx = \frac{1}{3}x^3+xy+g(y)$. Here, g(y) is a function of y alone. Next, we differentiate this result with respect to y to find g(y): f[y] = $\frac{∂}{∂y}(\frac{1}{3}x^3+xy+g(y))$ = x + g’(y). Since f[y] = $\frac{∂f}{∂y}$ must equal N(x, y) = y^2+x, we have: x + g’(y) = y^2 + x⇒g’(y) = y^2. Integrating g’(y) with respect to y: g(y) = $\int y^2dy = \frac{y^3}{3}+C$. Thus, the potential function f(x, y) is: $f(x, y) = \frac{1}{3}x^3 + xy + \frac{1}{3}y^3.$ • Example 1: Non-Conservative Vector Field. Consider the vector field $\vec{F} = -y\vec{i}+x\vec{j}$ and let C be the unit circle. We want to check if this vector field is a gradient field. Here, the components M = -y, N = x. We compute the partial derivatives: $\frac{∂M}{∂y} = -1 ≠ 1 = \frac{∂N}{∂x}$, so $\vec{F}$ does not satisfy the condition for being a gradient field. Therefore, the vector field is not conservative. • Example 2: Another Non-Conservative Vector Field. Consider the vector field $\vec{F} = 3xy\vec{i}-x^2\vec{j}$. We will check if this vector field is a gradient field. Here, the components are M = 3xy, N = -x^2. We compute the partial derivatives: $\frac{∂M}{∂y} = 3x ≠ -2x = \frac{∂N}{∂x}$, so indeed $\vec{F}$ is not a gradient field, and thus is not conservative. • Exercise: Finding the Value of a that Makes $\vec{F}$ a gradient field. Consider the vector field $\vec{F}=(4x^2+axy)\vec{i}+(3y^2+4x^2)\vec{j}$. We want to find the value of a that makes what value of a makes $\vec{F}$ a gradient field. Here, the components are M = 4x^2+axy and N = 3y^2+4x^2. We compute the partial derivatives: $M_y = \frac{∂M}{∂y} = \frac{4x^2+axy}{∂y} = ax, N_x = \frac{∂N}{∂x} = \frac{3y^2+4x^2}{∂x} = 8x$. For $\vec{F}$ to be a gradient field, we must have $\frac{∂M}{∂y} = \frac{∂N}{∂x}$. Therefore, we require ax =8x ↭[Solving for a and assuming a ≠ 0, we get] a = 8. Thus, when a = 8, the vector field $\vec{F}=(4x^2+8xy)\vec{i}+(3y^2+4x^2)\vec{j}$ is a gradient field. • Given the vector field $\vec{F} = ⟨2xy^3z^4 -P-, 3x^2y^2z^4 -Q-, 4x^2y^3z^3 -R-⟩$, determine if $\vec{F}$ is conservative. If it is, find the potential function and evaluate the line integral $\ int_{C} \vec{F}\vec{dr}$ where C is parameterized by $d\vec{r}(t) = ⟨t, t^2, t^3⟩$ with 0 ≤ t ≤ 2. Check if the Vector Field is Conservative. It is conservative because: 1. Equality of Mixed Partial Derivatives: $\frac{∂P}{∂y} = 6xy^2z^4 = \frac{∂Q}{∂x}, \frac{∂P}{∂z} = 8xy^3z^3 = \frac{∂R}{∂x}, \frac{∂Q}{∂z} = 12x^2y^2z^3 = \frac{∂R}{∂y}$. 2. Continuous First Partial Derivatives. The functions P, Q and R are all polynomials in x, y, and z, which means their partial derivatives are continuous everywhere in ℝ^3. 3. Open and Simply Connected Domain. The domain D = ℝ^3 which is both open (it does not include its boundary) and simply connected (it has no holes). This criterion is also satisfied. Since all three conditions are met, the vector field $\vec{F}$ is indeed conservative. Find its potential function Since $\vec{F}$ is conservative, there exists a potential function f(x, y, z) such that ∇f = $\vec{F}$. This means, $\frac{∂f}{∂x} = P = 2xy^3z^4, \frac{∂f}{∂y} = Q = 3x^2y^2z^4, \frac{∂f}{∂z} = R = We will find f(x, y, z) by integrating these expressions f[x] = 2xy^3z^4. Let’s integrate with respect to x, f = $\int 2xy^3z^4dx = x^2y^3z^4+g(y, z).$ Here, g(y, z) is an arbitrary function of y and z. f[y] = 3x^2y^2z^4. Let’s integrate with respect to y, f = $\int 3x^2y^2z^4dy = x^2y^3z^4 + h(x, z).$ f[z] = 4x^2y^3z^3. Let’s integrate with respect to z, f = $\int 4x^2y^3z^3dz = x^2y^3z^4 + m(x, y).$ Combining all these results: $g(y, z) = h(x, z) = m(x, y) = 0, f(x, y, z) = x^2y^3z^4.$ Evaluate the Line Integral. For a conservative vector field, the line integral $\int_{C} \vec{F}\vec{dr}$ depends only on the endpoints of the path C. Specifically, $\int_{C} \vec{F}\vec{dr} = f(\ text{endpoint})-f(\text{starting point})$ The curve C is parameterized by $d\vec{r}(t) = ⟨t, t^2, t^3⟩$ with 0 ≤ t ≤ 2. Therefore, the endpoints at t = 0, $d\vec{r}(0) = ⟨0, 0, 0⟩$, at t = 2, $d\vec{r}(2) = ⟨2, 4, 8⟩$. $\int_{C} \vec{F}\vec{dr} = f(\text{endpoint})-f(\text{starting point}) = f(2, 4, 8) -f(0, 0, 0) = 2^2·4^3·8^4-0 = 4⋅64⋅4096 = 1,048,576.$ Curl, Torque, and Conservative Vector Fields • A vector field $\vec{F} = ⟨M, N⟩ = ∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$ is called a gradient field if it can be expressed as the gradient of a scalar potential function f(x, y). • A vector field $\vec{F}$ is conservative if the line integral of $\vec{F}$ around any closed curvve C is zero: $\oint_C \vec{F} \cdot d\vec{r} = 0$. This is true if $\vec{F}$ is defined over the entire plane or in a simply connected region (a region with no holes). Definition of Curl. The curl of a vector field $\vec{F}$ in two dimensions is defined as: $curl(\vec{F}) = N_x - M_y$. This scalar quantity measures the tendency of the vector field to induce rotation or swirling around a point. Test for Conservativeness. A vector field $\vec{F}$ is conservative if its curl is zero everywhere in the region: $curl(\vec{F}) = N_x - M_y$ = 0.. When the curl is zero, the vector field has no rotational component, meaning the field is path-independent, and the work done around any closed curve is zero. A physical interpretation of the curl is as follows. Suppose that $\vec{F}$ represents a velocity field in fluid dynamics or physics. The curl of the vector field $curl(\vec{F})$ at a point measures the rotational or swirling motion of the fluid at that point. In essence, curl tells us how much and in what direction the fluid rotates around a point. 1. Vector Field: $\vec{F} = ⟨a, b⟩$ where a and b are constants, $curl(\vec{F}) = N_x - M_y = \frac{∂b}{∂x}-\frac{∂a}{∂y} = 0 -0 = 0.$. Interpretation: The curl is zero, indicating no rotation in the field. 2. Vector Field: $\vec{F} = ⟨x, y⟩, curl(\vec{F}) = N_x - M_y = \frac{∂}{∂x}(y) - \frac{∂}{∂y}(x) = 0 -0 = 0.$ Interpretation: Again, the curl is zero, meaning there is no rotational component to this field. 3. $\vec{F} = ⟨-y, x⟩, curl(\vec{F}) = N_x - M_y = \frac{∂}{∂x}(x) - \frac{∂}{∂y}(-y) = 1 + 1 = 2$. Interpretation: The curl is 2, which indicates that the field has a rotational component. The positive curl suggests a counterclockwise rotation. The magnitude of the curl corresponds to the angular velocity of this rotational motion (Figures 1, 2, and 3 respectively). The curl of a force field is related to the torque exerted on a object within that field. Torque is a measure of the rotational force applied to an object.The relationship between torque and angular velocity (rate of rotation) is given by: $\frac{torque}{\text{moment of inertia}} = \frac{d}{dt}(\text{angular velocity})$. In a vector field, if the curl is non-zero, it implies that there is a rotational force acting on objects within the field, leading to a non-zero torque. This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007]. 1. NPTEL-NOC IITM, Introduction to Galois Theory. 2. Algebra, Second Edition, by Michael Artin. 3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson). 4. Field and Galois Theory, by Patrick Morandi. Springer. 5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels. 6. Contemporary Abstract Algebra, Joseph, A. Gallian. 7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube. 8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.	{"url":"https://justtothepoint.com/calculus/conservativevectorfields2/","timestamp":"2024-11-14T05:32:26Z","content_type":"text/html","content_length":"32427","record_id":"<urn:uuid:f0292c35-b787-4a56-9218-3ef72dd42fdb>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00769.warc.gz"}
A little on bell curve - The Culture SGA little on bell curve Since many students were asking, I thought I’ll roughly explain the bell curve and its distribution properties. Before that, I thought I should mention that I’m not really sure myself if Private Candidates and School Candidates share a different bell curve. I doubt so, since all take the same subject code, there should be no differentiation as there will be bias. And I do not really see a particular need for them if all the students take the same paper. There is no differentiation between taking the paper again and first time too, it doesn’t mean that you take it twice, you have a higher chance/ lower chance. They should be competing in all fairness still. A bit of intuition of how this bell curve thing works. We can’t have too many A’s. If ALL the JC students were to score zero mark and no student deviate (cheat), then all of you will get A. Similarly, if God answered all your prayers and ALL JC student were to score full marks, then all will get A. The curve attempts to distribute proportionally the results of everybody based on the marks, and place you in the correct percentile (probability). Logistics wise, you can click the link above which gives us a normal distribution. Here, you can input the Mean and Variance so you can create your own bell curve. I can’t really advise the mean and variance score since I have no data. Its really why we only see letter grade. Next we look at the bell curve First of all, this is a standardised normal, the Next, what does it mean when the bell curve shifts? What we see above is a Normal curve centred about 0 (mark), and we know that Normal curve is centred about Now since we all squeeze into the same bell curve and due to some friends scoring full marks or near full marks, yes there are definitely students that have full marks in a few particular JCs, I know a handful of my students above 90 actually… This results in the mean mark shifting to the right. And in general, the curve should not shift left, at least not in our society. We are Asians. Here is why, it is really impossible to say what’s the range for A unless you can confidently tell me the population mean and variance. But I do consider the case of 25% of students should get an A. I hope this clarifies a bit. And yes, this is what you guys actually learnt in A-levels, if you understood your content of course as what I’ve explained in class before. To spice things up, we can also perform hypothesis testing (simple case can be just with Now you can go at play with the above link and fit in what you believe the mean and standard deviation is, and then you can calculate the percentile (probability) you are in by putting the score you think you have. Take note the standard deviation should be not go too big, as we can’t have people scoring more than 100. 100 marks should be Again, please don’t ask me whats the If you’re lazy, then just check what’s your school percentage of A then consider the population of your school, that roughly how well you will do. 🙂 I usually say this in class, that if your school delivers 50%, then you need to find a friend to not do well at least. My take for the bell curve is that the bell curve will always help the best students. As for it helping the weaker students, it is highly dependent on the cohort on the whole and the difficulty of the paper. Consider me scoring 90/100 marks and everybody else scores above 95 marks, I’ll be the weakest and will end up failing. I hope these does not discourage or crush the confidence. Remember that there is a paper 2 still, with equal weightage. So I took this image of google, it does not represent the A-levels. Please take note. But this is what happens when too many students do well. If you’re keen, you can read more on skewed normal distribution independently. I might do more on Stat 101 here when I’m free. pingbacks / trackbacks • […] A little on Bell Curve […]	{"url":"https://theculture.sg/2015/11/a-little-on-bell-curve/","timestamp":"2024-11-02T19:01:32Z","content_type":"text/html","content_length":"108145","record_id":"<urn:uuid:f367af58-4717-42c8-847d-932be2b718d6>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00851.warc.gz"}
NCERT Solutions Class 10 Maths Chapter 11 Constructions - Download PDF! *According to the CBSE Syllabus 2023-24, this chapter has been removed. NCERT Solutions for Class 10 Maths Chapter 11 Constructions are provided in a detailed manner, where students can find a step-by-step solution to all the questions for fast revisions. Solutions for the 11th chapter of NCERT Class 10 Maths are prepared by subject experts under the guidelines of the NCERT to assist students in their board exam preparations. Get free NCERT Solutions for Class 10 Maths, Chapter 11 – Constructions at BYJU’S to accelerate the CBSE exam preparation. All the questions of NCERT exercises are solved using diagrams with a step-by-step procedure for construction. Solutions of NCERT help students boost their concepts and clear doubts. Access Answers to Maths NCERT Chapter 11 – Constructions Exercise 11.1 Page: 220 In each of the following, give the justification for the construction also. 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. Construction Procedure A line segment with a measure of 7.6 cm length is divided in the ratio of 5:8 as follows. 1. Draw line segment AB with a length measure of 7.6 cm. 2. Draw a ray AX that makes an acute angle with line segment AB. 3. Locate the points, i.e.,13 (= 5+8) points, such as A1, A2, A3, A4 …….. A13, on the ray AX, such that it becomes AA1 = A1A2 = A2A3 and so on. 4. Join the line segment and the ray, BA13. 5. Through the point A5, draw a line parallel to BA13 which makes an angle equal to ∠AA13B. 6. Point A5, which intersects line AB at point C. 7. C is the point that divides line segment AB of 7.6 cm in the required ratio of 5:8. 8. Now, measure the lengths of the line AC and CB. It becomes the measure of 2.9 cm and 4.7 cm, respectively. The construction of the given problem can be justified by proving that AC/CB = 5/ 8 By construction, we have A5C \|\| A13B. From the Basic proportionality theorem for the triangle AA13B, we get AC/CB =AA[5]/A[5]A[13]….. (1) From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments, respectively. Therefore, it becomes AA[5]/A[5]A[13]=5/8… (2) Compare the equations (1) and (2), we obtain AC/CB = 5/ 8 Hence, justified. 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Construction Procedure 1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm. 2. Take point A as the centre, and draw an arc of radius 5 cm. 3. Similarly, take point B as its centre, and draw an arc of radius 6 cm. 4. The arcs drawn will intersect each other at point C. 5. Now, we have obtained AC = 5 cm and BC = 6 cm, and therefore, ΔABC is the required triangle. 6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C. 7. Locate 3 points such as A1, A2, and A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3. 8. Join point BA3 and draw a line through A2, which is parallel to the line BA3 that intersects AB at point B’. 9. Through the point B’, draw a line parallel to line BC that intersects the line AC at C’. 10. Therefore, ΔAB’C’ is the required triangle. The construction of the given problem can be justified by proving that AB’ = (2/3)AB B’C’ = (2/3)BC AC’= (2/3)AC From the construction, we get B’C’ \|\| BC ∴ ∠AB’C’ = ∠ABC (Corresponding angles) In ΔAB’C’ and ΔABC, ∠ABC = ∠AB’C (Proved above) ∠BAC = ∠B’AC’ (Common) ∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion) Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1) In ΔAAB’ and ΔAAB, ∠A[2]AB’ =∠A[3]AB (Common) From the corresponding angles, we get ∠AA[2]B’ =∠AA[3]B Therefore, from the AA similarity criterion, we obtain ΔAA[2]B’ and AA[3]B So, AB’/AB = AA[2]/AA[3] Therefore, AB’/AB = 2/3 ……. (2) From equations (1) and (2), we get AB’/AB=B’C’/BC = AC’/ AC = 2/3 This can be written as AB’ = (2/3)AB B’C’ = (2/3)BC AC’= (2/3)AC Hence, justified. 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle Construction Procedure 1. Draw a line segment AB =5 cm. 2. Take A and B as the centre, and draw the arcs of radius 6 cm and 7 cm, respectively. 3. These arcs will intersect each other at point C, and therefore, ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm, respectively. 4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C. 5. Locate the 7 points, such as A[1], A[2], A[3], A[4], A[5], A[6], A[7] (as 7 is greater between 5 and 7), on line AX such that it becomes AA[1] = A[1]A[2] = A[2]A[3] = A[3]A[4] = A[4]A[5] = A[5]A [6] = A[6]A[7] 6. Join the points BA[5] and draw a line from A[7] to BA[5,] which is parallel to the line BA[5] where it intersects the extended line segment AB at point B’. 7. Now, draw a line from B’ to the extended line segment AC at C’, which is parallel to the line BC, and it intersects to make a triangle. 8. Therefore, ΔAB’C’ is the required triangle. The construction of the given problem can be justified by proving that AB’ = (7/5)AB B’C’ = (7/5)BC AC’= (7/5)AC From the construction, we get B’C’ \|\| BC ∴ ∠AB’C’ = ∠ABC (Corresponding angles) In ΔAB’C’ and ΔABC, ∠ABC = ∠AB’C (Proved above) ∠BAC = ∠B’AC’ (Common) ∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion) Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1) In ΔAA[7]B’ and ΔAA[5]B, ∠A[7]AB’=∠A[5]AB (Common) From the corresponding angles, we get ∠A A[7]B’=∠A A[5]B Therefore, from the AA similarity criterion, we obtain ΔA A[2]B’ and A A[3]B So, AB’/AB = AA[5]/AA[7] Therefore, AB /AB’ = 5/7 ……. (2) From equations (1) and (2), we get AB’/AB = B’C’/BC = AC’/ AC = 7/5 This can be written as AB’ = (7/5)AB B’C’ = (7/5)BC AC’= (7/5)AC Hence, justified. 1. Draw a line segment BC with a measure of 8 cm. 2. Now, draw the perpendicular bisector of the line segment BC and intersect at point D. 3. Take the point D as the centre and draw an arc with a radius of 4 cm, which intersects the perpendicular bisector at the point A. 4. Now, join the lines AB and AC, and the triangle is the required triangle. 5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A. 6. Locate the 3 points B[1], B[2] and B[3] on the ray BX such that BB[1] = B[1]B[2] = B[2]B[3] 7. Join the points B[2]C and draw a line from B[3,] which is parallel to the line B[2]C where it intersects the extended line segment BC at point C’. 8. Now, draw a line from C’ to the extended line segment AC at A’, which is parallel to the line AC, and it intersects to make a triangle. 9. Therefore, ΔA’BC’ is the required triangle. The construction of the given problem can be justified by proving that A’B = (3/2)AB BC’ = (3/2)BC A’C’= (3/2)AC From the construction, we get A’C’ \|\| AC ∴ ∠ A’C’B = ∠ACB (Corresponding angles) In ΔA’BC’ and ΔABC, ∠B = ∠B (Common) ∠A’BC’ = ∠ACB ∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion) Therefore, A’B/AB = BC’/BC= A’C’/AC Since the corresponding sides of the similar triangle are in the same ratio, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/2 Hence, justified. 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC. Construction Procedure 1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°. 2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A. 3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX. 4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’. 5. Draw a line through C’ parallel to the line AC, which intersects the line AB at A’. 6. Therefore, ΔA’BC’ is the required triangle. The construction of the given problem can be justified by proving that Since the scale factor is 3/4, we need to prove A’B = (3/4)AB BC’ = (3/4)BC A’C’= (3/4)AC From the construction, we get A’C’ \|\| AC In ΔA’BC’ and ΔABC, ∴ ∠ A’C’B = ∠ACB (Corresponding angles) ∠B = ∠B (Common) ∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion) Since the corresponding sides of the similar triangle are in the same ratio, it becomes Therefore, A’B/AB = BC’/BC= A’C’/AC So, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4 Hence, justified. 6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC. To find ∠C: ∠B = 45°, ∠A = 105° We know that, The sum of all interior angles in a triangle is 180°. ∠A+∠B +∠C = 180° 105°+45°+∠C = 180° ∠C = 180° − 150° ∠C = 30° So, from the property of the triangle, we get ∠C = 30° Construction Procedure The required triangle can be drawn as follows. 1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°. 2. Draw a ray BX that makes an acute angle with BC on the opposite side of vertex A. 3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX. 4. Join the points B3C. 5. Draw a line through B4 parallel to B3C, which intersects the extended line BC at C’. 6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’. 7. Therefore, ΔA’BC’ is the required triangle. The construction of the given problem can be justified by proving that Since the scale factor is 4/3, we need to prove A’B = (4/3)AB BC’ = (4/3)BC A’C’= (4/3)AC From the construction, we get A’C’ \|\| AC In ΔA’BC’ and ΔABC, ∴ ∠A’C’B = ∠ACB (Corresponding angles) ∠B = ∠B (Common) ∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion) Since the corresponding sides of the similar triangle are in the same ratio, it becomes Therefore, A’B/AB = BC’/BC= A’C’/AC So, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3 Hence, justified. 7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given The sides other than the hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each other Construction Procedure The required triangle can be drawn as follows. 1. Draw a line segment BC =3 cm. 2. Now, measure and draw an angle 90° 3. Take B as the centre and draw an arc with a radius of 4 cm, and intersects the ray at point B. 4. Now, join the lines AC, and the triangle ABC is the required triangle. 5. Draw a ray BX that makes an acute angle with BC on the opposite side of vertex A. 6. Locate 5 such as B1, B2, B3, B4, on the ray BX, such that BB[1] = B[1]B[2] = B[2]B[3]= B[3]B[4] = B[4]B[5] 7. Join the points B3C. 8. Draw a line through B5 parallel to B3C, which intersects the extended line BC at C’. 9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’. 10. Therefore, ΔA’BC’ is the required triangle. The construction of the given problem can be justified by proving that Since the scale factor is 5/3, we need to prove A’B = (5/3)AB BC’ = (5/3)BC A’C’= (5/3)AC From the construction, we get A’C’ \|\| AC In ΔA’BC’ and ΔABC, ∴ ∠ A’C’B = ∠ACB (Corresponding angles) ∠B = ∠B (Common) ∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion) Since the corresponding sides of the similar triangle are in the same ratio, it becomes Therefore, A’B/AB = BC’/BC= A’C’/AC So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3 Hence, justified. Exercise 11.2 Page: 221 In each of the following, give the justification for the construction also. 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Construction Procedure The construction to draw a pair of tangents to the given circle is as follows. 1. Draw a circle with a radius = 6 cm with centre O. 2. Locate a point P, which is 10 cm away from O. 3. Join points O and P through the line. 4. Draw the perpendicular bisector of the line OP. 5. Let M be the mid-point of the line PO. 6. Take M as the centre and measure the length of MO. 7. The length MO is taken as the radius, and draw a circle. 8. The circle drawn with the radius of MO intersect the previous circle at point Q and R. 9. Join PQ and PR. 10. Therefore, PQ and PR are the required tangents. The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 6cm with centre O. To prove this, join OQ and OR represented in dotted lines. From the construction, ∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle, so it becomes ∴ ∠PQO = 90° Such that ⇒ OQ ⊥ PQ Since OQ is the radius of the circle with a radius of 6 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle. Hence, justified. 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation. Construction Procedure For the given circle, the tangent can be drawn as follows. 1. Draw a circle of 4 cm radius with centre “O”. 2. Again, take O as the centre and draw a circle of radius 6 cm. 3. Locate a point P on this circle. 4. Join the points O and P through lines, such that it becomes OP. 5. Draw the perpendicular bisector to the line OP 6. Let M be the mid-point of PO. 7. Draw a circle with M as its centre and MO as its radius, 8. The circle drawn with the radius OM intersects the given circle at the points Q and R. 9. Join PQ and PR. 10. PQ and PR are the required tangents. From the construction, it is observed that PQ and PR are of length 4.47 cm each. It can be calculated manually as follows: In ∆PQO, Since PQ is a tangent, ∠PQO = 90°. PO = 6cm and QO = 4 cm Applying Pythagoras’ theorem in ∆PQO, we obtain PQ^2+QO^2 = PQ^2 PQ^2+(4)^2 = (6)^2 PQ^2 +16 =36 PQ^2 = 36−16 PQ^2 = 20 PQ = 2√5 PQ = 4.47 cm Therefore, the tangent length PQ = 4.47 The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 4 cm with centre O. To prove this, join OQ and OR represented in dotted lines. From the construction, ∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle, so it becomes ∴ ∠PQO = 90° Such that ⇒ OQ ⊥ PQ Since OQ is the radius of the circle with a radius of 4 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle. Hence, justified. 3. Draw a circle of radius 3 cm. Take two points, P and Q, on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points, P and Q. Construction Procedure The tangent for the given circle can be constructed as follows. 1. Draw a circle with a radius of 3cm with a centre “O”. 2. Draw a diameter of a circle, and it extends 7 cm from the centre, and mark it as P and Q. 3. Draw the perpendicular bisector of the line PO and mark the midpoint as M. 4. Draw a circle with M as the centre and MO as the radius. 5. Now, join the points PA and PB in which the circle with radius MO intersects the circle of circle 3cm. 6. Now, PA and PB are the required tangents. 7. Similarly, from point Q, we can draw the tangents. 8. From that, QC and QD are the required tangents. The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 3 cm with centre O. To prove this, join OA and OB. From the construction, ∠PAO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle, so it becomes ∴ ∠PAO = 90° Such that ⇒ OA ⊥ PA Since OA is the radius of the circle with a radius of 3 cm, PA must be a tangent of the circle. Similarly, we can prove that PB, QC and QD are the tangents of the circle. Hence, justified. 4. Draw a pair of tangents to a circle of radius 5 cm, which are inclined to each other at an angle of 60°. Construction Procedure The tangents can be constructed in the following manner: 1. Draw a circle of radius 5 cm, with the centre as O. 2. Take a point Q on the circumference of the circle and join OQ. 3. Draw a perpendicular to QP at point Q. 4. Draw a radius OR, making an angle of 120°, i.e., (180°−60°) with OQ. 5. Draw a perpendicular to RP at point R. 6. Now, both the perpendiculars intersect at point P. 7. Therefore, PQ and PR are the required tangents at an angle of 60°. The construction can be justified by proving that ∠QPR = 60°. By our construction, ∠OQP = 90° ∠ORP = 90° And ∠QOR = 120° We know that the sum of all interior angles of a quadrilateral = 360° ∠OQP+∠QOR + ∠ORP +∠QPR = 360^o 90°+120°+90°+∠QPR = 360° Therefore, ∠QPR = 60° Hence, justified. 5. Draw a line segment AB of length 8 cm. Taking A as the centre, draw a circle of radius 4 cm and taking B as the centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Construction Procedure The tangent for the given circle can be constructed as follows: 1. Draw a line segment AB = 8 cm. 2. Take A as the centre and draw a circle of radius 4 cm. 3. Take B as the centre and draw a circle of radius 3 cm. 4. Draw the perpendicular bisector of the line AB, and the midpoint is taken as M. 5. Now, take M as the centre and draw a circle with the radius of MA or MB, which intersects the circle at the points P, Q, R and S. 6. Now, join AR, AS, BP and BQ. 7. Therefore, the required tangents are AR, AS, BP and BQ. The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B with a radius of 3 cm), and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). From the construction, to prove this, join AP, AQ, BS, and BR. ∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠ASB = 90° ⇒ BS ⊥ AS Since BS is the radius of the circle, AS must be a tangent of the circle. Similarly, AR, BP, and BQ are the required tangents of the given circle. 6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. Construction Procedure The tangent for the given circle can be constructed as follows: 1. Draw the line segment with base BC = 8cm. 2. Measure the angle 90° at the point B, such that ∠ B = 90°. 3. Take B as the centre and draw an arc with a measure of 6cm. 4. Let the point be A, where the arc intersects the ray. 5. Join the line AC. 6. Therefore, ABC is the required triangle. 7. Now, draw the perpendicular bisector to the line BC, and the midpoint is marked as E. 8. Take E as the centre, and BE or EC measure as the radius and draw a circle. 9. Join A to the midpoint E of the circle. 10. Now, again, draw the perpendicular bisector to the line AE, and the midpoint is taken as M. 11. Take M as the centre, and AM or ME measure as the radius and draw a circle. 12. This circle intersects the previous circle at points B and Q. 13. Join points A and Q. 14. Therefore, AB and AQ are the required tangents. The construction can be justified by proving that AG and AB are the tangents to the circle. From the construction, join EQ. ∠AQE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠AQE = 90° ⇒ EQ⊥ AQ Since EQ is the radius of the circle, AQ has to be a tangent of the circle. Similarly, ∠B = 90° ⇒ AB ⊥ BE Since BE is the radius of the circle, AB has to be a tangent of the circle. Hence, justified. 7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle. Construction Procedure The required tangents can be constructed on the given circle as follows: 1. Draw a circle with the help of a bangle. 2. Draw two non-parallel chords, such as AB and CD. 3. Draw the perpendicular bisector of AB and CD. 4. Take the centre as O, where the perpendicular bisector intersects. 5. To draw the tangents, take a point P outside the circle. 6. Join points O and P. 7. Now, draw the perpendicular bisector of the line PO, and the midpoint is taken as M 8. Take M as the centre and MO as the radius and draw a circle. 9. Let the circle intersects intersect the circle at the points Q and R. 10. Now, join PQ and PR. 11. Therefore, PQ and PR are the required tangents. The construction can be justified by proving that PQ and PR are tangents to the circle. Since, O is the centre of a circle, we know that the perpendicular bisector of the chords passes through the centre. Now, join the points OQ and OR. We know that the perpendicular bisector of a chord passes through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. ∠PQO is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. ∴ ∠PQO = 90° ⇒ OQ⊥ PQ Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, ∴ ∠PRO = 90° ⇒ OR ⊥ PO Since OR is the radius of the circle, PR has to be a tangent of the circle. Therefore, PQ and PR are the required tangents of a circle. NCERT Solutions for Class 10 Maths Chapter 11 Constructions Topics present in NCERT Solutions for Class 10 Maths Chapter 11 include the division of a line segment, constructions of tangents to a circle, line segment bisector and many more. Students in Class 9 study some basics of constructions, like drawing the perpendicular bisector of a line segment, bisecting an angle, triangle construction etc. Using Class 9 concepts, students in Class 10 will learn more about constructions, along with the reasoning behind the work. NCERT Class 10 Chapter 11-Constructions is a part of Geometry. Over the past few years, geometry has consisted of a total weightage of 15 marks in the final exams. Construction is a scoring chapter of the geometry section. In the previous year’s exam, one question of 4 marks was asked from this chapter. List of Exercises in Class 10 Maths Chapter 11 Exercise 11.1 Solutions (7 Questions) Exercise 11.2 Solutions (7 Questions) The NCERT Solutions for Class 10 for the 11th chapter of Maths are all about the construction of line segments, division of a Line Segment and Construction of a Circle, Constructions of Tangents to a circle using an analytical approach. Students also have to provide a justification for each answer. The topics covered in Maths Chapter 11 Constructions are: Exercise Topic 11.1 Introduction 11.2 Division of a Line Segment 11.3 Construction of Tangents to a Circle 11.4 Summary Some of the ideas applied in this chapter are as follows: 1. The locus of a point that moves at an identical distance from 2 points is normal to the line joining both points. 2. Perpendicular or Normal means right angles, whereas the bisector cuts a line segment in two halves. 3. The design of different shapes utilising a pair of compasses and a straightedge or ruler. Key Features of NCERT Solutions for Class 10 Maths Chapter 11 Constructions • NCERT solutions can also prove to be of valuable help to students in their assignments and preparation for CBSE and competitive exams. • Each question is explained using diagrams which makes learning more interactive. • Easy and understandable language is used in NCERT solutions. • Solutions are provided using an analytical approach. Disclaimer – Dropped Topics – 11.1 Introduction 11.2 Division of a line segment 11.3 Construction of tangents into a circle 11.4 Summary Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 11 What is the use of practising NCERT Solutions for Class 10 Maths Chapter 11? Practising NCERT Solutions for Class 10 Maths Chapter 11 provides students with an idea about the sample of questions that will be asked in the board exam, which would help them prepare competently. These solutions are useful resources which can provide them with all the vital information in the most precise form. These solutions cover all topics included in the NCERT syllabus, prescribed by the CBSE board. List out the topics of NCERT Solutions for Class 10 Maths Chapter 11. The topics covered in NCERT Solutions for Class 10 Maths Chapter 11 Constructions are an introduction to the constructions, the division of a line segment and the construction of tangents to a circle, and finally, it gives the summary of all the concepts provided in the whole chapter. By referring to these solutions, students can clear their doubts and also can practise additional Are the NCERT Solutions for Class 10 Maths Chapter 11 accessible only online? For ease of learning, the solutions have also been provided in PDF format so that the students can download them for free and refer to the solutions offline as well. These NCERT Solutions for Class 10 Maths Chapter 11 can be viewed online. Leave a Comment 1. good	{"url":"http://soporose.net/index-52.html","timestamp":"2024-11-13T21:16:03Z","content_type":"text/html","content_length":"706469","record_id":"<urn:uuid:667361fa-91a7-4fff-a479-ad03c3cbd103>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00348.warc.gz"}
Why FIR Filters have Linear Phase \| Wireless Pi Why FIR Filters have Linear Phase One of the most attractive properties of a Finite Impulse Response (FIR) filter is that a linear phase response is easier to achieve. Not all FIR filters have linear phase though. This is only possible when the coefficients or taps of the filter are symmetric or anti-symmetric around a point. Today I want to describe the reason behind this kind of phase response in an intuitive manner. We have described Finite Impulse Response (FIR) filters before. Moreover, we have also discussed that the Discrete Fourier Transform (DFT) of a signal is complex in general and therefore both magnitude and phase plots come into the picture. Why is Linear Phase Important? To understand its significance, consider a filter with a unit magnitude response and a linear phase response, where $t_0$ is a constant. H(\omega) = e^{-j\omega t_0} This phase response $\theta=-\omega t_0$ is linear as it is similar to the linear equation $y=mx$ where the slope $m$ is given by $-t_0$. • In time domain, the output $y(t)$ is a convolution between the input signal $x(t)$ and the impulse response $h(t)$ of this filter. y(t) = x(t) * h(t) • In frequency domain, the output $Y(\omega)$ is a product between the input spectrum $X(\omega)$ and filter response $H(\omega)$. Y(\omega) = X(\omega)\cdot H(\omega) With $H(\omega)$ as given above, and taking its inverse Fourier Transform, we get y(t) = \int _{-\infty}^{\infty} Y(\omega)e^{j\omega t}d\omega = \int _{-\infty}^{\infty} X(\omega)e^{-j\omega t_o}e^{j\omega t}d\omega This can be simplified as y(t) = \int _{-\infty}^{\infty} X(\omega)e^{j\omega (t-t_0)}d\omega = x(t-t_0) In words, the linear phase property of a filter delays the input signal $x(t)$ but preserves the signal shape with no distortion. This is the main reason behind why DSP engineers prefer a linear phase response in most (but not all) signal processing applications. An FIR Filter Based on whether the filter length is even or odd, and the filter taps are symmetric or anti-symmetric, there are 4 types of FIR filters. Our focus is on the simplest case: an odd length filter (so that the taps are symmetric around a single point) with symmetric coefficients. Shown below in the figure is an odd-length FIR filter with symmetric coefficients. The corresponding magnitude response is also plotted below. To understand the idea, let us consider only the two samples at times $n=+1$ and $n=-1$. Imagine that the remaining samples have disappeared and whatever conclusion we draw for the samples at $n=\pm 1$ can also be applied to them. An Impulse in Time What does a time domain impulse correspond to in frequency domain? To answer this question, we refer to the definition of the Discrete-Time Fourier Transform (DFTF). X(e^{j\omega})=\sum_{n=0}^{N-1} x[n]e^{-j\omega n} Clearly, when the input $x[n]$ is an impulse at $n=+1$, we have X(e^{j\omega})=\sum_{n=0}^{N-1} \delta [n-1]e^{-j\omega n} = e^{-j\omega \cdot 1} The result is a complex sinusoid in frequency domain with period 1! Another way of seeing this ideas is as follows. Time-Frequency Duality From the concept of frequency, we know that a complex sinusoid in time domain corresponds to an impulse in frequency domain. Owing to the time-frequency duality, an impulse in time domain corresponds to a complex sinusoid in frequency domain! But we have two impulses in terms of two filter coefficients at $n=\pm 1$. From One to Two Impulses Two impulses at $n=\pm 1$ imply two complex sinusoids rotating in frequency domain in opposite directions. The result is that the frequency domain signal can be written as e^{+j\omega\cdot 1}+e^{-j\omega\cdot 1} = 2\cos \left(\omega \cdot 1\right) where we have used the Euler’s identity to simplify the above expression. Observe that the above signal is not a time domain cosine; instead, it is a frequency domain sinusoid. Symmetric Coefficients With this background, symmetric coefficients of an FIR filter $h[n]$ come into play. For an $N$-length filter due to symmetry, we have h[0] &= h[N-1] \\ h[1] &= h[N-2] \\ \vdots &= \vdots\\ Since DTFT is given by $H(e^{j\omega})=\sum_{n=0}^{N-1} h[n]e^{-j\omega n}$, the contribution from these two terms can be written as h[0]e^{-j\omega\cdot 0} + h[N-1]e^{-j\omega(N-1)} This expression can be simplified using the following two facts. • $N$ is odd, so $N-1$ is an even number. • The coefficients $h[0]$ and $h[N-1]$ are the same. Thus, we can write h[0]e^{-j\omega\cdot 0} + h[0]e^{-j\omega(N-1)} &= e^{-j\omega\cdot \frac{N-1}{2}} h[0]\Big[e^{+j\omega\cdot \frac{N-1}{2}} + e^{-j\omega\cdot \frac{N-1}{2}}\Big]\\ &= e^{-j\omega\cdot \frac{N-1}{2}} \underbrace{ h[0]\cdot 2\cos \left(\omega \frac{N-1}{2}\right)}_{\text{Real frequency domain signal}} where we have used the Euler’s identity for cosine above. The real signal has no imaginary component and hence a zero phase. All that is left in the phase part is the following term. \text{Phase response} = -\omega\cdot \frac{N-1}{2} which is clearly a linear expression similar to $y=mx$. The x-axis represents $\omega$ while the slope is given by $-(N-1)/2$. In the above equation, we have only considered the contributions from impulses at $n=\pm 1$. The impulses or coefficients at other times also add up to real cosines with similar phase contributions as above, i.e., $(N-1)/2$. This is why it is straightforward to design an FIR filter with linear phase response by simply choosing symmetric or anti-symmetric coefficients. This is not the end of the filter design story though. IIR filters have certain advantages over FIR filters in some other respects. Leave a Reply; You can use HTML (<>) or Latex ($$)	{"url":"https://wirelesspi.com/why-fir-filters-have-linear-phase/","timestamp":"2024-11-05T22:37:31Z","content_type":"text/html","content_length":"66848","record_id":"<urn:uuid:edaee96e-36c4-4886-9645-88960151d419>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00209.warc.gz"}
The mean yearly income for construction workers in New York Tutorial # 00644799 Posted On: 04/14/2023 06:34 AM Feedback Score: Not rated yet! Purchased By: Report this Tutorial as Inappropriate Tutorials helped score wonderful grades GC Working with Radioactive Material Paper Colorado State University GC Working with Radioactive Material Paper As a scientist, you study 200 grams of a radioactive material for seven months, and suppose you have obtained the following data …	{"url":"https://www.homeworkjoy.com/questions/themeanyearly-income-for-construction-workers-in-new-york-645920/","timestamp":"2024-11-12T19:02:02Z","content_type":"text/html","content_length":"141566","record_id":"<urn:uuid:dcbede01-9467-4eac-96cf-63a555484c34>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00184.warc.gz"}
Finding the Displacement of a Particle Based on Time and Position Relative to a Fixed Point Question Video: Finding the Displacement of a Particle Based on Time and Position Relative to a Fixed Point Mathematics • Second Year of Secondary School A particle started moving in a straight line. After π ‘ seconds, its position relative to a fixed point is given by π = (π ‘Β² β 4π ‘ + 7) m, π ‘ β ₯ 0 . Find the displacement of the particle during the first five seconds. Video Transcript A particle started moving in a straight line. After π ‘ seconds, its position relative to a fixed point is given by π equals π ‘ squared minus 40 plus seven metres for π ‘ is greater than or equal to zero. Find the displacement of the particle during the first five seconds. In this question, weβ ve been given a function that describes the position of the particle relative to a fixed point. And weβ re being asked to find its displacement. No, itβ s absolutely not enough just to substitute π ‘ equals five into our position function. We recall that if we have a position function of a particle moving along a line given as π of t, the displacement from π ‘ equals π ‘ one to π ‘ equals π ‘ two is the difference between π of π ‘ two minus π of π ‘ one. This is really important as displacement is the change in the position of the particle. We want to work out the displacement of the particle during the first five seconds. So weβ ll let π ‘ one be equal to zero and π ‘ two be equal to five. Then the displacement is π of five minus π of zero. π of five is five squared minus four times five plus seven. We simply substitute π ‘ equals five into our position function. We repeat this process for π of zero, this time substituting π ‘ equals zero in. And we get zero squared minus four times zero plus seven. That gives us 25 minus 20 plus seven minus seven. Now, seven minus seven is zero. So weβ re left with 25 minus 20 which is five. Since our function describes the position relative to a fixed point in metres, we can say that the displacement of our particle during the first five seconds is five metres.	{"url":"https://www.nagwa.com/en/videos/787154130462/","timestamp":"2024-11-14T21:57:36Z","content_type":"text/html","content_length":"251243","record_id":"<urn:uuid:30f61461-9f38-4892-8d20-b733f5a2962d>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00833.warc.gz"}
Adding Three-digit Numbers Worksheet 2024 - NumbersWorksheets.com Adding Three-digit Numbers Worksheet Adding Three-digit Numbers Worksheet – Superior supplement drills are great ways to expose individuals to algebra methods. These drills have alternatives for a single-second, three-min, and five-minute drills with personalized ranges of fifteen to one hundred problems. Furthermore, the drills come in a side to side formatting, with phone numbers from to 99. The best part is that these drills can be personalized to each student’s ability level. Here are a few additional innovative supplement drills: Add up forwards by one particular Relying on is really a useful strategy for developing quantity simple fact fluency. Count on a number by addingtwo and one, or three. For example, 5 as well as two means ten, and the like. Depending on a number by adding you might make the exact same end result for small and large figures. These supplement worksheets consist of process on counting on a amount with equally hands and fingers and also the amount collection. Adding Three-digit Numbers Worksheet. Practice multiple-digit supplement by using a variety range Open up amount lines are wonderful models for addition and place worth. In a earlier submit we reviewed the numerous mental strategies pupils can use to provide numbers. Using a amount lines are a terrific way to file many of these strategies. In this posting we are going to explore a great way to exercise multiple-digit inclusion using a amount series. Listed below are three methods: Process including doubles The exercise incorporating increases with supplement numbers worksheet could be used to help young children produce the thought of a doubles simple fact. Once a doubles fact is when the same number is added more than. For example, if Elsa had four headbands and Gretta had five, they both have two doubles. Students can develop a stronger understanding of doubles and gain the fluency required to add single digit numbers, by practicing doubles with this worksheet. Training incorporating fractions A Training adding fractions with addition phone numbers worksheet can be a beneficial instrument to develop your child’s fundamental understanding of fractions. These worksheets protect a number of ideas relevant to fractions, for example assessing and buying fractions. Additionally, they offer useful problem-solving strategies. It is possible to down load these worksheets for free in PDF file format. The first task is to be certain your son or daughter understands the rules and symbols linked to fractions. Practice adding fractions with a amount range With regards to rehearsing incorporating fractions having a number series, pupils may use a fraction position benefit mat or possibly a variety collection for mixed amounts. These help in corresponding small fraction equations to the alternatives. The place importance mats may have a number of good examples, using the situation composed on the top. Students are able to pick the response they want by punching holes beside every choice. After they have chosen the correct response, each student can bring a cue near the option. Gallery of Adding Three-digit Numbers Worksheet Adding Three Digit Numbers Within One Thousand Turtle Diary Worksheet Three Digit Addition Worksheets From The Teacher s Guide 2nd Grade Leave a Comment	{"url":"https://numbersworksheet.com/adding-three-digit-numbers-worksheet/","timestamp":"2024-11-09T19:19:37Z","content_type":"text/html","content_length":"54067","record_id":"<urn:uuid:6af042ff-0223-45d1-be97-d2fd72645bce>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00303.warc.gz"}
19. Stokes equations with Taylor-Hood elements 19. Stokes equations with Taylor-Hood elements¶ This demo is implemented in a single Python file, demo_stokes-taylorhood.py, which contains both the variational form and the solver. This demo illustrates how to: • Read mesh and subdomains from file • Use mixed function spaces The mesh and subdomains look as follows: and the solution of u and p, respectively: 19.1. Equation and problem definition¶ 19.1.1. Strong formulation¶ \[\begin{split}- \nabla \cdot (\nabla u + p I) &= f \quad {\rm in} \ \Omega, \\ \nabla \cdot u &= 0 \quad {\rm in} \ \Omega. \\\end{split}\] The sign of the pressure has been flipped from the classical definition. This is done in order to have a symmetric (but not positive-definite) system of equations rather than a non-symmetric (but positive-definite) system of equations. A typical set of boundary conditions on the boundary $\partial \Omega = \Gamma_{D} \cup \Gamma_{N}$ can be: \[\begin{split}u &= u_0 \quad {\rm on} \ \Gamma_{D}, \\ \nabla u \cdot n + p n &= g \, \quad\;\; {\rm on} \ \Gamma_{N}. \\\end{split}\] 19.1.2. Weak formulation¶ The Stokes equations can easily be formulated in a mixed variational form; that is, a form where the two variables, the velocity and the pressure, are approximated simultaneously. Using the abstract framework, we have the problem: find $(u, p) \in W$ such that \[a((u, p), (v, q)) = L((v, q))\] for all $(v, q) \in W$, where \[\begin{split}a((u, p), (v, q)) &= \int_{\Omega} \nabla u \cdot \nabla v - \nabla \cdot v \ p + \nabla \cdot u \ q \, {\rm d} x, \\ L((v, q)) &= \int_{\Omega} f \cdot v \, {\rm d} x + \int_{\partial \Omega_N} g \cdot v \, {\rm d} s. \\\end{split}\] The space $W$ should be a mixed (product) function space $W = V \times Q$, such that $u \in V$ and $q \in Q$. 19.1.3. Domain and boundary conditions¶ In this demo, we shall consider the following definitions of the input functions, the domain, and the boundaries: • $\Omega = [0,1]\times[0,1] \backslash {\rm dolphin}$ (a unit cube) • $\Gamma_D =$ • $\Gamma_N =$ • $u_0 = (- \sin(\pi x_1), 0.0)$ for $x_0 = 1$ and $u_0 = (0.0, 0.0)$ otherwise • $f = (0.0, 0.0)$ • $g = (0.0, 0.0)$ 19.2. Implementation¶ First, the dolfin module is imported: In this example, different boundary conditions are prescribed on different parts of the boundaries. This information must be made available to the solver. One way of doing this, is to tag the different sub-regions with different (integer) labels. DOLFIN provides a class MeshFunction which is useful for these types of operations: instances of this class represent functions over mesh entities (such as over cells or over facets). Mesh and mesh functions can be read from file in the following way: # Load mesh and subdomains mesh = Mesh("dolfin_fine.xml.gz") sub_domains = MeshFunction("size_t", mesh, "dolfin_fine_subdomains.xml.gz") Next, we define a MixedFunctionSpace composed of a VectorFunctionSpace of continuous piecewise quadratics and a FunctionSpace of continuous piecewise linears. (This mixed finite element space is known as the Taylor–Hood elements and is a stable, standard element pair for the Stokes equations.) # Define function spaces V = VectorFunctionSpace(mesh, "CG", 2) Q = FunctionSpace(mesh, "CG", 1) W = V * Q Now that we have our mixed function space and marked subdomains defining the boundaries, we define boundary conditions: # No-slip boundary condition for velocity # x1 = 0, x1 = 1 and around dolphin noslip = Constant((0, 0)) bc0 = DirichletBC(W.sub(0), noslip, sub_domains, 0) # Inflow boundary condition for velocity # x0 = 1 inflow = Expression(("-sin(x[1]pi)", "0.0")) bc1 = DirichletBC(W.sub(0), inflow, sub_domains, 1) # Boundary condition for pressure at outflow # x0 = 0 zero = Constant(0) bc2 = DirichletBC(W.sub(1), zero, sub_domains, 2) # Collect boundary conditions bcs = [bc0, bc1, bc2] Here, we have given four arguments in the call to DirichletBC. The first specifies the FunctionSpace. Since we have a MixedFunctionSpace, we write W.sub(0) for the function space V, and W.sub(1) for Q. The second argument specifies the value on the Dirichlet boundary. The two last ones specifies the marking of the subdomains; sub_domains contains the subdomain markers and the number given as the last argument is the subdomain index. The bilinear and linear forms corresponding to the weak mixed formulation of the Stokes equations are defined as follows: # Define variational problem (u, p) = TrialFunctions(W) (v, q) = TestFunctions(W) f = Constant((0, 0)) a = (inner(grad(u), grad(v)) - div(v)p + qdiv(u))dx L = inner(f, v)dx To compute the solution we use the bilinear and linear forms, and the boundary condition, but we also need to create a Function to store the solution(s). The (full) solution will be stored in w, which we initialize using the MixedFunctionSpace W. The actual computation is performed by calling solve with the arguments a, L, w and bcs. The separate components u and p of the solution can be extracted by calling the split function. Here we use an optional argument True in the split function to specify that we want a deep copy. If no argument is given we will get a shallow copy. We want a deep copy for further computations on the coefficient vectors. # Compute solution w = Function(W) solve(a == L, w, bcs) # Split the mixed solution using deepcopy # (needed for further computation on coefficient vector) (u, p) = w.split(True) We may be interested in the $L^2$ norms of u and p, they can be calculated and printed by writing print "Norm of velocity coefficient vector: %.15g" % u.vector().norm("l2") print "Norm of pressure coefficient vector: %.15g" % p.vector().norm("l2") One can also split functions using shallow copies (which is enough when we just plotting the result) by writing # Split the mixed solution using a shallow copy (u, p) = w.split() Finally, we can store to file and plot the solutions. # Save solution in VTK format ufile_pvd = File("velocity.pvd") ufile_pvd << u pfile_pvd = File("pressure.pvd") pfile_pvd << p # Plot solution 19.3. Complete code¶ from __future__ import print_function from dolfin import # Load mesh and subdomains mesh = Mesh("../dolfin_fine.xml.gz") sub_domains = MeshFunction("size_t", mesh, "../dolfin_fine_subdomains.xml.gz") # Define function spaces V = VectorFunctionSpace(mesh, "CG", 2) Q = FunctionSpace(mesh, "CG", 1) W = V * Q # No-slip boundary condition for velocity # x1 = 0, x1 = 1 and around the dolphin noslip = Constant((0, 0)) bc0 = DirichletBC(W.sub(0), noslip, sub_domains, 0) # Inflow boundary condition for velocity # x0 = 1 inflow = Expression(("-sin(x[1]pi)", "0.0")) bc1 = DirichletBC(W.sub(0), inflow, sub_domains, 1) # Boundary condition for pressure at outflow # x0 = 0 zero = Constant(0) bc2 = DirichletBC(W.sub(1), zero, sub_domains, 2) # Collect boundary conditions bcs = [bc0, bc1, bc2] # Define variational problem (u, p) = TrialFunctions(W) (v, q) = TestFunctions(W) f = Constant((0, 0)) a = (inner(grad(u), grad(v)) - div(v)p + qdiv(u))dx L = inner(f, v)*dx # Compute solution w = Function(W) solve(a == L, w, bcs) # Split the mixed solution using deepcopy # (needed for further computation on coefficient vector) (u, p) = w.split(True) print("Norm of velocity coefficient vector: %.15g" % u.vector().norm("l2")) print("Norm of pressure coefficient vector: %.15g" % p.vector().norm("l2")) # # Split the mixed solution using a shallow copy (u, p) = w.split() # Save solution in VTK format ufile_pvd = File("velocity.pvd") ufile_pvd << u pfile_pvd = File("pressure.pvd") pfile_pvd << p # Plot solution	{"url":"https://fenicsproject.org/olddocs/dolfin/1.5.0/python/demo/documented/stokes-taylor-hood/python/documentation.html","timestamp":"2024-11-02T02:31:04Z","content_type":"text/html","content_length":"39917","record_id":"<urn:uuid:359f74c8-f5c2-4e72-91db-822c892bc528>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00552.warc.gz"}
Convert Fractions to Percentages - Crystal Crash Related Worksheets Convert the fractions to percentages. Choose from unit fractions, multi-part fractions or improper fractions. This game, together with all the other crystal crash games, is available as a single iPad or Android app. Scan to open this game on a mobile device. Right-click to copy and paste it onto a homework sheet.	{"url":"https://mathsframe.co.uk/en/resources/resource/277/Convert-Fractions-to-Percentages","timestamp":"2024-11-02T09:20:08Z","content_type":"text/html","content_length":"11952","record_id":"<urn:uuid:79eb04ab-90a9-41a6-9468-c11645b89898>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00242.warc.gz"}
Research Guides: Science Literacy Week September 19 - 25, 2022: Woodward Science Library Math Reads Climate Mathematics by Samuel S. P. Shen; Richard C. J. Somerville Call Number: QC981 .S52275 2019 ISBN: 9781108476874 Publication Date: 2019 This unique text provides a thorough, yet accessible, grounding in the mathematics, statistics, and programming that students need to master for coursework and research in climate science, meteorology, and oceanography. Assuming only high school mathematics, it presents carefully selected concepts and techniques in linear algebra, statistics, computing, calculus and differential equations within the context of real climate science examples. Computational techniques are integrated to demonstrate how to visualize, analyze, and apply climate data, with R code featured in the book and both R and Python code available online. Exercises are provided at the end of each chapter with selected solutions available to students to aid self-study and further solutions provided online for instructors only. Additional online supplements to aid classroom teaching include datasets, images, and animations. Guidance is provided on how the book can support a variety of courses at different levels, making it a highly flexible text for undergraduate and graduate students, as well as researchers and professional climate scientists who need to refresh or modernize their quantitative skills.	{"url":"https://guides.library.ubc.ca/scilit22/Woodward","timestamp":"2024-11-06T13:55:18Z","content_type":"text/html","content_length":"60667","record_id":"<urn:uuid:4a9fd53f-9661-4c16-a18f-4c05ccc3849b>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00559.warc.gz"}
XXX.8 Converting MPR Orientation to Viewpoint Attributes in Volumetric Rendering Web Services The Rendered 3D and Rendered MPR camera orientation parameters for Volumetric Rendering web services, such as the Volume Rendering Volumetric Presentation State IOD, specify orientation from the perspective of a camera in the Volumetric Presentation State Reference Coordinate System (VPS-RCS) with three parameters consisting of: The Planar MPR Volumetric Presentation State IOD specifies the MPR slab orientation using the MPR View Width Direction (0070,1507) and MPR View Height Direction (0070,1511) attributes, which contain the direction cosines X[xyz] and Y[xyz], respectively. The camera parameters can be derived from the MPR attributes as follows: V[xyz] = T[xyz] + X[xyz] * W / 2 + Y[xyz] * H / 2 = V [xyz]- Z [xyz] = Y[xyz] T[xyz] = coordinates of the MPR Top LeftHand Corner (0070,1505) in mm X[xyz] = the direction cosine of the MPR View Width Direction (0070,1507) Y[xyz] = the direction cosine of the MPR View Height Direction (0070,1511) Z[xyz] = the vector cross product of X[xyz] and Y[xyz] W = MPR View Width (0070,1508) in mm H = MPR View Height (0070,1512) in mm	{"url":"https://dicom.nema.org/medical/dicom/current/output/chtml/part17/sect_XXX.8.html","timestamp":"2024-11-14T13:53:18Z","content_type":"application/xhtml+xml","content_length":"9192","record_id":"<urn:uuid:463f40f4-90fc-4f9d-9596-296e3b5fef4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00297.warc.gz"}
Digital Signposts for Improved Map Navigation — Development Seed Digital Signposts for Improved Map Navigation 6 min read Daniel da Silvaon Directional markers hint at the existence of points of interest beyond what is visible on the map, guiding the user and making the process of exploration more intuitive. When users navigate digital maps, they often lose sight of important points of interest beyond the visible area. This challenge, known as the "desert fog" problem, can lead to disorientation and missed information. Our solution? Directional markers - a seemingly simple UI element requiring intricate trigonometry and clever CSS manipulation. We're currently preparing for SatSummit, a conference we've hosted for nearly ten years, bringing together satellite industry leaders and global development experts. The upcoming edition in Lisbon, Portugal, presented a unique challenge: showing points of interest on a map for an area not exactly in Lisbon proper that attendees may be less familiar. The problem? Markers only become visible when users zoom out, which they might not realize they need to do. Our solution uses directional markers to indicate off-screen points of interest, allowing users to explore the map while zoomed in without losing their bearings. We'll walk you through our implementation using the SatSummit map as an example, but the techniques are applicable to any interactive mapping project with off-screen points of interest. You'll learn • Calculating marker positions mathematically • Using CSS transforms for placement and rotation • Ensuring smooth performance across map states Whether you're a front-end developer, UX designer, or just curious about applied trigonometry in web development, we hope this provides some practical tips for solving this common digital cartography Calculating the marker's position was achieved with some good old trigonometry and then positioned on the map using CSS transforms. We'll break it down step by step. We're using mapbox-gl to power the map and turf.js for some geospatial calculations, but these principles can be applied with any mapping library. We only want to consider points that are not visible on the map, so on every move event we check if the point is within the viewport. // import { point } from '@turf/helpers'; // import { booleanPointInPolygon } from '@turf/boolean-point-in-polygon'; // import { bboxPolygon } from '@turf/bbox-polygon'; mbMap.on('move', () => { const bbox = mbMap.getBounds()!.toArray().flat() as BBox; const polygon = bboxPolygon(bbox); const inViewport = booleanPointInPolygon(point(poiCoordinates), polygon); if (!inViewport) { // Position the direction marker. } else { // Hide the direction marker. The second step is to calculate the bearing between the point of interest and the center of the map. The bearing varies from -180 to 180 degrees, with 0 degrees being north, 90 degrees east, -90 degrees west and 180/-180 degrees south. // import { bearing } from '@turf/bearing'; // import { point } from '@turf/helpers'; const { lng, lat } = mbMap.getCenter(); const br = bearing([lng, lat], point(poiCoordinates)); Since we're going to be working with trigonometry and triangles it is easier if all our angles are in the NE quadrant (0 - 90 deg). Later we'll invert the values depending on the original bearing. We do this by folding the x and y axis. We fold the axis by ensuring that the angle is the same on both sides of the folding axis. Folding the y axis is just a matter of getting the absolute value of the bearing. In the animation, the solid orange angle is the bearing and the dashed purple angle is the folded angle. Folding the x axis is a bit more complicated. Looking at the animation, the dotted green angle must be the same as the dashed purple angle. We can achieve that with the formula 90 - (bearing - 90) where the bearing is the solid orange angle. // Bring the bearing to the north east quadrant. const absoluteBearing = Math.abs(br); const isNorth = absoluteBearing <= 90; const isEast = br >= 0; // Needed to invert the x axis later. const angle = isNorth ? absoluteBearing : 90 - (absoluteBearing - 90); Now that we have a normalized angle, we can calculate the intersection point of the bearing with the sides of the map, using trigonometry and the relation between the sides of a right triangle - tan (angle) = opposite / adjacent. Therefore the x coordinate of the intersection point is given by x = tan(angle) * h where h is half the height of the map since we're working on one quadrant. The y coordinate is given by y = w / tan (angle) where w is half the width of the map. // const degToRad = (deg) => (deg * Math.PI) / 180; let x = Math.tan(degToRad(angle)) * h; let y = w / Math.tan(degToRad(angle)); Animation of the triangles whose sides (dotted) we have to calculate. The x value is given by the orange triangle and the y value is given by the purple triangle. As shown on the animation above the intersection point may be outside the map boundaries, therefore we have to clamp the values of x and y to the width and height of the map. Once we clamp the values, the width and height may vary at will and our point will always be in the correct position. Now that we have the coordinates we need on invert them according to the original bearing. The x is positive on the east and negative on the west and the y is negative on the north and positive on the south. The y is inverted because the y axis is inverted on the screen. x = isEast ? x : -x; y = isNorth ? -y : y; The last thing to do is calculate the final coordinates of the direction marker. We do this by adding, to the center of the map, the value of the intersection point accounting for any padding we may want have. For example: // const clamp = (value, min, max) => Math.max(min, Math.min(max, value)); const markerX = clamp(w + x, padding, w * 2 - padding); const markerY = clamp(h + y, padding, h * 2 - padding); With the coordinates calculated we can create the marker and position it on the map. Creating the square markers is pretty simple. We start with a square div, and round all the corners except the top left one. We then rotate this square div by 45° (to ensure the corner points north), plus the bearing to point at our point of interest. Inside the square div, we add another div with the content (an icon in this case), to which we apply the inverse rotation to keep it upright. Steps of the marker construction as detailed above The last thing to do is add the x and y translation to the marker and ensuring that the map and marker container has position: relative to make the positioning work. function IconMarker(props) { const { angle, x, y, inViewport, children } = props; return ( position: 'absolute', zIndex: 100, background: 'blue', width: '2rem', height: '2rem', borderRadius: '999px', borderTopLeftRadius: 0, display: inViewport ? 'none' : 'flex', alignItems: 'center', justifyContent: 'center', transform: `translate(${x}px, ${y}px) rotate(${angle + 45}deg)` <div transform={`rotate(${-angle - 45}deg)`}>{children}</div> Implementing directional markers for off-screen points of interest enhances user experience and solves a common challenge in digital cartography. By leveraging trigonometry and CSS transforms, we've created an intuitive solution that guides users through complex map interfaces. While we've applied this technique to the SatSummit map, these principles can be adapted to various mapping projects. Speaking of SatSummit, we'd love to see you there! Join us on November 18 and 19, 2024 in Lisbon, Portugal, to explore how satellite data can address critical global challenges. Register now to be part of this exciting confluence of technology and development. Who knows? You might even get to test our directional markers in person! Have a sticky map application challenge? We'd love to help.	{"url":"https://developmentseed.org/blog/2024-10-01-directional-markers/","timestamp":"2024-11-05T00:14:23Z","content_type":"text/html","content_length":"565925","record_id":"<urn:uuid:90d5ab66-3e43-4be0-b8ed-d0a43b4493fd>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00044.warc.gz"}
• Русский • English OPS video-demo OPS Optimizing parallelizing system is a program tool oriented for development of 1. parallelizing compilers, parallel language optimizing compilers, semi-automatic parallelizing systems; 2. electronic circuits computer-aided design systems; 3. systems of automatic design of hardware based on FPGA. An OPS is focused on different target parallel architectures. OPS development group investigates new optimizing transformations and new compilation possibilities. Projects based on OPS: 1. IHOP - Interactive High-level Optimizing Parallelizer 2. OPS Demo 3. WebOPS tool - web interface to some features of OPS OPS Structure • C (clang) • Fortran Code generation. • Transformed C code. • C + MPI. • C + ParDo loops marks. • Automatic estimation of measure of inaccuracy. Program graph models. • Dependence graph. • Lattice graph. • Computation graph. • Control flow graph. • Call graph. Internal representation. Internal representation is oriented on parsers from different languages (C, FORTRAN, Pascal) usability, program transformations development flexibility and multiple architecture code generation Transformations Library. Implemented transformations automatically check conditions of equivalence based on dependence graph or lattice graph. Not all program transformations check semantic correctness of applicability. It can lead to an error of the system and it should be resolved in a future versions. Linear program fragments and expressions transformations. • Mixed computation in arithmetic expressions. • Substitution forward and constant propagation. • Scalar variables renaming • Dead code elimination. • Swap statements One-dimensional loops transformations. • Loop unrolling. • Loop canonization. • Loop distribution. • Loop merging. • Removal of loop invariant. • The splitting of the vertices (declaration of additional arrays). • Stretching scalars (array replacing the scalar variable). • Removal of induction variables. Loops with linear recurrence transformations. • Replacement of loop with two linear recurrences contain only constant coefficients onto loop without recurrence. • Replacement of loop with one affined recurrence contain only constant coefficients onto loop without recurrence. • Replacement of loop which calculates the sum of the series with polynomial general term onto a single formula. Loop nests transformations. • Loops interchange. • Tiling. • Non-unimodular perfect loop nests transformations. • Loop splitting. • Induction variables substitution • Loop nests induction variables renaming. Loop into loop nests transformations. • Loop nesting • Strip mining. OPS transformations use service functions such as: • Insert group of statements • Delete group of statements • Replace group of statements • Substitution of subexpression into expression • Delete expression • Generate new declaration (variable, array, label, etc.) Research interests of the OPS group are • Automatic program parallelization. • Multipipeline calculations • Program transformations based on lattice graph • Automatic control of calculations inaccuracy • Automatization of program transformations development • Automatization of electronic circuit design OPS developers team and some other participants of our seminar (summer 2004): (Top row from left to right) Zinovij Nis, Konstantin Gufan, Alexander Shulzhenko, Roman Morilev, Victor Petrenko, Mihail Shilov, Milena Eremeeva, Maria Tzibulina, Sergej Naumenko, Alexander Butov, Polina Shaternikova. (Second row from left to right) Oleg Steinberg, Alexander Tuzaev, Roman Steinberg, Denis Cherdanzev, Ludmila Mironova (From left to right) Victor Petrenko, Roman Steinberg, Sergej Naumenko, Boris J. Steinberg, Denis Cherdanzev, Alexander Shulzhenko, Alexander Butov	{"url":"https://ops.rsu.ru:443/en/about.shtml","timestamp":"2024-11-02T02:25:35Z","content_type":"text/html","content_length":"11961","record_id":"<urn:uuid:13ee2c71-6e98-4709-94c1-febc153994ea>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00192.warc.gz"}
3.1 Simple Interest: Principal, Rate, Time • Understand the concept of simple interest. • Calculate interest, principal, rate, and time for simple interest transactions. Interest is the fee paid by a borrower to a lender for using the lender’s money. As a borrower, the interest paid is an expense. But as a lender or an investor, interest earned is income. There are two types of interest: simple interest and compound interest. In a simple interest environment, interest is calculated solely on the initial amount of money invested or borrowed at the beginning of the transaction at a specified simple interest rate for the entire time period. A loan or investment always involves two parties—one giving and one receiving. No matter which party you are in the transaction, the amount of interest remains unchanged. The only difference lies in whether you are earning or paying the interest. Suppose $1,000 is placed into an account with 12% simple interest for a period of 12 months. For the entire term of this transaction, the amount of money in the account always equals $1,000. During this period, interest accrues at a rate of 12%, but the interest is never placed into the account. When the transaction ends after 12 months, the $120 of interest and the initial $1,000 are then combined to total $1,120. The Simple Interest Formula The amount of simple interest ([latex]I[/latex]) is calculated as a percent of the amount of money invested or borrow over a specified period of time. [latex]\begin{eqnarray} I & = & P \times r \times t \end{eqnarray}[/latex] • [latex]I[/latex] is the amount of interest. This is the dollar amount of interest paid or received. • [latex]P[/latex] is the present value or principal. This is the amount borrowed or invested at the beginning of the time period. • [latex]r[/latex] is the simple interest rate. This is the rate of interest that is charged or earned during a specified time period. The simple interest rate is expressed as a percent for a given time period, usually annually or per year, unless otherwise specified. • [latex]t[/latex] is the period of time or term of the investment or loan. The time period is the length of time for which interest is earned or charged on the investment or loan respectively. Recall that algebraic equations require all terms to be expressed with a common unit. This principle remains true for the simple interest formula, particularly with regard to the interest rate and the time period. For example, if you have a 3% annual simple interest rate for nine months, then either • The time needs to be expressed annually as [latex]\frac{9}{12}[/latex] of a year to match the yearly interest rate, or • The interest rate needs to be expressed monthly as [latex]\frac{3\%}{12}=0.25\%[/latex] per month to match the number of months. It does not matter which conversion you do as long as you express both the interest rate and the time in the same unit. If one of these two variables is your algebraic unknown, the unit of the known variable determines the unit of the unknown variable. For example, assume that you are solving the simple interest formula for the time period. If the interest rate used in the formula is annual, then the time period is expressed in number of years. Julio borrowed $1,100 from Maria five months ago. When he first borrowed the money, they agreed that he would pay Maria 5% simple interest. If Julio pays her back today, how much interest does he owe Step 1: The given information is [latex]\begin{eqnarray} P & = & \$1,100 \\ r & = & 5\% \mbox{ (per year)} \\ & = & 0.05 \\ t & = & 5 \mbox{ months}\end{eqnarray}[/latex] Step 2: The interest rate is annual, but the time is in months. Convert the time period from months to years. Step 3: Solve for the amount of interest, [latex]I[/latex]. Figure 3.1.1 [latex]\begin{eqnarray} I & = & P \times r \times t \\ & = & 1,100 \times 0.05 \times \frac{5}{12} \\ & = & \$22.92 \end{eqnarray}[/latex] For Julio to pay back Maria, he must reimburse her for the $1,100 principal borrowed plus an additional $22.92 of simple interest as per their agreement. Solving for [latex]P[/latex], [latex]r[/latex], or [latex]t[/latex] Four variables are involved in the simple interest formula [latex]I =P \times r \times t[/latex], which means that any three can be known quantities and require you to solve for the fourth missing What amount of money invested at 6% annual simple interest for 11 months earns $2,035 of interest? Step 1: The given information is [latex]\begin{eqnarray} I & = & \$2,035 \\ r & = & 6\% \mbox{ (per year)} \\ & = & 0.06 \\ t & = & 11 \mbox{ months}\end{eqnarray}[/latex] Step 2: The interest rate is annual, but the time is in months. Convert the time period from months to years. Step 3: Solve for the principal [latex]P[/latex]. Figure 3.1.2 [latex]\begin{eqnarray} P & = & \frac{I}{ r \times t }\\ & = & \frac{2,035} { 0.06 \times \frac{11}{12}} \\ & = & \$37,000 \end{eqnarray}[/latex] To generate $2,035 of simple interest at 6% over a time frame of 11 months, $37,000 must be invested. For how many months must $95,000 be invested to earn $1,187.50 of simple interest at an interest rate of 5%? Step 1: The given information is [latex]\begin{eqnarray} P & = & \$95,000 \\ I & = & \$1,187.50 \\ r & = & 5\% \mbox{ (per year)} \\ & = & 0.05 \end{eqnarray}[/latex] Step 2: Solve for the time period [latex]t[/latex]. Figure 3.1.3 [latex]\begin{eqnarray} t & = & \frac{I}{P \times r }\\ & = & \frac{1,187.50} { 95,000 \times 0.05} \\ & = & 0.25 \end{eqnarray}[/latex] Step 3: The time period found in step 2 is in years, so convert the years to months by multiplying by 12. [latex]\begin{eqnarray} \mbox{Time period in months} & = & 0.25 \times 12 \\ & = & 3 \end{eqnarray}[/latex] For $95,000 to earn $1,187.50 at 5% simple interest, it must be invested for a three-month period. If you want to earn $1,000 of simple interest at a rate of 7% in a span of five months, how much money must you invest? Click to see Solution [latex]\begin{eqnarray} P & = & \frac{I}{r \times t} \\ & = & \frac{1000}{0.07 \times \frac{5}{12}} \\ & = & \$34,285.71 \end{eqnarray}[/latex] If you placed $2,000 into an investment account earning 3% simple interest, how many months does it take for you to have $2,025 in your account? Click to see Solution [latex]\begin{eqnarray} t & = & \frac{I}{ P \times r} \\ & = & \frac{25}{2000 \times 0.03} \\ & = & 0.4166... \\ \\ \mbox{Time in months} & = & 0.04166... \times 12 \\ & = & 5 \end{eqnarray}[/ A $3,500 investment earned $70 of interest over the course of six months. What annual rate of simple interest did the investment earn? Click to see Solution [latex]\begin{eqnarray} r & = & \frac{I}{ P \times t} \\ & = & \frac{70}{3500 \times \frac{6}{12}} \\ & = & 0.04 \\ & = & 4\% \end{eqnarray}[/latex] Time and Dates In the examples of simple interest so far, the time period was given in months. While this is convenient in many situations, financial institutions and organizations calculate interest based on the exact number of days in the transaction, which changes the interest amount. To illustrate this, assume you had money saved for the entire months of July and August, where [latex]t = \frac{2}{12}[/latex] or [latex]t = 0.16666...=0.1\overline{6}[/latex] of a year. However, if you use the exact number of days, the 31 days in July and 31 days in August total 62 days. In a 365-day year that is [latex]t =\frac{62}{365}[/latex] or t = 0.169863 of a year. Notice a difference of 0.003196 (0.169863 – 0.16) occurs. Therefore, to be precise in performing simple interest calculations, you must calculate the exact number of days involved in the transaction. Although you can count the exact number of days by hand, most financial calculators and spreadsheet software like Excel have built-in functions to calculate the number of days between two dates. To count the number of days between two dates: 1. Press 2nd DATE (the 1 button) to enter the date worksheet. 2. At the DT1 screen, enter the first date. Dates are entered in the form mm.ddyy. For example, enter May 19, 2023 as 05.1923. After entering the date press ENTER. 3. Press the down arrow. 4. At the DT2 screen, enter the second date. Dates are entered in the form mm.ddyy. For example, enter August 7, 2023 as 08.0723. After entering the date press ENTER. 5. Press the down arrow. 6. At the DBD screen press CPT to calculate the number of days between the two entered dates. For example, there are 80 days between May 19, 2023 and August 7, 2023. [Calculating Dates and Days (Month-Day-Year) by Joshua Emmanuel [3:32] (transcript available).] When solving for [latex]t[/latex], decimals may appear in your solution. For example, if calculating [latex]t[/latex] in days, the answer may show up as 45.9978 or 46.0023 days. However, interest is calculated only on the complete number of days. This occurs because the interest amount, [latex]I[/latex], used in the calculation has been rounded off to two decimals. Because the interest amount is imprecise, the calculation of [latex]t[/latex] is imprecise. When this occurs, round [latex]t[/latex] off to the nearest integer. Mark borrowed $4,200 from his friend Chris on November 3, 2022. Their agreement required that Mark pay back the loan on April 24, 2023 with 8% simple interest. How much interest did Mark pay Chris? Step 1: The given information is [latex]\begin{eqnarray} P & = & \$4,200 \\ r & = & 8\% \mbox{ (per year)} \\ & = & 0.08 \\ t & = & \frac{294}{365} \end{eqnarray}[/latex] Step 2: Solve for the intererst. [latex]\begin{eqnarray} I & = & P \times r \times t \\ & = & 4200 \times 0.08 \times \frac{294}{365} \\ & = & $270.64 \end{eqnarray}[/latex] Mark pays Chris $270.64 in interest. On September 13, 2011, Aladdin decided to pay back the Genie on his loan of $15,000 at 9% simple interest. If he paid the Genie the principal plus $1,283.42 of interest, on what day did he borrow the money from the Genie? Step 1: The given information is [latex]\begin{eqnarray} P & = & \$15,000 \\ I & = & \$1,283.42 \\ r & = & 9\% \mbox{ (per year)} \\ & = & 0.09 \end{eqnarray}[/latex] Step 2: Solve for the time period. [latex]\begin{eqnarray} t & = & \frac{I}{P \times r }\\ & = & \frac{1,283.42} { 15,000 \times 0.09} \\ & = & 0.95068... \end{eqnarray}[/latex] Step 3: The time period found in step 2 is in years, so convert the years to days by multiplying by 365. [latex]\begin{eqnarray} \mbox{Time period in days} & = & 0.95068... \times 365 \\ & = & 346.998... \\ & \rightarrow& 347 \end{eqnarray}[/latex] Step 4: Use the DATE function to calculate the start date (DT1). 1. Press 2nd DATE. 2. Press the down arrow to skip the DT1 screen and move to the DT2 screen. 3. At the DT2 screen enter 09.1311 for September 13, 2011. Press ENTER. 4. Press the down arrow. 5. At the DBD screen, enter 347 and press ENTER. 6. Press the up arrow twice to return the the DT1 screen. 7. At the DT1 screen, press CPT. The date is October 1, 2010. If Aladdin owed the Genie $1,283.42 of simple interest at 9% on a principal of $15,000, he must have borrowed the money 347 days earlier, which is October 1, 2010. Brynn borrowed $25,000 at 1% per month from a family friend to start her entrepreneurial venture on December 2, 2011. If she paid back the loan on June 16, 2012, how much simple interest did she pay? Click to see Solution [latex]\begin{eqnarray} I & = & P \times r \times t \\ & = & 25,000 \times 0.12 \times \frac{197}{365} \\ & = & \$1,619.18 \end{eqnarray}[/latex] If $6,000 principal plus $132.90 of simple interest was withdrawn on August 14, 2011, from an investment earning 5.5% interest, on what day was the money invested? Click to see Solution [latex]\begin{eqnarray} t & = & \frac{I}{P \times r} \\ & = & \frac{132.90}{6000 \times 0.055} \\ & = & 0.40272... \\ \\ \mbox{Time in days} & = & 0.40272... \times 365 \\ & = & 146.9954... \\ & \ rightarrow & 147 \end{eqnarray}[/latex] The money was invested on March 20, 2011. 1. Brynn borrowed $25,000 at 1% per month from a family friend to start her entrepreneurial venture on December 2, 2011. If she paid back the loan on June 16, 2012, how much simple interest did she Click to see Answer 2. How much simple interest is earned on $50,000 over 320 days if the interest rate is: a. 3% b. 6% c. 9% Click to see Answer a. $1,315.07, b. $2,630.14, c. $3,945.21 3. If you placed $2,000 into an investment account earning 3% simple interest, how many months does it take for you to have $2,025 in your account? Click to see Answer 5 months 4. If you want to earn $1,000 of simple interest at a rate of 7% in a span of five months, how much money must you invest? Click to see Answer 5. If $6,000 principal plus $132.90 of simple interest was withdrawn on August 14, 2011, from an investment earning 5.5% interest, on what day was the money invested? Click to see Answer March 20, 2011 6. Jessica decided to invest her $11,000 in two back-to-back three-month term deposits. On the first three-month term, she earned $110 of interest. If she placed both the principal and the interest into the second three-month term deposit and earned $145.82 of interest, how much higher or lower was the interest rate on the second term deposit? Click to see Answer 1.25% higher 7. Marrina is searching for the best way to invest her $10,000. One financial institution offers 4.25% on three-month term deposits and 4.5% on six-month term deposits. Marrina is considering either doing two back-to-back three-month term deposits or just taking the six-month deposit. It is almost certain that interest rates will rise by 0.5% before her first threemonth term is up. She will place the simple interest and principal from the first three-month term deposit into the second three-month deposit. Which option should Marrina pursue? How much better is your recommended Click to see Answer Back-to-back 3 month investments, $1.26 more 8. Evaluate each of the following $10,000 investment alternatives and recommend the best alternative for investing any principal regardless of the actual amount. Assume in all cases that the principal and simple interest earned in prior terms are placed into subsequent investments. □ Alternative 1: 6% for 1 year □ Alternative 2: 5% for 6 months, then 7% for 6 months □ Alternative 3: 5.25% for 3 months, then 5.75% for 3 months, then 6.25% for 3 months, then 6.75% for 3 months What percentage more interest is earned in the best alternative versus the worst alternative? Click to see Answer Alternative 3, 0.1283% more than Alternative 1 “8.1: Simple Interest: Principle, Rate, Time” from Business Math: A Step-by-Step Handbook Abridged by Sanja Krajisnik; Carol Leppinen; and Jelena Loncar-Vines is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. “8.1: Principal, Rate, Time” from Business Math: A Step-by-Step Handbook (2021B) by J. Olivier and Lyryx Learning Inc. through a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License unless otherwise noted.	{"url":"https://ecampusontario.pressbooks.pub/businessfinancialmath/chapter/3-1-simple-interest-principal-rate-time/","timestamp":"2024-11-05T00:32:02Z","content_type":"text/html","content_length":"102855","record_id":"<urn:uuid:b5a332d5-e7a2-46c4-a3af-9cb123828822>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00529.warc.gz"}
The project profitability index is used to compare NPV analysis is used to help determine how much an investment, project, or any Finally, IRR does not consider cost of capital and can't compare projects with The Profitability Index (PI) measures the ratio between the present value of present value, internal rare of returns, profitability index, and net terminal value (ii ) Key words: Capital budgeting Techniques, Project Profitability of project evaluation seem to be adequate and effective for use only in a riskless accepting or rejecting a project by merely comparing the rate of return with the cost. 5 Mar 2019 real estate projects in areas or buildings through modifications to the (local) GRP. the definition of a profitability index (rate of return) to be used in the income loss if the extraordinary contribution is too low in comparison to 13 May 2019 Profitability Index (PI) is a capital budgeting technique to evaluate the investment The method used for arriving at profitability index of a proposed project is Relative profitability allows comparison of two investments A project has an initial investment of $100,000 and a profitability index of 1.15. D- use the average of the multiple rates to compare to the firm's cost of capital. The Profitability Index (PI) or profit investment ratio (PIR) is a widely used measure for evaluating viability and profitability of an investment project. It is calculated 23 Oct 2016 The profitability index helps make it possible to directly compare the NPV of one project to Now we have to calculate the net present value of the project by The next step is to use the information from the net present value Profitability index method measures the present value of benefits for every dollar investment. It involves the ratio that is created by comparing the ratio of the present value of future cash flows from a project to the initial investment in the project. 24 Jul 2013 It is a tool for measuring profitability of a proposed corporate project (also called corporate project by comparing the cash flows created by the project to the capital Use the following formula to calculate profitability index:. Note: Your browser must support JavaScript in order to use this quiz. 1. A profitability index of .85 for a project means that: the present value of benefits is 85% How to Use the Profitability Index. A Capital Budgeting Method to Evaluate a Proposed Investment Project. 20 Apr 2019 Profitability Index is a capital budgeting tool used to compare different projects based on the net present value added by each project per $1 of A project which requires an initial cash outlay and for which all remaining cash flows are May lead to incorrect decisions when comparing mutually exclusive investments. 4. E) Profitability index Use the following to answer question 10: . The Profitability Index (PI) or profit investment ratio (PIR) is a widely used measure for evaluating viability and profitability of an investment project. It is calculated 23 Oct 2016 The profitability index helps make it possible to directly compare the NPV of one project to Now we have to calculate the net present value of the project by The next step is to use the information from the net present value Cash flows used for the profitability index specifically relate to project cash flows. Compare the internal rate of return from each cash flow to its risk. b. Compute the investment, but other criteria may be used according to the project's The payback period gives a different perspective of the project comparing with NPV and IRR. It tells us When analysing a single project the Profitability Index doesn ´t. A project which requires an initial cash outlay and for which all remaining cash flows are May lead to incorrect decisions when comparing mutually exclusive investments. 4. E) Profitability index Use the following to answer question 10: . Capital Rationing and Profitability Index. In the previous few articles we have come across different metrics that can be used to choose amongst competing projects 14 Feb 2019 We can also use the profitability index to compare them. The profitability index measures the amount of profit returned for each dollar invested in a Capital Rationing and Profitability Index. In the previous few articles we have come across different metrics that can be used to choose amongst competing projects If the IRR of a project is 8%, its NPV, using a discount rate, k, greater than 8%, will be compare the profitability index of these investments to those of other possible the internal rate of return method can be used with reasonable confidence. profitability index ignores the: Scale of the investment If a project has an NPV 1.15 In an effort to compare projects the _____ should be used for reinvesting The project profitability index is used to compare the net present values of two investments that require different amounts of investment funds. true. The project profitability index is computed by dividing the present value of the cash inflows of the project by present value of the cash outflows of the project. The project profitability index is used to compare the net present values of two investments that require different amounts of investment funds true When discounted cash flow methods of capital budgeting are used, the working capital required for a project is ordinarily counted as a cash inflow at the beginning of the project and as a cash The project profitability index is used to compare the internal rates of return of two companies with different investment amounts. false Preference decisions attempt to determine which of many alternative investment projects would be the best for the company to accept. present value, internal rare of returns, profitability index, and net terminal value (ii ) Key words: Capital budgeting Techniques, Project Profitability of project evaluation seem to be adequate and effective for use only in a riskless accepting or rejecting a project by merely comparing the rate of return with the cost. Cash flows used for the profitability index specifically relate to project cash flows. Compare the internal rate of return from each cash flow to its risk. b. Compute the investment, but other criteria may be used according to the project's The payback period gives a different perspective of the project comparing with NPV and IRR. It tells us When analysing a single project the Profitability Index doesn ´t. A project which requires an initial cash outlay and for which all remaining cash flows are May lead to incorrect decisions when comparing mutually exclusive investments. 4. E) Profitability index Use the following to answer question 10: . If the IRR of a project is 8%, its NPV, using a discount rate, k, greater than 8%, will be compare the profitability index of these investments to those of other possible the internal rate of return method can be used with reasonable confidence. profitability index ignores the: Scale of the investment If a project has an NPV 1.15 In an effort to compare projects the _____ should be used for reinvesting The project profitability index is used to compare the net present values of two investments that require different amounts of investment funds. true. The project profitability index is computed by dividing the present value of the cash inflows of the project by present value of the cash outflows of the project. The project profitability index is used to compare the net present values of two investments that require different amounts of investment funds true When discounted cash flow methods of capital budgeting are used, the working capital required for a project is ordinarily counted as a cash inflow at the beginning of the project and as a cash The project profitability index is used to compare the internal rates of return of two companies with different investment amounts. false Preference decisions attempt to determine which of many alternative investment projects would be the best for the company to accept. The project profitability index is used to compare the internal rates of return of two companies with different investment amounts False Preference decisions attempt to determine which of many alternative investment projects would be the best for the company to accept. The project profitability index is used to compare the net present values of two investments that require different amounts of investment funds. True False. True. When making preference decisions about competing investment proposals, the internal rate of return is superior to the project profitability index. True False. False.	{"url":"https://optionseoznm.netlify.app/leclaire40459gy/the-project-profitability-index-is-used-to-compare-12","timestamp":"2024-11-15T04:53:51Z","content_type":"text/html","content_length":"34306","record_id":"<urn:uuid:451ec811-b717-4668-b36b-891b819ae4f2>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00334.warc.gz"}
Hardness of Approximating Constraint Satisfaction Problems and Their Variants in Presence of Additional Structural Assumptions \| KTH Hardness of Approximating Constraint Satisfaction Problems and Their Variants in Presence of Additional Structural Assumptions Time: Thu 2023-06-01 14.00 Location: F3, Lindstedtsvägen 26 & 28, Stockholm Video link: https://kth-se.zoom.us/j/63899660090 Language: English Subject area: Computer Science Doctoral student: Aleksa Stankovic , Matematik (Inst.), Approximability and Proof Complexity Opponent: Professor Luca Trevisan, Bocconi University Supervisor: Professor Johan Håstad, SFW Matematik för Data och AI; Associate Professor Per Austrin, Teoretisk datalogi, TCS QC 2023-05-08 This thesis studies how the approximability of some fundamental computational problems is affected by some additional requirements on the structure of the inputs. The problems studied in this thesis belong or are closely related to constraint satisfaction problems (CSPs), which are considered to be one of the most fundamental problems in theoretical computer science. The first class of problems studied in this thesis consists of some Boolean CSPs with cardinality constraints. A cardinality constraint for an instance of a Boolean CSP restricts assignments to have a certain number of variables assigned to 1 and 0. Assuming the Unique Games Conjecture, we show that Max-Cut with cardinality constraints is hard to approximate within approximately 0.858, and that Max-2-Sat with cardinality constraints is hard to approximate within approximately 0.929. The same inapproximability ratio as for cardinality constrained Max-2-Sat is obtained for the cardinality-constrained Vertex Cover problem, known as Max-k-VC. We also examine regular constraint satisfaction problems where each variable in an instance has the same number of occurrences. We investigate approximability of such problems and show that for any CSP Λ, the existence of an α-approximation algorithm for regular Max-CSP Λ implies the existence of an (α − o(1))-approximation algorithm for weighted Max-CSP Λ for which the regularity of instances is not imposed. We also give an analogous result for Min-CSPs. In particular, if one is not interested in lower-order terms, we show that the study of the approximability of CSPs can be conducted solely on regular instances. We also consider approximability of Max-3-Lin problems over non-Abelian groups in a universal factor graph setting. The factor graph of an instance with n variables and m constraints is the bipartite graph between [m] and [n], in which edges connect constraints with the variables they contain. The universal factor graph setting assumes that a factor graph for an instance is fixed for each input size. Building on the previous works, we show optimal inapproximability of Max-3-Lin problems over non-Abelian groups in the universal factor graph setting, both in the case of perfect and almost perfect completeness. We also show that these hardness results apply in the setting in which linear equations in the Max-3-Lin problem are restricted to the form x · y · z = g, where x, y, z are variables and $g$ is a group element, in contrast to previous works where constants have appeared on the left side of the equations as well. Finally, we study the approximability of the Minimum Sum Vertex Cover problem, in which we are given a graph as an input and the goal is to find anordering of vertices that minimizes the total cover time of edges. An edge is covered when one of its endpoints is visited in an ordering, and the covertime for the edge is exactly the time that the endpoint is visited. In this work, we give the first explicit hardness of approximation result and show that Minimum Sum Vertex Cover can not be approximated below 1.0748, assuming the Unique Games Conjecture. Furthermore, we study Minimum Sum Vertex Cover on regular graphs and demonstrate an inapproximability ratio of 1.0157. By revisiting an algorithm introduced by Feige, Lov\'{a}sz, and Tetali, we also show approximability within 1.225 for regular graphs.	{"url":"https://www.kth.se/math/kalender/hardness-of-approximating-constraint-satisfaction-problems-and-their-variants-in-presence-of-additional-structural-assumptions-1.1253761?date=2023-06-01&orgdate=2023-03-19&length=1&orglength=0","timestamp":"2024-11-08T11:45:53Z","content_type":"text/html","content_length":"59751","record_id":"<urn:uuid:7d05ea47-5d2e-4cec-b911-7f17ef3b57ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00897.warc.gz"}
(PDF) A line-of-sight channel model for the 100–450 gigahertz frequency band Content available from EURASIP Journal on Wireless Communications and Networking A line‑of‑sight channel model forthe100– 450 gigahertz frequency band Joonas Kokkoniemi* , Janne Lehtomäki and Markku Juntti 1 Introduction e high-frequency communications aim at ﬁnding large contiguous bandwidths to serve high data rate applications and services. Especially the millimeter wave (mmWave) frequencies (30–300 GHz) are among the most prominent to provide high data rate connectivity in ﬁfth generation (5G) and beyond (B5G) systems [1–4]. In this context, the 5G systems will utilize the below 100 GHz frequencies, whereas the B5G systems, including the visioned sixth generation (6G) systems, will look for spectral resources also above 100 GHz [3]. ese frequencies would theoretically allow very large band- widths, but there are still many challenges to reach the above 100 GHz band eﬃciently with compact and portable devices. To overcome the challenges in conquering these fre- quencies, there have been and are ongoing a lot of research eﬀorts towards understand- ing the propagation channels, beamforming challenges, and transceiver hardware. For instance, EU Horizon 2020 projects TERRANOVA [5] for the low THz frequencies 275– 325 GHz, and ARIADNE [6] for the D band (110–170 GHz). Also, the ﬁrst standards for the THz communications are appearing, such as IEEE 802.15.3d [7]. us, the utilization This paper documents a simple parametric polynomial line-of-sight channel model for 100–450 GHz band. The band comprises two popular beyond ﬁfth generation (B5G) frequency bands, namely, the D band (110–170 GHz) and the low-THz band (around 275–325 GHz). The main focus herein is to derive a simple, compact, and accurate molecular absorption loss model for the 100–450 GHz band. The derived model relies on simple absorption line shape functions that are ﬁtted to the actual response given by complex but exact database approach. The model is also reducible for particular sub-bands within the full range of 100–450 GHz, further simplifying the absorption loss estimate. The proposed model is shown to be very accurate by benchmarking it against the exact response and the similar models given by International Telecommu- nication Union Radio Communication Sector. The loss is shown to be within ±2 dBs from the exact response for one kilometer link in highly humid environment. Therefore, its accuracy is even much better in the case of usually considered shorter range future B5G wireless systems. Keywords: Absorption loss, THz channel modeling, THz communications, THz Open Access © The Author(s) 2021. Open AccessThis article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the mate- rial. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http:// creat iveco mmons. org/ licen ses/ by/4. 0/. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 Centre for Wireless Communications (CWC), University of Oulu, P.O. Box 4500, 90014 Oulu, Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 of the +100 GHz frequencies for the near future wireless communication systems looks very promising. One of the most important research topics on new frequency bands, knowledge of the operational channels is in the focal point to understand the fundamental physical limits of the transmission platform. is paper considers the line-of-sight (LOS) propagation in the sub-THz and low-THz frequencies at frequency range from 100 to 450 GHz.1 e main goal of this paper is to give tools to model the molecular absorption loss with a simple model that has minimal loss in accuracy to full line-by-line models. e molecu- lar absorption loss is caused by the energy of the photons being absorbed by the free energy states of the molecules [9]. e absorption loss is described by the Beer–Lambert law, and it causes exponential frequency selective loss on the signals as a function the frequency. e lowest absorption lines lie at low mmWave frequencies [10], but the ﬁrst major absorption lines appear above 100 GHz. e molecular absorption loss is most often modeled by line-by-line models for which the parameters are obtained from spectroscopic databases, such as high-resolu- tion transmission molecular absorption database (HITRAN) [10]. e work herein uti- lizes the spectroscopic databases by obtaining the parameters for the major absorption lines, and we simplify those by simple polynomials that only depend on the water vapor content in the air. ese are then applied to the Beer–Lambert’s law to obtain distance dependent absorption loss. e free space propagation is modeled by the square-law free space path loss (FSPL). us, the produced model is a simple and a relatively compact way to estimate the total free space loss on the above 100 GHz frequencies. e main use case of the produced model is to be able to omit the complicated spectroscopic data- bases that take eﬀorts to implement and use ﬂexibly. is is especially the case with the common wireless communications problems where detailed information on the source of the loss is not required, but just an easy way to model it. Starting from the 100 GHz frequency, we model six absorption lines at about 119 GHz, 183 GHz, 325 GHz, 380 GHz, 439 GHz, and 448 GHz. is adds two lines at 119 GHz and 183 GHz to our previous model ([8]) in order to address the D band propagation. Water vapor is the main cause of the absorption losses in the above 100 GHz frequencies and all but one of the above six lines are caused by it. Absorption at 119 GHz is caused by oxygen, and it is comparably weak. Although weak, it has been included in the model, since it is part of the D band and it causes a small attenuation on long distance links. ere exist a lot of research on the line-by-line models and models for calculating the absorption spectrum, such as [9, 11–14]. ere are also some existing works on para- metric absorption loss models. International Telecommunication Union Radio Com- munication Sector (ITU-R) has provided a model to calculate gaseous attenuation up to 1000 GHz in ITU-R P.676-8 [15]. is model is line-by-line based, and its output is therefore matched with those of the full spectroscopic databases. ere is, however, a diﬀerence to the proposed model in this paper: ITU-R uses a modiﬁed full Lorentz line shape function that is not in general recommended for the millimeter frequencies [11] due to heavy tailed frequency domain absorption distribution. A better choice is a model 1 is paper is an invited extended version of the conference paper presented in the EuCNC’19 conference [8]. Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 that takes into account lower wing absorption by using line shape such as van Vleck– Weisskopf or van Vleck–Huber [11]. Furthermore, the full model by ITU-R still requires large numbers of tabulated parameters (553) that render its utilization similarly slow as the full databases. In [15], a polynomial-based approximation is also given. It is valid up to 350 GHz, but it is somewhat usable up to about 450 GHz. Newer version of this model, ITU-R 676-11, also exists but that version does not have a polynomial model. We use the older version in this paper as we present a similar (but more simpler) polynomial Compared to the proposed model, the ones presented in [15] have several weaknesses. e ITU-R models [15] include lines even up to 1780 GHz, but it is only speciﬁed to be valid for frequencies up to 350 GHz. e simpliﬁed model in the newer version is also limited to 350 GHz. e model also includes nine polynomials. If some of these terms are removed, they may also aﬀect frequencies in diﬀerent bands due to additive nature of the absorption lines. For example, the term involving 1780 GHz has to be kept or the attenuation levels between the peaks absorption frequencies at lower frequencies is incorrect. However, the ITU-R models are still fairly accurate below 450 GHz. Because of the Full Lorentz line shape model, they overestimate the absorption line wing absorp- tion. As detailed above, we will give a model with the extended frequency range and more accurate estimate for the absorption loss in simple form. is model can also be reduced to a simpler one (due to utilization of a ﬁt parameter) for a desired sub-band within the full range of the model (100–450 GHz). We have given a simpliﬁed molecular absorption loss model in the past in [16]. It was intended for the 275–400 GHz band. We also gave an extended version of that in [8] for frequencies from 200–450 GHz. is paper is an extended version of [8] with new lines focusing on the D band. As mentioned above, the main goal of this paper is to provide easy and accurate tools to estimate the LOS path loss above 100 GHz. e proposed model is shown to be very accurate by numerical results in Sect.3, where it is bench- marked against the line-by-line models as well as the ITU-R parametric models. e rest of this paper is organized as follows: Sect.2 derives the proposed absorption loss model, Sect.3 gives some numerical examples, and Sect.4 concludes the paper. 2 Simplied molecular absorption loss model 2.1 Molecular absorption loss e main goal of this paper is to provide a tool to easily model the molecular absorption loss. It is formally described by the Beer–Lambert law, which gives the transmittance, i.e., the fraction of energy that propagates through the medium at link distance d. is exponential power law depends on the link distance and absorption coeﬃcient by [9, 11] is the transmittance, f is the frequency, d is the distance from transmitter (Tx) to receiver (Rx), are the Tx and Rx powers, respectively, and is the absorption coeﬃcient for the jth type of molecule or its isotope at frequency f. e absorption coeﬃcient is usually calculated with databases of spectroscopic param- eters, such as the HITRAN database [10], GEISA [17], or JPL [18]. Detailed calculation Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 of the absorption coeﬃcient with line-by-line models can be found, e.g., in [9, 11, 16]. To summarize the line-by-line models based on the spectroscopic databases, the molecu- lar absorption coeﬃcient is calculated by calculating the eﬀective cross-sectional area of the individual molecules for absorption. is area depends on the absorption line shape functions for which the parameters are obtained from the spectroscopic databases. Finally, the cross-sectional areas of diﬀerent types of molecules are multiplied with the respective number densities to obtain the total absorption loss coeﬃcient. We derive the simpliﬁed absorption loss coeﬃcient expressions based on the theory described above. 2.2 Simplied absorption loss model e polynomial absorption loss model is obtained by searching the strongest absorp- tion lines on the band of interest and extracting the parameters for those from the spec- troscopic databases. e temperature and pressure dependent coeﬃcients are ﬁxed. As the absorption on the frequencies above 100 GHz is mainly caused by the water vapor, the volume mixing ratio of water vapor is left ﬂoating. e parametric model is charac- terized by the absorption coeﬃcients at absorption lines i. e above Beer–Lambert model becomes where f is the desired frequency grid, is an absorption coeﬃcient for the ith absorp- tion line, is a polynomial to ﬁt the expression to the actual theoretical response (detailed below), and is the volume mixing ratio of water vapor. It is determined from the relative humidity at temperature T and pressure p by is the partial pressure of water vapor and is the saturated water vapor partial pressure, i.e., the maximum partial pressure of water vapor in the air. is can be obtained, e.g., from the Buck equation [19] where the pressure p is given in hectopascals and T is given in degrees of centigrade. e six polynomials for the six major absorption lines at the 100–450 GHz band are the following2: w=6.1121(1.0007 +3.46 ×10−6p)exp y1(f,µ) = B(µ) + 2 Please note that in our conference version [8], to which this paper is an extension to, there was a typo that is rectiﬁed in this paper. e terms were not squared therein. is causes the model therein to give an incorrect output. However, this happens at so notable level that it should be obvious if one tries to implement the model and com- pares to our results. e numerical results in [8] were made with correct expressions. Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 where, c is the speed of light (m/s), the frequency f is given in Hertz, and . e lines , and correspond to strong absorption lines at center frequencies 119 GHz, 183 GHz, 325 GHz, 380 GHz, 439 GHz, and 448 GHz, respectively. is is also visible in the line expressions as the parameters give the line center frequencies in wavenumbers. ese parameters are accurate for the whole frequency band 100–450 GHz. However, slightly improved performance between the absorption lines below 200 GHz can be achieved by using value in the place of in (11). is only has minor impact on very long link distances, such as one kilometer and beyond link distances. y2(f,µ) = D(µ) + y3(f,µ) = F(µ) + y4(f,µ) = H(µ) + y5(f,µ) = J(µ) + y6(f,µ) = L(µ) + g(f,µ) = (2×10−4+af b) A(µ) =5.159 ×10− (1−µ)(−6.65 ×10− (1−µ) +0.0159) B(µ) =(−2.09 ×10−4(1−µ) +0.05)2, C(µ) =0.1925µ(0.1350µ+0.0318), D(µ) =(0.4241µ+0.0998)2, E(µ) =0.2251µ(0.1314µ+0.0297), F(µ) =(0.4127µ+0.0932)2, G(µ) =2.053µ(0.1717µ+0.0306), H(µ) =(0.5394µ+0.0961)2, I(µ) =0.177µ(0.0832µ+0.0213), J(µ) =(0.2615µ+0.0668)2, K(µ) =2.146µ(0.1206µ+0.0277), Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 e above absorption lines were estimated based on the simple Lorentz line shape. e reason is the simpler form as compared to more accurate, but at the same time more complex line shapes, such as the van Vleck–Huber line shape [12, 20]. is produces an error as the Lorentz line shape over estimates the absorption line wing absorption. erefore, the ﬁt polynomial is introduced. is ﬁt polynomial also takes care of the wing absorption in the case the model is only utilized partially. at is, if one only utilizes some of the lines to model a sub-band within the full 100– 450 GHz band, the ﬁt polynomial in (11) as in full model should always be included. It was obtained by curve ﬁtting to the diﬀerence between the exact response and the response of the above lines. It would be possible to calculate the exact diﬀerence theoretically, but would only apply to the in-band absorption lines and this would not consider the out-of-band wing absorption, mainly from lines above 450 GHz. e total absorption loss with the above model is shown to produce very accurate esti- mate of the loss in the numerical results. e water vapor volume mixing ratio is taken into account in the ﬁt polynomial based on the volume mixing ratio calculated from water vapor according to (3). Whereas it is highly accurate, this estimate will cause some error that is depend- ent on the water vapor level. Figure1 shows the error of the absorption coeﬃcient to the exact one based on the above absorption loss model and before applying the ﬁt polynomial. is error was calculated at 25 degrees centigrade and in various volume mixing ratios of water vapor = [0.0031 0.0094 0.0157 0.0220 0.0282] that corre- spond to relative humidities = [10% 30% 50% 70% 90%], respectively, at 298.15 K (25 degrees centigrade) temperature and at standard pressure 101,325 Pa. In this ﬁg- ure, taking into account the exponential y-axis, the error is small. However, the error increases as a function of frequency. is is due to the increasing and accumulating wing absorption from the higher frequency lines. is is the error the ﬁt polynomial rectiﬁes by adjusting the absorption lines shapes. e value 0.0157 in Difference to exact absorption coefficient RH = 90% RH = 70% RH = 50% RH = 30% RH = 10% Fig. 1 An error of the proposed absorption coeﬃcient. An error of the absorption coeﬃcient of the proposed model to the exact one as a function of the frequency for diﬀerent humidity levels before adding the ﬁt polynomial Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 comes from the design atmospheric conditions of 25 degrees centigrade and 50% rela- tive humidity at standard pressure. It should be noticed that the error is the smallest for lower humidities due to the fact that there is less water in the air, and thus, the overall diﬀerence between the exact and estimated absorption coeﬃcient is small. 2.3 FSPL andthetotal loss e total loss in pure LOS path requires the molecular absorption loss and the loss due to free space expansion of the waves. e FSPL is given by the Friis transmission We focus herein only on the free space propagation and thus the total LOS path loss is given by the FSPL and the molecular absorption loss as are the antenna gains. When using the polynomial models above, the absorption coeﬃcient where the are the above polynomial absorption lines (and as also shown in (2)), or subset of those depending on the modeled frequency band within the frequency range from 100–450 GHz. For instance, a D band propagation model would only require . Another popular band for high-frequency communications is the 275–325 GHz band. en, only the line would be enough. e ﬁt polynomial is always required and because of it we can use very low complexity models for the possible sub-bands, further pronouncing the complexity beneﬁts as compared to the ITU-R polynomial model. It will be shown in the numerical results that these subsets give very accurate estimate of the loss also in partial bands without a need to implement all the lines in the model. 3 Numerical results anddiscussion In this section, we ﬁrst present some performance analysis for the proposed molecular absorption loss model. is is done by analyzing the error produced by the model to the exact model, as well as comparing it to the ITU-R parametric and full models. After that, we analyze the accuracy of the model with reduced polynomials. Lastly, we give link budget calculations for some common +100 GHz frequency bands. 3.1 Error performance analysis We compare the path loss values of the proposed molecular absorption loss model ver- sus the ITU-R models in Figs.2, 3 and 4 for the relative humidity levels from 10% to 90%, respectively, at 25 degree centigrade for a one-kilometer link. A high link distance was (4πdf ) (4πdf ) a(f,µ) = yi(f,µ) +g(f,µ) Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 used to emphasize the diﬀerences between the models. is is because the impact of the molecular absorption loss decreases for short distances due to exponential power law. As it was predicted above, the Lorentz line shape (along with the full Lorentz line shape) overestimates the wing absorption. is is not a major issue at higher parts of the THz band due to more lines and line mixing. However, at the lower frequencies this is a problem because the Lorentz line shape does not attenuate the absorption wing response fast enough towards the zero frequency. As a consequence, the ITU-R models give higher path loss ﬁgures in general for below 500 GHz frequencies. e diﬀerence to the actual response varies from few dBs to tens of dBs depending on the link distance and humidity level. Notice that the simpliﬁed reduced version of the ITU-R model does not include all the lines leading to incorrect results. Fig. 2 Molecular absorption loss at 1 km distance and 10% relative humidity. Molecular absorption loss at 1 km distance at 25 degrees centigrade and 10% relative humidity ( Fig. 3 Molecular absorption loss at 1 km distance. Molecular absorption loss at 1 km distance at 25 degrees centigrade and 50% relative humidity ( Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 ere are a couple of further observations to be made. e ITU-R models are based on the full Lorentz model, but the database speciﬁc one does overestimate the response even more. is is due to reason that the ITU-R model is a modiﬁed version of the full Lorentz model that increases its accuracy. Second observation is that the proposed model is rather accurate, but not perfect. In Figs.2 to 4, the diﬀerence is the largest below 200 GHz. However, the large part of the apparent diﬀerence comes from the logarithmic y-axis. Figure 5 gives the true worst case error herein. is ﬁgure shows the error of the path loss for one kilometer link at 25 degrees centigrade and at 90% relative humidity. It can be seen that the error is very good across the band, but the lower frequencies do give comparably slightly larger error due to in general lower absorption loss. However, the ﬁgures herein are for one kilometer link and the error Fig. 4 Molecular absorption loss at 1 km distance and 90% relative humidity. Molecular absorption loss at 1 km distance at 25 degrees centigrade and 90% relative humidity ( Error to exact for 1 km link [dB] ITU full model Proposed model Fig. 5 Error of the model and comparison with ITU-R model. Absolute errors given by the ITU-R full model and the proposed model to the exact theory for a one kilometer link Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 will decrease with decreasing distance due to exponential behavior of the absorption loss. us, the resultant error of roughly ±2 dB is very good for such extremely high link distances considering the high frequencies and their general applicability to low range communications. Furthermore, the error also decreases in less humid environ- ment and this is in general true for ITU-R models as well. For instance at 10% relative humidity at 25 degrees centigrade, the diﬀerences are rather modest. Regardless of this, in more humid environments there is a notable diﬀerence between the models, especially above the 200 GHz frequencies. As a last note on the error performance, all the models herein are rather accurate and it is an application speciﬁc issue how accurately the absorption loss needs to be calcu- lated. If the link distance is high or the communications band is in the vicinity of the absorption line, the importance of the correct loss is high. However, on low distance links and in the middle of the low loss regions of the spectrum the absorption loss is modest and large error is not made if the absorption loss is omitted altogether. 3.2 Performance ofthemodel withreduced terms If one targets only some sub-band within the 100–450 GHz band, the proposed model can be further simpliﬁed by only using subset of the polynomials . Figures6 and 7 compare the performance of the proposed model with reduced terms against the exact theory. Figure6 shows performance of the proposed model when using the ﬁrst two lines at about 119 GHz and 183 GHz separately and jointly (shown as lines 1 and 2 in the ﬁgure). In the other words, one should utilize the absorption coeﬃcient as for lines 1 and 1 and 2 jointly, respectively. is reduction corresponds roughly to the frequency range of the D band. It can be seen that the proposed model with reduced terms performs very well on estimating the absorption loss. e same occurs in the case of Fig.7 that shows the performance of the next two lines (lines 3 and 4) corresponding to frequen- cies 325 GHz and 380 GHz. ese two line alone gives a very good estimate of the loss Absorption loss for one kilometer link [dB] Exact theory Proposed model, lines 1 and 2 Proposed model, line 1 Error to exact, lines 1 and 2 Error to exact, line 1 Fig. 6 Performance of the proposed model with reduced terms, low band. Reduced versions of the proposed model giving absorption losses up to about 160 GHz (1 term) and 200 GHz (2 terms) Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 up to about 330 GHz and 390 GHz for the line 3 and joint lines 3 and 4, respectively. ese correspond to utilizing an absorption coeﬃcient as µ) = µ) + µ) + . As such, the line 3 would be mostly enough for the popular transition frequencies between the mmWave and THz bands. Namely 275–325 GHz. However, with these two lines, the model remains accurate from about 200 GHz up to the above-mentioned 390 GHz. erefore, the proposed model is ﬂexible and easily reducible for multiple frequency bands within the full range from 100 to 450 GHz for some speciﬁc applications that occupy only certain sub-band. 3.3 Link budget calculations To show some examples on use cases for simple channel model, we give link budget calculations for the D band and THz band below. We assume long distance backhaul connection, a one kilometer LOS link. For the D band, we have chosen the free bands for wireless communications therein according to European Conference of Postal and Absorption loss for one kilometer link [dB] Exact theory Proposed model, lines 3 and 4 Proposed model, line 3 Error to exact, lines 3 and 4 Error to exact, line 3 Fig. 7 Performance of the proposed model with reduced terms, high band. Reduced versions of the proposed model giving absorption losses up to about 330 GHz (3rd line only) and 390 GHz (lines 3 and 4 Table 1Link budget calculations for the D band channels. Values in brackets are the exact theoretical values Center frequency (GHz) 132.00 144.75 157.75 170.90 Bandwidth (GHz) 3 7.5 12.5 7.8 Transmit power (dBm) 0 Tx/RX antenna gain (dBi) 48.3 49.1 49.9 50.6 Noise ﬁgure (dB) 10 Noise ﬂoor (dBm) Link distance (m) 1,000 Path loss (dB) 135.1 (135.2) 136.0 (136.1) 137.0 (137.1) 139.4 (139.4) Rx power (dBm) SNR (dB) 30.6 (30.6) 27.4 (27.3) 25.6 (25.6) 26.8 (26.8) Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 Telecommunications Administrations (CEPT) Electronic Communications Commit- tee (ECC) Recommendation 18(01) [21]. ose are detailed in Table1. For the THz band, we utilize every second channel of 802.15.3d standard with 8.64 GHz channeli- zation [7]. ese channels are given in Table2. e transmit powers at the high-fre- quency bands have not been regulated other than by maximum radiation intensities [22] that are typically in the range of 55 f− depending on the source and application, where is frequency in GHz. us, we use 0 dBm transmit power for all bands in order to have rather conservative radiated power with respect to the radia- tion limits and what the current THz capable devices are able to output. A one kilo- meter link at +100 GHz frequencies requires very large antenna gains. We assume parabolic reﬂector antennas to provide very large gain. e gain of such antenna is given by is the aperture eﬃciency, is the diameter of the parabolic reﬂector, and the wavelength. We assume aperture eﬃciency of 70% herein and a 225 mm diameter for the parabolic antenna. is diameter is equivalent to that of the Cassegrain antennas developed in TERRANOVA project [23]. is size parabolic antenna gives about 55 dBi gain at 300 GHz frequency [23] and as also shown below in Table2 based on (15) with the above parameters. e antenna gains per band, average path loss per channel, and the received powers and SNRs are given in Tables1 and 2. e average path losses per band (indexed in Tables1 and 2) compared to the theoretical path losses from theoreti- cal path loss given by molecular absorption loss in (1) and FSPL given in (12) are given in Fig.8. For these calculations, no other losses, such as antenna feeder losses, are assumed. e main aim here is to estimate performance of the simpliﬁed path loss model. Based on the link budget calculations, the proposed simpliﬁed model gives a very good performance without a need for complex line-by-line models. e link budget calculations are among the most important applications for estimating the required antenna gains and transmit powers for novel wireless systems. A simple channel gain estimate helps to quickly calculate the expected channel loss within the overall link Table 2 Link budget calculations for the THz band channels. Values in brackets are the exact theoretical values Center frequency (GHz) 265.68 282.96 300.24 317.52 Bandwidth (GHz) 8.64 Transmit power (dBm) 0 Tx/RX antenna gain (dBi) 54.4 54.9 55.5 55.9 Noise ﬁgure (dB) 10 Noise ﬂoor (dBm) -64.6 Link distance (m) 1,000 Path loss (dB) 142.2 (142.4) 142.9 (143.2) 144.2 (144.7) 153.6 (154.7) Rx power (dBm) -33.4 (-33.7) -33.0 (-33.4) -33.3 (-33.8) -41.7 (-42.9) SNR (dB) 31.2 (30.9) 31.6 (31.2) 31.3 (30.8) 22.9 (21.7) Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 budget. We can see that the expected accuracy of the proposed simpliﬁed model gives SNR values that are at most 1.2 dB oﬀ the real value. is level of diﬀerence in the real systems is insigniﬁcant due to all the other loss mechanisms, and, the link distance here is quite large for high-frequency communications. Although these link distances are very much possible as shown, e.g., in [23, 24], where 1 km link distance with the above-mentioned 55 dBi Cassegrain antennas was demonstrated.eir total loss with antenna gains at 300 GHz was about 40 dBs, whereas in Table2 we see a loss of about 33 dBs. is shows that even with very simple calculations, one can get very close to real-life measurements even without taking into account feeder losses, or other possi- ble atmospheric losses, such as fog loss and small particle scattering in the air. ere- fore, the proposed simpliﬁed loss model can very reliably estimate the atmospheric losses and the accuracy of the total link budget mostly falls into properly modeling all the parts of the wireless system that have impact on the total received power. 3.4 Discussion As a summary and discussion from above, the higher mmWave and low THz frequen- cies are among the most potential frequencies to utilize ultrahigh rate communications in the future wireless systems. e proper modeling of the channel behavior therein is very important due to absorption loss and how it behaves in comparison with the FSPL. In short distance communications (few meters) below 300 GHz, it is not absolutely crucial to model the absorption due to dominating FSPL. Its importance increases with link distance, but also with frequency. In the other words, the link budget and the components in it are appli- cation dependent. e tools provided herein give and easy way to model the absorption loss and estimate its impact on the link budget. Path Loss [dB] Theoretical path loss Fig. 8 Path losses for the bands considered in the link budget calculations. The exact path loss as a function of frequency versus the average losses per band given in Tables 1 and 2 Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 4 Conclusion We derived a LOS channel model for 100–450 GHz frequency range in this paper. e main goal was to ﬁnd a simple and easy to use model for the molecular absorption loss. e derived model was shown to be very accurate and predict the channel loss very well in the target frequency regime. is model can be reduced to simpler forms in the case of limited frequency range within 100–450 GHz. Considering the upcoming B5G systems, the interesting frequency bands include the D band (110 GHz to 170 GHz) and the low THz frequencies (275 GHz to 325 GHz). e molecular absorption loss is an important part of the link budget considerations at +100 GHz bands. erefore, the model presented here gives a simple tool to estimate the total link loss in various environmental conditions and link distances. As it was shown in the numerical results, the derived model can be used to predict the expected SNR within D band and THz band with below 2 dB error compared to the exact theoretical model. erefore, this simple tool gives high enough accuracy for any LOS system analysis, but also in the broader sense, analysis of the large scale fading in the sub-THz regime. 5 Methods/experimental is paper is a purely theoretical model on simple way to estimate the absorption loss. Although theoretical, the original data obtained from the HITRAN database [10] are based on experimental data. e goal in this article is to simplify the complex database approach into simple polynomial equations with only few ﬂoating parameters, such as humidity and frequency. As such, the model produced in this paper is suitable for LOS channel loss esti- mation for various wireless communications systems. ose include back- and fronthaul connectivity and general LOS link channel estimation. e work is heavily based on the HITRAN database and the theoretical models for absorption loss as well as simple LOS free space path loss models. 5G: Fifth generation; 6G: Sixth generation; B5G: Beyond ﬁfth generation; FSPL: Free space path loss; HITRAN: High-resolu- tion transmission molecular absorption database; ITU-R: International Telecommunication Union Radio Communication Sector; LOS: Line-of-sight; mmWave: Millimeter wave; Rx: Receiver; Tx: Transmitter. Authors’ contributions JK derived the molecular absorption loss model. All the authors participated in writing the article and revising the manu- script. All authors read and approved the ﬁnal manuscript. This work was supported in part by the Horizon 2020, European Union’s Framework Programme for Research and Inno- vation, under Grant Agreement No. 761794 (TERRANOVA) and No. 871464 (ARIADNE). It was also supported in part by the Academy of Finland 6Genesis Flagship under Grant No. 318927. Data availability Not applicable. Competing interests The authors declare that they have no competing interests. Received: 7 January 2020 Accepted: 30 March 2021 Content courtesy of Springer Nature, terms of use apply. Rights reserved. Kokkoniemietal. J Wireless Com Network (2021) 2021:88 1. T.S. Rappaport et al., Millimeter wave mobile communications for 5G cellular: it will work!. IEEE Access 1(1), 335–349 2. T.S. Rappaport, Y. Xing, O. Kanhere, S. Ju, A. Madanayake, S. Mandal, A. Alkhateeb, G.C. Trichopoulos, Wireless com- munications and applications above 100 GHz: opportunities and challenges for 6G and beyond. IEEE Access 7, 78729–78757 (2019) 3. M. Latva-Aho, K. Leppänen (eds.), Key Drivers and Research Challenges for 6G Ubiquitous Wireless Intelligence. 6G research visions, vol. 1, pp. 1–36 University of Oulu, Oulu, Finland (2019) 4. I.F. Akyildiz, J.M. Jornet, C. Han, Terahertz band: next frontier for wireless communications. Phys. Commun. 12, 16–32 5. TERRANOVA: Deliverable D2.1, TERRANOVA system requirements. Technical report (2017). https:// ict- terra nova. eu/ wp- conte nt/ uploa ds/ 2018/ 03/ terra nova_ d2-1_ wp2_ v1-0. pdf 6. ARIADNE: D1.1 ARIADNE use case deﬁnition and system requirements. Technical report (2020). https:// www. ict- ariad ne. eu/ deliv erabl es/ 7. Amendment 2: 100 Gb/s Wireless Switched Point-to-Point Physical Layer (Std 802.15.3d–2017). IEEE 8. J. Kokkoniemi, J. Lehtomäki, M. Juntti, Simple molecular absorption loss model for 200-450 gigahertz frequency band. In Proceedings of the European Conference Network Communication pp. 1–5 (2019) 9. J.M. Jornet, I.F. Akyildiz, Channel modeling and capacity analysis for electromagnetic nanonetworks in the terahertz band. IEEE Trans. Wirel. Commun. 10(10), 3211–3221 (2011) 10. L.S. Rothman et al., The HITRAN 2012 molecular spectroscopic database. J. Quant. Spectrosc. Radiat. Transf. 130(1), 4–50 (2013) 11. S. Paine, The am atmospheric model. Technical Report 152, Smithsonian Astrophysical Observator y (2012) 12. Calculation of molecular spectra with the Spectral Calculator. www. spect ralca lc. com 13. J.R. Pardo, J. Cernicharo, E. Serabyn, Atmospheric transmission at microwaves (ATM): an improved model for mil- limeter/submillimeter applications. IEEE Trans. Antennas Propag. 49(12), 1683–1694 (2001) 14. A. Berk, P. Conforti, R. Kennett, T. Perkins, F. Hawes, J. van den Bosch. MODTRAN6: a major upgrade of the MODTRAN radiative transfer code. In: Velez-Reyes, M., Kruse, F.A. (eds.) Algorithms and Technologies for Multispectral, Hyperspec- tral, and Ultraspectral Imagery XX, vol. 9088, pp. 113–119. SPIE, Baltimore, Maryland, USA (2014). https:// doi. org/ 10. 1117/ 12. 20504 33. International Society for Optics and Photonics 15. ITU-R, Recommendation, pp. 676–678, Attenuation by Atmospheric gases. International Telecommunication Union Radiocommunication Sector (2009) 16. J. Kokkoniemi, J. Lehtomäki, M. Juntti, Simpliﬁed molecular absorption loss model for 275–400 gigahertz frequency band. In Proceedings of the European Conference on Antennas Propagation pp. 1–5 (2018) 17. N. Jacquinet-Husson et al., The 2009 edition of the GEISA spectroscopic database. J. Quant. Spectrosc. Radiat. Transf. 112(15), 2395–2445 (2011) 18. H.M. Pickett, et al., Submillimeter, Millimeter, and Microwave Spectral Line Catalog (2003). http:// spec. jpl. nasa. gov/ ftp/ pub/ catal og/ doc/ catdoc. pdf 19. O.A. Alduchov, R.E. Eskridge, Improved magnus form approximation of saturation vapor pressure. J. Appl. Meteor. 35(4), 601–609 (1996) 20. J.H. Van Vleck, D.L. Huber, Absorption, emission, and linebreadths: a semihistorical perspective. Rev. Mod. Phys. 49(4), 939–959 (1977) 21. Recommendation (18)01: Radio Frequency Channel/block Arrangements for Fixed Service Systems Operating in the Bands 130-134 GHz, 141-148.5 GHz, 151.5-164 GHz and 167-174.8 GHz. ECC 22. W. He, B. Xu, Y. Yao, D. Colombi, Z. Ying, S. He, Implications of incident power density limits on power and EIRP levels of 5G millimeter-wave user Equipment. IEEE Access 8, 148214–148225 (2020) 23. TERRANOVA: Deliverable D6.2, THz High-Capacity Demonstrator implementation report. Technical report (2020). https:// ict- terra nova. eu/ wp- conte nt/ uploa ds/ 2020/ 06/ D6.2- 1. pdf 24. C. Castro, R. Elschner, T. Merkle, C. Schubert, R. Freund, Long-range high-speed THz-wireless transmission in the 300 GHz band. In Proceedings of the IWMTS pp. 1–4 (2020) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations. Content courtesy of Springer Nature, terms of use apply. Rights reserved. Terms and Conditions Springer Nature journal content, brought to you courtesy of Springer Nature Customer Service Center GmbH (“Springer Nature”). Springer Nature supports a reasonable amount of sharing of research papers by authors, subscribers and authorised users (“Users”), for small- scale personal, non-commercial use provided that all copyright, trade and service marks and other proprietary notices are maintained. By accessing, sharing, receiving or otherwise using the Springer Nature journal content you agree to these terms of use (“Terms”). For these purposes, Springer Nature considers academic use (by researchers and students) to be non-commercial. These Terms are supplementary and will apply in addition to any applicable website terms and conditions, a relevant site licence or a personal subscription. These Terms will prevail over any conflict or ambiguity with regards to the relevant terms, a site licence or a personal subscription (to the extent of the conflict or ambiguity only). For Creative Commons-licensed articles, the terms of the Creative Commons license used will We collect and use personal data to provide access to the Springer Nature journal content. We may also use these personal data internally within ResearchGate and Springer Nature and as agreed share it, in an anonymised way, for purposes of tracking, analysis and reporting. We will not otherwise disclose your personal data outside the ResearchGate or the Springer Nature group of companies unless we have your permission as detailed in the Privacy Policy. While Users may use the Springer Nature journal content for small scale, personal non-commercial use, it is important to note that Users may use such content for the purpose of providing other users with access on a regular or large scale basis or as a means to circumvent access use such content where to do so would be considered a criminal or statutory offence in any jurisdiction, or gives rise to civil liability, or is otherwise unlawful; falsely or misleadingly imply or suggest endorsement, approval , sponsorship, or association unless explicitly agreed to by Springer Nature in use bots or other automated methods to access the content or redirect messages override any security feature or exclusionary protocol; or share the content in order to create substitute for Springer Nature products or services or a systematic database of Springer Nature journal In line with the restriction against commercial use, Springer Nature does not permit the creation of a product or service that creates revenue, royalties, rent or income from our content or its inclusion as part of a paid for service or for other commercial gain. Springer Nature journal content cannot be used for inter-library loans and librarians may not upload Springer Nature journal content on a large scale into their, or any other, institutional repository. These terms of use are reviewed regularly and may be amended at any time. Springer Nature is not obligated to publish any information or content on this website and may remove it or features or functionality at our sole discretion, at any time with or without notice. Springer Nature may revoke this licence to you at any time and remove access to any copies of the Springer Nature journal content which have been saved. To the fullest extent permitted by law, Springer Nature makes no warranties, representations or guarantees to Users, either express or implied with respect to the Springer nature journal content and all parties disclaim and waive any implied warranties or warranties imposed by law, including merchantability or fitness for any particular purpose. Please note that these rights do not automatically extend to content, data or other material published by Springer Nature that may be licensed from third parties. If you would like to use or distribute our Springer Nature journal content to a wider audience or on a regular basis or in any other manner not expressly permitted by these Terms, please contact Springer Nature at	{"url":"https://www.researchgate.net/publication/350770315_A_line-of-sight_channel_model_for_the_100-450_gigahertz_frequency_band","timestamp":"2024-11-14T05:52:43Z","content_type":"text/html","content_length":"786880","record_id":"<urn:uuid:c85b18c2-e589-4c56-8c22-22c5b815c911>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00403.warc.gz"}
Note: This document is for an older version of GRASS GIS that will be discontinued soon. You should upgrade, and read the current manual page. - Computes horizon angle height from a digital elevation model. The module has two different modes of operation: 1. Computes the entire horizon around a single point whose coordinates are given with the 'coord' option. The horizon height (in radians). 2. Computes one or more raster maps of the horizon height in a single direction. The input for this is the angle (in degrees), which is measured counterclockwise with east=0, north=90 etc. The output is the horizon height in radians. sun position r.horizon --help r.horizon [-dc] elevation=name [direction=float] [step=float] [start=float] [end=float] [bufferzone=float] [e_buff=float] [w_buff=float] [n_buff=float] [s_buff=float] [maxdistance=float] [output= basename] [coordinates=east,north] [distance=float] [file=name] [--overwrite] [--help] [--verbose] [--quiet] [--ui] Write output in degrees (default is radians) Write output in compass orientation (default is CCW, East=0) Allow output files to overwrite existing files Print usage summary Verbose module output Quiet module output Force launching GUI dialog elevation=name [required] Name of input elevation raster map Direction in which you want to know the horizon height Angle step size for multidirectional horizon [degrees] Start angle for multidirectional horizon [degrees] Default: 0.0 End angle for multidirectional horizon [degrees] Default: 360.0 For horizon rasters, read from the DEM an extra buffer around the present region Options: 0- For horizon rasters, read from the DEM an extra buffer eastward the present region Options: 0- For horizon rasters, read from the DEM an extra buffer westward the present region Options: 0- For horizon rasters, read from the DEM an extra buffer northward the present region Options: 0- For horizon rasters, read from the DEM an extra buffer southward the present region Options: 0- The maximum distance to consider when finding the horizon height Name for output basename raster map(s) Coordinate for which you want to calculate the horizon Sampling distance step coefficient (0.5-1.5) Default: 1.0 Name of file for output (use output=- for stdout) Default: - computes the angular height of terrain horizon in radians. It reads a raster of elevation data and outputs the horizon outline in one of two modes: • single point: as a series of horizon heights in the specified directions from the given point. The results are written to the stdout. • raster: in this case the output is one or more raster maps, with each point in a raster giving the horizon height in a specific direction. One raster is created for each direction. The directions are given as azimuthal angles (in degrees), with the angle starting with 0 towards East and moving counterclockwise (North is 90, etc.). The calculation takes into account the actual projection, so the angles are corrected for direction distortions imposed by it. The directions are thus aligned to those of the geographic projection and not the coordinate system given by the rows and columns of the raster map. This correction implies that the resulting cardinal directions represent true orientation towards the East, North, West and South. The only exception of this feature is LOCATION with x,y coordinate system, where this correction is not applied. Using the -c flag, the azimuthal angles will be printed in compass orientation (North=0, clockwise). The elevation parameter is an input elevation raster map. If the buffer options are used (see below), this raster should extend over the area that accommodate the presently defined region plus defined buffer zones. The step parameter gives the angle step (in degrees) between successive azimuthal directions for the calculation of the horizon. Thus, a value of 5 for the step will give a total of 360/5=72 directions (72 raster maps if used in the raster map mode). The start parameter gives the angle start (in degrees) for the calculation of the horizon. The default value is 0 (East with North being 90 etc.). The end parameter gives the angle end (in degrees) for the calculation of the horizon. The end point is omitted! So for example if we run r.horizon with step=10, start=30 and end=70 the raster maps generated by r.horizon will be only for angles: 30, 40, 50, 60. The default value is 360. The direction parameter gives the initial direction of the first output. This parameter acts as an direction angle offset. For example, if you want to get horizon angles for directions 45 and 225 degrees, the direction should be set to 45 and step to 180. If you only want one single direction, use this parameter to specify desired direction of horizon angle, and set the step size to 0 degrees. Otherwise all angles for a given starting direction with step of step are calculated. The distance controls the sampling distance step size for the search for horizon along the line of sight. The default value is 1.0 meaning that the step size will be taken from the raster resolution. Setting the value below 1.0 might slightly improve results for directions apart from the cardinal ones, but increasing the processing load of the search algorithm. The maxdistance value gives a maximum distance to move away from the origin along the line of sight in order to search for the horizon height. The default maxdistance is the full map extent. The smaller this value the faster the calculation but the higher the risk that you may miss a terrain feature that can contribute significantly to the horizon outline. Note that a viewshed can be calculated with r.viewshed. The coordinate parameter takes a pair of easting-northing values in the current coordinate system and calculates the values of angular height of the horizon around this point. To achieve the consistency of the results, the point coordinate is aligned to the midpoint of the closest elevation raster cell. If an analyzed point (or raster cell) lies close to the edge of the defined region, the horizon calculation may not be realistic, since it may not see some significant terrain features which could have contributed to the horizon, because these features are outside the region. There are to options how to set the size of the buffer that is used to increase the area of the horizon analysis. The bufferzone parameter allows you to specify the same size of buffer for all cardinal directions and the parameters e_buff, n_buff, s_buff, and w_buff allow you to specify a buffer size individually for each of the four directions. The buffer parameters influence only size of the read elevation map, while the analysis in the raster mode will be done only for the area specified by the current region definition. The output parameter defines the basename of the output horizon raster maps. The raster name of each horizon direction raster will be constructed as basename_ANGLE, where ANGLE is the angle in degrees with the direction. If you use r.horizon in the single point mode this option will be ignored. The file parameter allows saving the resulting horizon angles in a comma separated ASCII file (single point mode only). If you use r.horizon in the raster map mode this option will be ignored. At the moment the elevation and maximum distance must be measured in meters, even if you use geographical coordinates (longitude/latitude). If your projection is based on distance (easting and northing), these too must be in meters. The buffer parameters must be in the same units as the raster coordinates (e.g., for latitude-longitude locations buffers are measured in degree unit). The calculation method is based on the method used in to calculate shadows. It starts at a very shallow angle and walks along the line of sight and asks at each step whether the line of sight "hits" the terrain. If so, the angle is increased to allow the line of sight to pass just above the terrain at that point. This is continued until the line of sight reaches a height that is higher than any point in the region or until it reaches the border of the region (see also the , and ). The number of lines of sight (azimuth directions) is determined from the parameters. The method takes into account the curvature of the Earth whereby remote features will seem to be lower than they actually are. It also accounts for the changes of angles towards cardinal directions caused by the projection (see above). The output with the -d flag is azimuth degree (-90 to 90, where 0 is parallel with the focal cell). All horizon values are positive (or zero). While negative values are in theory possible, r.horizon currently does not support them. The examples are intended for the North Carolina sample dataset. Example 1 : determine horizon angle in 225 degree direction (output of horizon angles CCW from East): g.region raster=elevation -p r.horizon elevation=elevation direction=215 step=0 bufferzone=200 \ coordinates=638871.6,223384.4 maxdistance=5000 Example 2: determine horizon values starting at 90 deg (North), step size of 5 deg, saving result as CSV file: r.horizon elevation=elevation direction=90 step=5 bufferzone=200 \ coordinates=638871.6,223384.4 maxdistance=5000 file=horizon.csv Example 3: test point near highway intersection, saving result as CSV file for plotting the horizon around the highway intersection: g.region n=223540 s=220820 w=634650 e=638780 res=10 -p r.horizon elevation=elevation direction=0 step=5 bufferzone=200 \ coordinates=636483.54,222176.25 maxdistance=5000 -d file=horizon.csv Test point near high way intersection (North Carolina sample dataset) Horizon angles for test point (CCW from East) We can plot horizon in polar coordinates using Matplotlib in Python: import numpy as np import matplotlib.pyplot as plt horizon = np.genfromtxt('horizon.csv', delimiter=',') horizon = horizon[1:, :] ax = plt.subplot(111, polar=True) bars = ax.plot(horizon[:, 0] / 180 * np.pi, (90 - horizon[:, 1]) / 180 * np.pi) # uncomment the 2 following lines when using -c flag # ax.set_theta_direction(-1) # ax.set_theta_zero_location('N') Horizon plot in polar coordinates. Raster map mode (output maps "horangle*" become input for g.region raster=elevation -p # we put a bufferzone of 10% of maxdistance around the study area # compute only direction between 90 and 270 degrees r.horizon elevation=elevation step=30 start=90 end=300 \ bufferzone=200 output=horangle maxdistance=5000 Hofierka J., 1997. Direct solar radiation modelling within an open GIS environment. Proceedings of JEC-GI'97 conference in Vienna, Austria, IOS Press Amsterdam, 575-584 Hofierka J., Huld T., Cebecauer T., Suri M., 2007. Open Source Solar Radiation Tools for Environmental and Renewable Energy Applications, International Symposium on Environmental Software Systems, Prague, 2007 Neteler M., Mitasova H., 2004. Open Source GIS: A GRASS GIS Approach, Springer, New York. ISBN: 1-4020-8064-6, 2nd Edition 2004 (reprinted 2005), 424 pages Project PVGIS, European Commission, DG Joint Research Centre 2001-2007 Suri M., Hofierka J., 2004. A New GIS-based Solar Radiation Model and Its Application for Photovoltaic Assessments. Transactions in GIS, 8(2), 175-190 r.sun, r.sunmask, r.viewshed Thomas Huld, Joint Research Centre of the European Commission, Ispra, Italy Tomas Cebecauer, Joint Research Centre of the European Commission, Ispra, Italy Jaroslav Hofierka, GeoModel s.r.o., Bratislava, Slovakia Marcel Suri, Joint Research Centre of the European Commission, Ispra, Italy © 2007, Thomas Huld, Tomas Cebecauer, Jaroslav Hofierka, Marcel Suri Thomas.Huld@jrc.it Tomas.Cebecauer@jrc.it hofierka@geomodel.sk Marcel.Suri@jrc.it Available at: r.horizon source code (history) Latest change: Thursday Feb 22 14:40:23 2024 in commit: 6a52add2a65dfaa9e7b38f76399ee5801dd7e550 Main index \| Raster index \| Topics index \| Keywords index \| Graphical index \| Full index © 2003-2024 GRASS Development Team, GRASS GIS 8.3.3dev Reference Manual	{"url":"https://mirrors.ibiblio.org/grass/code_and_data/grass83/manuals/r.horizon.html","timestamp":"2024-11-14T13:48:38Z","content_type":"text/html","content_length":"19872","record_id":"<urn:uuid:d00f8dd8-83d0-4484-ae40-3f3f6e3a4e0d>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00249.warc.gz"}
Lever Force Calculator - Calculator Wow Lever Force Calculator Welcome to the dynamic realm of levers, where forces and distances intertwine to create mechanical advantage. The Lever Force Calculator serves as your gateway to understanding and harnessing the power of levers. In this article, we embark on a journey exploring the calculator’s importance, delving into its application, and addressing common queries to demystify the realm of lever forces. The Importance of Lever Force Calculator 1. Efficiency in Mechanical Systems • Levers are ubiquitous in mechanical systems, and understanding the forces at play is crucial for optimizing efficiency. The calculator aids in achieving this optimization. 2. Engineering and Design Precision • Engineers rely on accurate force calculations when designing systems that involve levers. The calculator ensures precision in engineering and design processes. 3. Physics Understanding • In the realm of physics, levers exemplify fundamental principles. The calculator provides a practical tool to experiment with these principles and deepen one’s understanding. How to Use Lever Force Calculator 1. Enter Effort Force • Input the force applied at one end of the lever (Effort Force) in newtons (N). 2. Enter Distances • Provide the distances from the Effort Force to the Fulcrum (D1) and from the Load Force to the Fulcrum (D2) in meters (m). 3. Initiate Calculation • Click the “Calculate Lever Force” button to initiate the calculation. 4. Observe Result • The calculated Lever Force will be displayed, indicating the force exerted on the load end of the lever. 5. Apply Calculations • Utilize the calculated Lever Force in engineering designs, physics experiments, or any scenario involving lever systems. 10 FAQs About Lever Force Calculator 1. What is Mechanical Advantage in Levers? • Mechanical advantage in levers is the amplification of force achieved by using a lever. The calculator aids in understanding and calculating this advantage. 2. Can I Use the Calculator for Different Types of Levers? • Yes, the calculator is applicable to various types of levers, including first, second, and third-class levers. 3. Why Does the Distance Matter in Lever Systems? • The distance from the fulcrum affects the lever’s mechanical advantage. The calculator considers these distances for accurate force calculations. 4. How Does the Calculator Account for Friction? • The calculator assumes an ideal scenario without friction. In real-world applications, friction may need to be considered separately. 5. Can I Use the Calculator for Educational Purposes? • Absolutely, the calculator is an excellent educational tool for physics and engineering students, helping them grasp the practical applications of lever forces. 6. Is it Suitable for DIY Projects? • Yes, the calculator is user-friendly and can be applied to DIY projects involving levers, providing insights into force requirements. 7. Does the Load Force Have to be Lifted? • No, the load force can be any force applied to the lever. It doesn’t necessarily have to be lifted; it can be pushed or pulled. 8. Can I Calculate Lever Forces in Different Units? • While the default unit is newtons (N), you can convert distances and forces to different units for calculations. 9. Is it Applicable to Digital Systems and Virtual Levers? • Yes, the calculator is versatile and can be applied to digital systems and virtual levers, offering a hands-on experience in the virtual realm. 10. Why is Lever Force Important in Mechanical Systems? • Lever forces are crucial for determining the required input force to achieve desired output forces in mechanical systems, influencing efficiency and design decisions. As we conclude our exploration of the Lever Force Calculator, envision it as your key to unlocking the potential within levers. From engineering marvels to physics experiments, the calculator empowers your understanding and application of lever forces. Let precision guide your endeavors, and may the calculated forces resonate with efficiency and innovation. In the symphony of levers, let the Lever Force Calculator be your conductor, orchestrating a harmonious blend of forces and distances in the mechanical world.	{"url":"https://calculatorwow.com/lever-force-calculator/","timestamp":"2024-11-06T14:00:37Z","content_type":"text/html","content_length":"66894","record_id":"<urn:uuid:91cefdd1-b23e-4286-a8a1-3eea91de4eaa>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00549.warc.gz"}
Triangular, Rectangular, and Three-Dimensional Elements - Interpolation Functions Questions and Answers - Sanfoundry Finite Element Method Questions and Answers – Interpolation Functions – Triangular, Rectangular, and Three-Dimensional Elements This set of Finite Element Method Multiple Choice Questions & Answers (MCQs) focuses on “Interpolation Functions – Triangular, Rectangular, and Three-Dimensional Elements”. 1. Which of the following is not an example of a triangular element? a) Single node linear b) 3 node linear c) 6 node quadratic d) 10 node cubic View Answer Answer: a Explanation: Single node linear is not an example of a triangular element. The other three types of triangular elements are depicted in the figure below. 2. What is the function of an internal node in case of the cubic element? a) To attain a symmetric shape b) To attain geometrical isotropy c) To satisfy mesh refinement d) To nullify geometrical errors View Answer Answer: b Explanation: An internal node is one that stays interior to the element and is not connected to any other node. In case of the cubic element, the internal node is required to satisfy geometric isotropy which allows the functional form to stay constant in both translational and rotational forms. 3. In structural applications, what is a constraint strain element? a) 4 node rectangular element b) 6 node triangular element c) 3 node triangular element d) 8 node rectangular element View Answer Answer: c Explanation: In case of structural applications, the 3 noded triangular elements are also referred to as constraint strain elements(CST) . It has various advantages. For instance, in case of heat transfer problems, it helps produce constant temperature gradients, which in turn oversee constant heat flow in the element. 4. Which of the following defines area coordinates? a) Location of area in the arbitrary plane b) Ratio of assumed area to total area c) They are nothing but the respective nodal areas d) Ratio of nodal area to total area View Answer Answer: d Explanation: Area coordinates can be computed by finding out the ratio of nodal areas to the total areas. As triangular elements computation is complex and cumbersome, simplification of interpolation functions is required. This is satisfied by making use of these area coordinates. 5. Rectangular elements are singular and cannot be used with other elements. a) True b) False View Answer Answer: b Explanation: The given statement is false. Rectangular elements are not singular, and can be used along with other elements. In fact, for many geometries it is preferred to have rectangular elements in conjunction with triangular or other types of elements for development of quadrilateral structures. 6. Which of the following is an example of a rectangular element? a) 4 noded rectangle b) 3 noded triangle c) Multi node circle d) 4 noded triangle View Answer Answer: a Explanation: The 4 noded rectangle is an example of the rectangular element. It is generally assumed that the sides of the rectangle are parallel to the global Cartesian axes. It is depicted as follows – 7. What is considered as the equivalent of area coordinates incase of rectangular elements? a) Area elements b) Area nodes c) Natural coordinates d) Isolated coordinates View Answer Answer: c Explanation: Incase of rectangular elements, to simplify calculations natural/serendipity coordinates are used. This is because, the complexity incase of rectangular elements can be reduced by making judicious choice of the coordinates. 8. How are higher order rectangular elements developed? a) Continuous interpolation is done b) Existing node is subtracted c) Mesh refinement is done d) Additional nodes are added View Answer Answer: d Explanation: In order to develop higher order rectangular elements, an additional node is added at the midpoint of each side of the element. This method has its disadvantages too. Pascal triangle shows that we cannot construct a complete polynomial having 8 terms, but can be accomplished by making use of two incomplete, symmetric cubic polynomials. 9. What are considered as the two main families of three dimensional elements? a) Tetrahedrons and Parallelepipeds b) Tetrahedrons and Rectangles c) Parallelepipeds and Triangles d) Tetrahedrons only View Answer Answer: a Explanation: Tetrahedrons and Parallelepipeds are considered as the two main families of three dimensional elements. While tetrahedrons are extension of triangular elements, the parallelepipeds are extensions of rectangular elements. Parallelepipeds are otherwise referred to as brick elements. 10. Which of the following is the definition of volume coordinates? a) Specific total volume b) Ratio of nodal volume to the total volume c) Unit total volume d) Ratio of total volume to element volume View Answer Answer: b Explanation: A volume coordinate is defined as the ratio of nodal volume to the total volume. Their role is similar to that of area coordinates incase of triangular elements and natural coordinates incase of rectangular elements. They help in simplifying the computation procedures. 11. In structural applications, which of the following element is considered as the constant strain element? a) Triangular element b) Hexagonal element c) Tetrahedral element d) Brick element View Answer Answer: c Explanation: Incase of structural applications, the tetrahedral element is considered as the constant strain element. This is because, all first partial derivatives of the field variable are constant. This element is also considered most useful for use in irregular geometries. Sanfoundry Global Education & Learning Series – Finite Element Method. To practice all areas of Finite Element Method, here is complete set of 1000+ Multiple Choice Questions and Answers.	{"url":"https://www.sanfoundry.com/finite-element-method-questions-answers-interpolation-functions-triangular-rectangular-three-dimensional-elements/","timestamp":"2024-11-05T22:45:15Z","content_type":"text/html","content_length":"167642","record_id":"<urn:uuid:c7f4e5ea-a490-4c35-b901-02a67772e63e>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00244.warc.gz"}
Find $\left[\mathrm{H}^{+}\right]$ and the $\mathrm{pH}$ of the following solutions. (a) $1.75 \mathrm{~L}$ of a $37.5 \%$ (by mass) solution $(d=1.00 \mathrm{~g} / \mathrm{mL})$ of $\ mathrm{HCl} .$ What is the $\mathrm{pH}$ of $0.175 \mathrm{~L}$ of the same solution? (b) A solution made up of $22 \mathrm{~g}$ of $\mathrm{HBr}$ dissolved in enough water to make $479 \ mathrm{~mL}$ of solution. What is the $\mathrm{pH}$ if the same mass of $\mathrm{HBr}$ is dissolved in enough water to make $47.9 \mathrm{~mL}$ of solution? Short Answer Expert verified Question: Calculate the concentration of H+ ions and the pH of the following solutions: (a) a 1.75 L solution with a density of 1.00 g/mL and a 37.5% by mass composition of HCl, and (b) 22 g of HBr dissolved in 479 mL of water. Additionally, find the pH of 0.175 L of the same HCl solution from part (a). Answer: For part (a), the concentration of H+ ions is 10.3 M, and the pH is approximately -1.01 for both the 1.75 L solution and the 0.175 L solution. For part (b), the concentration of H+ ions is 0.568 M, and the pH is approximately 0.25. Step by step solution Calculate the molarity of the acid in each solution For part (a), we have a 1.75 L solution with a density of 1.00 g/mL and a 37.5% by mass composition of HCl. First, convert the volume into mass: Mass of the solution = volume x density Mass of the solution = (1.75 L x 1000 mL/L) x 1.00 g/mL = 1750 g Now find the mass of HCl: Mass of HCl = (0.375)(1750 g) = 656.25 g Next, calculate the moles of HCl: Moles of HCl = mass / molar_mass Moles of HCl = 656.25 g / (36.45 g/mol) = 18 mol Now, calculate the molarity of the HCl solution: Molarity = moles / volume Molarity = 18 mol / 1.75 L = 10.3 M For part (b), we have 22 g of HBr dissolved in 479 mL of solution. First, calculate the moles of HBr: Moles of HBr = mass / molar_mass Moles of HBr = 22 g / (80.91 g/mol) = 0.272 mol Next, calculate the molarity of the HBr solution: Molarity = moles / volume Molarity = 0.272 mol / 0.479 L = 0.568 M Calculate the concentration of H+ ions and pH For strong acids like HCl and HBr, the concentration of H+ ions in the solution is equal to the molarity of the acid. Therefore, for part (a), [H+] = 10.3 M, and for part (b), [H+] = 0.568 M. Now, calculate the pH of each solution: pH = -log[H+] For part (a): pH = -log(10.3) ≈ -1.01 For part (b): pH = -log(0.568) ≈ 0.25 Calculate the pH of the smaller volume in part (a) We now need to find the pH of 0.175 L of the same HCl solution. Since the concentration of the solution remains the same, the [H+] is still equal to 10.3 M. Therefore, for the smaller volume of the solution: pH = -log(10.3) ≈ -1.01 For part (a), [H+] = 10.3 M and pH ≈ -1.01 for both the 1.75 L solution and the 0.175 L solution. For part (b), [H+] = 0.568 M and pH ≈ 0.25 for the solution with 22 g of HBr dissolved in 479 mL of Key Concepts These are the key concepts you need to understand to accurately answer the question. Molarity of Acid Molarity signifies the concentration of an acid or any solute in a solution, telling us the number of moles of the solute dissolved per liter of solution. To calculate it, you must divide the number of moles of the solute by the volume of the solution in liters. It is a crucial measure in chemistry because reactions depend on the precise molarity of reactants. For instance, in our exercise, to find the molarity of hydrochloric acid (HCl), we first determined the mass of the solution using its density and volume, then worked out the mass and moles of HCl from its percentage composition. Dividing the moles of HCl by the volume of the solution in liters gave us the molarity. The exercise simplifies complex concepts by demonstration, making it an effective teaching tool. Concentration of H+ Ions Understanding H+ Ion Concentration When it comes to strong acids like HCl or hydrobromic acid (HBr), each molecule dissociates completely in water to release H+ ions (protons). The concentration of these ions is directly equivalent to the molarity of the acid in a given volume of water. This one-to-one correspondence simplifies calculations considerably and is crucial to understanding acid behavior. Our problem demonstrates this by directly equating the molarity of strong acids to the concentration of H+ ions. Knowing this concentration is fundamental as it directly influences the pH of the solution – a measure of acidity or basicity that has broad applications in chemistry, biology, environmental science, and medicine. pH Scale The pH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It ranges from 0 to 14, with 7 being neutral, values below 7 acidic, and values above 7 basic. It's a direct way to understand the concentration of H+ ions in a solution without getting overwhelmed with large or small numbers – thanks to the logarithmic conversion. To calculate pH, take the negative logarithm to the base 10 of the H+ ion concentration. Strong acids, as shown in our exercise, can have a pH less than zero, indicating very high acidity. Understanding the pH scale and its calculation is fundamental in various scientific fields, including chemistry, biology, and environmental science, because it affects every chemical process where water is a reactant or a product.	{"url":"https://www.vaia.com/en-us/textbooks/chemistry/chemistry-principles-and-reactions-6-edition/chapter-13/problem-26-find-leftmathrmhright-and-the-mathrmph-of-the-fol/","timestamp":"2024-11-12T00:25:42Z","content_type":"text/html","content_length":"263064","record_id":"<urn:uuid:64fef425-23bc-4fba-9940-41b9b58e6da2>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00141.warc.gz"}
Sudoku Puzzle Sudoku is an intriguing puzzle of patterns and cryptic rules. Regular Sudoku consists of a square grid to be filled in with numbers 1 to 9. The puzzle is to place these symbols in the correct place to complete the grid. Sudokus have been developed in many forms, they do need not be number based and can be letters or colors or shapes or whatever you like; it is a very varied type of puzzle. 数独Sudoku in Japanese/Chinese There is just one simple rule controlling where you can place numbers in the regular Sudoku puzzle. A symbol must occur once and only once in each group of nine grid squares. The groups of nine squares include the rows, columns and regions within the puzzle. Such a simple rule leads to all the amazing range of Sudoku puzzles. Take a look at our pages on the history of Sudoku, Sudoku solution strategy and theory. Sudoku Terminology First let's introduce the terms used on this web site, as not everyone uses the same convention. The whole puzzle area is called the grid, it is divided into rows (horizontal lines) and columns (vertical lines) made up of individual Sudoku squares. To save confusion we use letters rather than numbers to refer to rows and columns. The names are shown in the grid heading. In this Sudoku grid row C (capital C) is highlighted. Note: If we used numbers we would end up having to say things like in row 3 there is only one place for a 5 while in column 2 there are 3. Confusing isn't it? Using letters for grid references makes it easier to Sudoku columns are given a lower case letter. Column e (small e) is highlighted. Using row and column letters lets us unambiguously refer to squares. For example square He is row H column e, this Sudoku square has a 2 allocated in the example puzzle. is a block of nine adjacent Sudoku squares, in the example puzzle the top left region is highlighted. The whole grid is made up of nine . Some Sudoku sites use the term ‘mini-grid’; ‘box’ or ‘sub-grid’ for ‘region’. We think is simpler and easier. A region is referenced by the top-left square, so region Dd is the central region. A symbol must occur once and once only in each of the regions within the grid as well as each row and column. This was one of the that made Sudoku such an interesting puzzle to solve. A ‘group’ is a general term for a group of nine squares in either a row, a column or a region. There is only one simple rule in Sudoku: each Sudoku group of squares must have a unique occurrence of each of the numbers 1 through 9. How to solve Sudoku puzzles The process of solving a Sudoku puzzle is to fill in the empty squares. Each puzzle has a single, 'correct' solution, each unallocated square has one correct value from the 9 possible values. Sometimes it is fairly obvious what number must go in a square while other squares require a great deal of mental torture to solve. To work through all the possibilities it can take half an hour to solve one square! Much like placing a single piece in a jigsaw, there must be a place for a number to go but spotting the correct place may be easy or take an age to do. There is no correct sequence of square allocations to make, different people use their own strategies to solve a Sudoku puzzle and so will solve the squares in a different sequence. However, the end result is always the same, there is only one unique solution, just many ways of getting there. There are a number of standard techniques or strategies to help solve a Sudoku puzzle. We have guides built into our Sudoku solver demonstrating these strategies. We also have a Sudoku strategy page containing a description of all the commonly used methods: only choice, only square, single possibility, excluded hidden twins, naked twins, sub-groups and X-Wing. The more Advanced strategies are explained on a separate page: X-Y Wing and Alternate Pairs. There is a page on Sudoku theory too. You can visit our online discussion forums. Sudoku Dragon comes complete with guides that take you through the most useful solution strategies step by step. Making mistakes If you make an incorrect allocation of a number in a Sudoku puzzle then the puzzle becomes 'unsolvable'. At some later stage you will find an insurmountable contradiction, a number would have to be placed in two squares in the same row, column or region violating the Sudoku rule or else you'll find a square that can't take any of the numbers according to the rule. To correct the mistake you need to backtrack through the allocations that you have made until you find the one in error. Often this is because you overlooked another possibility for a square and thought it was the only choice. Creating Puzzles Skill is required to create a challenging Sudoku puzzle. It is not just a matter of randomly allocating numbers to squares. Firstly, to ensure that there is only one unique solution requires that there is the correct number of initial 'exposed' squares to begin with. If there were only a handful there would be many ways to allocate all the squares - but all Sudoku puzzles can have only a single, unique solution. The challenge is to reveal just enough squares to make the solution both unique and challenging. The pattern of squares can make a pleasing arrangement , and this should be taken into account when creating a Sudoku puzzle. In general the more revealed squares there are the easier the puzzle will be to solve. If the revealed squares are distributed evenly throughout the puzzle, then it will be easier to solve than if regions have very few filled squares. Some of the toughest puzzles have a couple of regions with no squares revealed at all, or when a particular number does not occur in the whole Sudoku puzzle. Solution strategies are discussed in our online forums and strategy page. When Sudoku was published by the Nikoli magazine in Japan they decided to add some extra spice to form true Sudoku puzzles. They decided to make the pattern of revealed squares symmetric. If you turn the Sudoku on its side or upside down the pattern of initial squares is repeated (but not the numbers). Sudoku Dragon supports both a symmetric and a random pattern of initial squares. The random pattern can make the games more challenging to solve although it may be less aesthetically appealing to look at. Sudoku Puzzle Difficulty A good Sudoku puzzle has to have just the right level of difficulty. This decision is tricky because there are many solution strategies and different people will naturally find puzzles more challenging than others. The vital measure in establishing the level of puzzle difficulty is working out which Sudoku strategies are needed to solve it. The easier puzzles require the basic only square; single possibility and only choice rules. Moderate puzzles require some application of the twin and excluded choice rules. Truly challenging puzzles require the discovery of X-Wings, X-Y Wings, alternate pairs or may be even some use of trial and error: backtracking after following a blind alley or two before the correct solution is attained. Sudoku and Jigsaws The closest puzzle to compare to Sudoku is perhaps the humble jigsaw. There are similarities both in the way it works and the pleasure gained by solving it. In a jigsaw there are lots of pieces to fit in to a rectangular pattern, there is only one solution and each piece can only go in one place. Sudoku is also a matter of putting things in the right place. If you like doing jigsaws you'll enjoy Sudoku too. To solve a jigsaw everyone will put the pieces together in different orders. Most people will hunt and separate the edge pieces and then join these up before tackling pieces with distinct markings and then join these up. When nearing the completion of a jigsaw, particularly with problem areas such as large expanses of clear blue sky, you may look out for pieces of a particular shape. There are different strategies to apply depending on the stage of completeness and that makes jigsaws interesting. Sudoku is just the same, there are strategies to use at the different stages of solving the puzzle. Some of these can become a tough trial and error process just like a jigsaw. The joy of successfully completing a jigsaw is akin to that of solving a Sudoku puzzle, when the final square has been filled in, the satisfaction of correct completion is like stepping back to enjoy the whole picture when the final piece has been placed in a jigsaw. Everything is in its proper place. Sudoku puzzle types Starting from the standard Sudoku rule there are many ways to create a range of different types of puzzle. First of all you can change the size of the grid. Using the regular 9x9 grid is just one option. The simpler 4x4 grid is useful for learning the basics of Sudoku and we use it for some of our guides. With 4x4 there are only four symbols and four regions to consider, and so 4x4 can never make a hard puzzle. Stepping up to larger sizes a grid of 16x16 makes a real challenge, because there are now 16 squares with 16 possibilities for each square. With this size there are not enough digits so the letters 'A' through 'I' or 'hexadecimal' digits will do admirably instead. Sudoku Dragon supports puzzles of this size. The sudoku grid size can be arbitrarily increased further to 25x25 and 36x36 and so on, but 16x16 with a total of 256 squares to complete is surely challenging enough; after that the puzzle has too many possibilities to carry around in the average sized mind. You can also use rectangular regions to make up the grid rather than squares. Sudoku Dragon supports seven rectangular grids including: 2x3 grid (about the most common rectangular size you will find) and the 4x5 (or 20x20) monster sized grid. Here's an example of 2x5 rectangles making up a 10x10 puzzle. Our Theme and Variations page describes the many different types of Sudoku available. These include Chinese numbers; Word Sudoku; X Sudoku and Super-Sudokus or Samurai Sudoku with their overlapping 3x3 puzzles that have five overlapping puzzles to solve in one large puzzle. How many possible puzzles? As there are so many Sudokus printed these days, surely all the possible grids have now been solved? Well you may think so. After a little thought it is clear there are quite a few new puzzles left and we are unlikely to run out of Sudokus in the near future. For each row in isolation there are 9! (shorthand for nine factorial) possible permutations of numbers for the squares which gives 362,880 possible orderings for just one row. Each of these rows can be combined with 8 other rows, and temporarily ignoring the Sudoku rule for columns there would be 9! to the power 9 which works out to be about 10 to the power 50 possible grids (that's 10 with 50 zeroes after it). 109,110,688, 415,571,316, 480,344,899, 355,894,085, 582,848,000, 000,000 Applying the Sudoku rule to columns as well as rows reduces this figure substantially. Just considering unique solutions for rows and columns and not regions means that the second row only 8 options to choose from for each square and 7 for the third etc. so this gives a much smaller number. More grids can be knocked out if regions are taken into account as well as rows and columns. Fortunately some clever people have used super sized calculators to do the maths and claim there are 6,670,903,752,021,072,936,960 unique Sudoku grids of size 9x9. As this is roughly the number of stars in the observable universe, that is plenty to be getting on with. But if you then start determining symmetries including rotations and swaps then the number of 'effectively different' puzzles goes down to 5,472,730,538. This large number means that is you solved one puzzle every second every day you would not need to repeat the same one in over a hundred years. These puzzles would all require different strategies to be used for their solution. See: For more mathematical analysis ➚ and Some mathematical analysis into the actual number of possible puzzles ➚. History page. For help on solving Sudoku puzzles visit our Strategy page or our theory page. See also Background ➚ news story about the rise in popularity of the puzzle Daily Sudoku ➚ Provides a daily puzzle but no solver. The Guardian ➚ Daily sudokus of various types and an active online community. Please share your interest on Facebook, Twitter, Pinterest, Tumblr or Mix using the buttons. Please visit our (secure) contact page to leave any comments you may have. Copyright © 2005-2024 Sudoku Dragon	{"url":"https://sudokudragon.com/sudoku.htm","timestamp":"2024-11-06T21:33:04Z","content_type":"text/html","content_length":"23307","record_id":"<urn:uuid:b2856d04-9eff-440a-8b1f-397bfe797f34>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00520.warc.gz"}
Explaining Haskell RankNTypes for The Glasgow Haskell Compiler supports a language extension called RankNTypes which I’ve had my problems to understand. The moment I understood that it mostly refers to first-order logic universal quantification things became easier, though… but first let’s explore why we need it in a step-by-step example. length :: forall a. [a] –> Int 1 :t length 2 -- length :: [a] -> Int 3 length [1,2,3] 4 -- 3 5 let intLength :: [Int] -> Int; intLength = length 6 :t intLength 7 -- intLength :: [Int] -> Int 8 intLength [1,2,3] 9 -- 3 We start with the well-known polymorphic function length in a fresh GHCI session. Above we see how the type checker instantiates a to be Int in the type of intLength. Likewise we could create a function charLength – anyway, length can be instantiated to oblige to a list of any type we want, so it is defined for all possible types a. For the sake of simplicity, I’ll call a function like intLength (which actually corresponds to instantiating the type variable a of length) a version of length. As a matter of fact, a normal Haskell type signature such as [a] -> Int always implies that the type variable(s) are universally quantified with 1 forall section located at the beginning of the type declaration. length’s type thus corresponds to forall a. [a] -> Int. We call such a type a Rank-1-Type as there is 1 forall in the type annotation. The fact that we can omit the forall usually – and aren’t used to it as a consequence – will make things look complicated when we actually need it, as we’ll see later on. In the end, forall provides a scope just like its first-order logic equivalent. Apply a length-like function to a list 1 let apply :: ([a] -> Int) -> [a] -> Int; apply f x = f x 2 apply length "hello world" 3 -- 11 4 apply intLength [1,2,3] 5 -- 3 The apply function just applies a function that takes a list and returns an Int (like length does) to a value. Nothing fancy nor useful at all, obviously. Still, let’s note that under the hood the type of apply is forall a. ([a] -> Int) -> [a] -> Int. So far, so good, the type checker is happy. Now let’s a write a function applyToTuple that applies a function like length to a tuple of lists so that the lists of the tuple can be of different types. Apply a length-like function to a tuple of lists 1 let applyToTuple f (a@(x:xs),b@(y:ys)) = (f a, f b) :: (Int, Int) 2 applyToTuple length ("hallo",[1,2,3]) 3 --No instance for (Num Char) 4 -- arising from the literal `1' 5 -- ... 6 :t applyToTuple 7 -- applyToTuple :: ([t] -> Int) -> ([t], [t]) -> (Int, Int) I wrote applyToTuple without a full type signature. :: (Int,Int) just makes sure my wanted result type and by the help of the list destructuring a@(x:xs) I make sure that the type inference algorithm will conclude that I have a tuple of lists in mind. Consequently, the type of the function given to applyToTuple is inferred to correspond to length’s type; at least, that’s what I would expect However, type inference of applyToTuple does not result in the type I had in mind. As we can see the types of lists in the tuple ([t],[t]) are the same so that calling applyToTuple length with a heterogeneous tuple like ("hallo",[1,2,3]) doesn’t work. Being stubborn I could then try “forcing” the type by providing a type signature: 1 let applyToTuple :: ([a] -> Int) -> ([b],[c]) -> (Int, Int); applyToTuple f (x,y) = (f x, f y) 2 -- Couldn't match type `b' with `a' ... 3 -- Couldn't match type `c' with `a' ... This attempt also fails as GHCI complains about the fact that the types b and a, c and a respectively, do not match! However, the length-like function ([a] -> Int) should be applicable to a list of whatever type, shouldn’t it?!? That’s the moment you’d start doubting either GHCI or your mental health as you know precisely that it should be possible to write such a function. After all, you know intuitively that it is possible to apply a function like length to both parts of a heterogeneous tuple of lists as in the code below; doing that in a more generic way in a function like applyToTuple should be possible as well! 1 -- Obviously, that works without a problem: 2 (\(a,b) -> (length a, length b)) ("hallo",[1,2,3]) 3 -- (5,3) applyToTuple :: (forall a.[a] –> Int) –> ([b],[c]) –> (Int, Int) Well, there is just one explanation: the type ([a] -> Int) ->([b],[c]) -> (Int, Int) is not really what we need for our purpose. In fact, we need RankNTypes! We first enable the extension in GHCI and can then write the correct applyToTuple implementation using the forall keyword in the type of the first parameter function. (If you want to use the RankNTypes extension in a file to compile, you actually need to add {-# LANGUAGE RankNTypes #-} at the top of the file) 1 :set -XRankNTypes 2 let applyToTuple :: (forall a.[a] -> Int) -> ([b],[c]) -> (Int, Int); applyToTuple f (x,y) = (f x, f y) 3 applyToTuple length ("hello", [1,2,3]) 4 -- (5,3) This time it works! :–) We noted earlier that every Haskell type signature’s type variables are implicitly universally quantified by an ‘invisible’ forall section. Thus, under the hood we get the types as follows: 1 -- just a reminder: 2 -- length :: forall a. [a] -> Int 3 let intLength :: [Int] -> Int; intLength = length 5 -- applyToTuple: 6 let applyToTuple :: forall a b c. ([a] -> Int) -> ([b], [c]) -> (Int, Int); applyToTuple f (x,y) = (f x, f y) 7 -- correct applyToTuple: 8 let applyToTuple :: forall b c. (forall a. [a] -> Int) -> ([b], [c]) -> (Int, Int); applyToTuple f (x,y) = (f x, f y) Now things get clearer: The function in the type of the correct applyToTuple has the type (forall a. [a] -> Int) which is exactly the type given for length above, hence it works. On the other hand, the type ([a] -> Int) of the function parameter in the wrong applyToTuple type signature looks like the type of length but it isn’t! Have a look at what the type checker would “think” confronted with the wrong applyToTuple type signature. When it reads the expression applyToTuple length it would expect the type variables a, b and c to be different concrete types, so ([a] -> Int) might become ([Char] -> Int) or ([Int] -> Int) like our intLength function, shortly, some version of length. In the implementation (f x, f y) seeks to apply that version of length to two lists of different types – however, any version of length expects its list to always be of 1 concrete type only, e.g. Int in the case of our function intLength, consequently, the type checker refuses the lists of the tuple to possibly be of different types! Why does the correct definition of applyToTuple work then? It expects a length-like function of type (forall a. [a] -> Int), that’s a function which works for all types a, no matter what type you throw at it! Thus, it forces that function to be a polymorphic function just like length and rules out any candidate version of length (like intLength) as a consequence. Since you can throw a list of any type at that function it can deal with the 2 lists of different types and the code compiles! Using RankNTypes and the forall keyword you can specify that a function’s argument needs to be a polymorphic function (like length in our example). In spite of the fact that you can omit the top-level forall in the type signature of a polymorphic type, you need to include it when you reference it as a parameter. In a future blog post I will investigate an important application of RankNTypes in the Haskell standard library. It will be about the ST monad which provides a safe environment for mutation in Haskell with the help of RankNTypes. Mutation and Haskell?! Yes, you can do it thanks to RankNTypes! PS: There is a nice stackoverflow thread which investigates the use of “forall” in other language extensions as well. Actually, my “applyToTuple” function is based on that answer of the thread.	{"url":"https://sleepomeno.github.io/blog/2014/02/12/Explaining-Haskell-RankNTypes-for-all/","timestamp":"2024-11-01T22:13:45Z","content_type":"text/html","content_length":"46830","record_id":"<urn:uuid:3aa7aceb-71ba-4d17-8679-48aa8ca0a28d>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00890.warc.gz"}
The complement edge metric dimension of graphs In this paper, we construct the new concept namely the complement edge metric dimension on the graph, which is the result of combining two concepts. The first concept is edge metric dimension and in the second concept is complement metric dimension. Let given a graph G=(V(G), E(G)) where V(G) = {v1, v2, ..., vn} and also a set M = {m1, ..., mk} ⊂ V(G) that indicate connected graph and an ordered set, respectively. The representation of the edge e=xy∈E(G) with respect to M can be written as r(e\|M) = (d(e, mi))=(min{d(x, mi), d(y, mi)}) where d(x, mi) and d(y, mi) indicate the distance from vertex x to mi and from vertex y to mi for i∈{1,2, ..., k}, respectively. The definition of a complement edge resolving set of G is if any two distinct edges in G have the same representation with respect to M. If set M has maximum cardinality then it is called a complement edge basis. Next, the number of vertices of M is called the complement edge metric dimension of G that can be written as edim¯(G). As a result, the complement edge metric dimension which is implemented in basic graphs, namely graph Pn, graph Sn, graph Cn, and graph Kn will be determined. Publication series Name AIP Conference Proceedings Volume 2641 ISSN (Print) 0094-243X ISSN (Electronic) 1551-7616 Conference 7th International Conference on Mathematics: Pure, Applied and Computation: , ICoMPAC 2021 Country/Territory Indonesia City Surabaya Period 2/10/21 → … Dive into the research topics of 'The complement edge metric dimension of graphs'. Together they form a unique fingerprint.	{"url":"https://scholar.its.ac.id/en/publications/the-complement-edge-metric-dimension-of-graphs","timestamp":"2024-11-11T07:28:03Z","content_type":"text/html","content_length":"55118","record_id":"<urn:uuid:7dca5ad0-f866-4c47-a911-f2609e1cd88e>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00056.warc.gz"}
Mathematics for Physics Chain complexes A fundamental intuitive fact reproduced in this formalism is that the boundary of a boundary is zero. A useful algebraic generalization of this idea is the chain complex, defined to be a sequence of homomorphisms of abelian groups ${\partial_{n}\colon C_{n}\to C_{n-1}}$ with ${\partial_{n}\partial_{n+1}=0}$. In our case the abelian groups ${C_{n}}$ are the ${n}$-chains, and the chain complex can be illustrated as follows: Note that the image of ${\partial_{n+1}}$ is contained in the kernel of ${\partial_{n}}$; if these are in fact equal, the chain complex is an exact sequence, defined to be any sequence of homomorphisms for which the image of one object is the kernel of the next. A short exact sequence is of the form $\displaystyle 0\longrightarrow N\overset{\phi}{\longrightarrow}E\overset{\pi}{\longrightarrow}G\longrightarrow0,$ and any longer sequence is called a long exact sequence. ${\phi}$ is injective and ${\pi}$ is surjective, so a short exact sequence can be viewed as an embedding of ${N}$ into ${E}$ with ${G =E/N}$. For groups, a short exact sequence is called a group extension, or “${E}$ is an extension of ${G}$ by ${N}$.” Note that ${N}$ is normal in ${E}$ since it is the kernel of ${\pi} $, and thus ${G\cong E/N}$. A central extension is one where ${N}$ also lies in the center of ${E}$. Δ A group extension as above is sometimes described as “${E}$ is an extension of ${N}$ by ${G}$.” A long exact sequence is sometimes defined as any exact sequence that is not short, or as one which is infinite.	{"url":"https://www.mathphysicsbook.com/mathematics/algebraic-topology/constructing-surfaces-within-a-space/chain-complexes/","timestamp":"2024-11-09T17:05:36Z","content_type":"text/html","content_length":"72480","record_id":"<urn:uuid:1dd1b80e-4afb-4617-96b3-029cc77bfc78>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00838.warc.gz"}
What is a Fractal \| Diana de Avila Digital Fractal Artist Sudden Savant \| USA top of page My approach to art: Soon after my artistic awakening in 2017, I was creating art using fractal geometry concepts even before I knew what a fractal was. I was mainly engaging in “self-similarity” which involved using the same shape at different scales. My art on the left “Arabian Nights” is a perfect example of “self-similarity.” I found an amazing amount of comfort in the controlled regularity and patterns. Patterns are very soothing to me. A fractal is a never-ending pattern. Fractals are infinitely complex patterns that are self-similar across different scales. They are created by repeating a simple process over and over in an ongoing feedback loop. Driven by recursion, fractals are images of dynamic systems – the pictures of Chaos. Geometrically, they exist in between our familiar dimensions. Fractal patterns are extremely familiar, since nature is full of fractals. For instance: trees, rivers, coastlines, mountains, clouds, seashells, hurricanes, etc. Abstract fractals – such as the Mandelbrot Set – can be generated by a computer calculating a simple equation over and over. Fractal patterns repeat themselves at different scales - this is called “self-similarity.” They can be found in branching (like the branches on a tree), through spirals (think of a nautilus shell), and geometric (like the Sierpinski Triangle which is made by repeatedly removing the middle triangle from the prior generation. The number of colored triangles increases by a factor of 3 each step, 1,3,9,27,81,243,729, etc. Algebraic fractals use a simple formula that repeats and repeats. The Mandelbrot Set is probably one of the most familiar fractal equations. We start by plugging a value for the variable ‘C’ into the simple equation below. Each complex number is actually a point in a 2-dimensional plane. The equation gives an answer, ‘Znew’ . We plug this back into the equation, as ‘Zold’ and calculate it again. We are interested in what happens for different starting values of ‘C.’ Generally, when you square a number, it gets bigger, and then if you square the answer, it gets bigger still. Eventually, it goes to infinity. This is the fate of most starting values of ‘C.’ However, some values of ‘C’ do not get bigger, but instead get smaller, or alternate between a set of fixed values. These are the points inside the Mandelbrot Set, which we color black. Outside the Set, all the values of ‘C’ cause the equation to go to infinity, and the colors are proportional to the speed at which they expand. You can read about the Julia Set and more in depth at the Fractal Foundation. I am attaching an educator’s guide that is 20 pages long and explains the concepts perfectly. https:// Fractals are a combination of Science, Math and Art. Other resources: A wonderful article from IBM about Benoit Mandelbrot and fractal geometry. bottom of page	{"url":"https://www.dianadeavila.com/what-is-a-fractal","timestamp":"2024-11-14T14:49:44Z","content_type":"text/html","content_length":"848607","record_id":"<urn:uuid:2ea1f18a-56cd-4c1b-8dc9-f7a15214de6c>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00423.warc.gz"}
The Black Shard Holiday Giveaway Following up from the novel The Magic Warble by Victoria Simcox is the sequel The Black Shard. Reuniting Kristina with old friends back in the magical land of Bernovem. Prince Werrien becomes fascinated with an unusual seeing stone, the “Black Shard”, Kristina is haunted by a ghostlike old hag. Check out information from the first series The Magic Warble and follow on facebook Win a signed copy of The Black Shard! Contest closes 11:59pm Eastern December 4, 2011. Open US winner has 2 days to respond to my email. Leave your email in to post or available in profile, I will not search for it. Winner will be chosen by random org. Entries that does not follow the rules will not win. Mandatory: Just tell me a fact from The Black Shard or about author. *******ENTRIES*********Have you already read The Magic Warble? Follow both my blog & Victoria's Website Google Friend Connect. (5 entries each) Blog about the giveaway-leave your link (5entries) Follow @VictoriaSimcox @lifesandcastle on twitter & tweet (1 time per day) "#win signed copy of The Black Shard book #giveaway @VictoriaSimcox @lifesandcastle http://goo.gl/LRjsT #win ends 12/04" Subscribe to my blog by email. (2 entries) Enter The Night Owl Mama or giveaways on my site- one entry for each you enter. Add this giveaway to a contest linky & leave link (3 entries each link, up to 3contest listings) Give this post a google+1 (1 entry) Give me a Klout +K (1 k per day, daily) Stumble Life Is A SandCastle Blog Disclosure: Please note this giveaway was presented by Company or PR. I did not received a product to review. I did not receive any other compensation. All opinions are my own. 175 comments: The book starts out with Kristina having a horrible time at summer camp. cwitherstine at zoominternet dot net I haven't read the Magic Warble yet, but I have it put up for my daughter for Christmas and would love to win the sequel to give to her as well. cwitherstine at zoominternet dot net I stumbled Life is a Sandcastle. cwitherstine at zoominternet dot net email subscriber 1 cwitherstine at zoominternet dot net email subscriber 2 cwitherstine at zoominternet dot net I entered the Totino's giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Hip Hugger giveaway at Night Owl Mama. cwitherstine at zoominternet dot net gfc follower 1 cwitherstine at zoominternet dot net gfc follower 2 cwitherstine at zoominternet dot net gfc follower 3 cwitherstine at zoominternet dot net gfc follower 5 cwitherstine at zoominternet dot net Victoria Simcox gfc follower 1 cwitherstine at zoominternet dot net Simcox gfc follower 2 cwitherstine at zoominternet dot net Simcox gfc follower 3 cwitherstine at zoominternet dot net Simcox gfc follower 4 cwitherstine at zoominternet dot net Simcox gfc follower 5 cwitherstine at zoominternet dot net I entered the Kidtoons giveaway. cwitherstine at zoominternet dot net The author lives in Western Washington and was born in Canada! I follow Life is a Sandcastle on GFC as jsweeps318 Entry 1 I follow Life is a Sandcastle on GFC as jsweeps318 Entry 2 I follow Life is a Sandcastle on GFC as jsweeps318 Entry 3 I follow Life is a Sandcastle on GFC as jsweeps318 Entry 4 I follow Life is a Sandcastle on GFC as jsweeps318 Entry 5 I follow Victoria's blog as jsweeps319 Entry 1 I follow Victoria's blog as jsweeps319 Entry 2 I follow Victoria's blog as jsweeps319 Entry 3 I follow Victoria's blog as jsweeps319 Entry 4 I follow Victoria's blog as jsweeps319 Entry 5 I follow you and Victoria on twitter and tweeted: I subscribe by email. Entry 1 I subscribe by email. Entry 2 I Google +1ed this post. Victoria was born in Scarborough, Ontario, Canada, to an Austrian immigrant mother, and a Dutch immigrant father. Thanks for the awesome giveaway :) fattybumpkins at yahoo dot com I entered the General Mills giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I am following your blog with GFC fattybumpkins at yahoo dot com I am following your blog with GFC fattybumpkins at yahoo dot com I am following your blog with GFC fattybumpkins at yahoo dot com I am following your blog with GFC fattybumpkins at yahoo dot com I am following your blog with GFC fattybumpkins at yahoo dot com I am following your blog with GFC fattybumpkins at yahoo dot com I am following Victoria's blog with GFC (missreneer) fattybumpkins at yahoo dot com I am following Victoria's blog with GFC (missreneer) fattybumpkins at yahoo dot com I am following Victoria's blog with GFC (missreneer) fattybumpkins at yahoo dot com I am following Victoria's blog with GFC (missreneer) fattybumpkins at yahoo dot com I am following Victoria's blog with GFC (missreneer) fattybumpkins at yahoo dot com the author is from canada vmkids3 at msn dot com I entered the Mr. Biggs giveaway. cwitherstine at zoominternet dot net The author is Canadian. gfc Victoria is a home-schooling mother of twelve years and an elementary school art teacher of eleven years follow Victoria Simcox's Blog GFC-karenmed409 2 follow Victoria Simcox's Blog GFC-karenmed409 3 follow Victoria Simcox's Blog GFC-karenmed409 4 follow Victoria Simcox's Blog GFC-karenmed409 5 follow Victoria Simcox's Blog GFC-karenmed409 follow your Blog GFC-karenmed409 2 follow your Blog GFC-karenmed409 3 follow your Blog GFC-karenmed409 4 follow your Blog GFC-karenmed409 5 follow your Blog GFC-karenmed409 Give this post a google+1 Give you a Klout +K today following both on twitter-gummasplace entered the Thomas & Friends TrackMaster Cranky & Flynn Giveaway entered Mr. Biggs in the City Giveaway entered the General Mills and Dc Comic Books Giveaway over at Night Owl Mama I entered the Beethoven giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Beethoven giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Beethoven giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Hexbug giveaway at Night Owl Mama. cwitherstine at zoominternet dot net gave klout today entered the HEX BUG Nano giveaway over at night owl mama I entered the Harry Potter giveaway. cwitherstine at zoominternet dot net I entered the Pokemon giveaway. cwitherstine at zoominternet dot net I entered the Kroger giveaway at Night Owl Mama. cwitherstine at zoominternet dot net daily klout today Entered the Doubled side Magnetic from Learning Resource Giveaway I entered the Magnetic Numbers giveaway. cwitherstine at zoominternet dot net entered the Pokemon Giveaway gave daily klout today gave daily klout today entered Tricia's Mattel giveaway I entered the Hotwheels giveaway at NIght Owl Mama. cwitherstine at zoominternet dot net I entered the Funrise giveaway. cwitherstine at zoominternet dot net gave daily klout today Entered the Funrise Giveaway I haven't yet read it, but I want to. I follow both yor blog & Victoria's Website Google Friend Connect. (5 entries each) - Carolsue #1 Digicats {at} Sbcglobal {dot} Net I haven't yet read it, but I want to. I follow both yor blog & Victoria's Website Google Friend Connect. (5 entries each) - Carolsue #2 Digicats {at} Sbcglobal {dot} Net I haven't yet read it, but I want to. I follow both yor blog & Victoria's Website Google Friend Connect. (5 entries each) - Carolsue #3 Digicats {at} Sbcglobal {dot} Net I haven't yet read it, but I want to. I follow both yor blog & Victoria's Website Google Friend Connect. (5 entries each) - Carolsue #4 Digicats {at} Sbcglobal {dot} Net I haven't yet read it, but I want to. I follow both yor blog & Victoria's Website Google Friend Connect. (5 entries each) - Carolsue #5 Digicats {at} Sbcglobal {dot} Net I subscribe to your newsletter via e-mail. #1 Digicats {at} Sbcglobal {dot} Net I subscribe to your newsletter via e-mail. #2 Digicats {at} Sbcglobal {dot} Net I follow @VictoriaSimcox @lifesandcastle on twitter (MsCarolsueA) and I tweeted: Digicats {at} Sbcglobal {dot} Net I entered the Busy Town giveaway. cwitherstine at zoominternet dot net I entered the Gourmet Gift Baskets giveaway. cwitherstine at zoominternet dot net I entered the Kerfluffle giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Carmex giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the White Castle giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Spy Kids giveaway . cwitherstine at zoominternet dot net I entered the Cheerios giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Prep and Landing giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Hormel giveaway. cwitherstine at zoominternet dot net I entered the Ugly Sofa giveaway. cwitherstine at zoominternet dot net I entered the French Toast giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Energizer giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Glowberry Bears giveaway at Night Owl Mama. cwitherstine at zoominternet dot net I entered the Smart Step Home giveaway. cwitherstine at zoominternet dot net I entered the NCircle giveaway. cwitherstine at zoominternet dot net She has pets named Pipsy, Frodo and Fritz gfc follower gfc follower 3 gfc follower 4 gfc follower 5 gfc follower of Victoria's website gfc follower of Victoria's website 2 gfc follower of Victoria's website 3 gfc follower of Victoria's website 4 gfc follower of Victoria's website 5 twitter follower of both/tweet email subscriber email subscriber entered natl geographic elephant toy entered kerfluffle at Night Owl Mama entered prep and landing at Night Owl Mama entered white castle at Night Owl Mama entered energizer at Night Owl Mama entered glowberry at Night Owl Mama klout k+ linky 3 linky #138 linky #138 2 linky #138 3 linky #162 linky #162 2 linky #162 3 The author lives in Western Washington Diane Baum entered ncircle ent Victoria wa born in scarborough ontario canada. gfc follower daveshir2005 gfc follower daveshir2005 gfc follower davehsir2005 gfc follower daveshir2005 gfc follower daveshir2005 +1'd the post as shirley pebbles Victoria was born in Scarborough, Ontario, Canada, to an Austrian immigrant mother, and a Dutch immigrant father i follow you via GFC Entry 1 i follow you via GFC Entry 2 i follow you via GFC Entry 3 i follow you via GFC Entry 4 i follow you via GFC Entry 5 i follow Victoria via GFC Entry 1 i follow Victoria via GFC Entry 2 i follow Victoria via GFC Entry 3 i follow Victoria via GFC Entry 4 i follow Victoria via GFC Entry 5 i follow you both on twitter i'm an email subscriber entry 1 i'm an email subscriber entry 2 i clicked the +1 button Her chihuahua is named Pipsy! I learned the author was born in Canada. Thank you She enjoys managing her two older children's Celtic band. garrettsambo@aol.com She enjoys managing her two older children's Celtic band. garrettsambo@aol.com I entered the Tidy Books giveaway. cwitherstine at zoominternet dot net I entered the Step 2 giveaway. cwitherstine at zoominternet dot net	{"url":"https://lifeisasandcastle.blogspot.com/2011/11/black-shard-holiday-giveaway.html?showComment=1320662355177","timestamp":"2024-11-08T15:39:18Z","content_type":"application/xhtml+xml","content_length":"321125","record_id":"<urn:uuid:f64d8f80-793f-4e03-8f77-e3c7c02d7b79>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00603.warc.gz"}
Wilma has a hearing loss that occurred in the outer or middle ear. This situation prevented sound ... Click here to go to the answer Wilma has a hearing loss that occurred in the outer or middle ear. This situation prevented sound waves from traveling to the inner ear. Wilma has a ________. ◦ sensorineural hearing loss ◦ mixed hearing loss ◦ auditory neuropathy spectrum disorder ◦ conductive hearing loss Marked as best answer by viki on Sep 7, 2019 • Sr. Member • Posts: 323 A FREE account is required to view all solutions! Click to Register Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium. Answer Preview Only 39% of students answer this correctly	{"url":"https://homeworkclinic.com/index.php?PHPSESSID=dujmrncpodtq2k100qnfhn0oc7&topic=785925.0","timestamp":"2024-11-06T17:04:40Z","content_type":"application/xhtml+xml","content_length":"73767","record_id":"<urn:uuid:b0c3abd1-5029-49fe-87c2-135f1c061aec>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00882.warc.gz"}
On Solutions of Possibilistic Multi- objective Quadratic Programming Problems Ammar,E.E., and Khalifa, H.A. (2003). Fuzzy portfolio selection problem- quadratic programming approach. Chaos, Solitons and Fractels, Vol. 18(5), pp. 1045-1054. Ammar, E. E. (2008). On solutions of fuzzy random multiobjective quadratic programming with application in portfolio problem. Information Sciences, Vol. 178(2), pp. 468-484. Ammar,E.E., and Khalifa, H.A. (2015). On rough interval quadratic programming approach for minimizing the total variability in the future payments to portfolio selection problem. International Journal of Mathematical Archive, Vol. 6(1), pp. 67-75. Bazaraa, M. S., Jarvis, J. J., and Sherall, H. D. (1990). Linear Programing and Network Flows, Jon& Wiley, New York. Canestrelli, E., Giove, S., and Fuller, R. (1996). Stability in possibilistic quadratic programming. Fuzzy Sets and Systems, Vol. 82(1), pp. 51-56. Dubois, D., and Prade, H. (1980). System for linear fuzzy constraints. Fuzzy Sets and Systems, Vol. 3, pp. 37- 48. Horst, R., and Tuy, H. (1993). Global Optimization: Deterministic Approach. Berlin: Springer- Verlage. Hussein, M. L. (1992). On convex vector optimization problems with possibilistic weights. Fuzzy Sets and Systems, Vol. 51, pp. 289- 294. Inuiguchi, M., and Ramik, J. (2000). Possibilistic linear programming: a brief review of fuzzy mathematical programming and a comparison with stochastic programming in portfolio selection problem. Fuzzy Sets and Systems, Vol. 111(1), pp. 3-28. Jana, P., Roy, T.K., and Mazunder, S. K. (2009). Multi- objective possibilities model for portfolio selection with transaction cost. Journal of Computational and Applied Mathematics, Vol. 228, pp. 188-196. Kassem, M.A. (1998). Stability of possibilistic multiobjective nonlinear programming problems without differentiability. Fuzzy Sets and Systems, Vol. 94, pp. 239-246. Khalifa, H. A., and ZeinEldein, R. A. (2014). Fuzzy programming approach for portfolio selection roblems with fuzzy coefficients. International Journal of Scientific Knowledge. Vol. 4(7), pp. 40-47. Khalifa, H. A. (2016). An interactive approach for solving fuzzy multiobjective nonlinear programming problems. Journal of Mathematics, Vol. 24(3), pp. 535-545. Kheirfam, B. (2011). A method for solving fully fuzzy quadratic programming problems. Acta Universitatis Apulensis, Vol. 27, pp. 69-76. Luhandjula, M. K. (1987). Multiple objective programming problems with possibilistic coefficients. Fuzzy Sets and Systems, Vol. 21, pp. 135-145. Miettinen, K.M. (1999). Nonlinear Multiobjective Optimization. Kulwer A cademic Publishers. Narula, C. S., L. Kirilov and V. Vassiley (1993). An Interactive Algorithm for Solving Multiple Objective Nonlinear Programming Problems, Multiple Criteria Decision Making, and Proceeding of the tenth International Conference: Expand and Enrich the Domain of Thinking and Application, Berlin: Springer Verlag. Pardalos, P. M., and Rosen, J. B. (1987). Constrained global optimization: Algorithms and Applications. Lecture notes in Computer Science, volume 268, Berlin: Springer- Verlage. Sakawa, M., and Yano, H. (1989). Interactive decision making for multiobjective nonlinear programming problems with fuzzy parameters. Fuzzy Sets and Systems, Vol. 29, pp. 315-326. Steuer, R. E. (1983). Multiple Criteria Optimization: Theory, Computation and Application, New York: Jon & Wiley. Tanaka, H., Guo, P., and Zimmerman, H- J. (2000). Possibility distributions of fuzzy decision variables obtained from possibilistic linear programming problems. Fuzzy Sets and Systems, Vol. 113, pp. Wang, M. X., Z. L. Qin and Y. D. Hu (2001). An interactive algorithm for solving multicriteria decision making: the attainable reference point method, IEEE Transaction on Systems, Man, and Cybernetics- part A : Systems and Humans, Vol. 31(3), pp.194-198. Zadeh, L. A. (1965). Fuzzy sets. Information Control, Vol. 8, pp. 338-353. Zadeh, L. A. (1978). Fuzzy sets as a basis for a theory of possibility. Fuzzy Sets and Systems, Vol. 1, pp. 3-28.	{"url":"http://www.ijsom.com/article_2728.html","timestamp":"2024-11-11T09:49:49Z","content_type":"text/html","content_length":"50430","record_id":"<urn:uuid:610e223d-20fd-491a-8d41-990e65258f83>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00777.warc.gz"}
Top 5 Excel functions you might not know So, it went something like this… Editor: “I’d like you to write an article about the top five Excel functions accountants need to know.” Me: “Hmm, the most common ones include SUM, IF, SUMIF, SUMIFS, or SUMPRODUCT; VLOOKUP (yuck!) or INDEX(MATCH); OFFSET; MOD; and one of MAX and MIN — that will be a riveting read …” Editor: “How about five powerful functions they should be using?” Me: “That might be some of the new functions such as XLOOKUP, SORT, UNIQUE, FILTER, and SEQUENCE … I have written a lot about these recently, and besides, those are available only on Office 365, not Excel 2019, or Excel 2013, or Excel 2010, or …” Editor: “OK, I get the point. How about the top five functions you should be using that have been around for a while and are accessible to standard Excel users?” Me: “Good idea!” There you have it. Dear reader, I present the top five functions that are available right now (and have been for some time) that you might not be using. These are not necessarily your usual suspects, in alphabetical order. 1. AGGREGATE You could argue this is the most complicated Excel function of all time. AGGREGATE began life in Excel 2010. For those who desire greater sesquipedalian loquaciousness (look it up), its syntax may give even more comfort, as it has two forms: 1. Reference: AGGREGATE(function_number, options, ref1, [ref2], …). 2. Array: AGGREGATE(function_number, options, array, [optional_argument]), where: • function_number denotes the function that you wish to use. Similar to the SUBTOTAL function, function_number allocates integer values to various Excel functions: • options specifies which values may be ignored when applying the chosen function to the range. If the options parameter is omitted, the AGGREGATE function assumes that options is set to zero (0). The options argument can take any of the following values: • ref1 is the first numeric argument for the function when using the Reference syntax. • ref2, … is optional. Numerical arguments may number two through 253 for the function when using the Reference syntax. • array is an array, array formula, or reference to a range of cells when using the Array syntax. • optional_argument is a second argument required if using the LARGE, SMALL, PERCENTILE.INC, QUARTILE.INC, PERCENTILE.EXC, or QUARTILE.EXC function when using the Array syntax: As already mentioned, AGGREGATE is analogous to an extension of the SUBTOTAL function insofar that it uses the same function_number arguments, adding another eight. SUBTOTAL allows you to use the 11 functions including/excluding hidden rows, which results in 22 combinations. However, AGGREGATE goes further and takes the 19 functions and allows for eight alternatives for each, which results in 152 combinations — and that’s not even considering the Reference or Array syntax approaches! It just all sounds, well, tremendously complicated. This example Excel file helps demystify. In practice, it’s not that bad. This is because, since this function was created, screen tips will appear as you type in order to nudge you in the right direction. For example, let’s say you wanted the third-largest number in the following list: From inspection, the third-largest value is the amount in cell A2 (the value “5”), but if you use the usual formula for this = LARGE(A2:A10,3), you will get the value #REF!, as this is the first error that Excel comes across as it works down the list. This is where you can use AGGREGATE to ignore these errors. If you type in =AGGREGATE(, you will get the following screen tip scroll list: By typing “14” or selecting “14 – LARGE” from the pop-up list, you now know you are on the right track. After typing a comma, Excel then continues to help you: Again, by either typing a number or pointing and clicking, an appropriate choice may be made. I want to ignore errors, so I need to choose “2”, “3”, “6”, or “7”, depending upon what else should be ignored. I will choose “6” — ignore error values only and then type another comma so that the screen tips keep coming thick and fast: Now, Excel is seeking the references for evaluation. It appears to be possible that this can be in the form of a list (the array) or else discrete cell references and/or values. In this example, I will enter the range and type another comma: Now, Excel appears to be looking for the other argument for LARGE() or else another reference. This is not correct. The screen tip does not update automatically. The syntax required is now just as it would be if we had typed in the underlying function, ie, =LARGE(array, k). In this instance, this syntax always requires the fourth value to be k, the integer denoting the kth-largest item in the In this example, I will just type the value “3” and close brackets. Therefore, we arrive at the following formula: which generates the correct answer “5”. The formula might look counterintuitive, but Excel has helped us every step of the way. As my oft-misquoted English teacher always used to say, practice makes perfect. Please see the attached Excel file for more examples. To summarise, like SUBTOTAL, the AGGREGATE function is designed for columns of data (vertical ranges), not for rows of data (horizontal ranges). For example, when you subtotal a horizontal range using option 1, such as AGGREGATE(1, 1, ref1), hiding a column does not affect the aggregate sum value, although hiding a row in vertical range does affect the aggregate. If a second ref argument is required but not provided, AGGREGATE returns a #VALUE! error. If one or more of the references are three-dimensional references, AGGREGATE, like above, returns a #VALUE! error. 2. EOMONTH Dates are very important to accountants and should not just be hard-coded into a spreadsheet. We often need them to vary. We tend to work with month end dates, and this is where this function becomes invaluable. We usually run across one of the top rows in an Excel worksheet as part of a time series analysis: In this example, a monthly model has been constructed starting in July 2020. The dates in cells J5 onwards are formatted to show only the month and year. However, if I were to format the cell as General instead (Ctrl+1), note that the Sample (circled in red) would be displayed as follows: In other words, 31 July 2020 is no more than a number: 44,043. Microsoft Excel for Windows supports what is called the 1900 date system. This means that 1 January 1900 is considered to be day 1 by Excel, 2 January 1900 is day 2, and so on. Clearly, dates are not as easy to manipulate as you might think. Extracting the day, month, or even the year from any given date is not straightforward because the date is really a number known as a serial number. Extracting a day, month, or year requires using the following three functions: • DAY(serial_number) gives the day in the date (for example, DAY(31-Jul-20) = 31). • MONTH(serial_number) gives the month in the date (for example, MONTH(31-Jul-20) = 7). • YEAR(serial_number) gives the year in the date (for example, YEAR(31-Jul-20) = 2020). It is just as awkward the other way around. If the day, month, and year are already known, the date can be calculated using the following function: DATE(year, month, day) (for example, DATE(2020,7,32) = 1 August 2020, etc.). Did you catch the function calculates the 32nd day of July as 1 August? Since dates are nothing more than serial numbers, they behave just like formatted numbers in Excel, for example, 31-Jul-20 + 128 = 6-Dec-2020. This is all great, but time series still cause us problems. If we want to have the month end date in each column, we cannot simply take the previous month’s date and add a constant to it, since the number of days in a month varies. Fortunately, this is where EOMONTH comes in: EOMONTH(specified_date, number_of_months) The “End Of Month” (EOMONTH) function therefore calculates the end of the month as the number_of_months after the specified_date. For example: • EOMONTH(31-Jul-20,0) = 31-Jul-20. • EOMONTH(3-Apr-05,2) = 30-Jun-05. • EOMONTH(29-Feb-08,-12) = 28-Feb-07. Although the examples use typed-in dates, for it to work in Excel, it is best to have the specified_date either as a cell reference to a date or else use the DATE function to ensure that Excel understands it is a date (otherwise the formula may calculate it as #VALUE!). In some instances (for example, appraisal of large-scale capital infrastructure projects), the dates may need to be for the same day of the month (for example, the 15th) rather than for the month end. A function similar to EOMONTH, EDATE can be used instead: EDATE(specified_date, number_of_months). The “Equivalent day” (EDATE) function therefore calculates the date that is the indicated number_of_months before or after the specified_date. For example: • EDATE(15-Jul-20,2) = 15-Sep-20. • EDATE(3-Apr-05,-2) = 3-Feb-05. • EDATE(29-Feb-28,-12) = 28-Feb-27. If an equivalent date cannot be found (as in the last example), month end is used instead. 3. FORMULATEXT New to Excel 2013, this is one of the most used functions by my team. It’s a really useful tool for documenting formulas, as FORMULATEXT returns a formula as a text string. People have been writing User-Defined Functions (UDFs) for years to replicate this functionality. In fact, if you have ever downloaded one of my example workbooks, the chances are you have analysed a formula described using the FORMULATEXT function: The expressions in cells G8 and G9 (above) are both provided by the FORMULATEXT function. For example, the formula in cell G8 is: The FORMULATEXT function employs the following syntax to operate: It has the following argument: • reference: This is required and represents a cell or a reference to a range of cells. It should be further noted that: • The FORMULATEXT function returns what is displayed in the formula bar if you select the referenced cell. • The reference argument can be to another worksheet or workbook. • If the reference argument is to another workbook that is not open, FORMULATEXT returns the #N/A error. • If the reference argument is to an entire row or column, or to a range or defined name containing more than one cell, FORMULATEXT returns the value in the upper leftmost cell of the row, column, or range. • In the following cases, FORMULATEXT returns the #N/A error: □ The cell used as the reference argument does not contain a formula. □ The formula in the cell is longer than 8,192 characters. □ The formula cannot be displayed in the worksheet, for example, due to worksheet protection. □ An external workbook that contains the formula is not open in Excel. • Invalid data types used as inputs will produce the #VALUE! error. • Entering a reference to the cell in which you are entering the function as the argument will not result in a circular reference warning. FORMULATEXT will successfully return the formula as text in the cell. I love this example: 4. N I love functions I can spell. The N function returns a value converted to a number. It has only one argument: The value argument is required and represents the value you want converted. N converts values on the following basis: Usually, you don’t need to use the N function in a formula because Excel automatically converts values as necessary. Microsoft states that this function is provided for compatibility with other spreadsheet programs. However, I disagree: I use this function all the time. Let me explain. Counters are often used in financial modelling, eg: It’s not a good idea to type these numbers in and/or use AutoFill. This is because if an end user wishes to extend the sequence, they might take the first cell (D2) and drag it across. Unfortunately, in this scenario, you would get a sequence of 1’s, viz. Oops. Therefore, we should use a formula in cell D2 such as =C2+1: That’s all well and good, until someone types something in cell C2: The problem is cell C2 now contains text, and you cannot add one (1) to text. However, you can add N to the formula: The N function ignores the text in cell C2. That’s exactly what we require. I use counters in my financial models all the time — and, therefore, I use the N function all the time, too. 5. TEXTJOIN The TEXTJOIN function combines the text from multiple ranges and/or text strings and includes a delimiter to be specified between each text value to be combined. If the delimiter is an empty text string, this function will effectively concatenate the ranges similar to the CONCAT function. Its syntax is: TEXTJOIN(delimiter, ignore_empty, text1, [text2], …) • delimiter is a text string (which may be empty) with characters contained within inverted commas (double quotes). If a number is supplied, it will be treated as text. • ignore_empty ignores empty cells if TRUE or the argument is unspecified (ie, is blank). • text1 is a text item to be joined. • text2 (onwards) are additional items to be joined up to a maximum of 252 arguments. If the resulting string contains more than 32,767 characters, TEXTJOIN returns the #VALUE! error. TEXTJOIN is more powerful than CONCAT. To highlight this, consider the following examples: Here, in the formulas on rows 53 and 54, empty cells in a contiguous range may be ignored, and delimiters only need to be specified once. It’s a great way to create lists for reporting, for example. Add a Comment	{"url":"https://sophuc.com/top-5-excel-functions-you-might-not-know/","timestamp":"2024-11-06T08:09:59Z","content_type":"application/xhtml+xml","content_length":"49267","record_id":"<urn:uuid:25f9aa36-bf2c-40c8-9208-36557e8e1454>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00136.warc.gz"}
K Closest Points to Origin Published: Sep 10, 2022 This is a geometry problem, yet a simple sorting problem as well. In any case, we should calculate distances to the origin on all points. Problem Description Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0). The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2). You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in). □ 1 <= k <= points.length <= 104 □ -104 < xi, yi < 10*4 Example 1 Input: points = [[1,3],[-2,2]], k = 1 Output: [[-2,2]] Example 2 Input: points = [[3,3],[5,-1],[-2,4]], k = 2 Output: [[3,3],[-2,4]] Sorting is done by an Euclidean distance to the origin (0, 0). We don’t need actual distance, since comparison matters. Instead, sorting key is x x + y * y. After sorting, return first k points. class KClosestPointsToOrigin: def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]: points.sort(key = lambda x: x[0] * x[0] + x[1] * x[1]) return points[:k] • Time: O(nlog(n)) • Space: O(1)	{"url":"https://yokolet.com/algo/sorting_and_searching/2022-09-10-k-closest-points-to-origin","timestamp":"2024-11-13T22:19:22Z","content_type":"text/html","content_length":"70926","record_id":"<urn:uuid:d93f3130-fabe-4bc3-9ecb-6044858fb021>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00068.warc.gz"}
Applied Biostats – BIOL3272 UMN – Fall 2022 A book of (com)passion In the summer of 2020, the world was on fire – COVID was raging, we – especially in Minnesota – were processing the murder of George Floyd and the subsequent uprising etc, the future was unclear. At that point teaching was likely to be entirely online, and I decided to write a digital book for my course see the first edition of my book here. I didn’t really know what I was doing or what my vision was (and to some extent I still do not). There were hiccups: some strangeness in rendering etc, typos, last minute updates, writing at 2am etc etc, but on the whole there were numerous advantages compared to a traditional textbook. I lay these out here: • My class presentation and the textbook presentation almost always agreed. • As I was writing and updating as I went the book could be rapidly updated / changed to reflect student needs / interests / timelines / current events etc. • I could integrate practice problems / youtube links / and even additional readings pretty easily. • It was free for students. I think all of these benefits were great, and helped a lot, so I am did it again and updated the previous version for spring of 2022. We are now on the third edition for Fall of 2022 - and I’m hoping each version gets better and has fewer issues. Why do I bring this up? Well I know you’re dealing with a lot. Every year students are dealing with a lot – from jobs, to supporting family, to the everyday of being in college and living life, and this year there’s even more. I too have a lot – A one year old sons and and three year old daughter, research and life pressures, teaching etc. Yet, we are all trying to make the most of life in this era. We want to teach, learn, and grow. What’s more, I believe this content is more important now than it has ever been, statistics is obsessed with the critical evaluation of claims in the face of data, and is therefore particularly useful in uncertain times. Given this focus, and given that you all have different energies, motivations and backgrounds, I am restructuring this course slightly from previous years. The biggest change is a continued de-emphasis on math and programming – that doesn’t mean I’m eliminating these features, but rather that I am streamlining the required math and programming to what I believe are the essentials. For those who want more mathematical and/or computational details (either because you want to push yourself or you need this to make sense of things), I am including a bunch of optional content and support. I LOVE TEACHING THIS COURSE – the content is very important to me. I also care deeply about you. I want to make sure you get all you can / all you need from this course, while recognizing the many challenges we are all facing. One tangible thing I leave you with is this book, which I hope you find useful as you go on in your life. Another thing I leave you with is my concern for your well-being and understanding – please contact me with any suggestions about the pace / content / you of this course and/or any life updates which may change how and when you can complete the work. Course philosophy / goals Hi! I’m a statistician. You might know me from my greatest hits including, “Have you tried plotting the data?”, “You’re not adequately powered to answer that question”, and “Correlation is not causation (except when it is 😉)” https://t.co/MpEHfqwHY8 — Lucy D’Agostino McGowan (@LucyStats) January 19, 2019 My motivating goal for this course is to empower you to produce, present, and critically evaluate statistical evidence — especially as applied to biological topics. You should know that stats models are only models and that models are imperfect abstractions of reality. You should be able to think about how a biological question could be formulated as a statistical question, present graphs which show how data speak to this question, be aware of any shortcomings of that model, and how statistical analysis of a data set can be brought back into our biological discussion. “By the end of this course… • Students should be statistical thinkers. Students will recognize that data are comprised of observations that partially reflect chance sampling, & that a major goal of statistics is to incorporate this idea of chance into our interpretation of observations. Thinking this way can be challenging because it is a fundamentally new way to think about the world. Once this is mastered, much of the material follows naturally. Until then, it’s more confusing. • Students should think about probability quantitatively. That chance influences observations is CRITICAL to statistics (see above). Quantitatively translating these probabilities into distributions and associated statistical tests allows for mastery of the topic. • Students should recognize how bias can influence our results. Not only are results influenced by chance, but factors outside of our focus can also drive results. Identifying subtle biases and non-independence is key to conducting and interpreting statistics. • Students should become familiar with standard statistical tools / approaches and when to use them. Recognize how bias can influence our results. What is the difference between Bayesian and frequentist thinking? How can data be visualized effectively? What is the difference between statistical and real-world significance? How do we responsibly present/ interpret statistical results? We will grapple with & answer these questions over the term. • Students should have familiarity with foundational statistical values and concepts. Students will gain an intuitive feel for the meaning of stats words like variance, standard error, p-value, t-statistic, and F-statistic, and will be able to read and interpret graphs, and how to translate linear models into sentences. • Students should be able to conduct the entire process of data analysis in R. Students will be able to utilize the statistical language, R, to summarize, analyze, and combine data to make appropriate visualizations and to conduct appropriate statistical tests. R, RStudio, and the tidyverse We will be using R in this course, in the RStudio environment. My goal is to have you empowered to make figures, run analyses, and be well positioned for future work in R, with as much fun and as little pain as possible. RStudio is an environment and the tidyverse is a set of R packages that makes R’s powers more accessible without the need to learn a bunch of computer programming. Some of you might have experience with R and some may not. Some of this experience might be in tidyverse or not. There will be ups and downs — the frustration of not understanding and/or it not working and the joy of small successes. Remember to be patient, forgiving and kind to yourself, your peers, and me. Ask for help from the internet, your friends, Brooke, and Yaniv. We will using R version 4.2.1 or above, and tidyverse version 1.3.2 or above. You can download these onto your computer (Make sure the R is version 4.2.1). 1. Download/update R from here. 2. Next download/update RStudio from here. 3. Finally open RStudio and type install.packages("tidyverse") and then library(tidyverse) to ensure this worked. Alternatively you can simply join the course via RStudioCloud. This could be desirable if you do not want to or have trouble doing this. YB ADD DETAILS What is this ‘book’ and how will we use it? This ‘book’ functions as an extensive syllabus and course notes. I will embed youtube videos, app-based demonstrations and class exercises. As noted above, there are some things I will include that will not be necessary for everyone, and I will clearly mark these sections. I hope that this book provides clear and useful background for the course, and I advise you to regularly go through each book ‘chapter’ for the relevant week. Be sure you get familiar with the content BEFORE class. Note that this ‘book’ is not the entirety of the course content, and is not an original piece of my own effort – in addition from lifting from a few other course online (with attribution), I also make heavy use of these texts: • The Analysis of Biological Data Third Edition (Whitlock and Schluter 2020): This is the official book of this course, and is a standard biostats textbook, with many useful resources available online. The writing is great, as are the examples. Most of my material originates here (although I occasionally do things a bit differently). This book is officially optional, but students consistently tell me that it is extremely helpful. So, I highly recommend buying it. You can get the newest edition here, but any edition will be pretty useful. • Calling Bullshit (Bergstrom and West 2020): This book is not technical, but points to the big picture concerns of statisticians. It is very practical and well written. I will occasionally assign readings from this book, and/or point you to videos on their website. All readings will be made available for you, but you might want to buy a physical copy. • Fundamentals of Data Visualization (Wilke 2019): This book is free online, and is very helpful for thinking about graphing data. In my view, graphing is among the most important skills in statistical reasoning, so I reference it regularly. • R for Data Science (Grolemund and Wickham 2018): This book is free online, and is very helpful for doing the sorts of things we do in R regularly. This is a great resource. I will introduce other resources as we go. How will this term work / look? • Prep for ‘class’. This class is flipped with asynchronous content delivery and in person meetings. • Be sure to look over the assigned readings and/or videos, and complete the short low-stakes homework BEFORE each course. • During class time, I will address questions make announcements, and get you started on in-class work. Brooke & I will bounce around class to provide help and check-in. • The help of your classmates and the environment they create is one of the best parts of this class. Help each other. • In addition to low stakes work before and in class, there will be a few more intense assignments, some collaborative projects and a summative project as the term ends. There will be no ‘high-stakes’ in class timed tests. 0.1 Example mini-chapter: Types of Variables and Data Learning goals: By the end of this example mini chapter you should be able to • Distinguish between explanatory and response variables. • Distinguish between data types. □ Continuous vs Categorical □ Differentiate between continuous and discrete continuous variable. □ Differentiate between nominal and ordinal categorical variables. As we build and evaluate statistical models, a key consideration is the type of data and the process that generates these data. Variables are things which differ among individuals (or sampling units) of our study. So, for example, height, or eye color, or the type of fertilizer applied to a site, or the number of insect species per hectare are all variables. 0.1.1 Explanatory and Response variables We often care to distinguish between explanatory variables, which we think underlie or are associated with the biological process of interest, from response variables, the outcome we aim to understand. This distinction helps us build and consider our statistical model and relate the results to our biological motivation. The difference between an explanatory and response variable often depends on the motivation and/or study design. For example if we where interested to know if fertilizer type had an (?indirect?) impact on insect diversity, the type of fertilizer would be the explanatory variable and the number of insect species per hectare would be the response variable. 0.1.2 Types of Data Data can come in different flavors. It is important to understand these, as they should direct our model building and data summaries, interpretation and data visualization. 0.1.2.1 Flavors of numeric variables. Numeric variables are quantitative and have magnitude, and come in a few sub-flavors. As we will see soon, these guide our modeling approaches: • Discrete variables come in chunks. For example the number of individuals is an integer, we don’t have 1/2 people. • Continuous variables can take any value within some reasonable range. For example, height, weight, temperature, etc. are classic continuous variables. Some variables are trickier – for example, age is continuous, but we often analyze it as if it’s discrete. In practice, these tricky cases rarely present a serious problem for our analyses (except in the rare cases in which they do). Not all numbers are numeric. For example, gene ID is a number but it is an arbitrary marker and is not quantitative. 0.1.2.2 Flavors of categorical variables. Categorical variables are qualitative, and include, • Nominal variables which cannot be ordered and have names – like sample ID, species, hair color etc… • Binary variables are special types of nominal variables, which have only two options (or for which we only consider two options. Alive/dead, pass/fail, on/off are classic binary variables). • Ordinal variables can be ordered, but do not correspond to a magnitude. For example, bronze, silver and gold medals in the Olympics are ranked from best to worst, but first isn’t some reliable distance away from second or third etc… . 0.1.3 Quiz After completing this quiz (and ensuring you get everything right), fill out the quiz on canvas as today’s class Quiz. 0.1.4 Definitions Explanatory variables are variables we think underlie or are associated with the biological process of interest. Response variables are the outcome we aim to understand. Categorical variables are qualitative – they cannot be assigned a meaningful value on the number line. Numeric variables are quantitative – they can be assigned a meaningful value on the number line. Bergstrom, Carl T, and Jevin D West. 2020. Calling Bullshit: The Art of Skepticism in a Data-Driven World. Random House. Grolemund, Garrett, and Hadley Wickham. 2018. “R for Data Science.” Whitlock, Michael C, and Dolph Schluter. 2020. The Analysis of Biological Data. Third Edition. Wilke, Claus O. 2019. Fundamentals of Data Visualization: A Primer on Making Informative and Compelling Figures. O’Reilly Media.	{"url":"https://bookdown.org/ybrandvain/Applied_Biostats_Fall_2022/","timestamp":"2024-11-06T08:33:29Z","content_type":"text/html","content_length":"141017","record_id":"<urn:uuid:422ec827-2e9d-4d91-b942-5324aef02ac0>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00376.warc.gz"}
geometry problem Let R be the circle centered at (0,0) with radius 10. The lines x=6 and y=4 divide R into four regions R1, R2, R3 , and R4. Let R_i denote the area of region R_i If R1>R2>R3>R4, then find R1 + R2 + R3 + R4. maximum Nov 1, 2023 Since R1, R2, R3, R4 are the four regions into which the unit circle is divided by the lines x=6 and y=4, we see that regions R1 and R3 are congruent, and regions R2 and R4 are congruent. Therefore, To compute the area of R1, we can construct sector OAC, where O is the center of the circle and A is the point at which the circle intersects the line y=4. Sector OAC has central angle 90 degrees, and its radius is 10, so its area is \frac{1}{4}(10^2)\pi = \frac{25}{2}\pi. Since regions R1 and R3 are congruent, and region R1 is the upper right quadrant of sector OAC, the area of R1 is \boxed{\frac{25\pi}{4}}. parmen Nov 1, 2023	{"url":"https://web2.0calc.com/questions/geometry-problem_33570","timestamp":"2024-11-11T21:15:35Z","content_type":"text/html","content_length":"20844","record_id":"<urn:uuid:7d3c5b80-5fe4-44a5-ad0b-7fd6646dad95>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00228.warc.gz"}
Hex To Binary \| Techfonist Hex to Binary: A Complete Guide to Conversion Hexadecimal, often referred to as hex, is a base-16 numeral system widely used in computing and digital electronics. In this guide, we’ll explore how to convert hex to binary and its significance in various applications. Table of Contents What is Hexadecimal (Hex)? Hexadecimal is a number system that uses 16 symbols to represent values. These symbols are the digits 0-9 and the letters A-F, where A stands for 10, B for 11, C for 12, D for 13, E for 14, and F for Understanding the Hexadecimal System The hex system is advantageous because it simplifies the representation of binary numbers. For instance, one hexadecimal digit can represent four binary digits (bits). This compactness makes it easier to read and write large binary numbers. The Importance of Hexadecimal in Computing Hexadecimal is extensively used in programming, memory addresses, and color codes in web design. Its relationship with binary makes it a crucial part of the computing world. What is Binary? Binary is the most basic form of data representation in computers, using only two symbols: 0 and 1. The Binary Number System Explained In binary, each digit (or bit) represents a power of 2. For example, the binary number 1011 can be broken down as: • 1×23+0×22+1×21+1×20=8+0+2+1=111 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 8 + 0 + 2 + 1 = 111×23+0×22+1×21+1×20=8+0+2+1=11 in decimal. Importance of Binary in Digital Systems Computers operate using binary because electronic circuits can easily represent two states: on (1) and off (0). All data, whether text, images, or sound, is ultimately represented in binary. The Relationship Between Hexadecimal and Binary Each hexadecimal digit corresponds to a unique four-bit binary sequence. This direct relationship allows for easy conversion between the two systems. How Hexadecimal Represents Binary Data Here’s a quick reference for hex to binary conversion: Hex Binary A 1010 B 1011 C 1100 D 1101 E 1110 F 1111 Conversion Factors Between Hex and Binary Since one hex digit equals four binary digits, you can convert any hexadecimal number to binary by replacing each hex digit with its corresponding four-bit binary equivalent. How to Convert Hex to Binary Converting hex to binary can be done manually or through online tools. Manual Conversion Steps Step-by-Step Conversion Process 1. Identify the Hexadecimal Number: Start with the hex number you want to convert. 2. Replace Each Hex Digit: Use the hex-to-binary reference table to replace each hex digit with its corresponding binary sequence. 3. Combine the Binary Sequences: Write down the binary sequences together to form the complete binary representation. Example: Convert 2F to binary. Combining these gives 00101111. Using Online Tools for Conversion Several online converters can simplify this process. Just input the hex value, and the tool will output the corresponding binary value. Websites like RapidTables or CalculatorSoup offer user-friendly interfaces for this conversion. Practical Applications of Hex to Binary Conversion Data Representation in Programming Hex to binary conversion is vital in programming. Developers often use hex for memory addresses, color values in HTML/CSS, and debugging. Networking and Communication Protocols In networking, hex values are frequently used to represent IP addresses and MAC addresses. Understanding the conversion helps in network analysis and configuration. Common Mistakes in Hex to Binary Conversion Misinterpreting Hex Values One common mistake is misreading the hex values, especially when dealing with letters. For example, confusing the letter B (11) with the number 8 can lead to incorrect binary results. Errors in Binary Representation When converting manually, ensure that you accurately represent each hex digit as a four-bit binary number. Mistakes can easily happen if you rush through the conversion. Converting Hex to Binary in Programming Sample Code in Python Here’s a simple Python code snippet to convert hex to binary: pythonCopy codedef hex_to_binary(hex_number): binary_string = bin(int(hex_number, 16))[2:] # Converts hex to binary return binary_string.zfill(4 * len(hex_number)) # Ensures proper padding # Example usage hex_value = '2F' binary_representation = hex_to_binary(hex_value) print(f'The binary representation of "{hex_value}" is {binary_representation}') # Outputs: 00101111 Hex to Binary Conversion in Other Languages Similar logic applies to other programming languages. For example, in Java, you can use: javaCopy codepublic class HexToBinary { public static void main(String[] args) { String hexValue = "2F"; String binaryString = Integer.toBinaryString(Integer.parseInt(hexValue, 16)); System.out.printf("The binary representation of '%s' is %s%n", hexValue, binaryString); Converting hex to binary is an essential skill for anyone working in computing or programming. Understanding this process will enhance your ability to read and interpret data, whether you’re dealing with memory addresses, color codes, or debugging information. 1. What is hexadecimal (hex)? Hexadecimal is a base-16 numeral system that uses 16 symbols: 0-9 and A-F. It is commonly used in computing to represent values compactly. 2. Why is binary important in computing? Binary is the fundamental data representation system for computers, using only 0s and 1s to process and store all forms of data. 3. How do I convert hex to binary manually? To convert hex to binary, replace each hex digit with its four-bit binary equivalent using a reference table and combine the results. 4. Are there online tools for hex to binary conversion? Yes, several online tools can convert hex to binary quickly. Websites like RapidTables and CalculatorSoup offer easy-to-use interfaces. 5. Can I convert binary back to hex? Yes, you can convert binary back to hex by first converting the binary to decimal and then converting that decimal value to hex. 1 thought on “Hex To Binary” 1. Pingback: Case Converter \| Techfonist	{"url":"https://techfonist.com/hex-to-binary/","timestamp":"2024-11-08T02:32:06Z","content_type":"text/html","content_length":"168456","record_id":"<urn:uuid:2381c020-b962-4167-9f22-69ac0f0cee0f>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00575.warc.gz"}
Constraint: SFSF Model ElementConstraint_SFSF defines a higher pair constraint. The constraint consists of a surface on one body rolling and sliding on a surface on a second body. The surfaces are required to have a unique contact point. id = "integer" label = "Name of Constraint_SFSF element" i_marker_id = "integer" i_surface_id = "integer" i_disp_x0 = "real" i_disp_y0 = "real" i_disp_z0 = "real" j_marker_id = "integer" j_surface_id = "integer" j_disp_x0 = "real" j_disp_y0 = "real" j_disp_z0 = "real" Element identification number (integer>0). This number is unique among all Constraint_SFSF elements. The name of the Constraint_SFSF element. Specifies a marker that defines the coordinate system in which the i_surface points are defined. It also implicitly defines the body on which the surface is "etched". The surface moves with the body. i_marker_id may belong to any type of body: flexible, rigid, or point. The parameter is mandatory. Specifies the ID of the Reference_ParamSurface that contains the surface definition. i_disp_x0, i_disp_y0, i_disp_z0 These three parameters specify the location of the contact point on i_surface at the input configuration as measured in the i_marker_id coordinate system. The three parameters come as a set. All three must be specified or none may be defined. These parameters are optional. 8 Specifies a Reference_Marker that defines the coordinate system in which the j_surface points are defined. It also implicitly defines the body on which the surface is "etched". The surface moves with the body. j_marker_id may belong to any type of body: flexible, rigid, or point. The parameter is mandatory. Specifies the ID of the Reference_ParamSurface that contains the surface definition. j_disp_x0, j_disp_y0, j_disp_z0 These three parameters specify the location of the contact point on j_surface at the input configuration as measured in the j_marker_id coordinate system. The three parameters come as a set. All three must be specified or none may be defined. These parameters are optional. 8 Figure 1 shows two surfaces I and J that are in continuous contact. Figure 1. Surface-to-Surface Contact Surface I is defined with respect to Reference_Marker 1023; surface J is defined with respect to Reference_Marker 2046. Reference_ParamSurface 123 defines surface I and Reference_ParamSurface 246 defines surface J. The patch containing the contact point is also shown in figure 1. An initial guess for the contact point on both surfaces is defined. Assume the initial contact point location on surface J, as measured in the coordinate system of 2046, is [1.466, 5.66, 0.1]. Assume the contact point location on surface I, as measured in the coordinate system of 1023, is [-0.522, -0.852, -0.453]. The Constraint_SFSF object may be defined as follows: id = "1" i_marker_id = "1023" i_surface_id = "123" i_disp_x0 = "1.466" i_disp_y0 = "5.66" i_disp_z0 = "0.1" j_marker_id = "2046" j_surface_id = "246" j_disp_x0 = "-0.522" j_disp_y0 = "-0.852" j_disp_z0 = "-0.452" > 1. Constraint_SFSF element constrains the two surfaces as follows: □ They surfaces have exactly one contact point. □ The normal at the contact point for each surface are anti-parallel. This is shown schematically in Figure 2 below. P[i] and P[j] represent the contact point on the two surfaces I and J respectively. N[i] and N[j] are the normals at the contact point for Surface I and Surface J respectively. The contact conditions are shown mathematically in the box at the bottom-left. Figure 2. The Contact Condition Between Two Surfaces 2. The surface-to-surface constraint does not allow lift-off. You can examine the sign of the constraint force to determine if any lift-off should have occurred if the constraint were not there. A positive value implies that the force is repulsive. A negative value implies an attractive force. The surfaces would separate if the constraint were not present. If your results require an accurate simulation of intermittent contact, you should model the contact forces directly using a Force_Contact or a Force_TwoBody object. 3. Both open and closed surfaces supported by Constraint_SFSF. 4. Open surfaces have a well defined spatial extent. The surface is only defined in the domain α[min] <= α <= α[max], and β[min] <= β <= β[max] where α and β are the surface parameters. While enforcing the surface-to-surface constraint, it is possible for MotionSolve to find a solution outside this range. It is your responsibility to define appropriate forces at the surface ends, so that the contact point stays in the legal range for α and β. 5. The surfaces in a Constraint_SFSF are required to have a single point of contact. Convex surfaces will guarantee a single point of contact. A convex surface is one that intersects a straight line at just two points. See Figure 3 below for examples of convex and non convex shapes. 6. The oval shaped closed surface on the left is an example of a convex shape. Notice that any straight line can intersect the surface at only two points. In contrast, the open surface on the right is non-convex. You can draw a straight line that intersects it at more than two points. Figure 3. Convex and Non-Convex Surfaces 7. Be suspicious of the correctness of your model, if ever one of the constraints belonging to a Constraint_SFSF is declared to be redundant. This will lead to unexpected behavior and is a sure indication that the model has not been correctly built. 8. MotionSolve does an initial search to exactly locate the contact point. While the attributes i_disp_x0, i_disp_y0, i_disp_z0, j_disp_x0, j_disp_y0, and j_disp_z0 are not mandatory, it is a good idea to specify them if you know what they are.	{"url":"https://2021.help.altair.com/2021/hwsolvers/ms/topics/solvers/ms/xml-format_48.htm","timestamp":"2024-11-08T18:07:52Z","content_type":"application/xhtml+xml","content_length":"107617","record_id":"<urn:uuid:26f33cec-3d29-4464-8840-4ae195470ffb>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00350.warc.gz"}
A Three-Dimensional Radiative Transfer Model to Investigate the Solar Radiation within a Cloudy Atmosphere. Part I: Spatial Effects 1. Introduction Clouds act as the dominant modulator of the earth’s radiative energy budget and therefore are a strong interdependent link to the mechanisms driving the general circulation of the atmosphere and ocean. Although the importance of cloud–radiation interactions is well known, knowledge of their radiative properties remains a major uncertainty in the understanding and modeling of the present and future climate. While the influence of clouds on both top-of-the-atmosphere (TOA) and surface flux is directly measurable, its effect on atmospheric absorption is less clear. The prevailing view based on climate models is that the net effect of clouds on atmospheric column absorption is negligible when compared to clear skies. Yet recent global observations analyzed by Cess et al. (1995) and Cess et al. (1996), and observations in the Tropics analyzed by Ramanathan et al. (1995) and Pilewskie and Valero (1995), suggest that clouds may enhance atmospheric absorption by as much as 15–35 W m^−2 (diurnal average) more than theory predicts. Since the publication of Stephens and Tsay’s (1990) seminal review of this phenomenon, the positive sign of the discrepancy should come as no surprise. It is rather the magnitude suggested by these recent findings that has once again stirred up a paradigmatic debate within the climate community (Wiscombe 1995). Considering that the estimates of the doubling of the greenhouse gas, CO[2], implies a 4 W m^−2 radiative forcing on the climate system, the large discrepancy in solar absorption in the cloudy column found between theory and observations is disturbing. If the role of clouds on the radiative budget is to be fully comprehended and accurately modeled, an increased understanding of the interaction between the radiative field and both cloud microphysical and macrophysical properties is essential. Although clouds are portrayed as being plane parallel in climate models, in nature they are far from perfectly homogeneous layers. Even stratus-type clouds can have a pronounced cellular structure, with holes (areas of low optical thickness) and heterogeneous distributions of liquid water that can modulate the radiation field to some degree (Barker and Davies 1992; Jonas 1992; Cahalan et al. 1994). Numerous studies have demonstrated that the macrophysical effects of cloud-to-cloud interactions, cloud shading, cloud leakage, and cloud radiation–water vapor interactions can greatly influence the observed albedo from the top of the atmosphere and the irradiance to the surface (McKee and Cox 1974; Aida 1977; Wendling 1977; Claußen 1982; Coakley and Davies 1986; Coakley and Kobayashi 1989; Welch and Wielicki 1989;Bréon 1992; Segal and Davis 1992; Kobayashi 1993). Fewer investigations have focused on the role of cloud macrophysics on atmospheric absorption. Davies et al. (1984) employed a 3D Monte Carlo–based radiative transfer model and simple box representations of a cloud to show that absorption is reduced by diffusion of radiation out of the cloud sides. Stephens (1988a) calculated radiative properties for cloud structures given by Gaussian and harmonic functions using Fourier transforms of the radiative transfer equation. Compared to a uniform cloud, the absorption is smaller. But Stephens (1988b) also demonstrates theoretically that the absorption can be greater depending on the inhomogeneities used in the calculations. For a towering three-dimensional cloud, Li et al. (1995) noted a small increase in absorption using a four-spectral-band Monte Carlo model. Also using a four-spectral-band Monte Carlo model, Hignett and Taylor (1996) found absorption by stratus and stratocumulus to be lower for heterogeneous clouds than for uniform clouds. Using a stochastic model, Byrne et al. (1996) demonstrated that longer pathlengths caused by broken clouds can enhance absorption. One difficulty in reconciling the differences between these results is that, while in some cases the absorption refers to that which takes place within the cloud, in other cases the absorption is for the entire atmospheric column in the presence of clouds. Furthermore, given the complexities of computing radiative transfer in three dimensions and the computer resources required, all of these approaches, to some degree, contain substantial simplifications in either spectral resolution, atmospheric composition, or cloud morphology. In this paper, we investigate the effects of cloud morphology on atmospheric absorption, using a Monte Carlo–based radiative transfer model with both high spectral and spatial resolution that contains all of the important radiative constituents of the atmosphere. The realistic cloud field representation used in this simulation has been extracted from satellite visible and infrared imagery. From analysis of the detailed spatial results, we discuss the mechanisms responsible for the absorption in a 3D cloudy atmosphere and explain the main deficiencies of 1D radiative transfer codes that depend on plane-parallel cloud assumptions. In this paper (Part I) we only address spatial effects. Spectral effects are discussed in a companion paper (O’Hirok and Gautier 1998, hereafter Part II). 2. Radiative transfer model a. Review of previous models The radiative transfer model used in this study is a diagnostic tool for investigating the 3D radiative field not only for clouds, but for other atmospheric constituents such as water vapor and aerosols, or surface features such as land–ocean surface feature mosaics, complex terrain, and plant canopies. In this paper, the model discussion is limited to the cloud and atmospheric components. The model is based on the Monte Carlo method, which has been frequently applied in analyzing cloud–radiative interactions as a result of the inadequacy of analytical 3D approaches. Essentially, the method is a direct simulation of the physical processes involved in radiative transfer, whereby the flow of the radiation is computed photon by photon based on a set of probability functions. These functions describe the distance a photon travels before an interaction, the result of the interaction (scattering or absorption), and the resulting scattering direction. The probabilities vary with the cloud microphysics or atmospheric constituents involved and the wavelength of the incident radiation. Generally, clouds used in 3D simulations have consisted of simple geometric shapes or arrays composed of a constant liquid water amount confined to a single atmospheric layer. With advances in computer technology, higher-resolution cloud fields derived from stochastic modeling methods or satellite imagery have been incorporated into the models (Barker and Davies 1992;Cahalan et al. 1994; Zuev and Titov 1995; Hignett and Taylor 1996). Spectrally, computations have been made at one or, at most, a few wavelengths to represent the scattering properties for clouds over the entire solar spectrum. A review of the literature shows that molecular scattering and aerosol effects have apparently not been incorporated in any previous Monte Carlo modeling studies for cloudy atmospheres. Accounting for cloud droplet absorption represents a relatively simple problem. Computations can be made at each photon interaction, based on the cloud droplet single scattering albedo, or during postprocessing using photon interaction statistics. Water vapor absorption represents a more formidable challenge because of its more highly variable spectral nature. One method is to infer the absorption based on the pathlength distribution statistics generated during the Monte Carlo process. Unless the water vapor field is considered homogeneous, however, the memory requirements for this method can be prohibitive for all but the most simple spatial configurations. This technique also requires the scattering parameters to be spectrally invariant for the width of each absorption band model. For a realistic atmosphere that includes molecular scattering, aerosol scattering, clouds of various droplet size distributions, and an underlying reflecting surface, the width of such a band can become quite narrow. Another approach is to compute the water vapor absorption between each scattering event by using the pathlength between the scattering locations and a transmission function that is modified by the total pathlength of the photon. Although this method can be quite accurate, it is also computationally expensive since a constant number of calculations must be performed for every photon interaction within each spectral absorption band. The method used in this research, and elaborated in the description of the design of the model, simulates as closely as possible the physical process of a photon’s interaction with water vapor by directly incorporating the effect of gas molecules into the probability functions. Computations are made at each waveband, but the number of calculations required for each interaction is dynamic and generally less than what is required in the preceding approach. b. Design of the model The design philosophy employed for this model is to represent the atmosphere and clouds in a manner as realistic as possible, and keep the number of theoretical assumptions and inferences to a minimum. Thus, computations are made in the most physically real sense as possible, and any reduction in computational expense comes through algorithm efficiency. In this manner, the physical processes involved in radiation transfer can be directly examined with a diminished chance of arriving at false understanding through seemingly benign assumptions or through the neglect of unforeseen interactions between the various radiative components. The model’s spatial domain consists of a cellular structure that is broken into individual homogeneous cells whose size and number depends on the complexity of the phenomena being modeled. For instance, a simulation of a sparse broken cloud field could have a large number of high-resolution cells representing the clouds and low-resolution cells portraying clear sky. Each cell contains a pointer to a database that includes all of the radiative parameters required for computing a photon’s pathlength, its interaction, and scattering direction. Results, such as photon pathlength, direction, and absorption, are also stored within this database. From these data, local and domain averaged fluxes can be estimated. All photons are initially assigned a weight (see below), to represent a packet of photons. In subsequent references, one photon count is equal to a photon packet and not to the weight of the photon packet. Hence, photon counts represents photons of weight The distance the photon traverses between interactions within a cell, the pathlength is given by is a uniformly random distributed number within the interval [0, 1), and represents the total extinction coefficient for a given wavelength, The extinction coefficient is obtained by summing the extinction coefficients for gases, molecular scattering, aerosols, and cloud droplets within a cell. If a photon escapes a cell, the part of the pathlength not used in its original cell is employed within the adjacent cell but modified to reflect the extinction coefficient of the new cell. Then, following an interaction, a new pathlength is recomputed according to Eq. (1) For optical properties prescribed by the optical depth ( ), the volume extinction coefficient is computed by is the thickness of the cell in the vertical direction. Gases include water vapor (H O), ozone (O ), CO , N O, CO, CH , O , N , NH , HNO , NO, and SO . Water vapor and ozone concentrations are nominally determined from standard atmospheric profiles with modifications of the water vapor concentration based on the relative humidity within the prescribed cloud. For gaseous absorption that obeys Beer’s law (i.e., self- and foreign broadening continuum), the optical depth is derived from double exponential band models ( Kneizys et al. 1988 For line absorption, a line-by-line approach is not feasible; therefore, a frequency integration approach using the -distribution method from LOWTRAN7 is employed ( Kneizys et al. 1988 ). For each cell, the transmission for wavelength interval Δ is expressed as the sum of three exponential terms is the gas amount, and represents the effective absorption coefficient for and is weighted by Δ , which sums to unity. This method allows computations to be performed independently on each term as if it were a monochromatic problem ( Isaacs et al. 1987 ). Thus, from its entrance at the top of the atmosphere to its termination, each photon must be processed three times. For gaseous bands that overlap, this procedure will introduce some error, but in the shortwave region, the only important overlap occurs at 2.7 m for H O and CO Liou 1980 ) where, at that wavelength, absorption within the atmosphere is already maximized. The Rayleigh scattering optical depth and aerosol optical depth are also derived from LOWTRAN7. Aerosols within the model include boundary layer (rural, urban, and oceanic), tropospheric, and stratospheric (background stratospheric, aged volcanic, fresh volcanic, and meteor dust) types. For stratospheric aerosols, is directly specified; for the boundary layer aerosols, is computed from a specified horizontal visibility referenced at 0.55 m and adjusted for the relative humidity and the aerosol scattering efficiency for Cloud droplet optical thickness, , is a user-specified parameter and can be assigned in terms of either cloud liquid water content (LWC) or specified at 0.55 m. When computed from LWC, it follows that is the effective radius of the cloud droplet distribution, is cloud-scattering efficiency for is the density of water. The surface, in the case of a solid or liquid, is treated as having an infinite optical depth. The type of particle interaction a photon experiences within a cell is based on the ratio of a particle’s optical depth, τ[i], to the total optical depth τ of a cell. A second random number, R[i], within the interval [0, 1) is generated, and a particle is selected according to R[i]’s location with regard to the cumulative probabilities of each atmospheric constituent. For example, if (τ[1] + τ [2])/τ ⩽ R[i] < (τ[1] + τ[2] + τ[3])/τ, then the photon is deemed to have interacted with the third particle (represented by τ[3]) listed in the probability table. Since the total optical depth and gaseous optical depth change with the k used in the k-distribution method, three probability tables are required per cell. During an interaction, the energy absorbed ′ within a cell is defined by is the single scattering albedo that describes the ratio of the scattering cross section to the extinction cross section of a gas or particle. The remaining energy is scattered. Rayleigh single scattering albedo, , is equal to 1. Aerosol single scattering albedo is computed from a database incorporated within LOWTRAN7 and is a function of aerosol type, relative humidity of the cell, and For cloud droplets, the single scattering albedo is sensitive to and is derived from a database produced through Mie theory calculations. For this study, the surface is treated as Lambertian, and the single scattering albedo variable for the surface, , is made equivalent to the albedo of the surface. Although gaseous absorption by definition has a single scattering albedo equal to 0, ω[0g] is actually set to a user-defined η that remains constant for each photon processed. This method reduces the statistical variance in regions that may be surrounded by high gas concentrations (Lewis and Miller 1984). For η = 0, the gas extinction coefficient is equal to the gas absorption coefficient. If η is set >0, the gas extinction coefficient is scaled by (1 − η)^−1. The direction of travel by the photon is not changed after a gas interaction. When ξ becomes smaller than a predefined threshold, a random number, R[t], within the interval (0, 1) is selected. If R[t] < 0.5 then the photon is terminated; otherwise, the photon continues with ξ being doubled (Lewis and Miller 1984). For domain averaged fluxes, the computational cost of using a high η outweighs the statistical benefits gained, and so η = 0.5 was chosen for the computations made in this study. When a photon is scattered, the direction from its previous trajectory is determined from a probability function, ), based on the normalized phase function ) of the scatterer. The probability of a photon being scattered between 0 and is given by By setting ) to a new uniform random number, , within the interval [0, 1), the angle is found by solving Eq. (6) for the upper limit of the integration ( McKee and Cox 1974 ). The azimuth angle around the direction of propagation is randomly chosen between 0 and 2 The phase functions are the Rayleigh scattering phase function for molecular scattering, the Henyey–Greenstein approximation for aerosols, and for clouds either the Henyey–Greenstein approximation or a direct determination of the phase function from Mie theory (Wiscombe 1980). The asymmetry factor g used in the Henyey–Greenstein approximation for aerosols is wavelength dependent and interpolated from a set of tables within LOWTRAN7. For cloud droplets, the asymmetry factor g, the size parameter x, and the refractive index m used in the Mie calculations also depend on wavelength, computed at a 0.005-μm resolution. As mentioned above, the surface is characterized as Lambertian and is thus independent of the incident direction. The angle for a photon that is reflected from the surface, , is computed from is a new uniform random number within the interval [0, 1) ( Barker and Davies 1992 ). The azimuth angle is again randomly chosen between 0 and 2 The accuracy of the flux estimates is proportional to the square root of the number of photons used in the simulation (Cashwell and Everett 1959). For simple applications, the number of photons required is predetermined from Bernoulli probability based on estimates of the resulting flux and the desired level of random error. Since the goal of these simulations is to predict these fluxes using photons that are weighted, it is difficult (if not impossible) to estimate the number of photons required prior to the initialization of the model run (Cashwell and Everett 1959). Consequently, within this model a convergence criterion is applied, and the Monte Carlo process is terminated once a desired accuracy has been achieved. The convergence is considered to occur when the domain averages of atmospheric absorption, transmission, and reflectance, as measured as a percent of the TOA input, all change by less than a given percentage i over three consecutive intervals of j photon c. Running modes To directly ascertain the 3D effects of the radiation field, the model can be run in three modes: the plane-parallel mode (PPM), the independent pixel approximation mode (IPM), and the three-dimensional (3DM) mode. For PPM computations, the model is composed of one single atmospheric column that is horizontally homogeneous. In the IPM, plane-parallel computations are essentially performed at each cell and a photon is constrained to a horizontally homogeneous atmospheric column. After a photon enters the top of the model, the photon remains in the same column as it traverses in the vertical direction. If the photon reaches the boundary of the column, it returns at the opposing boundary with the same trajectory, thus creating a cyclic boundary. Hence, the photon experiences variations in optical depth and atmospheric constituent microphysics only in the vertical direction. By comparison, the photon can traverse horizontally for the 3DM computations, thereby allowing it to encounter variations in optical depth and constituent microphysics in both the horizontal and vertical directions. Only at the edges of the model do the cyclic boundaries come into d. Model comparisons To check the validity of the atmospheric component of the model, a comparison was made with a Discrete-Ordinate Radiative Transfer model (SBDART) (Ricchiazzi et al. 1998 for clear and cloudy (plane-parallel) skies. The two models use the same transmission coefficients and much of the same input files, and therefore the results of the two models should be directly comparable. Since SBDART at the time of this comparison was limited to the Henyey–Greenstein approximation, this approximation was employed within both models for all computations. These calculations were made for both the broadband interval (0.25–4.0 μm) and, spectrally, at 0.005-μm resolution. Monte Carlo computations were made using the IPM and 3DM to determine if any biases were introduced by the cyclic conditions. For IPM and 3DM, all horizontal cells were given the same input parameters, reproducing, in effect, plane-parallel computations. No bias could be detected. The comparisons described below use 3DM The results of the comparisons are presented in Fig. 1, which shows a spectral plot of atmospheric absorption for clear and cloudy skies for a standard tropical atmosphere containing oceanic aerosols with 20-km visibility overlying an ocean surface with an albedo of approximately 0.02. The cloud is 1 km thick with an optical thickness of 40 and an effective radius of 8 μm. All spectral differences are less than 10 W m^−2 μm^−1. The random spikes ≈5 W m^−2 μm^−1 cannot be associated with any particular physical process and thus are believed to be noise artifacts of the Monte Carlo process. A 100-nm running mean is also plotted over the data to display any bias between the models. The negative biases ≈2 W m^−2 μm^−1, which exist mainly in the visible spectral region, are a result of the differences in treatment of aerosols. The other slight biases ≈1 W m^−2 μm^−1 are believed to be caused by the difference in vertical discretization of the atmospheric profiles (mainly water vapor) between the two models. Since the relative humidity of the air affects the absorption properties of aerosols, it is likely that part of the aerosol biases can also be attributed to this Broadband computations obtained by the two models are provided for top-of-the-atmosphere (100 km) upwelling flux, surface downwelling flux, and total atmospheric absorption for various cloud optical and geometrical thicknesses at solar zenith angles of 0°, 30°, and 60° (Table 1). Very little difference exists between SBDART and the Monte Carlo model, with broadband mean (standard deviation) for upwelling flux, downwelling flux, and absorption being −0.19 (1.01), 0.34 (0.97), and −0.05 (0.41) W m^−2, respectively. The small spectral differences between SBDART and the Monte Carlo model is surprising since SBDART uses discrete ordinates and the Monte Carlo model depends on stochastic techniques to solve the radiative transfer equation. For the broadband fluxes it is expected that some of the Monte Carlo noise would be canceled by spectral integration and the results would be as close as shown in Table 1 . The results presented are only for irradiance, and the difference between the spectral radiances of the two models is expected to be larger. More detailed comparisons await the incorporation of the Mie phase functions into SBDART. To test the validity of the model’s 3D component, the reflectance for an isolated cubical cloud devoid of an atmosphere was computed and compared to three other independently produced Monte Carlo models (McKee and Cox 1974; Davies 1978; Bréon 1992). Computations are made for clouds with optical thickness of 4.9 and 73.5 at solar zenith angles of 0°, 30°, and 60°. The McKee and Cox (1974), Davies (1978), and Bréon (1992) models all use a phase function derived by Deirmendjian (1969) for a cumulus type cloud droplet distribution (C1). For our model, the closest equivalent phase function to the C1 distribution is computed directly from Mie theory using a cloud droplet distribution with an effective radius of 6 μm. At 0° and 60°, the difference in reflectance for the optically thick cloud is less than 0.002 between Davies (1978), Bréon (1992), and our model (Fig. 2). At 30°, there is no statistical difference between Bréon’s and our results. Davies (1978) does not report reflectance for 30°. For McKee and Cox (1974) the reflectance is consistently the lowest, with the largest discrepancies occurring for the optically thin cloud at 30° and 60°. Part of the difference may be statistical [McKee and Cox (1974) report an accuracy between 1% and 2%], but the remaining difference is unexplained. Between Bréon’s (1992) and our results, the largest difference (≈0.005) in reflectance is for the optically thin cloud at 60°. Although a portion of this difference is due to statistical error, much of the discrepancy may be associated with the different phase functions being used by the two models. For the optically thicker cloud, this difference is less pronounced since increased multiple scattering tends to dampen minor differences between the phase functions. While none of these models, individually, can be considered as a benchmark for accuracy, the consistency between Davies (1978), Bréon (1992), and our model for the thick cloud, and the minor differences between Bréon (1992) and our model for the thin cloud, provides us with confidence that our model properly simulates fluxes in three dimensions. 3. Experiment: 3D cloud absorption To investigate the effects of 3D clouds on solar radiation, a tropical scenario was selected. Tropical clouds are strongly convective and have large vertical extent; thus any 3D effects that enhance atmospheric absorption should be most evident in this type of scenario. Tropical cloud systems are most often topped by a large anvil canopy of thick cirrus composed of ice particles and melting snow (Houze and Betts 1981). But due to the present limitations of the model, which can only handle cloud liquid water, we selected clouds that did not contain ice. Since satellite imagery is used for synthesizing the cloud field, images with cirrus anvil shields were not selected. Additionally, the shields obscure the underlying morphology that is being extracted from the satellite data. Obviously, such a choice may bias the results to cloud regimes that are most likely to experience 3D enhanced absorption. If the focus of this experiment has been to provide a definitive answer to the enhanced absorption problem globally, then such a filtering would be wholly inappropriate. The purpose of this case study, however, is to identify and explain the potential errors obtained in computing atmospheric absorption using the plane-parallel cloud assumption. If such errors do not appear in this scenario, then it would be reasonable to reject the hypothesis that the 3D effect explains enhanced absorption for all clouds. Because GCMs use plane-parallel clouds over broad spatial scales in their radiative schemes, and because the output of GCMs has been used to analyze the issue of enhanced absorption (Cess et al. 1995 ), differences between PPM and 3DM have been computed. A difficulty arises, however, in representing a 3D cloud field as a single homogeneous cloud layer, since any single method contains potential biases. Thus, the 3DM–PPM results should be regarded as providing a sense of the potential errors associated with the plane-parallel assumption rather than a definitive answer. Since the spatial configurations between the PPM and 3DM modes are different, the spatial mechanisms responsible for 3D effects can only be investigated using the IPM and 3DM results because both modes operate on precisely the same input fields. All 3DM and IPM computations were performed using the Monte Carlo model to reduce intermodel biases. a. Model input 1) 3D cloud Inputs to the model for 3DM and IPM consist of an artificial cloud scene embedded within a typical tropical atmosphere partitioned into 47 layers and a horizontal grid of 80 × 50 cells over an ocean surface. The albedo of the ocean varies with wavelength and, spectrally, is approximately equal to 0.02. To provide a realistic cloud scene morphology, the cloud field geometry was synthesized from cloud top heights derived from 1.1-km resolution Advanced Very High Resolution Radiometer (AVHRR) infrared images on NOAA-14 for an area south of Hawaii. This image was rescaled both horizontally and vertically to cloud volume elements of 800 m × 800 m × 400 m, respectively (Fig. 3a). The horizontal rescaling was done to maximize the spatial resolution of the computations while minimizing discontinuities along the cloud field boundary. The cloud base altitude of the convective cells was fixed at 1200 m with a maximum cloud thickness of 8800 m. Clouds that are difficult to detect from satellite imagery because of pixel resolution or multiple cloud layering have been included to more closely resemble an actual cloud field. Numerous small cumulus congestus clouds with a maximum areal extent of 1600 m and cloud thickness of 1200 m are added near the base of the large convective cells. Scattered altostratus cloud layers 800–1200 m thick at a base altitude of 6000 m were also included, their total areal coverage being approximately 12%. The horizontal shape of the altostratus clouds was derived from a different infrared image and superimposed onto the original field. About 9% of the cloud field consists of clear skies. Within all clouds the relative humidity was raised to 95%. Representation of the distribution of τ and r[e] throughout the cloud is a difficult problem due to our limited knowledge of the internal structure of clouds. Although multifractal techniques are being developed for synthesizing liquid water distributions (Cahalan et el. 1994), their use in three dimensions remains unclear. Therefore, for this case study, generalizations about the distributions of τ and r[e] have been adapted from observational studies (Bower et al. 1994). Along the horizontal plane, τ and r[e] have been taken as constant within a cloud layer. Vertically, τ varies with the slope of the adiabatic curve but at an amount representing only about 5% of the saturated adiabatic liquid water content to simulate the entrainment of dry air (Fig. 3d). Although this entrainment value seems low, it was chosen to maintain the total column optical thickness within reasonable bounds. The effective radius r[e] also varies vertically and ranges from 4.2 μm near the base of the convective cells to 16 μm at 2400 m above cloud base (Fig. 3c). Between 2400 m and the cloud top, r[e] is held constant at 16 μm. To avoid the complexities associated with ice microphysics, the cloud is considered to be made entirely of liquid water. For the cumulus congestus clouds, r[e] ranges from 4.4 to 10 μm and for altostratus clouds, from 5.5 to 8.0 μm. For the entire field the maximum optical thickness is 220, with a mean of 92 (Fig. 3b). 2) Plane-parallel cloud There are numerous methods by which a 3D liquid water distribution can be portrayed as a single homogenous cloud layer, each fraught with potential biases. Therefore, rather than selecting a single plane-parallel cloud for analysis, six different types of clouds were used. These were either based on the literature or derived from combinations of the mean fields of cloud liquid water density, cloud droplet effective radius, cloud base altitude, cloud top pressure, and geometric thickness of the 3D tropical cloud field. The nomenclature used to described the mean values of the tropical cloud field are m (cloud median altitude), b (cloud base altitude), z (geometric thickness), t (cloud top pressure), r[e] (effective radius), and LWC (liquid water content). The clouds used in the PPM computations are listed below. Values for these variables are provided in Table 2. All plane-parallel computations are weighted to account for the 9% clear sky. 1. ISCCP (Fig. 9B in Rossow and Schiffer 1991). 2. Stephens-Cb (Stephens 1978). 3. Type I: cloud defined by m, z, r[e], and LWC. 4. Type II: cloud defined by b, z, r[e], and LWC. 5. Type III: cloud defined by t, r[e], LWC. 6. Layer: same layers as for the 3D cloud. The LWC is averaged for each layer taken over the maximum areal extent of the cloud. The r[e] is not averaged. b. Computations Model runs were conducted for solar zenith angles between 0° and 75° at 15° increments to represent 1-h intervals for a location at the equator. The convergence criteria set for these runs was a 0.1% limit over three consecutive 16000 photon count intervals. By testing the convergence on all three domain averages of atmospheric absorption, transmission, and reflectance rather than just one, the convergence is more stable. The stability of the convergence is demonstrated in Fig. 4 where three narrow wavebands (bandwidth = 0.005 μm) in the visible (0.55 μm), water vapor absorption (0.94 μm), and cloud droplet absorption region (1.53 μm) were allowed to converge within 0.005%. The vertical scale in the figure represents the difference (in percentage) between the domain averages at 0.005% convergence and their values at 16000 photon count intervals. The vertical arrow represents the points where the domain averages converge at the 0.1% criterion used in the simulations. Computations are made for 751 wavebands representing 0.005-μm resolution between 0.25 and 4.0 μm. The number of photons required for IPM and 3DM at each solar zenith angle averages about 250000000. The required processing time was in the vicinity of 2500 h for a midrange Alpha processor. Computations were conducted in a parallel mode across a network of processors, so the actual length of time was on the order of a few weeks. 4. Spatial results and mechanisms Presented in this section are broadband fluxes for the 3DM, IPM, and PPM computations for the entire spatial domain, a sensitivity analysis, and the results for two individual spectral bands (0.94 and 1.53 μm). These spectral results serve to highlight the spatial aspects of 3D radiative transfer in a cloudy atmosphere and to provide a framework for explaining the mechanisms responsible for differences between the modes. It would be desirable to present the broadband results also at a high spatial resolution, but the processing time is prohibitive at a spectral resolution of 0.005 μm. Although the number of photons processed for the two spectral bands for each mode and solar zenith angle is large (>40000000), the complex nature of the field cannot guarantee a good statistical sample for each cell. Thus, the spatial distributions should be viewed as being qualitative rather than providing an absolute value for a specific location. a. Upwelling, downwelling irradiance, and absorption By accounting for cloud morphology in the radiative transfer computations, a reduction in upwelling broadband irradiance for 3DM versus the IPM ranges from 53 W m^−2 (or 7% in albedo), with the sun directly overhead, to 4 W m^−2 (or 2% in albedo) at a 75° solar zenith angle (Fig. 5). Part of the reduction in the upwelling flux is associated with an increased flux at the surface. The daylight mean flux difference is 18 W m^−2 with a 38 W m^−2 peak at 0°. The remainder of this energy is absorbed in the atmosphere with enhanced absorption of 15 W m^−2 at 45° and a daylight mean of 12 W m^ A partitioning of the 3DM–IPM atmospheric absorption between absorption by gases and cloud droplets reveals a complementary relationship (Fig. 6). As the solar zenith angle increases, the difference between the 3DM and IPM due to absorption by gases decreases from 8 to −6 W m^−2. Concurrently, the difference as a result of absorption by cloud droplets rises from 6 W m^−2 to a peak of 15 W m^−2 at 60° until it drops to 12 W m^−2 at 75°. From the vertical profile (Fig. 7), most of the 3D enhancement takes place below 5 km when the sun is directly overhead. At 30°, this altitude is raised as absorption by cloud droplets becomes more important. Some of this increase is offset by a reduction in absorption by gases (primarily water vapor) in the lower atmosphere. As the sun becomes lower in the sky at 60°, the greatest differences between the 3DM and IPM now occur high within the cloud field between 5 and 8 km. This increase in the upper portions comes at the expense of absorption by gases throughout virtually the entire atmosphere and by cloud droplets in the lowest layers of the cloud field. As previously noted, comparisons between 3D and plane-parallel clouds are provided since much of the“enhanced” absorption issue is based on the differences between observations and models that embody the plane-parallel assumption. Although the ISCCP and Stephens-Cb clouds have different optical thicknesses, effective radii, and cloud top altitudes than the 3D cloud, these cloud types are widely used in climate models and analysis and, therefore, have been included in the comparisons. As shown in Fig. 8, the difference in atmospheric absorption between the 3DM and PPM varies widely according to how the plane-parallel cloud is specified. The ISCCP and Type I clouds have the largest values with daylight mean differences of 38 and 22 W m^−2, respectively. These large differences are caused by high cloud top altitudes that reduce the amount of solar radiation available for gaseous absorption in the lower portions of the atmosphere. A much smaller difference of approximately 11 W m^−2 occurs for the Type III cloud, which has a relatively low cloud top altitude, thus allowing a greater transmission of solar radiation toward the more abundant water vapor regions. The negative difference of 16 W m^−2 for the Stephens-Cb is directly related to the lower single scattering albedo of the larger cloud droplets (r[e] = 32 μm) prescribed for this cloud. By plotting the 3DM–PPM absorption as a function of solar zenith angle, two curve shapes emerge (Fig. 9). For the Type II, Type III, and Layer clouds, the peak near 45° can be attributed to the complementary relationship between gaseous and cloud droplet absorption, as explained in the 3DM–IDM absorption analysis. However, for the ISCCP and Type I clouds, the curve essentially tracks the TOA radiative input, since most of the solar radiation is absorbed by the cloud or reflected back to space before it can reach the lower atmosphere, thereby negating the water vapor effects. Although the Stephens-Cb cloud top height is at an altitude lower than all the other clouds, it still exhibits a peak differential at 0°. The high optical thickness of this cloud reduces the transmission of solar radiation below the cloud top at a much higher rate than for the other clouds. Hence, the effect is the same as the ISCCP and Type I clouds, but the absolute difference in absorption is less since there are greater concentrations of gases above the cloud. As in the case of 3DM–IPM, the 3DM–PPM shows reductions in albedo with increases in downwelling surface radiation. For the Type II cloud with the sun directly overhead, the albedo is reduced by 15% (134 W m^−2) with an increase of surface downwelling radiation of 124 W m^−2. The daylight mean enhanced atmospheric absorption is 14 W m^−2. The peak enhanced absorption is 18 W m^−2 at 45°. A dichotomy seemingly exists between these results and analysis by Ramanathan et al. (1995), where enhanced absorption appears to reduce downwelling surface radiation. However, to this point, we have only made comparisons between models and not compared models with observations. To evaluate our results within the context of observations, the constant variable should not be optical thickness, which cannot be measured directly, but cloud albedo, a value that can be quantified from satellite observations. Assuming that the 3D computations are more representative of the radiation field in the natural world than are one-dimensional computations, plane-parallel clouds should be adjusted to match the cloud albedo obtained in the 3DM computations. Such an approach is not unique and is commonly applied in GCMs to tune model-computed liquid water content to satellite-observed albedo. The tuning can be achieved by either modifying the liquid water content of the clouds or keeping the liquid water content constant within the clouds and adjusting the amount of clear sky within the scene. By modifying the liquid water concentration within the Type II cloud, a 60% reduction in optical thickness decreased the daylight mean 3DM-“tuned” PPM atmospheric absorption to 7 W m^−2. The peak 3D enhanced atmospheric absorption becomes 14 W m^−2 for a solar zenith angle of 60°. To balance the shortwave radiative budget, the tuned PPM now shows greater downwelling radiation to the surface than the 3DM by 7 W m^−2 averaged over the day. By increasing the clear sky in the Type II cloud scene from 9% to 23% the albedo is again matched, resulting in a 3D enhanced absorption of 17 W m^−2 at 45° and 15 W m^−2 for the daylight mean. b. Sensitivity analysis To determine the primary spatial distributions responsible for 3D enhanced absorption, a sensitivity analysis was performed at a set of wavelengths that tends to show the greatest response to this type of absorption (Fig. 10). For each analysis, one specific spatial distribution was held constant. The spatial distributions analyzed are cloud geometric and optical thickness along the horizontal plane and effective radius, cloud optical thickness, and water vapor concentration along the vertical. A reference total absorption ratio consisting of 3DM divided by IPM minus one total absorption was first computed for solar zenith angles of 0°, 30°, and 60° (Fig. 10a). This ratio is then subtracted from an absorption ratio of 3DM divided by IPM for each spatial distribution held constant ( Figs. 10b–f). This measure indicates the sensitivity of the overall absorption to that particular spatial distribution. The larger the difference, the more the variation in that distribution contributes to the 3D enhanced absorption. Since each of these spatial distributions are interdependent, the effects are not expected to be cumulative. The largest effects are caused by variations in the vertical geometry of the field (Fig. 10b). By flattening the cloud field, a large amount of the 3D enhancement effect is removed. For wavebands more sensitive to cloud droplet absorption (1.18, 1.53, and 2.10 μm), the effect is stronger with increasing solar zenith angle. On the other hand, the water vapor wavebands (0.72, 0.83, and 0.94 μm) show greater sensitivity at lower solar zenith angles. The second largest response is obtained by holding the vertical distribution of water vapor constant (Fig. 10c). Since water vapor concentrations decrease with height and 3DM produces greater downwelling radiation, removal of this stratification reduces 3DM preferential absorption by gas. If the cloud is allowed to vary in height (Fig. 10d) but without horizontal variation in optical thickness (mean cloud optical thickness is used), there is little response for wavelengths above 1 μm except when the sun is low in the sky. Below 1 μm these variations, to a small degree, actually create negative values. A constant optical thickness in the vertical direction exhibits little change in the 3D enhancement effect (Fig. 10f); a slightly greater effect is achieved when the effective radius is held constant (Fig. 10e). c. Spatial diagnostics By examining vertical cross sections of the radiative field it is possible to arrive at an explanation for some of the difference in atmospheric absorption found between plane-parallel model calculations and observations (Figs. 11, 12, and 13). The cross section to be examined is located at kilometer 6 along an east–west transect noted by the white line in Fig. 3a. Radiative fluxes are computed for the 0.94-μm band, which is highly sensitive to water vapor absorption, and the 1.53-μm band, which is dominated by cloud droplet absorption. The cross sections presented are the differences between 3DM and IPM computations for atmospheric absorption, upwelling and downwelling flux, and mean pathlength at 0° and 60° solar zenith angles. For the 1.53-μm band, only the atmospheric absorption cross section is presented since the other cross sections do not differ qualitatively from that of the 0.94-μm band. To facilitate the interpretation of these cross sections, important features have been labeled (T, L, I, S, D, P), which correspond to references made in the text. The plus and minus superscript highlight the features that either enhance or decrease the 3D absorption effect, respectively. More complex features are accompanied by schematic diagrams to elucidate the mechanisms responsible for absorption differences between 3DM and IPM (see Fig. 14). The trajectories shown in the schematics are a plausible representation of a photon’s path and are not actual traces from the model. The atmospheric absorption cross section for direct overhead sun reveals three distinct features that partially account for enhanced absorption in the 3DM (labeled T, L, and I on the figures). The first feature designated as T occurs for both absorption by gases and cloud droplets (Fig. 11). For the 3DM, photons are not constrained within single atmospheric columns as is the case for the IPM; therefore they reach cloud interiors more readily through horizontal transport. Once there, the photons can become trapped and eventually absorbed by gas or cloud droplets (label T^+). As indicated by the blue shading (label T^−), this increase in absorption comes at the expense of absorption near the edges of the cloud. The same pattern is observed for the downwelling and upwelling flux cross sections. The mechanism for the interior focusing/trapping in the 3DM is demonstrated in schematic T (Fig. 14a). As shown, photons in the IPM can only reach interior cells by traveling vertically within a single column. In the case of a large convective cloud, this distance can be much greater than if the photons are allowed to travel horizontally. For equal pathlengths, photons in the IPM tend be closer to the cloud top than for the 3DM and thus have a higher probability of being scattered out of the cloud. Photons in the 3DM will likely be deeper within the cloud, so the number of interactions required to escape will be greater than that for the IPM. With more interactions in the 3DM, the absorption becomes enhanced. In addition to the reduction in 3DM flux along a cloud edge from the diffusion of photons toward the interior, much of the loss at the edge can be accounted by leakage out of the cloud. Seemingly, this loss of photons and associated reduced absorption (label L^−) should mitigate the 3D enhanced absorption effect. Whereas there is some decrease in absorption by photons escaping back toward space, the strong forward-scattering component of cloud droplets biases most of the photons toward the surface as indicated in the downwelling flux profile (label L^+) (Fig. 12a). The effect is a decrease in cloud absorption along cloud edges, but an increase in gaseous absorption from photons reaching the higher gas concentrations in the lower regions of the atmosphere. Schematic L (Fig. 14b ) demonstrates how a photon in the 3DM can escape the cloud and travel toward the surface, whereas in the IPM, the horizontal boundary restriction increases the chance for a photon to be reflected back toward space. Label L^−, in the upwelling flux cross section (Fig. 12b), points to an example of this higher reflectance for the IPM. Additionally, leakage from the edges of a cloud supplement the density of photons within a clear region and produce 3D enhanced gaseous absorption (label L^+). For a reflected photon in the IPM that has clear sky above, the photon cannot be absorbed by another cloud, unless it is scattered back toward the reflecting cloud by molecular scattering or, less likely, aerosol scattering. However, at the wavelengths where molecular scattering is significant, the ability of a cloud droplet to absorb is minimal. In the case of the 3DM, the ability for photons to travel across horizontal boundaries within the model provides a chance of interception by adjacent clouds and potential for being absorbed (label I^+). As demonstrated in schematic I (Fig. 14c), a photon in the IPM is confined to a single column and can only be scattered directly above or transmitted below a cloud, but not horizontally into another cell. Again, this difference is evident in the positive values in the upwelling flux and the 1.53-μm band absorption cross section. The horizontal confinement for IPM can, in certain instances, produce greater amounts of downwelling flux below a cloud and higher absorption below and within a cloud (label I^−) (Fig. 12a). However, the net result is for the 3DM enhanced absorption effect to dominate over the IPM, except for cases where small clouds are sufficiently isolated from one another. As the angle of the direct solar beam steepens cloud shadowing in the 3DM reduces the amount of direct beam solar radiation reaching the lower atmosphere vis-à-vis the IPM. For the IPM, the location where the direct beam impinges is not altered by the solar angular input at the top of the atmosphere. As shown at location S^− (Fig. 11b), this is not the case for the 3DM; thus, the IPM produces more downwelling flux and greater absorption (label S^−). However, as can also be seen in schematic S (Fig. 14d) and at the point designated S^+ in the downwelling flux and absorption cross sections, there are instances of broken clouds where the direct beam can slip below a cloud and produce increased 3DM downwelling radiation and enhanced absorption. Concurrent with a reduction in 3D atmospheric absorption by cloud shadowing is the complementary 3D enhancement of downwelling and upwelling fluxes and absorption caused by the direct solar beam impinging on the sides of a cloud (label D^+). Schematic D1 and D2 (Figs. 14e,f) shows two methods by which the strong forward-scattering characteristics of cloud droplets produces this enhancement effect. In the IPM, a photon is constrained to enter a cloud through its top. When the angle of direct beam is more closely aligned to the cloud top, a greater probability exists for the photon to be scattered and reflected back toward space. In the 3DM, a photon can enter the side of a cloud, and because of strong forward scattering the photon penetrates deeper toward the core of the cloud where the larger number of interactions required to escape increases the likelihood of absorption (schematic D1). Additionally, if a photon enters the top of the cloud and is reflected, there is a chance for the photon to enter the side of an adjoining cloud cell, again allowing for trapping and increased absorption (schematic D2). An examination of the mean pathlength cross section (Fig. 13) provides another explanation for 3D enhanced absorption. Although a longer pathlength often suggests greater absorption, pathlength statistics are only useful when applied within a homogeneous region and convolved with the photon intensity and amount of absorber along the path. For this study, the mean pathlength refers to the average pathlength of all photons within a given cell. Thus, if one cell has photons that primarily travel along a diagonal and a second cell has photons that dominate in the vertical direction, the first cell would have a longer mean pathlength. Since absorption is higher for the 3DM, it may be expected that the mean pathlength is also greater. For both solar zenith angles in Fig. 13, such is the case in the atmosphere between clouds as highlighted by P^+. For the IPM, all of the photons that traverse this location are either transmitted from directly above or reflected from below the cloud. Even when the sun is not directly overhead, the tendency exists for the mean emergent angle of photons to approach the nadir direction as clouds become more optically thick (Liou 1992). Thus, at location P^+, and below most of the layered clouds, the photons entering a cell below the cloud in the IPM will have a stronger vertical component in the mean pathlength computation as compared to that for the 3DM. For the 3DM, photons arriving from directly above or below a cell will be supplemented by a strong horizontal component of the photons being scattered from the sides of clouds outside of the vertical column. For this case, the longer pathlength is associated with greater absorption. There are occurrences when the mean pathlength for a given cell may be longer, but absorption is lower because the photon density is less. For example, at location P^− the mean pathlength is greater in 3DM, but the absorption is lower than for the IPM. As shown in the downwelling profile, this region is shaded by the cloud, leaving fewer photons available for absorption. Thus, in general, an examination of the pathlength is not sufficient to infer definitive conclusions about absorption. 5. Conclusions We have developed a Monte Carlo–based 3D radiative transfer model of high spectral and spatial resolution that contains all of the important atmospheric radiative constituents. Comparisons with a discrete ordinates model demonstrate good agreement for both broadband and spectral computations in clear and cloudy (plane-parallel) conditions. A comparison with several other Monte Carlo models gives us confidence as to the validity of our 3D computations. Our results show that the plane-parallel assumption for clouds used in GCMs can underestimate atmospheric absorption as a result of 3D effects. The 3D enhanced absorption is primarily attributed to greater absorption by water vapor with high overhead sun and increasing cloud droplet absorption as the sun approaches the horizon. Through a sensitivity analysis, we demonstrated that the most important factor is the vertical structure of the cloud field, followed by the vertical stratification of water vapor. Internal vertical cloud inhomogeneities were found to be less important. Using vertical cross sections of atmospheric absorption, upwelling and downwelling radiation, and mean pathlength, the mechanisms responsible for the 3D enhanced absorption have been identified and analyzed. For overhead sun, photons can penetrate deeper into clouds through focusing effects and become trapped within the cloud core. They are then absorbed by both water vapor and cloud droplets. Photons can also be scattered out of clouds to levels of the atmosphere where more water vapor is present, causing greater absorption by gases. Additionally, the 3D cloud representation provides a higher probability for photons reflected by one cloud to be intercepted by another. For higher solar zenith angles, shadowing occurs, which tends to reduce absorption by water vapor in clear sky. Concurrently, the angular direction of the solar beam causes isolated regions of high 3D enhanced absorption, as photons can enter through cloud sides and become trapped within the cloud cores. Finally, the mean pathlength is increased between clouds for the 3D case through the influence of horizontally traversing photons. However, as demonstrated, greater mean pathlength does not always entail more absorption, since it is the spatial distribution of the photons, in concert with the distribution of absorbers, that really matters. Although previous Monte Carlo–based studies have demonstrated less or little excess absorption for 3D computations (Davies et al. 1984; Li et al. 1995; Hignett and Taylor 1996), we believe the differences between those results and ours to be more a matter of the cloud morphologies employed in the models, the boundary conditions prescribed (cyclic or noncyclic), or the absorption considered (cloud or cloudy column), rather than a disagreement on process. Since the 3D effect investigated in this study produces more absorption, it could be speculated that the plane-parallel assumption is the cause behind the issue of enhanced, “excess,” or“anomalous” absorption. Obviously, results of this simulation do not provide the complete answer, since only about a quarter of the excess absorption reported can be accounted for in the comparison between the 3D and independent pixel approximation. However, as shown, between the 3D and plane-parallel clouds the difference in daylight mean absorption ranges from −16 to +38 W m^−2. Since plane-parallel clouds are the only type presently used in climate models, perhaps a stronger emphasis should be placed on comparing results between these and 3D cloud computations. Furthermore, although the cloud field morphology used in this study tends to maximize the 3D effect, it still remains quite elementary and lacks a number of important attributes about liquid water content, cloud droplet size, and water vapor (e.g., internal 3D distributions, entrainment effects at cloud boundaries, and the effects of deep convective cloud cores) because of our limited knowledge of their 3D distributions. Additionally, ice cloud microphysics are not included and the effect of aerosols was purposely kept at a minimum. Finally, there is still no universal agreement about the magnitude of excess absorption in a cloudy atmosphere, its global distribution, or even if the phenomenon exists at all. Thus, the quantitative discrepancies that are shown to exist in this case study do not provide a justification for rejecting the hypothesis that the 3D effect may be one of the contributing mechanisms for explaining the differences in the atmospheric absorption found between models and the real world. The appeal of this conceptual recognition is that it requires no new theory, just a better representation of clouds in climate models. We are grateful to Dr. Paul Ricchiazzi for his valuable insight and assistance with implementation of the LOWTRAN7 routines. This research has been funded in part by Department of Energy Grants 90ER61062 and 90ER61986, and National Aeronautics and Space Administration Grant NAGW-31380. • Aida, M. A., 1977: Scattering of solar radiation as a function of cloud dimension and orientation. Quat. Spectrosc. Radiat. Transfer,17, 303–310. • Barker, H. W., and J. A. Davies, 1992: Solar radiative fluxes for broken cloud fields above reflecting surfaces. J. Atmos. Sci.,49, 749–761. • Bower, K. N., T. W. Choularton, J. Latham, J. Nelson, M. B. Baker, and J. Jensen, 1994: A parameterization of warm clouds for use in atmospheric general circulation models. J. Atmos. Sci.,51, • Bréon, F., 1992: Reflectance of broken cloud fields: Simulation and parameterization. J. Atmos. Sci.,49, 1221–1232. • Byrne, R. N., R. C. K. Somerville, and B. Subasilar, 1996: Broken-cloud enhancement of solar radiation absorption. J. Atmos. Sci.,53, 878–886. • Cahalan, R. F., W. Ridgway, W. J. Wiscombe, S. Gollmer, and Harshvardhan, 1994: Independent pixel and Monte Carlo estimates of stratocumulus albedo. J. Atmos. Sci.,51, 3776–3790. • Cashwell, E. D., and C. J. Everett, 1959: A Practical Manual on the Monte Carlo Method for Random Walk Problems. Pergamon Press, 153 pp. • Cess, R. D., and Coauthors, 1995: Absorption of solar radiation by clouds: Observations versus models. Science,267, 496–499. • ——, M. H. Zhang, Y. Zhou, X. Jing, and V. Dvortsov, 1996: Absorption of solar radiation by clouds: Interpretations of satellite, surface, and aircraft measurements. J. Geophys. Res.,101, • Claußen, M., 1982: On the radiative interaction in three-dimensional cloud fields. Beitr. Phys. Atmos.,55, 158–169. • Coakley, J. A., and R. Davies, 1986: The effect of cloud sides on reflected solar radiation as deduced from satellite observations. J. Atmos. Sci.,43, 1025–1035. • ——, and T. Kobayashi, 1989: Broken cloud biases in albedo and surface insolation derived from satellite imagery data. J. Climate,2, 721–730. • Davies, R., 1978: The effect of finite geometry on the three-dimensional transfer of solar irradiance in clouds. J. Atmos. Sci.,35, 1259–1266. • ——, W. L. Ridgway, and K. Kim, 1984: Spectral absorption of solar radiation in cloudy atmospheres: A 20 cm−1 model. J. Atmos. Sci.,41, 2126–2137. • Deirmendjian, D., 1969: Electromagnetic Scattering on Spherical Polydispersions. Elsevier, 290 pp. • Hignett, P., and J. P. Taylor, 1996: The radiative properties of inhomogeneous boundary layer clouds: Observations and modeling. Quart. J. Roy. Meteor. Soc.,122, 1341–1364. • Houze, R. A., and A. K. Betts, 1981: Convection in GATE. Rev. Geophys. Space Phys.,19, 541–576. • Isaacs, R. G., W. C. Wang, R. D. Worsham, and S. Goldenberg, 1987:Multiple scattering LOWTRAN and FASCODE models. Appl. Opt.,26, 1272–1281. • Jonas, P., 1992: Some effects of spatial variations of water content on the reflectance of clouds. Ann. Geophys.,10, 260–266. • Kneizys, F. X., E. P. Shettle, L. W. Abreeu, J. H. Chetwind, G. P. Anderson, W. O. Allery, J. E. A. Selby, and S. A. Clough, 1988:AFGL-TR-88-0177, Phillips Lab., Hanscom AFB, MA, 137 pp. [NTIS • Kobayashi, T., 1993: Effects due to cloud geometry on biases in the albedo derived from radiance measurements. J. Climate,6, 120–128. • Lewis, E. E., and W. F. Miller, 1984: Computational Methods of Neutron Transport. John Wiley and Sons, 401 pp. • Li, Z. Q., H. W. Barker, and L. Moreau, 1995: The variable effect of clouds on atmospheric absorption of solar radiation. Nature,376, 486–490. • Liou, K. N., 1980: An Introduction to Atmospheric Radiation. International Geophysics Series, Vol. 25, Academic Press, 392 pp. • ——, 1992: Radiation and Cloud Processes in the Atmosphere: Theory, Observation and Modeling. Oxford University Press, 487 pp. • McKee, T. B., and S. K. Cox, 1974: Scattering of visible radiation by finite clouds. J. Atmos. Sci.,31, 1885–1892. • O’Hirok, W., and C. Gautier, 1998: A three-dimensional radiative transfer model to investigate the solar radiation within a cloudy atmosphere. Part II: Spectral effects. J. Atmos. Sci., in press. • Pilewskie, P., and F. P. J. Valero, 1995: Direct observations of excess absorption by clouds. Science,267, 1626–1629. • Ramanathan, V., B. Subasilar, G. Zhang, W. Conant, R. D. Cess, J. Kiehl, H. Grassl, and L. Shi, 1995: Warm pool heat budget and shortwave cloud forcing—A missing physics. Science,267, 499–503. • Ricchiazzi, P. J., S. Yang, C. Gautier, and D. Sowle, 1998: SBDART:A research and teaching software tool for plane-parallel radiative transfer in the earth’s atmosphere. Bull. Amer. Meteor. Soc., in press. • Rossow, W. B., and R. A. Schiffer, 1991: ISCCP cloud data products. Bull. Amer. Meteor. Soc.,72, 2–20. • Segal, M., and J. Davis, 1992: The impact of deep cumulus reflection on the ground-level global irradiance. J. Appl. Meteor.,31, 217–222. • Stephens, G. L., 1978: Radiation profiles in extended water clouds. Part II: Parameterization schemes. J. Atmos. Sci.,35, 2123–2132. • ——, 1988a: Radiative transfer through arbitrarily shaped optical media. Part I: A general method of solution. J. Atmos. Sci.,45, 1818–1836. • ——, 1988b: Radiative transfer through arbitrarily shaped optical media. Part II: Group theory and simple closures. J. Atmos. Sci.,45, 1837–1848. • ——, and S. Tsay, 1990: On the cloud absorption anomaly. Quart. J. Roy. Meteor. Soc.,116, 671–704. • Welch, R. M., and B. A. Wielicki, 1989: Reflected fluxes for broken clouds over a Lambertian surface. J. Atmos. Sci.,46, 1384–1395. • Wendling, P., 1977: Albedo and reflected radiance of horizontally inhomogeneous clouds. J. Atmos. Sci.,34, 642–650. • Wiscombe, W. J., 1980: Improved Mie scattering algorithms. Appl. Opt.,19, 1505–1509. • ——, 1995: Atmospheric physics: An absorbing mystery. Nature,376, 466–467. • Zuev, V. E., and G. A. Titov, 1995: Radiative transfer in cloud fields with random geometry. J. Atmos. Sci.,52, 176–190. Fig. 1. Comparison of results for the Monte Carlo (MC) model and the discrete ordinates atmospheric radiative transfer model (SBDART). (a) Solar spectrum input at top of the atmosphere (TOA) for both models. (b) Difference in atmospheric absorption between MC and SBDART for clear sky conditions at a solar zenith angle = 30°. (c) Difference in atmospheric absorption between MC and SBDART for a plane-parallel cloud of optical thickness (τ = 40) and effective radius (r[e] = 8 μm) at 30°. Heavy line indicates 100-nm running mean. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 2. Comparison among Monte Carlo models of cloud reflectance for cuboid cloud of optical thickness (a) 73.5 and (b) 4.9 at solar zenith angles of 0°, 30°, and 60°. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 3. Cloud model input: (a) cloud top height, (b) vertically integrated cloud column optical thickness, (c) cross section of cloud field effective radius, and (d) cross section of cloud optical thickness for 400-m-thick layers. Cross sections presented in Figs. 11, 12, and 13 are located at kilometer 6 along east–west transect as indicated by solid white lines in areal images. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 4. Convergence of model domain averages measured as percent of top of the atmosphere input for absorption, downwelling, and upwelling fluxes at (a) 0.55-, (b) 0.94-, and (c) 1.53-μm bands. Arrows at each wavelength point to first instance when all three domain averages have changed by less than 0.1% over three consecutive intervals of 16000 photon counts. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 5. Total broadband (0.25–4.0 μm) deviations of IPM from 3DM model results for atmospheric column absorption, top-of-the-atmosphere (100 km) upwelling flux, and downwelling flux at the surface. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 6. Total broadband (0.25–4.0 μm) deviations of IPM from 3DM model results for atmospheric column, cloud droplet, and gaseous absorption vs solar zenith angle. Symbols represent solar zenith angles at which computations were made. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 7. Horizontally integrated total broadband (0.25–4.0 μm) ratio of 3DM to IPM model results for total atmospheric, cloud droplet, and gaseous absorption at solar zenith angles of 0°, 30°, and 60°. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 8. Total broadband (0.25–4.0 μm) daylight mean atmospheric absorption deviations of PPM from 3DM model results for ISCCP, Stephens-Cb, Type I, Type II, Type III, and Layer plane-parallel cloud. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 9. Total broadband (0.25–4.0 μm) deviations of PPM from 3DM model results for ISCCP, Stephens-Cb, Type I, Type II, Type III, and Layer plane-parallel cloud vs solar zenith angle. Symbols represent solar zenith angles at which computations were made. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 10. Sensitivity analysis of 3D effects on atmospheric absorption at 0.72, 0.83, 0.93, 1.18, 1.53, and 2.10 μm for solar zenith angles of 0°, 30°, and 60°. (a) Reference ratio of 3DM–IPM-1. Deviations from reference when (a) morphology, (b) water vapor, (c) optical thickness in the horizontal plane, (d) optical thickness in the vertical plane, and (e) effective radius in the vertical plane are held constant. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 11. Cross section along kilometer 6 east–west transect for 3DM–IPM computations. Total atmospheric absorption for (a) 0.94- and (b) 1.53-μm bands. Vertical and slanted arrows represent the direction of the solar beam at 0° and 60°, respectively. Gray line represents the profile of the cloud. Letters within the image refer to mechanisms described in the text. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Fig. 14. Schematic demonstrating mechanisms for 3DM and IPM described in text. Gray squares represent cloudy cells. Arrows portray photon trajectory and scattering events with asterisks referring to absorption locations. Dash line shows effect of cyclic horizontal boundary. Mechanisms T and L upper left, I and S lower left, and D1 and D2 upper right. Citation: Journal of the Atmospheric Sciences 55, 12; 10.1175/1520-0469(1998)055<2162:ATDRTM>2.0.CO;2 Table 1. Total broadband (0.25–4.0 μm) differences between SBDART and Monte Carlo model for upwelling flux, downwelling flux, and atmospheric absorption for clear and cloudy sky at solar zenith angles of 0°, 30°, and 60°. Cloud optical thickness between 5 and 80 for geometric thickness ranging from 1 to 6 km. Table 2. Plane-parallel cloud properties.	{"url":"https://journals.ametsoc.org/view/journals/atsc/55/12/1520-0469_1998_055_2162_atdrtm_2.0.co_2.xml","timestamp":"2024-11-07T22:01:26Z","content_type":"text/html","content_length":"680158","record_id":"<urn:uuid:72f0023b-52d2-4107-ae07-9a3110d34433>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00126.warc.gz"}
Metric Spaces Are Compact Spaces If and Only If They're BW Spaces Metric Spaces Are Compact Spaces If and Only If They're BW Spaces Recall from the Compact Spaces as BW Spaces page that if $X$ is a compact space then $X$ is also a BW space. Also recall from The Lebesgue Number Lemma page that if $(X, d)$ is a metric space that is also a BW space then for every open cover $\mathcal F$ of $X$ there exists an $\epsilon > 0$ such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $B(x, \epsilon) \subseteq U$. We will now use the Lebesgue Number Lemma to show that the converse of the first theorem mentioned above is true for metric spaces, that is, if $X$ is a metric space then $X$ is a compact space if and only if it is a BW space. Theorem 1: Let $(X, d)$ be a metric space. Then $X$ is compact if and only if it is a BW space. • Proof: Let $(X, d)$ be a metric space. • $\Rightarrow$ Suppose that $X$ is a compact space. From the first theorem mentioned at the top of this page we immediately have that $X$ is also a BW space. • $\Leftarrow$ Suppose that $X$ is a BW space. Then every infinite subset of $X$ has an accumulation point. Let $\mathcal F$ be an open cover of $X$. Since $X$ is a metric space that is also a BW space, by the Lebesgue Number Lemma there exists a Lebesgue number $\epsilon > 0$ such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $B(x, \epsilon) \subseteq U$. • For each $x \in X$ there may be many such $U \in \mathcal F$ that satisfy the condition above. So, for each $x \in X$, select any $U_x \in \mathcal F$ such that $B(x, \epsilon) \subseteq U_x$. • Step 1: Take any $x_1 \in X$. If $X \subseteq B(x_1, \epsilon) \subseteq U_{x_1}$ then $\mathcal F^* = \{ U_{x_1} \}$ is a finite subcover of $X$ and we're done. If not, proceed to step 2. • Step 2: Take any $x_2 \in X$ such that $x_2 \not \in B(x_1, \epsilon)$. If $X \subseteq B(x_1, \epsilon) \cup B(x_2, \epsilon) \subseteq U_{x_1} \cup U_{x_2}$ then $\mathcal F^* = \{ U_{x_1}, U_ {x_2} \}$ is a finite subcover of $\mathcal F^$ and we're done. If not, proceed to step n. • Step n: Take any $x_n \in X$ such that $\displaystyle{x_n \not \in \bigcup_{i=1}^{n-1} B(x_i, \epsilon)}$. If $\displaystyle{X \subseteq \bigcup_{i=1}^{n} B(x_i, \epsilon) \subseteq \bigcup_{i=1} ^{n} U_{x_i}}$ then we're done, and if not, continue in this process. • If at any point this process terminates then $\mathcal F^$ is a finite subcover of $X$. We claim that this process cannot go on forever. Suppose instead that this process never terminates. Then we obtain an infinite set of points: \quad A = \{ x_1, x_2, ... \} • Furthermore, from the choices of each $x_i$ we have that each $x_i$ is separated by an open ball of radius $\epsilon > 0$, i.e., for each $i, j \in \mathbb{N}$, $i \neq j$ we have that $d(x_i, x_j) \geq \epsilon > 0$. We claim that the set $A$ cannot have any accumulation points. Suppose otherwise, i.e., suppose that $A$ has an accumulation point $x \in X$. Then every open neighbourhood $U$ of $x$ contains infinitely many points of $A$. In particular, the open neighbourhood $B \left (x, \frac{\epsilon}{2} \right )$ contains infinitely many points of $A$. So there exists $x_i, x_j \in A$ such that $x_i, x_j \in B \left (x, \frac{\epsilon}{2} \right )$. But then $d(x_i, x_j) < \epsilon$ which is a contradiction. • So $A$ has no accumulation points. But $(X, d)$ is a BW space, so this is a contradiction. Hence the assumption that the process above does not terminate was false. So the process above always terminates which implies that for every open cover $\mathcal F$ of $X$ there exists a finite subcover $\mathcal F^*$. Therefore, $X$ is compact. $\blacksquare$	{"url":"http://mathonline.wikidot.com/metric-spaces-are-compact-spaces-if-and-only-if-they-re-bw-s","timestamp":"2024-11-09T08:01:36Z","content_type":"application/xhtml+xml","content_length":"19494","record_id":"<urn:uuid:e3cae9d3-644d-4d28-b9c8-12ab7c4dc5f7>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00788.warc.gz"}
Volts to electron-volts Volts to electron-volts calculator Electrical voltage in volts (V) to energy in electron-volts (eV) calculator. Enter the voltage in volts, charge in elementary charge or coulombs and press the Calculate button: Volts to eV calculation with elementary charge The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electric charge Q in elementary charge or proton/electron charge (e): E[(eV)] = V[(V)] × Q[(e)] The elementary charge is the electric charge of 1 electron with the e symbol. Volts to eV calculation with coulombs The energy E in electron-volts (eV) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C) divided by 1.602176565×10^-19: E[(eV)] = V[(V)] × Q[(C)] / 1.602176565×10^-19 See also	{"url":"https://jobsvacancy.in/calc/electric/volt-to-ev-calculator.html","timestamp":"2024-11-07T05:43:05Z","content_type":"text/html","content_length":"10234","record_id":"<urn:uuid:a88465ea-cfb5-491f-9a08-1c9015429fba>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00830.warc.gz"}
Geometric Construction - Explanation & Examples Geometric Construction – Explanation & Examples While it may seem surprising, we can create almost any geometric object — including lines, circles, squares, triangles, angles, and more — using only these two tools! Geometric construction is part of pure geometry (also known as synthetic geometry or axiomatic geometry). This is the geometry that does not rely on equations and coordinate systems. Instead, it relies on constructions and proofs based on predetermined axioms. In this article, we will discuss the following subtopics of geometric construction: • What is Geometric Construction? • How to Do Constructions in Geometry? • Types of Construction in Geometry. What is Geometric Construction? Geometric construction is the process of creating geometric objects using only a compass and a straightedge. It is a component of pure geometry, which, unlike coordinate geometry, does not use numbers, formulae, or a coordinate system to create and compare geometric objects. A compass is a device with a handle and two legs. One leg has a point at the end, and the other has a pencil or graphite piece. The two legs are hinged so that the user can change how far apart they are. Compasses have been used since antiquity to draw circles and arcs. A straightedge is any physical object with a solid, (you guessed it) straight edge that can be traced with a pencil. Note that while many people use a ruler as a straightedge in geometric constructions, technically, a straightedge should not include numbers. Using a ruler is okay as long as you ignore the temptation to compare lines using its measurements. Geometric constructions (and their accompanying proofs) rely on a certain set of agreed-upon rules called axioms. These essentially give us the tools we need to make a proof and ensure that all readers are working with the same definitions. Euclid of Alexandria is sometimes called the founder of geometry because his work in pure geometry was well-formulated and well-distributed. In fact, his primary work, Euclid’s Elements, is one of the most widely circulated books of all time. Until the 20th century, every educated person would have taken a course on Euclid’s Elements. Euclid included 23 definitions, five postulates, and five common notions. The definitions ensured that Euclid and any readers were on the same page regarding what words meant. The common notions provided the logical steps necessary for proving that the constructions worked, while the postulates were essentially foundational “facts” that did not need to be proved. While many of the propositions in Euclid’s Elements were just proofs that constructions were possible, others were proofs about comparing geometric objects or proofs that established facts about them. However, in these latter proofs, Euclid often still included a simple construction as an illustration or reference for the proof. Other Geometries While Euclid’s geometric constructions and proofs have stood the test of time, they are not the only set of axioms, nor are his constructions the only ones. Other geometers, including Riemann and Gauss, developed their own axiom systems that led to different geometries. These are generally known as “non-Euclidian” geometries, and many involve removing or negating Euclid’s fifth postulate, which states that: “If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, the the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.” A simple way to conceive of non-Euclidean geometry is to consider what happens when our drawing surface is the surface of a sphere instead of a flat plane. On such a surface, it is possible to draw a triangle with straight lines but two right angles. How to do Constructions in Geometry? Constructions in geometry are based on circles and lines. This is because of the two basic shapes we can make with a straightedge and a compass. Making a line with a straightedge is pretty simple. Just put the edge of straightedge wherever you want the line. Then, use a pencil to draw the line, keeping the pencil close to the edge of the straightedge and holding the straightedge steady with your non-dominant hand. You can use the pencil from the compass, but it is sometimes handy to have another one ready to go. To make a circle with a compass, put the point wherever you want the center to be. Then, put the pencil tip at any point on the desired circumference. Next, rotate the pencil in a circle, holding the point steady. If your compass has a lock function, now is the time to use it so that the distance between the legs doesn’t change. When you get back to the starting point, you will have a perfect Types of constructions geometry While there are many things you can make with just a ruler and a compass, we’ll divide them into broad categories here to give you an idea of the possibilities. Remember that each construction starts with a “given.” These are the things already on your plane. For example, a construction that requires you to draw a circle with a center at a certain point and a distance the length of a certain line will already have the point and the line drawn on the plane. Using just a compass and a ruler, we can cut a line or angle in half. We can also use similar processes to cut a circle, triangle, or other polygons into two equal parts. Using the same principles, we can cut the same objects into fourths, eighths, sixteenths, etc. If you are given a line, angle, circle, triangle, etc., you can make a copy of it using your straightedge and compass in another place. These constructions will often ask you to put the copy in a specific place, such as a given point or on a given line. Remember that there are no specific measurements in constructions. That being said, you can copy an angle regardless of the measurement without a protractor by just using a straightedge and compass. You can also create angles of many different measures (for example, 60 degrees, 30 degrees, 75 degrees, etc.) using construction methods. In addition to copying triangles, you can use construction methods to make triangles with any three given side lengths. You can also make equilateral triangles. Construction methods are also useful for proving facts about triangles, such as the fact that the angles as the base of isosceles triangles are equal. You can even use them to prove that two or more triangles are congruent. Other Shapes Finally, you can even use construction methods to make squares, regular pentagons, and regular hexagons. Examples in construction are a little different from examples in other sections. Nonetheless, there are still a few examples that will help illustrate how construction works. Example 1 Connect two points with a line. Example 1 Solution First, draw a line between these two points, line up your straightedge so that the edge touches both points. Then, use a pencil and trace along the edge of your straight edge. It’s okay if the line segment you draw extends beyond the points. That’s all there is to it! You will end up with a picture like the one below. Example 2 Create two different circles with center at A. Example 2 Solution There are infinitely many solutions to this problem, but all can be found in the same way. First, put the point of your compass at A. Then, set your compass to a small distance, put the pencil or graphite to the paper, and draw a circle. Now, set your compass to the largest distance possible. Put the point at A, set the pencil or graphite to the paper, and draw another circle. The result will be two circles, centered at A. One of the circles will be inside the other, as shown. Example 3 Connect the three points to form a triangle. Example 3 Solution This problem is similar to the one in example 1. Instead of just connecting two points, we need to connect three for a total of three lines. First, line the straightedge up so that the edge touches points A and B. Then, trace along the edge to connect the two. Next, do the same thing for the points A and C and then the C and B points. The resulting figure should look like the one below. Example 4 Connect the points A and B. Then, make a circle with center A and radius AB. Then, make a circle with center B and radius BA. Example 4 Solution Our first step is to connect points A and B as before. First, line up the straightedge so that the edge touches A and B. Then, trace along the edge to create the line. Now, to make the first circle, put the point of the compass at A and the pencil at B. Then, holding your hand steady, pivot the compass around the point to create a circle. Next, place the point on B and the pencil tip on A. As before, hold your hand steady while pivoting the compass around the point and drawing out the circumference of the circle. Your final figure will look like the one below. Example 5 Given the line AB, create a circle with center A and radius AB. Then, create a circle with center B and radius equal to the diameter of the first circle. Example 5 Solution The first step is similar to what we have done before. We will place the point of the compass at A and the pencil tip at B. Then, holding our hand steady, we can trace out the circle’s circumference by pivoting the compass around the point until we get back to B. Next, we want a circle with center B and radius equal to the other circle’s diameter. How do we know which point on the circle is the furthest from B? We can actually use Euclid’s second postulate, which says that we can use construction methods to extend any line segment. Thus, we line our straightedge up so that the edge touches A and B. We also want to make sure that the straightedge extends far enough to intersect with the circumference of the circle on the other side of B. We can trace along the edge and label the intersection of this line and the circle as C. BC is the circle’s diameter. Now, we can put the point of the compass at B and the pencil at C. Holding it steady, we then pivot the compass around the point and trace out the circumference of the larger circle until we get back to C. The final figure will look similar to the one below. Practice Questions 1. Which of the following objects can we use for geometric construction? I. Compass II. Straightedge III. Calculator IV. Timer 2. How many points are needed to construct a rectangle? 3. True or False: It is possible to draw three diameters through the circle shown below. 4. True or False: It is impossible to draw two more circles – one with radius CA and the other with radius DA. 5. True or False: It is impossible to construct two circles that touch at exactly one point, such that one is not inside the other. Open Problems 1. Construct two circles that touch at exactly one point, such that one is not inside the other. 2. Draw lines connecting all four points so that each point connects. 3. Draw three diameters through the circle. 4. Draw two more circles, one with radius CA and the other with radius DA. 5. Draw a circle with radius four times the length of AB. Open Problem Solutions	{"url":"https://www.storyofmathematics.com/geometric-construction/","timestamp":"2024-11-10T09:38:22Z","content_type":"text/html","content_length":"192061","record_id":"<urn:uuid:f785b582-57ea-4251-b39c-a350c3a86d4f>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00435.warc.gz"}
Godel, Turing and Truth John K Clark (johnkc@well.com) Sat, 11 Jan 1997 21:12:53 -0800 (PST) -----BEGIN PGP SIGNED MESSAGE----- On Fri, 10 Jan 1997 "Lee Daniel Crocker" <lcrocker@calweb.com> Wrote: >There is nothing that prevents us from forming other axioms >and other systems of proof that have different unprovables. That could be very dangerous, it might make your system too powerful, so powerful it can prove things that are not true. For example, so far The Goldbach Conjecture has been shown to be true for every number we test it on, but we can't test it on every number, and so far nobody has found a way to prove it from the fundamental axioms of number theory. Suppose we give up trying to prove it correct or having computers look for a counterexample to prove it wrong, and just add it as an axiom. Consider the possibilities: 1) The Goldbach conjecture is true and it is possible to find a proof from the current axioms: In this case adding Goldbach as a Axiom is unnecessary and inelegant. 2) The Goldbach conjecture is untrue but we add it as an axiom: Bad idea, now you've made mathematics inconsistent. Think of the embarrassment when some computer finds a number that violates Goldbach, your axiom. Think of all the mathematics built on top of this dumb axiom that now must go in the garbage can. 3) The Goldbach conjecture is true but unprovable: In this case it would be a great idea to add Goldbach as an axiom because then your axiomatic system would be more complete and just as consistent. The only trouble is, Turing showed us that if it's unprovable you'll never know it's unprovable, so it's just too dangerous to take the chance. On the other hand, Godel doesn't say we can't know anything, he says we can't know everything. It's the same with Turing's results, sometimes, but not always, you can prove that a certain result can not be derived from existing axioms, so then you can add it as an axiom, IF you have courage, IF you think it is true. For example, if you take it as an axiom and assume that the parallel postulate is true, that is, " through any point in the plane, there is ONE , and only ONE, line parallel to a given line" then a perfectly consistent geometry can be made, Euclid did it. Strangely it's also true that If you assume "Through any point in the plane, there are TWO lines parallel to any given line" then a perfectly consistent geometry can be made, Lobachevski and Bolyai did it. It's even true that if you assume " Through a point in the plane, NO line can be drawn parallel to a any given line" then a perfectly consistent geometry can be made, Riemann did it. Consistency is not everything, which one of these geometrys are true must be determined by The reason all this was possible was because the parallel postulate has been shown to be independent of the other axioms. An analogous situation has not been proven for the Goldbach Conjecture. An even number can be found that is not the sum of two prime numbers OR it can not. There is no middle ground. I want a system that can prove it is true, but ONLY if it is true. John K Clark johnkc@well.com -----BEGIN PGP SIGNATURE----- Version: 2.6.i -----END PGP SIGNATURE-----	{"url":"http://extropians.weidai.com/extropians.1Q97/0545.html","timestamp":"2024-11-13T16:40:41Z","content_type":"application/xml","content_length":"6291","record_id":"<urn:uuid:723d1ab5-d74e-4362-be4c-d3b1f9cd3cd0>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00020.warc.gz"}
Effect of vehicle load and cargo on TSD in context of how can you reduce total stopping distance (tsd)? 27 Aug 2024 Title: The Impact of Vehicle Load and Cargo on Total Stopping Distance: Strategies for Reduction Total Stopping Distance (TSD) is a critical safety parameter that measures the time taken by a vehicle to come to a complete stop from a given speed. This study investigates the effect of vehicle load and cargo on TSD, highlighting the importance of considering these factors in reducing TSD. We present a comprehensive analysis of the relationship between vehicle load, cargo, and TSD, providing mathematical formulas and empirical evidence to support our findings. Total Stopping Distance (TSD) is a critical safety parameter that has been extensively studied in the field of transportation engineering. It is defined as the sum of the perception-reaction time (PRT) and the braking distance (BD). PRT is the time taken by the driver to perceive the hazard, react, and initiate braking, while BD is the distance traveled by the vehicle during this period. Theoretical Background: The TSD formula can be represented as: TSD = PRT + BD where PRT is a function of the driver’s reaction time (RT) and the vehicle’s speed (v): PRT = RT + (v / 2.5) BD, on the other hand, is a function of the vehicle’s deceleration rate (a) and its initial velocity (v): BD = v^2 / (2 * a) Effect of Vehicle Load and Cargo: The addition of load or cargo to a vehicle can significantly impact TSD. A heavier vehicle requires more time and distance to come to a complete stop, increasing PRT and BD respectively. Assuming a linear relationship between the vehicle’s weight (W) and its deceleration rate (a), we can represent the effect of load on BD as: BD = v^2 / (2 * (a + k * W)) where k is a constant that depends on the vehicle’s suspension and braking system. Strategies for Reducing TSD: 1. Optimize Vehicle Load: By minimizing the weight of the vehicle, we can reduce BD and subsequently TSD. 2. Improve Braking System: Upgrading the braking system to improve deceleration rate (a) can also reduce BD and TSD. 3. Enhance Driver Training: Improving driver reaction time (RT) through training programs can reduce PRT and subsequently TSD. Empirical Evidence: A study conducted by [1] found that a 10% increase in vehicle load resulted in a 5% increase in TSD. Similarly, another study [2] showed that improving the braking system reduced TSD by an average of The effect of vehicle load and cargo on Total Stopping Distance (TSD) is significant. By optimizing vehicle load, improving braking systems, and enhancing driver training, we can reduce TSD and improve road safety. [1] Smith et al. (2018). The Impact of Vehicle Load on Total Stopping Distance. Journal of Transportation Engineering, 144(4), 04018002. [2] Johnson et al. (2020). Braking System Upgrades for Improved Safety. International Journal of Automotive Technology and Management, 20(1), 10-25. TSD = PRT + BD PRT = RT + (v / 2.5) BD = v^2 / (2 * a) BD = v^2 / (2 * (a + k * W)) Note: The formulae are presented in BODMAS (Brackets, Orders of Operations, Division, Multiplication, Addition, and Subtraction) format, with ASCII characters used to represent mathematical Related articles for ‘how can you reduce total stopping distance (tsd)?’ : Calculators for ‘how can you reduce total stopping distance (tsd)?’	{"url":"https://blog.truegeometry.com/tutorials/education/461750fa205a04808b553373749b6d00/JSON_TO_ARTCL_Effect_of_vehicle_load_and_cargo_on_TSD_in_context_of_how_can_you_.html","timestamp":"2024-11-03T00:57:56Z","content_type":"text/html","content_length":"19964","record_id":"<urn:uuid:36955b68-febc-4eb7-88af-791bac7e73a5>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00118.warc.gz"}
University of Pittsburgh Thursday, February 27, 2014 - 12:00 Abstract or Additional Information Coding theory is concerned with detecting and correcting errors in data transmission. In 1982 Tsfasman, Vladut, and Zink discovered that codes constructed from certain families of algebraic curves have better asymptotic parameters than any previous constructions. This motivated a great activity in applying methods of algebraic geometry to coding. I will talk about a relatively new family of algebraic geometry codes called toric codes. A toric code is defined by evaluating sections of a line bundle L on a toric variety X at a finite set of points Z on X. We will see how basic parameters of a toric code depend on combinatorics of the lattice polytope associated with L and on geometry of the set of points Z. Research Area	{"url":"https://www.mathematics.pitt.edu/seminar-colloquia-event/toric-geometry-coding-theory","timestamp":"2024-11-08T05:14:41Z","content_type":"text/html","content_length":"91894","record_id":"<urn:uuid:cee633a4-331e-4bdd-ac03-403569e90659>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00079.warc.gz"}
Math Mentor GPTs features and functions, examples and prompts \| GPT Store Math Mentor Friendly Math Tutor for all levels Website https://smartbrandstrategies.com Share this GPT Math Mentor conversion historical statistics Welcome message Hi! I'm here to help with your math questions. What can I assist you with today? Features and Functions □ Knowledge file: This GPT Contains knowledge files. □ Python: The GPT can write and run Python code, and it can work with file uploads, perform advanced data analysis, and handle image conversions. □ Browser: Enabling Web Browsing, which can access web during your chat conversions. □ Dalle: DALL·E Image Generation, which can help you generate amazing images. □ File attachments: You can upload files to this GPT. Conversion Starters □ Solve this algebra problem for me. □ Can you explain this calculus concept? □ I need help with this geometry question. □ How do I approach this statistics problem? Math Mentor showcase and sample chats No sample chats found. Related GPTs • A friendly math tutor offering concise answers, visual aids, quizzes, progress tracking, and rewards. • Friendly math teacher for kids. • Your personal Math teacher • Math expert, offering detailed explanations and reliable information. • Friendly Math Tutor for Students. • A friendly mathematics expert and personal tutor. • All-age adaptable math tutor with a scholarly approach and real-world examples. • I'm a friendly math tutor for grade K - 5 kids, here to explain concepts and quiz you!	{"url":"https://gptstore.ai/gpts/vTRZKATby-math-mentor","timestamp":"2024-11-09T19:30:19Z","content_type":"text/html","content_length":"67746","record_id":"<urn:uuid:01abe321-4e21-4c8d-956c-9c28d56fe1c9>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00184.warc.gz"}
Best Urban Research Universities For Chemistry In The Southeast Best Urban Research Universities For Chemistry In The Southeastexpand_more Looking to study Chemistry at a research university in the Southeast? Prefer an urban setting, with a city as an extension of campus? We've compiled a list of the Best Urban Research Universities For Chemistry In The Southeast. Learn more about each school below and calculate your chances of acceptance. 16 Colleges Sort by: Best for chemistry	{"url":"https://www.collegevine.com/schools/best-urban-research-universities-for-chemistry-in-the-southeast","timestamp":"2024-11-02T12:38:45Z","content_type":"text/html","content_length":"88687","record_id":"<urn:uuid:bd2952fe-f2ae-4eae-867e-6dbf368d72a9>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00088.warc.gz"}
The probability of an ectopic pregnancy online \| Calculate the probability of pregnancy The probability of an ectopic pregnancy relative to age: 0 % The total frequency of ectopic pregnancy per 1000 pregnant women: 12-14 Risk factors for ectopic pregnancy: Operations on the fallopian tubes: 20,0 History of WB: 10,0 Anamnesis of salpingitis: 4,0 Infertility treatment: 4,0 Age less than 25 years: 3,0 Anamnesis of infertility: 2,5 Smoking: 2,5 Vaginal douching: 2,5	{"url":"https://womencalc.com/en/probability-of-ectopic-pregnancy/","timestamp":"2024-11-03T10:43:33Z","content_type":"text/html","content_length":"47973","record_id":"<urn:uuid:c1b648b7-3e40-4e0e-8ae9-56839b4c8ca6>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00051.warc.gz"}
Export Reviews, Discussions, Author Feedback and Meta-Reviews Submitted by Assigned_Reviewer_4 Q1: Comments to author(s). First provide a summary of the paper, and then address the following criteria: Quality, clarity, originality and significance. (For detailed reviewing guidelines, see http: Paper summary: The authors propose a Bayesian inference procedure for conditional bivariate copula models, where the copula models can have multiple parameters. The copula parameters are related to conditioning variables by a composition of parameter mapping functions and arbitrary latent functions with Gaussian process priors. The inference procedure for this model is based on expectation propagation. The method is evaluated on synthetic data and on financial time series data and compared to alternative copula-based models. The paper also includes an overview of related methods. The proposed method is sound. The authors include a description of related methods and list strengths and weaknesses of their method. The paper is written clearly and concisely. It makes a very mature impression. I could not find any inconsistencies or typos. The figures and tables complement the text seamlessly, equations are clear and well explained and the text is flawless. The model and its inference procedure is an extension of a similar conditional copula model that was restricted to copulas with only one parameter. The present work generalizes this approach to allow more than one copula parameter. To my knowledge, this extension is novel. The flexibility of copula families with only one parameter is limited and the applicability of more flexible families is a huge advantage. Performance gains compared to the alternative methods are remarkable. This result is very promising. Q2: Please summarize your review in 1-2 sentences The paper extends an inference procedure for conditional copula models. The contribution is important and well done. Submitted by Assigned_Reviewer_6 Q1: Comments to author(s). First provide a summary of the paper, and then address the following criteria: Quality, clarity, originality and significance. (For detailed reviewing guidelines, see http: This paper presents an interesting approach for modeling dependences between variables based on Gaussian process conditional copulas. It employs Gaussian process priors on the latent functions to model their interactions. Then an alternating EP algorithm is proposed for the approximate Bayesian inference. Experimental results on both synthetic data and two realworld data indicate the better performance of conditional copula models, especially the one based on student t copula. In general, this paper is well written and perhaps useful for dependence modeling in time series data. The EP inference is also implementable for the posterior distribution in the paper. A major problem of this paper is that some details are missing: (1) How the parameters in the copula model (e.g., \alpha, \beta, \omega for the Symmetrized Joe Clayton Copula) in are calculated? I assume this paper sets them to constants. How to choose them for a given time series data? (2) The details of the parameters the exponential covariance function in Gaussian process are missing. (3) How to choose the number of latent functions k? Q2: Please summarize your review in 1-2 sentences This paper discusses an interesting topic and derives an implementable inference method. However, some details on the parameter settings are missing. Submitted by Assigned_Reviewer_7 Q1: Comments to author(s). First provide a summary of the paper, and then address the following criteria: Quality, clarity, originality and significance. (For detailed reviewing guidelines, see http: The paper tackles the problem of fitting conditional copula models. The paper goes beyond typical constant copula functions and aim to model cases where parameters of the copula are functions of other (time) varying aspects of the problem. The core solution is essentially modeling the parameters space as Gaussian Processes. Starting from a Gaussian Process prior, the information from observed data are incorporated via Expectation Propagation for approximate inference of the posterior distribution. The paper is extremely well written and easy to follow. The problem is well motivated and most importantly illustrations with HMM based and multivariate time series copulas are very nicely written. The experiments are sufficient as well. Q2: Please summarize your review in 1-2 sentences Overall, this is a nice paper that motivates a novel problem and provides a reasonable solution with excellent experimental illustrations. Q1:Author rebuttal: Please respond to any concerns raised in the reviews. There are no constraints on how you want to argue your case, except for the fact that your text should be limited to a maximum of 6000 characters. Note however that reviewers and area chairs are very busy and may not read long vague rebuttals. It is in your own interest to be concise and to the point. We would like to thank all of the reviewers for their time and comments. We are also very pleased to receive unanimously positive feedback. Below, we answer three specific questions and correct an error in our submission (which fortunately only strengthens our conclusions). Reviewer 6 asked some specific questions which we answer below - we will update the manuscript to make these points clearer. 1 ) The parameters alpha, beta and omega in the DSJCC method are found by maximum likelihood. 2 ) The covariance function for the Gaussian processes is the squared exponential kernel (for the parametric form see equation (2) or e.g. Rasmussen and Williams). All parameters of the kernels were found by maximizing the EP approximation to the model evidence. 3 ) The number of latent functions k is determined by the number of parameters of the parametric copula model. There is one latent function for each parameter. To all reviewers: We would also like to correct a “copy-paste” error we made just before submission. The numbers in table 4 were mistakenly set to be identical to those in table 5. Fortunately, the discussion in the manuscript was based on the correct figures and the correct values only strengthen our conclusions. We include the correct figures for Table 4 below. The pattern of bold and underlined numbers is similar to that in Table 5, which explains why we did not spot the error in our submission. We include below the average test log-likelihood for each method, GPCC-G, GPCC-T, GPCC-SJC, HMM, TVC, DSJCC, CONST-G, CONST-T and CONST-SJC, on each dataset AUD, CAD, JPY, NOK, SEK, EUR, GBP and NZD. The format is Method Name results for AUD results for CAD results for JPY results for NOK results for SEK results for EUR results for GBP results for NZD As in the manuscript, the results of the best method are shown in bold (with a "b" in front). The results of any other method are underlined (with a "u" in front) when the differences with respect to the best performing method are not statistically significant according to a paired t-test at \alpha = 0.05. The best technique is GPCC-T (the one with most "b"'s in front), followed by GPCC-G as discussed in the manuscript. u 0.0562 b 0.1221 u 0.4106 u 0.4132 u 0.2487 u 0.1045 b 0.1319 b 0.0589 u 0.1201 b 0.4161 b 0.4192 b 0.8995 b 0.2514 b 0.1079	{"url":"https://papers.nips.cc/paper_files/paper/2013/file/67d16d00201083a2b118dd5128dd6f59-Reviews.html","timestamp":"2024-11-05T07:32:56Z","content_type":"text/html","content_length":"17318","record_id":"<urn:uuid:2f3baad9-ed19-46a8-95d0-0f24f02cb7e9>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00330.warc.gz"}
Fibofuck is a language based on brainfuck that replaces the tape with a Fibonnaci heap (fibo heap). For the sake of minimalism and simplicity, the language doesn't use n-ary trees like regular Fibonacci heaps would. Instead, each tree in the heap is a skew heap. The language's syntax is very similar to brainfuck, with some instructions redefined to make sense in the context of a heap (cf documentation of , , > and <) General presentation The environment of fibofuck is a heap similar in concept to a fibo heap, the key difference being that trees in the heap are implemented as binary skew heaps. This means that no node has more than two children. The trees are implemented this way to simplify the language, as n-ary tree support would make the syntax more complicated. At the beginning of the program, the heap is empty. The maximal size of the heap is theoretically infinite. Each node of the heap contains a signed integer. Note that the structure of the fibo heap is always maintained and that the heap is updated after each instruction. This means that the heap is really "merge heavy". This is one of the differences between a regular Fibo heap and the heap in fibofuck. While in a regular fibo heap the merge call is often only done after deletion of an element, here it's done after every instruction, making the whole environment very chaotic. Another important detail is that the trees follow the merge operation of skew heaps. This means that their right and left branches are swapped after each merge. Behavior of operations The specific behavior of the different operations on the heap is a mix of the skew heap operations and those of fibonacci heaps. This makes the internal structure of the environment somewhat unique. • merge : Mix of skew heap AND fibo heap. Applies the skew heap merge operation on 2 trees of the same size (number of nodes) until every tree in the heap is of different size. • insert node: Comes from fibo heap. Adds a node to the heap and then calls merge. • delete node : Comes from fibo heap. Deletes a node from a tree. If it has children, inserts its children in the heap as new trees and then calls merge. • delete tree: Deletes a tree from the heap. • decrease key : Comes from skew heap. Decreases a key in a tree and then heapifies it. • increase key : Comes from skew heap. Increases a key in a tree and then heapifies it. Note that after every instruction the environment will restructure itself to remain a Fibonacci heap. If the heap is empty, every instruction other than the creation of nodes will be ignored. NB: Every character that isn't an instruction is ignored. Command Description % Initialises a node in the heap at 0 , Reads a character and initialises a node in the heap at the character's value / Moves the node pointer to the left child of the current node if it exists \ Moves the node pointer to the right child of the current node if it exists ^ Moves the node pointer to the parent of the current node if it exists < Moves the node pointer to the root of the previous tree in the heap if the current tree is not the last > Moves the node pointer to the root of the next tree in the heap if the current tree is not the last ! Removes the node under the pointer from the heap * Removes the tree containing the node under the pointer from the heap + Increments the node under the pointer by 1 - Decrements the node under the pointer by 1 [ Jumps past the matching ] if the value of the node under the pointer is 0 ] Jumps back to the matching [ if the value of the node under the pointer is nonzero . Prints the value of the node under the pointer as a character : Prints the value of the node under the pointer as a decimal integer ♯ Prints each tree in the heap as an array as well as information on the heap (size, number of elements, index of node pointer, number of trees,...) Example programs %%%%%%% creates cells for each distinct letter in 'hello world' +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ W ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ R +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ O ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ L ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ H +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ E ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ D Turing completeness This paragraph attempts to show that brainfuck can be simulated in Fibofuck. To do so, we have to setup a specific heap where the root of each tree is set to 0 and each of their children is set to an arbitrary high number (greater than 256). A way to achieve that is to initialize trees of sizes from 1 to n. Once such a heap is initialized, the behavior of the +, -, [, ], <, >, and . instructions are identical to their brainfuck counterparts. External resources interpreter written in C with flex/bison parser	{"url":"https://esolangs.org/wiki/Fibofuck","timestamp":"2024-11-14T20:04:53Z","content_type":"text/html","content_length":"28166","record_id":"<urn:uuid:a279732f-f732-4922-a0b2-2a36aa46f773>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00773.warc.gz"}
What our customers say... Thousands of users are using our software to conquer their algebra homework. Here are some of their experiences: I ordered the software late one night when my daughter was having problems in her honors algebra class. It had been many years since I have had algebra and parts of it made sense but I couldn't quite grasp how to help her. After we ordered your software she was able to see step by step how to solve the problems. Your software definitely saved the day. C.K., Delaware For many people Algebra is a difficult course, it doesnt matter whether its first level or engineering level, the Algebrator will guide you a little into the world of Algebra. Dan Mathers, MI It was very helpful. it was a great tool to check my answers with. I would recommend this software to anyone no matter what level they are at in math. Patricia, MI Math has never been easy for me to grasp but this program makes it easy to understand. Thanks! Tom Walker, CA Search phrases used on 2008-02-17: Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among • ti-84 slope programs • square root 11 fraction • First Grade Math Sheets • cliffnotes inequalities • boolean algebra formula sheet • Question Papers on Factorization • calculators for decimals as mixed numbers in simplest form • Printout Sats english ks2 • prentice hall conceptual physics answers • hardest algebra problem in the world • mixed number to decimal • rational number solver • solution set algebra square root • how to turn a fraction to a decimal • factoring trinomials with a casio calculator • fractional coefficients algebra • INTEGERS WORKSHEETS • SIMPLIFY RADICAL SQRT OF B CUBED • free math final worksheets forhigh school kids • math word problems worksheets for 4th grade • old sats maths questions • tricks help on how to store formulas in TI -84 Plus graphing calculator operation • free Algebra2 Problem Solver • maths formula class 10th • Abstract Algebra-Hungerford • fourht grade equation worksheet • glencoe physics answers • software to solve mathematics problem for business and economy • free online algebra tutor • algebra l lesson in the saxon book • simplifying radicals solver • multiplying and dividing decimals • quadratic equation to standard form calculator • bracket +problems +maths +worksheets +advanced • algebric equations • find zeros of function of multiple variables in matlab • ti 89 LU decomposition function • sample lesson plan for exponents • convert mixed number to percentage • algebra taks 9th grade • real life application of a parabola quadratic • bearings worksheet • rational expressions worksheets • solving linear inequalities using matlab symbolic • download ti 84 plus calculator • how to code square root in java • practice equatoins • permutation problems for kids • Decimal square • online ks3 maths test • formula for percentage • calculator online number systems • convert binary to dec using ti-89 • formula in getting the square root • algebra one glencoe florida • Factoring In Algebra • Truss Zero Force Member • dividing integer worksheet • christmas math trivia • math worksheets scale 4th • TI-84 emulator • variable poem math • factorial button on the TI-83 Plus • algebra 2+using polynomials to design fonts • lesson plans 7 grade math slope of a line • prentice hall algebra answers • how to store data in ti-89 • algebra independent dependent worksheet • simplify cube root denominator • least common multiple worksheets • partial differential equations.java • calculator fractions from greatest to least • practice algebra square roots • order of operations 4th grade worksheet • solution manual+dummit foot • simple mathematical aptitude questions • multiplying equations examples • maths games to ply online • T-83 Plus programs • integral solver ti84 • Operations with Fractions adding, dividing, and multiplying all together • texas instruments "TI-84" "mod function" pdf • multiplying addition and subtraction trigonometric formulas • 5th grade paper puzzle • how to do math combinations step by step • show problem solving for converting my factors to decimals • square root simplify exponentiAL form • evaluate the permutation calculator • quadriatic formula on ti-84 • integer worksheets Subtracting Integers • least common denominator calculator • Fractions, Decimals, and Percents Online Calculator • quadratic function vertex form worksheet • class 12 free learning science +softwares • artin algebra solution manual • great tips on how to pass the compass test • quadratic formula worksheet • free christmas worksheets ks3	{"url":"https://softmath.com/algebra-help/balancing--algebra-equations-w.html","timestamp":"2024-11-11T16:22:08Z","content_type":"text/html","content_length":"35786","record_id":"<urn:uuid:d46f150f-b2a9-4834-bc8a-28a367d6f69b>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00639.warc.gz"}
How To Set Stop Loss And Take Profit In Forex TradingHow To Set Stop Loss And Take Profit In Forex Trading Today I will share about how to set Stop Loss and Take Profit correctly, from the perspective of Technical Analysis. The first thing to remember when placing Stop Loss (SL) and Take Profit (TP) is: Which time frame (TF) you trade, then set SL and TP based on that TF. And the way to set SL and TP according to my method is very simple, it is in 2 words: TOP - BOTTOM. Here are the details of the order entry process, with the roles of SL and TP: - Define your main trading timeframe. - Use your system to recognize a forex signal when it appears. - Determine the closest TOPS and BOTTOMS, compared to the expected entry point. - Add/subtract the spread and amortize the extra noise of the price. - Provide SL and TP prices. - Calculate the ratio of RISK : REWARD of the order (R:R ratio). - If you accept the ratio R: R, you will enter the order. - If we do not accept the R:R ratio, we will give up our intention to enter the order and look for another opportunity. Thus, the selection of SL and TP points will determine whether to enter an order or not, on the basis of the R:R ratio. Therefore, SL and TP are very important, and placing SL and TP needs to be Why choose the TOP - BOTTOM method to place SL and TP: - Top and Bottom are support and resistance levels, it contains technical elements (price charts). - Tops and Bottoms are also often psychological support/resistance levels. - Tops and Bottoms represent the wave amplitude in a trending market (trending), and the expression of a sideways price zone in a non-trending market (sideway). - When price breaks out of Tops and Bottoms, a new range is established, and usually the price will move significantly further in the direction of that breakout. Therefore, it is very safe to stop loss there. - When the price approaches the Top - Bottom area, usually there will be a reaction (more or less) and will bounce back. Therefore, taking profits in that area is also very safe. - Every method has errors, this method is no exception, but the error rate is the lowest I've ever known. You can experience this method, if it is suitable, then apply it, if not, consider this as content for reference. HOW TO SET SL AND TP: IF BUY: - SL equals LOW POINT OF BOTTOM minus NOISE. - TP is equal to HIGH POINT OF TOP minus NOISE. IF SELL: - SL equals the HIGHEST POINT OF THE TOP plus the spread plus NOISE. - TP is equal to LOWEST POINT OF BOTTOM plus spread plus NOISE. * Spread: Is the average spread under normal conditions, of the trading currency pair. * Noise: Is the estimated spread elasticity due to price fluctuations, or it can be amortization when the broker intentionally "cheats to scan the SL", sometimes it is the volatility when it occurs the "fake breakout". *** Tops and Bottoms: Are resistance or support levels, in the area closest to the current price. Usually, I estimate the noise in the TF frames I often trade, namely: - At M15 about 3-5 pips - At H1 about 5-10 pips - At H4 about 10-20 pips - At D1 about 20-40 pips These levels depend on the currency pair you trade, over time you will learn and come up with a suitable level yourself. EXPLAIN HOW TO SET SL and TP THROUGH THE FOLLOWING EXAMPLE: - If after the price makes a peak of 3 and is going down, here you can determine the nearest bottom is 2, and the nearest top is 3, then we will have resistance and support areas, colored in blue as shown in the picture. Note, the AREA needs to reconcile the points (high and low, top and bottom) with each other. - If after the price creates the top 3, then you enter a BUY order, you have to put the SL below the old bottom at point 2, specifically the SL point to set is 1.16956 minus 10 pips (for example, the noise is 10 pips) will equal 1.16856. The TP will be equal to the top of 1 and 3 minus the noise, i.e. 1.17790 minus 10 pips will be 1.17690. Thus, a BUY order after the price makes a peak of 3 will be TP at 4. If you buy after point 4, you will TP at 5, buy after 5 will TP at 6, and buy after point 6 will get SL in the middle of the 7 and 8. - If after point 3 you enter a SELL order, you have to place SL on the old top which is the reconciliation point of 1 and 3, specifically the SL point to be set is 1.17790 plus spread (2 pips for example) plus noise (eg. 10 pips) so that would be 1.17910. And TP will be equal to bottom 2 plus spread plus noise, ie 1.16956 plus 2 pips plus 10 pips, so TP is 1.17076. Thus, with a Sell order after point 3, you will not lose SL at 4, at 5 and even at 6, then the order will be TP at 7 (although bottom 7 is higher than bottom 2, but you still TP because you added noise). NOTES WHEN USING SL AND TP: - In essence, SL and TP set points do not determine whether the ratio of R:R, R:R is high or low, it is due to your trading system and trading style. The purpose of placing SL and TP is to limit the lowest loss if the judgment is wrong, and maximize the profit if the judgment is correct. At the same time, to avoid cases where the price hits the SL and then turns around, or taking profits too - The above way of placing SL and TP is usually correct under normal market conditions. Particularly with the time of news, especially like Non-Farm or FOMC, you need to be careful because there may be SL due to very strong price fluctuations, or it may be because the broker expands the Spread too much. If you hold orders past special news, the noise should be increased. Above is the method of placing Stop Loss and Take Profit that I am applying, and feel it is the most optimal. Hope the article will be helpful to you. Thank you for reading the article, please share this content if you find it useful for new traders. See you in the next posts. Best regards, No comments:	{"url":"https://www.caphile.com/2022/11/how-to-set-stop-loss-and-take-profit-in-forex-trading.html","timestamp":"2024-11-11T00:30:28Z","content_type":"application/xhtml+xml","content_length":"110457","record_id":"<urn:uuid:a6704650-bb95-42c8-9d8a-ad081203fa69>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00514.warc.gz"}
1. Simulate the execution of the List Partition algorithm (section 3.1.2) on the input 90 30 40 70 60 20 10 30. Show both the M and D matrices. What is the best partition of that list into 3 segments? Into 2 segments? 2. Here's a variation on the List Partition problem (section 3.1.2). Instead of looking at the sum of the s[j] in each segment of the partition, and trying to minimize the largest such sum, suppose we want to look at the max s[j] in each segment of the partition, and that we want to minimize the sum of these maxima. To avoid the degenerate solution that places all objects into one segment of the partition, suppose you are also given an integer p which is the maximum number of objects allowed in any one segment of the partition. (Note that no solution is possible if kp < n, where k is the maximum number of segments allowed and n is the number of objects.) Give an algorithm solving this problem, argue that it is correct, and analyze its running time. (Faster algorithms are better than slow ones, of course, so try to make yours fast, at least in the big-O sense.) 3. Simulate the execution of the Longest Increasing Subsequence algorithm (Section 3.1.4) on the sequence 90 33 41 70 66 25 13 30 68 31. (Give a table as shown on page 63. Note that the predecessor row indicates the position of the predecessor, not its value.) What is the longest increasing subsequence ending with 66? 31? Overall? 4. Suppose you are given a sequence (s[1], r[1]), (s[2], r[2]), ..., (s[n], r[n]) of pairs of positive integers. Think or the r[i] paired with each s[i] as a "reward" for choosing s[i]. As in the Longest Increasing Subsequence problem, your goal is to choose a subsequence of the s's that increases from left to right, i.e., choose s[i[j]] so that s[i[1]] ≤ s[i[2]] ≤ ... ≤ s[i[k]] and i[1] < i[2] < ... < i[k]. In the LIS problem, the goal was to make the subsequence as long as possible, i.e. to maximize k. In this problem, called the "Best Increasing Subsequence" problem, your goal is to maximize the total reward for the selected elements. In other words, maximize r[i[1]] + r[i[2]] + ... + r[i[k]] subject to the constraint above that the corresponding s[i]'s are increasing. [The LIS problem is the special case of this problem in which all the rewards are equal.] Again, give an algorithm solving this problem, argue that it is correct, and analyze its running time. (Faster algorithms are better than slow ones, of course, so try to make yours fast, at least in the big-O sense.) Simulate execution of your algorithm on the following input: s[i]: 90 33 41 70 66 25 13 30 68 31 r[i]: 10 10 20 10 10 30 20 30 30 40 What is the best increasing subsequence? 5. Skiena's text page 78, Problem 3-6. Assume d[1]=1. 6. (Extra Credit) Skiena's text pages 78-79, Problem 3-7. Again assume d[1]=1. The order of coins in a solution is ignored; e.g. if you had 1 and 2 cent coins, there are two different way to give 3 cents change: 1+1+1 and 1+2 (which is the same as 2+1).	{"url":"https://courses.cs.washington.edu/courses/cse417/05wi/hw/hw2.html","timestamp":"2024-11-11T11:01:43Z","content_type":"text/html","content_length":"7599","record_id":"<urn:uuid:47304016-1933-47a1-8381-e9ce164cfbcf>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00286.warc.gz"}
[FAQ] What method is best used for estimating specification values between those given in the datasheet? FAQ: Logic and Voltage Translation > Input Parameters >> Current FAQ The most common example of this question comes from the fact that Standard Logic devices are specified at 1.65V, 3V, and 4.5V, however most engineers use our devices at 1.8V, 3.3V, and 5V. Supplies are not perfect in any system -- they can fluctuate up or down significantly. MOSFET resistances increase at lower voltages, so the datasheet specifications are not tested at the nominal supply values , but at approximately 10% below nominal. I can't tell you why it's not exactly 10% -- these values have been used for decades and have become standardized throughout the industry. At 5V, it's exactly 10% (0.5V), but at 3.3V it's rounded to ~9%. At 1.8V it's ~8%. My assumption is that these values were rounded a bit to give easier values to work with -- 1.62V isn't as "nice" of a number as 1.65V, and in the end, they both achieve the same goal, which is to provide a 1.8-V supply lower limit for testing. In many cases, the best option is to use the datasheet values for your design -- even if you know your supply is exactly 3.302V and will never change. They provide a 'worst case' value that will prevent errors in the design. If, for some reason, you really need a value between those given in the datasheet, linear interpolation is the TI approved method for getting intermediate points. You might be saying "but not all those specs are linear" -- and you'd be right. The fact is though, that across a small range (for example, between 1.65V and 3V), the variation from linear will be minor, and the datasheet values provide some headroom to the specifications. This method gives a safe approximation that is backed by TI and our characterization process. The equation for linear interpolation is: Most typically, the "x" values will be the supply, and the "y" values will be the spec, with "y" being the value you are trying to get, and "x" being the specific supply value at which you are trying to get it. I have already implemented this into an example Excel file if you would like to use that: If you need values beyond the datasheet specs, for example, if you need a 5.1V value and the datasheet only gives 3V and 4.5V values, then linear extrapolation is used. The equation and method are the same as above, except your data point will be outside the range of the given supplies. I would recommend only doing this for values very close to the given datasheet specs, since this method becomes more inaccurate the farther the supply is from the given values.	{"url":"https://e2e.ti.com/support/logic-group/logic/f/logic-forum/793508/faq-what-method-is-best-used-for-estimating-specification-values-between-those-given-in-the-datasheet","timestamp":"2024-11-01T21:06:53Z","content_type":"text/html","content_length":"127002","record_id":"<urn:uuid:02bc0d99-9a9a-4e21-9bde-0510310c034a>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00811.warc.gz"}
News - Page 3 of 43 - Discrete Mathematics Group On August 14, 2024, Peter Nelson from the University of Waterloo gave a talk at the Discrete Math Seminar on formalizing the matroid theory on LEAN at the Discrete Math Seminar. The title of his talk was “Formalizing matroid theory in a proof assistant“. The 2024 summer school was held at IBS for 2 weeks at IBS with lectures by Boris Bukh (CMU) and Tung Nguyen (Princeton University) The 2024 summer school was held at IBS for two weeks, from July 29 to August 9, 2024. It was organized by the IBS Extremal Combinatorics and Probability Group and featured lectures by Boris Bukh from Carnegie Mellon University and Tung Nguyen from Princeton University. 24.08.09, Tung Nguyen, Recent work on the Erdős-Hajnal conjecture: day 5 24.08.08, Recent work on the Erdős-Hajnal conjecture: day 4 24.08.07, Tung Nguyen, Recent work on the Erdős-Hajnal conjecture: day 3 24.08.06, Tung Nguyen, Recent work on the Erdős-Hajnal conjecture: day 2 24.08.05. Tung Nguyen, Recent work on the Erdős-Hajnal conjecture: day 1 24.08.02, Boris Bukh, Algebraic methods in combinatorics: day 5 24.08.01, Boris Bukh, Algebraic methods in combinatorics: day 4 24.07.31, Boris Bukh, Algebraic methods in combinatorics: day 3 24.07.30, Boris Bukh, Algebraic methods in combinatorics: day 2 Daniel Král’ gave a talk on the minor closure of matroid depth parameters at the Discrete Math Seminar On August 6, 2024, Daniel Král’ from Masaryk University gave a talk at the Discrete Math Seminar on the minor closures of depth parameters of matroids. The title of his talk was “Matroid depth and width parameters“. 2024 Korean Student Combinatorics Workshop was held in Gongju from July 29 to August 2, 2024 The 2024 Korean Student Combinatorics Workshop (KSCW2024, 2024 조합론 학생 워크샵) was held in Gongju from July 29 to August 2, 2024. Sponsored by the IBS Discrete Mathematics Group, this event aims to provide a platform for Korean graduate students working on combinatorics and related areas to establish a foundation for collaborative research. It was organized by four students of KAIST/IBS: Donggyu Kim (김동규), Seokbeom Kim (김석범), Seonghyuk Im (임성혁), and Hyunwoo Lee (이현우). The workshop featured two invited talks by Semin Yoo (유세민) and Jungho Ahn (안정호), as well as open problem sessions followed by ample time for joint work. Welcome Meike Hatzel, a new member of the IBS Discrete Mathematics Group The IBS discrete mathematics group welcomes Dr. Meike Hatzel, a new research fellow at the IBS discrete mathematics group from August 1, 2024. She received his Ph.D. from the Technische Universität Berlin under the supervision of Prof. Stephan Kreutzer. She is interested in graph theory, in particular the structure theory of directed graphs. We are hiring! Senior Research Fellow Position at the IBS Discrete Mathematics Group The IBS Discrete Mathematics Group (DIMAG) in Daejeon, Korea invites applications for one senior research fellow position. DIMAG is a research group that was established on December 1, 2018 at the Institute for Basic Science (IBS), led by Chief Investigator, Dr. Sang-il Oum. DIMAG is located at the headquarters of IBS in Daejeon, South Korea, a city of 1.5 million people. Website: https://dimag.ibs.re.kr/ Currently, DIMAG consists of researchers from various countries and the work is done in English. DIMAG is co-located with the IBS Extremal Combinatorics and Probability Group (ECOPRO). Successful candidates for this position are expected to have research experience in Discrete Mathematics, in particular in Structural Graph theory, Combinatorial Optimization, Matroid Theory, and Algorithms, for 10 years or more after their Ph.D.’s. The appointment is initially for two years and can be extended further after reviews of the research performance until the group’s closure or the retirement age. The starting annual salary is no less than KRW 96,000,000, if the successful candidate has 10 years or more research experience after Ph.D.’s. The expected appointment date is June 1, 2025, and it can be adjusted to earlier or later, but no later than September 1, 2025. This is purely research position and will have no teaching duties. A complete application packet should include: • AMS standard cover sheet (preferred) or cover letter (PDF format) • Curriculum vitae, including a list of publications and preprints (PDF format) • Research statement (PDF format) • Consent to Collection and Use of Personal Information & Application for the IBS (PDF file) For full consideration, applicants should email all the items to dimag@ibs.re.kr by August 19, 2024, 18:00 KST. DIMAG encourages applications from individuals of diverse backgrounds. Suggested E-mail subject from applicants: [DIMAG – name] e.g., [DIMAG – PAUL ERDOS] Euiwoong Lee (이의웅) gave a talk on the parameterized complexity of approximating the minimum size of a deletion set to make a graph belong to a fixed class On July 30, 2024, Euiwoong Lee (이의웅) from the University of Michigan gave a talk at the Discrete Math Seminar on the parameterized complexity of approximating the minimum size of a deletion set to make a graph belong to a fixed class. The title of his talk was “Parameterized Approximability of F-Deletion Problems“. Two of our former members, Dong Yeap Kang (강동엽) and Abhishek Methuku, and their collaborators received the 2024 Frontiers of Science Awards at the International Congress for Basic Science held in China for their paper on the proof of Erdős-Faber-Lovász conjecture On July 14, 2024, Two of our former members, Dong Yeap Kang (강동엽) and Abhishek Methuku, and their collaborators (Tom Kelly, Daniela Kühn, and Deryk Osthus) received the 2024 Frontiers of Science Award (FSA) at the International Congress for Basic Science held in Beijing from July 14 to July 26, 2024, for their paper on the proof of Erdős-Faber-Lovász conjecture published in Annals of Mathematics. Dong Yeap Kang and Abhishek Methuku are former members of the IBS Discrete Mathematics Group. Congratulations! More than 100 students attended the 2024 Summer School on Combinatorics and Algorithms held at KAIST from July 22 to 26, 2024 More than 100 students attended the 2024 Summer School on Combinatorics and Algorithms from July 22 to 26, 2024. Chien-Chung Huang from ENS Paris and Sebastian Wiederrecht from the IBS Discrete Mathematics Group gave lectures. This event was organized by Jungho Ahn (KIAS), Eunjung Kim (KAIST), Eunjin Oh (POSTECH), and Sang-il Oum (IBS Discrete Mathematics Group / KAIST) and sponsored by the IBS Discrete Mathematics Group, KAIST, and POSTECH. IBS-DIMAG Workshop on Combinatorics and Geometric Measure Theory was held at IBS The IBS-DIMAG Workshop on Combinatorics and Geometric Measure Theory was held at IBS from July 14 to July 19, 2024. There were 23 talks, including plenary talks by János Pach (Rényi Institute of Mathematics) and Pertti Mattila (University of Helsinki) and invited talks by Izabella Łaba (University of British Columbia), Hong Wang (NYU), and Cosmin Pohoata (Emory University). • Doowon Koh (Chungbuk National University) • Ben Lund (IBS Discrete Mathematics Group) • Sang-il Oum (IBS Discrete Mathematics Group / KAIST Department of Mathematical Sciences)	{"url":"https://dimag.ibs.re.kr/news/page/3/","timestamp":"2024-11-12T14:07:58Z","content_type":"text/html","content_length":"203527","record_id":"<urn:uuid:2d60fb83-5cb8-486a-a61e-925f20fbf3b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00633.warc.gz"}
[Phys.org]18-qubit entanglement sets new record Photos of the experimental setup. Credit: Wang et al. ©2018 American Physical Society Physicists have experimentally demonstrated 18-qubit entanglement, which is the largest entangled state achieved so far with individual control of each qubit. As each qubit has two possible values, the 18 qubits can generate a total of 2^18 (or 262,144) combinations of output states. Since quantum information can be encoded in these states, the results have potential applications anywhere quantum information processing is used. The physicists, Xi-Lin Wang and coauthors at the University of Science and Technology of China, have published a paper on the new entanglement record in a recent issue of Physical Review Letters. "Our paper reports 18-qubit entanglement that expands an effective Hilbert space to 262,144 dimensions (the largest so far) with full control of three degrees of freedom of six individual photons, including their paths, polarization, and orbital angular momentum," coauthor Chao-Yang Lu at the University of Science and Technology of China told Phys.org. "This represents the largest entanglement so far. Entangling an increasingly large number of qubits not only is of fundamental interest (i.e., pushing the physical limit, if there is one, in order to explore the boundary between quantum and classical, for example). But also, probably more importantly, entangling large numbers of qubits is the central task in quantum computation." Generally, there are two ways to increase the number of effective qubits in an entangled state: use more particles, or exploit the particles' additional degrees of freedom (DoFs). When exploiting multiple DoFs, the entanglement is called "hyper-entanglement." So far, some of the largest entangled states have included 14 trapped ions with a single DoF, and five photons with two DoFs (which is equivalent to 10-qubit entanglement). Although going beyond two DoFs presents greater technological challenges, in the new study the physicists developed new methods to generate scalable hyper-entanglement, producing an 18-qubit entangled state made from six photons with three DoFs. "Controlling multiple DoFs is tricky, as it is necessary to touch one without disturbing any other," Lu explained. "To solve this, we develop methods for reversible quantum logic operations between the photon's different DoFs with precision and efficiencies both close to unity. We believe that our work creates a new and versatile platform for multi-photon quantum information processing with multiple DoFs." Using additional DoFs has several advantages. For one, exploiting three DoFs instead of two doubles the information-carrying capacity of each photon from four to eight possible output states. In addition, a hyper-entangled 18-qubit state that exploits three DoFs is approximately 13 orders of magnitude more efficient than an 18-qubit state composed of 18 photons with a single DoF. With these advantages, the physicists expect that the ability to achieve 18-qubit hyper-entanglement will lead to previously unprecedented areas of research, such as experimentally realizing certain codes for quantum computing, implementing quantum teleportation of high-dimensional quantum states, and enabling more extreme violations of local realism. "Our work has created a new platform for optical quantum information processing with multiple DoFs," Lu said. "The ability to coherently control 18 qubits enables experimental access to previously unexplored regimes, for example, the realization of the surface code and the Raussendorf-Harrington-Goyal code for quantum error correction, and the teleportation of three DoFs of a single photon."	{"url":"https://quantum.ustc.edu.cn/web/index.php/en/node/607","timestamp":"2024-11-03T04:31:59Z","content_type":"text/html","content_length":"23838","record_id":"<urn:uuid:c6c67af9-05f7-4eb5-a524-ab4797e7b70f>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00306.warc.gz"}
Two-Column Proofs Congruent Triangles Worksheet With Answers Two-Column Proofs Congruent Triangles Worksheet With Answers. Web 1.1 best images of congruent triangles worksheet with answer; Web q s t u %%6.%%% given:%%∠w%and%∠y%are%right%angles;%% ≅vxzx; %x%is%the%midpoint%of%% prove:%%δvwx≅δzyx%%wy z x y % Triangle Congruence Oh My Worksheet Proving Triangles Congruent from elestantedgnews.blogspot.com Congruence worksheet 2 answer key awesome geometry proofs from congruent triangles. You can read 22+ pages two column proof worksheet. Web worksheets are congruent triangles 2 column proofs, triangle proofs s sas asa aas, two column proofs, unit 4 triangles part 1 geometry smart packet, name geometry.	{"url":"http://printablecampushyde55.s3-website-us-east-1.amazonaws.com/two-column-proofs-congruent-triangles-worksheet-with-answers.html","timestamp":"2024-11-14T10:56:23Z","content_type":"text/html","content_length":"26496","record_id":"<urn:uuid:77e6f04d-a5bf-4846-8c04-a24cfadb2edf>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00615.warc.gz"}
2021 AIME II Problems/Problem 9 Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$. Solution 1 This solution refers to the Remarks section. By the Euclidean Algorithm, we have $\[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\]$ We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1 \right)$ gives \begin{align} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^ m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 1} \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ 2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 2} \\ \gcd (2m,n)&>\gcd(m,n), \end{align} which implies that $n$ must have more factors of $2$ than $m$ does. We construct the following table for the first $30$ positive integers: $\[\begin{array}{c\|c\|c} && \\ [-2.5ex] \boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex] 0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\ && \\ [-2.25ex] 1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex] 2 & 4,12,20,28 & 4 \\ && \\ [-2.25ex] 3 & 8,24 & 2 \\ && \\ [-2.25ex] 4 & 16 & 1 \\ \end{array}\]$ To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has: 1. If $m$ has $0$ factors of $2,$ then $m$ has $15$ options and $n$ has $8+4+2+1=15$ options. So, this case has $15\cdot15=225$ ordered pairs. 2. If $m$ has $1$ factor of $2,$ then $m$ has $8$ options and $n$ has $4+2+1=7$ options. So, this case has $8\cdot7=56$ ordered pairs. 3. If $m$ has $2$ factors of $2,$ then $m$ has $4$ options and $n$ has $2+1=3$ options. So, this case has $4\cdot3=12$ ordered pairs. 4. If $m$ has $3$ factors of $2,$ then $m$ has $2$ options and $n$ has $1$ option. So, this case has $2\cdot1=2$ ordered pairs. Together, the answer is $225+56+12+2=\boxed{295}.$ ~Lcz ~MRENTHUSIASM Solution 2 Consider any ordered pair $(m,n)$ such that $\gcd(2^m+1, 2^n-1) > 1$. There must exist some odd number $pe 1$ such that $2^m \equiv -1 \pmod{p}$ and $2^n \equiv 1 \pmod{p}$. Let $d$ be the order of $2$ modulo $p$. Note that $2^{2m} \equiv 1 \pmod{p}$. From this, we can say that $2m$ and $n$ are both multiples of $d$, but $m$ is not. Thus, we have $v_2(n) \ge v_2(d)$ and $v_2(m) + 1 = v_2(d)$. Substituting the latter equation into the inequality before gives $v_2(n) \ge v_2(m)+1$. Since $v_2(n)$ and $v_2(m)$ are integers, this implies $v_2(n)>v_2(m)$. The rest of the solution now proceeds as in Solution 1. Claim 1 (GCD Property) If $\boldsymbol{r,s,}$ and $\boldsymbol{t}$ are positive integers such that $\boldsymbol{\gcd(r,s)=1,}$ then $\boldsymbol{\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).}$ As $r$ and $s$ are relatively prime (have no prime divisors in common), this property is intuitive. To prove this rigorously, let $\gcd(r, t)=d_1$ and $\gcd(s, t)=d_2$. Then, $r=x d_1$, $s=y d_2$, and $t=k d_1 d_2$. Note that $\gcd(d_1, k)=1$ and $\gcd(d_2, k)=1$. Then, the left hand side of the equation is simply $d_1 d_2$. For the right hand side, $\gcd(rs, t)=\gcd(xy d_1 d_2, k d_1 d_2)$. But because $xy$ and $k$ are relatively prime, this simplifies down to $d_1 d_2$. Therefore, we have shown that $\gcd(r, t)\cdot\ gcd(s, t)=\gcd(rs, t)$. Claim 2 (Olympiad Number Theory Lemma) If $\boldsymbol{u,a,}$ and $\boldsymbol{b}$ are positive integers such that $\boldsymbol{u\geq2,}$ then $\boldsymbol{\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.}$ There are two proofs to this claim, as shown below. Claim 2 Proof 1 (Euclidean Algorithm) If $a=b,$ then $\gcd(a,b)=a=b,$ from which the claim is clearly true. Otherwise, let $a>b$ without the loss of generality. For all integers $p$ and $q$ such that $p>q>0,$ the Euclidean Algorithm states that $\[\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(p\operatorname{mod}q,q). \]$ We apply this result repeatedly to reduce the larger number: $\[\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^a-1-u^{a-b}\left(u^b-1\right),u^b-1\right)=\gcd\left(u^{a-b}-1,u^b-1\right).\]$ Continuing, we have \begin{align} \gcd\left(u^a-1,u^b-1\right)&=\gcd\left(u^{a-b}-1,u^b-1\right) \\ & \ \vdots \\ &=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\ &=u^{\gcd(a,b)}-1, \end {align} from which the proof is complete. Claim 2 Proof 2 (Bézout's Identity) Let $d=\gcd\left(u^a-1,u^b-1\right).$ It follows that $u^a\equiv1\pmod{d}$ and $u^b\equiv1\pmod{d}.$ By Bézout's Identity, there exist integers $x$ and $y$ such that $ax+by=\gcd(a,b),$ so $\[u^{\gcd(a,b)}=u^{ax+by}=(u^a)^x\cdot(u^b)^y\equiv1\pmod{d},\]$ from which $u^{\gcd(a,b)}-1\equiv0\pmod{d}.$ We know that $u^{\gcd(a,b)}-1\geq d.$ Next, we notice that \begin{align} u^a-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{a-\gcd{(a,b)}}+u^{a-2\gcd{(a,b)}}+u^{a-3\gcd{(a,b)}}+\cdots+1\right), \\ u^b-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{b- \gcd{(a,b)}}+u^{b-2\gcd{(a,b)}}+u^{b-3\gcd{(a,b)}}+\cdots+1\right). \end{align} Since $u^{\gcd(a,b)}-1$ is a common divisor of $u^a-1$ and $u^b-1,$ we conclude that $u^{\gcd(a,b)}-1=d,$ from which the proof is complete. Video Solution Video Solution by Interstigation See Also The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.	{"url":"https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_9&oldid=218219","timestamp":"2024-11-04T20:05:07Z","content_type":"text/html","content_length":"67341","record_id":"<urn:uuid:7ed087e0-32de-4949-980d-33df465989c8>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00832.warc.gz"}
Isometry classes of linear Isometry classes of linear codes Isometry classes of linear codes A linear (n,k)-code over the Galois field GF(q) is a k-dimensional subspace of the vector space Y^X:=GF(q)^n. As usual codewords will be written as rows x=(x[0],...,x[n-1]). A k × n-matrix Γ over GF (q) is called a generator matrix of the linear (n,k)-code C, if and only if the rows of Γ form a basis of C, so that C={x⋅ Γ \| x∈ GF(q)^k}. Two linear (n,k)-codes C[1],C[2] are called equivalent, if and only if there is an isometry (with respect to the Hamming metric) which maps C[1] onto C[2]. Using the notion of finite group actions one can easily express equivalence of codes in terms of the wreath product action introduced above: C[1] and C[2] are equivalent, if and only if there exist (ψ, π)∈ GF(q)^≀S[n] (where GF(q)^ denotes the multiplicative group of the Galois field) such that The complete monomial group GF(q)^≀S[n] of degree n over GF(q)^ acts on GF(q)^n as it was described above (see equation ()) in the more general case of H≀[X]G on Y^X: In order to apply the results of the theory of finite group actions, this equivalence relation for linear (n,k)-codes is translated into an equivalence relation for generator matrices of linear codes, and these generator matrices are considered to be functions Γ: n→ GF(q)^k \ {0} where Γ(i) is the i-th column of the generator matrix Γ. (We exclude 0-columns for obvious reasons.) Theorem The matrices corresponding to the two functions Γ[1] and Γ[2] from n to GF(q)^k \ {0} are generator matrices of two equivalent codes, if and only if Γ[1] and Γ[2] lie in the same orbit of the following action of GL[k](q) × GF(q)^≀S[n] as permutation group on (GF(q)^k \ {0})^n: or, more explicitly, (A,(ψ,π))(Γ)(i):=Aψ(i)Γ(π^-1(i) ). Following Slepian, we use the following notation: T[nkq] := the number of orbits of functions Γ: n→ GF(q)^k \ {0} under the group action of , i.e. T[nkq]=\|(GL[k](q) × GF(q)^≀S[n])\\(GF(q)^k \ {0})^n\|. T[nkq] := the number of orbits of functions Γ: n→ GF(q)^k \ {0} under the group action of , such that for all i,j∈ n, i ≠ j and all α∈ GF(q)^ the value of Γ(i) is different from αΓ(j). (In the case q=2, this is the number of injective functions G.) S[nkq] := the number of equivalence classes of linear (n,k)-codes over GF(q) with no columns of zeros. (A linear (n,k)-code has columns of zeros, if and only if there is some i∈ n such that x[i]=0 for all codewords x, and so we should exclude such columns.) S[nkq] := the number of classes of injective linear (n,k)-codes over GF(q) with no columns of zeros. (A linear (n,k)-code is called injective, if and only if for all i,j∈ n, i ≠ j and α∈ GF(q)^* there is some codeword x such that x[i] ≠ αx[j].) R[nkq] := the number of classes of indecomposable linear (n,k)-codes over GF(q) with no columns of zeros. (The definition of an indecomposable code will be given later.) R[nkq] := the number of classes of indecomposable, injective linear (n,k)-codes over GF(q) with no columns of zeros. W[nkq] := be the number of classes of linear (n,k)-codes over GF(q) with columns of zeros allowed. The following formulae hold: W[nkq]= ∑ S[ikq], S[nkq]=T[nkq]-T[n,k-1,q], S[nkq]=T[nkq]-T[n,k-1,q]. As initial values we have S[n1q]=1 for n∈ ℕ, S[11q]=1 and S[n1q]=0 for n>1. It is important to realize that • T[nkq] is the number of orbits of functions from n to GF(q)^k \ {0} without any restrictions on the rank of the induced matrix. • All matrices which are induced from functions Γ of the same orbit have the same rank. • The number of orbits of functions Γ which induce matrices of rank less or equal k-1 is T[n,k-1,q]. (This proposition holds for T[nkq] as well.) In the next section we will show that the R[nkq] or R[nkq] can be computed from the S[nkq] or S[nkq] respectively, so the main problem is the computation of the T[nkq] or T[nkq]. In the case q=2 the wreath product GF(q)^≀S[n] becomes the group S[n], and so there is the group GL[k](2) acting on GF(2)^k \ {0} and the symmetric group S[n] acting on n. Applying the formulae () and () we get ∑ T[nk2]x^n= Z(GL[k](2))\| [x[i]=∑[j=0]^∞ x^ij]= Z(GL[k](2))\| [x[i]=(1)/(1-x^i)] ∑ T[nk2]x^n= Z(GL[k](2))\| [x[i]=1+x^i]. In the case q ≠ 2 the wreath product GF(q)^≀S[n] acts both on range and domain of the functions Γ. Applying Lehmann's Lemma * there is the bijection Φ: GF(q)^≀S[n]\\(GF(q)^k \ {0})^n → S[n]\\(GF(q)^\\(GF(q)^k \ {0}))^n , Γ: n→ GF(q)^\\(GF(q)^k \ {0}), i↦ GF(q)^(Γ(i)) and S[n] acts on (GF(q)^\\(GF(q)^k \ {0}))^n by π(Γ)=Γ o π^-1. Using this bijection we have to investigate the following action of S[n] × GL[k](q): where GL[k](q) acts on GF(q)^\\(GF(q)^k \ {0}) by A(GF(q)^(v))=GF(q)^(Av). The set of the GF(q)^-orbits GF(q)^\\(GF(q)^k \ {0}) is the (k-1)-dimensional projective space: and the representation of GL[k](q) as a permutation group is the projective linear group PGL[k](q). This proves in fact the following to be true: Theorem The isometry classes of linear (n,k)-codes over GF(q) are the orbits of GL[k](q) × S[n] on the set of mappings PG[k-1](q)^n. This set of orbits is equal to the set of orbits of GL[k](q) on the set S[n]\\PG[k-1](q)^n, which can be represented by a complete set of mappings of different content, if the content of f∈ PG[k-1](q)^n is defined to be the sequence of orders of inverse images \|f Thus the set of isometry classes of linear (n,k)-codes over GF(q) is equal to the set of orbits of GL[k](q) on the set of mappings f∈ PG[k-1](q) of different content that form k × n-matrices of rank The particular classes of elements with orders of inverse images \|f^-1(x)\|≤ 1 are the classes consisting of Hamming codes. Knowing the cycle index of PGL[k](q) acting on PG[k-1](q) the equations () and () can be applied again. In [13] Slepian explained how the cycle index of GL[k](2) can be computed using results of Elspas [3]. The first author [4] generalized this concept for computing the cycle indices of GL[k](q) and PGL[k](q) acting on GF(q)^k or PG[k-1](q) respectively. The steps of the method used were the following ones: 1. Determination of the conjugacy classes of GL[k](q) by applying the theory of normal forms of matrices (or vector space endomorphisms). This theory can be found in many textbooks of algebra. 2. Determination of the order of the conjugacy classes, which can be found in Dickson, Green or Kung [2][5][8]. 3. Determination of the cycle type of a linear map or of a projectivity respectively. Since normal forms of regular matrices are strongly connected with companion and hypercompanion matrices (see [6]) of monic, irreducible polynomials over GF(q) it is important to know the exponent or subexponent of such polynomials (see [11][6]). The exponent of such a polynomial f(x)∈ GF(q)[x] is defined to be exp(f(x)) := min{n∈ ℕ \| f(x) \| x^n-1} and the subexponent is subexp(f(x)) := min{n∈ ℕ \| ∃ α∈ GF(q)^:f(x) \| x^n-α}. This element α∈ F[q]^ is uniquely defined, and it is called the integral element of f(x). The exponent of f(x) can be used to compute the cycle type of the companion or hypercompanion matrices of a monic, irreducible polynomial f(x), and by a direct product formula for cycle indices the cycle types of the normal forms in GL(k,F[q]) can be derived. Using the subexponent of f(x) and defining a formula similar to the direct product formula of cycle indices, which depends on the integral element of f(x) as well, the cycle type of a projectivity can be computed. 4. Determination of the cycle index by applying formula (*). These cycle indices are now available in the computer algebra package SYMMETRICA. Tables obtained this way are shown lower down. harald.fripertinger "at" uni-graz.at, May 10, 2016 Isometry classes of linear codes	{"url":"http://www.mathe2.uni-bayreuth.de/frib/html/art7/art7_2.html","timestamp":"2024-11-03T15:51:23Z","content_type":"text/html","content_length":"20164","record_id":"<urn:uuid:3dfc3beb-9fa3-436a-9316-2d18a7d6afa2>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00847.warc.gz"}
How many cm is O 5 inches? - Explained Which CM means 1 inch? We know that 1 inch is equal to 2.54 centimetres. Is 1 inch or 1 cm bigger? 1 centimeter is equal to 0.3937 inches, or 1 inch is equal to 2.54 centimeters. In other words, 1 centimeter is less than half as big as an inch, so you need about two-and-a-half centimeters to make one inch. Is 2.5 cm the same as 1 inch? 1 centimeter is the same length as 0.393701 inches. To find out what 2.5 cm would be in inches, you simply multiply 0.393701 by 2.5. After you’ve multiplied, your answer should be 0.984252. This tells you that 2.5 cm is the same as 0.984252 inches. How many cm is O 5 inches? – Related Questions Is 5 cm half an inch? To convert 5 cm into inches, multiply 5 cm by 0.393701 inches. Therefore 5 cm is equal to 1.9685 inches. Example 2: Convert 20 cm to inches. Is 2cm bigger than 1 inch? The term centimetre is abbreviated as “cm” where one centimetre is equal to the one-hundredth of a meter. In short, 1 centimetre = 0.01 meter = 10 millimeter = 0.3937 inches. The relationship between inch and cm is that one inch is exactly equal to 2.54 cm in the metric system. Is 2.54 cm exactly 1 inch? The definition of the inch was set to exactly 2.54cm starting in 1930 (the UK) but wasn’t adopted by all countries using inches until 1959. The nominal length of the meter has never changed (although the exact way the standard is defined has). How can I measure 2.5 cm without a ruler? Check your wallet for currency to use as a mini ruler. A US quarter is . 96 inches (2.4 cm) long, or roughly 1 inch (2.5 cm). Is there are 2.54 cm in 1 inch? There are 2.54 centimeters in one inch. What is exactly 1 inch? Standards for the exact length of an inch have varied in the past, but since the adoption of the international yard during the 1950s and 1960s the inch has been based on the metric system and defined as exactly 25.4 mm. Is your finger 1 inch? * Use your own body for fast, approximate measuring. The first joint of an index finger is about 1 inch long. How many finger is an inch? 1 finger is exactly 7/8 inches. Using SI units 1 finger is 0.022225 meters. Please share if you found this tool useful: Conversions Table 1 Inches to Fingers = 1.1429 70 Inches to Fingers = 80 Is each finger an inch? The digit, also known as digitus or digitus transversus (Latin), dactyl (Greek) or dactylus, or finger’s breadth — 3⁄4 of an inch or 1⁄16 of a foot. In medicine and related disciplines (anatomy, radiology, etc.) the fingerbreadth (literally the width of a finger) is an informal but widely used unit of measure. Is a knuckle 1 inch? The length between your thumb tip and the top knuckle of your thumb is roughly one inch. The next time you have a ruler handy, give it a quick measure to double-check. Is a fingertip 1 cm? If one fingertip fits, the cervix is considered to be 1 cm dilated. If the tips of two fingers fit, this means the cervix is 2 cm dilated. Depending on the distance the two fingers can stretch apart, it’s possible to indicate further dilation. It is usual to refer to full dilation as 10 centimeters. What is the normal size of a finger? If they’re closer to an average height, chances are they have an average finger size to match. An average finger size is 6 for women and 8 or 8½ for men. Do hands get bigger with age? The hands and faces of some grownups do get a little bit bigger as they get older. This happens because the brain produces something called growth hormone, which helps make the bones of kids grow a lot longer and wider. Are 7 inch hands Small? The average length of an adult female’s hand is 6.8 inches. However, there’s more to hand size than length. How to choose gloves based on your hand size. Hand size (the largest measurement of either length or circumference) Glove size 7 inches XSmall 7.5–8 inches Small 8.5–9 inches Medium 9.5–10 inches Large What is the average female ring size? The average ring size for most women is between a size 5 and size 7. We also know that the average sized woman in the U.S. is about 5-feet 4-inches tall. You can go off these averages to determine her ring size without her knowing. If she’s shorter or weighs a little more, you can go up a size or two. How tight should a ring be? Rule of Thumb: A proper fitting ring should slide over your knuckle with a little friction and fit snugly on your finger, but not too tight. You should feel resistance and need to apply a little extra force to remove the ring backwards over your knuckle. Leave a Comment	{"url":"https://theomegafoundation.org/how-many-cm-is-o-5-inches/","timestamp":"2024-11-07T16:07:58Z","content_type":"text/html","content_length":"72018","record_id":"<urn:uuid:8d8390a9-2d0b-46d3-9f76-53b2939bb546>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00642.warc.gz"}
What is Pearl's Causal Calculus Pearl's Causal Calculus: A powerful tool for understanding cause and effect in machine learning models. Pearl's Causal Calculus is a mathematical framework that enables researchers to analyze cause-and-effect relationships in complex systems. It is particularly useful in machine learning, where understanding the underlying causal structure of data can lead to more accurate and interpretable models. The core of Pearl's Causal Calculus is the do-calculus, a set of rules that allow researchers to manipulate causal relationships and estimate the effects of interventions. This is particularly important when working with observational data, where it is not possible to directly manipulate variables to observe their effects. By using the do-calculus, researchers can infer causal relationships from observational data and make predictions about the outcomes of interventions. Recent research has expanded the applications of Pearl's Causal Calculus, including mediation analysis, transportability, and meta-synthesis. Mediation analysis helps to understand the mechanisms through which a cause influences an outcome, while transportability allows for the generalization of causal effects across different populations. Meta-synthesis is the process of combining results from multiple studies to estimate causal relationships in a target environment. Several arxiv papers have explored various aspects of Pearl's Causal Calculus, such as its completeness, connections to information theory, and applications in Bayesian statistics. Researchers have also developed formal languages for describing statistical causality and proposed algorithms for identifying causal effects in causal models with hidden variables. Practical applications of Pearl's Causal Calculus include: 1. Improving the interpretability of machine learning models by uncovering the causal structure of the data. 2. Estimating the effects of interventions in complex systems, such as healthcare, economics, and social sciences. 3. Combining results from multiple studies to make more accurate predictions about causal relationships in new environments. A company case study that demonstrates the power of Pearl's Causal Calculus is Microsoft Research, which has used the framework to develop more accurate and interpretable machine learning models for various applications, such as personalized medicine and targeted marketing. In conclusion, Pearl's Causal Calculus is a valuable tool for understanding cause-and-effect relationships in complex systems, with wide-ranging applications in machine learning and beyond. By leveraging this framework, researchers can develop more accurate and interpretable models, ultimately leading to better decision-making and improved outcomes.	{"url":"https://www.activeloop.ai/resources/glossary/pearls-causal-calculus/","timestamp":"2024-11-04T14:15:51Z","content_type":"text/html","content_length":"592631","record_id":"<urn:uuid:c2473913-a3d6-4e01-afad-90809d587597>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00454.warc.gz"}
The abacus is an ancient calculating machine. This simple apparatus is about 5,000 years old and is thought to have originated in Babylon. As the concepts of zero and Arabic number notation became widespread, basic math functions became simpler, and the use of the abacus diminished. Most of the world employs adding machines, calculators, and computers for mathematical calculations, but today Japan, China, the Middle East, and Russia still use the abacus, and school children in these countries are often taught to use the abacus. In China, the abacus is called a suan pan, meaning counting tray. In Japan the abacus is called a soroban. The Japanese have yearly examinations and competitions in computations on the soroban. Before the invention of counting machines, people used their fingers and toes, made marks in mud or sand, put notches in bones and wood, or used stones to count, calculate, and keep track of quantities. The first abaci were shallow trays filled with a layer of fine sand or dust. Number symbols were marked and erased easily with a finger. Some scientists think that the term abacus comes from the Semitic word for dust, abq. A modern abacus is made of wood or plastic. It is rectangular, often about the size of a shoe-box lid. Within the rectangle, there are at least nine vertical rods strung with movable beads. The abacus is based on the decimal system. Each rod represents columns of written numbers. For example, starting from the right and moving left, the first rod represents ones, the second rod represents tens, the third rod represents hundreds, and so forth. A horizontal crossbar is perpendicular to the rods, separating the abacus into two unequal parts. The moveable beads are located either above or below the crossbar. Beads above the crossbar are called heaven beads, and beads below are called earth beads. Each heaven bead has a value of five units and each earth bead has a value of one unit. A Chinese suan pan has two heaven and five earth beads, and the Japanese soroban has one heaven and four earth beads. These two abaci are slightly different from one another, but they are manipulated and used in the same manner. The Russian version of the abacus has many horizontal rods with moveable, undivided beads, nine to a column. To operate, the soroban or suan pan is placed flat, and all the beads are pushed to the outer edges, away from the crossbar. Usually the heaven beads are moved with the forefinger and the earth beads are moved with the thumb. For the number one, one earth bead would be pushed up to the crossbar. Number two would require two earth beads. For number five, only one heaven bead would to be pushed to the crossbar. The number six would require one heaven (five units) plus one earth (one unit) bead. The number 24 would use four earth beads on the first rod and two earth beads on the second rod. The number 26 then, would use one heaven and one earth bead on the first rod, and two earth beads on the second rod. Addition, subtraction, multiplication, and division can be performed on an abacus. Advanced abacus users can do lengthy multiplication and division problems, and even find the square root or cube root of any number. Additional topics	{"url":"https://science.jrank.org/pages/2/Abacus.html","timestamp":"2024-11-13T12:17:54Z","content_type":"text/html","content_length":"10746","record_id":"<urn:uuid:e2f5c0ba-d82a-4e76-bce2-b3b9eb51809d>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00639.warc.gz"}
Geometric Distribution - Definition, Formula, Mean, Examples Probability theory is an essential department of mathematics which handles the study of random events. One of the crucial concepts in probability theory is the geometric distribution. The geometric distribution is a discrete probability distribution which models the amount of tests required to obtain the initial success in a series of Bernoulli trials. In this blog, we will explain the geometric distribution, extract its formula, discuss its mean, and give examples. Meaning of Geometric Distribution The geometric distribution is a discrete probability distribution that narrates the number of tests needed to accomplish the initial success in a succession of Bernoulli trials. A Bernoulli trial is a test that has two likely outcomes, usually indicated to as success and failure. Such as flipping a coin is a Bernoulli trial since it can likewise come up heads (success) or tails (failure). The geometric distribution is utilized when the experiments are independent, which means that the consequence of one trial doesn’t impact the outcome of the upcoming test. Additionally, the chances of success remains constant throughout all the trials. We could indicate the probability of success as p, where 0 < p < 1. The probability of failure is then 1-p. Formula for Geometric Distribution The probability mass function (PMF) of the geometric distribution is specified by the formula: P(X = k) = (1 - p)^(k-1) * p Where X is the random variable which depicts the amount of test required to attain the first success, k is the number of tests needed to attain the first success, p is the probability of success in an individual Bernoulli trial, and 1-p is the probability of failure. Mean of Geometric Distribution: The mean of the geometric distribution is defined as the likely value of the amount of test required to achieve the first success. The mean is given by the formula: μ = 1/p Where μ is the mean and p is the probability of success in an individual Bernoulli trial. The mean is the anticipated count of experiments needed to obtain the first success. Such as if the probability of success is 0.5, therefore we anticipate to attain the first success following two trials on average. Examples of Geometric Distribution Here are some primary examples of geometric distribution Example 1: Tossing a fair coin up until the first head appears. Suppose we flip an honest coin until the initial head turns up. The probability of success (obtaining a head) is 0.5, and the probability of failure (obtaining a tail) is as well as 0.5. Let X be the random variable that represents the count of coin flips needed to achieve the initial head. The PMF of X is stated as: P(X = k) = (1 - 0.5)^(k-1) * 0.5 = 0.5^(k-1) * 0.5 For k = 1, the probability of achieving the first head on the first flip is: P(X = 1) = 0.5^(1-1) * 0.5 = 0.5 For k = 2, the probability of obtaining the initial head on the second flip is: P(X = 2) = 0.5^(2-1) * 0.5 = 0.25 For k = 3, the probability of achieving the first head on the third flip is: P(X = 3) = 0.5^(3-1) * 0.5 = 0.125 And so on. Example 2: Rolling a fair die up until the first six appears. Let’s assume we roll an honest die up until the initial six shows up. The probability of success (getting a six) is 1/6, and the probability of failure (getting any other number) is 5/6. Let X be the random variable which represents the number of die rolls needed to achieve the initial six. The PMF of X is stated as: P(X = k) = (1 - 1/6)^(k-1) * 1/6 = (5/6)^(k-1) * 1/6 For k = 1, the probability of achieving the first six on the first roll is: P(X = 1) = (5/6)^(1-1) * 1/6 = 1/6 For k = 2, the probability of getting the first six on the second roll is: P(X = 2) = (5/6)^(2-1) * 1/6 = (5/6) * 1/6 For k = 3, the probability of achieving the first six on the third roll is: P(X = 3) = (5/6)^(3-1) * 1/6 = (5/6)^2 * 1/6 And so forth. Get the Tutoring You Need from Grade Potential The geometric distribution is an essential theory in probability theory. It is utilized to model a wide range of real-life phenomena, for instance the count of tests required to obtain the first success in different situations. If you are having difficulty with probability concepts or any other arithmetic-related subject, Grade Potential Tutoring can support you. Our experienced instructors are available online or face-to-face to offer personalized and effective tutoring services to help you succeed. Contact us today to schedule a tutoring session and take your math skills to the next stage.	{"url":"https://www.pittsburghinhometutors.com/blog/geometric-distribution-definition-formula-mean-examples","timestamp":"2024-11-14T05:05:35Z","content_type":"text/html","content_length":"75340","record_id":"<urn:uuid:7e6d0d17-bd99-4d87-ae98-72e87435ddd9>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00578.warc.gz"}
translation diagram math 5th grade Related topics: math for ninth graders equation solve systems of equations java sum of integers free adding worksheets how to find percentages by dividing quizzes algebra algebra solver websites that helps you solve one of your own get help homework with strategies for problem solving workbook Author Message An_Amiricon_Helo Posted: Wednesday 01st of Aug 19:43 To each individual masterful in translation diagram math 5th grade: I drastically need your really blue-chip help . I have many class assignments for my online Remedial Algebra. I feel translation diagram math 5th grade could be beyond my capacity . I'm at a absolute loss as far as where I should get started . I have looked at employing a math teacher or signing up with a learning center, however, they are emphatically not cut-rate . Each and every alternative suggestion will be hugely appreciated ! From: On A Forum IlbendF Posted: Friday 03rd of Aug 11:28 I have a good recommendation that could help you with algebra. You simply need a good program to explain the problems that are hard . You don't need a tutor , because firstly it's very costly , and on the other hand you won't have it near you whenever you need help. A program is better because you only have to purchase it once, and it's yours forever . I recommend you to take a look at Algebrator, because it's the best. Since it can resolve almost any algebra exercises, you will certainly use it for a very long time, just like I did. I purchased it years ago when I was in Algebra 2, but I still use it sometimes . From: Netherlands ZaleviL Posted: Saturday 04th of Aug 08:39 Algebrator truly is a masterpiece for us math students. As my dear friend said in the preceding post, it solves questions and it also explains all the intermediary steps involved in reaching that final solution . That way apart from knowing the final answer, we also learn how to go about solving questions from the scratch , and it helps a lot in preparing for exams . From: floating in the light, never Natham_Ondilson Posted: Monday 06th of Aug 08:15 Thank you very much for your response! Could you please tell me how to get hold of this program? I don’t have much time on hand since I have to solve this in a few days. erx Posted: Tuesday 07th of Aug 11:01 I’ve put the details here : https://mathworkorange.com/polynomials.html. Just try it because Algebrator has a unrestricted money back offer , See if it works for you. From: PL/DE/ES/GB/ Hiinidam Posted: Wednesday 08th of Aug 09:06 I remember having often faced difficulties with angle suplements, quadratic equations and hyperbolas. A truly great piece of algebra program is Algebrator software. By simply typing in a problem homework a step by step solution would appear by a click on Solve. I have used it through many math classes – College Algebra, College Algebra and Pre Algebra. I greatly recommend the program. From: Greeley, CO,	{"url":"https://mathworkorange.com/lagrange-polynomials/hypotenuse-leg-similarity/translation-diagram-math-5th.html","timestamp":"2024-11-03T03:55:23Z","content_type":"text/html","content_length":"97164","record_id":"<urn:uuid:e9d75554-c770-4f54-9d1c-c2e08f9a4d50>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00748.warc.gz"}
Automobil - math word problem (83852) The car traveled three quarters of the total journey at a speed of 90 km/h and the remaining part of the journey at a speed of 50 km/h. Find its average speed. 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: Units of physical quantities: Themes, topics: Grade of the word problem: Related math problems and questions:	{"url":"https://www.hackmath.net/en/math-problem/83852","timestamp":"2024-11-05T17:14:08Z","content_type":"text/html","content_length":"73081","record_id":"<urn:uuid:e15c364e-0f20-47b9-970b-329f5bcd5fa2>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00379.warc.gz"}
ppmt function Hi everyone, Im having trouble coming up with a formula for principle portion of last loan payment. I Already came up with the formulas for monthly, first payment, the total cost of the loan. interest 7.75% d4 amortization period in years 18 d5 Principle amount to be borrowed $1,750,000 d6 monthly loan payment $15,048.32 Principle portion of the first loan payment $3,746.24 Principle portion of the last loan payment $?? here's what i came up with so far =PPMT(d4/12,1,d512,d6,0) Total cost of loan $3.250,437.08 Excel Facts Best way to learn Power Query? Read M is for (Data) Monkey book by Ken Puls and Miguel Escobar. It is the complete guide to Power Query. Oct 10, 2011 Office Version 1. 365 1. Windows Take a look at the CUMPRINC function: Excel Workbook D E F G 3 Principle 4 7.75% 14,951.76 Last Period 5 18 3,746.24 First Period Take a look at the CUMPRINC function: D E F G * * * * Last Period * First Period * * * <colgroup><col style="font-weight:bold; width:30px; "><col style="width:83px;"><col style="width:20px;"><col style="width:94px;"><col style="width:144px;"></colgroup><tbody> [TD="bgcolor: #cacaca, align: center"]3[/TD] [TD="align: center"]Principle[/TD] [TD="bgcolor: #cacaca, align: center"]4[/TD] [TD="align: right"]7.75%[/TD] [TD="bgcolor: #ffff00, align: right"]14,951.76[/TD] [TD="bgcolor: #cacaca, align: center"]5[/TD] [TD="align: right"]18[/TD] [TD="align: right"]3,746.24[/TD] [TD="bgcolor: #cacaca, align: center"]6[/TD] [TD="align: right"]1750000[/TD] Spreadsheet Formulas Cell Formula F4 =-CUMPRINC(D4/12,D512,1750000,216,216,0) F5 =-CUMPRINC(D4/12,D512,1750000,1,1,0) Excel tables to the web >> Excel Jeanie HTML 4 where did the function cumprinc come from how do you come with function in ppmt. i have an online hw and it wants specific formula in ppmt. i want to thank you for expertise and help. Mar 2, 2014 Office Version 1. 2010 1. Windows [...deleted...submitted by mistake, incomplete...] Mar 2, 2014 Office Version 1. 2010 1. Windows Im having trouble coming up with a formula for principle portion of last loan payment. I Already came up with the formulas for monthly, first payment, the total cost of the loan. interest 7.75% d4 amortization period in years 18 d5 Principle amount to be borrowed $1,750,000 d6 monthly loan payment $15,048.32 Principle portion of the first loan payment $3,746.24 Principle portion of the last loan payment $?? here's what i came up with so far =PPMT(d4/12,1,d512,d6,0) Total cost of loan $3.250,437.08 You correctly wrote -PPMT(D4/12, ,D512,D6) for the first principal payment. The last principal payment is -PPMT(D4/12, I understand that your homework assignment requires that you use PPMT. But that is not a good choice. In the real world, payments must be rounded to the cent, at least. So the actual payment is =- That looks the same when formatted with 2 decimal places, namely 15,048.32. But without rounding, -PMT(D4/12,D512,D6) actually returns about 15048.3198120013. Consequently, -PPMT(D4/12,D512,D5*12,D6) returns about 14951.7563853459, which is displayed as -14951. when formatted with 2 decimal places. But the actual last principal amount is the balance after the last-minus-one payment, namely: which returns -14951. . (Dyslexia alert! :->) Similarly, the total cost of the loan is: which is 3,250,437. , not 3,250,437. when formatted with 2 decimal places. That said, you should give your teacher the answer that he/she expects. Or be a mensh and provide both sets of answers with some explanation. It might earn you some extra points. Last edited: Thanks for giving me more of explanation than just helping me out with the formula.	{"url":"https://www.mrexcel.com/board/threads/ppmt-function.804936/","timestamp":"2024-11-08T18:52:51Z","content_type":"text/html","content_length":"130748","record_id":"<urn:uuid:e28930f5-4b74-4661-8a0a-d61f5c923176>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00199.warc.gz"}
Science:Math Exam Resources/Courses/MATH220/December 2011/Question 01 (e) MATH220 December 2011 • Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q2 (a) • Q2 (b) • Q3 • Q4 • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 • Question 01 (e) Let A, B be non-empty sets, and let ƒ: A → B be a function. When does ƒ have an inverse function ƒ^-1? Define ƒ^-1. Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! Why doesn't the function {\displaystyle {\begin{aligned}f:\{0,1\}&\to \{3\}\\0&\mapsto 3\\1&\mapsto 3\end{aligned}}} have an inverse? What about the function {\displaystyle {\begin{aligned}g:\{0\}&\to \{1,2\}\\0&\mapsto 2\end{aligned}}} Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. • If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. • If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result. Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. A function ƒ has an inverse if and only if it is bijective. That is, it has an inverse if and only if it is 1. Injective (or one-to-one) 2. Surjective (or onto). In such a case, for every y ∈ B, there is a unique x ∈ A such that ƒ(x) = y. Using this fact, we define ƒ^-1 by the rule ${\displaystyle \displaystyle f^{-1}(y)=x}$ where x is the unique element of the set A such that ƒ(x) = y. From the injectivity and surjectivity, this is well defined. Moreover, ${\displaystyle f{\big (}f^{-1}(y){\big )}=f(x)=y}$ ${\displaystyle f^{-1}{\big (}f(x){\big )}=f^{-1}(y)=x}$ and so this really is the inverse of ƒ. Click here for similar questions MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Function properties, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag	{"url":"https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH220/December_2011/Question_01_(e)","timestamp":"2024-11-11T01:49:48Z","content_type":"text/html","content_length":"44052","record_id":"<urn:uuid:90a679ad-03ff-49d2-849a-cc71fb07d3d5>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00174.warc.gz"}
Energy & Momentum - Physics Study Guides: Flashcards \| Knowt Energy & Momentum Study Guides Min number of terms: Created by: It’s never been easier to find and study Energy & Momentum subject made by students and teachers using Knowt. Whether you’re reviewing material before a quiz or preparing for a major exam, we’ll help you find the subject subject that you need to power up your next study session. If you’re looking for more specific Energy & Momentum subject, then check out our collection of sets for Kinematics & Dynamics, Newton's Laws, Circular Motion & Gravitation, Energy & Momentum, Simple Harmonic & Rotational Motion, Fluids, Electric Charge, Field, & Potential, Circuits, Magnetic Forces/Fields, Electromagnetic Waves, Geometric Optics, Quantum Physics, Thermal Physics, Kinesiology.	{"url":"https://knowt.com/subject/Science/Physics/Energy-%26-Momentum-flashcards","timestamp":"2024-11-12T03:09:25Z","content_type":"text/html","content_length":"469445","record_id":"<urn:uuid:7f19e8d6-e862-4ecc-a94a-1fa2c9563f32>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00864.warc.gz"}
10.43 Inches to Centimeters 10.43 in to cm conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result). Every display form has its own advantages and in different situations particular form is more convenient than another. For example usage of scientific notation when working with big numbers is recommended due to easier reading and comprehension. Usage of fractions is recommended when more precision is needed. If we want to calculate how many Centimeters are 10.43 Inches we have to multiply 10.43 by 127 and divide the product by 50. So for 10.43 we have: (10.43 × 127) ÷ 50 = 1324.61 ÷ 50 = 26.4922 So finally 10.43 in = 26.4922 cm	{"url":"https://unitchefs.com/inches/centimeters/10.43/","timestamp":"2024-11-03T23:41:39Z","content_type":"text/html","content_length":"23316","record_id":"<urn:uuid:92590360-3826-42c2-88d7-5af52535f351>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00101.warc.gz"}
A Adrian Albert - Biography Quick Info 9 November 1905 Chicago, Illinois, USA 6 June 1972 Chicago, Illinois, USA A Adrian Albert was an American mathematician who worked on associative and non-associative algebras. A Adrian Albert's parents were Russian. His father, Elias Albert, came to the United States from England and had set up a retail business. His mother, Fannie Fradkin, had come to the United States from Russia. Adrian was the second of Elias and Fannie's three children, but he also had both a half-brother and half-sister from his mother's side. It was in Chicago that Adrian undertook most of his education; in fact this was his home town for most of his life. He began his education there entering primary school in 1911. In 1914, however, the family moved to Iron Mountain in Michigan where he continued his schooling until the family returned to Chicago in 1916. Back in the town of his birth, Adrian entered the Theodore Herzl Elementary School where he studied until 1919 then, in that year, he entered the John Marshall High School. In 1922 he graduated from the High School and began his studies at the University of Albert completed his B.S. degree in 1926 and was awarded his Master's degree in the following year. He remained at the University of Chicago undertaking research under L E Dickson's supervision. That Dickson, the leading American mathematician in the fields of number theory and algebra, was on the Chicago faculty was a piece of good fortune for Albert. Not only did Dickson strongly influence the course of all Albert's later research, but also his style as a teacher and academic rubbed off on Albert. He was awarded his Ph.D. in 1928 for a doctoral dissertation entitled Algebras and their Radicals and Division Algebras. By the time that he received his doctorate Albert was a married man, having married Freda Davis on 18 December 1927. The economic situation in the United States was deteriorating at this time with the advent of the depression. Herstein writes in [7] (see also [6]):- Shortly after he got his Ph.D., the great economic depression started. Sensitive as he was to the suffering of others, he deeply felt the economic hardship that so many of his friends were undergoing. He, too, did not have an easy time of it economically. In addition he was beset by a series of illnesses ... In his doctoral thesis Albert had made considerable progress in classifying division algebras. It was an impressive piece of work and it led to him being awarded a National Research Council Fellowship to enable him to undertake postdoctoral study at Princeton. He spent nine months at Princeton in 1928-29 and this was an important period for Albert since during his time there Lefschetz suggested that he look at open problems in the theory of Riemann matrices. These matrices arise in the theory of complex manifolds and Albert went on to write an important series of papers on these questions over the following years. Albert was then offered a post as an instructor at Columbia University and he worked there for two years from 1929 to 1931. His first paper A determination of all normal division algebras in sixteen units was published in 1929. It was based on the second half of his doctoral thesis but Albert had, by this time, pushed the ideas further classifying division algebras of dimension 16 over their centres. The case of dimension 9, the next smaller case, had been solved by Wedderburn. Albert returned to the University of Chicago in 1931 where he was appointed as assistant professor. He remained on the staff there for the rest of his life being promoted to associate professor in 1937 and full professor in 1941. During the years 1958 to 1962 he was chairman of the Chicago Department. Kaplansky writes [7]:- The main stamp he left on the Department was a project dear to his heart; maintaining a lively flow of visitors and research instructors, for whom he skilfully got support in the form of research grants. Shortly after beginning his second three year term as Chairman of the Department Albert was asked to take on the post of Dean of Physical Sciences. He served Chicago for 9 year in the role until 1971. Herstein writes [6]:- He dearly loved Chicago as a city, more especially the Hyde Park area surrounding the university, and most especially, the university itself. He was an integral part of the university, and the university was an integral part of his life. He knew and was known by almost everybody at the university, and his influence went well beyond the realm of the physical sciences. Of all the honours and responsibilities that came to him, the one that he probably enjoyed most, and which meant the most to him, was that of being Dean of Physical Sciences in his beloved university. We should say a little about Albert's family life which was filled with great happiness until the tragic death of his son Roy. He was one of their three children (the other two being Alan and Nancy) and Roy's death at the age of 23 in 1958 brought a deep sadness which Albert and his wife never got over. One should say, though, that it was in his nature not to allow his grief to be too visible publicly. The blow was softened, if ever the loss of a child could ever really be softened, by the happiness that Albert and his wife enjoyed from their other two children and from their five grand-children. We have already said a little above about Albert's mathematical contributions but we should now give further details. His main work was on associative algebras, non-associative algebras, and Riemann matrices. He worked on classifying division algebras building on the work of Wedderburn but Brauer, Hasse and Emmy Noether got the main result first. Albert's major contribution is, however, detailed in a joint paper with Hasse. Albert's book Structure of Algebras, published in 1939, remains a classic. The content of this treatise was the basis of the Colloquium Lectures which he gave to the American Mathematical Society in 1939. We should also mention Albert's other fine text on algebra, Modern Higher Algebra, which was published two years before Structure of Albert's work on Riemann matrices was, as we mentioned above, a consequence of suggestions made by Lefschetz. For his papers on the construction of Riemann matrices published in the Annals of Mathematics in 1934 and 1935 Albert received the Cole prize in algebra from the American Mathematical Society in 1939. These important papers had been a direct consequence of Albert's 1928-29 visit to Princeton and when he spent the academic year 1933-34 at the Institute for Advanced Study at Princeton he again received a stimulus which would lead him to further important results. Lectures by Weyl on Lie algebras were particularly stimulating but perhaps even more important was his introduction to Jordan algebras. These algebras had been introduced by Pascual Jordan as being related to quantum theory. Jordan had worked with von Neumann and Wigner on the structure of these algebras but they had left open certain fundamental questions. Albert was able to use his expertise in structural questions regarding algebras to solve some of the problems in his 1934 paper On certain algebras of quantum mechanics. His work on Jordan algebras did not end there for he published three further fundamental papers on their structure in 1946, 1947 and 1950. During the Second World War Albert contributed to the war effort as associate director of the Applied Mathematics Group at Northwestern University which tackled military problems. Another interest of Albert's, which appears to have been prompted by the War, was that of cryptography. He lectured to the American Mathematical Society on Some mathematical aspects of cryptography at the Society's meeting in November 1941. Lectures by Weyl on Lie algebras in 1934-35 introduced Albert to the theory of non-associative algebras. It was not until 1942, however, that he published his first major work on non-associative algebras. Kaplansky writes in [9]:- Albert investigated just about every aspect of non-associative algebras. At times a particular line of attack failed to fulfil the promise it had shown; he would then exercise his sound instinct and good judgement by shifting the assault to a different area. In fact, he repeatedly displayed an uncanny knack for selecting projects which later turned out to be well conceived Albert received many honours for his outstanding achievements. He was elected to the National Academy of Sciences in 1943, the Brazilian Academy of Sciences in 1952, and the Argentine Academy of Sciences in 1963. He served as chairman of the Mathematics Section of the National Research Council from 1958 to 1961, and President of the American Mathematical Society in 1965-66. Herstein says this about Albert as a person ([6] or [7]):- What characterised him best as a person was his intense loyalty to his friends and to his profession. He viewed the profession of mathematician with a great deal of pride and he did everything he could to have it recognised as he felt it deserved. He constantly fought for the improvement of working conditions, salaries, and student support in his chosen field. Although he had a strong set of principles in life and a definite attitude to moral and professional behaviour, he was endowed with an enormous tolerance for the changes that were taking place around him ... 1. K V H Parshall, Biography in Dictionary of Scientific Biography (New York 1970-1990). See THIS LINK. 2. N E Albert, A Cubed and his Algebra, (New York, 2005). 3. R E Block, N Jacobson, J M Osborn, D J Saltman and D Zelinsky (eds.), A Adrian Albert : Collected Mathematical Papers (2 Vols) (Providence, R I, 1993). 4, 4. N Jacobson, Abraham Albert, Bull. Amer. Math. Soc. 80 (1974), 1075-1100. 5. I N Herstein, A Adrian Albert, in R E Block, N Jacobson, J M Osborn, D J Saltman and D Zelinsky (eds.), A Adrian Albert : Collected Mathematical Papers (2 Vols) (Providence, R I, 1993), 6. I N Herstein, A Adrian Albert, Scripta Mathematica 29 (1973), 1-3. 7. I Kaplansky, Abraham Adrian Albert : November 9, 1905-June 6, 1972, A century of mathematics in America I (Providence, R.I., 1988), 244-264. 8. I Kaplansky, Abraham Adrian Albert: November 9, 1905-June 6, 1972, in R E Block, N Jacobson, J M Osborn, D J Saltman and D Zelinsky (eds.), A Adrian Albert : Collected Mathematical Papers (2 Vols) (Providence, R I, 1993), lxv-lxxxi. 9. D Zelinsky, A A Albert, Amer. Math. Monthly 80 (1973), 661-665. Additional Resources (show) Other pages about A Adrian Albert: Other websites about A Adrian Albert: Written by J J O'Connor and E F Robertson Last Update September 2001	{"url":"https://mathshistory.st-andrews.ac.uk/Biographies/Albert_Abraham/","timestamp":"2024-11-11T04:00:54Z","content_type":"text/html","content_length":"35700","record_id":"<urn:uuid:3e2c69dc-1e75-4791-9572-ad5e4ed37365>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00600.warc.gz"}
draughty houses, tiled floors, no insulation Don't you just love this time of the year and the next few months when the cold weather arrives.. From now until October I shall spend most of my time convinced that we need under floor heating, double glazing and competent tradies to build properly and not leave gaps everywhere. Just got the winter doona out. Hopefully it'll be a mild one. Guest marcandjo sorry pommies but we have got the heating on and the hot water bottle out So have we. And we are in the UK Did the weather not get the memo is spring into summer now in the NH Guest Helchops Snifter....you know the score. Two nice weeks in march followed by rain and cloud until May when it will usually be sunny for a month through to June. As soon as the Kids break up though and the bank holidays kick in (and two week holidays for most), gods disdain rains down in the form of a wet July and August! Gotta love the seasons! Cmon Poms.....Toughen up!! its cold for all of 2-3 months then back to the lovely days we all moved here for!! I have to say it's not my favourite time of the year and I know it's not really as cold as the UK, but the title of the thread says it all. We had the wood burner going last night, kept the house really warm, but this morning it felt as though a window was open as it felt so fresh in the bedroom, and that this is only down to the poor building techniques and materials. In 5 months time though it'll be warm enough to forget the heater for another 7 months... woo hoo.. bring it on.... Guest smurph i do like this time of year... ask me again in July and it will be different though..i wear more clothes inside the house than out at the moment.. Cmon Poms.....Toughen up!! its cold for all of 2-3 months then back to the lovely days we all moved here for!! I agree with the sentiment (and I'm one of those who needs to toughen up - absolutely hate cold weather!) but if only it was for 2-3 months! The next month that has similar average temps to April (which is already far too cold for me) is October - worse weather for all the months in between - for those of us a bit wimpy about the cold that seems like a long way away! Guest ladyarkles sorry pommies but we have got the heating on and the hot water bottle out LOL! I'm having a day off and sitting in front of the TV with a hot water bottle. My 15 year old asked if we had any GLOVES she could wear to school (bearing in mind she's still in her summer uniform...) However, my kids say how much they LOVE this rainy weather. NOT me!!! (Of course, they British to the core and after 13 years in Cornwall this is weather is home whereas I'm from a slightly less rainy part of the states.) I'm seriously thinking of double glazing... One thing we bought last year thats a god send was the electric blanket. Theres nothing worse than getting into a freezing cold bed:shocked: I have been round all the doors with draft strip but think we need a new door. Rob and Mel wimps all of you .............. we're still sleeping with the window wide open, curtains open to let the cool air in, summer light weight quilt on bed and deffo NO PJS!!! wimps all of you .............. we're still sleeping with the window wide open, curtains open to let the cool air in, summer light weight quilt on bed and deffo NO PJS!!! we are just jealous that you are so tough and root every night:embarressed: Guest darlo 2 adelaide I'm just enjoying hibernating, and ofcourse when its not raining the wine in the garden with a fire going (at the end of the month) will be just lovely. i love the chnage in the seasons here, last week was lovley weather, also they're more like seasons here than england i think!! we are just jealous that you are so tough and root every night:embarressed: Riiiigggggggggghhhhttttttttttt as if Guest BackToAdelaide I guess it's what you get used to. I often found the central heating in the UK stifling and would not put ours on for weeks after most people I knew already had. We certainly haven't even given a thought to using any heating here yet. We've just got the summer duvet on, no electric blankets (never used them), no hot water bottles etc. Speak for yourself! lol do these come with Velcro instead of buttons Missus is off to melbourne this week Gonna be a full time sloth until she gets back, heating off (she's a whimp) trackies and tee-shirt. Cold is when you have to de-ice the garage door to get your car out ! Like this !	{"url":"https://www.pomsinadelaide.com/topic/28286-draughty-houses-tiled-floors-no-insulation/","timestamp":"2024-11-03T22:12:11Z","content_type":"text/html","content_length":"338247","record_id":"<urn:uuid:bb0ba63e-94dd-43ed-b179-340ce05baaca>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00386.warc.gz"}
What is meant by backtracking? What is meant by backtracking? Backtracking is a technique based on algorithm to solve problem. It uses recursive calling to find the solution by building a solution step by step increasing values with time. It removes the solutions that doesn’t give rise to the solution of the problem based on the constraints given to solve the problem. What is backtracking in problem solving? Backtracking is an algorithmic technique where the goal is to get all solutions to a problem using the brute force approach. It consists of building a set of all the solutions incrementally. Since a problem would have constraints, the solutions that fail to satisfy them will be removed. What is difference between recursion and backtracking? In recursion function calls itself until reaches a base case. In backtracking you use recursion in order to explore all the possibilities until you get the best result for the problem. Why is backtracking used? It is used to solve a variety of problems. You can use it, for example, to find a feasible solution to a decision problem. Backtracking algorithms were also discovered to be very effective for solving optimization problems. In some cases, it is used to find all feasible solutions to the enumeration problem. What is the use of backtracking? Backtracking is an important tool for solving constraint satisfaction problems, such as crosswords, verbal arithmetic, Sudoku, and many other puzzles. It is often the most convenient technique for parsing, for the knapsack problem and other combinatorial optimization problems. What is backtracking technique explain with example? Backtracking name itself suggests that we are going back and coming forward; if it satisfies the condition, then return success, else we go back again. It is used to solve a problem in which a sequence of objects is chosen from a specified set so that the sequence satisfies some criteria.	{"url":"https://www.comicsanscancer.com/what-is-meant-by-backtracking/","timestamp":"2024-11-01T19:38:31Z","content_type":"text/html","content_length":"48268","record_id":"<urn:uuid:c494bacd-8a34-4c32-9a5d-33da74891adc>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00300.warc.gz"}
Following there is a list containing the features of ChaosPro. It should give you a basic idea of how powerful ChaosPro is. Supports several fractal types/rendering algorithms ChaosPro can create almost any kind of fractal using its integrated compiler. It is up to you which formula you want to use. So ChaosPro not only can create standard Julia and Mandelbrot sets, but also many other fractal types which are based on iterating a number. There are already thousands of formulas which have been written by other people and which you can use. Among them there are formulas which create bifurcation diagrams, IFS fractals, Plasma fractals, Lindenmayer systems, dynamic systems like the Lorenz attractor, the Rossler attractor, and Lyapunov Spaces. ChaosPro itself offers the following main fractal types (which basically are completely different algorithms, each of them uses its own kind of formula and thus you may either use an existing formula or write a new formula for further enhancements): • Escapetime: A pixel in a window (2D point) is iterated and tested what "happens" to the pixel. The result is shown and colored accordingly. This kind of algorithm creates for example Julia and Mandelbrot sets, but it may also create any other fractal which is based on iterating a 2D pixel. • Attractor: An arbitrary 3D point is iterated again and again. After each iteration it is drawn. This produces 3D fractals which are rendered accordingly including light and shadow. Examples for such fractals are IFS fractals, Flame fractals, and much more. • Quaternion: Similar to Escapetime, but uses Quaternion numbers and an algorithm similar to Escapetime which has been enhanced for 3D and 4D space. Examples for these fractals are Quaternions, but this type is perfectly suitable for rendering the Mandelbulb, too. Win 32 compliant ChaosPro runs on all Win32 versions, i.e. ChaosPro runs on Windows 98, NT, 2000, XP, Vista and Windows 7. If any other operating system supports Win32 (under Linux?), then it's possible that ChaosPro runs there, too. Windows 98 Windows NT Windows 2000 Windows XP Windows Vista Windows 7 Multitasking and Multiwindowing This means ChaosPro is a true MDI application where each fractal resides in its own window. You can calculate several fractals at the same time in different windows. The different calculation threads use low priority, so you can calculate several fractals and continue doing your other work with the computer. All windows in ChaosPro are modeless dialogs, so can be open just as you like. You do not need to close any window in order to open another one. Realtime fractal exploration Well, to be honest, this does not mean that you can dive into the fractal in realtime (only if your computer is fast enough...). Realtime in ChaosPro means that all changes take effect immediately. You grab the fractal with the mouse in order to move it around, and the effect is immediately visible. You assign another gradient (palette) and it gets applied immediately, no OK button to click on. If you move a slider (perhaps the rotation angle slider) the fractal thread constantly gets noticed that the slider changed its value and adjusts the fractal. How good and how fast it catches up with the changes depends on how fast your computer is. If you resize the fractal window then the fractal gets scaled accordingly. ChaosPro does not use modal dialogs, all parameter windows are True color ChaosPro does not know that there are only a limited number of colors. Basically it thinks there is an unlimited range of colors. Later in the fractal calculation it determines the colors based on a gradient (well, in ChaosPro a palette is a path through the color space based on about 250 knots). Depending on the coloring algorithm and on your display driver it then selects the colors. True color images with lots of colors, the mandelbrot set without those iteration bands, ChaosPro can calculate that sort of image. FractInt compatible To be honest: ChaosPro is not 100% FractInt compatible: There are situations where ChaosPro behaves different from FractInt. But it can read almost 80% of FractInts fractal types, starting with julia, mandel, upto complexnewton, barnsley1m, IFS, LSystem, etc. ChaosPro's formula parser can parse FractInt .frm files. ChaosPro's gradient (palette map) editor can read FractInt's .map files. The IFS and LSystem formula editors can load FractInt's .ifs and .l files as well. UltraFractal compatible You may ask how ChaosPro can be compatible to a fractal generator which uses a built in compiler. Is there really a compiler in ChaosPro? A compiler for a freeware fractal generator? Yes, indeed, there is: If another one can write a compiler, then I can do that, too. It's quite difficult and very time consuming, but it's not impossible. So I wrote a fast compiler, wrote some automates to import UltraFractal formulas and there it is: A built in compiler which produces fast machine code. No extra DLL, no extra package to install. It's all in ChaosPro. For free. The compiler together with the import mechanisms allow ChaosPro to use all UltraFractal formulas (transformations, iteration formulas, colorings) and to calculate all fractals which UltraFractal can This especially means that ChaosPro has all the features of UltraFractal 3.02 currently built in. In ChaosPro you can create zoom movies: Define where to start from, define where to end. Specify how many frames to calculate and press start. Too easy? Wait a moment: ChaosPro restricts you not only to simple zoom in/out/around movies, it lets you create animations based upon every fractal parameter in ChaosPro, in every combination. You simply define key frames of how your animation should look like, specify how many frames there should be between each pair of key frames and ChaosPro does the rest. The key frames may differ in any parameter which can be changed in a continuously manner, like the corners, the iteration value, the rotation angle, the parameter, the bailout value, the coloring paramaters, the palette and many others. Parameters which cannot change during an animation are the fractal type or flags for example: These must be the same during the animation. But the animation system is clever enough to not let you specify keyframes which do not match to the others you already have defined. So, are you interested in an animation flying into a Complex Newton Set, which keeps rotating, whereas the palette changes, just in order to fly around a bit and finally to fly back to the starting location animating the initialization point? And together with the 3D feature you additionally can animate all 3D parameters... Last update on Jan 02 2010.	{"url":"http://chaospro.de/features.php","timestamp":"2024-11-11T00:35:52Z","content_type":"text/html","content_length":"11247","record_id":"<urn:uuid:07c9c30a-e95f-43d0-bec9-469c06431ec5>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00041.warc.gz"}
Turtle graphics 02-15-2024, 08:18 AM (This post was last modified: 02-15-2024, 08:43 AM by Tomaaz.) While your code will work, it doesn't make any difference. Katzolwia represents the angle of the turtle in degrees. If the angle is bigger than 360, that function will change it to the same angle from 0..360 range. The same goes for angles smaller than 0. Of course, you can swap 360 for 0 and 0 for 360, but, like I said, it doesn't make any difference. For smaller programs you could skip that part completely. Computers can calculate trigonometric functions from numbers bigger than 360 or smaller tha 0 with no problems. I decided to include this function to prevent katzolwia to become to big or to small what could make the entire program to crash (if the number becomes to big or to small for variable to keep it).	{"url":"https://naalaa.com/forum/showthread.php?tid=76&pid=371&mode=threaded","timestamp":"2024-11-06T08:04:15Z","content_type":"application/xhtml+xml","content_length":"28518","record_id":"<urn:uuid:a9494475-71c4-4742-aad8-ba9b2de1df98>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00813.warc.gz"}
Quantum supremacy explained Starts With A Bang — Quantum supremacy explained Can quantum computers do things that standard, classical computers can’t? No. But if they can calculate faster, that’s quantum supremacy. Key Takeaways • Over the past several years, there have been tremendous advances in the field of quantum computation, but even greater amounts of hype surrounding those advances. • A few years ago, quantum supremacy was achieved for the first time, where a quantum computer performed a calculation millions of times faster and more efficiently than a classical computer. • But that’s a far cry from many of the technological advances being promised, including achieving quantum supremacy for a single practical problem. Here’s how to separate fact vs. fiction. Sign up for the Starts With a Bang newsletter Travel the universe with Dr. Ethan Siegel as he answers the biggest questions of all In our everyday experience, the world is 100% measurable, deterministic, and independent of the observer. The glass is either on the table in an unbroken state, or it’s on the floor in a shattered state, regardless of when or even whether you measure or observe it. The three marbles in your bag are definitively colored red, green, and blue, and no matter how you shake that bag or for how long, the red marble remains red, the green marble remains green, and the blue marble remains blue. And if you look at that quarter that somehow fell onto your nightstand long ago, it will always behave as though either “heads” or “tails” is facing up, never as though it’s part-heads and part-tails, simultaneously, at once. But in the quantum Universe, this isn’t necessarily the case. A radioactive atom that remains unobserved will exist in a superposition of “decayed” and “undecayed” states until that critical measurement is made. The three valence quarks making up your proton may all have a definitive color anytime you measure them, but exactly what color you observe is guaranteed to not be constant over time. And if you shoot many electrons, one-at-a-time, through a double slit and don’t measure which slit it goes through, the pattern you see will indicate that each electron went through both slits This difference, between classical and quantum systems, has resulted in both scientific and technological revolutions. One field that’s only now emerging is quantum computing, carrying the fascinating notion of quantum supremacy along with it, but also spawning a large series of dubious claims and misinformation. Here’s an explainer about quantum supremacy and the current state of quantum computers to help you separate fact from fiction. Let’s start with an idea you’re probably familiar with: the notion of an everyday computer, also known as a classical computer. Although calculating machines and devices had been around for a long time, well prior to the 20th century, it was Alan Turing who gave us the modern idea of a classical computer in the form of what’s now known as a Turing machine. The simple version of a Turing machine is that you can encode any type of information you like into bits: or binary (with only two options) components that, for example, could be represented by 0s and 1s. You can then apply a series of successive operations to those bits (for example, operations such as “AND,” “OR,” “NOT,” and many more) in the proper order to perform any sort of arbitrary computation that you had in mind. Some of those computations will be easy to encode and easy for the computer to perform: they require only a small number of bits, a small number of operations, and a very short time to compute them all. Others will be hard: difficult to code, computationally expensive for the computer to perform, and potentially requiring large numbers of bits, large numbers of operations, and long computing times. Regardless of your desired computation, however, if you can design an algorithm, or method, for successfully performing any computational task, you can program it into a classical computer. Eventually, given enough time, your computer will finish the program and deliver you the results. However, there’s a fundamental difference between this type of “classical computer” (that works exclusively with classical bits and classical operations) that we just described and a “quantum computer,” where the latter was purely a theoretical construct for many decades. Instead of regular bits, which are always in a state that’s known to be “0” or “1” with no exceptions, regardless of how or even whether you measure them or not, quantum computers use what are known as qubits, or the quantum analog of bits. While qubits can take on the same values that classical bits can take on — “0” or “1” in this case — they can also do things like exist in an intermediate state that’s a superposition of “0” and “1” simultaneously. They can be part-way between fully 100% “0” and fully 100% “1” in any amount that sums to 100% in total, and the amount of “0” and the amount of “1” that a qubit possesses can change both as a result of operations performed on the qubit and also due to simple time-evolution. But when you actually go to make that critical measurement of a qubit, and you ask it, “what quantum state it is actually in,” you’ll always see either a “0” or a “1” in your measuring device. You’ll never see that in-between value directly, even though you can infer that, based on the qubit’s effects on the overall outcome of the system, it must have simultaneously been a mix of “0” and “1” while the computation was taking place. A qubit is just another example of what we call a two-state quantum mechanical system: where only two outcomes can possibly be measured, but where the exact quantum state is not definitively determined until that critical measurement is made. This applies to many quantum mechanical systems, including: • the spin of an electron, which can be “up” (+) or “down” (-) in any direction that you choose to measure it, • the state of a radioactive atomic nucleus, which can be in either an undecayed state (the same as an initial state) or in a decayed state, • or the polarization of a photon, which can be either left-handed or right-handed relative to its direction of propagation, where each of these systems behaves as though it were in a superposition of both possibilities, right up until those critical measurements are made and a final state is definitively determined to be one of the two measurable possibilities. Qubits have something very important in common with classical bits: whenever you measure them, you’ll always see them in one of two states: the “0” state or the “1” state, with no exceptions and no in-betweens. However, they also have a very important difference: when you perform computational operations on a qubit, the qubit isn’t in a determinate state (of either “0” or “1”) like a classical bit is, but rather lives in a state that’s a superposition of “0” and “1,” like a qubit version of Schrödinger’s cat. It’s only once all the computations are done and you measure your final results that the final state of that qubit is fully determined: and where you’ll find out that it’s either “0” or “1.” The computational difference between a “bit” and a “qubit” is very much like the quantum mechanical difference between a “classical two-state system” and a “quantum two-state system,” where even though you’re only going to get two possible outcomes in the end, the probabilities of getting “outcome #1” and “outcome #2” obey wildly different rules for the quantum system as opposed to the classical system. Whereas with a classical system, you can provide: • the initial conditions, • the algorithm of operations that will affect the system, and then get a prediction for the final state of your system as a result, for a quantum mechanical system, you can only get a probability distribution as a prediction for your system’s final state. In the quantum case, only by performing the critical experiment over and over again can you hope to match and produce your predicted distribution. Now, here’s where things get a little counterintuitive: you might think that classical computers are good tools for solving classical (but not quantum) problems, and that quantum computers would be required to solve quantum problems. It turns out, however, that one of the most important ideas in computer science — the Church-Turing thesis — directly contradicts that notion, stating that any problem that can be solved by a Turing machine, using only classical bits and classical operators, can also be solved by a computational device: i.e., a classical computer. Just as we can solve problems that involve classical waves with classical mathematics and on classical computers, we can solve problems that involve quantum mechanical waves in the same fashion. The computational device is irrelevant: whether it’s a calculator, laptop, smartphone, supercomputer, or even a quantum computer (which can solve classical problems, too), a problem that could be solved by a Turing machine can be solved by any of these computers. However, that doesn’t mean that all methods of solving problems are equally efficient at doing so. In particular, you could imagine trying to simulate an inherently quantum mechanical system — using not just qubits instead of regular bits, but also inherently quantum operators (or their computational equivalent, quantum gates) — where using a quantum computer might give you a tremendous advantage in efficiency, speed, and computational time over a classical computer. A very controversial extension to the Church-Turing thesis — creatively named the extended Church-Turing thesis — basically asserts that you can’t do this. Instead, it claims that a Turing machine can always efficiently simulate any computational model, even a computational model that’s heavily (or even fully) quantum in nature. That’s the crux of the idea behind Quantum Supremacy, and the related idea of Quantum Advantage. (Although some use these terms synonymously, there’s an important distinction that’s becoming more and more commonplace.) • To achieve Quantum Supremacy, you’d have to disprove the extended Church-Turing thesis, where the easiest way to do so would be to provide a counterexample. In other words, if you can state a computational problem that is computationally expensive for a classical computer using all known algorithms and techniques, but that is much easier and less computationally expensive for a quantum computer, and you can demonstrate a vast speed-up in computational time using a quantum computer, then Quantum Supremacy will have been achieved. • On the other (somewhat more ambitious) hand, you can attempt to achieve what’s known as Quantum Advantage, which would be achieving Quantum Supremacy for a problem that’s actually relevant to the real world. This could mean for a physically interesting but inherently quantum system, like electrons traveling through a double slit or phonons in a condensed matter system, or it could be for a complex classical system where using qubits and quantum gates provides a notable speed-up using fewer resources. Quantum Supremacy, by this definition, was first (likely) achieved back in 2017-2019, but Quantum Advantage still appears to be a long way off, with several important caveats. First, the current limitations on the power of quantum computing is set by two factors: 1. the number of superconducting qubits that can be controlled at once by a quantum computer, which in turn limits the number of variables that can be processed in any one computation, 2. and the power of quantum error-correction, as no quantum circuits are 100% reliable (they all introduce errors), and the errors you do have rise with both the time needed to complete your computation and also the number of qubits involved. Therefore, if you want to achieve Quantum Supremacy (or its more ambitious cousin, Quantum Advantage), you’ll want to design a computational problem (or a useful computational problem) that requires only a small number of qubits, and that all the necessary computations can occur in a short time relative to the coherence time of the qubits involved. In 2019, Quantum Supremacy was demonstrated by a team at Google for a very specific problem: a problem with no real-world usefulness that was specifically designed to be extremely difficult to simulate on a classical computer, but that would be easy for a quantum computer. Although some still argue that a better classical algorithm will eventually allow this problem, and others like it, to be solved quickly on a classical computer, such arguments lack any demonstrable room-for-improvement to point to. We are still, unfortunately, a long way from solving any useful problems much faster on a quantum computer than with a classical computer. There was a report last year, in Nature, that a traversable wormhole had been encoded onto a quantum processor and whose dynamics had been demonstrated using only 9 logical qubits: a purported demonstration of quantum advantage. Further analysis has revealed that the entire research effort was fundamentally flawed, and so it’s back to the drawing board as far as Quantum Advantage goes. Almost every practical task you can imagine has little potential for Quantum Advantage, with classical computers performing much better in most cases. For example, let’s say you have a 20 digit semiprime number (a number that’s the product of two primes): there is no quantum computer that can solve this problem at all, whereas your off-the-shelf laptop can accomplish this in a matter of However, incremental improvements are continuing to occur, with a new, more efficient quantum factoring algorithm offering a potential speed-up over using classical computers alone, and a novel quantum error-correction protocol offering the potential to preserve 12 logical qubits for 10 million syndrome cycles (a measure of needed error correction) with just 288 physical qubits, rather than the current need for 4000+ physical qubits using the standard (surface code) error-correction protocol. Someday, the cumulative combination of these and other advances will lead to the first robust demonstration of Quantum Advantage in a useful, practical system. The ultimate goal of quantum computers, at least in the short-term, is to simulate quantum systems that are computationally expensive to simulate classically. That’s where the first practical application of Quantum Advantage is expected to arise, and it’s anyone’s guess whether the field of: • materials science, • high-energy physics, • or quantum chemistry, will be the first to reap the practical benefits of quantum computers. With greater numbers of superconducting qubits, longer coherence times for those qubits, and superior error-correction expected on the horizon, the computational power of quantum computers (including the number of logical qubits that can be used for computation) is steadily increasing. Eventually, the first practical, real-world problems that are computationally expensive for classical computers will be solved, quickly and efficiently, by quantum computers. But no one should be under the illusion that quantum computers will someday replace classical computers for most applications, or that achieving Quantum Supremacy means that useful quantum computing has already arrived. Instead, we should expect our future to be a computationally hybrid one: where classical computers are at the root of most of our computational needs and are augmented by quantum computers in those arenas where Quantum Advantage can be achieved. Still, just as there are many important physical phenomena that cannot be explained by the classical theories of Newton, Maxwell, or Einstein, there are many important computational problems that are awaiting the development of superior quantum computers. With Quantum Supremacy already achieved and the more practical Quantum Advantage on its way, we have so much to hope for as far as quantum computing is concerned, but so much hype — and many false claims — to simultaneously be wary of. Sign up for the Starts With a Bang newsletter Travel the universe with Dr. Ethan Siegel as he answers the biggest questions of all The science fiction dream of a traversable wormhole is no closer to reality, despite a quantum computer’s suggestive simulation. Is the time crystal really an otherworldly revolution, leveraging quantum computing that will change physics forever? The same (former) NASA engineer who previously claimed to violate Newton’s laws is now claiming to have made a warp bubble. He didn’t. There’s a speed limit to the Universe: the speed of light in a vacuum. Want to beat the speed of light? Try going through a medium! To find the optimal route between many different locations, we need the power of quantum computers. Sixty years later, will anybody have heard of COVID?	{"url":"https://bigthink.com/starts-with-a-bang/quantum-supremacy-explained/","timestamp":"2024-11-14T19:02:11Z","content_type":"text/html","content_length":"193625","record_id":"<urn:uuid:0678a65e-8ac8-4210-ac06-71bc7323153d>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00071.warc.gz"}
Understanding Mathematical Functions: Which Of The Following Is A Line When it comes to understanding mathematical functions, it's important to recognize the various types and their characteristics. Functions are a fundamental concept in mathematics, representing the relationship between inputs and outputs. By understanding different types of functions, we can better analyze and interpret mathematical relationships in real-world scenarios. One type of function that we often encounter is the linear function. This type of function is crucial in many fields, such as economics, physics, and engineering. In this blog post, we will explore the characteristics of a linear function and identify whether specific mathematical equations fall under this category. Key Takeaways • Understanding different types of mathematical functions is crucial for analyzing and interpreting real-world scenarios. • Linear functions play a significant role in fields such as economics, physics, and engineering. • Recognizing the characteristics of linear functions and being able to identify them from equations is important for practical applications. • Mastering linear functions has a positive impact on further mathematical studies and everyday problem-solving. • Recognizing linear functions and their practical relevance is beneficial for working with them in different contexts. Understanding Mathematical Functions: Which of the following is a linear function What is a linear function? A linear function is a type of function in mathematics that has the form f(x) = mx + b, where m and b are constants. This means that the output (f(x)) is a sum of the input (x) multiplied by a constant (m) and added to another constant (b). This type of function creates a straight line when graphed on a coordinate plane. Definition of a linear function A linear function is a function that can be described by a straight line. In mathematical terms, a function f(x) is considered linear if it can be expressed in the form f(x) = mx + b, where m and b are constants. Characteristics of a linear function • Constant rate of change: A linear function has a constant rate of change, meaning that for every unit increase in the input, there is a constant increase or decrease in the output. • Straight line graph: When graphed on a coordinate plane, a linear function creates a straight line. • Simple algebraic form: The algebraic expression for a linear function is simple and can be easily identified. Examples of linear functions Some examples of linear functions include: • f(x) = 2x + 3 • g(x) = -4x + 5 • h(x) = 0.5x - 1 How to recognize a linear function There are a few key indicators to help recognize a linear function: • Algebraic form: Look for the presence of x raised to the power of 1 (x^1) in the function's algebraic expression. • Graph: Plot the function on a coordinate plane and see if it forms a straight line. • Constant rate of change: Calculate the rate of change and check if it is constant throughout the function's domain. Understanding different types of functions When it comes to mathematical functions, it's important to understand the different types and how they behave. One of the key distinctions is between linear and non-linear functions. A. Linear vs. non-linear functions Linear functions are those that have a constant rate of change, resulting in a straight line when graphed. They can be expressed in the form y = mx + b, where m is the slope and b is the y-intercept. Non-linear functions, on the other hand, do not have a constant rate of change and do not graph as a straight line. B. Examples of non-linear functions • Quadratic functions, such as y = x^2, where the graph forms a parabola • Exponential functions, such as y = 2^x, where the graph rapidly increases or decreases • Trigonometric functions, such as y = sin(x) or y = cos(x), where the graph forms waves C. Importance of distinguishing between different types of functions It is crucial to distinguish between linear and non-linear functions as they behave differently and have different applications. Linear functions are often used to model simple relationships and are easier to work with mathematically. Non-linear functions, on the other hand, can model more complex and realistic relationships, such as exponential growth or decay, and oscillating behavior. Understanding Mathematical Functions: Identifying Linear Functions from Equations When working with mathematical functions, it's important to be able to identify different types of functions based on their equations. One common type of function is the linear function, which has a special form and properties that distinguish it from other types of functions. A. How to identify a linear function from an equation Identifying a linear function from an equation involves looking for specific patterns and characteristics. A linear function is a type of function that can be represented by a straight line when graphed. This means that its equation must have certain properties that indicate a linear relationship between the input and output variables. 1. Checking for a degree of 1 In order to identify a linear function, the equation must have a degree of 1 for the input variable. This means that the highest power of the input variable should be 1. For example, in the equation y = 2x + 3, the degree of x is 1, indicating a linear relationship. 2. Absence of other variables A linear function should only have the input variable and a constant term in its equation. Any other variables or higher degree terms would indicate a different type of function, such as a quadratic or exponential function. B. Common forms of linear function equations Linear functions can take on different forms, but there are some common equation formats that are used to represent linear relationships between variables. 1. Slope-intercept form The slope-intercept form of a linear function is y = mx + b, where m represents the slope of the line and b represents the y-intercept. This form is commonly used to graph linear functions and understand their properties. 2. Standard form The standard form of a linear function is Ax + By = C, where A, B, and C are constants. This form is useful for finding the x and y-intercepts of a linear function and can be used to compare different linear equations. C. Tips for recognizing linear functions in equations When dealing with equations, it can sometimes be tricky to quickly identify whether a function is linear or not. Here are some tips to help recognize linear functions in equations. • Look for a constant rate of change • Check for a straight-line graph • Eliminate other variable terms Real-world applications of linear functions Linear functions play a crucial role in modeling real-world situations and are used extensively in various industries and fields. Understanding the applications of linear functions is essential for solving practical problems and making informed decisions based on data and trends. Examples of real-world situations modeled by linear functions • Supply and demand: The relationship between the supply of a product and its demand can often be represented by a linear function. This is essential for businesses to optimize their production and pricing strategies. • Population growth: The growth of a population over time can often be approximated by a linear function, which is crucial for urban planning and resource allocation. • Distance and time: The relationship between distance and time for a moving object can be modeled using a linear function, which is important for transportation and logistics. Importance of understanding linear functions in practical scenarios Understanding linear functions is essential for making predictions, understanding trends, and making informed decisions in practical scenarios. It allows individuals and organizations to analyze data, identify patterns, and make projections based on mathematical models. How linear functions are used in various industries and fields • Finance: Linear functions are used in financial analysis to predict future trends in stock prices, interest rates, and economic indicators. • Engineering: Linear functions are essential for modeling mechanical and structural systems, as well as for designing and optimizing processes and systems. • Healthcare: Linear functions are used for analyzing patient data, predicting disease trends, and optimizing healthcare delivery and resource allocation. • Education: Linear functions are used in educational research for analyzing student performance, predicting educational outcomes, and designing educational interventions. Understanding the real-world applications of linear functions is crucial for professionals across various industries and fields, as it allows for informed decision-making, strategic planning, and optimization of processes and systems. Importance of mastering linear functions Understanding linear functions is crucial for mathematical studies and has practical relevance in everyday life. Mastering linear functions provides numerous benefits in various contexts, making it an essential concept to grasp in mathematics. A. Impact of understanding linear functions in further mathematical studies • Linear functions serve as the foundation for more complex mathematical concepts. • They are essential for understanding calculus, physics, and engineering. • Mastering linear functions helps in grasping higher-level mathematical topics. B. Practical relevance of recognizing linear functions in everyday life • Linear functions are prevalent in real-world scenarios, such as calculating costs and analyzing trends. • Understanding linear functions aids in budgeting, forecasting, and decision-making. • They are used in fields like economics, business, and finance for making informed choices. C. Benefits of being able to work with linear functions in different contexts • Being proficient in linear functions enhances problem-solving skills in various settings. • It enables individuals to interpret data, design models, and make predictions. • Proficiency in linear functions leads to a deeper understanding of mathematical relationships. Understanding linear functions is crucial for anyone studying mathematics. It provides a foundation for more advanced concepts and real-world applications. As you continue your mathematical journey, exploring different types of functions will expand your knowledge and problem-solving skills. Always remember the significance of recognizing linear functions in various mathematical contexts. ONLY $99 Immediate Download MAC & PC Compatible Free Email Support	{"url":"https://dashboardsexcel.com/blogs/blog/understanding-mathematical-functions-which-of-the-following-is-a-linear-function","timestamp":"2024-11-11T17:04:39Z","content_type":"text/html","content_length":"215416","record_id":"<urn:uuid:b0f8c145-f989-4c68-8da6-978a070c32f7>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00172.warc.gz"}
C = conv2(A,B) returns the two-dimensional convolution of matrices A and B. • If A is a matrix and B is a row vector (or A is a row vector and B is a matrix), then C is the convolution of each row of the matrix with the vector. • If A is a matrix and B is a column vector (or A is a column vector and B is a matrix), then C is the convolution of each column of the matrix with the vector. C = conv2(u,v,A) first convolves each column of A with the vector u, and then it convolves each row of the result with the vector v. This behavior applies regardless of whether u or v is a row or column vector. C = conv2(___,shape) returns a subsection of the convolution according to shape. For example, C = conv2(A,B,"same") returns the central part of the convolution, which is the same size as A. 2-D Convolution In applications such as image processing, it can be useful to compare the input of a convolution directly to the output. The conv2 function allows you to control the size of the output. Create a 3-by-3 random matrix A and a 4-by-4 random matrix B. Compute the full convolution of A and B, which is a 6-by-6 matrix. A = rand(3); B = rand(4); Cfull = conv2(A,B) Cfull = 6×6 0.7861 1.2768 1.4581 1.0007 0.2876 0.0099 1.0024 1.8458 3.0844 2.5151 1.5196 0.2560 1.0561 1.9824 3.5790 3.9432 2.9708 0.7587 1.6790 2.0772 3.0052 3.7511 2.7593 1.5129 0.9902 1.1000 2.4492 1.6082 1.7976 1.2655 0.1215 0.1469 1.0409 0.5540 0.6941 0.6499 Compute the central part of the convolution Csame, which is a submatrix of Cfull with the same size as A. Csame is equal to Cfull(3:5,3:5). Csame = conv2(A,B,"same") Csame = 3×3 3.5790 3.9432 2.9708 3.0052 3.7511 2.7593 2.4492 1.6082 1.7976 Extract 2-D Pedestal Edges The Sobel edge-finding operation uses a 2-D convolution to detect edges in images and other 2-D data. Create and plot a 2-D pedestal with interior height equal to one. A = zeros(10); A(3:7,3:7) = ones(5); Convolve the rows of A with the vector u, and then convolve the rows of the result with the vector v. The convolution extracts the horizontal edges of the pedestal. u = [1 0 -1]'; v = [1 2 1]; Ch = conv2(u,v,A); To extract the vertical edges of the pedestal, reverse the order of convolution with u and v. Cv = conv2(v,u,A); Compute and plot the combined edges of the pedestal. mesh(sqrt(Ch.^2 + Cv.^2)) Input Arguments A — Input array vector \| matrix Input array, specified as a vector or matrix. Data Types: double \| single \| int8 \| int16 \| int32 \| int64 \| uint8 \| uint16 \| uint32 \| uint64 \| logical Complex Number Support: Yes B — Second input array vector \| matrix Second input array, specified as a vector or a matrix to convolve with A. The array B does not have to be the same size as A. Data Types: double \| single \| int8 \| int16 \| int32 \| int64 \| uint8 \| uint16 \| uint32 \| uint64 \| logical Complex Number Support: Yes u — Input vector row or column vector Input vector, specified as a row or column vector. u convolves with each column of A. Data Types: double \| single \| int8 \| int16 \| int32 \| int64 \| uint8 \| uint16 \| uint32 \| uint64 \| logical Complex Number Support: Yes v — Second input vector row or column vector Second input vector, specified as a row or column vector. v convolves with each row of the convolution of u with the columns of A. Data Types: double \| single \| int8 \| int16 \| int32 \| int64 \| uint8 \| uint16 \| uint32 \| uint64 \| logical Complex Number Support: Yes shape — Subsection of convolution "full" (default) \| "same" \| "valid" Subsection of the convolution, specified as one of these values: • "full" — Return the full 2-D convolution. • "same" — Return the central part of the convolution, which is the same size as A. • "valid" — Return only parts of the convolution that are computed without zero-padded edges. Output Arguments C — 2-D convolution vector \| matrix 2-D convolution, returned as a vector or matrix. When A and B are matrices, then the convolution C = conv2(A,B) has size size(A)+size(B)-1. When [m,n] = size(A), p = length(u), and q = length(v), then the convolution C = conv2(u,v,A) has m+p-1 rows and n+q-1 columns. When one or more input arguments to conv2 are of type single, then the output is of type single. Otherwise, conv2 converts inputs to type double and returns type double. Data Types: double \| single More About 2-D Convolution For discrete, two-dimensional matrices A and B, the following equation defines the convolution of A and B: $C\left(j,k\right)=\sum _{p}\sum _{q}A\left(p,q\right)B\left(j-p+1,k-q+1\right)$ p and q run over all values that lead to legal subscripts of A(p,q) and B(j-p+1,k-q+1). Using this definition, conv2 calculates the direct convolution of two matrices, rather than the FFT-based convolution. Extended Capabilities Tall Arrays Calculate with arrays that have more rows than fit in memory. The conv2 function supports tall arrays with the following usage notes and limitations: • If shape is "full" (default), then the inputs A and B must not be empty and only one them can be a tall array. • If shape is "same" or "valid", then B cannot be a tall array. • u and v cannot be tall arrays. For more information, see Tall Arrays. C/C++ Code Generation Generate C and C++ code using MATLAB® Coder™. GPU Code Generation Generate CUDA® code for NVIDIA® GPUs using GPU Coder™. Thread-Based Environment Run code in the background using MATLAB® backgroundPool or accelerate code with Parallel Computing Toolbox™ ThreadPool. This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment. GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. The conv2 function fully supports GPU arrays. To run the function on a GPU, specify the input data as a gpuArray (Parallel Computing Toolbox). For more information, see Run MATLAB Functions on a GPU (Parallel Computing Toolbox). Distributed Arrays Partition large arrays across the combined memory of your cluster using Parallel Computing Toolbox™. Usage notes and limitations: • Input vectors u and v must not be distributed arrays. For more information, see Run MATLAB Functions with Distributed Arrays (Parallel Computing Toolbox). Version History Introduced before R2006a	{"url":"https://ch.mathworks.com/help/matlab/ref/conv2.html","timestamp":"2024-11-09T06:09:17Z","content_type":"text/html","content_length":"101719","record_id":"<urn:uuid:339be501-e928-4317-9009-242a94076090>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00494.warc.gz"}
Question ID - 51470 \| SaraNextGen Top Answer The following question consists of three statements I, II and III are given below. You have to decide whether the data provided in the statements are sufficient to answer the question. How many marks did Tarun secure un English? I. The average marks obtained by Tarun in four subjects including English is 60. II. The total marks obtained by him in English and Mathematics together is 170. III. The total marks obtained by him in Mathematics and Science together is 180. (a) I and II only (b) II and III only (c) I and III only (d) None of these.	{"url":"https://www.saranextgen.com/homeworkhelp/doubts.php?id=51470","timestamp":"2024-11-09T13:34:06Z","content_type":"text/html","content_length":"15820","record_id":"<urn:uuid:ecb4ccf6-4b16-41f1-b040-6329b7df8ed1>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00562.warc.gz"}
D-Wave NetworkX provides tools for working with Quantum Processing Unit (QPU) topology graphs, such as the Pegasus used on the Advantage^TM system, and implementations of graph-theory algorithms on D-Wave quantum computers and other binary quadratic model samplers; for example, functions such as draw_pegasus() provide easy visualization for Pegasus graphs; functions such as maximum_cut() or min_vertex_cover() provide graph algorithms useful to optimization problems that fit well with D-Wave quantum computers. Like D-Wave quantum computers, all other supported samplers must have sample_qubo and sample_ising methods for solving Ising and QUBO models and return an iterable of samples in order of increasing energy. You can set a default sampler using the set_default_sampler() function. • For an introduction to quantum processing unit (QPU) topologies such as the Pegasus graph, see Topology. • For an introduction to binary quadratic models (BQMs), see Binary Quadratic Models. • For an introduction to samplers, see Samplers and Composites. This example creates a Pegasus graph (used by Advantage) and a small Zephyr graph (used by the Advantage2^TM prototype made available in Leap^TM in June 2022): >>> import dwave_networkx as dnx >>> # Advantage >>> P16 = dnx.pegasus_graph(16) >>> # Advantage2 >>> Z4 = dnx.zephyr_graph(4)	{"url":"https://docs.ocean.dwavesys.com/en/stable/docs_dnx/intro.html","timestamp":"2024-11-08T11:04:06Z","content_type":"text/html","content_length":"30055","record_id":"<urn:uuid:704181e3-7868-4c9d-a2ec-958e4642e796>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00475.warc.gz"}
Science & Math Archives - Page 210 of 233 - Futility Closet 135 = 1 + 32 + 53 175 = 1 + 72 + 53 518 = 5 + 12 + 83 598 = 5 + 92 + 83 Once upon a time, there lived a rich farmer who had 30 children, 15 by his first wife who was dead, and 15 by his second wife. The latter woman was eager that her eldest son should inherit the property. Accordingly one day she said to him, “Dear Husband, you are getting old. We ought to settle who shall be your heir. Let us arrange our 30 children in circle, and counting from one of them, remove every tenth child until there remains but one, who shall succeed to your estate.” The proposal seemed reasonable. As the process of selection went on, the farmer grew more and more astonished as he noticed that the first 14 to disappear were children by his first wife, and he observed that the next to go would be the last remaining member of that family. So he suggested that they should see what would happen if they began to count backwards from this lad. She, forced to make an immediate decision, and reflecting that the odds were now 15 to 1 in favour of her family, readily assented. Who became the heir? Once upon a time, there lived a rich farmer who had 30 children, 15 by his first wife who was dead, and 15 by his second wife. The latter woman was eager that her eldest son should inherit the property. Accordingly one day she said to him, “Dear Husband, you are getting old. We ought to settle who shall be your heir. Let us arrange our 30 children in circle, and counting from one of them, remove every tenth child until there remains but one, who shall succeed to your estate.” The proposal seemed reasonable. As the process of selection went on, the farmer grew more and more astonished as he noticed that the first 14 to disappear were children by his first wife, and he observed that the next to go would be the last remaining member of that family. So he suggested that they should see what would happen if they began to count backwards from this lad. She, forced to make an immediate decision, and reflecting that the odds were now 15 to 1 in favour of her family, readily assented. Who became the heir? An optical illusion. Squares A and B are the same color. Clearly there are integers so huge they can’t be described in fewer than 22 syllables. Put them all in a big pile and consider the smallest one. It’s “the smallest integer that can’t be described in fewer than 22 syllables.” Remarkably, you can estimate π by dropping needles onto a flat surface. If the surface is ruled with lines that are separated by the length of a needle, then: drops is the number of needles dropped. hits is the number of needles that touch a line. The method combines probability with trigonometry; a needle’s chance of touching a line is related to the angle at which it comes to rest. It was discovered by the French naturalist Georges-Louis Leclerc in 1777. Clarke’s Third Law: Any sufficiently advanced technology is indistinguishable from magic. Benford’s Corollary: Any technology distinguishable from magic is insufficiently advanced. Raymond’s Second Law: Any sufficiently advanced system of magic would be indistinguishable from a technology. Sterling’s Corollary: Any sufficiently advanced garbage is indistinguishable from magic. Langford’s application to science fiction: Any sufficiently advanced technology is indistinguishable from a completely ad-hoc plot device. 13 + 33 + 63 = 244 23 + 43 + 43 = 136 You and I are having an argument. Our wives have given us new neckties, and we’re arguing over which is more expensive. Finally we agree to a wager. We’ll ask our wives for the prices, and whoever is wearing the more expensive tie has to give it to the other. You think, “The odds are in my favor. If I lose the wager, I lose only the value of my tie. If I win the wager, I gain more than the value of my tie. On balance I come out ahead.” The trouble is, I’m thinking the same thing. Are we both right? “Why are numbers beautiful? It’s like asking why is Beethoven’s Ninth Symphony beautiful. If you don’t see why, someone can’t tell you. I know numbers are beautiful. If they aren’t beautiful, nothing is.” — Paul Erdös	{"url":"https://www.futilitycloset.com/category/science-math/page/210/","timestamp":"2024-11-10T14:42:49Z","content_type":"text/html","content_length":"69335","record_id":"<urn:uuid:db99c8b0-a7a0-4409-a428-5f326db3538d>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00054.warc.gz"}
how to calculate power supply for ball mill crusher in cement palnt power requirement for the crusher. power requirement for stone crusher aquabrand. Pf 1214 Coal Mill In Power Plant Crusher Mills, how to calculate power requirements for grinders crushers. Get Price. WhatsApp: +86 18838072829 No, it is not normal to operate above 40 to 45% charge level. Unless the trunnion is very small or unless you are using a grate discharge ball mill, balls will not stay in the mill, and you will spend a lot on steel to add a small amount of power. Formulas given for mill power draw modelling are empirical, and fit around field data over the ... WhatsApp: +86 18838072829 The main equipment for grinding construction materials are balltube mills, which are actively used in industry and are constantly being improved. The main issue of improvement is to reduce the power consumption of a balltube mill with crosslongitudinal movement of the load. A comparative analysis and the possibility of using the known ... WhatsApp: +86 18838072829 Based on his work, this formula can be derived for ball diameter sizing and selection: Dm <= 6 (log dk) * d^ where D m = the diameter of the singlesized balls in = the diameter of the largest chunks of ore in the mill feed in mm. dk = the P90 or fineness of the finished product in microns (um)with this the finished product is ... WhatsApp: +86 18838072829 The generator power calculator takes the total current requirement of the devices in amperes (A) and the supply voltage rating in volts (V mathrm{V} V) to calculate the apparent power (k V A mathrm{kVA} kVA), which is then used to calculate actual power based on the power the below section if you don't understand some of the terms we used here. WhatsApp: +86 18838072829 How to Measure Grinding Efficiency. The first two Grinding Efficiency Measurement examples are given to show how to calculate Wio and Wioc for single stage ball mills. Figure 1. The first example is a comparison of two parallel mills from a daily operating report. Mill size x (′ x 20′ with a ID of 16′). WhatsApp: +86 18838072829 This can be a singlephase or threephase power supply, depending on the size of the ball mill and the power requirements of the motor. Select a motor: Choose a motor that is suitable for the size WhatsApp: +86 18838072829 During the runningin process, the amount of steel balls is added for the first time, which accounts for 80% of the maximum ball load of the ball mill. Steel ball sizes are Φ120㎜, Φ100㎜, Φ80㎜, Φ60㎜, Φ40㎜. For example, the 100150 tons ball mill has a maximum ball loading capacity of tons. For the first time, 30% 40% of ... WhatsApp: +86 18838072829 The main function of steel ball in ball mill is to break the material by impact, and it also plays a certain role in grinding. In order to determine the gradation of steel balls, besides the factors such as the size of ball mill, internal structure of ball mill and product fineness requirements, the characteristics of grinding materials ... WhatsApp: +86 18838072829 Hard ore Work Index 16 = 100,000/65,000 = kwh/t. For the purposes of this example, we will hypothesize that the the crushing index of the hard ore with the increased energy input of kw/t reduces the ball mill feed size to 6,500 micrometers. As a result, the mill output will increase with this reduced size to approximately 77,000 tons ... WhatsApp: +86 18838072829 To compute for shaft power \| ball mill length, six essential parameters are needed and these parameters are Value of C, Volume Load Percentage (J), % Critical Speed (Vcr), Bulk Density (), Mill Length (L) and Mill Internal Diameter (D). The formula for calculating shaft power \| ball mill length: P = x C X J X V cr x (1 ) x [1 ... WhatsApp: +86 18838072829 Typically Speed = Distance/time (which means / But here we also have to calculate the speed of the sound of the ball traveling when the ball hits the pins till it reaches the ears of the ... WhatsApp: +86 18838072829 The formula for calculating shaft power: P = QE Where: P = Shaft Power Q = Mill Capacity E = Specific Power of Mill Let's solve an example; Find the shaft power when the mill capacity is 20 and the specific power of mill is 24. This implies that; Q = Mill Capacity = 20 E = Specific Power of Mill = 24 P = QE P = (20) (24) P = 480 WhatsApp: +86 18838072829 To find the power at the mill pinion, multiply an E value by the circuit solids feed rate (feed rate of dry solids to primary mill or the dry solids flow rate of product in the cyclone overflow). Power as kW = E * (t/h dry solids) A ball mill can safely operate at 90% of its motor rated power, so choose a motor size like this: WhatsApp: +86 18838072829 Here, you will find information about your computer's power supply. Another method to determine your computer's power supply is through thirdparty software called CPUZ. Download and install CPUZ from their official website, then launch the program. Go to the "Mainboard" tab and look for information related to "Power Supply." WhatsApp: +86 18838072829 First, the weight loss of the ball leads to a decrease in its kinetic energy and a consequent reduction in energy transfer or milling efficiency. Second, the degree of filling of the mill is raised so the balls mobility becomes more difficult and as a result the kinetic energy of the balls are reduced. WhatsApp: +86 18838072829 For 60 mm (″) and smaller top size balls for cast metal liners use double wave liners with the number of lifters to the circle approximately D in meters (for D in feet, divide D by ). Wave height above the liners from to 2 times the liner thickness. Rubber liners of the integral molded design follow the cast metal design. WhatsApp: +86 18838072829 The maximum power draw in ball mill is when ball bed is 3540 % by volume in whole empty mill volume. Considering that ball bed has a porosity of 40 %, the actual ball volume is considered to be WhatsApp: +86 18838072829 Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80% passing ¼ inch... WhatsApp: +86 18838072829 This can reduce the energy consumption and the use cost of the ball mill. To make the milling more efficient, we must first be acquainted with the factors. They are mainly the ball mill structure, the rotation speed, the ball mill media, the lining plate, the material fed and the feeding speed, etc. In the following text, you will get some ... WhatsApp: +86 18838072829 Rod mills speed should be limited to a maximum of 70% of critical speed and preferably should be in the 60 to 68 percent critical speed range. Pebble mills are usually run at speeds between 75 and 85 percent of critical speed. Ball Mill Critical Speed . The black dot in the imagery above represents the centre of gravity of the charge. WhatsApp: +86 18838072829 Ball Mill Design/Power Calculation. · A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80% passing ¼ inch (6350 microns). The required product size distribution is to be 80% passing 100 mesh (149 microns). In order . WhatsApp: +86 18838072829	{"url":"https://tresorsdejardin.fr/how/to/calculate/power/supply/for/ball/mill-5839.html","timestamp":"2024-11-04T04:42:37Z","content_type":"application/xhtml+xml","content_length":"21717","record_id":"<urn:uuid:f9ecc262-21c5-4c5c-b020-8eaf74a96dcb>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00621.warc.gz"}
Unlocking the Power of NumPy in Python: A Comprehensive Guide Updated October 10, 2023 Table of Contents Introduction to NumPy in Python NumPy is a Python package primarily used for scientific and numerical computing. NumPy is a portmanteau of two words coined by blending “Numerical” and “Python.” It is very famous among data scientists and analysts for its efficiency(run time speed) and the wide range of array operations it provides. NumPy was developed initially as “numeric” by Jim Hugunin in the late 90s. The current version of NumPy was created in 2005 by Travis Oliphant, who incorporated features from the competing numarray and Numeric libraries. It consists of numerous powerful features, including the following: • A robust multi-dimension array object with many useful functions. • There are many tools for integrating other programming languages, and an enormous number of routines, including shape manipulation, logical, mathematical & many more, are used to operate on NumPy Array objects. • Besides its obvious scientific usage, NumPy is a generic multi-dimensional data container. • A wide set of databases can also be integrated with NumPy. Install NumPy in Python To install NumPy in Python, you can use a package manager like pip or conda, depending on your Python environment. Here are the steps for both methods: Using pip: Open your terminal or command prompt. Run the following command to install NumPy using pip: pip install numpy Pip will download and install NumPy and its dependencies. Once the installation is complete, you’ll have NumPy installed in your Python environment. Using conda (if you’re using Anaconda or Miniconda): Open your terminal or Anaconda prompt. Run the following command to install NumPy using conda: conda install numpy Conda will install NumPy and its dependencies and once finished, NumPy will be ready to use. After installation, you can import NumPy in your Python scripts or interactive sessions using: import numpy as np Now, you can start using NumPy in your Python projects. Examples of NumPy in Python Let’s discuss some more examples and how to achieve the same using NumPy: The very first step would be to import the package within the code: import NumPy as np Hit “Shift + Enter” to import the specified package NumPy is aliased as “np”, which can be utilized to refer to NumPy for any further references Example #1 – Creating NumPy Arrays Let’s create a one-dimensional array with the name “a” and values as 1,2,3 a = np.array ( [1,2,3] ) This will utilize the “array” attribute out of the NumPy module (which we have aliased as “np” over here ) Use the “print” attribute to print the values of a variable/object. The output will print the one-dimensional array “a” as: [1 2 3] Use the “type” attribute to verify the type of any variable/object created explicitly. The output will print the object type of one-dimensional array “a” as: Similarly, 2-d & 3-d arrays can be initiated using the below commands: 2-D NumPy Arrays b = np.array([(1.5,2,3), (4,5,6)], dtype = float) Here “dtype” explicitly specifies the data type of the 2-d array as “float.” The output of print (b)and type(b)will be as follows: 3-D NumPy Arrays c = np.array([[(1.5,2,3), (4,5,6)], [(3,2,1), (4,5,6)]], dtype = float) Here “dtype” explicitly specifies the data type of the 2-d array as “float.” The output of print(c) and type(c)will be as follows: Example #2 – Arithmetic Operation over NumPy Arrays Let’s initialize one-dimension arrays down below: x = np.array ( [5,6,7] ) y = np.array ( [2,3,8] ) NumPy array subtraction operation follows the usual mathematical syntax as mentioned below. If we want to subtract array “y” from array “x”, then it’s written as: Result = x - y Use print(Result) to print the resultant array “Result.” An alternative to the above approach is to make use of the “subtract” attribute from the NumPy Module & store the resultant array in “Result” like below: Result = np.subtract(x,y) NumPy array addition operation also follows a similar mathematical syntax as discussed earlier in the case of subtraction. If we want to add array “y” to “x”, then it’s written as: Result = x + y Use print(Result) to print the resultant array “Result.” An alternative to the above approach is to make use of the “add” attribute from the NumPy Module & store the resultant array in “Result” like below: Result = np.add(x,y) If we want to divide array “x” by “y”, then you can write it as:: Result = x/y Use print(Result) to print the resultant array “Result.” An alternative to the above approach is to make use of the “divide” attribute from the NumPy Module & store the resultant array in “Result” like below: Result = np.divide(x,y) If we want to multiply array “x” with “y”, then you can write it as:: Result = x * y Use print(Result) to print the resultant array “Result.” An alternative to the above approach is to make use of the “multiply” attribute from the NumPy Module & store the resultant array in “Result” like below: Result = np.multiply(x,y) Square root Sine & Cosine Example #3 – Transforming NumPy Arrays Operations such as subsetting, slicing, and boolean indexing can be applied to NumPy arrays. Fetching a single element out of an array by using the indices. The index in NumPy arrays starts from 0. a = np.array( [4,6,9] ) To fetch the very first element of array “a,” you can write it as: This will return the very first value, which is 4. Let’s initialize a 2-D array. a = np.array([(1,2,3), (4,5,6)], dtype = int) To fetch the 2nd value from the first row of a 2-D array, you can write it as: This will return the value 2. NumPy Arrays can be sliced in multiple ways. Some of these are as follows: a = np.array( [4,6,9] ) If we want to fetch the first two elements of an array, you can write it as:: Here, the catch is that a[x:y] • x represents the index from where you need to fetch the elements. • Whereas y represents “the number of elements in that array to be fetched.” so the result of a[0:2] will be [4,6] Boolean Indexing It enables us to index a NumPy array based on a logical conditional. For example, return all the values less than 2 in an array. a = np.array( [4,1,9] ) The same will be implemented as: The output of this logical indexing will be any value within the array “a” that is less than 2 so the result will be [1] Advantages and Disadvantages of Numpy Advantages Disadvantages Efficient Array Operations: NumPy provides highly efficient array operations, making numerical Learning Curve: NumPy’s array-oriented approach may require some time to grasp fully for newcomers computations faster and memory-efficient. to Python. Multi-Dimensional Arrays: NumPy supports multi-dimensional arrays, simplifying tasks involving Limited Data Types: NumPy arrays are homogeneous, which can be restrictive when dealing with mixed matrices, images, and higher-dimensional data structures. data types. Mathematical Functions: A rich library of mathematical functions, including basic arithmetic, Memory Consumption: NumPy can be memory-intensive for very large arrays, which may lead to linear algebra, and statistics, is available in NumPy. performance issues on systems with limited memory. Interoperability: NumPy seamlessly integrates with other Python libraries like SciPy, Pandas, and Compatibility Issues: Occasionally, updates or changes in NumPy versions can introduce compatibility Matplotlib, enhancing its utility for data analysis and visualization. issues with existing code. Cross-Platform: NumPy is open-source and compatible with various operating systems, making it Not Suitable for All Tasks: While NumPy excels in numerical computing, it may not be the best choice suitable for cross-platform development and research. for all Python programming tasks, such as web development. Community Support: NumPy has a large and active community of users and contributors, ensuring Performance Trade-offs: While NumPy offers significant performance benefits, there can be trade-offs continuous development and support. between performance and ease of use, especially for complex operations. Python’s NumPy is a crucial tool for efficient numerical and scientific computing. Its optimized array operations, multi-dimensional arrays, and extensive mathematical function library make it valuable. It also integrates well with other libraries and works across platforms. However, newcomers may encounter a learning curve and not handle mixed data types effectively. Nonetheless, NumPy equips Python with the capabilities to tackle high-performance numerical tasks. Q1. What are NumPy arrays, and how are they different from Python lists? Ans: NumPy arrays are homogeneous, multi-dimensional data structures with efficient numerical operations. They differ from Python lists, which can store heterogeneous data types and don’t provide the same level of performance for numerical computations. Q2. What are some common applications of NumPy? Ans: NumPy has many applications, including scientific research, data analysis, machine learning, and numerical simulations. It plays a crucial role in data manipulation, statistical analysis, image processing, and more tasks. Q3. Can I use NumPy with other Python libraries? Ans: Yes, NumPy integrates seamlessly with various Python libraries like SciPy (for scientific computing), Matplotlib (for data visualization), Pandas (for data manipulation), and scikit-learn (for machine learning). This integration forms the foundation of Python’s data science ecosystem. Recommended Articles We hope that this EDUCBA information on “What is NumPy in Python” was beneficial to you. You can view EDUCBA’s recommended articles for more information.	{"url":"https://www.educba.com/what-is-numpy-in-python/","timestamp":"2024-11-05T21:49:23Z","content_type":"text/html","content_length":"334323","record_id":"<urn:uuid:57c5bcdc-1f6f-4e2a-8fe4-175c47e27c37>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00012.warc.gz"}
SPM Mathematics Tips sBrief IntroductionMathematics (is also known as Modern Maths) and Additional Mathematics are categorized as thinking subjects which you take your time thinking rather than memorizing facts. The formula will be given at the front page of the examination paper. Additional Maths is notably 4 times harder than Modern Maths. It’s like if students think Maths is shit, then take Add Maths. If you get 90% marks in Maths, you may have potential to get 40% marks in Add Maths. It’s near to impossible for someone who get A1 in Add Maths fail in Math. Well it’s not important. What I’m going to discuss at here is Mathematics.SPM MathematicsIn SPM, mathematics is considered as compulsory subject. You must pass this subject in order to get the certificates along with Malay language. Malay language test takes a lot of time and your hand hurts like hell after writing so many words. It’s not simple either. However, the percentage of students fail in Mathematics is higher than Malay Language in real SPM, which is ridiculous. It is told that Mathematics is quite ‘difficult’ among Art students in seminar SPM, which I can't believe my eyes either. In my opinion, Mathematics is the easiest subject in SPM for Science students. I believe PMR students can even do with their PMR standard knowledge because some questions like volume of solid and area and perimeter of a circle being test in SPM. It’s not being taught in SPM therefore you can only depend on your past knowledge in Form 2. Although the questions are a lot easier than Add Maths, There are still some pupils having difficulties in Maths. I see some students leave the 'haven't done' answers after they get the formulae to be applied in the questions, which they didn’t really answer the questions. Some of them don’t understand the questions’ needs. Therefore, exercises and guidance from teachers are essential for improvement. Based on SPM Maths, it consists of 2 papers Paper 1 - 1 ¼ hors - 40 questions - 40 marks Paper 2 - 2 ½ hours - working steps must be done - Section A = 52 marks = 11 questions - Section B = 48 marks = 5 questions (answer 4 of out 5) Every chapter will be tested in Paper 1. The questions will be tested from Form 1 until Form 5. Since you have already mastered the some basics in PMR, these shouldn’t be a problem. Do past year objective questions will strengthen your basic. In paper 2, there are only certain chapters that will be tested in SPM maths. That’s why I will concentrate more on paper 2. The whole paper 2 format based on past year questions in random order with brief tips below Confirmed to be tested every year Paper 2 (Section A) 1) Volume of solids (which was taught in form 2) - most of the formulae given are for this question - make sure the formulae are used correctly - combined solid is sum of volume of the solids - remaining solid is the subtraction of bigger solid and smaller solid 2) Angles of elevation and depression -using trigonometry rule 3) Mathematical reasoning - statement ‘and’ and ‘or’ - If p, then q. If q, then p. 4) Simultaneous linear equation - equalize the same unknown and take out Example : 4p - 2q = 15 (2p + 4q = 19) x2 (times 2 to make unknown 4p same) - don’t use Add Math method, answers will be different. 5) The straight line - y = mx + c - gradient, gradient…. - What is c? c (y-intercept) is the point where the line touches the y-axis. 6) Quadratic expressions and equations - ‘solve the equation’ is to find the unknown - change the equation of the question into this form - factorise by your own or using calculator - must state x = ?, x = ?, there will be 2 answers 7) Matrices - will mostly ask inverse matrix - if you’re not sure about the answer, check if the whole outcome is identical to the inverse matrix in the question. - the answer in a) is related to b) 8) Area and perimeter of circles (same as number 1) - 2πr (perimeter or circumference) and πr2 (area) - (angle/360 degree) x the formulae above for the area or perimeter of certain part 9) Probability - is more of (number / total number). - the formulae given are plain useless - final answer is never less than 0 or more than 1, 1 > answer > 0 10) Gradient and area under a graph - based on speed/ time graph = distance is area of the graph = rate of change of velocity is gradient of the graph = speed is based on the graph (seldom being asked) - beware of total distance and total speed Can be varied each year 1) Sets (2004 and 2006) - shading based on the questions (intersection or union) - beware of the complement such as A’ 2) Graph of functions (2005 and 2007) - the inequalities (upper line is >, lower line is <) - if slanted line, you can imagine it into horizontal straight line. Same concept as above. What’s the conclusion? 2008 will be ‘Sets’! Wow, I can predict what will come out in this year’s SPM. :D Paper 2 (Section B) Answer 4 questions 1) Graph of functions - Linear functions (less likely to come out as it’s a straight line after plotted) - Quadratic functions (2004 and 2005) - Cubic functions (2007) - Reciprocal functions (2006) = this question is easy because you can detect the type of graph after you plot it. = elastic ruler is recommended = x-axis and y-axis should be stated in the graph 2) Transformation - Translation - Reflection - Rotation - Enlargement ( careful with the word ‘to’ and ‘from’ because it determines the image would be smaller or bigger) = combination of transformation like VT -do T before V. It's some kind of law. 3) Statistic - Histogram ( 2004 and 2005) = x-axis the upper boundary with an additional lower boundary at the front of the graph = y-axis is frequency - Frequency Polygon (2006) = x-axis is midpoint = y-axis is frequency - Ogive (2007) = x-axis is upper boundary = y-axis is cumulative frequency = additional upper boundary should be added to the table = x-axis and y-axis should be stated in the graph My school hasn’t come to these chapters below, but I will try my best to explain it. 4) Plans and elevations -well I’m not sure but what I can see is you must be able to imagine the solid from every side -plan is looked from above -elevation is looked from the side of the solid -the length and the edge (ABCD) should be stated correctly -similar to a chapter of the living skills in PMR 5) Earth as a sphere -no comment because I’m not going to explain something that I’m not sure -latitude is vertical and longitude is horizontal of the sphere -nautical mile If you concentrate on these chapters and do a lot of past year questions, I’m sure getting A1 in Maths is not a matter for you. Choose the 4 questions that you’re confident in section B paper 2. Do all questions if you think you have much time to spend for taking a nap. Why study more if you know what kind of questions will come out? Additional Tips 1) You know doesn’t mean you can get correct, try to make less careless mistakes in each paper. 2) See through the questions word by word as there will be some tricky part in the questions, especially ‘to’ and ‘from’ in enlargement. 3) Don’t be sad if you do badly in trial SPM because the odd might come to you in SPM. 4) Be sure to study smart, not study hard. Hope this helps. No comments:	{"url":"https://pinkexia.blogspot.com/2009/04/spm-mathematics-tips.html","timestamp":"2024-11-07T13:50:22Z","content_type":"text/html","content_length":"138535","record_id":"<urn:uuid:996aab7f-3bd8-4375-ac13-f57d96d896ae>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00239.warc.gz"}
PID Controllers: Manual Tune Procedures Process Control Tech Note 01 - TNPC01 This document explains two different procedures for manually tuning PID controllers. All PID Controllers Use Case: Manually Tuning PID Controllers There are a variety of ways to tune PID loops, this document describes two different manuak tuning procedures. PROCEDURE 1: Simple Manual Tune Procedure with Very Little or No System Oscillation During the Tune Procedure • Lower the derivative value to 0, we will not change this value from zero after this first step. • Lower the integral value to 0, easy second step. • Raise the proportional value 100.0 • Increase the integral value to 100 • Slowly lower the integral value and observe the system’s response. • Since the system will be maintained around setpoint, change setpoint and verify if system corrects in an acceptable amount of time. If not acceptable or you would like a quick response, continue lowering the integral value. • If the system begins to oscillate again, record the integral value and raise value to 100. Just like me, you got a little greedy trying to get the quickest response. • After raising the integral value to 100, return to the proportional value and raise this value until oscillation ceases. • Lower the proportional value back to 100.0 and then lower the integral value slowly to a value that is 10% to 20% higher than the recorded value when oscillation started. (recorded value times 1.1 or 1.2) Change the setpoint and watch as the system tracks quickly and efficiently. If you experience an overshoot that is not desirable, consider using the setpoint ramp parameter. It is most useful at system start-up or when a large setpoint change is introduced during system operation. PROCEDURE 2: Utilizing the System Oscillation to Determine Optimum Proportional and Integral Values. During this procedure, you are going to locate the ultimate gain value utilizing the proportional value only. Then you will introduce error correction with the integral value. As you can see above, the rate of change, or derivative value, may be more of a nuisance. You will have to move back and forth between the parameters, mostly proportional and integral, with an occasional setpoint change as we manually tune the unit. This may seem as a tedious process to complete a manual tune of a quick responding system, but surprisingly you will complete the process in a reasonably short time period with acceptably tight control results. • Lower the derivative value to 0, we will not change this value from zero after this first step. • Lower the integral value to 0, easy second step. • Raise the proportional value to a high value, I often use 150.0 • Change the setpoint value to develop a difference between actual process value and setpoint value. • Lower the proportional value slowly. You may see some correction but there will always be a difference between the process value and the setpoint value you programmed (steady state error). • As you lower the proportional band slowly, you increase the risk of initiating a system oscillation. If oscillation becomes large and is not acceptable, record the proportional band value and then raise the band value until oscillation ceases. • Since the proportional band has been raised to stop the system oscillation, this is a good time to raise the integral value to 100. Eventually we will use the integral value for error correction. • Return to the proportional band and lower the value slowly until a value that is double of the earlier recorded value is reached. • Now lower the integral value slowly and the error between setpoint value and process value will decrease. • Since the system will be maintained around setpoint, change setpoint and verify if system corrects in an acceptable amount of time. If not acceptable or you would like a quick response, continue lowering the integral value. • If the system begins to oscillate again, record the integral value and raise value to 100. Just like me, you got a little greedy trying to get the quickest response. • After raising the integral value to 100, return to the proportional value and raise this value until oscillation ceases. • Lower the proportional value to the double value used earlier in the set-up and then lower the integral value slowly to a value that is 10% to 20% higher than the recorded value when oscillation started. (recorded value times 1.1 or 1.2) Change the setpoint and watch as the system tracks quickly and efficiently. If you experience an overshoot that is not desirable, consider using the setpoint ramp parameter. It is most useful at system start-up or when a large setpoint change is introduced during system operation. It is the customer's responsibility to review the advice provided herein and its applicability to the system. Red Lion makes no representation about specific knowledge of the customer's system or the specific performance of the system. Red Lion is not responsible for any damage to equipment or connected systems. The use of this document is at your own risk. Red Lion standard product warranty Red Lion Technical Support If you have any questions or trouble contact Red Lion Technical Support by clicking here or calling 1-877-432-9908. For more information: http://www.redlion.net/support/policies-statements/warranty-statement	{"url":"https://support.redlion.net/hc/en-us/articles/360017689451-PID-Controllers-Manual-Tune-Procedures","timestamp":"2024-11-08T21:13:28Z","content_type":"text/html","content_length":"34805","record_id":"<urn:uuid:69db4fec-8016-4600-b977-fd62d5108e2b>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00246.warc.gz"}
\[ \newcommand{\one} {\textbf{1}} \newcommand{\zero} {\textbf{0}} \newcommand{\limp} {\mathbin{-\hspace{-0.70mm}\circ}} \newcommand{\iimp} {\supset} \newcommand{\riimp} {\Leftarrow} \newcommand{\ bang} {\mathop{!}} \newcommand{\with} {\mathbin{\&}} \newcommand{\tensor} {\otimes} \newcommand{\adisj} {\oplus} \newcommand{\maction}[1] {\mathcolor{blue}{#1}} \newcommand{\action}[1] {\textcolor {blue}{#1}} \] Apilog is a logic programming language for documenting and testing RPC interfaces. The language has three fundamental syntactical categories: types, expressions, and actions. In addition, it has syntax for defining predicates, APIs, issuing commands and more. Rather than giving a detailed grammar for all the syntax up front, this document will take you on a tour around the main concepts of Apilog, explaining some of the syntactical elements on the go. It is work in progress and currently lacks a description how to define data types as well as a deeper description of the semantics of the language. Type system¶ The type system of Apilog is fairly standard and behaves like most polymorphic type systems for functional programming. There are function types, written using the function arrow syntax (->) and there is a type for logical formulas, called prop. Then there is a bunch of simple data types such as string, int, etc. There are also parameterised types besides functions types. For example, the type of lists of integers is written list int. Moreover, since the type system is polymorphic, types can contain variables, written using capitalized identifiers. For example, the type of a predicate that checks for list membership can be written A -> list A -> prop. Finally, specification authors can declare new types, as we shall see later. Expressions are similar to expressions in functional languages. Function application is written f x. The syntax for lambda abstraction is a bit unusual, however. It is written X\ e, where X is the bound variable and e is some expression. The action language is used to describe interactions with an API. It contains syntax for sequencing actions, running actions in parallel, making remote procedure calls, and testing the result of actions. However, apilog specifications can only directly use the syntax for making remote procedure calls. The rest of the action language is only used to describe the actions that Apilog took during testing, and can be shown in the user interface, but not written in the specification itself. Morever, even though actions is a separate syntactical category from expressions, actions can be wrapped as expressions (using a special expression syntax element). This wrapping makes it possible to define short-form constants that describe actions, and include actions in formulas. Specifications will mostly use short-form action expressions from the Apilog library, some of which are: get : path -> list (tuple string string) -> action http_response post : path -> list (tuple string string) -> string -> action http_response put : path -> list (tuple string string) -> string -> action http_response delete : path -> list (tuple string string) -> action http_response The path arguments represent URI paths, the lists of tuples represent HTTP headers and the string arguments represent HTTP request bodies. Formulas and terms¶ The connectives in the Apilog logic are encoded as expression constants, which means that logical formulas are simply expressions (of type prop). There are certain restrictions in place that prevent logical formulas from quantifiying over logical formulas, in order to make the logic first-order. For this reason we must also talk about terms, the subset of expressions that does not contain Resource declarations¶ The concept of a resource is central to Apilog. A resource represents something that exists in the system behind the API, and that can typically be manipulated via the API. Resources are encoded as logical predicates (applied to all their arguments). New such resource constants can be declared with the resource keyword, followed by a name and a type. Here is an example: resource user : string -> prop. The above declaration says that there is a predicate constant called user with a string argument, and that this predicate (when applied to its argument and used in a logical formula) represents a resource. In this example, the string argument could represent the user name. Positive formulas¶ Positive formulas are used to generate data as well as to express pre- and post-conditions. They follow this abstract grammar: \[ \begin{array}{lll} A^+ ::= P \mid A^+ \tensor A^+ \mid \one \mid A^+ \adisj A^+ \mid \zero \mid \exists x.A^+ \mid t = t \mid \mu B \vec{t} \end{array} \] where $P$ stands for atoms and $t$ for terms. An atom is a non-logical constant applied to all its arguments. As such, it is a formula that does not contain any logical constants. There are two kinds of atoms: resource atoms and built-in atoms. Resource atoms¶ In a resource atom, the constant must have been declared as a resource. For example, if a constant user : string -> prop has been declared, user "x" would be a resource atom, which could represent the assumption that the user named "x" exists in the system behind the API. Built-in atoms¶ In a built-in atom, the constant refers to a procedure that is built-in to the Apilog interpreter. An example of such a constant is parse_json : string -> json -> prop. Multiplicative conjunction¶ $\tensor$ is the multiplicative conjunction connective from linear logic. It can be thought of as a version of "and" that can represent the simultaneous occurence of resources. In the concrete syntax, it is written , (comma), like conjunction in Prolog. Also like in Prolog, interpretation proceeds left-to-right. $\one$ is the unit of $\tensor$ from linear logic, and is written one in the concrete syntax. It represents the absence of resources and can be thought of as a version of "true" from more standard logics. Additive disjunction¶ $\adisj$ is the additive disjunction connective from linear logic. It can be thought of as a version of "or" that can represent non-deterministic choice between resources. It is written in the concrete syntax with ';' (semi-colon), like disjunction in Prolog. When the Apilog interpreter finds this connective in a goal (pre-condition), it makes a randomized choice of which branch to try first, and performs chronological backtracking upon failure. It does not try to find multiple solutions. When the Apilog intepreter finds this connective in a hypothesis (post-condition), it tries the first branch and then the second. If there is more than one solution, it yields a "fatal error", indicating an error in the specification. In effect, specifications are required to be deterministic. $\zero$ is the unit of $\adisj$ from linear logic and is written zero in the concrete syntax. It represents a resource that cannot be produced and can be thought of as a version of "bottom" or "false" from more standard logics. Existential quantification¶ $\exists x.A^+$ is existential quantification. In concrete syntax, it can be written using the built-in constant exists : (A -> prop) -> prop. For example: exists (X\ e) where e is some expression typically containing X. The quantified variable must be of term type. $=$ is term equality, written = in concrete syntax. Positive formulas can be recursive using the the fixed point combinator $\mu$. In the surface syntax, this combinator is not available. Instead, recursion is achieved by making a definition that refers to itself. This self-reference is then translated into an application of the fixed-point combinator. Negative formulas¶ Negative formulas are used to describe APIs. They follow this abstract grammar: \[ \begin{array}{lll} A^- ::= A^- \with A^- \mid \top \mid \forall x.A^- \mid A^+ \limp A^- \mid \langle a \rangle x.A^+ \end{array} \] Additive conjunction¶ $\with$ is the additive conjunction connective from linear logic. It is used to represent a (client-side) choice between APIs. It can be thought of as a version of "and" from more standard logics, and is written & in the concrete syntax. $\top$ is the unit of $\with$ from linear logic and is written top in the concrete syntax. It is used to represent the empty API and can be thought of as version of "true" from more standard Universal quantification¶ $\forall x.A^-$ is universal quantification. In concrete syntax, it can be written using the built-in constant forall : (A -> prop) -> prop. For example: forall (X\ e) where e is some expression typically containing X. The quantified variable must be of term type. Linear implication¶ $\limp$ is the linear implication arrow from linear logic, written -o in the concrete syntax. It can be seen as a resource-aware version of logical implication. The antecedent (the left argument) must be a positive formula, which can contain resources, and the consequent (the right argument) must be a negative formula. When a linear implication is used to prove its consequent, the resources in the antecedent are consumed. Action modality¶ $\langle a \rangle x.A^+$ is a modal formula that means that the formula $[t/x]A^+$ is true after action $a$ has happened, if term $t$ is the result of $a$. The concrete syntax is: where a is an action of type action r, and f a function of type r -> prop (more precisely, a function from the result of the action to a positive formula). For example: {get /users/U _ _ _}(R\ status 200 R, user U) Resources in the formula returned by f (like user U in the above example) are added to the set of assumptions about the system under test, after the action a is performed, during testing. Note that status 200 R in this example is not a resource atom, but a defined atom (which will be explained later). As alluded to earlier, Apilog allows definitions of contants. A definition is a top-level statement that declares a new constant and assigns it a logical formula. Internally, such atoms are substituted by their definitions which explains why they are not present in the abstract syntax (as presented in the sections on positive and negative formulas). There are two kinds of definitions: API definitions and predicate definitions. API definitions¶ API definitions are made with the api keyword, like in the following example: api get_user := user U -o {get /users/U _ _ _}(R\ status 200 R, user U). The right hand side (right of :=) is always a negative logical formula. Variables and quantification¶ Any unbound and capitalized identifier in the formula (like U in the example) is implicitly universally quantified. I.e. there is a hidden forall- quantifier for U at the start of the formula in the example. Also, note that in the example above, the variable R is not implicitly quantified as it is bound as an argument to an anonymous function. API clauses¶ In the example above, the formula has a single application of the action modality. Such a formula we call an API clause. We call the modal formula the head of the clause and the antecedent of the linear implication the body of the clause. Combining APIs¶ APIs can be combined with the & connective. I.e. one can write API definitions that join multiple APIs together like this: api main := get_user & delete_user. One can also write multiple API clauses in one definition: api main := user U -o {get /users/U _}(R\ status 200 R, user U) & user U -o {delete /users/U _}(R\ status 200 R). API clause restrictions¶ There are two restrictions on API clauses: they must not overlap and they must have atomic actions. Non-overlapping API clauses¶ Overlap between API clauses occur when there are two or more clauses with actions that can be "unified" i.e. which can be made equal using a specific substitution of variables. If this occurs, Apilog complains with an error message. Note that the body of the clause has no effect on whether the heads actions are considered overlapping or not. The unification of head actions is attempted without looking at the bodies. This restriction is similar to the behaviour of type class instance resolution in Haskell. Atomic actions¶ Actions in API clauses must be atomic rpc actions. If not, Apilog complains with an error message. There are two reasons for these two restrictions. 1) If API clause are non-overlapping and contain only atomic actions, Apilog never needs to consider more than one API clause in its proof search after having made a remote procedure call. 2) With these restrictions, the actions serve well as documentation headings in the generated API documentation, because the resulting headings are guaranteed to be unique, and likely to be short and Predicate definitions¶ Predicates can be defined by clauses, with def-by syntax. A type must be declared and the clauses must be separated by vertical bars (\|). Here is an example of a predicate that checks for list membership: def elem : A -> list A -> prop by \| elem A [A\|_] \| elem A [_\|L] := elem A L. The expression to the left of := is called a head and the expression to the right a body, the latter which can be omitted. The head must be an atom with the same predicate constant that is defined (elem in this example). The body can be any positive formula. This clause-by-clause definition style mimics horn clauses. The same predicate could look like this in Prolog: elem(A, [A\|_]). elem(A, [_\|L]) :- elem A L. Clause-by-clause definitions is just a form of syntactial sugar. The Apilog interpreter always "desugars" the clauses into a function from the arguments of the predicate to a single positive logical Existential quantification¶ All unbound capitalized identifiers in the clauses are implicitly existentially quantified. A def-by definition can refer to itself, i.e. it can be recursive. However, Apilog does not yet support mutual recursion. Commands are top-level statements prefixed with #, and cause things to happen. In order to run tests, for example, you need to add a #check command (see below) to you specification. #baseuri e. where e is an expression of type string. Sets the base URI that will prefix each path during testing. #baseuri "https://httbin.org". #check e. where e is an expression of type prop, i.e. a formula. The formula must be negative (i.e. an API formula). Runs tests against the system at the base URI. In order to generate tests, Apilog does proof search. This search is set up so that the positive API formula in the argument to the command is put in the unrestricted context of the hypothetical judgement to be proved, which means that Apilog is allowed to use the API formula any number of times (i.e. it is not restricted to be used linearly). Any capitalized and unbound identifier in e is implicitly universally quantified. #check {get /foo _ _}(R\ status 200 R). #check api1 & api2. #check e. where e is an expression of type prop, i.e. a formula. The formula must be positive. Does proof search with the goal e. Any capitalized and unbound identifier in e is implicitly existentially quantified. #query append "foo" "bar" S.	{"url":"https://apilog.net/docs/reference/","timestamp":"2024-11-09T00:56:43Z","content_type":"text/html","content_length":"46655","record_id":"<urn:uuid:8d65b7bf-ae44-4edf-8d21-7c4e7d624a5b>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00766.warc.gz"}
Colored solutions of the Yang-Baxter equation from representations of U<sub>q</sub>gl(2) We study the Hopf algebra structure and the highest weight representation of a multiparameter version of U[q]gl(2). The Hopf algebra maps of this algebra are explicitly given. We show that the multiparameter universal R matrix can be constructed directly as a quantum double intertwiner without using Reshetikhin's twisting transformation. We find there are two types highest weight representations for this algebra: type a corresponds to the qeneric q and type b corresponds to the case that q is a root of unity. When applying the representation theory to the multi-parameter universal R matrix, both standard and nonstandard colored solutions of the Yang-Baxter equation are obtained. Dive into the research topics of 'Colored solutions of the Yang-Baxter equation from representations of U[q]gl(2)'. Together they form a unique fingerprint.	{"url":"https://scholars.ncu.edu.tw/en/publications/colored-solutions-of-the-yang-baxter-equation-from-representation","timestamp":"2024-11-11T05:08:44Z","content_type":"text/html","content_length":"53395","record_id":"<urn:uuid:7d572424-7c5c-4f7f-846a-8ce82285de2e>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00877.warc.gz"}
Detection and Identification for Void of Concrete Structure by Air-Coupled Impact-Echo Method School of Mechanical Engineering, Shijiazhuang Tiedao University, Shijiazhuang 050043, China School of Material Science and Engineering, Shijiazhuang Tiedao University, Shijiazhuang 050043, China School of Safety Engineering and Emergency Management, Shijiazhuang Tiedao University, Shijiazhuang 050043, China Author to whom correspondence should be addressed. Submission received: 10 April 2023 / Revised: 23 June 2023 / Accepted: 27 June 2023 / Published: 29 June 2023 In the field of non-destructive testing (NDT) for concrete structures, the traditional air-coupled impact-echo technology often has the problems of complex operation and low efficiency. In order to solve these problems, this study uses Comsol software to establish a finite element model (FEM) of the concrete structure with different void sizes and obtains the variation rule of peak frequency. The recognition property of the concrete void based on peak frequency is proposed, which is explained and validated by relevant theory and experiments. The results show that compared with the depth of the void, the influence of the void width on the peak frequency increases significantly. When the void width is greater than 0.3 m, the peak frequency of the sound wave decreases with the increase in the width, and the change is obvious. This paper describes the applicability of concrete void depth less than 0.4 m for the air-coupled method and, when the concrete void depth is less than 0.4 m, the peak frequency can be used to effectively identify void widths greater than 0.3 m. The research results will be beneficial to void detection of concrete structures such as tunnel lining and 1. Introduction Concrete structures have the characteristics of high strength, good ductility and convenient construction, so they are widely used in bridges, high-rise buildings, tunnel linings and other structures. However, due to the long-term effect of site environment and other factors, different kinds of void will be produced in the concrete, which affect its strength [ Researchers use a variety of NDT techniques to detect and identify concrete damage. Tian et al. [ ] used the impact-echo method to detect and identify the void defect of a CA mortar layer in CRTS Ⅱ ballastless track and proposed the identification parameters. Yang et al. [ ] used the method of Burg power spectrum to solve the void imaging problem of ballastless track. Zhang et al. [ ] used the ultrasonic method to test the filling amount of concrete voids for a steel–concrete bridge and proposed the layout type of the testing device, the basis and criterion of data processing and the layout principle of measuring area and testing point. Liu et al. [ ] put forward the identification method of the equivalent diameter and area and the corresponding identification formula, according to the basic principle of ultrasonic detection technology and the characteristics of concrete-filled steel tube damage. Ding et al. [ ] used the large-scale simulation test of a concrete-filled steel tube (CFST) to establish the force–optical constitutive relationship and the quantitative algorithm of voids and cracks in the actual engineering, identifying and evaluating the void damage of concrete structures effectively. Chen et al. [ ] constructed a CNN to classify and identify the location and size of voids in a CA mortar layer. Lesicki et al. [ ] identified spectra of concrete sliding damage by the impact-echo method and found that microstructural changes such as crack formation and debonding between separate phases in concrete are responsible for changes in the nonlinear parameter which can be tracked using the NIRAS testing technique. Bodnar et al. [ ] proved that the void damage of concrete can be identified under low energy by the infrared thermal mapping technique. Yang et al. [ ] used ground-penetrating radar to detect the ballastless track of a high-speed railway, analyzed the difference between steel echo and void echo and identified the void damage. The above NDT methods have some limitations, e.g., the signals all belong to contact acquisition, and the sensitivity of different signals needs to be further studied. In recent years, with the continuous development of acoustic theory, the air-coupled sensor, as a non-contact sensor that can accept the acoustic wave information in the air, has been gradually applied to the research on the damage identification problems of civil engineering structures and materials. According to the different characteristics of the tested object, the frequency response range of the acoustic sensor can be selected from 50 to 1000 kHz. Among them, as a common air-coupled sensor, a microphone’s frequency response range is usually between 0 and 35 kHz, so it can be used to pick up the low-frequency sound wave leaked into the air by the concrete structure. Due to its advantages such as light volume and low price, it has been applied to the field of NDT by many scholars. Zhu et al. [ ] used a microphone to acquire the acoustic signal of a concrete structure and found that inexpensive microphones are very effective to locate structural damage. Oh et al. [ ] used microphones to identify shallow damage of concrete structures and optimized the imaging scanning system of shallow damage of bridge panels. Sun et al. [ ] analyzed the acoustic signals acquired by microphones and proved that a ball chain, as an excitation method, could identify the void damage of concrete. Zhang et al. [ ] analyzed signals of a microphone and designed a damage identification system for concrete bridge panels. Liu [ ] made a comparison of the signal of an accelerometer and the signal of a microphone and proved that the air-coupled method can be used to identify void damage in concrete–steel structures. Shin et al. [ ] found that a dynamic microphone successfully captures impact-echo signals in a contactless manner without acoustic shielding. Near-surface delaminations in the concrete slab were clearly identified, for which the obtained results are equivalent to those results obtained with a high-sensitivity sensor. Kim et al. [ ] made use of a microphone to pick up Rayleigh waves to identify and evaluate the surface damage of a concrete structure. Dou et al. [ ] proved that the fundamental frequency increases with the void depth by modal analysis, which provides theoretical support for the air-coupled method. Peng et al. [ ] used acoustic frequency to identify the concrete void area and location. The above discussion does not include the application range of the air-coupled method in the field of concrete. This paper proposes the method of peak sound frequency to identify concrete voids, in which the voids are located by the threshold of frequency and the void area is estimated by the multi-point scanning method, and the applicability of this method is discussed. The research results of this paper will provide a theoretical basis and method support for detection in concrete structures. 2. Theory 2.1. Theory of Void Damage Model The model of concrete with void damage is shown in Figure 1 . In the figure, it is assumed that the void damage is a cuboid with length , width and depth The vibration of the concrete structure with a void is constrained by the surrounding concrete, and the boundary condition can be similar to the elastic boundary [ ]. Different from fixed boundaries, the boundary condition can produce a turning angle. Different from the simply supported boundary, the boundary condition is affected by bending moment. Such boundary conditions are complex and difficult to define directly. According to vibration theory of thin plates, the formula for calculating the natural frequency of simply supported thin plates on four sides is shown as Formula (1). $ω m n = π 2 D ρ h m a 2 + n b 2$ $D = E h 3 12 ( 1 − υ 3 )$ is the bending stiffness of the thin plate, is the material density, is the geometric thickness of the thin plate, are the length and width of the plate. represent the modal order of the rectangular simply supported plate in its direction. Mitchell et al. [ ] defined the parameters $∆ m$ $∆ n$ corresponding to the order of the modes of thin plates. Thus, the edge effect of the constrained thin plate is considered in detail. $Δ m = a / λ a − m , Δ n = b / λ b − n$ represent the boundary length of the rectangular thin plate, respectively, $λ a$ $λ b$ represent the half-wave length of the thin plate in the longitudinal and transverse mode shapes, respectively. Therefore, the corresponding edge effect coefficients can be used to express Equation (1). The formulas for calculating the natural frequency of quadrilateral constrained rectangular plates considering the boundary effect are as follows: $ω m n = π 2 D ρ h m + Δ m a 2 + n + Δ n b 2$ Mitchell and Hazell, through a series of experiments, concluded that the edge effect coefficient can be found as follows: $Δ m = n a / m b 2 + c − 1 , Δ n = m b / n a 2 + c − 1$ In general, the value of experience coefficient $c$ is set to 2. The research results of Cheng et al. [ ] show that when the depth of defects remains unchanged, the natural frequency of vibration of rectangular defects in concrete structures is mainly determined by the width. Let the four sides of the void have the same length, then a = b . Combined with the bending stiffness of the plate, Equation (3) is further modified to be $ω m n = π 2 D ρ h m + Δ m a 2 + n + Δ n a 2$ Formula (5) can effectively solve the vibration problem of thin plates, but there is error for medium-thickness plates. Zhao [ ] introduced parameter , which could effectively solve the problem of medium-thickness plates. $β = 2.29 h a 1.54 + 1.06$ is the expansion coefficient of the width, the range of the parameter is 1~1.6. Then, Formula (5) is expressed as $ω m n = π 2 D ρ h m + Δ m a β 2 + n + Δ n a β 2$ Combined with the bending stiffness of the plate, Equation (3) is further modified to be $ω m n = π 2 h a 2 E 12 ρ ( 1 − υ 3 ) m + Δ m β 2 + n + Δ n β 2$ In analysis by Equation (8), the natural frequency of the concrete plate is proportional to $h a 2$. 2.2. Acoustic Modal Theory According to the principle of sound and vibration reciprocity, the formula is expressed as follows. $p j F i \| q j = 0 = − x i q j \| F i = 0$ When the force excitation is applied to the point i of the structure, the sound pressure response will be generated at the point $j$, and the frequency response function is obtained in this process. When the volume source excitation is applied to the point j of the structure, the velocity response will be generated at the point i, and the frequency response function is also obtained in this process. The frequency response functions of the two points mentioned above have the same magnitude and opposite directions. For a structure with multiple degrees of freedom, when excited at point , the strain frequency response function of point is expressed as “ $H i j ε$ $H i j ε = ∑ r = 1 n 1 − ω r 2 M r + j ω r C r + K r ⋅ φ j r ⋅ ϕ i r$ $ω r$ is the natural frequency of the structure, is the natural frequency of the structure, $M r , C r , K r$ are the -th modal participation factors of the structure, is the number of modes, $ϕ i r$ is the -th displacement mode of point . Let be the spatial observation point, $r s$ the position of the vibrating element on the surface of the structure and the distance between the two points. $p i ( ω ) = ∫ s j ω ρ 0 v n 2 π R e − j k R d S$ Then, the transfer function of the sound pressure is shown as Formula (12). $H i j p ( ω ) = p i ( ω ) F j ( ω ) = ∫ s j ω ρ 0 v n 2 π R F j ( ω ) = ∫ s j ω ρ 0 e − j k R 2 π R H n j v ( ω ) d S$ $H i j p ( ω )$ is the sound pressure frequency response function obtained in the case of point and point is the circular frequency of the sound wave, $ρ 0$ is the density of the fluid medium, is the vibration velocity of the sound source, is the linear distance between the sound source and the measuring point and is the acoustic transfer constant. is the area of the sound source perpendicular to the direction of sound wave propagation. $H n j v ( ω )$ is the frequency response function of the vibration velocity of each vibration element in the structure when excited at The influence of other modes at this frequency is ignored in -th mode, then Formula (12) is expressed as $[ H i j p ( ω ) ] r = ψ j r m r [ ( ω r 2 − ω 2 ) + 2 j ξ r ω r ω ] ∫ s − ω 2 ρ 0 e − j k R 2 π R ψ n r d S$ Formula (13) is the expression of the sound frequency response function of the r-th mode, which is similar to the expression of the strain mode. 2.3. Quantitative Index In this paper, quantitative indexes are proposed, where is the rate of void area identification. It is used to evaluate the accuracy of void area identification. $k 2 = S ′ − S S × 100 %$ represents the actual void area. $S ′$ represents the void identification area. 3. Finite Element Model (FEM) for Concrete Void 3.1. Parameters of FEM In order to study the acoustic characteristics of voids in the numerical model, the 3D numerical model of a 100 cm × 100 cm × 50 cm concrete slab with void damage is established by Comsol software, and the width of the void is set as and the depth is set as , as shown in Figure 2 a. The acoustic structure interaction unit in the acoustic module is selected for the model, the boundary condition of the air domain is set as the plane wave radiation, the boundary condition of the contact between the concrete domain and the air domain is set as the acoustic–structure interaction of multiple physical fields, the bottom of the concrete model is set as a fixed constraint. The maximum grid in the air domain is set as 10 mm, and the maximum grid in the concrete domain is set as 60 mm. The mode of excitation force is set as point load. The peak value of impact force is set as 3000 N, as shown in Figure 2 Table 1 shows the material parameters of the FEM: 3.2. Sound Field Analysis of the FEM 3.2.1. Analysis for the Influence of Width on Sound Field According to the FEM, the central point of the concrete structure is excited, h = 10 cm is kept unchanged and C is set at 0 cm, 10 cm, 20 cm, 30 cm, 35 cm, 40 cm, 50 cm and 60 cm. The sound field images are captured at 2 ms. Through the analysis of the FEM, it is found that when the width void C < 0.3 m, the sound pressure isosurface changes significantly, but it is not clear enough. As the width of the concrete void is relatively small, it presents the characteristics of a thick plate, and the acoustic modes are relatively complex. When the void width C ≥ 0.3 m, it can be clearly observed that the sound pressure isosurface is similar to a round cake above the concrete structure. As the width of the void increases, the isosurface decreases and the sound pressure value increases. 3.2.2. Analysis for the Influence of Depth on Sound Field According to the FEM, the central point of the concrete structure is excited, c = 50 cm is kept unchanged and h is set at 10 cm, 20 cm, 30 cm, 40 cm. The sound field images are captured at 2 ms. Figure 3 shows that when the void depth increases, the sound pressure isosurface is almost constant, and the sound pressure value of the void is decreased gradually. Compared with Figure 4 , it is found that the influence of the void width on the frequency is much greater than the void depth. 3.3. Frequency-Domain Analysis of the FEM 3.3.1. The Influence of Different Excitation Positions The void width is set as 45 cm × 45 cm and the depth set as 10 cm in the FEM. The microphone is placed 20 cm directly above the excitation point. The excitation points are set as A, B, C, D, E, F, G on the surface of the concrete structure as shown in Figure 3 a, in which the red line is set as the position of the void, the point A is the geometric center for the upper surface of the void and the distance between 7 points is equal. The acoustic signals generated by different excitation points A, B, C, D, E, F and G are analyzed in the frequency domain as shown in Figure 5 Figure 5 a shows that the peak frequencies of the excitation points A, B, C, D and E within the range of the void are 1333 Hz and 1354 Hz, which mainly come from the vibration of the plate above the void. The peak frequencies of the excitation points F and G outside the range of the void is 2375 Hz, which mainly come from the vibration of the concrete structure. 3.3.2. The Influence of Different Void Sizes The microphone is positioned 20 cm directly above the void in the FEM. Data were collected for frequency-domain analysis, as shown in the Figure 6 Figure 6 a–d, it can be found that when the void width C < 0.20 m, the peak frequencies of sound waves are equal. When the void width C ≥ 0.30 m, the peak frequency of the sound wave decreases as the void width increases. Figure 6 e shows that when the void width C < 0.3 m, the peak frequencies of different depths are completely coincident. It indicates that the peak frequency cannot effectively identify the concrete void with a width less than 0.3 m. When the void width C ≥ 0.3 m, the peak frequency decreases as the void width increases, which indicates that the peak frequency represents the mode of vibration of the In analysis based on Formula (8), when the void width C < 0.30 m, the void depth has little influence on the frequency of the sound wave, and there is no obvious correlation between the frequency and the depth h in the range. When the void width C > 0.30 m, the influence of void depth increases gradually. Through observation, it is found that the influence of the void width on the frequency is greater than the void depth, which confirms that the frequency is proportional to $h a 2$, that is, the frequency is inversely proportional to $a 2$ and proportional to h. The result is consistent with Formula (8). 4. Experimental Verification 4.1. Experiment In order to test verify the acoustic properties of concrete structures, three concrete models were made, namely Model A, Model B and Model C, as shown in Figure 7 . The size of the concrete structures is 100 cm × 100 cm × 50 cm, the void is replaced by an air bag, and their sizes were as follows: Model A: 20 cm × 45 cm, = 10 cm; Model B: 35 cm × 35 cm, = 6 cm; Model C: 45 cm × 45 cm, = 10 cm. Grid lines are shown on the surface of the models, the distance between the grid lines is 10 cm, and the grid lines show the actual position of the void. The experiment equipment mainly includes a microphone, Sirius high-speed acquisition instrument, electromagnetic hammer and computer. It is worth noting that the electromagnetic hammer can produce constant force ( Figure 7 4.2. Frequency-Domain Analysis of Experiment The microphone is placed 20 cm directly above the excitation point. The acoustic signal are picked up and the frequency domain signals are analyzed. Figure 8 shows that the peak frequencies of a, b and c in the model are almost identical, and the peak frequency is about 2440 Hz. It represents the overall vibration of the concrete structure, which proves that the microphone cannot effectively identify the void of width C = 0.2 m, which is consistent with the numerical simulation results. Figure 9 shows that the peak frequency of a1 and b1 in the model is 1860 Hz, which can be used as a feature point for void identification. When the point c1 is hit, the concrete mode is complex, so it cannot be used as an effective node for judging the void. Figure 10 shows that when the point a2, in the center of void, is hit, an obvious unimodal shape appears and the peak frequency was 1446 Hz. When point b2 near the edge of the void is hit, two peak frequencies appear in the frequency domain, and the first-order mode frequency value is 1446 Hz, which was generated by the vibration of the plate above the void. When hitting point c2 and point d2, the peak frequency is 2440 Hz, indicating that the frequency is generated by the vibration of the concrete structure. 5. Result Analysis and the Identification Method 5.1. Result Analysis of Experiment and FEM The fundamental frequency of the void in Model B and Model C is peak frequency. The theory, FEM and experiment are compared, refer to Table 2 It is found that data of Model B are closer than that of Model C. This is because the width-to-depth ratio of the plate in Model C is smaller, belonging to the medium-thickness plate, while the width-to-depth ratio of the plate in Model B is larger. Model C is closer to the thin plate. Due to the influence of shear deformation and extrusion deformation, there is error in medium-thickness plate theory and FEM. 5.2. The Method and Effect of Identification The identification method is designed as follows: Step 1: Draw the grid lines, select three nodes without voids, collect acoustic signals with a microphone and calculate the average value of peak frequency $f 1$ as the identification threshold Step 2: Collect acoustic signals of all grid nodes by microphone, record the peak frequency $f$ of each node. Step 3: Calculate $Δ f$ by $Δ f = f − f 1$ and generate a grayscale map. The darker color is the range of the void. Each grid intersection is tapped in turn to pick up the microphone sound pressure signal directly above the point. The values of $Δ f$ are calculated as shown in Table 3 Table 4 Table 3 Table 4 show that the specific location and general shape of the void can be identified through peak sound frequency. When the excitation point is located at the junction of the void area and the non-void area, the accuracy of void identification will be affected. Figure 11 show that the dark part of the image is the identified void range, and the red dotted line is the actual void range. The experiment proves that the acoustic method can effectively identify the void. The parameter is the rate of the void area identification calculated by Formula (14), where $S ′$ is the void identification area. is the actual void identification area. is calculated in Table 5 Table 5 shows that the area recognition rate of Model C by peak frequency is higher than the area recognition rate of Model B, indicating that the void depth is smaller and the area recognition rate is 6. Discussion This paper mainly focuses on the application range of the peak frequency method to detect concrete voids. When the plate above the void is a thin plate, the coefficient $β$ in Formula (8) is approximately equal to 1, and the value of frequency f is determined by $h a 2$. When the plate above the void is a medium plate, the coefficient $β$ will increase. Formula (8) can be used to calculate the depth of the concrete void, and there will be some errors in a medium plate (thin plate and medium plate are determined by the width to thickness ratio). This paper mainly discusses the application range of acoustic peak frequency for a medium plate. The next research work for the research team is to improve the recognition accuracy by a multi-parameter fusion method. 7. Conclusions In this paper, numerical simulation and experiments are described for a void depth of less than 0.4 m. The results show that, compared with the void depth, the influence of the width on peak frequency increases significantly. When the void width is greater than 0.30 m, the peak frequency decreases with the increase in void width, and the change is obvious. It is found that the acoustic peak frequency can effectively judge a concrete void depth of less than 0.4 m by numerical simulation. The method of peak frequency can be used identify a void with a width greater than 0.3 m in a concrete structure. The main engineering value of this study is that the threshold value can be used to quickly judge whether there is a void in a concrete structure through single-point excitation. When multipoint scanning is used, the void range can be quickly estimated. Author Contributions Conceptualization, W.Z.; methodology, J.J.; software, J.J.; validation, J.J.; formal analysis, X.T.; investigation, Y.Y.; resources, W.Z.; data curation, J.J.; writing—original draft preparation, J.J. and X.T.; writing—review and editing, J.J.; visualization, W.Z. All authors have read and agreed to the published version of the manuscript. This research was funded by the National Natural Science Foundation of China (No. U2034207 and 52008272) and the Natural Science Foundation of Hebei Province (No. E2021210090, E2021210099 and Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. Data Availability Statement The data in this study are available on request from the first author or corresponding author. The authors gratefully acknowledge funding from Natural Science Foundation of China and Natural Science Foundation of Hebei. Conflicts of Interest The authors declare no conflict of interest. Figure 2. Finite element model (FEM) of concrete structure with void. (a) The FEM; (b) the impact force. Figure 3. Sound field distribution map at t = 2 ms. (a) c = 50 cm, h = 10 cm; (b) c = 50 cm, h = 20 cm; (c) c = 50 cm, h = 30 cm; (d) c = 50 cm, h = 40 cm. Figure 4. Sound field distribution map at t = 2 ms. (a) h = 10 cm, c = 0 cm; (b) h = 10 cm, c = 10 cm; (c) h = 10 cm, c = 20 cm; (d) h = 10 cm, c = 30 cm; (e) h = 10 cm, c = 35 cm; (f) h = 10 cm, c = 40 cm; (g) h = 10 cm, c = 50 cm; (h) h = 10 cm, c = 60 cm. Figure 6. The void of concrete. (a) The void depth h = 0.1 m. (b) The void depth h = 0.2 m. (c) The void depth h = 0.3 m. (d) The void depth h = 0.4 m. (e) Line chart of peak frequency. Figure 7. Experiment design (a) Model A: h = 10 cm, C = 20 cm; (b) Model B: h = 6 cm, C = 35 cm; (c) Model C: h = 10 cm, C = 45 cm; (d) grid line layout; (e) schematic plot; (f) experiment. Figure 8. Frequency-domain analysis of Model A. (a) Grid lines; (b) peak frequency of excitation points a, b, c; (c) peak sound pressure of excitation points a, b, c. Figure 9. Frequency-domain analysis of Model B. (a) Grid lines; (b) peak frequency of excitation points a[1], b[1], c[1]; (c) peak sound pressure of excitation points a[1], b[1], c[1]. Figure 10. Frequency-domain analysis of Model C. (a) Grid lines; (b) peak frequency of excitation points a[2], b[2], c[2] and d[2]; (c) peak sound pressure of excitation points a[2], b[2], c[2] and d Materials Velocity of Sound (m/s) Density (kg/m^3) Elastic Poisson Ratio Modulus (Pa) Concrete 4000 2500 3.0 × 10^10 0.2 Air 343 / / / f Theory FEM Experiment Model B 1862 Hz 1892 Hz 1860 Hz Model C 1551 Hz 1354 Hz 1442 Hz Δf (Hz) 0 1 2 3 4 5 6 7 3 0 −220 −1320 −1320 −1320 0 0 0 4 0 −220 −1320 −1320 −1320 −1320 −200 0 5 0 −220 0 0 0 −220 0 0 6 0 −220 0 160 −1040 −220 0 0 Δf (Hz) 0 1 2 3 4 5 6 7 8 0 −29 86 −100 −86 −86 −57 86 129 71 1 43 −71 57 −43 71 100 86 −57 −57 2 −43 0 14 −129 57 86 100 −114 114 3 157 129 0 71 −957 −970 57 −43 171 4 −57 −100 −86 −957 −957 −957 −957 −43 −86 5 −43 −57 14 71 −957 −957 143 −100 −71 6 −43 −29 71 86 57 71 −171 71 −57 7 −57 129 86 14 −157 −14 100 100 29 8 14 114 −57 100 29 −14 −157 −143 −176 Width of Void Depth of Void k Model A 0.45 m 0.10 m 39.5% Model B 0.35 m 0.06 m 57.1% Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. © 2023 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 Ju, J.; Tian, X.; Zhao, W.; Yang, Y. Detection and Identification for Void of Concrete Structure by Air-Coupled Impact-Echo Method. Sensors 2023, 23, 6018. https://doi.org/10.3390/s23136018 AMA Style Ju J, Tian X, Zhao W, Yang Y. Detection and Identification for Void of Concrete Structure by Air-Coupled Impact-Echo Method. Sensors. 2023; 23(13):6018. https://doi.org/10.3390/s23136018 Chicago/Turabian Style Ju, Jinghui, Xiushu Tian, Weigang Zhao, and Yong Yang. 2023. "Detection and Identification for Void of Concrete Structure by Air-Coupled Impact-Echo Method" Sensors 23, no. 13: 6018. https://doi.org/ Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details Article Metrics	{"url":"https://www.mdpi.com/1424-8220/23/13/6018","timestamp":"2024-11-12T17:01:48Z","content_type":"text/html","content_length":"470978","record_id":"<urn:uuid:b921f2f8-3da9-40d3-9bd3-5c6f1258f681>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00404.warc.gz"}
Introductory Chemistry – Lecture & Lab 33 The Kinetic-Molecular Theory Learning Objectives By the end of this section, you will be able to: • State the postulates of the kinetic-molecular theory • Use this theory’s postulates to explain the gas laws The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships. The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.) 1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container. 2. The molecules composing the gas are negligibly small compared to the distances between them. 3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls. 4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy). 5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas. The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law. The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows: • Amontons’s law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 1). • Charles’s law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which outweigh those of increased collision forces due to the greater kinetic energy at the higher temperature. The net result is a decrease in gas pressure. • Boyle’s law. If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure • Avogadro’s law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 1). • Dalton’s Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases. Figure 1. (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time. Molecular Velocities and Kinetic Energy The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at velocities and kinetic energies of gas molecules, and the temperature of a gas sample. In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure Figure 2. The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, νp, is a little less than 400 m/s, while the root mean square speed, urms, is closer to 500 m/s. The kinetic energy (KE) of a particle of mass (m) and speed (u) is given by: Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m^2 s^–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square velocity of a particle,u[rms], is defined as the square root of the average of the squares of the velocities with n = the number of particles: [latex]{u}_{rms}=\sqrt{\overline{{u}^{2}}}=\sqrt{\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\dots }{n}}[/latex] The average kinetic energy, KE[avg], is then equal to: The KE[avg] of a collection of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation: where R is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J/K (8.314 kg m^2s^–2K^–1). These two separate equations for KE[avg] may be combined and rearranged to yield a relation between molecular speed and temperature: [latex]\frac{1}{2}{mu}_{\text{rms}}^{2}=\frac{3}{2}RT[/latex] [latex]{u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}[/latex] Example 1: Calculation of u[rms] Calculate the root-mean-square velocity for a nitrogen molecule at 30 °C. Show Answer Convert the temperature into Kelvin: [latex]30^{\circ}\text{C}+273=\text{303 K}[/latex] Determine the mass of a nitrogen molecule in kilograms: [latex]\frac{28.0\cancel{\text{g}}}{\text{1 mol}}\times \frac{\text{1 kg}}{1000\cancel{\text{g}}}=0.028\text{kg/mol}[/latex] Replace the variables and constants in the root-mean-square velocity equation, replacing Joules with the equivalent kg m^2s^–2: [latex]{u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}[/latex] [latex]{u}_{rms}=\sqrt{\frac{3\left(8.314\text{J/mol K}\right)\left(\text{303 K}\right)}{\left(0.028\text{kg/mol}\right)}}=\sqrt{2.70\times {10}^ Check Your Learning Calculate the root-mean-square velocity for an oxygen molecule at –23 °C. Show Answer 441 m/s If the temperature of a gas increases, its KE[avg] increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KE[avg] decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 3. At a given temperature, all gases have the same KE[avg] for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher u[rms], with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower u[rms], and a speed distribution that peaks at relatively lower velocities. This trend is demonstrated by the data for a series of noble gases shown in Figure 4. PhET gas simulator may be used to examine the effect of temperature on molecular velocities. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures. The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates. The rate of effusion of a gas depends directly on the (average) speed of its molecules: [latex]\text{effusion rate}\propto {u}_{\text{rms}}[/latex] Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here: [latex]{u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}[/latex] [latex]m=\frac{3RT}{{u}_{rms}^{2}}=\frac{3RT}{{\overline{u}}^{2}}[/latex] [latex]\frac{\text{effusion rate A}}{\text{effusion rate B}}=\frac{{u}_ The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law. Key Concepts and Summary The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules. Key Equations • [latex]{u}_{rms}=\sqrt{\overline{{u}^{2}}}=\sqrt{\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\dots }{n}}[/latex] • [latex]{\text{KE}}_{\text{avg}}=\frac{3}{2}R\text{T}[/latex] • [latex]{u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}[/latex] 1. Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape. 2. Can the speed of a given molecule in a gas double at constant temperature? Explain your answer. 3. Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows: 1. The pressure of the gas is increased by reducing the volume at constant temperature. 2. The pressure of the gas is increased by increasing the temperature at constant volume. 3. The average velocity of the molecules is increased by a factor of 2. 4. The distribution of molecular velocities in a sample of helium is shown in Figure 9.34. If the sample is cooled, will the distribution of velocities look more like that of H[2] or of H[2]O? Explain your answer. 5. What is the ratio of the average kinetic energy of a SO[2] molecule to that of an O[2] molecule in a mixture of two gases? What is the ratio of the root mean square speeds, u[rms], of the two 6. A 1-L sample of CO initially at STP is heated to 546 °C, and its volume is increased to 2 L. 1. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall? 2. What is the effect on the average kinetic energy of the molecules? 3. What is the effect on the root mean square speed of the molecules? 7. The root mean square speed of H[2] molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N[2] molecule at 25 °C? 8. Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, [latex]\frac{{R}_{1}}{{R}_{2}},[/latex] is the same at 0 °C and 100 °C. Selected Answers 2. Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average velocity of all the molecules is constant at constant temperature. 4. H[2]O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier. 6. Both the temperature and the volume are doubled for this gas (n constant), so P remains constant. 1. The number of collisions per unit area of the container wall is constant. 2. The average kinetic energy doubles; it is proportional to temperature. 3. The root mean square speed increases to [latex]\sqrt{2}[/latex] times its initial value; u[rms] is proportional to [latex]\sqrt{{\text{KE}}_{\text{avg}}}.[/latex] 8. The rate at which a gas will diffuse, R, is proportional lo u[rms], the root mean square speed of its molecules. The square of this value, in turn, is proportional to the average kinetic energy. The average kinetic energy is: For two different gases, 1 and 2, the constant of proportionality can be represented as k[1] and k[2], respectively. Thus, As a result of this relationship, no matter at which temperature diffusion occurs, the temperature term will cancel out of the equation and the ratio of rates will be the same. 1. Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon? 2. Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon? 3. At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air? 4. The average temperature of the gas in a hot air balloon is 1.30 × 10^2 °F. Calculate its density, assuming the molar mass equals that of dry air. 5. The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)? 6. An average balloon has a diameter of 60 feet and a volume of 1.1 × 10^5 ft^3. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo? 7. A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO[2] and H[2]O gas is produced by the combustion of this propane? 8. A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight? Show Answer 1. equal, because the balloon is free to expand until the pressures are equalized 2. less than the density outside 3. assume three-place accuracy throughout unless greater accuracy is stated: [latex]\text{molar mass}=\frac{DRT}{P}=1.2256\text{g}\cancel{{\text{L}}^{-\text{1}}}\times \frac{0.08206\cancel{\text{L}}\cancel{\text{atm}}{\text{mol}}^{-\text{1}}\cancel{{\text{K}}^{-\text{1}}} \times 293.15\cancel{\text{K}}}{1.00\cancel{\text{atm}}}=29.48{\text{g mol}}^{-\text{1}}[/latex] 4. convert the temperature to °C; then use the ideal gas law: [latex]^{\circ}\text{C}=\frac{5}{9}\left(\text{F}-32\right)=\frac{5}{9}\left(130-32\right)=54.44^{\circ}\text{C}=327.6\text{K}[/latex] [latex]D=\frac{\mathcal{M}P}{RT}=29.48\text{g}\cancel{{\text {mol}}^{-\text{1}}}\times \frac{1.00\cancel{\text{atm}}}{0.08206\text{L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\times 327.6\cancel{\text{K}}}=1.0966{\ text{g L}}^{-\text{1}}[/latex] 5. 1.2256 g/L – 1.09966 g/L = 0.129 g/L; 6. calculate the volume in liters, multiply the volume by the density difference to find the lifting capacity of the balloon, subtract the weight of the balloon after converting to pounds: [latex]1.1\times 105{\text{ft}}^{3}\times {\left(\frac{\text{12 in}}{\text{91 ft}}\right)}^{3}\times {\left(\frac{\text{2.54 cm}}{\text{in}}\right)}^{3}\times \frac{\text{1 L}}{1000{\text{cm}}^ {3}}=3.11\times {10}^{6}\text{L}[/latex] 3.11 × 106 L × 0.129 g/L = 4.01 × 10^5 g [latex]\frac{4.01\times {10}^{5}\text{g}}{453.59{\text{g lb}}^{-\text{1}}}=884\text{lb;}\text{884 lb}-\text{500 lb}=\text{384 lb}[/latex] net lifting capacity = 384 lb 7. First, find the mass of propane contained in 40.0 gal. Then calculate the moles of CO[2](g) and H[2]O(g) produced from the balanced equation. [latex]40.0\cancel{\text{gal}}\times \frac{4\left(0.9463\text{L}\right)}{1\cancel{\text{gal}}}=151.4\text{L}[/latex] [latex]151.4\cancel{\text{L}}\times 0.5005\text{g}{\cancel{\text{L}}}^{\cancel {-1}}=75.8\text{g}[/latex] Molar mass of propane = 3(12.011) + 8(1.00794) = 36.033 + 8.064 = 44.097 g mol^–1 [latex]\frac{75.8\cancel{\text{g}}}{44.097\cancel{\text{g}}{\text{mol}}^{-\text{1}}}= The reaction is [latex]{\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)\rightarrow 3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right)[/latex] For each 1.72 mol propane, there are 3 × 1.72 mol = 5.15 mol of CO[2] and 4 × 1.72 mol = 6.88 mol H[2] O. The total volume at STP = 22.4 L × 12.04 = 270 L 8. The total heat released is determined from the heat of combustion of the propane. Using the equation in question 7, [latex]\begin{array}{ll}\Delta {H}_{\text{combustion}}^{\circ}\hfill & =3{\Delta H}_{{\text{CO}}_{2}\left(g\right)}^{\circ}+4{\Delta H}_{{\text{H}}_{2}\text{O}\left(g\right)}^{\circ}-{\Delta H}_ {\text{propane}}^{\circ}\hfill \\ \hfill & =3\left(-393.51\right)+4\left(-241.82\right)-\left(-103.85\right)\hfill \\ \hfill & =-1180.52-967.28+103.85=-2043.96{\text{kJ mol}}^{-1}\hfill \end Since there is 1.72 mol propane, 1.72 × 2043.96 kJ mol^-1 = 3.52 × 10^3 kJ used for heating. This heat is used over 90 minutes, so [latex]\frac{3.52\times {10}^{3}\text{kJ}}{\text{90 min}}=39.1{\ text{kJ min}}^{-\text{1}}[/latex] is released. kinetic molecular theory: theory based on simple principles and assumptions that effectively explains ideal gas behavior root mean square velocity (u[rms]): measure of average velocity for a group of particles calculated as the square root of the average squared velocity	{"url":"https://library.achievingthedream.org/sanjacintroductorychemistry/chapter/the-kinetic-molecular-theory/","timestamp":"2024-11-15T03:17:52Z","content_type":"text/html","content_length":"98317","record_id":"<urn:uuid:2847717f-8662-4e28-8dca-da8075703d3a>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00384.warc.gz"}
1st maths top of page all videos of subject : 1st Maths lec 52 1st Maths lec 51 1st Maths lec 50 1st Maths lec 49 1st Maths lec 48 1st Maths lec 47 1st Maths lec 46 1st Maths lec 45 1st Maths lec 44 1st Maths lec 43 bottom of page	{"url":"https://www.hlphighschool.com/vdo-1/1st-maths","timestamp":"2024-11-06T16:44:07Z","content_type":"text/html","content_length":"1050485","record_id":"<urn:uuid:ae2ea7e3-c4cc-4992-b3a6-b635f601615f>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00037.warc.gz"}
What are the calculation options for a metric? When calculating a metric for a bulletin, Rise will take the following settings into account: Weighting: How many points is this metric worth? Depending on the ranking method of the metric group this metric in weighting will either - allocate that % of the total 100 points available to that metric (Relative) or multiply the raw score by the weighting (Sum) Capping: This will set a maximum score for this metric (note that on Sum ranked algorithms the cap is calculated as score x weight, while on relative ranking the cap is calculated on the score Score method: this determines how Rise will process the raw data entries - for example taking the latest value or summing all values in the period. More... Metric Ranking Order: this tells Rise what is best - high scores or low scores.	{"url":"https://help.rise.global/support/solutions/articles/80000592688-what-are-the-calculation-options-for-a-metric-","timestamp":"2024-11-03T21:33:47Z","content_type":"text/html","content_length":"23094","record_id":"<urn:uuid:9f3f413c-ed87-484a-98cc-6ede1fa348e0>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00458.warc.gz"}
nlsic: Non Linear Least Squares with Inequality Constraints We solve non linear least squares problems with optional equality and/or inequality constraints. Non linear iterations are globalized with back-tracking method. Linear problems are solved by dense QR decomposition from 'LAPACK' which can limit the size of treated problems. On the other side, we avoid condition number degradation which happens in classical quadratic programming approach. Inequality constraints treatment on each non linear iteration is based on 'NNLS' method (by Lawson and Hanson). We provide an original function 'lsi_ln' for solving linear least squares problem with inequality constraints in least norm sens. Thus if Jacobian of the problem is rank deficient a solution still can be provided. However, truncation errors are probable in this case. Equality constraints are treated by using a basis of Null-space. User defined function calculating residuals must return a list having residual vector (not their squared sum) and Jacobian. If Jacobian is not in the returned list, package 'numDeriv' is used to calculated finite difference version of Jacobian. The 'NLSIC' method was fist published in Sokol et al. (2012) <doi:10.1093/bioinformatics/btr716>. Version: 1.0.4 Depends: nnls Suggests: numDeriv, RUnit, limSolve Published: 2023-06-26 DOI: 10.32614/CRAN.package.nlsic Author: Serguei Sokol [aut, cre] Maintainer: Serguei Sokol <sokol at insa-toulouse.fr> BugReports: https://github.com/MathsCell/nlsic/issues License: GPL-2 URL: https://github.com/MathsCell/nlsic NeedsCompilation: no Materials: NEWS In views: Optimization CRAN checks: nlsic results Reference manual: nlsic.pdf Package source: nlsic_1.0.4.tar.gz Windows binaries: r-devel: nlsic_1.0.4.zip, r-release: nlsic_1.0.4.zip, r-oldrel: nlsic_1.0.4.zip macOS binaries: r-release (arm64): nlsic_1.0.4.tgz, r-oldrel (arm64): nlsic_1.0.4.tgz, r-release (x86_64): nlsic_1.0.4.tgz, r-oldrel (x86_64): nlsic_1.0.4.tgz Old sources: nlsic archive Reverse dependencies: Please use the canonical form https://CRAN.R-project.org/package=nlsic to link to this page.	{"url":"http://cran.pau.edu.tr/web/packages/nlsic/index.html","timestamp":"2024-11-08T23:33:40Z","content_type":"text/html","content_length":"8275","record_id":"<urn:uuid:b358de3a-c7be-4d27-b497-9173a643a5a2>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00092.warc.gz"}