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https://www.groundai.com/project/convergence-rates-of-smooth-message-passing-with-rounding-in-entropy-regularized-map-inference/ | 1 INTRODUCTION
## Abstract
Maximum a posteriori (MAP) inference is a fundamental computational paradigm for statistical inference. In the setting of graphical models, MAP inference entails solving a combinatorial optimization problem to find the most likely configuration of the discrete-valued model. Linear programming (LP) relaxations in the Sherali-Adams hierarchy are widely used to attempt to solve this problem, and smooth message passing algorithms have been proposed to solve regularized versions of these LPs with great success. This paper leverages recent work in entropy-regularized LPs to analyze convergence rates of a class of edge-based smooth message passing algorithms to -optimality in the relaxation. With an appropriately chosen regularization constant, we present a theoretical guarantee on the number of iterations sufficient to recover the true integral MAP solution when the LP is tight and the solution is unique.
\runningtitle
Convergence Rates of Smooth Message Passing with Rounding
## 1 Introduction
Undirected graphical models are a central modeling formalism in machine learning, providing a compact and powerful way to model dependencies between variables. Here we focus on the important class of discrete-valued pairwise models. Inference in discrete-valued graphical models has applications in many areas including computer vision, statistical physics, information theory, and genome research (Antonucci et al., 2014; Wainwright and Jordan, 2008; Mezard and Montanari, 2009).
We focus on the problem of identifying a configuration of all variables that has highest probability, termed maximum a posteriori (MAP) inference. This problem has an extensive literature across multiple communities, where it is described by various names, including energy minimization (Kappes et al., 2013) and constraint satisfaction (Schiex et al., 1995). In the binary case, the MAP problem is sometimes described as quadratic-pseudo Boolean optimization (Hammer et al., 1984) and it is known to be NP-hard to compute exactly (Kolmogorov and Zabin, 2004; Cooper, 1990) or even to approximate (Dagum and Luby, 1993). Consequently, much work has attempted to identify settings where polynomial-time methods are feasible. We call such settings “tractable” and the methods “efficient.” A general framework for obtaining tractable methodology involves “relaxation”—the MAP problem is formulated as an integer linear program (ILP) and is then relaxed to a linear program (LP). If the vertex at which the LP achieves optimality is integral, then it provides an exact solution to the original problem. In this case we say that the LP is tight. If the LP is performed over the convex hull of all integral assignments, otherwise known as the marginal polytope , then it will always be tight. Inference over the marginal polytope is generally intractable because it requires exponentially many constraints to enforce global consistency.
A popular workaround is to relax the marginal polytope to the local polytope (Wainwright and Jordan, 2008). Instead of enforcing global consistency, the local polytope enforces consistency only over pairs of variables, thus yielding pseudo-marginals which are pairwise consistent but may not correspond to any true global distribution. The number of constraints needed to specify the local polytope is linear in the number of edges. More generally, Sherali and Adams (1990) introduced a series of successively tighter relaxations of the marginal polytope, or convex hull, while retaining control on the number of constraints. However, even with these relaxations, it has been observed that standard LP solvers do not scale well (Yanover et al., 2006), motivating the study of solvers that exploit the structure of the problem, such as message passing algorithms.
Of particular interest to this paper are smooth message passing algorithms, i.e. algorithms derived from regularized versions of the relaxed LP (Meshi et al., 2012; Savchynskyy et al., 2011, 2012; Hazan and Shashua, 2008; Ravikumar et al., 2010). These regularized LPs conduce to efficient optimization in practice and have the special property that their fixed points are unique and optimal; however, this comes at the cost of solving an approximation of the true MAP problem and, without rounding, they do not recover integral solutions in general. Non-asymptotic convergence rates to the optimal regularized function value have been studied (Meshi et al., 2012), but guarantees on the number of iterations sufficient to recover the optimal integral assignment of the true MAP problem have not been considered to our knowledge.
In this work we provide a sharp analysis of the entropy-regularized MAP inference problem with Sherali-Adams relaxations. We first characterize the approximation error of the regularized LP in distance, based on new results on entropy-regularized LPs (Weed, 2018). We then analyze an edge-based smooth message passing algorithm, modified from the algorithms described in Werner (2007) and Ravikumar et al. (2010). We prove a rate of convergence of iterates in distance. Combining the approximation error and convergence results, we present a guarantee on the number of iterations sufficient to recover of the true integral MAP assignment using a standard vertex rounding scheme when the LP relaxation is tight and the solution is unique.
## 2 Related Work
The idea of entropy regularization to aid optimization in inference problems is well studied. It is well known that solving a scaled and entropy-regularized linear program over the marginal polytope yields the scaled Gibbs free energy, intimately related to the log partition function, when the temperature parameter equals one (Wainwright and Jordan, 2008). As the temperature parameter is driven to zero, the calculation of the free energy reduces to the value of the MAP problem. However, this problem is intractable due to the difficulty of both computing the exact entropy and characterizing the marginal polytope (Deza and Laurent, 2009). Therefore, there has been much work in trying to turn this observation into tractable inference algorithms. The standard Bethe approximation instead minimizes an approximation of the true entropy (Bethe, 1935). It was show by Yedidia et al. (2003) that fixed points of the loopy belief propagation correspond to its stationary points, but still the optimization problem resulting from this approximation is non-convex and convergence is not always guaranteed.
To alleviate convergence issues, much work has considered convexifying the free energy problem leading to classes of convergent convex belief propagation often derived directly from convex regularizers (Meshi et al., 2009; Heskes, 2006; Hazan and Shashua, 2008; Johnson and Willsky, 2008; Savchynskyy et al., 2012). For instance, Weiss et al. (2007) proposed a general convexified belief propagation and explored some sufficient conditions that enable heuristically recovering the MAP solution of the LP via a convex sum-product variant. However, the approximation error was still unclear and non-asymptotic convergence rates were not considered. A number of algorithms have also been proposed to directly optimize the unregularized LP relaxation often with only asymptotic convergence guarantees such as block-coordinate methods (Werner, 2007; Globerson and Jaakkola, 2008; Kovalevsky and Koval, 1975; Tourani et al., 2018; Kappes et al., 2013) and tree-reweighted message passing (Wainwright et al., 2005; Kolmogorov, 2006). The relationship between the regularized and unregularized problems can equivalently be viewed as applying a soft-max to the dual objective typically considered in the latter to recover that of the former (Nesterov, 2005; Sontag et al., 2011). Many other convergent methods exist such as augmented Lagrangian (Martins et al., 2011; Meshi and Globerson, 2011), bundle (Kappes et al., 2012), and steepest descent (Schwing et al., 2012, 2014) approaches, but again they are difficult to compare without rates.
Most closely related to our work is recent work in convergence analysis of certain smoothed message passing algorithms that aim to solve the regularized LP objective. Savchynskyy et al. (2011) proposed an accelerated gradient method that achieves convergence to the optimal regularized dual objective value. Convergence of the primal iterates was only shown asymptotically. Meshi et al. (2012) considered a general dual coordinate minimization algorithm based on the entropy-regularized MAP objective. They proved upper bounds on the rate of convergence to the optimal regularized dual objective value; however, closeness to the true MAP assignment was not formally characterized. Furthermore, convergence in the dual objective value again does not make it easy to determine when the true MAP assignment can be recovered. Meshi et al. (2015) later studied the benefits of adding a quadratic term to the LP objective instead and proved similar guarantees. Ravikumar et al. (2010) also considered entropic and quadratic regularization, using a proximal minimization scheme with inner and outer loops. They additionally provided rounding guarantees to recover true primal solutions. However, as noted by the authors, the inexact calculation of the inner loop prevents a convergence rate analysis once combined with the outer loop. Additionally, rates on the inner loop convergence were not addressed.
The approach of this paper can be understood as the bridging the gap between Meshi et al. (2012) and Ravikumar et al. (2010). Our first contribution is a characterization of the approximation error of the entropy-regularized MAP inference problem. We then study an edge-based message passing algorithm that solves the regularized LP, which is essentially a smoothed max-sum diffusion (Werner, 2007) or the inner loop of the proximal steps of Ravikumar et al. (2010). For our main contribution, we provide non-asymptotic guarantees to the integral MAP assignment for this message passing algorithm when the LP is tight and the solution is unique. To our knowledge, this is the first analysis with rates guaranteeing recovery of the true MAP assignment for smooth methods.
## 3 Background
We denote the -dimensional probability simplex as . The set of joint distributions which give rise to is defined as For any two vectors or matrices and having the same number of elements, we use to denote the dot product, i.e. elementwise multiplication then sum over all elements. We use to denote the sum of absolute values of the elements of . The Bregman divergence between with respect to a strictly convex function is We will consider the Bregman divergence with respect to the negative entropy , where need not be a distribution. When is a distribution, this corresponds to the Kullback-Leibler (KL) divergence. The Bregman projection with respect to of onto the set is defined as . The Hellinger distance between is defined as , where is the -norm. We denote the square of the Hellinger distance by . We will often deal with marginal vectors which are ordered collections of joint and marginal distributions in the form of matrices and vectors, respectively.
### 3.1 Pairwise Models
For a set of vertices, , and edges , a pairwise graphical model, , is a Markov random field that represents the joint distribution of variables , taking on values from the set of states . We assume that each vertex has at least one edge. For pairwise models, the joint distribution can be written as a function of doubletons and singletons: We wish to find maximum a posteriori (MAP) estimates of this model. That is, we consider the integer program:
maxxV∈χn∑i∈Vθi(xi)+∑ij∈Eθij(xi,xj). (Int)
The maximization in (Int) can be written as a linear program by defining a marginal vector over variable vertices and variable edges . The vector represents the marginal distribution probabilities on vertex while the matrix represents the joint distribution probabilities shared between vertices and . We follow the notation of Globerson and Jaakkola (2008) and denote indexing into the vector and matrix variables with parentheses, e.g. for . The set of marginal vectors that are valid probability distributions is known as the marginal polytope and is defined as
M\lx@stackreldef.=⎧⎪ ⎪⎨⎪ ⎪⎩μ : ∃ P,PXi(xi)=μi(xi), ∀i,xiPXi,Xj(xi,xj)=μij(xi,xj),∀ij,xi,xj⎫⎪ ⎪⎬⎪ ⎪⎭ (1)
We can think of as the set of mean parameters of the model for which there exists a globally consistent distribution . We abuse notation slightly and dually view as a potential “vector.” The edge matrix is indexed as , indicating the element at the th row and th column. The vertex vector is indexed as , indicating the th element. The MAP problem in (Int) can be shown to be equivalent to the following LP (Wainwright and Jordan, 2008):
max⟨θ,μ⟩ s.t. μ∈M
where .
The number of constraints in is unfortunately superpolynomial (Sontag, 2010). This motivates considering relaxations of the marginal polytope to outer polytopes that involve fewer constraints. For example, the local outer polytope is obtained by enforcing consistency only on edges and vertices:
L2\lx@stackreldef.={μ≥0 : μi∈Σd∀i∈Vμij∈Ud(μi,μj)∀ij∈E} (2)
Relaxations of higher orders have also been studied, in particular by Sherali and Adams (1990) who introduced a hierarchy of polytopes by enforcing consistency on joint distributions of increasing order up to : . The corresponding Sherali-Adams LP relaxation of order is then
max⟨θ,μ⟩ s.t. μ∈Lm, (LP)
where . Because is an outer polytope of , we no longer have that the solution to (LP) recovers the true MAP solution of (Int) in general. However if the solution to (LP) is integral, then recovers the optimal solution of the true MAP problem. In this case, we say is tight.
## 4 Entropy-Regularized Map
In this section, we present our first main technical contribution, characterizing the approximation error in the entropy-regularized MAP problem for Sherali-Adams relaxations. In contrast to solving the exact (LP), we aim to solve the entropy-regularized LP:
min⟨C,μ⟩−1ηH(μ) s.t. μ∈Lm, (Reg)
where and . The hyperparameter adjusts the level of regularization. Denote by the solution of (Reg) where we omit the reference to to alleviate notation. In addition to their extensive history in inference problems, entropy-regularized LPs have arisen in a number of other fields to aid optimization when standard LP solvers are insufficient. For example, recent work in optimal transport has relied on entropy regularization to derive alternating projection algorithms (Cuturi, 2013; Benamou et al., 2015) which admit almost linear time convergence guarantees in the size of the cost matrix (Altschuler et al., 2017). Some of our theoretical results draw inspiration from these works.
### 4.1 Approximation Error
When is tight and the solution is unique, we show that approximate solutions from solving (Reg) are not necessarily detrimental because we can apply standard vertex rounding schemes to yield consistent integral solutions. It was shown by Cominetti and San Martín (1994), and later refined by Weed (2018), that the approximation error of general entropy-regularized linear programs converges to zero at an exponential rate in . Furthermore, it is possible to determine how large should be chosen in order for rounding to exactly recover the optimal solution to (Int). The result is summarized in the following extension of Theorem 1 of Weed (2018)1.
###### Theorem 1.
Let , , be the set of vertices of , and the set of optimal vertices with respect to . Let be the smallest gap in objective value between an optimal vertex and any suboptimal vertex of . Suppose is tight and . If , the following rounded solution is a MAP assignment:
(round(μ∗η))i:=argmaxx∈χ(μ∗η)i(x)
###### Proof.
Define , where denotes an all-ones vector with the same dimensions as . If then , the set of optimal vertices of with respect to , satisfies and . If ; and , then . Let . If , and then . And therefore, by Corollary 9 of Weed (2018) . Since is assumed to be tight and contains a single integral vertex , the last equation implies . ∎
Consequently, since and 2, we have:
###### Corollary 1.
If is tight, , and , the rounded solution is a MAP assignment.
In general the dependence of on suggested by Theorem 1 is not improvable (Weed, 2018). Nevertheless, when and , since all vertices in have entries equal to either or —see Padberg (1989) or Theorem 3 of Weller et al. (2016)—if the entries of are all integral, we have , thus yielding a more concrete guarantee. The disadvantage of choosing exorbitantly large is that efficient computation of solutions often becomes more difficult in practice (Weed, 2018; Benamou et al., 2015; Altschuler et al., 2017). Thus, in practice, there exists a trade-off between computation time and approximation error that is controlled by . We will provide a precise theoretical characterization of the trade-off in Section 6. In our guarantees, multiplying by a constant (and therefore multiplying by ) is equivalent to multiplying by the same value.
### 4.2 Equivalent Bregman Projection
The objective (Reg) can be interpreted as a Bregman projection. This interpretation has been explored by Ravikumar et al. (2010) as a basis for proximal updates and also Benamou et al. (2015) for the optimal transport problem. The objective is equivalent to
minDΦ(μ,exp(−ηC)) s.t. μ∈Lm, (Proj)
where . The derivation, based on a mirror descent step can be found in the appendix. The projection, however, cannot be computed in closed form in general due to the complex geometry of .
Ravikumar et al. (2010) proposed using the Bregman method (Bregman, 1966), which has been applied in many fields to solve difficult constrained problems (Benamou et al., 2015; Goldstein and Osher, 2009; Osher et al., 2005, 2010), to compute for the inner loop calculation of their proximal algorithm. While the outer loop proximal algorithm can be shown to converge at least linearly, the inner loop rate was not analyzed and the constants (possibly dependent on dimension) were not made clear. Furthermore, the Bregman method is in general inexact, which makes the approximation and the effect on the outer loop unclear (Liu and Ihler, 2013).
## 5 Smooth Message Passing
We are interested in analyzing a class of algorithms closely inspired by max-sum diffusion (MSD) as presented by Werner (2007) and the proximal updates of Ravikumar et al. (2010) to solve (Proj) over the polytope. We describe it in detail here, with a few minor modifications and variations to facilitate theoretical analysis. In , the constraints occur only over edges between vertices3. Given an edge , we must enforce the constraints prescribed by (2), which is the intersection of the following sets:
1 Xij→i={μ : μij\mathbbm1=μi} 2 Xij,i={μ : μ⊤i\mathbbm1=1, \mathbbm1⊤μij\mathbbm1=1} 3 Xij→j={μ : μ⊤ij\mathbbm1=μj} 4 Xij,j={μ : μ⊤j\mathbbm1=1, \mathbbm1⊤μij\mathbbm1=1}.
The normalization of the joint distribution in 2 and 4 is actually a redundant constraint, but it facilitates analysis as we demonstrate in Section 6. For each of these affine constraints, we can compute the Bregman projections in closed form with simple multiplicative updates.
###### Proposition 1.
For a given edge , the closed-form solutions of the Bregman projections for each of the above individual constraints are given below.
1. Left consistency: If , then for all , and .
2. Left normalization: If , then for all , and .
3. Right consistency: If , then for all , and .
4. Right normalization: If , then for all , and .
These update rules are similar to a number of algorithms throughout the literature on LP relaxations. Notably, they can be viewed as a smoothed version of MSD (Werner, 2007; Kovalevsky and Koval, 1975) in that the updates enforce agreement between variables on the edges and vertices. Nearly identical smoothed updates were also initially proposed by Ravikumar et al. (2010). As in MSD, it is common for message passing schemes derived from LP relaxations to operate on dual objective instead. We presented the primal view here as the Bregman projections lend semantic meaning to the updates and ultimately the stopping conditions in the algorithms. An equivalent dual view is presented in Appendix C.1.
Based on these update rules, we formally outline the algorithms we wish to analyze, which we call edge-based message passing (EMP) for convenience. We consider two variants: EMP-cyclic (Algorithmic 1), which cyclically applies the updates to each edge in each iteration and EMP-greedy (Algorithmic 2), which applies a single projection update to only the edge with the greatest constraint violation in each iteration. We emphasize that these algorithms are not fundamentally new, but our analysis in the next section is our main contribution. EMP-cyclic is the Bregman method, almost exactly the inner loop proposed by Ravikumar et al. (2010). In both variants, is defined as the normalized value of . The GreedyEdge operation in EMP-greedy is defined as
These procedures are then repeated again until the stopping criterion is met, which is that is -close to satisfying the constraint that the joint distributions sum to the marginals for all edges. Both algorithms also conclude with a rounding operation. Any fixed point of EMP must correspond to an optimal (see details in appendix). Computationally, EMP-greedy requires a search over the edges to identify the greatest constraint violation, which can be efficiently implemented using a max-heap (Nutini et al., 2015).
## 6 Theoretical Analysis
We now present our main contribution, a theoretical analysis of EMP-cyclic and EMP-greedy. This result combines two aspects. First, we present a convergence guarantee on the number of iterations sufficient to solve (Proj), satisfying the constraints with error in distance. We note that, in finite iterations, the pseudo-marginals of EMP are not primal feasible in general due to this -error. We then combine this result with our guarantee on the approximation error in Theorem 1 to show a bound on the number of iterations sufficient to recover the true integral MAP assignment by rounding, assuming the LP is tight and the solution is unique. This holds with sufficient iterations and a sufficiently large regularization constant even though the pseudo-marginals may not be primal feasible. We emphasize that these theorems are a departure from usual convergence rates in the literature (Meshi et al., 2012, 2015). Prior work has guaranteed convergence in objective value to the optimum of the regularized objective (Proj), making it unclear whether the optimal MAP assignment can be recovered, e.g. by rounding. We address this ambiguity in our results.
We begin with the upper bound iterations to obtain -close solutions, which is the result of two facts which we show. The first is that the updates in Proposition 1 monotonically improve a Lyapunov (potential) function by an amount proportional to the constraint violation as measured via the Hellinger distance. The second is that the difference between the initial and optimal values of the Lyapunov function is bounded.
Let denote the maximum degree of graph and define:
S \lx@stackreldef.=∑ij∈E⎡⎣log∑xi,xj∈χe−ηCij(xi,xj)+∑xi,xj∈χηd2Cij(xi,xj)⎤⎦ +∑i∈V[log∑x∈χe−ηCi(x)+∑x∈χηdCi(x)].
###### Theorem 2.
For any , EMP is guaranteed to satisfy and for all in iterations for EMP-cyclic and iterations for EMP-greedy.
Here, . In this theorem, we give our guarantee in terms of distance rather than function value convergence. As we will see, this is significant, allowing us to relate this result to Theorem 1 in order to derive the main result. The proof is similar in style to Altschuler et al. (2017). We leave the full proof for EMP-cyclic for the appendix due to a need to handle tedious edge cases, but we state several intermediate results and sketch the proof for EMP-greedy for intuition as it reveals possibly how similar message passing algorithms can be analyzed. We first introduce a Lyapunov function written in terms of dual variables , indexed by the edges and vertices to which they belong in . We denote the iteration-indexed dual variables as . For a given edge , constraints enforcing row and column consistency correspond to , respectively. Normalizing constraints correspond to . The Lyapunov function, , is shown in Figure 3.
We note that maximizing over satisfies all constraints and yields the solution to (Proj) by first-order optimality conditions. We now present a result that establishes the monotone improvement in due to the updates in Proposition 1.
###### Lemma 1.
For a given edge , let and denote the updated primal and dual variables after a projection from one of 14 in Proposition 1. We have the following improvements on . If is equal to:
1. , then
2. , then
3. , then
4. , then .
This result shows that improves monotonically after each of the four updates in Proposition 1. Furthermore, at every update, improves by twice the squared Hellinger distance of the constraint violation between the joint and the marginals.
###### Lemma 2.
Let , denote the maximizers of . The difference in function value between the optimal value of and the first iteration value is upper bounded
L(λ∗,ξ∗)−L(λ(1),ξ(1))≤S0.
Turning to Theorem 2, the result is obtained by observing that as long as the constraints are violated by an amount (i.e., the algorithm has not terminated), then the Lyapunov function must improve by a known positive amount at each iteration. We provide a proof sketch for EMP-greedy.
###### Proof Sketch of Theorem 2 for EMP-greedy.
We now show how to combine the results of Lemma 1 and Lemma 2 to obtain Theorem 2. Let be the first iteration such that the termination condition in Algorithm 2 holds with respect to some . Then, for any satisfying , we have that selects such that either or .
Without loss of generality, suppose . Therefore, we have
ϵ24≤14∥μij\mathbbm1−μi∥21≤2h2(μij\mathbbm1,μi),
where again denotes the squared Hellinger distance and the last inequality is the Hellinger inequality. Since and are normalized for each iteration, this inequality is valid. Thus, improves by when occurs and by a non-negative amount when occurs by Lemma 1. Therefore, we can guarantee improvement of at least each iteration. Since the optimality gap is at most by Lemma 2, this means the algorithm must terminate in iterations.∎
We now turn to our main theoretical result. We combine our approximation and iteration convergence guarantees to fully characterize the convergence of EMP for to the optimal MAP assignment when the relaxation is tight and the solution is unique.
###### Theorem 3.
Let , and . If is tight and , the EMP algorithm returns a MAP assignment after iterations for EMP-cyclic and after iterations for EMP-greedy.
When is integral, , yielding a bound of all known parameters. The main technical challenge in producing this result is to relate the termination condition of EMP to the distance between and (the MAP assignment), as this may lie outside the polytope . It does not suffice to provide convergence guarantees in function value as the goal of MAP inference is to produce integral assignments. The proof proceeds in two steps. First we show that is the entropy-regularized solution to objective over a “slack” polytope . Where the slack vector corresponds to the constraint violations of . We use this characterization to “project” onto a nearby feasible point . Second, we can use the properties of the primal objective to bound and . The proof is in the appendix.
## 7 Numerical Experiments
We illustrate our theoretical results in a practical application of the EMP algorithms. Ravikumar et al. (2010) already gave empirical evidence that the basic EMP-cyclic is competitive with standard solvers. Therefore, the objective of these experiments is to understand how graph and algorithm properties affect approximation (Theorem 1) and convergence (Theorem 2). We consider the family of multi-label Potts models (Wainwright et al., 2005) with labels on . For each trial, the cost vector is , and
Cij(xi,xj) ={βijxi=xj0otherwise∀ij,xi,xj
where the parameters are random and . The graphs considered are structured as grids (Globerson and Jaakkola, 2008; Ravikumar et al., 2010; Erdogdu et al., 2017) and as Erdős-Rényi random graphs with edge probability . To evaluate recovery of the optimal MAP assignment, we first solved each graph with the ECOS LP solver (Domahidi et al., 2013) and selected graphs that were tight. Solving the LP to find the ground-truth was the main computational bottleneck. Further details can be found in Appendix E.
#### Approximation
In Figure 4, we evaluate the effect of regularization and graph size on the quality of the nearly converged solution from EMP for over iterations on grids. The box-plots indicate that large choices of often yield the exact MAP solution (cyan and purple). Moderate choices still yield competitive solutions but not optimal for larger graphs (orange and green). Low choices generally give poor solutions with high spread for all graph sizes (red and blue).
#### Convergence
We then investigate the effects of regularization on convergence for both variants. Figure 5 illustrates the distance of the rounded solution to the optimal MAP solution over projection steps on grids of size . EMP-greedy converges sharply and varying regularization has less of an effect on its convergence rate. Finally, in Figure 6, we look at Erdős-Rényi random graphs to observe the effect of the graph structure for both variants. We considered degree-limited random graphs with and . The figure shows convergence over projection steps for graphs of size . For both variants, the convergence rate deteriorates for higher degrees.
## 8 Conclusion
In this paper, we investigated the approximation effects of entropy regularization on MAP inference objectives. We combined these approximation guarantees with a convergence analysis of an edge-based message passing algorithm that solves the regularized objective to derive guarantees on the number of iterations sufficient to recover the true MAP assignment. We also showed empirically the effect of regularization and graph propertise on both the approximation and convergence. In future work, we wish to extend the analyses and proof techniques to higher order polytopes and general block-coordinate minimization algorithms.
#### Acknowledgements
We thank the anonymous reviewers and Marco Pavone for their invaluable feedback.
## Appendix A Bregman Projection Derivation
The objective (Reg) can be equivalently interpreted as a Bregman projection. This interpretation has been explored by Ravikumar et al. (2010) as a basis for proximal updates and also Benamou et al. (2015) for the optimal transport problem. Here, we review the transformation because it is central to the algorithm of Ravikumar et al. (2010), upon which our main theoretical results are based.
By definition of the Bregman projection with respect to the negative entropy, , we have
DΦ(μ,\mathbbm1) =+⟨μ,logμ−\mathbbm1⟩−⟨log\mathbbm1,μ−\mathbbm1⟩ =−H(μ)
where is a vector of ones of the same size as the marginal vector and denotes the two sides are equal up to a constant. Substituting this into (Reg) and multiplying through by yields the objective:
minη⟨C,μ⟩+DΦ(μ,\mathbbm1) s.t. μ∈Lm.
Note the similarity to a projected mirror descent update over starting from (Nemirovsky and Yudin, 1983; Bubeck, 2015). Using this insight and performing a single gradient update in the dual, we can transform the problem into a single Bregman projection of the vector. The unprojected marginal vector satisfies
∇Φ(μ′)=∇Φ(\mathbbm1)−ηC,
where is the dual map and is the inverse dual map. We have and the solution to the mirror descent update is . Therefore it is sufficient to solve the following Bregman projection problem:
minDΦ(μ,exp(−ηC)) s.t. μ∈Lm
The projection, however, cannot be computed in closed form due to the complex geometry of . Sinkhorn-like algorithms such as those used in Cuturi (2013) are unavailable because the transportation polytopes are dependent on variables and which are also involved in the projection operation.
## Appendix B Derivation of EMP Update Rules
We present the derivations of the update rules similar to Ravikumar et al. (2010) for a given edge based on the Bregman projections onto the individual constraint sets , , , . We refer the reader to Ravikumar et al. (2010) for the original algorithm and derivation. We derive only the first two projections; the last two can be found by exchanging the indices.
• For the projection , where
Xij→i ={μ : μij\mathbbm1=μi},
there are no constraints on any edges or vertices other than and . Therefore, , . Similarly, , .
The Lagrangian of the projection is given in terms of primal variables and dual variables :
L(μ′,α) =∑xi,xjμ′ij(xi,xj)(logμ′ij(xi,xj)μij(xi,xj)−1+α(xi))+∑xiμ′i(xi)(logμ′i(xi)μi(xi)−1−α(xi)).
By the first-order optimality condition, the primal solution in terms of the dual variables is
μ′ij(xi,xj) =μij(xi,xj)e−α(xi) μ′i(xi) =μi(xi)eα(xi).
Substituting this solution back in to the Lagrangian, we have
L(α)=−∑xi,xjμij(xi,xj)e−α(xi)−∑xiμi(xi)eα(xi).
Again, by the first-order optimality condition, the dual solution is
α∗(xi)=12log∑xjμij(xi,xj)μi(xi).
Substituting this value for into the primal solution yields the desired result.
• Again, for the projection onto
Xij,i ={μ : μ⊤i\mathbbm1=1, \mathbbm1⊤μij\mathbbm1=1},
only and are affected. enforces that the variables and each sum to one. It is well known and easy to show that the Bregman projection with respect to the negative entropy is simply the and normalized by their sums. This normalization can also be written as a multiplicative update of the same form by observing that
μ′ij(xi,xj) =μ′ij(xi,xj)e−ξ∗ij μ′i(xi) =μ′i(xi)e−ξ∗i,
where and . Again, these can be derived via the Lagrangian.
## Appendix C Extensions of EMP
### c.1 Dual EMP
We may also equivalently interpret the multiplicative updates in Algorithm 1 and Algorithm 2 as additive updates of the dual variables. The dual interpretation is consistent with past work in dual MAP algorithms (Sontag et al., 2011) and may be more practical to avoid numerical issues in implementation. Instead of tracking the primal variables , we track a sum of the dual variables with for each vertex and edge. Enforcing consistency between a given joint distribution and its marginals in 1 yields updated dual variable sums
ζ′ij(xi,xj)←ζij(xi,xj)−α∗(xi) ζ′i(xi)←ζi(xi)+α∗(xi),
where again | 2021-01-17 21:29:46 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9056931138038635, "perplexity": 538.8872583424125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00061.warc.gz"} |
https://homework.cpm.org/category/MN/textbook/cc2mn/chapter/7/lesson/7.2.7/problem/7-138 | Home > CC2MN > Chapter 7 > Lesson 7.2.7 > Problem7-138
7-138.
Rewrite each fraction below as an equivalent fraction, as a decimal, and as a percent.
1. $\frac { 6 } { 18 }$
1. $\frac { 7 } { 20 }$
${\frac{7}{20}=\frac{14}{40}=0.35=35\%}$
1. $\frac { 9 } { 10 }$
1. $\frac { 4 } { 25 }$
${\frac{4}{25}=\frac{12}{75}=0.16=16\%}$ | 2020-10-01 13:08:55 | {"extraction_info": {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5909161567687988, "perplexity": 7325.550680883642}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402131412.93/warc/CC-MAIN-20201001112433-20201001142433-00345.warc.gz"} |
https://stats.stackexchange.com/questions/linked/15287?sort=hot&pagesize=50 | 39 views
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https://planetmath.org/proofofsamplingtheorem | # proof of sampling theorem
## 0.1 Set-up
Let $w>0$ be the (two-sided) bandwidth. The variable $\xi$ below will denote frequency, and the variable $t$ will denote time. (Both $w$ and $\xi$ are measured in .)
Consider the space of functions:
$\mathcal{H}^{\prime}=\{g\in\mathbf{L}^{2}(\mathbb{R})\colon g(\xi)=0\text{ for% almost all }\lvert\xi\rvert>w/2\}$
which is clearly seen to be a complex Hilbert space with the usual inner product for $\mathbf{L}^{2}(\mathbb{R})$.
Let $\mathcal{F}$ denote the Fourier transform on $\mathbf{L}^{2}(\mathbb{R})$, which is a unitary transform by Plancherel’s theorem. So,
$\mathcal{H}=\{f\in\mathbf{L}^{2}(\mathbb{R})\colon(\mathcal{F}f)(\xi)=0\text{ % for almost all }\lvert\xi\rvert>w/2\}=\mathcal{F}^{-1}\mathcal{H}^{\prime}$
is also a Hilbert space.
## 0.2 Computation of orthonormal basis
One orthonormal basis for $\mathcal{H}^{\prime}$ consists of the usual Fourier functions on the interval $[-w/2,w/2]$, extended to be zero on $\mathbb{R}\setminus[-w/2,w/2]$:
$\phi_{n}(\xi)=\begin{cases}\frac{1}{\sqrt{w}}e^{-2\pi in\xi/w}\,,&\lvert\xi% \rvert\leq w/2\\ 0\,,&\lvert\xi\rvert>w/2\,,\end{cases}\quad n\in\mathbb{Z}\,.$
Mapping these by $\mathcal{F}^{-1}$ produces an orthonormal basis for $\mathcal{H}$:
$\displaystyle(\mathcal{F}^{-1}\phi_{n})(t)$ $\displaystyle=\frac{1}{\sqrt{w}}\int_{-w/2}^{w/2}e^{-2\pi in\xi/w}\,e^{2\pi i% \xi t}\,d\xi$ $\displaystyle=\frac{1}{\sqrt{w}}\int_{-w/2}^{w/2}e^{2\pi i\xi(t-n/w)}\,d\xi$ $\displaystyle=\frac{1}{\sqrt{w}}\bigl{(}w\,\operatorname{sinc}(w(t-n/w))\bigr{% )}=\sqrt{w}\,\operatorname{sinc}(wt-n)\,,$
where we have used the fact that the Fourier transform of $t\mapsto w\,\operatorname{sinc}(wt)$ (normalized sinc function) is the rectangular box function of bandwidth $w$, and vice versa.
## 0.3 Expansion by orthonormal basis
Given $f\in\mathcal{H}$, let $g=\mathcal{F}f\in\mathcal{H}^{\prime}$. We can expand $g$ in a Fourier series with respect to the $\{\phi_{n}\}$:
$g(\xi)=\sum_{n\in\mathbb{Z}}\langle g,\phi_{n}\rangle\,\phi_{n}(\xi)\,,$
with the infinite sum converging in $\mathbf{L}^{2}(\mathbb{R})$. Taking $\mathcal{F}^{-1}$ of both sides, we obtain:
$f(t)=(\mathcal{F}^{-1}g)(t)=\sum_{n\in\mathbb{Z}}\langle g,\phi_{n}\rangle\,(% \mathcal{F}^{-1}\phi_{n})(t)=\sqrt{w}\sum_{n\in\mathbb{Z}}\langle g,\phi_{n}% \rangle\,\operatorname{sinc}(wt-n)\,.$
Moreover,
$\langle g,\phi_{n}\rangle=\frac{1}{\sqrt{w}}\int_{-\infty}^{\infty}g(\xi)\,e^{% 2\pi in\xi/w}\,d\xi=\frac{1}{\sqrt{w}}(\mathcal{F}^{-1}g)\Bigl{(}\frac{n}{w}% \Bigr{)}=\frac{1}{\sqrt{w}}f\Bigl{(}\frac{n}{w}\Bigr{)}\,.$
(Since $g$ is also in $\mathbf{L}^{1}$, its inverse Fourier transform $\mathcal{F}^{-1}g$ is a continuous function. Provided that we modify $f$ on a set of measure zero, we can assume that $f=\mathcal{F}^{-1}(\mathcal{F}f)=\mathcal{F}^{-1}g$ is continuous. So it is legal to talk about the pointwise values $f(n/w)$.)
## 0.4 Result
Hence, we arrive at the representation:
$f(t)=\sum_{n\in\mathbb{Z}}f\Bigl{(}\frac{n}{w}\Bigr{)}\,\operatorname{sinc}(wt% -n)\,,$
thereby reconstructing any $f\in\mathcal{H}$ — a square-integrable band-limited function — from its samples at every time period of length $1/w$.
## 0.5 Uniform and absolute convergence of series
The infinite series for $f$ converges in $\mathbf{L}^{2}$ by construction, but in fact it also converges uniformly and absolutely. To see this, first note that by the Cauchy-Schwarz inequality,
$\sum_{n\in\mathbb{Z}}\lvert f(\tfrac{n}{w})\rvert\,\lvert\operatorname{sinc}(% wt-n)\rvert\leq\Bigl{(}\sum_{n\in\mathbb{Z}}\lvert f(\tfrac{n}{w})\rvert^{2}% \Bigr{)}^{1/2}\Bigl{(}\sum_{n\in\mathbb{Z}}\operatorname{sinc}^{2}(wt-n)\Bigr{% )}^{1/2}\,.$
The series $\sum_{n}\lvert f(n/w)\rvert^{2}$ converges by Parseval’s theorem ($w^{-1/2}f(n/w)$ are the Fourier coefficients of $g$). Also, the series $\sum_{n}\operatorname{sinc}^{2}(wt-n)$ is uniformly bounded for all $t\in\mathbb{R}$. To prove this, it suffices to restrict to $t$ bounded inside $[0,1/w]$ as the function $t\mapsto\sum_{n}\operatorname{sinc}^{2}(wt-n)$ is $1/w$-periodic; and then it becomes an easy estimate using the fact that $\sum_{n}n^{-2}<\infty$. It follows that the series $\sum_{n}\lvert f(n/w)\rvert\lvert\operatorname{sinc}(wt-n)\rvert$ is uniformly bounded for all $t$, and its tail tends to zero uniformly in $t$.
## 0.6 Illustrations
• http://aux.planetmath.org/files/objects/8650/sample.pyPython program to produce the three figures
Title proof of sampling theorem ProofOfSamplingTheorem 2013-03-22 16:28:57 2013-03-22 16:28:57 stevecheng (10074) stevecheng (10074) 13 stevecheng (10074) Proof msc 42A38 msc 94A20 PlancherelsTheorem | 2018-11-18 08:44:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 62, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9921930432319641, "perplexity": 232.63555686023886}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039744320.70/warc/CC-MAIN-20181118073231-20181118095231-00115.warc.gz"} |
https://gamedev.stackexchange.com/questions/146636/a-pathfinding-gridless-with-circles | # A* Pathfinding Gridless With Circles [closed]
I found that A* is an algorithm for grid (or tiled) maps. I want to use it or some alternative of it to my game. The problem is that I have no grid. I have canvas and GameObjects could be at position { x: 23.91324512, y: 131.12334253461 }. How would you do this?
And another question, I have element with stepSize (concretely called movementSpeed), and getPosition().getX(), getPosition().getY() AND getRadius() which returns the radius of a circle.
How would you do A* algorithm with radius?
Btw I am implementing it in Java, so maybe some Java snippet would be helpful.
EDIT
As you may suggest me to create a virtual grid, how do I check if I can move to the next node when I have my getRadius() and all the obstacles getRadiuses
## closed as unclear what you're asking by BlueRaja - Danny Pflughoeft, Engineer, Gnemlock, Alexandre Vaillancourt♦, Tyyppi_77Aug 4 '17 at 12:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• I suggest you be a lot clearer about what you are asking, or this question is going to be closed shortly. Supply a picture if necessary to explain what you want. – Engineer Aug 1 '17 at 21:33
• "A* is an algorithm for grid/tiled maps" is a common misconception. A* works for any planning problem that you can describe as a series of discrete options & state transitions. Those transitions do not need to form a regular grid. It's very common to implement A* and similar pathfinding algorithms over continuous space by first discretizing that space into a navigation mesh ("navmesh"), and treating the movement from (anywhere within) one convex polygon to (anywhere within) a neighbouring convex polygon as such a discrete transition. – DMGregory Aug 1 '17 at 22:17
• This appears to be the XY problem. Please describe the issue you're trying to solve first, then your attempted solution. – Alexandre Vaillancourt Aug 2 '17 at 10:01 | 2019-09-18 12:02:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.256744921207428, "perplexity": 1130.7538454011212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00458.warc.gz"} |
https://encyclopediaofmath.org/index.php?title=Splitting_field_of_a_polynomial&diff=next&oldid=16268 | # Difference between revisions of "Splitting field of a polynomial"
The smallest field containing all roots of that polynomial. More exactly, an extension $L$ of a field $K$ is called the splitting field of a polynomial $f$ over the field $K$ if $f$ decomposes over $L$ into linear factors:
$$f=a_0(x-a_1)\ldots(x-a_n)$$
and if $L=K(a_1,\ldots,a_n)$ (see Extension of a field). A splitting field exists for any polynomial $f\in K[x]$, and it is defined uniquely up to an isomorphism that is the identity on $K$. It follows from the definition that a splitting field is a finite algebraic extension of $K$.
Examples. The field of complex numbers $\mathbf C$ serves as the splitting field of the polynomial $x^2+1$ over the field $\mathbf R$ of real numbers. Any finite field $\operatorname{GF}(q)$, where $q=p^n$, is the splitting field of the polynomial $x^q-x$ over the prime subfield $\operatorname{GF}(p)\subset\operatorname{GF}(q)$. | 2021-01-27 16:46:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8504858613014221, "perplexity": 57.10028177887086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00031.warc.gz"} |
https://astarmathsandphysics.com/university-maths-notes/matrices-and-linear-algebra/4329-reflections-and-rotatiins-in-rn-are-one-to-one-and-onto.html | ## Reflections and Rotations in Rn are One to One and Onto
A linear transformation is one to one and onto if and only if it has an inverse.
We can represent a rotation in
$\mathbb{R}^n$
by a matrix
$R( \theta )$
where
$R$
means rotation and
$\theta$
means a rotation anticlockwise through an angle
$\theta$
.
The inverse to this transformation is the rotation
$R(- \theta )$
, the rotation about the same axis through the same angle in the opposite direction. Hence rotations in
$\mathbb{R}^n$
are one to one and onto.
Reflections in
$\mathbb{R}^n$
are also one to one and onto since the inverse of any reflection represented by a matrix
$q ( \theta )$
is
$q (- \theta )$
Since reflections are self inverse, they are also one to one and onto. | 2018-01-18 08:01:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7213630080223083, "perplexity": 389.62369906230987}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00776.warc.gz"} |
https://www.amplifiedparts.com/sets-by-amplifier/bogner/duende | # Tubes for the Bogner Duende
### The Bogner Duende amplifier requires the following components:
× 5
12AX7
Russian made reissue of the original Tungsol. The Ultimate 12AX7. Big, warm, and musical. High Gain, ultra-low microphonics, and superb linearity with a dynamic 3-D sound. Available with gold pins.
Russian made reissue of the original Tungsol. The Ultimate 12AX7. Big, warm, and musical. High Gain, ultra-low microphonics, and superb linearity with a dynamic 3-D sound. Available with gold pins.
The JJ Electronic 12AX7 / ECC83 has quickly become a modern classic in preamp tubes. Its rugged design and rich harmonics make this tube an obvious choice for musicians who need a reliable and high quality tube at a reasonable price. The balanced tone of the JJ 12AX7 works well in any amp.
The JJ Electronic 12AX7 / ECC83 has quickly become a modern classic in preamp tubes. Its rugged design and rich harmonics make this tube an obvious choice for musicians who need a reliable and high quality tube at a reasonable price. The balanced tone of the JJ 12AX7 works well in any amp.
A high performance, long plate version of the ECC83 / 12AX7. Like all JJ preamp tubes, these are built into a rugged package using the spiral filament and robust plate structure resulting in low microphonics and high reliability. The long plate version of this tube differs from the standard version with its stronger and fuller sound. This tube will work in any 12AX7 or ECC83 position.
A high performance, long plate version of the ECC83 / 12AX7. Like all JJ preamp tubes, these are built into a rugged package using the spiral filament and robust plate structure resulting in low microphonics and high reliability. The long plate version of this tube differs from the standard version with its stronger and fuller sound. This tube will work in any 12AX7 or ECC83 position.
Genalex Gold Lion ECC83/B759 (12AX7) dual triode preamp tube with gold pins. Original Genalex Gold Lion ECC83/B759 tubes are considered among the best 12AX7 types ever made. They are hard to find and when they are available, they are expensive. They exhibit a large detailed soundstage that is sought after by discerning audiophiles who demand the highest quality sound. The reissue Genalex Gold Lion ECC83/B759 recreates the sound signature of the original. This tube features balanced sections and gold plated pins.
Genalex Gold Lion ECC83/B759 (12AX7) dual triode preamp tube with gold pins. Original Genalex Gold Lion ECC83/B759 tubes are considered among the best 12AX7 types ever made. They are hard to find and when they are available, they are expensive. They exhibit a large detailed soundstage that is sought after by discerning audiophiles who demand the highest quality sound. The reissue Genalex Gold Lion ECC83/B759 recreates the sound signature of the original. This tube features balanced sections and gold plated pins.
Miniature high-mu twin triode each section of which has an individual cathode connection - especially suited for use in resistance-coupled voltage amplifiers, phase inverters, multivibrators, and industrial-control circuits.
Russian-made Mullard Reissue 12AX7 dual triode. Large sound stage with excellent detail. Factory selected for low noise and microphonics. Great for hi-fi amps.
9 pin miniature preamp tube (Amplification Factor =100). Very good gain, tight deep bass, fat mids and silky top end with overall definition and brightness. Selected with balanced systems and recommended for guitar amps to improve overall response.
Peavey 'Super 7' 12AX7
This Russian tube is full bodied and bright with a more straightforward presentation. It packs a lot of bite with its higher-than-normal gain. In overdrive, it offers a lot of crunchy fizz. High gain preamplifier tube with ultra low noise and detailed, musical tone. Unique spiral filament eliminates filament to cathode induced hum common on amplifiers that use AC on the filaments.
9 pin miniature preamp tube (Amplification Factor = 100). This Russian tube is full bodied and well balanced. It has a high gain with a very smooth presentation. In overdrive, it is not quite as defined as the 12AX7-S-JJ, but it offers more crunchy fizz.
9 pin miniature preamp tube (Amplification Factor = 100). This Russian tube has strong lows and bright highs. Its lows demand attention when playing full bar chords. In overdrive, it offers a lot of crunchy fizz.
Sovtek 12AX7WB dual triode.
Miniature high-mu twin triode each section of which as an individual cathode connection suited for use in resistance-coupled voltage amplifiers, phase inverters, multivibrators, andindustrial-control circuits where high voltage gain is desired.
Miniature, high-mu, twin triode primarily designed for use in low-level stages of high-gain audio-frequency amplifiers. Use of the 7025 in this application is advantageous because of its low hum output.
A dual triode that can be used in place of most 12AX7 circuits. The rugged build of this tube and reduced gain (30% less than a 12AX7) can improve headroom and give better control in high gain stages allowing users to more easily dial in sweet spots. The JJ 5751 features a smooth and balanced tone and a detailed response with low noise and low microphonics.
A new take on the classic 12AX7 preamp tube. While it uses the same rugged build and thick glass as other JJ preamp tubes, this tube differs in construction from the typical 12AX7 in that it features a medium-long plate, which provides a dynamic and open sound with a rich forward sounding mid range. This tube offers low noise and low microphonics.
Sovtek 12AX7WC dual triode preamp tube. Made in Russia.
Tung Sol 5751 dual triode preamp tube. Made in Russia. NOS specs call the 5751 a “special 12AX7” having a high-mu amplification factor of 70.
Low-noise Tube for Best Clean Tone! The TAD Selected 5751 is Tube Amp Doctor's first choice for noticeably improved clarity, smoothness, and a true vintage tone character for most guitar amps!
Warm, clean tone with creamy overdrive! PREMIUM selected quality for the most demanding amp stages. This tube works greatly for warm clean tones and creamy overdrive. We recommend this tube for the 1st gain stage and as input tube. It upgrades 12AX7A and ECC83 tubes and replaces E83CC tubes. The HIGHGRADE Selection is the tip-of-the-top quality at the lowest noise (microphony, hum, hiss) level. Typically only 0.5-2% out of the whole production of this tube meet the HIGHGRADE requirements.
$15.95 × 5$26.95 × 5
$12.95 × 5$22.50 × 5
$15.75 × 5$24.95 × 5
$24.75 × 5$43.95 × 5
$59.95 × 5$18.95 × 5
$19.35 × 5$20.50 × 5
$13.95 × 5$16.95 × 5
$14.75 × 5$16.75 × 5
$49.95 × 5$59.95 × 5
$14.95 × 5$13.95 × 5
$14.95 × 5$26.75 × 5
$24.95 × 5$30.95 × 5
× 1
5AR4
The YJR is a direct plug-in replacement adapter for use in most amplifiers that use 5AR4/GZ34, 5U4, 5Y3 or similar full-wave rectifier tubes. The YJR converts your amp's vacuum tube rectifier to a solid state rectifier, reducing tube sag for a tighter sound and feel. The tube rectifier can easily be swapped back in when sag is desired. This Yellow Jacket® is a solid state device and as such does not come with vacuum tubes.
• Converts most audio amplifiers which use a vacuum tube rectifier 5AR4/GZ34, 5U4 or 5Y3 to a solid state device.
• Safe for all common amplifiers and transformers.
All Yellow Jacket® converters have a one year warranty against manufacturer defects.
Chinese 5AR4 Vacuum Tube
The JJ Electronic GZ34 / 5AR4 is rugged rectifier at a reasonable price. JJ has a reputation for building sturdy tubes and this one is no exception. Users report that the JJ GZ34 / 5AR4 is a reliable rectifier in their Vox, Fender, and hifi amplifiers.
Octal rectifier tube (Max DC output current = 250 mA)
Tube Amp Doctor 5AR4 Rectifier, Premium Selected. A faithful reproduction of the Mullard/VALVO GZ34. Pure power and the typical sag is the distinguishing characteristic of the most popular guitar amps of the 1960s guitar heroes till today.
Solid State Rectifier Direct plug-in replacement for 5AR4, 5U4 and 5Y3 rectifier vacuum tubes in amplifiers with center tapped secondary power transformer. Use when less 'sag' is desired in the power supply; just switch back to the tube rectifier when 'sag' is desired.
Premium directly heated heavy duty rectifier Diode. Re-issue of the most popular rectifier valve ever built by Mullard. Helps extend the life of the other valves in the amplifier by allowing them to heat up before plate voltage is applied.
• Low internal voltage drop & controlled heater warm up time
• Excellent choice for both Hifi and instrument amplification
• Replacement/upgrade for all 5AR4/GZ34 types
Tung Sol 5AR4 rectifier tube. Made in Russia. Premium directly heated heavy duty rectifier Diode. Helps extend the life of the other valves in the amplifier by allowing them to heat up before plate voltage is applied. Tung-Sol tubes reflect the best in vintage construction with unsurpassed tone and reliability.
• Low internal voltage drop & controlled heater warm up time
• Excellent choice for both Hifi and instrument amplification
• Replacement/upgrade for all 5AR4/GZ34 types
Genalex Gold Lion U77 / GZ34 full-wave rectifier tube. Made in Russia.
$14.95 × 1$18.54 × 1
$17.75 × 1$21.95 × 1
$32.50 × 1$9.95 × 1
$39.95 × 1$34.85 × 1
42.95 × 1 × 2 6V6 Please make a selection above. Mullard 6V6GT tube. Made in Russia. This valve is built to stand the high voltages encountered in the Fender Deluxe Reverb and other guitar amplifiers that ran the valves above the published specification. The smooth and balanced frequency response also makes it suitable for hi-fi use. Genalex Gold Lion 6V6GT / CV511 beam-power tube. Made in Russia. This 6V6 has the standard rugged construction we’ve come to expect from JJ which allows it handle much higher plate voltages than the typical 6V6 tube. The JJ 6V6 can even be used in place of a 6L6 in some amplifiers. Overall this tube has a warm and balanced tone with incredible separation and response. These tubes can cover a spectrum of sounds and styles all within the same amplifier. Totally Tweed! The preferred OEM tube of '50s-era Fender Tweed Champ and Deluxe amplifiers The Tungsol 6V6 has a geometry designed to safely handle the higher voltages used in guitar amps " plus heavier plate and grid materials. The result.. better mids and bottom while keeping the smooth top of the classic 6V6's. The Tungsol 6V6 breaks up evenly from low E to high up the neck. Blues players will LOVE these tubes for the way they sing! Octal power tube (Max Plate Watts = 14W) Beam power tetrode with a specially developed cathode coating, careful alignment of the grid, tri-alloy plate material for flawless performance up to 475 volts. Perfect for high plate voltage amplifiers like the Fender Deluxe Reverb. This 6V6 has the standard rugged construction we’ve come to expect from JJ which allows it handle much higher plate voltages than the typical 6V6 tube. The JJ 6V6 can even be used in place of a 6L6 in some amplifiers. Overall this tube has a warm and balanced tone with incredible separation and response. These tubes can cover a spectrum of sounds and styles all within the same amplifier. This 6V6 has the standard rugged construction we’ve come to expect from JJ which allows it handle much higher plate voltages than the typical 6V6 tube. The JJ 6V6 can even be used in place of a 6L6 in some amplifiers. Overall this tube has a warm and balanced tone with incredible separation and response. These tubes can cover a spectrum of sounds and styles all within the same amplifier. The 6V6 is a low-output power tube (12-18 watts) often used in Fender® amplifiers, including the Champ®, Deluxe™, Dual Professional, Vibro-King®, Princeton®, Vibrolux® and others. It has a very warm, round and soft response with rich harmonics, and distorts easily. One of the best choices ever produced for rehearsal and recording amplifiers. Our most premium design, for long life and professional performance. The choice for many pro players who use Groove Tubes in their favorite amps for recording and rehearsal; fully captures the spirit and sound of tube amps from the '50s and '60s to the present. Octal power tube (Max Plate Watts = 14W) Beam power tetrode with a specially developed cathode coating, careful alignment of the grid, tri-alloy plate material for flawless performance up to 475 volts. Perfect for high plate voltage amplifiers like the Fender Deluxe Reverb. Totally Tweed! The preferred OEM tube of '50s-era Fender Tweed Champ and Deluxe amplifiers The Tungsol 6V6 has a geometry designed to safely handle the higher voltages used in guitar amps " plus heavier plate and grid materials. The result.. better mids and bottom while keeping the smooth top of the classic 6V6's. The Tungsol 6V6 breaks up evenly from low E to high up the neck. Blues players will LOVE these tubes for the way they sing! Mullard 6V6GT tube. Made in Russia. This valve is built to stand the high voltages encountered in the Fender Deluxe Reverb and other guitar amplifiers that ran the valves above the published specification. The smooth and balanced frequency response also makes it suitable for hi-fi use. Genalex Gold Lion 6V6GT / CV511 beam-power tube. Made in Russia.59.00 × 1
$69.90 × 1$36.30 × 1
$60.00 × 1$42.90 × 1
$38.80 × 1$16.65 × 2
$54.95 × 1$19.95 × 2
$28.50 × 2$28.50 × 2
$33.95 × 2 Total price for all components:$0.00
12AX7: Not selected 5AR4: Not selected 6V6: Not selected | 2021-10-24 02:12:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1816309094429016, "perplexity": 12589.838273630708}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585837.82/warc/CC-MAIN-20211024015104-20211024045104-00537.warc.gz"} |
https://tt.gsusigmanu.org/5069-interstellar-movie-what-is-the-ldquoportalrdquo-to-th.html | Interstellar movie: What is the “portal” to the other galaxy?
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I recently watched Interstellar with some friends and we didn't come to the same conclusion. In my opinion, the portal they use to go to the other galaxy (to visit the three planets) is not a black hole. My friends tell me it is.
(warning, contains spoiler!!)
To me, it's not because of these points:
• When they are in front of the portal, they can see the other galaxy, which is not the case when Coop goes to the black hole at the end
• When they go to that portal, they "flow" in a tube like channel. Again, that's not the case when Coop goes to the black hole at the end.
• When Coop goes to that black hole, he goes through some sort of light dust that hits his ship, which is not the case when they go to the portal.
• When the robot (I forgot his name) goes in the black hole, his purpose is to collect some quantic data. If the initial portal was also a black hole, he would have made that collection at that time.
• When they goes to the portal, they seems "confident", but when Coop goes to that black hole, it's for killing himself. If the portal was a black hole, they would behave more identical.
So I believe it's not a black-hole, but what is it exactly (maybe not in scientific terms, but in Interstellar's terms)? A Wormhole?
Yes, it is a wormhole indeed. This has been indicated quite clearly in the movie as well, when Dr. Romily explains with a pen and paper to Dr. Cooper. The explanation goes like this :
Imagine a sheet of paper to be 2-D space, then a line joining two points on the sheet of paper is the shortest distance possible to reach that point, but if due to some disturbance, the space is bent ( achieved by folding the sheet of paper), you can pierce a hole in the sheet after aligning the two points together. That is a 2-D wormhole (which is a circle indeed). So what happens when you consider a 3-D wormhole? It becomes a sphere.
Now, you can see the other end of a 2-D wormhole (which is effectively the other hole in the sheet visible from the first hole, and one can see beyond the hole in the other direction as well). Same happens with the 3-D spherical wormhole where you can see to the other side of the wormhole as well.
Now, the means to bend spacetime : Space time can be bent by having a very big mass placed inside the space time (read Einstien's General Relativity ). So, it is safe to assume that the wormhole is a space time disturbance created due to a massive object, but it is not a blackhole.
In addition to the previous answer about Romily's explanation, the black hole is just a black hole at the center of the star system they're exploring. It's the remnants of the large star, not a wormhole. When they first come through the wormhole in the movie they're fairly far from the "Gargantua". That's why it takes them some time to get to Miller's planet that is orbiting the black hole. Hope this helps. :)
Exploring the Philosophy of 'Interstellar': Why Is the Universe Like This?
'Interstellar,' the new science fiction film, is rightly praised for its adherence to the real science of black holes, wormholes, extra dimensions and time travel. Moreover, promoting frontier concepts of cosmology builds public appreciation of real science. Successful entertainment need not be all escapist and unscientific like, say, magic ('Harry Potter') and fantasy ('Transformers,' 'X-men').
It goes deeper. Can grasping the exotic nature of reality approach bigger issues of meaning and purpose (if there be any)? What's it all about? Perhaps the philosophy of 'Interstellar,' as well as its science, is worth exploring.
The cutting-edge science comes from Kip Thorne, Feynman Professor of Theoretical Physics, Emeritus, at Caltech, a global authority on gravitation, general relativity and "the warped side of the universe," as he likes to say. Kip is an executive producer of the film and his new book, appropriately, is The Science of Interstellar.
Kip stresses "gravitational anomalies" as both changing the course of astronomy and playing a major role in the film. "Gravitational anomalies are a big deal, today and in the past," he says. "And we're still trying to figure out what's going on." In astronomy, gravitational anomalies in Mercury's orbit were explained by Einstein's general relativity, in rapid rotations of galaxies and stars by "dark matter," and in the shocking accelerating expansion of the universe by "dark energy." In the film, gravitational anomalies are fields in a fifth dimension (called "The Bulk"), which are then harnessed to save humankind.
The science of 'Interstellar' is founded on black holes, which are unimaginably strange, converting all their energy-matter (from a collapsing star) into warped space-time and trapping light so that it cannot escape (that's why they're "black"). Black holes give insight into higher dimensions and wormholes -- those theoretical tunnels that link one part of the universe to another and serve as the central plot device in 'Interstellar.'
But can black holes go further? Can they give us insight into how the universe began, and what lies in the far future? Can black holes connect to other universes? We can never see them but the "light" of black holes brings us "closer to truth."
Some say science can only answer "How questions," not "Why questions." "Why" is the domain of philosophy. So what's the philosophy of 'Interstellar'? What is it about a universe that features black holes and perhaps extra dimensions and wormholes? What does it mean, if anything, that black holes help structure galaxies such that stars and planets are stable for billions of years? What does it mean, if anything, that a universe with black holes is congenial to life and mind?
"Some people have speculated that there's only one way that things could work, because there's only one manner in which all the fundamental physical laws could hang together in a logical, self-consistent way," Kip says. "But there is considerable evidence in the last decade or two that's not the case."
Here are five possibilities:
1. There is no meaning. The universe, as Bertram Russell put it, "is just there, and that's all." Reality is a "brute fact" with no explanation.
2. There is "Only One Way" that the laws of nature can be -- the so-called "Theory of Everything." (Cosmologist Max Tegmark suggests reality is mathematical.)
3. There are multiple universes, perhaps an infinite number of universes, so that anything that can happen must happen, including us. Because only if we exist can we ask why we exist ("selection bias"). We think we are special when we are not.
4. There is a "ground of being" that is a "supreme conscious being" from which all being comes -- perhaps the personal God of Abrahamic religions (Judaism, Christianity, Islam), perhaps the cosmic consciousness of Eastern religions.
5. There is a "ground of being" that drives (or is) a kind of teleology in reality (which the atheistic philosopher Thomas Nagel suggests), an impersonal directionality in the universe that reflects the foundations of reality. (The atheistic physicist Paul Davies suggests that the universe is "about" something. The philosopher John Leslie suggests that "value" is fundamental.)
I asked Kip to reflect on a reality that features black holes, wormholes, extra dimensions and time travel.
"I have no great wisdom on this," he said. Maybe, I thought, that's the greatest wisdom.
So while I admire the science of 'Interstellar,' I appreciate even more how it provokes us to ask the "Big Why Question." Why is the universe like this?
Robert Lawrence Kuhn is the creator, writer and host of Closer To Truth, the public television series and website archive of over 4,000 videos of leading scientist and philosophers.
1. There's a solar-powered drone that stays up for decades.
Depends on design, but this is probably possible. The solar-powered Opportunity Mars rover has been going strong for ten years, and it's on Mars, which is further away from the sun than we are.
(It seems really unlikely that Cooper would be able to hack into the drone though. Did all world governments use the same guidance programs, accessible via short-range wi-fi?)
Milky Way Galaxy Line Drawing / The milky way is the galaxy that contains our solar system, with the name describing the galaxy's appearance from earth:
300x168 - The milky way (like other spiral galaxies) is surrounded by a large halo region which contains globular clusters, large clouds of hydrogen gas, and a huge mass of the mysterious dark matter.
612x383 - Distributed in the halos around the milky way and other galaxies.
1600x1155 - Learn how to draw the easy, step by step way while having fun and building skills and confidence.
479x612 - Our sun lies near a small, partial arm called the orion arm, or orion spur.
661x672 - The milky way facts say that it is the galaxy that contains our solar system.
1024x1159 - Titanic mergers between galaxies or interstellar winds from catastrophic supernovas can also have smashed clouds together into stars, she explained.
848x798 - The beijing researchers thought forces from the milky way's spinning.
1600x1155 - You can learn how to color with markers, color pencils and much more.
This film provides examples of:
• 20 Minutes into the Future: SSTO VTOLs have been built, as well as cryogenic sleep pods, Uterine Replicators and conversant AI Robot Buddies. They're just not common because of the Second Dust Bowl (which also has robotic harvester machines), meaning most people drive 2014 cars, beaten to hell by decades of wear and tear.
• Ace Pilot: Cooper. At first only an Informed Attribute, he later proves this to be true when docking onto the wildly rotating Endurance without losing consciousness and when pulling off the Spaceship Slingshot Stunt around the black hole.
• Actually Pretty Funny: The NASA officials laugh when Cooper asks for assurances that they're leaving, and not in the trunk of a car.
• Adaptation Distillation: TARS has less dialogue in the novelization, and several of his funny lines and scenes with Cooper are cut or shortened, making CASE's comment about TARS being the much more talkative one almost into an Informed Attribute.
• The novelization, while naturally very close to the movie, provides more insight into several of the characters, such as Mann's deception and what ultimately happened to characters like Tom and Edmunds.
• One of the more common complaints about the characters in the movie is that, in the film, Cooper occasionally comes across as blatantly caring more for his daughter than his son, and not thinking about Tom nearly as much as he does Murphy. The novelization, while still emphasizing Murphy more (justifiably, since she is more important to the plot than Tom), evens it out better.
• Several characters occasionally address/refer to Cooper as "Coop."
• Cooper addresses TARS as "Slick" multiple times.
• Cooper, and some others, refer to his daughter Murphy as "Murph"
• Cooper is alarmed when TARS states that his Honesty setting is set to 90%, but TARS explains that Brutal Honesty isn't always preferable. Ultimately Cooper trusts the robots and they never betray him or the mission. This also allows him to joke about the astronauts being "slaves for [his] robot colony".
• We even get a scene that almost mirrors events of 2001, where CASE gets a message to relay to Cooper containing information that would hinder the mission he's perfectly capable of editing the message to remove the ending, but instead he shows it to Cooper and company in full.
• Cooper suffers this twice. First, Dr. Mann cracks Cooper's visor and takes the emergency oxygen packs out of his suit, leaving him to die. Later, Cooper is teleported out of the black hole and floats in free space near Saturn, only to be rescued by one of NASA's scouting Rangers.
• Humanity is suffering this on Earth. The reason why the Blight will eventually kill every human on Earth isn't starvation rather, its uncontrollable growth will reduce the level of oxygen to the point where everyone will suffocate to death. It's already starting to happen when Murph is an adult.
• Dylan Thomas' poem "Do Not Go Gentle Into That Good Night".
• "90 percent" (referring to TARS's honesty setting, with the explanation that Brutal Honesty is not always the way to go, and the subsequent agreement among the crew to stick by this). At the end, Coop rebuilds him and ups his honesty to 95%.
• Kip Thorne, one of the world's leading experts in astrophysics, put an amazing amount of research into the film, but this film is the first place many viewers will have seen it. And some of it &mdash the Time Travel in particular &mdash is bleeding-edge theoretical, and may be disproved in the future &mdash especially as Thorne was actually able to use Double Negative's (the CGI effects company that worked on the film) resources to make significant theoretical advances. Of particular note is a disagreement between Thorne and Roberto Totta over those advances.
• There is at least one major aspect of the film that is pure artistic license. A planet orbiting a black hole's accretion disk would not be habitable by humans for numerous reasons. For starters, while accretion disks do give off radiation, very little of it is visible light of the sort that humans would need most of it is deadly X-rays and gamma rays. The planet would also be at risk of being torn apart by the black hole's tidal force (that is, the difference in the pull of gravity on different sides of the planet).
• The conditions on Mann's planet are the exact opposite of what they should be. Chlorine is rather heavy and should have drifted to the surface instead of staying in the higher elevations. The film does imply a plausible reason, as Mann was lying about these conditions and no one caught the mistake .
• Despite being a common misconception, land plants don't generate the world's oxygen, which is mostly generated by algae in the ocean. And it would be fairly easy to concentrate oxygen from the atmosphere to supply people with it, even if the atmospheric oxygen content were to significantly drop (which it wouldn't do on the order of decades, but centuries to thousands of years).
• While it makes for a very interesting scene, frozen clouds, as they are displayed, are just not very likely, namely due to the implication that they were actual clouds formed from water vapor or similar substances. It is scientifically possible for ice mountains to form like that, but it's unlikely that they would take the shape of clouds in the process.
• The incredible time dilation of Miller's planet may seem to be an exaggeration &mdash indeed, for a non-rotating (Schwarzschild) black hole, such dilation is impossible. But for a rotating (Kerr) hole, as described in Kip Thorne's The Science of Interstellar, the time dilation level in the film is actually possible &mdash if the conditions are just right. The black hole must be spinning near the speed of light for the effects to be as strong as in the film. The real problem with the black hole comes from its visual appearance in the film. As a fast-rotating Kerr black hole, Gargantua would look highly asymmetric due to its spin in real life, looking very different from how it does in the film note It is asymmetric in the film, but only very slightly, so you need to know to look to spot it look at how the accretion disc hugs the edge of the disc more closely on one side than the other. . Thorne, who would otherwise never agree to unscientific elements in the film, had to concede this small but noticeable inaccuracy to Nolan so that audiences would not be confused by an asymmetric black hole.
• There is no realistic way the solid surface of the water planet would be that smooth if there are tidal and rotational forces as powerful as what there would need to be for waves that big. The probe debris being that close together is also unlikely, especially since it is all invisible a moment before it.
• Several of the orbital maneuvers performed by Cooper defy real-life orbital mechanics. Of particular note is the sequence where Cooper attempts to dock with the damaged Endurance, where several such inconsistencies appear in rapid succession.
• Objects spin around their centers of mass. After it's damaged by Mann's failed attempt to enter , the Endurance is no longer symmetrical, and would not spin around its former center, where the central hatch is located. Therefore, Cooper would not be able to dock it, no matter how fast his module is spinning around its own center of mass.
• It's conceivable the Endurance was designed for situations like this, considering how catastrophic such a situation would be, plus simple modularity. The ship could have ballast in all the modules that could be ejected as needed. Or even, perhaps, large hidden masses in each of the modules that can be moved toward or away from the center.
• The accident causes the Endurance's orbit to decay enough to cause it to enter the planet's stratosphere. This would require an external force to act against the ship's direction of travel. No such force was applied in the movie. And assuming the Endurance's orbit was indeed sufficiently high to be stable before the accident, it would then need to have been slowed down significantly in order to fall so quickly into the stratosphere (again, no such force was applied during the accident).
• In order to enter the stratosphere at all, the Endurance would have needed to be in such a low orbit that it would already be grazing the stratosphere. This in itself would have made the orbit decay and destroy the ship well before anyone returned to it.
• A small boost from Cooper's ship is apparently enough to not only deflect the Endurance out of its sub-orbital trajectory but enough to get it to interplanetary velocity by mistake.
• The sequence when Cooper docks with the out-of-control Endurance is evocative of a emergency that occurred during Neil Armstrong and David Scott's 1966 Gemini 8 mission, when a thruster failure while practicing docking and rendezvous caused the spacecraft to spin violently out of control, which Armstrong was able to recover from. It also harkens back to the Soviet mission Soyuz T-13, in which spacecraft rotation had to be matched to dock with a derelict Salyut 7.
• Mann's fate is evocative of (although not nearly as graphic as) the horrific 1983 Byford Dolphin diving bell incident.
• The genuine recollections of Dust Bowl survivors.
• Cooper had to leave his family for possible dead as he ventured out to find a suitable planet.
• Mann is described multiple times as "the best of us" and "remarkable". However, by the time the Endurance crew reaches his planet and awakens him, all those years by himself, believing that he would die, have broken him and caused him to Go Mad from the Isolation .
• Professor Brand becomes this to everyone who learns the truth&mdashbut especially Murph, Cooper, and his own daughter Amelia&mdashwhen it comes out that he was lying to nearly everyone about "Plan A." He believes that everyone still on Earth is doomed. Earth does eventually succumb to the dust bowl and part of humanity is saved by Murph, but it's not thanks to him .
• Judging by the way Dr. Mann is described as the best that NASA has to offer, and the fact that he went on a possible suicide mission in the first place, he must have been a respected and idolized astronaut. In spite of it all, he makes a false report that endangers the entire human race just to save his own skin .
• Possibly Cooper to Tom later on in the film, due to the stress of Tom not seeing his father for years and losing his firstborn child. When Murph tries to convince him to follow her advice based on something having to do with their father, Tom notes that their grandfather was the person who really raised him, not Coop.
• Briefly Coop to Murph as well, when, after years of already resenting him for leaving, she learns that Plan A is a sham and comes to fear that her father knew about it and abandoned her, Tom, and the rest of humanity to die. However, she eventually realizes that her father was her "ghost" all along and was giving her the info she needs to save humanity, and he becomes a Rebuilt Pedestal as a result. However, she does move on from waiting for her dad after this when he does return, she's glad to see him but tells him that she has filled her life by surrounding herself with a family that love her, and that he should seize the opportunity that being still young gives him.
• The wristwatch Cooper leaves for Murph as a Memento MacGuffin he later uses it to communicate the gravity equations to her. Before that (or after that depending on the moment you focus on in the movie), both the dust from the storm and the books in Murph's room.
• Cooper (from the future) turns out to have been Murph's "ghost" the whole time.
• Both Cooper and Murph know Morse code, which is important later when Cooper communicates with Murph via books and wristwatch.
• Murph's shared penchant with her father for scientific pursuit and her talents as a Child Prodigy result in Professor Brand taking her under his wing, and she grows up to work with NASA just like her dad.
• While it is only implied in the film proper, the official novelization and official prequel comic both confirm that Mann betrayed his Robot Buddy, KIPP , years before the beginning of the movie. Since KIPP found enough data to prove that Mann's planet was uninhabitable, Mann had him prepare a "hypothetical" set of perfect data for his planet that he used to create forged data, shut KIPP down before he could send any of his findings to Earth, and booby-trapped him to explode if anyone tries to access the archives that would reveal his deception. He did all this to a robot who was programmed to be his loyal helper and companion .
• He also betrays the entire Endurance crew who comes to rescue him. Once Cooper decides to return to Earth (but still help the others get settled on Mann's planet before he leaves), Mann resolves to kill him and Make It Look Like an Accident to the rest of the crew so he can still keep them as companions. When that fails (thanks to Cooper surviving and Romilly dying by inadvertently setting off the bomb), Mann decides, "Screw This, I'm Outta Here!" and tries to steal the Endurance and abandon all the remaining crew on his planet .
• On a less personal level, he also betrays the Lazarus mission (which he led) and all his fellow humans by succumbing to his cowardice and falsifying the data about his planet, which puts the entire Endurance mission at risk and almost destroys the possibility of either Plan A or Plan B&mdashand thus, the future of the human race .
• NASA's secret base is within a night's drive of Cooper's farm.
• Try calculating the odds of finding a lone astronaut floating in space just in the nick of time - but also note that the bulk beings who brought Cooper into the Tesseract also created the wormhole in the first place. Spitting him out right where the spacers would find him instead of just dropping him off inside the station itself is just them granting present-day humans their Willing Suspension of Disbelief.
• The Ranger is a VTOL Space Plane with a fighter jet aesthetic that chariots our heroes wherever they need to go.
• The bulkier and less sleek-looking Lander, designed to haul heavy scientific equipment, proves its own worth when Dr. Mann steals the Ranger and Brand uses the Lander to rescue Cooper .
• Subverted with Laura Miller. She is dead by the time the Endurance crew reaches her planet however, due to severe time dilation there, Brand speculates that, by this planet's time, she only arrived a couple of hours before they did, and probably only died minutes before they landed .
• Played straight with Wolf Edmunds. Once the crew gets through the wormhole, they learn that Edmunds&mdashwho was sending the "thumbs-up" signal for his planet&mdashstopped transmitting three years prior (i.e. a year before the Endurance left Earth). When Brand reaches his planet at the end of the movie, she discovers that this is because a rockslide crushed his habitat, including his beacon and his stasis pod, and killed him, meaning that, regardless of which order the crew visited the planets, she never had any chance of seeing him again. Still, Brand was right to believe in her love for Edmunds, since his planet is the only one of the three that is actually habitable .
• TARS, the sarcastic robot.
• Cooper, the sarcastic human. Not surprisingly, leads to some good old-fashioned Snark-to-Snark Combat between the two.
• The original twelve astronauts were chosen based on their skills and their lack of any familial connections. Everyone fails to consider that this means they would have nothing to lose by falsifying the data and claiming their world is habitable when it isn't .
• Dr. Brand looks slightly puzzled but mostly straight-faced as Cooper exchanges goodbyes with TARS when the latter detaches and drops into the black hole ("See you on the other side, Cooper" "See you there, slick!"). It's not until CASE announces that Ranger 2 (Coop's ship) is about to be detached that she panics and pleads for him not to do it. only to get the 90% rule response from Cooper before he detaches despite her protest.
• The wormhole is represented as an instantaneous passage between two points in space separated by millions of lightyears, yet it is possible to see through the "aperture" from multiple angles due to its spherical shape in physical space, allowing probes to scan the far galaxy in all directions to identify potential new worlds for humans to settle. Later when the Endurance passes through, it takes a noticeable amount of time to make the trip, and while inside the crew witnesses a prolonged passage of other stars and galaxies on the way. Also, the ship's controls do nothing inside, because it's not physical space.
• Miller's planet at first appears relatively earth-like and normal, but two big differences play a key role in the plot: time passes at approximately 1/60,000th of the Earth's rate, due to the extreme bending of space that occurs as a result of the planet orbiting right outside the event horizon of a black hole, and the surface (which is a uniform ocean only two feet deep across the entire planet) is regularly swept with literal mile high tidal waves, again due to the planet's close proximity to the black hole. It's also uniformly well lit like a slightly overcast day on Earth, but the planet has no sun - that light is caused by the black hole again, due to light bending around it from all directions.
• The black hole, both inside and out. On the outside it appears as a blacker than black sphere surrounded by a nimbus of light from all directions, which makes sense - the gravity causes such extreme distortions in local space that all light is literally wrapped around it. On the inside, Cooper flies through a stream of particles that steadily increase in size and speed which destroy his ship, forcing him to eject, which lands him inside the Tessaract. Also, his ship's systems mostly fail and he has no control because like the wormhole, the inside of the black hole is not physical space.
• At the baseball game, numerous people somehow fail to spot a gigantic dust cloud until it's almost on top of them.
• The waves on the water planet are only visible when the plot requires them to be, despite being so massive they can be seen from orbit. Also, when the crew is digging around in the water, there is no way the probe debris would be that close together with waves that powerful tossing it around. Finally, when the crew is looking for the probe wreckage, they proclaim they should be right on top of it when their robot steps on it and digs it out. at which point the camera pans to reveal that they're surrounded by brightly-painted pieces of metal they should have been able to see with ease. It also floats to the surface right after they proclaim they can't see the wreck. Though it's possible the water being drawn up into the advancing wave exposed them as the water receded.
• Cooper, through his bond with his family.
• Brand becomes this as well after the visit to Miller's Planet, which at first gets dismissed by the other characters.
• Quite a bit towards the "ghost"'s identity:
• Murph says that she calls the mysterious happenings in her bedroom the work of a "ghost" not because she's afraid of it, but because it feels like a person. Turns out she's right, and it is a person. namely, her father reaching out to her from the future .
• After being woken up by Cooper's nightmare in the beginning, Murphy even mentions "I thought you were the ghost", and it turns out he was the ghost all along.
• While leaving home, Cooper says to Murph, "Once you're a parent, you're the ghost of your children's future". Cooper is Murph's ghost in his future, and is able to use this to play a vital role in her future.
• The wormhole. How is a hole drawn on paper? A circle. What's the three-dimensional form of a circle? A sphere. This is also a hint that They don't really understand humanity that well Cooper mentions that he would have expected a wormhole to be, well, a hole.
• When Cooper enters the black hole with TARS, he deduces that the "aliens" have spread time into physical dimensions as snippets of Murph's timeline so that it's understandable to Cooper and makes it easier for him to communicate with Murph. The sympathy that he feels from this, and the fact that he deduces it so easily, makes him think that the creators of the black hole are futuristic humans who chose Murph (and him to a lesser extent) to be bookmarks and catalysts for humanity's exploration of space.
• Dr. Mann, to the point of lying about his planet's habitability, attempting to kill Cooper, setting a trap which kills Romilly, and docking his Ranger with Endurance while ignoring all warnings Cooper and Brand give him. Also, he gave in to selfish temptation from the isolation, as he knew that he would be never be rescued if his world was not habitable &mdash meaning all the Endurance's colony supplies would have been good for was keeping him alive a few more years.
• Downplayed for Romilly. He's stiff and awkward after 23 years of isolation on the Endurance (he spends a good portion of that time in cryosleep, but it isn't enough), but he's still sane and sober enough to continue functioning mostly-normally and remains helpful to the crew up until his death .
• Cooper's attempt to dock the Lander with the Endurance, while the ship is spinning out of control and falling into the atmosphere. It's a suicidally-risky maneuver, but if they lose the Endurance, they lose any remaining chance of saving humanity.
• Dr Mann tries to smash Cooper's helmet with his own. When Cooper points out he has an equal chance of cracking his own helmet, Mann replies that a 50/50 chance are the best odds he's had in years.
• Played straight with most of the astronauts of the Lazarus Mission, three-quarters of whom traveled to planets that they found to be completely uninhabitable, and thus are doomed to never be rescued and die there.
• Defied by Dr. Mann, who refuses to die alone on his planet and endangers the entire mission just to survive.
• Attempted but subverted by both Cooper and TARS, who let themselves fall into Gargantua so Brand and CASE can continue, but are saved by the Bulk Beings, who first place them in a tesseract that allows Cooper to make minimal contact with Murph in the past and give her TARS's data to save humanity, and then, after that's completed, send them back through the wormhole to a location where they can be rescued by the Plan A habitat.
• The crew agrees to land on Miller's planet, as she has been transmitting positive signals for years. However, as the planet is within Gargantua's gravity distortion, time there runs extremely slow, to the tune of one hour to every seven years outside, so they won't have long to assess viability. It never occurs to them that Miller likewise couldn't have been there very long by her planet's time&mdasha couple of hours at most&mdashand thus her signal is repeating because she simply hasn't had long enough to transmit anything else. Cooper is furious, and Brand admits she screwed this up after the fact.
• The same can be said of both Brand and Doyle while they're on Miller's planet:
• Despite Cooper issuing four warningsthat the next wave was approaching, which Brand can plainly see, she still insists on trying to recover the data recorder, and doesn't seem to realize how pointless it is until she falls and gets pinned beneath the wreckage. This makes her indirectly partially responsible for Doyle's death, since he sends CASE to save her and has to manually override the shuttle's outside hatch. mere seconds before being swept away by the tsunami .
• However, Doyle is also partially responsible for this after CASE gets Brand back to the Ranger and they both jump in, Doyle doesn't immediately jump in after them, and instead spends a few seconds gawking at the giant wave about to crash down on them. By the time he turns back to try to get in the Ranger, it's too late, and Cooper has had to forcibly close the outer hatch so the water doesn't get in and kill them all, trapping Doyle outside and causing him to be killed by the tidal wave .
• Miller who probably died when her ship was hit by the Giant Wall of Watery Doom.
• Also Edmunds, Brand's former significant other , who seems to have perished in a rock slide many years prior to her arrival (almost certainly when he stopped transmitting, which was a year before the Endurance mission leaves Earth) .
• Joseph Cooper is almost always referred to as Cooper or by his nickname "Coop" rather than his first name. His son even nicknames his grandson "Coop" in his honor.
• Wolf Edmunds and Laura Miller only get their first or full names stated once or twice, and Mann's first name is All There in the Script otherwise, they are referred to by their last names, which is Truth in Television (usually) for space flights.
• Cooper reveals that his wife died of a brain tumour in the opening portions. The fact that doctors were not able to save her is implied to be a big part of his motivation to go on the mission.
• Amelia is revealed to have one in Edmunds . He may or may not be alive, but she's partially motivated by the chance of seeing him again. He's confirmed to be dead in the epilogue .
• Brand finds out Cooper is about to detach himself from the Endurance to ensure her safe onward travel to Edmunds' planet, and distraughtly cries that he told her there were enough resources for both of them to make it. He responds with "We agreed, didn't we? 90% [honesty]."
• When Brand is making her somewhat bizarre speech after returning from Miller's Planet, she is repeating a lot of the words Cooper said to Donald on the porch of his house when explaining why he is going to leave on the mission to another galaxy.
• The Endurance, which besides its obvious implications championing the human psyche's resilience, is also the name of Ernest Shackleton expedition's ship. Bonus points for Mann's planet being an Antarctica-like Single-Biome Planet, with the Endurance orbiting it while the crew go off to search for Mann.
• Amelia Brand, the only female astronaut on the Endurance, is a callback to Amelia Earhart: the first female aviator to fly across the Pacific solo. At the end of the movie, Brand also successfully lands on an inhabitable planet &mdash and her whereabouts to the general public are unknown.
• Murph Cooper, whose name's meaning is blatantly discussed when Cooper explains that Murphy's Law actually means that anything could happen. Against all odds, Cooper ends up in a pocket dimension when he passes through Gargantua. He manages to transmit critical gravitational data TARS gleaned from the black hole to Murph through her room, so she can solve Brand's equation and save humanity.
• Dr. Hugh Mann, who eschews principle for base survival instincts when he realizes that his planet is unsustainable.
• The Lazarus Project, in a round-about way. Meaning to be about humanity's capacity to come back from the brink (existential "death"), it also fits in that Dr. Mann, the sole surviving member of the space-bound part of the project, is awoken from his cryo-pod, which was set to keep him asleep until it terminally malfunctioned or someone came-literally "awakening the dead". This one also gets a Lampshade Hanging &mdash Cooper, concerned, notes that "Lazarus had to die first", as well as quipping to Mann that he "rose like Lazarus from the grave" when they opened his hibernation chamber (as Mann was not expecting to be woken up ever again).
• Professor Brand, shortly after being introduced, informs the protagonist and the audience about the history that has lead the Earth to such a dire situation and NASA's attempts to save humanity.
• Romilly, from describing how a wormhole looks and behaves (from without andwithin), to stating numerous times that information cannot be garnered from a black hole, fits the bill of Mr. Exposition Scientific Advisor.
• In the final chapter, when Cooper wakes up in the hospital, the doctor gives him Infodumps about where he is and what happened to him.
• For Cooper: Donald and Tom. Donald dies somewhere in the 23-year-Time Skip while the crew was at Miller's planet, and (according to the novelization) Tom dies over a decade before Cooper reaches the human space colony at Saturn .
• For Brand: Her father and Edmunds (her former lover). Professor Brand dies while the crew is on the way to Mann's planet, and at that point, Amelia (and the entire crew) have been unable to send messages out for a long time. Edmunds died three years before the Endurance crew crosses through the wormhole (when his signal stopped transmitting) due to a rockslide that crushed his pod while he was in cryosleep .
• Part of the reason that the lone woman on the mission, Amelia Brand, is included is that she's Professor Brand's daughter. However, she is also a legitimately vital part of the crew in her own right as the biologist on board, she's the expert on and in charge of caring for the fertilized eggs of Plan B. She's the only human member of the crew to make it to Edmunds's habitable planet (with only the non-human CASE there to help her) and is heavily implied to have started enacting Plan B there, which would have been much more difficult for any of the other crew members to do.
• Likewise, Murph only gets to meet her eventual-mentor Professor Brand, who brings her to NASA to be educated, thanks to him knowing her father. However, the Professor is still quick to notice her intelligence after meeting her, and specifically seeks her out on her own merits after her father is gone to take her under his wing.
• The early parts of the film are spliced in with interviews with elderly people living in a "Dust Bowl". Given that they're speaking in English, with US accents, the viewer may well assume that they are talking about the USA's Great Depression-era Dust Bowl. note In real life, this is actually what these are real interviews from a documentary about the Dust Bowl are used for this movie. The ending scenes reveal that those interviews were part of a museum exhibit on Cooper Station about life on earth in the first part of the film.
• When Cooper is in the tesseract and realizes that he's Murph's "ghost", we get this for the various gravity anomalies seen in the first act, which are shown to have been caused by him using gravity to interact with Murph's bedroom. We even see some of the same scenes again, such as the binary dust lines, with this new context.
• Tom's first child, Jesse, dies from lung disease.
• Discussed by Murphy on her death bed. She sends Cooper away so he won't have to see her die .
• Professor Brand to Murph, as he wishes to cultivate her genius she grows up to work as an astrophysicist in NASA under his tutelage. Since her father and his daughter both left on the mission, they become like a surrogate father and daughter to each other. Her image of him is completely dashed when he confesses on his deathbed that he had always intended to restart the human race at the expense of Earth's current inhabitants.
• Also, Donald, who is charged of taking care of Murph and Tom while Coop is away. Tom particularly seems to regard him as his primary father figure after years of not hearing from Coop .
• TARS is Properly Paranoid about Dr. Mann and prepares accordingly, but never thinks to voice this concern to his fellow crew at any point. Had he done so, Romilly may have survived and the Endurance wouldn't have been badly damaged by Mann's failed docking attempt . Based on the way robot programming often works, the fact that that they do this on their own initiative, and the fact that CASE (who also knows about it, since the robots are in constant contact with each other) reveals it as soon as it becomes relevant to the current conversation, it's possible they don't bring it up beforehand because nobody asks them about it.
• When Cooper and Brand try to warn Mann not to open the hatch of his vessel to the Endurance because the robots have turned off the automatic locking and thus he would be propelled into space as soon as he does so, their calls just consist of them repeatedly telling him not to open the inner hatch without explaining why, failing to convey the danger he's in. Had they started explaining the danger as soon as they call him, he might have listened to them.
• Cooper's love for his daughter enables him to literally transcend time and space to give her younger self the information required to save humanity .
• Brand believes in her love for Edmunds. She makes an unscientific speech about that topic and is bashed for it. She turns out to be right.
• The giant dust clouds were created on location using large fans to blow cellulose-based synthetic dust through the air.
• The film's spacecrafts, as well as the robot companions, are almost entirely physical models/miniatures, and pretty much every scene (except for exterior space shots, of course) was shot in a real location.
• On Miller's planet, there is no life and nothing to sustain it, because not enough time has passed there to allow evolution to do its thing, and also likely because the violent tidal waves would make it quite difficult for complex life to survive there.
• NASA chose Lazarus mission crew members with no strong attachments to leave behind. One of them turns out to have no higher priority than his own survival and is willing to jeopardize the entire human race for a better shot at living.
• Earth society might be slowly crumbling and the people starving due to the blight, but that doesn't mean all aspects of civilization have vanished. Ignoring the NASA facility which operates in secret (and where apparently you can still buy drinks in old-fashioned cups with plastic lids and straws), people are still driving cars and trucks (suggesting fuel sources still exist), there is still electricity and there is still some form of internet.
• Dr. Mann sets one of these as a booby trap in his robot, KIPP , set to go off if anyone tries to access KIPP's archives (which would reveal the real data, which contradicts the stuff that Mann forged) . The bomb ends up killing Romilly .
• TARS pretends to start a destruct sequence while Cooper is tinkering with his settings.
• To Star Wars &mdash Protagonist in two-seater spaceplane with robot buddy in the back.
• From the robots shaped like monoliths to the score, the film is a love letter to 2001: A Space Odyssey. Even some of the dialog from Dr. Mann is oddly similar to that of HAL's. It's a bit of foreshadowing that Mann is going to be the character that will cause problems later on, rather than TARS and CASE that audiences would be expecting. Then there's an In-Universe example of the wise-cracking robot making a joke about blowing the team out of the airlock.
• Mann trying to enter the airlock with a vessel that can't seal with the entrance is also similar to Dave Bowman's dilemma in 2001. However even though Mann has his space helmet, unlike Bowman he dies because he can't seal the airlock properly before opening the inner hatch.
• Among the books in Murph's collection is Stephen King's The Stand &mdash kind of relevant, given the post-apocalypse of Earth.
• Early in the film, Cooper refers to his son Tom by the nickname Servo.
• In the movie Contact, McConaughey’s character gives Jodie Foster’s character a compass before she goes on her space voyage, and tells her it might just save her life (which it eventually does). The same actor in a similar movie performs the same gift-giving act with a similar gift that turns out to have similar plot results.
• Both of the Robot Buddies on the Endurance are named after characters from famous science-fiction novels. TARS is named after Tars Tarkas from John Carter of Mars, and CASE is named after Henry Case from Neuromancer.
• The scene of Cooper's launch (partly replaced by him driving his pick-up with launching sound effects) is a homage to Solaris (1972) (where the launching of the cosmonaut is totally replaced and symbolized by long shots from a car on a Japanese highway). Also the oceanic planet, of course.
• Donald's (John Lithgow's) comment "I want a hotdog" calls to mind the conversation about hotdogs in 2010: The Year We Make Contact, in which Lithgow played one of the characters.
• Romilly's explanation of travel through a wormhole - folding a piece of paper in half and pushing a pen through it - mirrors Weir's explanation of the gravity drive in Event Horizon.
• Like in 3001: The Final Odyssey, the protagonist is found floating near Saturn.
• A lot of the movie draws from Kip Thorne's theories of wormholes and portrays them accurately. Kip Thorne is an executive producer on the film. Moreover, the portrayal of other physics is also accurate, at least in comparison to most sci-fi movies/tv shows. Gravity doesn't suck, space isn't air and isn't noisy, and the engines are fired only when they need to be. Even the insane time dilation on the ocean planet is plausible &mdash Thorne worked out that the time dilation could exist if the planet were deep inside the black hole's gravity well and if the black hole were spinning extremely fast. Really the only thing one could get worked up over is how much fuel the spacecraft are using and the existence of the "ice clouds". note The ice clouds are the one big concession of scientific accuracy that Thorne made to Nolan when the movie was being written &mdash originally, the script had a spacecraft going faster than the speed of light!
• The film's black hole is often cited as the most realistic depiction of a black hole ever, as it was based on the actual equations which model black holes. It should be noted that lower-quality renderings have been made before for scientific papers and the like, but Interstellar is likely the first time such an equation-based rendering has been made at a photorealistic level of fidelity.
• To acquire inspiration for real-world space travel, Christopher Nolan invited former astronaut Marsha Ivins to the set.
• Mann has a pretty bad case of it, though it's more the isolation that has taken its toll on him.
• Romilly gets a mild case when he starts to fret about how thin the spaceship hull is, with all that nothingness behind it. Cooper soothes Rom by giving him his recorded nature sounds to listen to.
• The Endurance slingshots the black hole in order to get up to the speed needed for the onward flight to Edmunds' planet.
• Before the arrival at the wormhole, they make a flyby of Mars, which could have slightly given it velocity.
• The film acts as this towards Inception. Time passing by at bizarre speeds, a father being separated from his children in a faraway location, a Tragic Villain who almost screws up the team's goals , Nolan being in the Sci-Fi genre again, etc. Of course, the main difference is that the scale is much grander than Inception's smaller focus. Even lampshaded by Nolan himself in a couple of interviews.
• The film is a more practical and straightforward version of 1997's Contact, which also deals with an otherworldly message that triggers a drive for space exploration. In Interstellar's case, they actually have the logistics for space travel and Cooper speculates that the messengers are advanced futuristic humans , whereas in Contact, the beings basically provide the means and are extraterrestrial .
• First by 2 years, when the Endurance travels from Earth to Saturn to enter the wormhole.
• Next by 23 years, 4 months, and 8 days, which is how long half the crew spends on Miller's planet in "real" time (though, thanks to Time Dilation, from their perspective, they're only there for a few hours).
• Lastly, by more or less 60 years. Unlike the previous example, we don't see the direct effects of this until the denouement. At this point, Cooper is told he is chronologically 124 years old, but biologically, he is still only 40-ish.
• Murph is played by Mackenzie Foy as a child, Jessica Chastain in her mid-30s and Ellen Burstyn as an old woman.
• Tom is played by Timothée Chalamet as a teen and Casey Affleck as an adult.
• Brand should have listened to Cooper and hurried back to the Ranger, especially since she can see that the next wave is practically on top of them. Instead, she insists on trying to retrieve the data log anyway and Doyle dies saving her .
• Doyle as well in the same situation, since he freezes up and gawks at the approaching wave multiple times instead of continuing onward. The first time, Cooper has to tell him to keep moving, and the second time, it costs him his life even after Brand and CASE have already gotten into the Ranger, Doyle doesn't immediately jump in after them, but still keeps staring at the wave, and gets locked out of the Ranger and swept away .
• Mann really should have listened to the three separate warnings telling him not to open the airlock.
• The trailers also depict Mann's pod exploding, debris spewing from Endurance, and the waves of Miller's Planet.
• The Endurance crew comes to rescue him and, as he puts it, "literally raised [him] from the dead". In fact, since Mann's planet was uninhabitable, they shouldn't have even come there in the first place, and only do so because Mann faked his data to make it look like the planet could sustain human life. Mann repays them by 1) trying to kill Cooper to keep him from going home, 2) actually killing Romilly (indirectly) when Mann's booby trap in KIPP blows him up, and 3) attempting to steal the Endurance and continue the mission on his own while leaving Cooper, Brand, TARS, and CASE marooned on his planet.
• This also applies to his treatment of his Robot Buddy, KIPP . The robots are programmed to be loyal to their human masters, and the entire Endurance crew appreciates and values their robots, TARS and CASE, forming good friendships with them. By contrast, the novelization and prequel comic reveal that Mann threatened to shut KIPP down multiple times before actually doing so to keep him from transmitting their true findings to NASA (which is just KIPP doing his job), and then rigged him with a bomb to explode if anyone tries to access the real data in the archives .
• Cooper's relationship with his son Tom, as well as the daughter-in-law and grandson whom he never meets, seems to be thrown aside in favor of his bond with his daughter. Given the time skips, it's probable that he dies of old age and that his descendants are in the room with the rest of Murphy's family in the end . In any event, he isn't seen or mentioned in the finale in the movie, though the novelization expands on it by confirming that Tom died over a decade before Cooper arrives on the station .
• Seeing as it would have been physically impossible to evacuate everyone on earth, not without building thousands if not hundreds of thousands of ships the size of the colony, the finale avoids the issue of how many had to be left behind &mdash and, indeed, whether any group other than Americans even made it off the planet.
• The Indian Drone is collected in an exciting scene and then that subplot is shooed aside. Presumably, they stripped it for parts. The film never explains why an Indian Drone has been flying over the US for 10 years, and if it's meant to tie in with the gravitational anomalies, it isn't made clear or elaborated on at all.
20.6 Interstellar Matter around the Sun
We want to conclude our discussion of interstellar matter by asking how this material is organized in our immediate neighborhood. As we discussed above, orbiting X-ray observatories have shown that the Galaxy is full of bubbles of hot, X-ray-emitting gas. They also revealed a diffuse background of X-rays that appears to fill the entire sky from our perspective (Figure 20.19). While some of this emission comes from the interaction of the solar wind with the interstellar medium, a majority of it comes from beyond the solar system. The natural explanation for why there is X-ray-emitting gas all around us is that the Sun is itself inside one of the bubbles. We therefore call our “neighborhood” the Local Hot Bubble, or Local Bubble for short. The Local Bubble is much less dense—an average of approximately 0.01 atoms per cm 3 —than the average interstellar density of about 1 atom per cm 3 . This local gas has a temperature of about a million degrees, just like the gas in the other superbubbles that spread throughout our Galaxy, but because there is so little hot material, this high temperature does not affect the stars or planets in the area in any way.
What caused the Local Bubble to form? Scientists are not entirely sure, but the leading candidate is winds from stars and supernova explosions. In a nearby region in the direction of the constellations Scorpius and Centaurus, a lot of star formation took place about 15 million years ago. The most massive of these stars evolved very quickly until they produced strong winds, and some ended their lives by exploding. These processes filled the region around the Sun with hot gas, driving away cooler, denser gas. The rim of this expanding superbubble reached the Sun about 7.6 million years ago and now lies more than 200 light-years past the Sun in the general direction of the constellations of Orion, Perseus, and Auriga.
A few clouds of interstellar matter do exist within the Local Bubble. The Sun itself seems to have entered a cloud about 10,000 years ago. This cloud is warm (with a temperature of about 7000 K) and has a density of 0.3 hydrogen atom per cm 3 —higher than most of the Local Bubble but still so tenuous that it is also referred to as Local Fluff (Figure 20.20). (Aren’t these astronomical names fun sometimes?)
While this is a pretty thin cloud, we estimate that it contributes 50 to 100 times more particles than the solar wind to the diffuse material between the planets in our solar system. These interstellar particles have been detected and their numbers counted by the spacecraft traveling between the planets. Perhaps someday, scientists will devise a way to collect them without destroying them and to return them to Earth, so that we can touch—or at least study in our laboratories—these messengers from distant stars.
Artifacts In Space
All of a sudden, we have spacecraft and objects both coming into our solar system and leaving for interstellar space. This is highly unusual, and very intriguing.
The departing spacecraft is Voyager 2, which launched in 1977 and has traveled spaceward some 11 billion miles. It has now officially left the heliosphere, the protective bubble of particles and magnetic fields created by the sun. In this it follows Voyager I – which left our solar system in 2012 — and managers of the two craft have reason to think they can travel until they cross the half-century mark.
This is taking place the same time that scientists are puzzling over the nature of a cigar-shaped object that flew into the solar system from interstellar space last year.
Nobody knows what the object – called Oumuamua, Hawaiian for “first messenger,” or “scout” – really is. The more likely possibilities of it being a comet or an solar system asteroid have been found to be inconsistent with some observed properties of the visitor, and this has led some senior scientists to even hypothesize that it just might be an alien probe.
The likelihood may be small, but it was substantial enough for Harvard University Astronomy Department Chairman Avi Loeb to co-author a paper presenting the possibility. In the Astrophysical Journal Letters, Loeb and postdoc Shmuel Bialy wrote that the object “may be a fully operational probe sent intentionally to Earth vicinity by an alien civilization.”
They also say the object has some characteristics of a “lightsail of artificial origins,” rather like the one that Loeb is working on as chairman of the Breakthrough Starshot advisory committee. The well-funded private effort is hoping to develop ways to send a fleet of tiny lightsail probes to the star system nearest to us, Alpha Centauri.
Put the two phenomenon together — the coming into our solar system and the going out — and you have a pathway into the world of alien “artifacts,” products of civilizations near and far. They are the kind of “technosignatures,” the potential or actual handwork of intelligent beings, that NASA is now interested in learning about more.
We know this because during a fall conference in Houston convened by NASA at the request of members of Congress, scientists were brought together to discuss many different kinds of potential signs of intelligent extraterrestrial life. While artifacts were one of many topics discussed, the term carries a quite magnetic pedigree.
So far, that meaning is of course fictional, or a misreading of actual features. There is perhaps most famously the monoliths from the movie “2001: A Space Odyssey” and then the myriad sightings of alien spacecraft that turn out to be anything but that.
And then there’s the “Face on Mars.”
The original image taken by Viking 1 looked somewhat like a human face. The feature, found in the region where the highlands meet the northern plains of Mars, was subsequently broadly popularized as a potential “alien artifact,” with even a major motion picture.
So many people were convinced that an image had been sculpted on the surface of Mars that NASA ultimately put out a substantial release in 2001 to make clear that the face was actually a mountain.
That was after the Mars Global Surveyor orbiter determined that the “face” was created by unusual reflections in an otherwise ordinary Martian mountain.
So alien artifacts surely and properly have a steep hill to climb before they can be taken at all seriously.
But does that mean they shouldn’t be taken seriously at all? Loeb clearly says no, that they are a potential source of important and compelling science, even if they turn out to be unusual but natural phenomena.
And then there’s the question raised in the Houston “technosignatures” conference: What actually is meant by an artifact?
Longtime SETI scientist and advocate Jill Tarter, for instance, wondered if the signatures of intelligent civilizations could be imprinted on neutrinos. She said that a leak of the radioactive isotope tritium, which has a short 12-year half-life, could also signal the presence of advanced life because (unless it’s near a supernova) it would have to come quite recently from a nuclear reactor.
Taking it further, she and others argued that artifacts of intelligent life would include many atmospheric and planetary changes that could only be accomplished by intelligent beings. For instance, the presence of unnatural pollutants such as chloroflurocarbons ( CFC s) or sulfur hexafluoride (SF6) in an exoplanet atmosphere would, in this view, be an “artifact” of civilization.
Back, now, to Voyager 2, which is for sure an extraterrestrial artifact.
Voyager 2 was launched by NASA in August, 1977 to study the outer planets. Part of the larger Voyager program, it was launched 16 days before its twin, Voyager 1, on a trajectory that took longer to reach Jupiter and Saturn but enabled further encounters with Uranus and Neptune.
Both have traveled far their original destinations. The spacecraft were built to last five years and conduct close-up studies of Jupiter and Saturn. With the spacecraft holding up despite the rigors, additional flybys of the two outermost giant planets, Uranus and Neptune, proved possible, and then the Voyagers were directed to interstellar space.
Their five-year lifespans have stretched to 41 years, making Voyager 2 NASA’s longest running mission ever.
At the on-going American Geophysical Union annual meeting, NASA project manager Suzanne Dodd said she believed that Voyager 2 can keep functioning for 5 to 10 more years in this new region of space, though not with all its instruments operating.
The greatest concerns about keeping the probes operating, she said, involve power and temperature. The nuclear-powered Voyager 2 loses about 4 watts of power a year, and mission scientists have to shut off systems to keep instruments operating.
Voyager 2 is very cold — about 3.6 degrees Celsius and close to the freezing point of hydrazine — leading to concerns about the probe’s thruster that uses this fuel. Dodd says she’s set a personal goal of keeping at least one of the Voyagers going until 2027, making it a 50-year mission.
The cameras for both probes are no longer on. But before the camera on Voyager 1 was decommissioned, it took the iconic “Pale Blue Dot” picture of the Earth.
In preparation for the potentially deep space travels for the Voyager spacecrafts, both were fitted with a greeting for any intelligent life that might be encountered.
The message is carried by a phonograph record – -a 12-inch gold-plated copper disk containing sounds and images selected to show the diversity of life and culture on Earth. The contents of the record were selected for NASA by a committee chaired by space scientist and popularizer Carl Sagan. He and his associates assembled 115 images and a variety of natural sounds to give a sense of what Earth and Earthlings are like.
So are the Voyagers now artifacts from our civilization, messengers awaiting discovery by some distant beings?
Perhaps. But they actually have not even left the solar system, and won’t be leaving anytime soon. They are in what is considered interstellar space, but the boundary of our solar system is beyond the outer edge of the Oort Cloud, a collection of small objects that are still under the influence of the sun’s gravity.
The width of the Oort Cloud is not known precisely, but it is estimated to begin at about 1,000 astronomical units (AU) from the sun and to extend to about 100,000 AU. One AU is the distance from the sun to Earth. It will take about 300 years for Voyager 2 to reach the inner edge of the Oort Cloud and possibly 30,000 years to fly beyond it.
Astronomers have long predicted that objects from other solar systems get shot out into space and arrive in our system.
The first identified interstellar object to visit our solar system, Oumuamua, was discovered in late 2017 by the University of Hawaii’s Pan-STARRS1 telescope as part of a NASA effort to search for and track asteroids and comets in Earth’s neighborhood.
While originally classified as a comet, observations revealed no signs of cometary activity after it was slingshotted around the sun at a remarkable 196,000 miles per hour.
Oumuamua seems to be a dark red highly-elongated metallic or rocky object that (at last analysis) is somewhere between 400 and 100 meters long and is unlike anything normally found in the solar system. Researchers hypothesize that the shape and size suggest that the object has been wandering through the Milky Way, unattached to any star system, for hundreds of millions of years.
Immediately after its discovery, telescopes around the world were called into action to measure the object’s trajectory, brightness and color. Combining the images from several large telescopes, a team of astronomers led by Karen Meech of the Institute for Astronomy in Hawaii found that Oumuamua varies in brightness by a factor of 10 as it spins on its axis every 7.3 hours.
No known asteroid or comet from our solar system varies so widely in brightness, with such a large ratio between length and width. The most elongated objects we have seen to date are no more than three times longer than they are wide.
“This unusually big variation in brightness means that the object is highly elongated: about ten times as long as it is wide, with a complex, convoluted shape,” said Meech. “We also found that it had a reddish color, similar to objects in the outer solar system, and confirmed that it is completely inert, without the faintest hint of dust around it.”
Oumuamua is headed out of the solar system now, so it’s unlikely more will be learned about it. And with its odd shape and features, it clearly remains something of a mystery.
And that’s where Harvard’s Avi Loeb comes in. Especially due to the remarkably fast speed with which Oumuamua entered the solar system, he argues that a probe sent by intelligent others cannot be ruled out, that science must be open minded.
“There is data on the orbit of this object for which there is no other explanation” than that it is the product of intelligent others,” he has said. “The approach I take to the subject is purely scientific and evidence-based.”
Others strongly disagree. But the views of the chairman of the Harvard astronomy department are nonetheless an intriguing part of the story.
Sign-up to get the latest in news, events, and opportunities from the NASA Astrobiology Program.
YMEhkTxs3_E
With the highly-touted Passengers due out this fall, this earlier take on the "interstellar ark" film is worth a look. Dennis Quaid and Ben Foster star as two inhabitants of a ship sent on a 123-year voyage to another planet, who wake up to discover that many of the 60,000 others have devolved into a mutated form of human life. The movie suffers from budgetary and script problems, but it has some effective twists and the idea of the effects of deep space travel on human beings is a compelling one. Not a great movie by any means, but I could think of worse ways to pass some time.
Interstellar (2014)
Christopher Nolan’s past masterpiece include Batman’s “Dark Knight Trilogy” and “Inception”, and these days whenever the director releases a movie it automatically attracts levels of high expectation. So it should come as no shock that people expected great things from Nolan’s space epic film entitled “Interstellar.” Were those expectations met? Well, it definitely looks REALLY pretty, and will have obvious appeal to those people interested in Einstein’s theory of general relativity, wormholes, and all sorts of other theoretical physics related stuff. Others, however, have called it a “magnificent folly” or an “awe-inspiring mess”.
“Interstellar” is about the Earth’s impending destruction and an attempt by humankind’s remaining scientists to launch a rocket to another galaxy in the hopes of finding a new planet for Earth’s population to colonize and start a new life. One of the mission’s crew, NASA’s top space pilot Cooper (Matthew McConaughey) is afraid of the toll this mission will take on his family’s lifespan and the fact his 10-year-old daughter Murphy (Mackenzie Foy) may die from old age before her father ever returns. Nevertheless, Cooper, Dr. Brandt (Anne Hathaway) and a handful of explorers venture off into an unknown galaxy in Earth’s last ditch effort to save the human race before the planet we once called home becomes extinct.
“Interstellar” start off slow despite some amusing parental antics from McConaughey, then slowly get more interesting as they reach space, when things get emotional and heart breaking before eventually getting really weird. Throughout the movie, McConaughey, Hathaway and Michael Caine present some truly moving emotional scenes, and the journey into space wields a lot of great ideas and compelling conflicts. Overall, “Interstellar” has the cast, the music and the visuals to truly live up to Nolan’s well-earned reputation for excellence. One issue, however, is that the space voyage goes to too many places, and takes too long to get there while the finale results in more unanswered questions than many may care to spend the time contemplating.
Director: Christopher Nolan
Writers: Jonathan Nolan, Christopher Nolan
Music: Hans Zimmer
Budget: $165 million Box Office:$675.1 million
IMDB Rating: 8.6
Rotten Tomatoes: 71% | 2022-09-30 13:01:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3772493302822113, "perplexity": 2261.9748921129353}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335469.40/warc/CC-MAIN-20220930113830-20220930143830-00086.warc.gz"} |
http://www.mathblogging.org/posts/?type=post&filter0=blog&modifier0=topic&value0=Community | X
# Posts
### November 23, 2014
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This simple fraction game has students estimation the portion of a circle shown by the given fraction. It then rewards them for their accuracy.
### November 22, 2014
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A worksheet to accompany the problem shown on The Curiosity Show (link below).
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By using each digit from 1492 exactly once, how high can you build the positive integers?
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Are you being cheated on pizza?
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A 7-year-old can understand this problem which completely baffles mathematicians.
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People say that a vector is "a quantity with both magnitude and direction." What does that mean? This approach to vectors is too confusing. In this brief introduction allow me to demonstrate a more natural - and meaningful - approach to vectors. They're easy - and don't need to be made confusing.
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When will I use this? Meet professionals from a number of exciting fields who use mathematics in their jobs every day.
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XaoS is an interactive fractalzoomer. It allows the user to continuously zoom in or out of a fractal in a fluid, continuous motion. This capability makes XaoS great for exploring fractals, and it’s fun! Fractal zoomers on the iPad tend not to be free. This software is for Mac, Windows, Linux and is free. There is also a simplified online version.
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There are some great visualisations of mathematical concepts all over the net. This is a nice collection.
### November 21, 2014
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A couple of years ago, I committed to giving a series of lectures (a sort of mini-course) on measure theory at the Universidad de El Salvador. Not a big deal except for two challenges. (1) Teaching at a primarily undergraduate … Continue reading →
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This could form an interesting investigation for students studing the parabola. How can you use a parabola to multiply two numbers? Find out how. The YouTube description includes an online demonstration (Desmos) and there is a GeoGebra version.
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The shortest route for a spider to reach a fly seems obvious in this rectangular room, but Rob shows that it might be trickier than it seems.
### November 19, 2014
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Interactive to explore venn diagrams. Drag the numbers into the appropriate circle.
### November 18, 2014
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Art of Problem Solving {webinar_buttons.png} Enter the webinar room Ethnomathematics-RecordingMangahigh-RecordingRecording Full recording (text, voice, video)
### November 16, 2014
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La eclítica y el zodíaco, la nueva píldora de matemáticas donde mostramos como la invención del zodíaco griego se hizo en base a la eclíptica del Sol.
### November 14, 2014
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Marcus Du Sautoy explains a connection between Logarithms and Primes, and explains the connection between digital sound and Trigonometry.
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A ~3 min summary about integration and differentiation by well known mathematics presenter Marcus Du Sautoy.
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updated axiom
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### November 10, 2014
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A stunning animated slideshow which explores sequences, fractals, Fibonacci sequence, the golden ratio, Pascal's Triangle, and more. The text at the bottom explains each concept very clearly. The maths of Nature, of the real world, is emphasised. The text says 'It seems that the Land of Mathematics is even more curious and mysterious than Wonderland. Only that it is real...'
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A dynamic image which shows the creation of a parabola as the locus of a point equidistant from the focus and directrix. A gif file.
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A gif file that shows how an ellipse is created using 2 foci
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A YouTube playlist demonstrating the construction of four special quadrilaterals from paper rectangles. Rhombus, trapezium (trapezoid), parallelogram, kite. (Click through for the playlist.)
### November 09, 2014
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Experience a dynamic zoom into the Mandelbrot set, a famous and beautiful fractal. This resource is a 'featured picture' from Wikipedia, meaning members of the community have identified it as one of the finest images on the English Wikipedia.
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Leyland Numbers and Leyland Primes with Professor Ed Copeland.
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How to draw a perfect circle. Learn this simple technique to help you draw a near perfect circle freehand. | 2014-11-23 11:56:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26204583048820496, "perplexity": 2858.832597772893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416400379466.35/warc/CC-MAIN-20141119123259-00135-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://stacks.math.columbia.edu/tag/08PV | ## 91.5 Constructing a resolution
In the Noetherian finite type case we can construct a “small” simplicial resolution for finite type ring maps.
Lemma 91.5.1. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. Let $\mathcal{A}$ be the category of $A$-algebra maps $C \to B$. Let $n \geq 0$ and let $P_\bullet$ be a simplicial object of $\mathcal{A}$ such that
1. $P_\bullet \to B$ is a trivial Kan fibration of simplicial sets,
2. $P_ k$ is finite type over $A$ for $k \leq n$,
3. $P_\bullet = \text{cosk}_ n \text{sk}_ n P_\bullet$ as simplicial objects of $\mathcal{A}$.
Then $P_{n + 1}$ is a finite type $A$-algebra.
Proof. Although the proof we give of this lemma is straightforward, it is a bit messy. To clarify the idea we explain what happens for low $n$ before giving the proof in general. For example, if $n = 0$, then (3) means that $P_1 = P_0 \times _ B P_0$. Since the ring map $P_0 \to B$ is surjective, this is of finite type over $A$ by More on Algebra, Lemma 15.5.1.
If $n = 1$, then (3) means that
$P_2 = \{ (f_0, f_1, f_2) \in P_1^3 \mid d_0f_0 = d_0f_1,\ d_1f_0 = d_0f_2,\ d_1f_1 = d_1f_2 \}$
where the equalities take place in $P_0$. Observe that the triple
$(d_0f_0, d_1f_0, d_1f_1) = (d_0f_1, d_0f_2, d_1f_2)$
is an element of the fibre product $P_0 \times _ B P_0 \times _ B P_0$ over $B$ because the maps $d_ i : P_1 \to P_0$ are morphisms over $B$. Thus we get a map
$\psi : P_2 \longrightarrow P_0 \times _ B P_0 \times _ B P_0$
The fibre of $\psi$ over an element $(g_0, g_1, g_2) \in P_0 \times _ B P_0 \times _ B P_0$ is the set of triples $(f_0, f_1, f_2)$ of $1$-simplices with $(d_0, d_1)(f_0) = (g_0, g_1)$, $(d_0, d_1)(f_1) = (g_0, g_2)$, and $(d_0, d_1)(f_2) = (g_1, g_2)$. As $P_\bullet \to B$ is a trivial Kan fibration the map $(d_0, d_1) : P_1 \to P_0 \times _ B P_0$ is surjective. Thus we see that $P_2$ fits into the cartesian diagram
$\xymatrix{ P_2 \ar[d] \ar[r] & P_1^3 \ar[d] \\ P_0 \times _ B P_0 \times _ B P_0 \ar[r] & (P_0 \times _ B P_0)^3 }$
By More on Algebra, Lemma 15.5.2 we conclude. The general case is similar, but requires a bit more notation.
The case $n > 1$. By Simplicial, Lemma 14.19.14 the condition $P_\bullet = \text{cosk}_ n \text{sk}_ n P_\bullet$ implies the same thing is true in the category of simplicial $A$-algebras and hence in the category of sets (as the forgetful functor from $A$-algebras to sets commutes with limits). Thus
$P_{n + 1} = \mathop{\mathrm{Mor}}\nolimits (\Delta [n + 1], P_\bullet ) = \mathop{\mathrm{Mor}}\nolimits (\text{sk}_ n \Delta [n + 1], \text{sk}_ n P_\bullet )$
by Simplicial, Lemma 14.11.3 and Equation (14.19.0.1). We will prove by induction on $1 \leq k < m \leq n + 1$ that the ring
$Q_{k, m} = \mathop{\mathrm{Mor}}\nolimits (\text{sk}_ k \Delta [m], \text{sk}_ k P_\bullet )$
is of finite type over $A$. The case $k = 1$, $1 < m \leq n + 1$ is entirely similar to the discussion above in the case $n = 1$. Namely, there is a cartesian diagram
$\xymatrix{ Q_{1, m} \ar[d] \ar[r] & P_1^ N \ar[d] \\ P_0 \times _ B \ldots \times _ B P_0 \ar[r] & (P_0 \times _ B P_0)^ N }$
where $N = {m + 1 \choose 2}$. We conclude as before.
Let $1 \leq k_0 \leq n$ and assume $Q_{k, m}$ is of finite type over $A$ for all $1 \leq k \leq k_0$ and $k < m \leq n + 1$. For $k_0 + 1 < m \leq n + 1$ we claim there is a cartesian square
$\xymatrix{ Q_{k_0 + 1, m} \ar[d] \ar[r] & P_{k_0 + 1}^ N \ar[d] \\ Q_{k_0, m} \ar[r] & Q_{k_0, k_0 + 1}^ N }$
where $N$ is the number of nondegenerate $(k_0 + 1)$-simplices of $\Delta [m]$. Namely, to see this is true, think of an element of $Q_{k_0 + 1, m}$ as a function $f$ from the $(k_0 + 1)$-skeleton of $\Delta [m]$ to $P_\bullet$. We can restrict $f$ to the $k_0$-skeleton which gives the left vertical map of the diagram. We can also restrict to each nondegenerate $(k_0 + 1)$-simplex which gives the top horizontal arrow. Moreover, to give such an $f$ is the same thing as giving its restriction to $k_0$-skeleton and to each nondegenerate $(k_0 + 1)$-face, provided these agree on the overlap, and this is exactly the content of the diagram. Moreover, the fact that $P_\bullet \to B$ is a trivial Kan fibration implies that the map
$P_{k_0} \to Q_{k_0, k_0 + 1} = \mathop{\mathrm{Mor}}\nolimits (\partial \Delta [k_0 + 1], P_\bullet )$
is surjective as every map $\partial \Delta [k_0 + 1] \to B$ can be extended to $\Delta [k_0 + 1] \to B$ for $k_0 \geq 1$ (small argument about constant simplicial sets omitted). Since by induction hypothesis the rings $Q_{k_0, m}$, $Q_{k_0, k_0 + 1}$ are finite type $A$-algebras, so is $Q_{k_0 + 1, m}$ by More on Algebra, Lemma 15.5.2 once more. $\square$
Proposition 91.5.2. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. There exists a simplicial $A$-algebra $P_\bullet$ with an augmentation $\epsilon : P_\bullet \to B$ such that each $P_ n$ is a polynomial algebra of finite type over $A$ and such that $\epsilon$ is a trivial Kan fibration of simplicial sets.
Proof. Let $\mathcal{A}$ be the category of $A$-algebra maps $C \to B$. In this proof our simplicial objects and skeleton and coskeleton functors will be taken in this category.
Choose a polynomial algebra $P_0$ of finite type over $A$ and a surjection $P_0 \to B$. As a first approximation we take $P_\bullet = \text{cosk}_0(P_0)$. In other words, $P_\bullet$ is the simplicial $A$-algebra with terms $P_ n = P_0 \times _ A \ldots \times _ A P_0$. (In the final paragraph of the proof this simplicial object will be denoted $P^0_\bullet$.) By Simplicial, Lemma 14.32.3 the map $P_\bullet \to B$ is a trivial Kan fibration of simplicial sets. Also, observe that $P_\bullet = \text{cosk}_0 \text{sk}_0 P_\bullet$.
Suppose for some $n \geq 0$ we have constructed $P_\bullet$ (in the final paragraph of the proof this will be $P^ n_\bullet$) such that
1. $P_\bullet \to B$ is a trivial Kan fibration of simplicial sets,
2. $P_ k$ is a finitely generated polynomial algebra for $0 \leq k \leq n$, and
3. $P_\bullet = \text{cosk}_ n \text{sk}_ n P_\bullet$
By Lemma 91.5.1 we can find a finitely generated polynomial algebra $Q$ over $A$ and a surjection $Q \to P_{n + 1}$. Since $P_ n$ is a polynomial algebra the $A$-algebra maps $s_ i : P_ n \to P_{n + 1}$ lift to maps $s'_ i : P_ n \to Q$. Set $d'_ j : Q \to P_ n$ equal to the composition of $Q \to P_{n + 1}$ and $d_ j : P_{n + 1} \to P_ n$. We obtain a truncated simplicial object $P'_\bullet$ of $\mathcal{A}$ by setting $P'_ k = P_ k$ for $k \leq n$ and $P'_{n + 1} = Q$ and morphisms $d'_ i = d_ i$ and $s'_ i = s_ i$ in degrees $k \leq n - 1$ and using the morphisms $d'_ j$ and $s'_ i$ in degree $n$. Extend this to a full simplicial object $P'_\bullet$ of $\mathcal{A}$ using $\text{cosk}_{n + 1}$. By functoriality of the coskeleton functors there is a morphism $P'_\bullet \to P_\bullet$ of simplicial objects extending the given morphism of $(n + 1)$-truncated simplicial objects. (This morphism will be denoted $P^{n + 1}_\bullet \to P^ n_\bullet$ in the final paragraph of the proof.)
Note that conditions (b) and (c) are satisfied for $P'_\bullet$ with $n$ replaced by $n + 1$. We claim the map $P'_\bullet \to P_\bullet$ satisfies assumptions (1), (2), (3), and (4) of Simplicial, Lemmas 14.32.1 with $n + 1$ instead of $n$. Conditions (1) and (2) hold by construction. By Simplicial, Lemma 14.19.14 we see that we have $P_\bullet = \text{cosk}_{n + 1}\text{sk}_{n + 1}P_\bullet$ and $P'_\bullet = \text{cosk}_{n + 1}\text{sk}_{n + 1}P'_\bullet$ not only in $\mathcal{A}$ but also in the category of $A$-algebras, whence in the category of sets (as the forgetful functor from $A$-algebras to sets commutes with all limits). This proves (3) and (4). Thus the lemma applies and $P'_\bullet \to P_\bullet$ is a trivial Kan fibration. By Simplicial, Lemma 14.30.4 we conclude that $P'_\bullet \to B$ is a trivial Kan fibration and (a) holds as well.
To finish the proof we take the inverse limit $P_\bullet = \mathop{\mathrm{lim}}\nolimits P^ n_\bullet$ of the sequence of simplicial algebras
$\ldots \to P^2_\bullet \to P^1_\bullet \to P^0_\bullet$
constructed above. The map $P_\bullet \to B$ is a trivial Kan fibration by Simplicial, Lemma 14.30.5. However, the construction above stabilizes in each degree to a fixed finitely generated polynomial algebra as desired. $\square$
Lemma 91.5.3. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. Let $\pi$, $\underline{B}$ be as in (91.4.0.1). If $\mathcal{F}$ is an $\underline{B}$-module such that $\mathcal{F}(P, \alpha )$ is a finite $B$-module for all $\alpha : P = A[x_1, \ldots , x_ n] \to B$, then the cohomology modules of $L\pi _!(\mathcal{F})$ are finite $B$-modules.
Proof. By Lemma 91.4.1 and Proposition 91.5.2 we can compute $L\pi _!(\mathcal{F})$ by a complex constructed out of the values of $\mathcal{F}$ on finite type polynomial algebras. $\square$
Lemma 91.5.4. Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. Then $H^ n(L_{B/A})$ is a finite $B$-module for all $n \in \mathbf{Z}$.
Remark 91.5.5 (Resolutions). Let $A \to B$ be any ring map. Let us call an augmented simplicial $A$-algebra $\epsilon : P_\bullet \to B$ a resolution of $B$ over $A$ if each $P_ n$ is a polynomial algebra and $\epsilon$ is a trivial Kan fibration of simplicial sets. If $P_\bullet \to B$ is an augmentation of a simplicial $A$-algebra with each $P_ n$ a polynomial algebra surjecting onto $B$, then the following are equivalent
1. $\epsilon : P_\bullet \to B$ is a resolution of $B$ over $A$,
2. $\epsilon : P_\bullet \to B$ is a quasi-isomorphism on associated complexes,
3. $\epsilon : P_\bullet \to B$ induces a homotopy equivalence of simplicial sets.
To see this use Simplicial, Lemmas 14.30.8, 14.31.9, and 14.31.8. A resolution $P_\bullet$ of $B$ over $A$ gives a cosimplicial object $U_\bullet$ of $\mathcal{C}_{B/A}$ as in Cohomology on Sites, Lemma 21.39.7 and it follows that
$L\pi _!\mathcal{F} = \mathcal{F}(P_\bullet )$
functorially in $\mathcal{F}$, see Lemma 91.4.1. The (formal part of the) proof of Proposition 91.5.2 shows that resolutions exist. We also have seen in the first proof of Lemma 91.4.2 that the standard resolution of $B$ over $A$ is a resolution (so that this terminology doesn't lead to a conflict). However, the argument in the proof of Proposition 91.5.2 shows the existence of resolutions without appealing to the simplicial computations in Simplicial, Section 14.34. Moreover, for any choice of resolution we have a canonical isomorphism
$L_{B/A} = \Omega _{P_\bullet /A} \otimes _{P_\bullet , \epsilon } B$
in $D(B)$ by Lemma 91.4.3. The freedom to choose an arbitrary resolution can be quite useful.
Lemma 91.5.6. Let $A \to B$ be a ring map. Let $\pi$, $\mathcal{O}$, $\underline{B}$ be as in (91.4.0.1). For any $\mathcal{O}$-module $\mathcal{F}$ we have
$L\pi _!(\mathcal{F}) = L\pi _!(Li^*\mathcal{F}) = L\pi _!(\mathcal{F} \otimes _\mathcal {O}^\mathbf {L} \underline{B})$
in $D(\textit{Ab})$.
Proof. It suffices to verify the assumptions of Cohomology on Sites, Lemma 21.39.12 hold for $\mathcal{O} \to \underline{B}$ on $\mathcal{C}_{B/A}$. We will use the results of Remark 91.5.5 without further mention. Choose a resolution $P_\bullet$ of $B$ over $A$ to get a suitable cosimplicial object $U_\bullet$ of $\mathcal{C}_{B/A}$. Since $P_\bullet \to B$ induces a quasi-isomorphism on associated complexes of abelian groups we see that $L\pi _!\mathcal{O} = B$. On the other hand $L\pi _!\underline{B}$ is computed by $\underline{B}(U_\bullet ) = B$. This verifies the second assumption of Cohomology on Sites, Lemma 21.39.12 and we are done with the proof. $\square$
Lemma 91.5.7. Let $A \to B$ be a ring map. Let $\pi$, $\mathcal{O}$, $\underline{B}$ be as in (91.4.0.1). We have
$L\pi _!(\mathcal{O}) = L\pi _!(\underline{B}) = B \quad \text{and}\quad L_{B/A} = L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) = L\pi _!(\Omega _{\mathcal{O}/A})$
in $D(\textit{Ab})$.
Proof. This is just an application of Lemma 91.5.6 (and the first equality on the right is Lemma 91.4.3). $\square$
Here is a special case of the fundamental triangle that is easy to prove.
Lemma 91.5.8. Let $A \to B \to C$ be ring maps. If $B$ is a polynomial algebra over $A$, then there is a distinguished triangle $L_{B/A} \otimes _ B^\mathbf {L} C \to L_{C/A} \to L_{C/B} \to L_{B/A} \otimes _ B^\mathbf {L} C[1]$ in $D(C)$.
Proof. We will use the observations of Remark 91.5.5 without further mention. Choose a resolution $\epsilon : P_\bullet \to C$ of $C$ over $B$ (for example the standard resolution). Since $B$ is a polynomial algebra over $A$ we see that $P_\bullet$ is also a resolution of $C$ over $A$. Hence $L_{C/A}$ is computed by $\Omega _{P_\bullet /A} \otimes _{P_\bullet , \epsilon } C$ and $L_{C/B}$ is computed by $\Omega _{P_\bullet /B} \otimes _{P_\bullet , \epsilon } C$. Since for each $n$ we have the short exact sequence $0 \to \Omega _{B/A} \otimes _ B P_ n \to \Omega _{P_ n/A} \to \Omega _{P_ n/B}$ (Algebra, Lemma 10.138.9) and since $L_{B/A} = \Omega _{B/A}[0]$ (Lemma 91.4.7) we obtain the result. $\square$
Example 91.5.9. Let $A \to B$ be a ring map. In this example we will construct an “explicit” resolution $P_\bullet$ of $B$ over $A$ of length $2$. To do this we follow the procedure of the proof of Proposition 91.5.2, see also the discussion in Remark 91.5.5.
We choose a surjection $P_0 = A[u_ i] \to B$ where $u_ i$ is a set of variables. Choose generators $f_ t \in P_0$, $t \in T$ of the ideal $\mathop{\mathrm{Ker}}(P_0 \to B)$. We choose $P_1 = A[u_ i, x_ t]$ with face maps $d_0$ and $d_1$ the unique $A$-algebra maps with $d_ j(u_ i) = u_ i$ and $d_0(x_ t) = 0$ and $d_1(x_ t) = f_ t$. The map $s_0 : P_0 \to P_1$ is the unique $A$-algebra map with $s_0(u_ i) = u_ i$. It is clear that
$P_1 \xrightarrow {d_0 - d_1} P_0 \to B \to 0$
is exact, in particular the map $(d_0, d_1) : P_1 \to P_0 \times _ B P_0$ is surjective. Thus, if $P_\bullet$ denotes the $1$-truncated simplicial $A$-algebra given by $P_0$, $P_1$, $d_0$, $d_1$, and $s_0$, then the augmentation $\text{cosk}_1(P_\bullet ) \to B$ is a trivial Kan fibration. The next step of the procedure in the proof of Proposition 91.5.2 is to choose a polynomial algebra $P_2$ and a surjection
$P_2 \longrightarrow \text{cosk}_1(P_\bullet )_2$
Recall that
$\text{cosk}_1(P_\bullet )_2 = \{ (g_0, g_1, g_2) \in P_1^3 \mid d_0(g_0) = d_0(g_1), d_1(g_0) = d_0(g_2), d_1(g_1) = d_1(g_2)\}$
Thinking of $g_ i \in P_1$ as a polynomial in $x_ t$ the conditions are
$g_0(0) = g_1(0),\quad g_0(f_ t) = g_2(0),\quad g_1(f_ t) = g_2(f_ t)$
Thus $\text{cosk}_1(P_\bullet )_2$ contains the elements $y_ t = (x_ t, x_ t, f_ t)$ and $z_ t = (0, x_ t, x_ t)$. Every element $G$ in $\text{cosk}_1(P_\bullet )_2$ is of the form $G = H + (0, 0, g)$ where $H$ is in the image of $A[u_ i, y_ t, z_ t] \to \text{cosk}_1(P_\bullet )_2$. Here $g \in P_1$ is a polynomial with vanishing constant term such that $g(f_ t) = 0$ in $P_0$. Observe that
1. $g = x_ t x_{t'} - f_ t x_{t'}$ and
2. $g = \sum r_ t x_ t$ with $r_ t \in P_0$ if $\sum r_ t f_ t = 0$ in $P_0$
are elements of $P_1$ of the desired form. Let
$Rel = \mathop{\mathrm{Ker}}(\bigoplus \nolimits _{t \in T} P_0 \longrightarrow P_0),\quad (r_ t) \longmapsto \sum r_ tf_ t$
We set $P_2 = A[u_ i, y_ t, z_ t, v_ r, w_{t, t'}]$ where $r = (r_ t) \in Rel$, with map
$P_2 \longrightarrow \text{cosk}_1(P_\bullet )_2$
given by $y_ t \mapsto (x_ t, x_ t, f_ t)$, $z_ t \mapsto (0, x_ t, x_ t)$, $v_ r \mapsto (0, 0, \sum r_ t x_ t)$, and $w_{t, t'} \mapsto (0, 0, x_ t x_{t'} - f_ t x_{t'})$. A calculation (omitted) shows that this map is surjective. Our choice of the map displayed above determines the maps $d_0, d_1, d_2 : P_2 \to P_1$. Finally, the procedure in the proof of Proposition 91.5.2 tells us to choose the maps $s_0, s_1 : P_1 \to P_2$ lifting the two maps $P_1 \to \text{cosk}_1(P_\bullet )_2$. It is clear that we can take $s_ i$ to be the unique $A$-algebra maps determined by $s_0(x_ t) = y_ t$ and $s_1(x_ t) = z_ t$.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2023-02-03 08:04:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9795289635658264, "perplexity": 100.21939696737944}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00085.warc.gz"} |
http://www.maths.kisogo.com/index.php?title=Monotonicity_of_the_integral_of_non-negative_extended-real-valued_measurable_functions_with_respect_to_a_measure | # Monotonicity of the integral of non-negative extended-real-valued measurable functions with respect to a measure
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Stub, needs review, linking to from other pages, a bit of a measure-theory shuffle
## Statement
Let [ilmath](X,\mathcal{A},\mu)[/ilmath] be a measure space and let [ilmath]f,g\in[/ilmath][ilmath]\mathcal{M}_{\bar{\mathbb{R} }_{\ge 0} }(\mathcal{A})[/ilmath][Note 1], then[1]:
• if [ilmath]f\le g[/ilmath] - i.e.: [ilmath]\forall x\in X[f(x)\le g(x)][/ilmath] - then:
• [ilmath]\int f\ \mathrm{d}\mu\le\int g\ \mathrm{d}\mu[/ilmath]
## Proof
Recall the definition of the integral of a non-negative numerical functionAuthor's note:[Note 2]:
• $\int f\ \mathrm{d}\mu:\eq\text{Sup}\left(\left\{I_\mu(s)\in[0,+\infty]\subseteq\bar{\mathbb{R} }\ \Big\vert\ s\in\mathcal{E}_{\ge 0}(\mathcal{A})\wedge s\le f\right\} \right)$, where:
• [ilmath]s\le f[/ilmath] is an abuse of notation for [ilmath]\forall x\in X[s(x)\le f(x)][/ilmath] and
• [ilmath]s\in\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath] means [ilmath]s\in[/ilmath][[Set of all simple functions over a set|[ilmath]\mathcal{E}(\mathcal{A})[/ilmath] and [ilmath]\forall x\in X[s(x)\ge 0][/ilmath]
• Let [ilmath]f,g\in\mathcal{M}_{\bar{\mathbb{R} }_{\ge 0} }(\mathcal{A}) [/ilmath] be given and suppose [ilmath]f\le g[/ilmath]
• Define [ilmath]A:\eq\left\{r\in\mathcal{E}_{\ge 0}(\mathcal{A})\ \Big\vert\ r\le f\right\} [/ilmath]
• Define [ilmath]B:\eq\left\{s\in\mathcal{E}_{\ge 0}(\mathcal{A})\ \Big\vert\ s\le g\right\} [/ilmath]
• Lemma: we claim that [ilmath]A\subseteq B[/ilmath], using the implies-subset relation we see [ilmath]\big(A\subseteq B\big)\iff\big(\forall a\in A[a\in B]\big)[/ilmath]
• Let [ilmath]a\in A[/ilmath] be given, notice that by definition [ilmath]A\subseteq\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath], so [ilmath]a\in\mathcal{E}_{\ge 0}(\mathcal{A}) [/ilmath]
• Then [ilmath]a\le f[/ilmath], or [ilmath]\forall x\in X[a(x)\le f(x)][/ilmath]
• But [ilmath]f\le g[/ilmath] by hypothesis, i.e. [ilmath]\forall x\in X[f(x)\le g(x)][/ilmath], so we see [ilmath]\forall x\in X[a(x)\le g(x)][/ilmath] of [ilmath]a\le g[/ilmath]
• We see that both:
1. [ilmath]a\in\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath] - of which all elements of [ilmath]B[/ilmath] are too
2. [ilmath]a\le g[/ilmath] - the definition of a member of [ilmath]\mathcal{E}_{\ge 0}(\mathcal{A})[/ilmath] to be in [ilmath]B[/ilmath]
• So [ilmath]a\in B[/ilmath]
• Since [ilmath]a\in A[/ilmath] was arbitrary we have shown this for all [ilmath]a\in A[/ilmath] - as required to prove the lemma
• Define [ilmath]C:\eq\{I_\mu(r)\in[0,+\infty]\subseteq\bar{\mathbb{R} }\ \big\vert\ r\in A\} [/ilmath] and
• Define [ilmath]D:\eq\{I_\mu(s)\in[0,+\infty]\subseteq\bar{\mathbb{R} }\ \big\vert\ s\in B\} [/ilmath]
Grade: C
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The gist is all here, but it should link to the simple functions, their integral, the monotonicity of supremum, so forth. It's not quite finished basically. Alec (talk) 19:41, 14 April 2017 (UTC)
## Notes
1. functions which map to the extended real values are sometimes called numerical functions
2. Right now this is at integral of a positive function (measure theory), this is technically incorrect, but the information can be found here. | 2023-03-22 19:24:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9727169275283813, "perplexity": 13812.88359865087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00154.warc.gz"} |
https://www.techwhiff.com/issue/why-do-people-with-low-iron-have-pale-skin-and-no-energy--189281 | # Why do people with low iron have pale skin and no energy?
###### Question:
Why do people with low iron have pale skin and no energy?
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A situation in which the quantity of output supplied is greater than the quantity... | 2022-10-04 10:05:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2843892276287079, "perplexity": 2681.628244834736}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337490.6/warc/CC-MAIN-20221004085909-20221004115909-00258.warc.gz"} |
http://tex.stackexchange.com/questions/27144/documentclass-undefined-control-sequence | # \documentclass: undefined control sequence
There's probably a ton of questions about this already but I've been spending all day feeling overwhelmed with the TeX learning curve.
I'm trying to learn LuaTeX (by the way, if you know of a good tutorial I'd love to know about it). I have downloaded TeX Live 2011 for Windows yesterday. I have a LaTeX document with working syntax which renders properly. When I switch the typeset setting to "ConTeXt (LuaTeX)", I get "Undefined control sequence" for the first line, "\documentclass{report}". What am I doing wrong?
-
You don't say what your editing environment is, but you are trying to compile a LaTeX document using ConTeXt (which is not LuaLaTeX). So you need to compile your file with LuaLaTeX, and then things should work. – Alan Munn Sep 1 '11 at 3:26
You might use LuaTeX and not LuaLaTeX. Also, IMHO, there is not that much difference for a beginner between LuaLaTeX and the normal (pdf)LaTeX. LuaLaTeX is "just" a pdfLaTeX compiler with Lua (a scripting language) support. AFAIK there are some differences in font and language support, but the rest is just LaTeX. – Martin Scharrer Sep 1 '11 at 6:49
@Martin: glifchits asked a question about using system fonts earlier, so this is why he's getting to use LuaTeX. – ℝaphink Sep 1 '11 at 8:09
## 2 Answers
I hope this answer can clarify a few basic concepts for you.
TeX TeX is a (macro) programming language invented by Knuth. It is a Turing complete language. That is, theoretically you can do almost anything you can do with other languages, though not necessarily easily.
TeX Engine A program that can interpret the TeX language. It is like the perl program for the Perl language. The first TeX engine was written by Knuth. Nowadays, popular engines include pdftex, xetex, luatex. pdftex is much faster than the other two though the other two provide extra functionality, especially for using opentype fonts. luatex provides much more than fonts.
LaTeX and ConTeXt These are formats. A format is just a set of TeX macros (well, not exactly so simple). They define a lot of macros for you to use. For example, LaTeX defines the \documentclass and \begin{document} macros while ConTeXt defines \starttext. It will take a much longer paragraph to answer what there really is in a format. For now, you may just think the program lualatex is a shortcut for starting the luatex engine, inputting the LaTeX macro packages (i.e., the format), and then processing your input. (I know this is not the exact explanation, but I believe it helps a beginner to understand).
LuaLaTeX, XeLaTeX, pdfLaTeX, ... these are all just shortcuts for using the LaTeX format with a TeX engine.
In the question, you seems to think ConTeXt and LuaTeX are the same. This is not true. ConTeXt Mark IV requires LuaTeX as an engine to process this format. Just like many LaTeX packages and the forthcoming LaTeX3 format requires an engine with e-TeX features.
Please correct me if there's anything wrong with this answer.
-
Thank you for your helpful comments. This does help me understand things a lot better. – glifchits Sep 1 '11 at 23:31
Most likely, you typed luatex to compile a LaTeX document. The proper command to use to compile LateX code with LuaTeX is lualatex.
-
The real confusion I've been having stems from that you suggested I use LuaTeX since I installed TeX Live. However, (as you confirmed on my other question) TeXworks doesn't have LuaTeX. And to be specific, I don't manually type any commands when I compile my document. So I guess I should ask you, how do I typeset a document in LuaTeX using TeXworks? Does it mean I have to download, install, and configure (or any combination of those actions) LuaTeX within TeXworks on my own? Also, while I've got a giant question going, is LuaTeX and LuaLaTeX different? – glifchits Sep 1 '11 at 23:42 Right. I think @YanZhou's answer clears up most of these points. If it still doesn't answer everything, you should open a new question. – ℝaphink Sep 1 '11 at 23:44 | 2013-05-25 05:51:39 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9839068651199341, "perplexity": 2039.8423261423227}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705559639/warc/CC-MAIN-20130516115919-00024-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://web2.0calc.com/questions/solving-rational-equations_1 | +0
# Solving Rational Equations
0
206
2
+771
$${1\over x(x-4)}+{5-2x\over x^2-3x-4}={5\over x(x+1)}$$
Solve this rational equation.
AdamTaurus Sep 12, 2017
Sort:
### 2+0 Answers
#1
0
Solve for x:
1/(x (x - 4)) + (5 - 2 x)/(x^2 - 3 x - 4) = 5/(x (x + 1))
Hint: | Write the left hand side as a single fraction.
Bring 1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) together using the common denominator x (x - 4) (x + 1):
(-2 x^2 + 6 x + 1)/(x (x - 4) (x + 1)) = 5/(x (x + 1))
Hint: | Multiply both sides by a polynomial to clear fractions.
Cross multiply:
x (x + 1) (-2 x^2 + 6 x + 1) = 5 x (x - 4) (x + 1)
Hint: | Write the quartic polynomial on the left hand side in standard form.
Expand out terms of the left hand side:
-2 x^4 + 4 x^3 + 7 x^2 + x = 5 x (x - 4) (x + 1)
Hint: | Write the cubic polynomial on the right hand side in standard form.
Expand out terms of the right hand side:
-2 x^4 + 4 x^3 + 7 x^2 + x = 5 x^3 - 15 x^2 - 20 x
Hint: | Move everything to the left hand side.
Subtract 5 x^3 - 15 x^2 - 20 x from both sides:
-2 x^4 - x^3 + 22 x^2 + 21 x = 0
Hint: | Factor the left hand side.
The left hand side factors into a product with five terms:
-x (x + 1) (x + 3) (2 x - 7) = 0
Hint: | Multiply both sides by a constant to simplify the equation.
Multiply both sides by -1:
x (x + 1) (x + 3) (2 x - 7) = 0
Hint: | Find the roots of each term in the product separately.
Split into four equations:
x = 0 or x + 1 = 0 or x + 3 = 0 or 2 x - 7 = 0
Hint: | Look at the second equation: Solve for x.
Subtract 1 from both sides:
x = 0 or x = -1 or x + 3 = 0 or 2 x - 7 = 0
Hint: | Look at the third equation: Solve for x.
Subtract 3 from both sides:
x = 0 or x = -1 or x = -3 or 2 x - 7 = 0
Hint: | Look at the fourth equation: Isolate terms with x to the left hand side.
Add 7 to both sides:
x = 0 or x = -1 or x = -3 or 2 x = 7
Hint: | Solve for x.
Divide both sides by 2:
x = 0 or x = -1 or x = -3 or x = 7/2
Hint: | Now test that these solutions are correct by substituting into the original equation.
Check the solution x = -3.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ (5 - 2 (-3))/(-4 - 3 (-3) + (-3)^2) + 1/((-4 - 3) (-3)) = 5/6
5/(x (x + 1)) ⇒ -5/(3 (1 - 3)) = 5/6:
So this solution is correct
Hint: | Check the solution x = -1.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ (5 - 2 (-1))/(-4 - 3 (-1) + (-1)^2) + 1/((-4 - 1) (-1)) = ∞^~
5/(x (x + 1)) ⇒ -5/(1 - 1) = ∞^~:
So this solution is incorrect
Hint: | Check the solution x = 0.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ 1/((0 - 4) 0) + (5 - 2 0)/(-4 - 3 0 + 0^2) = ∞^~
5/(x (x + 1)) ⇒ 5/(0 (1 + 0)) = ∞^~:
So this solution is incorrect
Hint: | Check the solution x = 7/2.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ 1/(1/2 (7/2 - 4) 7) + (5 - (2 7)/2)/(-4 - (3 7)/2 + (7/2)^2) = 20/63
5/(x (x + 1)) ⇒ 5/(7/2 (1 + 7/2)) = 20/63:
So this solution is correct
Hint: | Gather any correct solutions.
The solutions are:
x = -3 or x = 7/2
Guest Sep 13, 2017
#2
+7048
+1
$$\frac{1}{x(x-4)}+\frac{5-2x}{x^2-3x-4}=\frac{5}{x(x+1)}$$
Factor the x2 - 3x - 4 . What two numbers add to -3 and multiply to -4 ? +1 and -4 .
$$\frac{1}{x(x-4)}+\frac{5-2x}{(x+1)(x-4)}=\frac{5}{x(x+1)}$$
Multiply the first fraction by $$\frac{x+1}{x+1}$$ .
$$\frac{1(x+1)}{x(x-4)(x+1)}+\frac{5-2x}{(x+1)(x-4)}=\frac{5}{x(x+1)}$$
Multiply the middle fraction by $$\frac{x}{x}$$ .
$$\frac{1(x+1)}{x(x-4)(x+1)}+\frac{(5-2x)x}{(x+1)(x-4)x}=\frac{5}{x(x+1)}$$
Multiply the last fraction by $$\frac{x-4}{x-4}$$ .
$$\frac{1(x+1)}{x(x-4)(x+1)}+\frac{(5-2x)x}{(x+1)(x-4)x}=\frac{5(x-4)}{x(x+1)(x-4)}$$
Now we have a common denominator. Let's multiply both sides by this denominator...but first we must say that x ≠ 0 , x ≠ 4 , and x ≠ -1 , because these values cause a zero in the denominator of the original equation.
1(x + 1) + (5 - 2x)x = 5(x - 4) When x ≠ 0 , x ≠ 4 , and x ≠ -1 .
Multiply out the parenthesees and simplify.
x + 1 + 5x - 2x2 = 5x - 20
-2x2 + x + 21 = 0
Solve for x with the quadratic formula.
$$x = {-1 \pm \sqrt{1^2-4(-2)(21)} \over 2(-2)} \\~\\ x = {-1 \pm 13\over -4} \\~\\ x=-\frac{12}{4}=-3\qquad\text{ or }\qquad x=\frac{-14}{-4}=\frac{7}{2}$$
hectictar Sep 13, 2017
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We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. Privacy Policy | 2018-05-21 11:01:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6636176109313965, "perplexity": 1377.0093598226304}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864063.9/warc/CC-MAIN-20180521102758-20180521122758-00132.warc.gz"} |
http://en.wikipedia.org/wiki/User_talk:NativeForeigner | # User talk:NativeForeigner
Jump to: navigation, search
Note: Archives are below in template as well. New archives will appear in header.
Header ripped off from Anonymous Dissident (Thanks)
Please, be my guest, and whack me with a large trout if the situation demands it.
This user replies where he likes, and is inconsistent in that respect.
$\lnot \mathrm{TB}$ Please refrain from using the dreaded Template:Talkback on this page multiple times in the same discussion (I'll have it watched after the first template)
Vote!
Formerly Redskunk (talk · contribs)
## The recent SPI
Um.. I don't want to turn out malicious or anything, but, has Atlantictire been sanctioned at all for all this? -- Director (talk) 11:32, 25 March 2014 (UTC)
Yup. Bbb extended the block out to ten days duration, if it happens again it will likely (almost certainly) be indef. NativeForeigner Talk 21:16, 25 March 2014 (UTC)
After all the "gestapo", "antisemitic crank", "bigot", etc. + the three attack socks? I always thought if you're a multiple sockpuppeteer you're indeffed on the spot. -- Director (talk) 08:19, 26 March 2014 (UTC)
It's generally left to the discretion of the blocking admin. I'd talk to @Bbb23: about this. NativeForeigner Talk 18:13, 26 March 2014 (UTC)
## The Bugle: Issue XCVI, March 2014
Your Military History Newsletter
The Bugle is published by the Military history WikiProject. To receive it on your talk page, please join the project or sign up here.
If you are a project member who does not want delivery, please remove your name from this page. Your editors, Ian Rose (talk) and Nick-D (talk) 12:04, 26 March 2014 (UTC)
## Puzzling sock block
NativeForeigner, I'm puzzled by your block of Electrostatic345345 (talk · contribs) for being a sockpuppet of Mittybark111 (talk · contribs). I don't see any mention of this user account at Wikipedia:Sockpuppet investigations/Mittybark111/Archive. If this is a sockpuppet case, shouldn't it be documented better? RockMagnetist (talk) 15:50, 26 March 2014 (UTC)
Ah, I see why you would be confused. I ran a checkuser on the case located at Wikipedia:Sockpuppet_investigations/207.255.205.112#Clerk.2C_CheckUser.2C_and.2For_patrolling_admin_comments. A clerk should merge the case to the Mittybark111 case at some point in the next day or two. NativeForeigner Talk 18:11, 26 March 2014 (UTC)
Whoah - Mittybark111 is one busy troublemaker. Thanks for cleaning up that mess. RockMagnetist (talk) 18:36, 26 March 2014 (UTC)
## User:Daveandbusters1345
Daveandbusters1345 was indeffed as one of the ceiling fan puppets. An IP left this message on my talk page, and I blocked the IP for one year for block evasion and being a confirmed proxy server. Then, when I probed further, I saw a range of IPs editing Archie Karas. My guess is Daveandbusters was using the proxy server to post to my talk page as the other edits of that IP and of the other IPs seem unrelated to the subject matter of the sock puppetry. Does anything further need to be done? Thanks.--Bbb23 (talk) 22:26, 26 March 2014 (UTC)
I've locked the account and globally blocked the proxy. --Rschen7754 22:35, 26 March 2014 (UTC)
## technically clean meat
In this edit to a sockpuppet investigation, you said "Some of these acoutns coule be meat but are technically clean." The mental image that comes to mind is Kosher spam (food), which of course does not exist. Thanks for the laugh. Oh, and I do understand what you were really trying to say. davidwr/(talk)/(contribs) 02:37, 29 March 2014 (UTC)
## Teaching wikipedia in comm 106i s14
Hi there
I am a professor in the communication department planning on a wikipedia editing assignment with my 80 studemt class this quarter. As campus ambassador, can you help me? I am reading up on the instructir and syllabus materials, but it would be great to meet and talk about advice or support to students as they undertake their projects for the quarter.
Might you be free to meet this week?
Thank you, Lilly Irani
## WikiCup 2014 March newsletter
A quick update as we are half way through round two of this year's competition. WikiCup newcomer Godot13 (submissions) (Pool E) leads, having produced a massive set of featured pictures for Silver certificate (United States), an article also brought to featured list status. Former finalist Adam Cuerden (submissions) (Pool G) is in second, which he owes mostly to his work with historical images, including a number of images from Urania's Mirror, an article also brought to good status. 2010 champion Sturmvogel_66 (submissions) (Pool C) is third overall, thanks to contributions relating to naval history, including the newly featured Japanese battleship Nagato. Cliftonian (submissions), who currently leads Pool A and is sixth overall, takes the title for the highest scoring individual article of the competition so far, with the top importance featured article Ian Smith.
With 26 people having already scored over 100 points, it is likely that well over 100 points will be needed to secure a place in round 3. Recent years have required 123 (2013), 65 (2012), 41 (2011) and 100 (2010). Remember that only 64 will progress to round 3 at the end of April. Invitations for collaborative writing efforts or any other discussion of potentially interesting work is always welcome on the WikiCup talk page; if two or more WikiCup competitors have done significant work on an article, all can claim points equally. If you are concerned that your nomination—whether it is at good article candidates, a featured process, or anywhere else—will not receive the necessary reviews, please list it on Wikipedia:WikiCup/Reviews. If you want to help out with the WikiCup, please do your bit to help keep down the review backlogs! Questions are welcome on Wikipedia talk:WikiCup, and the judges are reachable on their talk pages or by email. Good luck! If you wish to start or stop receiving this newsletter, please feel free to add or remove yourself from Wikipedia:WikiCup/Newsletter/Send. J Milburn (talkemail), The ed17 (talkemail) and Miyagawa (talkemail) 22:55, 31 March 2014 (UTC)
## Wikipedia:Sockpuppet investigations/Richard Daft
Hi there. I can forgive you for groaning when you see that heading. First, I'd like to thank you for your recent work on the case when you uncovered several latent accounts. Secondly, though, I wonder if you can give me some advice.
As you know Daft is a long-term abuser of the site who focuses on WP:CRIC by disrupting its articles and making personal attacks on several of its members. There is one exception, User:Johnlp, whom Daft sees as an "ally" of some kind although there is no real reciprocation of that, except in as much as Johnlp endeavours to be all things to all men and tries to see every point of view. The difficulty is that Johnlp believes Daft should be able to use User talk:Johnlp to express his views and he has resisted efforts by those of us trying to remove Daft's edits by saying that it is "his page" and he will decide what is displayed there.
I believe that under the terms of WP:BMB, Johnlp has no say in this and should stand aside when Daft's posts are removed from the talk page. Better still, I believe the page should be semi-protected to stop Daft using it. I have gone to WP:Requests for page protection but am not confident of that process as I'm proposing another user's talk page. I have also reported Daft to WP:Long-term abuse/Richard Daft.
Can you advise me on these steps? Am I going about things the right way and is there anything else you can suggest? Thanks very much. HCCC14 (talk) 18:27, 1 April 2014 (UTC)
Looks good. Only further step is AN or ANI but I'd generally try to avoid that unless absolutely necessary. NativeForeigner Talk 19:42, 2 April 2014 (UTC)
## SockPuppet Investigation Immediately closed by an administrator when the accused requested him to close it- Need Immediate attention
I had opened a sock puppet investigation on two users Shriram and Lihaas on India General election page- Wikipedia:Sockpuppet_investigations/Shriram. One of them suddently made a request to another Administrator ( RequestMadeHere ) to close the investigation and the page was immediately closed.
Excerpt- User:JamesBWatson, I would think canvassing around for his view is turning disruptive. (Wikipedia:Sockpuppet investigations/Shriram) How about a topic ban?Lihaas (talk) 14:44, 10 April 2014 (UTC)
The immediate closure of topic looks suspicious. Please do the necessary.
Thanks Soorejmg (talk) 15:52, 10 April 2014 (UTC)
I tend to agree with JamesBWatson. The two users agree, but that does not make them sockpuppets. NativeForeigner Talk 18:36, 10 April 2014 (UTC)
Soorejmg has sent this message to six different administrators. See User talk:Soorejmg#Canvassing. The editor who uses the pseudonym "JamesBWatson" (talk) 19:30, 10 April 2014 (UTC)
## Wikipedia:Sockpuppet investigations/TimeQueen32
Hi! Can I have an update on the status of this SPI? In particular I'm wondering why no blocks have yet been made if there is a paid editing ring at work here, as you alluded to in the SPI. I realise you may be busy offline, and that there might be additional checks going on behind the scenes, but the case has been open for quite awhile now without many updates. ThemFromSpace 17:09, 15 April 2014 (UTC)
It's really ambiguous. There are a couple groups of two accounts but they are stale. There is also some geographic grouping that could suggest meat. But hte groups are across four continents (as I recall). It might be best to block on behavior. I Relisted so that another checkuser could examine it, but as you can tell that hasn't occurred. I'll be much less busy at the end of the week, but no guarantees on actually blocking anything on CU evidence: this is not clear cut technically. NativeForeigner Talk 19:07, 15 April 2014 (UTC)
Thanks for the update! ThemFromSpace 15:11, 16 April 2014 (UTC)
## The Bugle: Issue XCVII, April 2014
Your Military History Newsletter Project news: From the editors; awards and honours; contest results Articles: Last month's new Featured and A-Class content Book reviews: Hawkeye7 looks at Britain's War Machine Interview: Article writers discuss German military history
The Bugle is published by the Military history WikiProject. To receive it on your talk page, please join the project or sign up here.
If you are a project member who does not want delivery, please remove your name from this page. Your editors, Ian Rose (talk) and Nick-D (talk) 14:30, 20 April 2014 (UTC) | 2014-04-24 14:00:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2156580537557602, "perplexity": 4231.391496520637}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00159-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/194112-need-help-logs-print.html | # Need help with logs
• Dec 12th 2011, 11:22 AM
MagisterPS3
Need help with logs
Write log p / square root of q in terms of log p and log q.
Solve the equation - 4 ^(3x + 1) = 7
Thanks
• Dec 12th 2011, 11:25 AM
MagisterPS3
Re: Need help with logs
That isnt a minus 4
• Dec 12th 2011, 11:29 AM
emakarov
Re: Need help with logs
Quote:
Originally Posted by MagisterPS3
Write log p / square root of q in terms of log p and log q.
I am assuming it's $\log(p/\sqrt{q})$, not $(\log p)/\sqrt{q}$. In plain text, one could write log(p) / sqrt(q). Have you tried applying different logarithm identities?
Quote:
Originally Posted by MagisterPS3
Solve the equation - 4 ^(3x + 1) = 7
Have you seen examples of solutions of such equations?
• Dec 12th 2011, 11:38 AM
MagisterPS3
Re: Need help with logs
I have seen examples of both but im not sure how to put in terms of a letter...
• Dec 12th 2011, 11:44 AM
Plato
Re: Need help with logs
Quote:
Originally Posted by MagisterPS3
I have seen examples of both but im not sure how to put in terms of a letter...
Here is an example to follow:
$\log(a^2\sqrt[3]b)=2\log(a)+\tfrac{1}{3}\log(b).$
We add logs for multiplication, what is done for division?
• Dec 12th 2011, 11:45 AM
emakarov
Re: Need help with logs
For the first problem, try applying some of the four identities in the link from post #3. For the second problem, multiply both sides by -1, apply $\log_4$ to both sides and use the fact that $\log_4\left(4^y\right)=y$.
• Dec 12th 2011, 11:51 AM
MagisterPS3
Re: Need help with logs
Thanks very much guys I got it. :) | 2017-10-19 00:20:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8832883238792419, "perplexity": 1904.5527254345006}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823168.74/warc/CC-MAIN-20171018233539-20171019013539-00478.warc.gz"} |
https://itectec.com/matlab/matlab-smooth-line-of-semilog-plot/ | MATLAB: Smooth line of semilog plot
curve fittingsemilogsmooth curvespline
Hi all,
Is there any effective and easy-understandable way to make a smoothly best-fitted line to the data on semilogx plot? The plot with spline seems to be effective way, however, it is uneasy to understand for me. Could somebody guide me to know how to smooth the semilogx plot? My data are
x = [5 15 25 35 45 55 65 75 85 95 105 135 155 165];
y = [0 0 7.5 30 47.5 62.5 75 85 90 92.5 95 97.5 100 100];
plot this with the semilogx and find the smooth curve fitting them through.
Thank you so much.
x = [5 15 25 35 45 55 65 75 85 95 105 135 155 165];y = [0 0 7.5 30 47.5 62.5 75 85 90 92.5 95 97.5 100 100];logistic_fcn = @(b,x) b(1) ./ (1 + exp(-b(2).*(x-b(3))));NRCF = @(b) norm(y - logistic_fcn(b,x));B = fminsearch(NRCF, [100; 0.1; 50]);xi = linspace(min(x), max(x));yi = logistic_fcn(B,xi);figure(1)semilogx(x, y, 'pg')hold onplot(xi, yi)hold offgrid | 2021-05-12 18:31:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5893257856369019, "perplexity": 1810.9865972309235}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989766.27/warc/CC-MAIN-20210512162538-20210512192538-00476.warc.gz"} |
https://bt.gateoverflow.in/281/gate2015-45 | # GATE2015-45
Match the microorganism in $\text{Group I}$ with their fermentation products in $\text{Group II}$.
$\begin{array}{|l|l|l|l|} \hline & \textbf{Group I} && \textbf{Group II} \\ \hline P. & \textit{Leuconostoc mesenteroides} & 1 .& \text{Cobalamin} \\ \hline Q. & \textit{Rhizopus oryzae} & 2. & \text{Sorbose} \\ \hline R. & \textit{Gluconobacter suboxydans} & 3. & \text{Dextran} \\ \hline S. & \textit{Streptomyces olivaceus} & 4. & \text{Lactic acid} \\ \hline && 5. & \text{Butanol} \\ \hline \end{array}$
1. $\text{P-5, Q-4, R-2, S-1}$
2. $\text{P-5, Q-3, R-2, S-4}$
3. $\text{P-3, Q-4, R-1, S-2}$
4. $\text{P-3, Q-4, R-2, S-1}$ | 2021-01-21 10:43:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9286773800849915, "perplexity": 7719.733879285625}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00328.warc.gz"} |
https://lmb.univ-fcomte.fr/My-favourite-branch-groups?lang=fr | My favourite branch groups.
par - publié le
Jared White
(LMB)
Branch groups are certain groups of tree automorphisms that have a track record
of providing counterexamples to famous conjectures. They come with a beauti-
ful self-similar structure. In the talk I will describe two of the most well-known
examples, the Grigorchuk group , and the Gupta-Sidki 3-group. Time permit-
ting, I will briefly describe how they can be used to solve a problem about the
second duals of group algebras. | 2021-11-27 18:10:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5831627249717712, "perplexity": 1370.4910768356915}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00609.warc.gz"} |
https://proofwiki.org/wiki/257 | # 257
Previous ... Next
## Number
$257$ (two hundred and fifty-seven) is:
The $55$th prime number
The $4$th Fermat number and Fermat prime after $3$, $5$, $17$:
$257 = 2^{\left({2^3}\right)} + 1 = 2^8 + 1$
The $4$th Sierpiński number of the first kind after $2$, $5$, $28$:
$257 = 4^4 + 1$
The $7$th prime number of the form $n^2 + 1$ after $2$, $5$, $17$, $37$, $101$, $197$:
$257 = 16^2 + 1$
The $3$rd (and largest known) prime Sierpiński number of the first kind after $2$, $5$:
$257 = 4^4 + 1$
The $6$th balanced prime after $5$, $53$, $157$, $173$, $211$:
$257 = \dfrac {251 + 263} 2$ | 2019-05-23 11:17:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9749646186828613, "perplexity": 1600.9846988546183}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257243.19/warc/CC-MAIN-20190523103802-20190523125802-00010.warc.gz"} |
https://stats.stackexchange.com/questions/429000/assign-an-error-to-the-parameters-of-map-estimate | # Assign an error to the parameters of MAP estimate
Through a MCMC Gibbs sampler I obtain $$M$$ chains of the parameters vector $$\mathbf{\theta}$$, meaning that each component of $$\mathbf{\theta}$$ is the value of one parameter at a given iteration.
Since I know the likelihood and the prior I found the maximum a posteriori estimate:
$$\hat{\mathbf{\theta}}=\underset{\mathbf{\theta}^{(1)},\,...,\,\mathbf{\theta}^{(M)}}{\operatorname{argmax}}\,p(\mathbf{\theta}),$$
I'd like to have an error on each component of $$\hat{\mathbf{\theta}}$$.
One other estimate of the best parameters is the mode or the mean of MCMC chains for each parameter, for which I can consider the standard deviation as an error, but this is not what I'm interested in.
Since we are treating $$\theta$$ as vector of random variables i think the term error is thematically wrong, indeed one of the strength of Bayesian methods is the richness of information granted about $$\theta$$ through it's posterior distribution. For MCMC sampling methods (of which the Gibbs sampler is one kind) the posterior distribution of paramters $$\theta$$ can generally be obtained in this manner:
2. Keep every x sample of $$\theta$$. Since the chain is correlated in time we only keep every xth value, this concept is known as thinning. The value of x is generally a trade-off between time and accuracy.
3. The kept samples of $$\theta$$ describes it's posterior distribution for which any moment can be calculated if so desired. | 2021-06-17 21:52:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9080538749694824, "perplexity": 357.9903869482401}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00043.warc.gz"} |
https://ardlussaestate.com/4gz3874z/interior-closure-boundary-examples-b48b38 | # interior closure boundary examples
If both Aand its complement is in nite, then arguing as above we see that it has empty interior and its closure is X. /Filter /FlateDecode Arcwise connected sets. Interior, Closure, Exterior and Boundary Interior, Closure, Exterior and Boundary Example Let A = [0;1] [(2;3). endobj Interior, closure, and boundary We wish to develop some basic geometric concepts in metric spaces which make precise certain intuitive ideas centered on the themes of “interior” and “boundary” of a subset of a metric space. 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Proof. Contents. For all of the sets below, determine (without proof) the interior, boundary, and closure of each set. stream Bounded, compact sets. /CA 0.25 /Pattern 15 0 R 2 0 obj Ω = { ( x , y ) | x 2 + y 2 ≤ 1 } {\displaystyle \Omega =\ { (x,y)|x^ {2}+y^ {2}\leq 1\}} is the disk's surrounding circle: ∂ Ω = { ( x , y ) | x 2 + y 2 = 1 } << A closed interval [a;b] ⊆R is a closed set since the set Rr[a;b] = (−∞;a)∪(b;+∞) is open in R. 5.3 Example. An external flow example would be airflow over an airplane wing. 5.2 Example. >> >> "���J��m>�ZE7�������@���|��-�M�䇗{���lhmx:�d��� �ϻX����:��T�{�~��ý z��N (In t A ) " ! /CA 0.4 /CA 0.7 /pgfprgb [ /Pattern /DeviceRGB ] /Parent 1 0 R %���� /F42 32 0 R /ca 0.6 One warning must be given. /Parent 1 0 R /F48 53 0 R Dense, nowhere dense set. A . A set A⊆Xis a closed set if the set XrAis open. The post office marks the [boundary] between the two municipalities. /pgf@ca.4 << /pgf@CA.4 << >> The set of boundary points is called the boundary of A and is denoted by ! b) Given that U is the set of interior points of S, evaluate U closure. is called open if is called closed if Lemma. Derived set. >> or U= RrS where S⊂R is a finite set. b(A). De–nition Theclosureof A, denoted A , is the smallest closed set containing A Perfect set. /CA 0.3 Interior, Closure, Boundary 5.1 Definition. Content: 00:00 Page 46: Interior, closure, boundary: definition, and first examples… We made a [boundary] of trees at the back of our… /Resources 67 0 R This topology course is frying my brain. /ca 0.6 endobj A set whose elements are points. For all of the sets below, determine (without proof) the interior, boundary, and closure of each set. << Precision perimeter Eclosure 0.182 ft. 939.46 ft. 1 5,176 Side Length (ft.) Latitude Departure degree minutes AB S 6 15 W 189.53 -188.403 -20.634 BC S 29 38 E 175.18 -152.268 86.617 CD N 81 18 W 197.78 29.916 -195.504 (a) A point in the interior of A is called an interior point of A. 9 >> /ProcSet [ /PDF /Text ] For each of the following subsets of R2, decide whether it is open, closed, both or neither. https://goo.gl/JQ8Nys Finding the Interior, Exterior, and Boundary of a Set Topology /pgf@ca.3 << gJ�����d���ki(��G���$ngbo��Z*.kh�d�����,�O���{����e��8�[4,M],����������_����;���$��������geg"�ge�&bfgc%bff���_�&�NN;�_=������,�J x LV�؛�[�������U��s3\Tah�$��f�u�b��� ���3)��e�x�|S�J4Ƀ�m��ړ�gL����|�|qą's��3�V�+zH�Oer�J�2;:��&�D��z_cXf���RIt+:6��3��9٠x� �t��u�|���E ��,�bL�@8��"驣��>�/�/!��n���e�H�����"�4z�dՌ�9�4. 3 0 obj /F59 23 0 R Classify It As Open, Closed, Or Neither Open Nor Closed. Let S be an arbitrary set in the real line R.. A point b R is called boundary point of S if every non-empty neighborhood of b intersects S and the complement of S.The set of all boundary points of S is called the boundary of S, denoted by bd(S). Some Basic De nitions Open Set: A set S ˆC is open if every z 0 2S there exists r >0 such that B(z 0;r) ˆS. A Comparison of the Interior and Closure of a Set in a Topological Space Example 1 Recall from The Interior Points of Sets in a Topological Space page that if$(X, \tau)$is a topological space and$A \subseteq X$then a point$a \in A$is said to be an interior point of$A$if there exists a$U \in \tau$with$a \in U$such that: >> Show transcribed image text. 7 0 obj /Subtype /Type1 /Count 8 ies: a theoretical line that marks the limit of an area of land Merriam Webster’s Dictionary of Law. /ca 0.4 >> Basic Theorems Regarding the Closure of Sets in a Topological Space; A Comparison of the Interior and Closure of a Set in a Topological Space; 2.5. Some of these examples, or similar ones, will be discussed in detail in the lectures. The boundary of Ais de ned as the set @A= A\X A. /Resources 65 0 R /ca 0.2 Lecture 2 Open Set and Interior Let X ⊆ Rn be a nonempty set Def. A topology on a set X is a collection τ of subsets of X, satisfying the following axioms: (1) The empty set and X are in τ (2) The union of any collection of sets in τ is also in τ (3) The intersection of any finite number of sets in τ is also in τ. boundary translation in English-Chinese dictionary. >> 1996. boundary I /pgf@ca.7 << 14 0 obj /F54 42 0 R /ColorSpace 14 0 R /F129 49 0 R Exercise: Show that a set S is an open set if and only if every point of S is an interior point. /pgf@ca.6 << >> >> 23) and compact (Sec. Note the difference between a boundary point and an accumulation point. Exterior points: If a point is not an interior point or boundary point of S, it is an exterior point of S. Lecture 2 Open and Closed set. >> bdy G= cl G\cl Gc. /FontBBox [ -350 -309 1543 1127 ] 11 0 obj << /Parent 1 0 R << Solutions to Examples 3 1. Videos for the course MTH 427/527 Introduction to General Topology at the University at Buffalo. /Type /Page /Parent 1 0 R Proof. /Type /Pages bwboundaries also descends into the outermost objects (parents) and traces their children (objects completely enclosed by the parents). Remark: The interior, exterior, and boundary of a set comprise a partition of the set. ��������9�L-M\��5�����vf�D�����ߔ�����T�T��oL��l~����],M T�?��� Wy#[ ���?��l-m~����5 ��.T��N�F6��Y:KXz L-]L,�K��¥]�l,M���m ��fg /Kids [ 3 0 R 4 0 R 5 0 R 6 0 R 7 0 R 8 0 R 9 0 R 10 0 R ] Selecting the analysis type. �06l��}g �i���X%ﭟ0| YC��m�. /F61 40 0 R endobj /Contents 57 0 R Example of a set whose boundary is not equal to the boundary of its closure. >> 03. Point set. 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Example 3.3. /CharSet (\057A\057B\057C\057E\057F\057G\057H\057I\057L\057M\057O\057P\057Q\057S\057T\057U\057a\057b\057bar\057c\057comma\057d\057e\057eight\057f\057ff\057fi\057five\057four\057g\057h\057hyphen\057i\057l\057m\057n\057nine\057o\057one\057p\057period\057r\057s\057seven\057six\057slash\057t\057three\057two\057u\057x\057y\057z\057zero) {\displaystyle \mathbb {R} ^ {2}} , the boundary of a closed disk. 4 0 obj Definition. /Filter /FlateDecode >> p������>#�gff�N�������L���/ /ca 0.3 The Boundary of a Set in a Topological Space; The Boundary of a Set in a Topological Space Examples 1; The Boundary of Any Set is Closed in a Topological Space You should change all open balls to open disks. >> /ca 0.4 Find its closure, interior and boundary in each case. Proposition 5.20. /Length 1969 of A nor an interior point of X \ A . 3 min read. Ask Question Asked 6 years, 7 months ago. By using our services, you agree to our use of cookies. Distinguishing between fundamentally different spaces lies at the heart of the subject of topology, and it will occupy much of our time. /Contents 66 0 R /ca 0.7 Cookies help us deliver our services. I could continue to stare at definitions, but some human interaction would be a lot more helpful. The set B is alsoa closed set. If anyone could explain interior and closure sets like I'm a five year old, and be prepared for dumb follow-up questions, I would really appreciate it. /pgf@CA0.7 << Interior and Boundary Points of a Set in a Metric Space. is open iff is closed. ����t���9������^m��-/,��USg�o,�� /MediaBox [ 0 0 612 792 ] for all z with kz − xk < r, we have z ∈ X Def. We give some examples based on the sets collected below. Topology (on a set). endobj /ca 1 Some examples. example. /Type /Page >> /Type /Page 1 Interior, closure, and boundary Recall the de nitions of interior and closure from Homework #7. Examples of … Derived Set, Closure, Interior, and Boundary We have the following definitions: • Let A be a set of real numbers. Suppose T ˆE satis es S ˆT ˆS. R R R R R ? One example is the Berlin Wall, which was built in 1961 by Soviet controlled East Germany to contain the portion of the city that had been given over to America, England, and France to administer. Interior, closure and boundary: examples Theorem 2.6 { Interior, closure and boundary One has A \@A= ? Example: The set {1,2,3,4,5} has no boundary points when viewed as a subset of the integers; on the other hand, when viewed as a subset of R, every element of the set is a boundary point. endobj 3.) Interior point. Theorem: A set A ⊂ X is closed in X iff A contains all of its boundary points. 9/20 . The Boundary of a Set. >> endobj /MediaBox [ 0 0 612 792 ] Closure of a set. /Type /FontDescriptor endobj A relic boundary is one that no longer functions but can still be detected on the cultural landscape. /Resources 76 0 R /F33 28 0 R 5 0 obj /Annots [ 61 0 R ] xڌ�S�'߲5Z�m۶]�eۿ��e��m�6��l����>߾�}��;�ae��2x�9��XQ�^��� ao�B����C$����ށ^�jc�D�����CN.�0r���3r��p00�3�01q��I� NaS"�Dr #՟ f"*����.��F�i������o�����������?12Fv�ΞDrD���F&֖D�D�����SXL������������7q;SQ{[[���3�?i�Y:L\�~2�G��v��v^���Yڙ�� #2uuT��ttH��߿�c� "&"�#��Ă�G�s�����Fv�>^�DfF6� K3������ @��� endstream /pgf@ca0.25 << Def. Examples of … As a consequence closed sets in the Zariski … /ca 0.8 (c)For E = R with the usual metric, give examples of subsets A;B ˆR such that A\B 6= A \B and (A[B) 6= A [B . /pgf@ca0 << endobj /Type /Page /ExtGState 17 0 R A . 12 0 obj Pro ve that for an y set A in a topological space we ha ve ! 1.4.1. 16 0 obj << >> /pgf@ca0.6 << General topology (Harrap, 1967). /CapHeight 696 Some of these examples, or similar ones, will be discussed in detail in the lectures. If A= [ 1;1] ( 1;1) inside of X= R2, then @A= A int(A) consists of points (x;y) on the edge of the unit square: it is equal to (f 1;1g [ 1;1]) [ ([ 1;1] f 1;1g); as you should check (from our earlier determination of the closure and interior of A). Latitudes and Departures - Example 22 EEEclosure L D 0.079 0.16322 0.182 ft. The closure of D is. De nition 1.1. >> /CA 0.6 The same area represented by a raster data model consists of several grid cells. << (By the way, a closed set need not have any boundary points at all: in $\Bbb R$ the only examples of this phenomenon are the closed sets $\varnothing$ and $\Bbb R$, but in more general topological spaces there can be many sets that are simultaneously open and closed and which therefore have empty boundary.) /MediaBox [ 0 0 612 792 ] a nite complement, it is open, so its interior is itself, but the only closed set containing it is X, so its boundary is equal to XnA. • The complement of A is the set C(A) := R \ A. Perfect set. 3 0 obj Our current model is internal and the fluid is bound by the pipe walls. /FontFile 20 0 R >> is open iff is closed. Interior and Boundary Points ofa Region in the Plane x1 x2 0 c a B 1.4. Since the boundary of a set is closed, ∂∂S=∂∂∂S{\displaystyle \partial \partial S=\partial \partial \partial S}for any set S. Thus, the algorithms implemented for vector data models are not valid for raster data models. Interior points of regions in space (R3). /Annots [ 68 0 R 69 0 R 70 0 R 71 0 R 72 0 R 73 0 R 74 0 R ] Interior and boundary points in space or R3. E X E R C IS E 1.1.1 . Set N of all natural numbers: No interior point. /Length2 19976 /pgf@CA0.8 << Interior, exterior and boundary points. /Descent -206 �� ��C]��R�����1^,"L),���>�xih�@I9G��ʾ�8�1�Q54r�mz�o��Ȑ����l5_�1����^����m ͑�,�W�T�h�.��Z��U�~�i7+��n-�:���}=4=vx9$��=��5�b�I�������63�a�Ųh�\�y��3�V>ڥ��H����ve%6��~�E�prA����VD��_���B��0F9��MW�.����Q1�&���b��:;=TNH��#)o _ۈ}J)^?N�N��u��Ez��v|�UQz���AڡD�o���jaw.�:E�VB ���2��|����2[D2�� /MediaBox [ 0 0 612 792 ] f1g f1g [0;1) (0;1) [0;1] f0;1g (0;1)[(1;2) (0;1)[(1;2) [0;2] f0;1;2g [0;1][f2g (0;1) [0;1][f2g f0;1;2g Z ? /CA 0 /CA 0.5 Table of Contents. /MediaBox [ 0 0 612 792 ] D = fz 2C : jzj 1g, the closed unit disc. From Wikibooks, open books for an open world < Real AnalysisReal Analysis. k = boundary(P) specifies points (x,y) or (x,y,z) in the columns of matrix P. example. %PDF-1.3 Please Subscribe here, thank you!!! Examples. These are boundaries that define our family and make it distinctive from other families. << >> Theorems. endobj /pgf@ca0.8 << Expert Answer 100% (1 rating) Previous question Next question Transcribed Image Text from this Question. For a general metric space, the closed ball ˜Br(x0): = {x ∈ X: d(x, x0) ≤ r} may be larger than the closure of a ball, ¯ Br(x0). /Length 20633 /Flags 4 /FirstChar 27 /LastChar 124 Its interioris the set of all points that satisfyx2+ y2+ z2 1, while its closure is x2+ y2+ z2= 1. 18), connected (Sec. 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Recall that U∈T Zaif either U= ? /FontName /KLNYWQ+Cyklop-Regular Interior and Boundary Points of a Set in a Metric Space. Bounded, compact sets. S = fz 2C : jzj= 1g, the unit circle. If fF /Contents 62 0 R This post is for a video which is the first in a three-part series. a is an interior point of M, because there is an ε-neighbourhood of a which is a subset of M. In any space, the interior of the empty set is the empty set. >> x��Z[oG~ϯط��x���(B���R��Hx0aV�M�4R|�ٙ��dl'i���Y��9���1��X����>��=x&X�%1ְ��2�R�gUu��:������{�Z}��ë�{��D1Yq�� �w+��Q J��t$���r�|�L����|��WBz������f5_�&F��A֯�X5�� �O����U�ăg�U�P�Z75�0g���DD �L��O�1r1?�/$�E��.F��j7x9a�n����$2C�����t+ƈ��y�Uf��|�ey��8?����/���L�R��q|��d�Ex�Ə����y�wǔ��Fa���a��lhE5�ra��$� �#�[Qb��>����l�ش��J&:c_чpU��}�(������rC�ȱg�ӿf���5�A�s�MF��x%�#̧��Va�e�y�3�+�LITbq/�lkS��Q�?���>{8�2m��Ža$����EE�Vױ�-��RDF^�Z�RC������P In the space of rational numbers with the usual topology (the subspace topology of R), the boundary of (-\infty, a), where a is irrational, is empty. >> 8. /Parent 1 0 R endobj /Parent 1 0 R Interior, exterior and boundary points. zPressure inlet boundary is treated as loss-free transition from stagnation to inlet conditions. << The closure of a set also depends upon in which space we are taking the closure. /Length1 980 The interior and exterior are both open, and the boundary is closed. 1 0 obj << I thought that U closure=[0,2] c) Give an example of a set S of real numbers such that if U is the set of interior points of S, then U closure DOES NOT equal S closure This one I was not sure about, but here is my example: S=(0,3)U(5,6) S closure=[0,3]U[5,6] If we let X be a space with the discrete metric, {d(x, x) = 0, d(x, y) = 1, x ≠ y. /Type /Page endobj Table of Contents. A . /F45 37 0 R /Font << >> b) Given that U is the set of interior points of S, evaluate U closure. Consider the subset A= Q R. Posted on December 2, 2020 by December 2, 2020 by example. /XHeight 510 8 0 obj If has discrete metric, 2. /Resources 63 0 R Interior and Boundary Points of a Set in a Metric Space. /ca 0.3 /pgf@ca0.7 << /F31 18 0 R R 2. /CA 0.4 Ob viously Aø = A % ! Find the interior of each set. Example 3.2. A vector x0 is an interior point of the set X, if there is a ball B(x0,r) contained entirely in the set X Def. For example, if X is the set of rational numbers, with the usual relative topology induced by the Euclidean space R, and if S = {q in Q : q 2 > 2, q > 0}, then S is closed in Q, and the closure of S in Q is S; however, the closure of S in the Euclidean space R is the set of all real numbers greater than or … >> a nite complement, it is open, so its interior is itself, but the only closed set containing it is X, so its boundary is equal to XnA. 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Our time, when these boundaries are, examples of … Please Subscribe here thank! | 2021-10-25 19:36:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7331698536872864, "perplexity": 1927.398518986281}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587767.18/warc/CC-MAIN-20211025185311-20211025215311-00142.warc.gz"} |
https://idrissi.eu/en/tags/operads/ | ##### The Voronov Product of Operads– Sep 22, 2016 #math #operads #swiss-cheese
My first real post in a while! It turns out that writing an actual paper (cf. previous blog post) takes a lot of time and effort. Who knew?
The Voronov product of operads is an operation introduced by Voronov in his paper The Swiss-cheese operad (he just called it “the product”). It combines an operad and a multiplicative operad to yield a new colored operad; the main example I know is the homology of the Swiss-cheese operad. This is a construction that I use in my preprint Swiss-Cheese operad and Drinfeld center, where as far as I know I coined the name “Voronov product” – I haven’t seen this operation at all outside of Voronov’s paper. I wanted to advertise it a bit because I find it quite interesting and I’m eager to see what people can do with it.
The purpose of this post is to record the definition of $$\infty$$-operads and explain why it works like that. For this I’m using Lurie’s definition of $$\infty$$-operads; there is also a definition by Cisinski–Moerdijk–Weiss using dendroidal sets, about which I might talk later.
Indeed, the definition on an $$\infty$$-operad is a bit mysterious taken “as-is” – see [HA, §2.1.1.10]. My goal is to explain how to reach this definition, mostly for my own sake. Most of what follows is taken either from the book Higher Algebra, the $$n$$Lab, or the semester-long workshop about hgiher category theory in Lille in 2015.
This post is about something somewhat weird I noticed about infinitesimal bimodules over operads and their relationships with some $$E_n$$ operads. I don’t know if it’s something significant, and I’d definitely be interested to hear more about it. | 2020-04-06 05:22:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7867717742919922, "perplexity": 638.2750686996085}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371618784.58/warc/CC-MAIN-20200406035448-20200406065948-00021.warc.gz"} |
https://en.wikipedia.org/wiki/Maxwell_bridge | # Maxwell bridge
A Maxwell-Wien bridge
A Maxwell bridge is a modification to a Wheatstone bridge used to measure an unknown inductance (usually of low Q value) in terms of calibrated resistance and inductance or resistance and capacitance. When the calibrated components are a parallel resistor and capacitor, the bridge is known as a Maxwell-Wien bridge. It is named for James C. Maxwell, who first described it in 1873.
It uses the principle that the positive phase angle of an inductive impedance can be compensated by the negative phase angle of a capacitive impedance when put in the opposite arm and the circuit is at resonance; i.e., no potential difference across the detector (an AC voltmeter or ammeter)) and hence no current flowing through it. The unknown inductance then becomes known in terms of this capacitance.
With reference to the picture, in a typical application ${\displaystyle R_{1}}$ and ${\displaystyle R_{4}}$ are known fixed entities, and ${\displaystyle R_{2}}$ and ${\displaystyle C_{2}}$ are known variable entities. ${\displaystyle R_{2}}$ and ${\displaystyle C_{2}}$ are adjusted until the bridge is balanced.
${\displaystyle R_{3}}$ and ${\displaystyle L_{3}}$ can then be calculated based on the values of the other components:
{\displaystyle {\begin{aligned}R_{3}&={\frac {R_{1}\cdot R_{4}}{R_{2}}}\\L_{3}&=R_{1}\cdot R_{4}\cdot C_{2}\end{aligned}}}
To avoid the difficulties associated with determining the precise value of a variable capacitance, sometimes a fixed-value capacitor will be installed and more than one resistor will be made variable. It cannot be used for the measurement of high Q values. It is also unsuited for the coils with low Q values, less than one, because of balance convergence problem. Its use is limited to the measurement of low Q values from 1 to 10.
${\displaystyle Q={\frac {\omega L}{R}}}$
The frequency of the AC current used to assess the unknown inductor should match the frequency of the circuit the inductor will be used in - the impedance and therefore the assigned inductance of the component varies with frequency. For ideal inductors, this relationship is linear, so that the inductance value at an arbitrary frequency can be calculated from the inductance value measured at some reference frequency. Unfortunately, for real components, this relationship is not linear, and using a derived or calculated value in place of a measured one can lead to serious inaccuracies.
A practical issue in construction of the bridge is mutual inductance: two inductors in propinquity will give rise to mutual induction: when the magnetic field of one intersects the coil of the other, it will reinforce the magnetic field in that other coil, and vice versa, distorting the inductance of both coils. To minimize mutual inductance, orient the inductors with their axes perpendicular to each other, and separate them as far as is practical. Similarly, the nearby presence of electric motors, chokes and transformers (like that in the power supply for the bridge!) may induce mutual inductance in the circuit components, so locate the circuit remotely from any of these.
The frequency dependence of inductance values gives rise to other constraints on this type of bridge: the calibration frequency must be well below the lesser of the self-resonance frequency of the inductor and the self-resonance frequency of the capacitor, Fr < min(Lsrf,Csrf)/10. Before those limits are approached, the ESR of the capacitor will likely have significant effect, and have to be explicitly modeled.
For ferromagnetic core inductors, there are additional constraints. There is a minimum magnetization current required to magnetize the core of an inductor, so the current in the inductor branches of the circuit must exceed the minimum, but must not be so great as to saturate the core of either inductor.
The additional complexity of using a Maxwell-Wien bridge over simpler bridge types[ambiguous] is warranted in circumstances where either the mutual inductance between the load and the known bridge entities, or stray electromagnetic interference, distorts the measurement results. The capacitive reactance in the bridge will exactly oppose the inductive reactance of the load when the bridge is balanced, allowing the load's resistance and reactance to be reliably determined. | 2019-10-21 14:46:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6262059211730957, "perplexity": 676.2224812918079}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987779528.82/warc/CC-MAIN-20191021143945-20191021171445-00446.warc.gz"} |
https://www.projectrhea.org/rhea/index.php/ECE_PhD_QE_CNSIP_2015_Problem1.2 | Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
Contents
Solution 1
$P((Z(t)=0) = P(Z(0)=0, N(t)=Even) + P(Z(0)=1, N(t)=Odd)\\ = pP( N(t)=Even) + (1-p)P( N(t)=Odd)\\ =p\sum_{m=0,1, 2, ...}P(N(t) = 2m)+ (1-p)\sum_{n=0,1,2,...}P(N(t)=2n-1)\\ =p\sum_{m=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^2m + (1-p)\sum_{n=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^{2n-1}\\ =p\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2} + (1-p)\cdot\frac{\lambda t}{1+\lambda t}\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ =\frac{p+\lambda t}{1+2\lambda t}$
$P((Z(t)=1) = 1 - P((Z(t)=0) = \frac{1+\lambda t - p}{1+2\lambda t}$
Solution 2
$P(Z(t)=0)=P(Z(t)=0|N(t)=even)P(N(t)=even)+P(Z(t)=0|N(t)=odd)P(N(t)=odd)$
Note that $\{Z(t)=0|N(t)=odd\}=\{Z(0)=1\}$ and $\{Z(t)=0|N(t)=even\}=\{Z(0)=0\}$, therefore,
$P(Z(t)=0)=P(Z(0)=0)P(N(t)=even)+P(Z(0)=1)P(N(t)=odd)\\ =p\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k}+(1-p)\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k+1}\\ =\frac{p}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}+ \frac{(1-p)\lambda t}{(1+\lambda t)^2}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ =\frac{p+\lambda t}{1+2\lambda t}\\ P(Z(t)=1) = 1- P(Z(t)=0) = 1-\frac{p+\lambda t}{1+2\lambda t} = \frac{1-p+\lambda t}{1+2\lambda t}\\ P(Z(t)=k)=\left\{ \begin{array}{cc} \frac{p+\lambda t}{1+2\lambda t}, k =0 \\ \frac{1-p+\lambda t}{1+2\lambda t}, k =1\\ 0, else \end{array} \right.$
The solution is correct. However the else case is not necessary. K can only be 0 or 1.
Solution 3
We know that $Z(t)$ can only take on the values 0 and 1, so we set out to find the probability that $Z(t)$ = 0; if we subtract this probability from 1, we will have found the probability that $Z(t) = 1$, and thus we will have described the entire pmf. We also know that if $Z(0)$ = 0, $Z(t)$ must be equal to 0 if $N(t)$ is even (i.e., if $k$ is even). Similarly, if $Z(0)\neq 0, Z(t)$ must be equal to 0 if $k$ is odd. As such, we can write the expression
$P(Z(t) = 0) = P(Z(0) = 0, \,N(t)\,\,is\,\,even) + P(Z(0) = 1, \,N(t)\,\,is\,\,odd) = P(Z(0) = 0)P(N(t)\,\,is\,\,even) + P(Z(0) = 1)P(N(t)\,\,is\,\,odd) \\= p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1}.$
We now recall that the sum of an infinite geometric series can be expressed as
$\sum_{k = 0}^\infty ar^k = \frac{a}{1-r}.$
We can use this to simplify the preceding equation:
$P(Z(t) = 0) = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1} \\ = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\frac{\lambda t}{1 + \lambda t}\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j} \\ = p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2}.$
We first combine terms:
$P(Z(t) = 0) =p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} \\ = \frac{p}{(1+\lambda t)\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)}.$
Then we simplify the denominator:
$P(Z(t) = 0) = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + \lambda t}{(1+\lambda t)^2 - (\lambda t)^2} \\ = \frac{p + \lambda t}{1 + 2\lambda t}.$
Now that we have found $P(Z(t) = 0)$, we can easily find $P(Z(t) = 1)$ by subtracting our result from 1:
$P(Z(t) = 1) = 1 - P(Z(t) = 0) \\ = 1 - \frac{p + \lambda t}{1 + 2\lambda t}\\ = \frac{1+2\lambda t}{1+ 2\lambda t} - \frac{p + \lambda t}{1 + 2\lambda t} \\ = \frac{1 + \lambda t-p}{1 + 2\lambda t}.$
Similar Problem
Find the mean function $\mu(t)$ and covariance function $C_{zz}(t_1,t_2)$ of the process $Z(t)$.
ECE462 Survivor
Seraj Dosenbach | 2019-01-21 18:55:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.917730450630188, "perplexity": 177.99425339311182}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583804001.73/warc/CC-MAIN-20190121172846-20190121194846-00382.warc.gz"} |
http://mathhelpforum.com/algebra/184384-parallel-perpendicular-lines.html | # Math Help - Parallel and perpendicular lines?
1. ## Parallel and perpendicular lines?
Write an equation of the line through the point that is parallel to the given line and perpendicular.
Point (7/8,3/4) line 5x+3y=0
which I turned into y=-5/3x
Then I plug into
(y-3/4)= -5/3(x-7/8) This is for parallel
3(3/4)=-5(x-7/8) Did I set this up correct
My second question is
(2,5) x=4 how would I set it up if the slope is undefined?
2. ## Re: Parallel and perpendicular lines?
Originally Posted by homeylova223
Write an equation of the line through the point that is parallel to the given line and perpendicular.
(2,5) x=4
Given the line $x=4$ and point $(2,5)$
one answer is $y=5$ and the other is $x=2$.
Which is which?
3. ## Re: Parallel and perpendicular lines?
Hmm I guess x=2 would be the parallel and y=5 perpendicular
4. ## Re: Parallel and perpendicular lines?
Originally Posted by homeylova223
Hmm I guess x=2 would be the parallel and y=5 perpendicular
Correct.
5. ## Re: Parallel and perpendicular lines?
Would this be parallel for my first question
3(y-3/4)=-5(x-7/8)
3y-2.25=-5x+4.375
3y=-5x+6.625
y=-5/3x+6.625/3 ?
6. ## Re: Parallel and perpendicular lines?
Originally Posted by homeylova223
Write an equation of the line through the point that is parallel to the given line and perpendicular.
Point (7/8,3/4) line 5x+3y=0
which I turned into y=-5/3x
Then I plug into
(y-3/4)= -5/3(x-7/8) This is for parallel
3(3/4)=-5(x-7/8) Did I set this up correct
[snip]
Yes. | 2014-07-28 21:01:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7001123428344727, "perplexity": 1958.1475130671831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510261958.8/warc/CC-MAIN-20140728011741-00333-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://dml.cz/handle/10338.dmlcz/134062?show=full | # Article
Title: Cantor-Bernstein theorem for lattices (English) Author: Jakubík, Ján Language: English Journal: Mathematica Bohemica ISSN: 0862-7959 Volume: 127 Issue: 3 Year: 2002 Pages: 463-471 Summary lang: English . Category: math . Summary: This paper is a continuation of a previous author’s article; the result is now extended to the case when the lattice under consideration need not have the least element. (English) Keyword: lattice Keyword: direct product decomposition Keyword: Cantor-Bernstein Theorem MSC: 06B05 idZBL: Zbl 1007.06005 idMR: MR1931330 . Date available: 2009-09-24T22:03:46Z Last updated: 2012-06-18 Stable URL: http://hdl.handle.net/10338.dmlcz/134062 . Reference: [1] A. De Simone, D. Mundici, M. Navara: A Cantor-Bernstein theorem for $\sigma$-complete $MV$-algebras.(to appear). Reference: [2] J. Jakubík: Cantor-Bernstein theorem for lattice ordered groups.Czechoslovak Math. J. 22 (1972), 159–175. MR 0297666 Reference: [3] J. Jakubík: On complete lattice ordered groups with strong units.Czechoslovak Math. J. 46 (1996), 221–230. MR 1388611 Reference: [4] J. Jakubík: Convex isomorphisms of archimedean lattice ordered groups.Mathware and Soft Computing 5 (1998), 49–56. MR 1632739 Reference: [5] J. Jakubík: Cantor-Bernstein theorem for $MV$-algebras.Czechoslovak Math. J. 49 (1999), 517–526. MR 1708370 Reference: [6] J. Jakubík: Direct product decompositions of infinitely distributive lattices.Math. Bohem. 125 (2000), 341–354. MR 1790125 Reference: [7] J. Jakubík: On orthogonally $\sigma$-complete lattice ordered groups.(to appear). MR 1940067 Reference: [8] J. Jakubík: Convex mappings of archimedean $MV$-algebras.(to appear). MR 1864107 Reference: [9] J. Jakubík: A theorem of Cantor-Bernstein type for orthogonally $\sigma$-complete pseudo $MV$-algebras.(Submitted). Reference: [10] R. Sikorski: A generalization of theorem of Banach and Cantor-Bernstein.Coll. Math. 1 (1948), 140–144. MR 0027264 Reference: [11] R. Sikorski: Boolean Algebras.Second Edition, Springer, Berlin, 1964. Zbl 0123.01303, MR 0126393 Reference: [12] E. C. Smith, jr., A. Tarski: Higher degrees of distributivity and completeness in Boolean algebras.Trans. Amer. Math. Soc. 84 (1957), 230–257. MR 0084466 Reference: [13] A. Tarski: Cardinal Algebras.New York, 1949. Zbl 0041.34502 .
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Partner of | 2019-07-17 06:16:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48993152379989624, "perplexity": 6669.036594175617}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00366.warc.gz"} |
https://tla.msr-inria.inria.fr/tlaps/content/Documentation/Tutorial/Practical_hints.html | ## Hints on using TLAPS effectively
The TLA+ proof system is designed to check the validity of claims as independently as possible of specific proof back-ends. We believe that users should concentrate on writing proofs in terms of their particular applications, not in terms of the capabilities of a particular proof system. In particular, TLAPS invokes its back-ends with some default setup for automatic proof, and we try to make it hard for users to change this default setup. Expert users of back-end provers may be frustrated because they may have to develop proofs somewhat further than what would be necessary with a fine-tuned tactic script. The main payoff of limited access to the nitty-gritty details of provers is greater clarity of the resulting proofs. They are also easier to maintain across minor changes of the specification or new releases of the TLA prover.
On some occasions users will encounter situations where the prover cannot prove an "obvious" proof obligation. Here are a few hints on what to try to make the proof go through. Your additions to this list are welcome.
### Control the size of formulas and expressions
Our provers are currently not good at making abstractions that humans understand immediately. They are easily confused by moderately big proof obligations and are just as likely to work on a top-level conjunction as on a set construction buried deeply inside the formula. This can cause back-ends to become very slow or even unable to complete seemingly trivial steps. While we intend to improve the back-ends in this respect, you can help them by using local definitions in proofs and hiding these definitions in order to keep expressions small. (Keep in mind that definitions introduced in a proof are usable by default and must be hidden explicitly, unlike definitions in specifications, which must be explicitly USEd.)
Here is a contrived example:
This kind of problem typically arises when reasoning about LET expressions, which are silently expanded by the proof manager. In a proof, introduce local definitions corresponding to the LET (using copy and paste from the specification), show that the overall expression equals the body of the LET, establish the necessary facts about these locally defined operators, and HIDE the definitions afterwards.
### Avoid "circular" (sets of) equations
Rewriting is one effective way to reason about equations, and it underlies the automatic proof methods used by the Isabelle back-end. The basic idea is to orient equalities such that the expressions on the left-hand side are systematically replaced by the right-hand sides. However, if the set of equations contains cycles as in
then rewriting may never terminate. Isabelle employs some (incomplete) heuristics to detect such cycles and will refuse to rewrite equations that it determines to be circular. This usually leads to its inability to infer anything about these equations. If circularity is not detected, it may cause Isabelle to enter an infinite loop. The suggested remedy is again to introduce local definitions that are hidden to break the loops.
As a concrete example consider the following proof snippet:
One possible workaround is as follows:
### Using set extensionality
The theorem of set extensionality asserts that two sets are equal if they contain the same elements:
This theorem is defined in the standard module TLAPS and can be proved automatically in TLAPS. Nevertheless, it is sometimes necessary to appeal to that theorem for proving that two set expressions are equal. Making set extensionality a part of the background theory would be counter-productive, since it could be applied to many occurrences of the equality symbol. When necessary, it can be added explicitly in a BY clause for the SMT backend. For Isabelle proofs, TLAPS defines a specific pragma IsaWithSetExtensionality that instructs Isabelle to try applying the set extensionality rule for proving equality of sets.
Consider a definition such as
In order to prove a property Q(foo), you will typically prove the two following assertions:
(a) \E x \in S : P(x)
(b) \A x \in S : P(x) => Q(x)
In some cases, assertion (b) can be trivial and need not be shown explicitly. Reasoning about an unbounded CHOOSE expression is analogous.
Remember that CHOOSE always denotes some value, even if P(x) holds for no x \in S (in particular, if S = {}), in which case the CHOOSE expression is fixed, but arbitrary. In practice, CHOOSE expressions usually arise when condition (a) is satisfied. Should you have designed your property to work even if the domain of the CHOOSE is empty, property Q must be trivial in that case, and you can structure your proof as follows:
A frequent TLA+ idiom is to define a "null" value by writing
The laws of set theory ensure that no set is universal, hence there exists an x that is not an element of set Value, ensuring condition (a) above. The theorem NoSetContainsEverything in the standard module TLAPS can be used to prove this condition.
The SMT backend may fail to prove obligations involving several CHOOSE expressions. In particular, the axioms for determinacy of CHOOSE stating
may not be available to the SMT solver.
### Help Zenon and Isabelle When Reasoning About Records
and were trying to prove
from facts that included
Zenon failed on the proof and Isabelle proved it only after a long time. (In fact, we originally stopped the proof because it was taking so long.) However, Zenon proved it instantly when we added mb.type = "1b" to the BY statement's list of facts. The provers are reluctant to try finding relations of the form record.field = value. They often need help.
The SMT backend should not require similar help for reasoning about records.
### Divide and Conquer
When the provers can't prove something that you think is obvious, it's usually because it isn't true. You can easily spend hours looking at a proof obligation without noticing a tiny mistake. The best way to find a mistake is by breaking the proof into simpler steps. Continuing to do this on the step or steps whose proof fails will eventually lead you to discover the problem – usually a missing hypothesis or a mistake in a formula. When you correct the mistake in the original proof step, the prover will usually be able to prove it.
### Don't Reinvent Mathematics
We expect that most people who use TLAPS will do so because they want to verify properties of an algorithm or system. We have therefore not devoted our limited resources to building libraries of mathematical results. If you want to create such libraries, we would welcome your help. However, if you are concerned with an algorithm or system, you should not be spending your time proving basic mathematical facts. Instead, you should assert the mathematical theorems you need as assumptions or theorems.
Asserting facts is dangerous, because it's easy to make a mistake and assert something false, making your entire proof unsound. Fortunately, you can use the TLC model checker to avoid such mistakes. For example, our example correctness proof of Euclid's algorithm uses this assumption
TLC cannot check this assumption because it can't evaluate a quantification over an infinite set. However, you can tell TLC to replace the definition of Nat with
(In the Toolbox, use the Definition Override section of the model's Advanced Options page.) TLC quickly verifies this assumption. (TLC checks each ASSUME; to add an assumption that you don't want TLC to check, make it an AXIOM.)
This kind of checking is almost certain to catch an error in expressing a fundamentally correct mathematical result – except when the only counterexamples are infinite. Fortunately, this is rarely the case when the result is needed for reasoning about an algorithm or system.
### It's Easier to Prove Something if it's True
Before trying to prove a property of an algorithm or system, try to check it with TLC. Even if TLC cannot check a large enough model to catch all errors, running it on a small model can still catch many simple errors. You will save a lot of time if you let TLC find these errors instead of discovering them while writing the proof. | 2021-12-07 14:05:11 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.844662070274353, "perplexity": 815.9893536060072}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00255.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-5x-5-10#419607 | # How do you solve 5x - 5 + -10?
May 8, 2017
It is $5 x - 15$
#### Explanation:
You can add numerical values simply
$5 x - 5 - 10$
which is
$5 x - 15$ | 2022-01-25 05:38:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.528389573097229, "perplexity": 6909.708674544281}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304760.30/warc/CC-MAIN-20220125035839-20220125065839-00065.warc.gz"} |
https://www.physicsforums.com/threads/double-slit-momentum.925425/ | # B Double Slit & Momentum
1. Sep 13, 2017
### fanieh
How true is this? Do you believe that in the double slit experiment, what is subsequently measured at the detection screen is actually the particle’s momentum?? How does this differ to the normal double slit explanation of the particle being deflected left or right of the slit, etc.?
http://www.users.csbsju.edu/~frioux/diffraction/s00897040748a1.pdf
"In Marcella’s quantum mechanical analysis of the double slit experiment, what is subsequently measured at the detection screen is actually the particle’s momentum. In other words, the well-known diffraction pattern created by the double-slit geometry is the particle’s momentum distribution in the plane of the detection screen. Therefore, to calculate the diffraction pattern one needs a momentum wave function, and this is obtained by a Fourier transform of eq 1 into momentum space"
2. Sep 14, 2017
### Simon Bridge
There is no difference between this momentum description and saying a particle is deflected. The use of this approach is that "momentum" is well defined while "particle deflection" is, at leats in this context, not.
Marcella's paper, Quantum interference at slits ( https://arxiv.org/abs/quant-ph/0703126 ), has other problems.
See followup paper: Rothman and Bouhm, Quantum interference at slits revisited ( https://arxiv.org/abs/1009.2408)
3. Sep 14, 2017
### fanieh
Say. Can the wave properties of any particles or fields be detected without using any material detector? Can anyone give an example proving the wave properties is inherent in the particles or fields? Thanks.
4. Sep 14, 2017
### DrChinese
As a matter of fact, they can. The example I like is using polarizers in front of each of the 2 slits. When the polarizers are parallel, there IS interference. When the polarizers are perpendicular, there is NO interference.
The only relevant variable here is the relative orientation of the 2 polarizers. Each individual polarizer's absolute orientation makes no difference. This is effectively a situation in which there is no material detecting the which path information.
5. Sep 14, 2017
### vanhees71
Well, the with-way information is simply in the polarization state of the particles. If you have, e.g., photons and the relative orientation of the polarization filters is $\pi/2$, then you have no interference at all, because the polarization states of the photons runing through slit 1 are perpendicular to thouse of slit 2. If the relative orientation is $0$ (i.e., parallel oriented polarizers) you have full interference contrast. For all relative angles between these extrems you get interference patterns with less than the maximal contrast.
6. Sep 14, 2017
### fanieh
The first paper was shared by Bill Hobba for over 50 times in the PF archives.. he summarized it thus:
"Here is the double slit experiments explanation without using that myth: http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf Whats going on is this. Each slit is a position measurement. After the slit it has a definite position so by the Heisenberg uncertainly relation it momentum is unknown. It's kinetic energy is still the same so the magnitude of its velocity doesn't change - its scattered in an unpredictable direction. When you have two slits the wavefunction is a superposition of the the wavefunction at each slit and when you work through the math as detailed in the above link you get interference. Thanks Bill"
Reference https://www.physicsforums.com/threa...lit-experiment-questions.799522/#post-5020416
I'd like to know. Is this momentum context of the explanation the consensus or 100% followed by all physicists.. but then Simon shared the second paper which gives the explanation as somewhat misleading.. so I'd like feedback from super experts on this. Thanks.
7. Sep 14, 2017
### Mentz114
Yes this is fine for light, which can be in physical superposition.
How does this help with 'matter waves' ?
8. Sep 14, 2017
### fanieh
I have polarized sunglasses so I know what it can do able to suppress all horizontal light waves. If you just put one polarizer in any slit in the double slit.. won't it cause a darkening of the interference pattern? But putting for example the polarized sunglasses in one of the slits would suppress all horizontal photons in that slit.. won't this create a which way path for the horizontally polarized photon or electron?
9. Sep 14, 2017
### Staff: Mentor
Yes, but not for the vertically polarized photons; these still have both paths available so will interfere.
10. Sep 14, 2017
### fanieh
It is said if you have two perpendicular polarizer.. there is no interference pattern because it can tell which path the photon passes (because one of them will pass it 100% and the other 0%). But if the photon has polarization photon oriented 45 degrees.. won't it pass thru both slits and polarization without giving any which way path?
11. Sep 14, 2017
### Mentz114
The polarizers argument is true but overcomplicated.
The general principle is that the two parts of the superposition must remain coherent for interference to happen.
If anything happens on one path that is different from the other - the coherence is lost and there is no interference.
12. Sep 15, 2017
### fanieh
Polarizer can prove the wave part of is in the particle being emitted and not in the detector as hidden pattern. See this New Scientist article shared by a science advisor here:
https://www.newscientist.com/article/mg21128241-700-beyond-space-time-welcome-to-phase-space/
"Smolin’s hunch is that we will find ourselves in a place where space-time and momentum space meet: an eight-dimensional phase space that represents all possible values of position, time, energy and momentum. In relativity, what one observer views as space, another views as time and vice versa, because ultimately they are two sides of a single coin – a unified space-time. Likewise, in Smolin’s picture of quantum gravity, what one observer sees as space-time another sees as momentum space, and the two are unified in a higher-dimensional phase space that is absolute and invariant to all observers. With relativity bumped up another level, it will be goodbye to both space-time and momentum space, and hello phase space."
Some question whether having objective phase space means there is a hidden interference guide in an object. This can give another view of how the double slit experiment can work. Using this particular phase space duplex-space perspective, one can see an entirely different explanation for the very famous Young's double slit experiment from the era of the classical mechanics paradigm. The conventional, single-space explanation (the old space and time explanation) saw the result as the interference of the light waves entering the two parallel slits. In that model, the slit structure itself contributes nothing but the two, parallel gap openings. This phase space duplex-space perspective says the slit structure itself, without the light waves, already has an Phase space substance interference pattern existing around the slit regions of the physical space structure. The model is that it is this objective phase space pattern that guides the light into its maxima and minima ordinary space intensity locations behind the slits.
Polarizer can help refute the idea. Anyway. If you are hit with electron beam (with electron matter wave).. and you use a polarizer.. would it filter some of the electron matter wave too? Or does polarizer only work for photons?
13. Sep 15, 2017
### vanhees71
Not true! As a single person I'm a tiny fraction of physicsts, but not 0% I think this paper is a didactical desaster and should not be used, and I told so in this forums several times. I can't help that nobody seems to take this criticism seriously. At least there is a published paper in Eur. J. Phys. also criticizing this paper, and this should be taken seriously.
14. Sep 15, 2017
### fanieh
But note that for all relative angles between these extremes of parallel and perpendicular, how can you get interference patterns when the polarizer can tell where the photon passes thru because it can make the pattern brighter.. for example. the right slit with the photon say 25 degree orientation closer to one of the polarizer vertical or horizontal.. do you get what I mean..i planned to draw to illustrate it but I think you get the idea...
15. Sep 15, 2017
### fanieh
Let's say you put the 2 polarizers outside the 2 slits of the double slit as DrChinese tried. The one on left is oriented vertical.. while the one on right is oriented 60 degrees. Now assume you have photon passing thru the slit with polarization of 25 degrees.. It's closer to the left (25 degrees) than the right polarizer which has 35 degrees from the nearest horizontal. Therefore the right polarizer would be brighter.
1. Wouldn't this produce a which way path to the 25 degrees photons producing lack of intereference for the 25 degree photons?
2. If yes. Does it mean only 30 degree photon can cause interference (since this is middle to the vertical and 60 degree oriented polarizer)?
3. In double slit, in one at a time photon or electron emission.. what degree of polarization does it use?
Thanks a lot!
16. Sep 15, 2017
### vanhees71
As I said, then you get interference patterns with reduced contrast. The point is that the polarization states for not perpendicular to each other oriented polarizers are not orthonal, i.e., $\langle \psi_1|\psi_2 \rangle \propto \cos \alpha \neq 0$, and thus you get some interference but with some function of the relative angle $\alpha$ of the polarizer orientation.
17. Sep 15, 2017
### Staff: Mentor
The incident photons have no polarization before they encounter the filter (because we haven't yet measured it). Thus, no matter what the angle of the filter, every photon has a 50% chance of passing the filter, and both polarizers are equally "bright".
18. Sep 15, 2017
### vanhees71
The point is not, how many photons run through the slits but whether there is interference between the two alternative paths. If the filters are in 90-degree relative orientation you have no interference, if they are in 0 (or 180)-degree relative orientation you have interference with full contrast. For any angle in between you have interference with reduced contrast.
19. Sep 15, 2017
### fanieh
You mean before measurement, the polarization is also in superposition just like the position or momentum? I thought only the position and other observables don't have values before measurement. I didn't know even the polarization didn't have definite degrees before measurement (?) Why is polarization not an observable in QM?
20. Sep 15, 2017
### vanhees71
Of course, polarization is an observable in QM. The standard basis are the helicity states ($\pm 1$ for photons), corresponding to left- and right-circular polarized modes of the em. field. | 2018-01-18 08:47:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.555306077003479, "perplexity": 1022.7457049941667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00507.warc.gz"} |
https://www.intechopen.com/online-first/optical-phase-modulation-techniques | Open access peer-reviewed chapter - ONLINE FIRST
Optical Phase-Modulation Techniques
By Ramón José Pérez Menéndez
Submitted: February 10th 2019Reviewed: October 31st 2019Published: December 9th 2019
DOI: 10.5772/intechopen.90343
Abstract
Optical phase-modulation technique is a very powerful tool used in a wide variety of high performance photonic systems. Fiber-optic sensors and gyroscopes, integrated-optics sensors, or high-performance photonic integrated circuits are some examples of photonic systems where the optical phase-modulation technique can be efficiently applied. In time, such a photonic system can be integrated as the core part of some specific applications like biosensors, 5G advanced optical communication devices, gyroscopes, or high-performance computation devices. In this work, the main optical phase-modulation techniques are analyzed. Also, a study of the most significant applications of this technique is made, relating it to the most appropriate type in each case.
Keywords
• optical phase-modulation
• electro-optic phase modulators
• sinusoidal phase-modulation
• square-wave phase-modulation
• triangular phase-modulation
• serrodyne phase-modulation
• phase-locking technique
• phase-locked-loop
• optical gyroscopes
1. Introduction
Optical interferometry constitutes an important technique used in a high number of measurement processes for multiple physical magnitudes and quantitative phenomenon [1]. Particularly, fiber-optic waveguides can act as very useful and efficient transmission medium for light guidance in a large group of interferometry-based sensor devices. On the other hand, the study of subject of interferometer fiber-optic sensors has received an extensive treatment in the literature [2, 3]. In this article, a wide analysis of optical phase-modulation is made focused on the optical gyroscopes as the main referenced application. A simple open-loop configuration of interferometric fiber-optic gyroscope based on the Sagnac effect is shown in Figure 1. This kind of gyro is based on the Sagnac-effect within an open optical path realized by a N-turn fiber-optic coil when two independent counter-propagating light modes are externally introduced from a broadband laser source through its two ends, respectively. This causes that an interference pattern between the CW and CCW light beams to be collected in a photo-detector with a phase shift given by the following equation, Ref. [4]:
ϕS=2πLDλ0c0ΩE1
where L and D are length and diameter of fiber-optic sensing coil, respectively, λ0 and c0 are wavelength and speed of light source in vacuum, respectively, and Ω is the rotation rate. Figure 1 clearly shows that this Interferometer-Fiber-Optic-Gyro (IFOG) has a passive configuration because the laser source is located externally to the sensing coil. In this system, the two counter-propagating light beams travel through the core of a conventional single-mode optical fiber (SMF) by total-internal-reflection phenomenon. As the core diameter of such an optical fiber is only about 8 μm, the spot size of the interference signal can only be coupled to a small area at the end of the fiber loop, for example on the small detection area of a photo-detector. So that, this interference signal affects only one or two interference fringes whose intensity can be evaluated by the following expression:
Iϕ=I01+cosϕE2
being I(ϕ) the output optical signal of interferometer, I0 the amplitude of each of two CW and CCW counter-propagating beams and ϕthe optical phase-difference between them.
Figure 2 represents the variation of light intensity along a single interference fringe as a function of ϕ. Notice the output intensity noise produced when the phase difference is detected with a phase error Δϕ. Phase noise sources of this gyro, their influence on output signal and solutions are exhaustively treated in Refs. [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16].
However, the simple and raw gyro configuration (laser-source, beam-splitter, fiber-coil sensor and photo-detector) showed in Figure 1 is not effective in practice mainly due to its inability to reduce phase errors. Thus, the decrease in phase error can be effectively achieved by a phase-modulation process of CW and CCW optical waves by adding an optical phase-modulator in the path of the waves entering the optical fiber sensing coil as it is shown in the arrangement of Figure 3. Phase error decreasing is mainly achieved by electronic filtering within phase-sensitive demodulation circuits (PSD).
In this scheme, the immediate consequence of the application of a phase modulation to CW and CCW waves is the need to have an electronic demodulation system for the optical phase (commonly called PSD). This electronic phase-sensitive demodulation system must be located at the electric output of photo-detector. This way, by a demodulation process of electrical output signal of photo-detector, the Sagnac phase shift can be retrieved as calculated from Eq. (1) when a rotation-rate Ω is applied to the whole system. Also, as showed in Figure 3, other elements like a second beam splitter, a polarizer and an optical filter are needed to complete the system. The main function of the electro-optical phase modulator is to provide a controlled phase shift which will be added to Sagnac phase shift produced by the rotation onto the system. This way, the signal detected by photo-detector can be demodulated with some ease to recover by electronic means the Ω rotation-rate value which affects the whole system.
A more advanced design is achieved by closing the measurement loop by means of a feedback signal becoming into the scheme so-called IFOG closed loop configuration. The general scheme of a closed loop IFOG is depicted on Figure 4. In this scheme, the output signal of demodulator circuit passes through a servo amplifier which drives a phase transducer placed in the interferometer path. Then, the whole system works under the phase-nulling principle. This means that the total phase shift becomes equal to zero because the phase transducer introduces a non-reciprocal phase shift that is equal, by in the opposite sign, to that generated by Sagnac phase shift induced by rotation. The output of the system is then the output of the phase transducer.
The main advantage of this configuration is the insensitivity to the laser source amplitude variations and the electronic circuitry gain because the system is always operated at zero total phase shift. Other design alternatives are possible, and so instead of using a fiber optic coil as a sensor it can be used a ring resonator integrated in a silicon waveguide, Ref. [17].
For open-loop configuration square-wave bias and sinusoidal phase modulation are usually applied while for closed-loop configuration, sinusoidal or square-wave bias and serrodyne feedback phase modulations are frequently used. In the following, a particular study of all these types of optical phase modulation will be made.
2. Square-wave optical phase modulation
One of the first attempts to apply the principle of phase modulation to CW-CCW optical waves in an optical gyroscope can be seen in Ref. [18]. In this case, a sinusoidal-wave phase modulation is applied mainly due first to the ease of finding fast bulk phase modulators in lithium niobate (LiNbO3), Ref. [19], and also reliable electronic sine wave oscillators. However, square-wave is frequently used as bias phase modulation because it allows periodically shift the working point of the gyro to each one of ±π/2 constant values, respectively. This last is due that when the central working point of the gyro is either +π/2 or −π/2, its sensitivity reach a maximum value, as it can be seen on Figure 5. Then, when the system is not subjected to rotation (Ω = 0), the output response is a pectinate-shaped curve with a constant value. But, when the system is subjected to a non-zero rotation rate, the interferometric response output curve is also a Square-Wave whose peak-to-peak amplitude is proportional to the value of the rotation speed. The latter can also be also checked by observing in detail the Figure 6. In this figure, τ is the transit time of the CW and CCW optical waves over the fiber-coil length and ϕSis the Sagnac phase shift caused between them by rotation.
For this purpose, one phase modulator is located at the end of fiber coil, as represented in the scheme of Figure 4. Thus, the calculation of effective phase shift induced by the phase modulation process between CW and CCW optical waves at the output of fiber coil after their respective roundtrip can be expressed as follows:
Δϕt=ϕCCWtϕCWt=ϕtϕtτE3
In Eq. (3), ϕtrepresents the time waveform of applied phase modulation and τ is, again, the transit time around the fiber coil which, in time, can be calculated as:
τ=nLcE4
here, n is the effective refractive index of fiber, L is the total length of fiber coil and c is the vacuum speed of light. For obtaining the result of Eq. (3) it has been taken into account that CW and CCW waves enter the fiber coil at opposite ends. Then, in the case of square-wave phase modulation as represented in Figure 6, analytics gives two constant phase values, namely, ±π/2 (continuous black wave) when no-rotation is applied to the system. However, when a non-zero rotation is applied, a phase difference equal to ϕSSagnac phase shift must be added to phase difference applied by the external phase modulation process. The explanation of blue (continuous) and red (dotted) output response curves of photo-detector is as follows. When rotation rate is equal to zero, the projection of the points of the input square waveform (continuous black) on the response curve of the interferometer gets a pectinate-shaped output waveform (continuous blue curve). However, when a non-zero rotation rate, a pectinate square-wave is obtained (dotted red curve). In this last case, as seen in the inset of Figure 6, the peak-to-peak square-wave value is very close proportional to rotation rate and can be evaluated as:
ΔIt=I0sinϕSKϕS=KΩΩE5
where the approximation can be justified because the value of the sine-function can be approximated by its argument when it is less than π/6 in absolute value.
Figure 6 shows a closed-loop block-diagram scheme of an IFOG model with square-wave BIAS and serrodyne FEEDBACK phase modulations. In this case it is necessary to use two different phase modulators, first one to apply the square wave BIAS and second one for the serrodyne FEEDBACK phase modulation. See the complete description of this block-diagram on Ref. [20].
Gyro designs with square-wave kind of phase modulation can be seen in Refs. [21, 22, 23, 24]. Last three engineered by Chinese authors utilize square waves staggered by sections (four-, five- or six-points phase modulation, respectively) and their main advantage is that all these schemes allow to improve the accuracy and scale factor of the gyro.
3. Sinusoidal optical phase modulation
The basic idea to apply the sinusoidal bias phase modulation to an IFOG configuration is that the amplitude of the first harmonic component of interferometer output signal contains information of the ϕSSagnac phase shift induced by rotation. In particular, this amplitude can be considered approximately linearly proportional to the absolute value of rotation rate that affects the system. This fact will be analytically derived next. Suppose first a simple open-loop IFOG configuration, like that represented on Figure 3. Then, a sinusoidal bias phase modulation is applied to phase modulator, like the supplied by an electric sine oscillator working at fm frequency and amplitude ϕ0in the following form:
ϕmt=ϕ0sinωmtE6
The phase difference between CW and CCW waves induced by this bias phase modulation will be:
Δϕmt=ϕCCWtϕCWt=ϕmtϕmtτΔϕmt=2ϕ0sinωmτ2cosωmtτ2ωmτ=π2ϕ0cosωmtτ2=ϕmsinωmtE7
here 2ϕ0=ϕm. As it can be seen from this equation, the maximum value of phase difference modulation for a given value of ϕ0amplitude will be reached when the ωmτ=πcondition to be accomplished. This condition is reached when the frequency fm equals the value:
fm=12τE8
referred as proper frequency of the system, here τ, the transit time. Then, under these conditions, when a rotation with ϕSSagnac phase shift affects the system, the total phase difference between CW and CCW waves will be:
Δϕt=Δϕmt+ϕS=ϕmsinωmt+ϕSE9
Therefore, the interference signal can be obtained by the following calculation:
IΔϕ=I01+cosΔϕ=I01+cosϕmsinωmt+ϕS=I01+J0ϕm+20J2nϕmcos2nωmtcosϕS20J2n1ϕmsin2n1ωmtsinϕSE10
here Jnis the Bessel function of first kind and n-order. From Eq. (10) it can be observed that interference signal is the sum of three terms: a constant first term that does not depend on the frequency of modulation, a second term that includes the factor cosϕSmultiplied by one infinite sum of even harmonics and a third term that includes the factor sinϕSmultiplied by one infinite sum of odd harmonics. When ϕmvalue is adjusted to be 1.84, then the J1term reach its maximum value, namely, J11.84=0.5815. It has also been shown that the amplitude of higher-order odd harmonics (3rd, 5th and successive) are getting smaller and smaller so their contribution can be neglected. Therefore, when a rotation rate affects the system, an electronic selective filtering of odd harmonics of interference signal is convenient to isolate and extract the data of rotation speed. Particularly, the first harmonic is the most convenient to recover, so that an electronic band-pass filtering at fundamental frequency of modulation should be added as phase-sensitive demodulator. This way, a successive low-pass filtering located after the band-pass one could retrieve the first harmonic amplitude, which is proportional to ϕSSagnac phase shift and, in turn, to rotation rate value.
Figure 7 represents the analytic sinusoidal phase modulation process. As it can be seen, when no rotation is applied, the interference signal (solid, blue curve) also contains even harmonics since the sinϕSfactor is canceled. However, when rotation is applied, both even and odd harmonics appear (dotted, red curve) in the interference output signal since the sinϕSand cosϕSare not canceled.
Figure 8 represents the block-diagram of a closed-loop IFOG configuration with sinusoidal BIAS and serrodyne FEED-BACK phase modulations, see Ref. [30]. The main novelty of this design is the structure of phase modulation feedback chain. In this case, one FET transistor (2N3848) is added on feedback branch of integrator OPAMP (block #7 on Figure 8). This block generates a linear ramp voltage Vγ on its output, and this ramp resets each one time-period driving by Vgate voltage. In this way, a resultant serrodyne-wave voltage is easily generated at the output of integrator circuit, obtaining finally the same intended sawtooth-wave voltage on feedback phase modulation chain as reported on previous designs. The output signal of the photodetector, in photocurrent form, is proportional to the light intensity at its optical input. This photocurrent signal is converted to voltage with a trans-impedance amplifier that is placed at the input of demodulation circuit. The demodulation circuit (PSD) takes the task of extracting the information of the ϕSSagnac phase shift induced by rotation. The corresponding voltage signal at its output (VS) scales as sine-function of ϕS. The PI controller realizes an integration of VS signal in time-domain, so that a voltage signal (Vγ) is obtained; this signal scales almost linearly with the time. This latter signal is filtered by means of a low-pass filter so that the corresponding output signal () is a DC voltage value that is more accurately proportional to the gyroscope rotation-rate Ω (since the following approximation is fulfilled in the working range: sinϕSϕS). Therefore, the analog output voltage signal constitutes the measurement of the rotation rate of the system. The control system, as a whole, acts as the principle of phase nulling. The phase-nulling process consists of generating a phase displacement (ϕm=ϕbias+ϕf) in such a way that the ϕfphase-difference associated with the voltage output signal (Vf) is equal and with opposite sign with regard to the Sagnac phase-shift induced by the rotation rate, i.e., ϕf=ϕS. To achieve this, the feedback phase modulation circuit holds a sample of the output signal . Note that this voltage signal is obtained at the end of low pass filter (Block #6 on Figure 8) and is proportional to rotation-rate Ω. An integration operation is needed for obtaining a linear ramp voltage to apply on phase modulator. Then, it integrates and inverts this signal by means of an operational integrator-inverter circuit, turning this signal into the following form:
Vf=1RC0tVΩdtE11
This way, the time variation of Vf voltage signal is a linear ramp, being its slope proportional to the rotation rate of the system (). Figure 8 represents clearly the optical and electronic subsystems of the gyroscope, including the feedback phase-modulation and bias phase-modulation circuits for getting phase-nulling process, both applied together to PM (phase-modulator). Referring now to Figure 8, the total voltage signal applied to PM will be:
Vm=Vbias+VfE12
Therefore, the output signal of the phase modulator will be the sum of the phase-difference signals associated with the Vbias and Vf voltages. In terms of phase differences, this is expressed as ϕm=ϕbias+ϕf. Then, the error signal at the output of the comparator ( voltage) tends to be nulled in average-time, due to the phase cancelation (since the average-time of the reference bias phase-modulation ϕbiasis 0).
Sinusoidal phase modulation has been used on either open-loop (Refs. [18, 25, 26]) or closed-loop (Refs. [27, 28, 29, 30]) IFOG configurations. In both cases, the PSD block (phase-sensitive-demodulation) must contain two selectively adjusted filtering circuits. First one is a low-pass filter with high enough cut-off frequency to filter the first harmonic component. Second one is a selective band-pass-filter to filter the component of first harmonic of interference signal, see also Ref. [30].
4. Serrodyne optical phase modulation
As announced at the beginning of this work, serrodyne-wave optical phase modulation is frequently used in closed loop IFOG schemes to configure one feedback signal which is able to cancel the ϕSSagnac phase shift induced by rotation. The justification for this is that the serrodyne-wave is the only one that produces a constant phase difference when applied to phase modulator (PM) in a gyro. This last can be checked observing the Figure 9(b). During a time span equal to (Tτ) the phase-difference between CW and CCW waves remains constant with a value equal to ϕm0Tτ, being τ the transit time of light around the fiber coil. In addition, a constant value of 2 π is usually taken as the ϕm0amplitude of phase modulation in most part of designs. This way, by adjusting appropriately the period T (or the frequency) of serrodyne-wave, the resulting value of phase-difference can be exactly matched with the ϕSSagnac phase shift to achieve the phase cancelation by means of a specific feed-back circuit located on the way of feed-back signal, see for example Refs. [30, 31, 32, 33].
For a proper operation of feedback circuit, it is essential that the falling edge (reset time) of sawtooth-wave be as fast as possible (ideally instantaneous), Ref. [31]. Since that serrodyne- (or sawtooth-) referred waveform is a periodic waveform that accomplishes Dirichlet conditions in the (0,T) interval, it is susceptible to be developed in a Fourier series (Ref. [34]) such as the one following next:
E13
This result shows that the series only contains sine terms because it refers to an odd function. On the other hand that result is very useful for filtering design purposes as it can be seen on simulated plots represented in Figure 10. Here, a successive sums containing the harmonics: first one (red-curve), the first and second ones (green-curve), the first, second and third ones (blue-curve), the first, second, third and fourth ones (cyan-curve) and finally the first, second, third, fourth and fifth ones (black-curve) are represented. The more terms are taken from the sum series, the better the approximation will be to the perfect sawtooth waveform.
In order to realize the serrodyne-wave phase modulation an Voltage-Controlled-Oscillator (VCO) circuit must be designed. The condition that this circuit must comply is:
ϕS=V2πTτE14
so that the frequency f=1Tof serrodyne-wave should be adjusted depending on the value of the ϕSSagnac phase shift, i.e., the more be ϕS, the more will be the frequency of serrodyne and, then, the lower the value of its period T. Several circuits have been designed to meet this condition. One of these circuits has been represented in Figure 11 and is described in Ref. [35]. Other VCO circuit for serrodyne-wave generation has already been explained above for FEED-BACK phase modulation, see Figure 8 and related Ref. [30].
Although the serrodyne wave is the one that produces the best results for the feed-back phase modulation purpose, other similar waves have been also proposed. For example, symmetric triangular-wave represented in Figure 12 can also perform the same function. Since it is an odd function, its development in Fourier series only contains the odd harmonics, then, it can be expressed in the following way:
E15
Figure 13 represents the three first harmonic sums (red, green, blue) of symmetric triangular-wave (black). So that taking ϕm0=π/2in Eq. (15), the first three harmonic terms can be written as follows:
E16
E17
E18
This way, the first three harmonic Fourier sums can be expressed, respectively, as:
E19
E20
E21
Then, the conclusion is that when the approximation of first three harmonic Fourier sum of symmetrical triangular-wave is taken in gyro phase modulation (blue and green curves in Figure 14), its effective phase-difference (red curve in Figure 14) can be computed. In this case, one switching circuit is needed.
5. Conclusions
Square, sinusoidal, serrodyne and symmetric triangular waveforms can be used in phase modulation processes for optical Sagnac interferometer gyros. For open-loop gyro schemes only one square-wave or sinusoidPM) to retrieve the Sagnac phase shift induced by rotation. However, for closed-loop scheme gyros two waveforms are needed, the first one (square-wave or sinusoidal-wave) for the bias phase modulation and the second one (serrodyne-wave or triangular-wave) for the feed-back phase modulation aiming the phase cancellation (phase-nulling). In the closed-loop scheme, the output signal of the phase-sensitive-demodulator (PSD) circuit passes through a servo-amplifier which drives a phase-shifter transducer placed in the interferometer path. Then, the phase transducer introduces a non-reciprocal phase shift that is equal, by in the opposite sign, to Sagnac phase shift induced by rotation. Thus, the output of the system is the output of the phase transducer. Closed-loop gyro configuration is advantageous with regard the open-loop one because a better accuracy (sensitivity) and scale-factor stability of the gyro are achieved.
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Ramón José Pérez Menéndez (December 9th 2019). Optical Phase-Modulation Techniques [Online First], IntechOpen, DOI: 10.5772/intechopen.90343. Available from: | 2020-01-23 09:49:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.691349983215332, "perplexity": 1260.495823020551}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00340.warc.gz"} |
http://mathoverflow.net/questions/78870/how-to-calculate-such-sum-of-product-of-binomial-coefficient | How to calculate such sum of product of binomial coefficient?
..I wonder if the following formula can be calculated?
$\sum_{k=0}^m {m \choose k} {2k \choose n}$
-
Look at the FAQ - even if such a thing can be calculated why do we care and what is your motivation? In particular, have you looked at various methods that are already available (combinatorial methods, generating functions etc.) and see if they work? – Somnath Basu Oct 23 '11 at 5:12
I am pretty sure this is a hw question as I could do it in 5 minutes. – John Jiang Oct 23 '11 at 6:02
The generating function in $n$ is $((1+t)^2+1)^m$. The case m=n is http://oeis.org/A006139. | 2014-03-10 15:58:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8443050384521484, "perplexity": 459.06949931374743}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010869716/warc/CC-MAIN-20140305091429-00023-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://socratic.org/questions/thermochemistry-calculations | Thermochemistry Calculations?
If the specific heat of water is 4.184 J/g.C degrees, how much energy is needed to raise the temperature of a 50 gram sample of water 12 degrees C?
May 8, 2018
$3000 J$ or $3 k J$
Explanation:
Use $q = m c \Delta T$ for equations like these.
$q$ = Energy/heat
$m$ = Mass
$c$ = Specific Heat
$\Delta T$ = Change in temperature.
Insert your values, and you are given:
q = "50g" * "4.184J/g°C" * 12°C
Our grams and °C cancel out:
$q = 50 \cdot \text{4.184J} \cdot 12$
$q = 2510.4 J$
After applying sig figs, we get 3000J, or 3kJ of energy. | 2021-10-28 21:08:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4690469205379486, "perplexity": 5620.2529128629885}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588526.57/warc/CC-MAIN-20211028193601-20211028223601-00153.warc.gz"} |
https://proofwiki.org/wiki/Definition:Oscillation/Real_Space | # Definition:Oscillation/Real Space
## Definition
Let $X$ and $Y$ be real sets.
Let $f: X \to Y$ be a real function.
### Oscillation on a Set
Let $A \subseteq X$ be any non-empty subset $A$ of $X$.
The oscillation of $f$ on (or over) $A$ is defined as:
$\displaystyle \map {\omega_f} A := \sup_{x, y \mathop \in A} \size {\map f x - \map f y}$
where the supremum is taken in the extended real numbers $\overline \R$.
### Oscillation at a Point
Let $x \in X$.
#### Definition 1
Let $\mathcal N_x$ be the set of open subset neighborhoods of $x$.
The oscillation of $f$ at $x$ is defined as:
$\displaystyle \omega_f \left({x}\right) := \inf_{U \mathop \in \mathcal N_x} \omega_f \left({U \cap X}\right)$
where $\omega_f \left({U \cap X}\right)$ denotes the oscillation of $f$ on $U \cap X$.
#### Definition 2
The oscillation of $f$ at $x$ is defined as:
$\displaystyle \omega_f \left({x}\right) := \inf \left\{{\omega_f \left({\left({x - \epsilon \,.\,.\, x + \epsilon}\right) \cap X}\right): \epsilon \in \R_{>0}}\right\}$
where $\omega_f \left({\left({x - \epsilon \,.\,.\, x + \epsilon}\right) \cap X}\right)$ denotes the oscillation of $f$ on $\left({x - \epsilon \,.\,.\, x + \epsilon}\right) \cap X$.
#### Definition 3
The oscillation of $f$ at $x$ is defined as:
$\displaystyle \omega_f \left({x}\right) := \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right) \cap X}\right)$
where $\omega_f \left({\left({x - h \,.\,.\, x + h}\right) \cap X}\right)$ denotes the oscillation of $f$ on $\left({x - h \,.\,.\, x + h}\right) \cap X$. | 2020-05-26 20:15:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999691367149353, "perplexity": 255.72704964590693}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00542.warc.gz"} |
https://www.physicsforums.com/threads/how-an-em-wave-propogates.149695/ | # How an em wave propogates?
#### rakeshbs
How an em wave propogates??
i understand that an em wave can be produced due to an oscillating electric field or oscillating magnetic field... but how does this wave move forward at the speed of light??
Related Classical Physics News on Phys.org
#### jtbell
Mentor
Loosely speaking, when the electric and magnetic fields at some point change, it causes the fields at nearby points to change also, but with a time delay that depends on the distance. This is like when a point on a stretched string moves, it causes nearby points to move also, but with a time delay.
Mathematically speaking, the electric and magnetic fields each obey the classical differential wave equation,
<oops... see robphy's posting below for the correct equation >
where $c = 1 / \sqrt{\epsilon_0 \mu_0}$
This can be proved from Maxwell's equations for the electric and magnetic fields, as was first done by Maxwell himself.
Last edited:
#### Gib Z
Homework Helper
i Think you wanted a visual interpretation rather than mathematical? I like to this of a photon as a sort of charged particle, which is attracted to its own electromagnetic fields that it generates. The fields induce the creation of another one is front, the photon is attracted and propagates. This helps you remember that light is a particle and a wave, but shouldn't be taken too seriously.
#### Integral
Staff Emeritus
Gold Member
I do not think that you analogy is a good one. It simply does not capture the mechanism well at all. Please reread JtBells explanation. It would be difficult to come up with a better one.
When you have acquired a better understanding of Mathematics you will be able to appreciate the formal mathematical statement. Meanwhile be careful about building incorrect visualizations as you will find that they can become a barrier to gaining a correct understanding.
#### robphy
Homework Helper
Gold Member
$$\left( \frac {\partial^2 \vec E}{\partial x^2} + \frac {\partial^2 \vec E}{\partial y^2} + \frac {\partial^2 \vec E}{\partial z^2}\right) = {\color{red} \frac {1}{c^2}} \frac {\partial^{\color{red}2} \vec E}{\partial t^{\color{red}2}}$$
#### tim_lou
the thing is, it doesn't really have to oscillate; electromagnetic wave just have to satisfy the equation robphy posted and the physical situation one is in.
notice that any twice differentiable equation in the form of:
$$E_i=f(x_i-ct) + g(x_i+ct)$$
satisfy the wave equation.
the sinusoidal wave is just one simple solution of the wave equation.
consider:
$$E_y=\cos\left[ \frac{2\pi}{\lambda}(x+ct) \right]$$
which satisfy the wave equation. you can visualize the electric field pointing in the y-direction changes as x or time varies.
Last edited: | 2019-11-13 10:20:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.726924479007721, "perplexity": 591.7884214378937}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496667177.24/warc/CC-MAIN-20191113090217-20191113114217-00339.warc.gz"} |
https://next.docs.getdbt.com/docs/faqs/failed-prod-run | If you're using dbt Cloud, we recommend setting up email and Slack notifications (Account Settings > Notifications) for any failed runs. Then, debug these runs the same way you would debug any runs in development. | 2022-08-08 23:24:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17348399758338928, "perplexity": 11879.964713144662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00051.warc.gz"} |
http://www.apmonitor.com/wiki/index.php/Main/Arrays | Main
## APMonitor Summation with Vectors
Variable or object arrays are defined by square brackets with a range of integers and separated by a colon as variable[index 1:index 2]. Arrays may be used to define multiple equations or connections on one line. Any line with an array is processed sequentially from the lowest to the highest index. The model parser creates and processes the arrays as if they were written sequential in non-array form as shown in the example.
### Higher dimensional arrays
Arrays with more than one dimension are allowed. The array indices are separated by brackets as var[i][j][k]. For operations on matrices, the precedence of operations is determined by the number of colons separating the vector indices. Matrix elements with fewer colon separators are executed first. For example, a set of 24 intermediate variables posed as:
• x[1:2][1::3][1:::4] = 1
results in the following set of equations:
• x[1][1][1] = 1
• x[2][1][1] = 1
• x[1][2][1] = 1
• x[2][2][1] = 1
• x[1][3][1] = 1
• x[2][3][1] = 1
• x[1][1][2] = 1
• etc...
### Array Index Consistency
When processing the arrays, the parser checks for array size consistency. An error with an appropriate message is returned if the vector indeces are of different dimension.
### Examples
! Method #1: Summation with arrays Model array Constants n = 5 End Constants Parameters p[1:n] = 1 End Parameters Variables sum End Variables Intermediates z[1] = p[1] z[2:n] = z[1:n-1] + p[2:n] End Intermediates Equations sum = z[n] End Equations End Model ! Method #2: Summation with the sum object Objects z = sum(5) End Objects Connections p[1:n] = z.x[1:n] y = z.y End Connections Model Constants n = 5 End Constants Parameters p[1:n] = 1 End Parameters Variables y sum End Variables Equations sum = y End Equations End Model ! Method #3: Summation without arrays Model array Parameters p[1] = 1 p[2] = 1 p[3] = 1 p[4] = 1 p[5] = 1 End Parameters Variables sum End Variables Intermediates z[1] = p[1] z[2] = z[1] + p[2] z[3] = z[2] + p[3] z[4] = z[3] + p[4] z[5] = z[4] + p[5] End Intermediates Equations sum = z[5] End Equations End Model
An additional example is a matrix summation where there are two indices of the parameter matrix p.
! Matrix Summation Model Parameters p[1:10][1::5] = 1 End Parameters Variables x End Variables Intermediates ! sum the rows n[0][1:5] = 0 n[1:10][1::5] = n[0:9][1::5] + p[1:10][1::5] ! sum the columns that are summation of rows m[0] = 0 m[1:5] = m[0:4] + n[10][1:5] End Intermediates Equations ! solution = 50 x = m[5] End Equations End Model
#### Arrays in Python GEKKO
Multi-dimensional arrays are defined in Python GEKKO with the m.Array() function or with list comprehensions.
from gekko import GEKKO
m = GEKKO()
ni = 3 # number of rows
nj = 2 # number of columns
# best method: use m.Array function
x = m.Array(m.Var,(ni,nj))
m.Equations([x[i][j]==i*j+1 for i in range(ni) for j in range(nj)])
# another way: list comprehensions
y = [[m.Var() for j in range(nj)] for i in range(ni)]
for i in range(ni):
for j in range(nj):
m.Equation(x[i][j]**2==y[i][j])
# summation
z = m.Var()
m.Equation(z==sum([sum([x[i][j] for i in range(ni)]) for j in range(nj)]))
m.solve()
print('x:')
print(x)
print('y=x**2:')
print(y)
print('z')
print(z.value) | 2019-05-21 04:01:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6604320406913757, "perplexity": 7184.5911541198175}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256215.47/warc/CC-MAIN-20190521022141-20190521044141-00071.warc.gz"} |
https://chemistry.stackexchange.com/questions/115514/is-there-a-quantitative-test-analysis-technique-that-can-determine-if-some-chlor | # Is there a quantitative test/analysis technique that can determine if some chloroform has degraded into phosgene?
Before I go to use any chemical that I've had in storage for a while, I try to check if it's degraded into something else (e.g.: checking the $$\ce{H2O}$$ content of $$\ce{H2O2}$$, checking for $$\ce{H2O2}$$ in $$\ce{Et2O}$$, etc.). This can usually be done using the appropriate test strip.
I have a 3/4 full bottle of chloroform that's been sitting on a shelf for about a year (probably less, actually), in an amber bottle to restrict the amount of light that can get to the chloroform. I'm aware that chloroform can break down into phosgene, and I know there are some ways to test for this, but I was hoping for a quick quantitative test that can be done without any hassle, but I can't seem to find anything out there. The Wikipedia page for chloroform states:
Suspected samples can be tested for phosgene using filter paper (treated with 5% diphenylamine, 5% dimethylaminobenzaldehyde in ethanol, and then dried), which turns yellow in phosgene vapor.
I don't have either of the diphenylamine or the dimethylaminobenzaldehyde, and would prefer not to have to purchase them just for this simple test (I don't have any use for either chemical currently). Does anyone know if there is a simple quantitative test that accomplishes the same thing? Preferably a test strip.
• Phosgene reacts with ethanol. Thus, if your chloroform is ethanol protected then your risk is minimal. – Mathew Mahindaratne May 19 at 8:03
Not a quantitative test but a good proof of phosgene. You could use a pH paper strip. Chloroform should show a neutral pH, but if there is phosgene/HCl the paper will show acidity.
• I see - That seems simple enough! Thank you – Justin Jun 7 at 9:18
I admire your safety precautions. Like we always check for peroxide content in older ether bottles, we should also test older chloroform bottles, specifically those of alkene-preserved, for the presence of phosgene (Ref.1). The C&E News letter recommended use of filter paper strips, wetted with 5% diphenylamine, 5% dimethylaminobenzaldehyde and then dried, which turn yellow in phosgene vapor (a qualitative test). Note that as test strips age they become light yellow/brown, they will still show a significant change to bright yellow upon detection of phosgene (Wooley Lab).
There are also Draeger tubes available for phosgene testing, but they are kind of expensive (e.g., check on Shopcross.com). The Draeger tube test uses the general reaction of an aromatic amine with phosgene to obtain the corresponding aromatic isocyanate (I believe), which has long been known. However, their aromatic amine turns red in the presence of phosgene. It has been suggested that such reactions proceed thru the acid salt of the amine such as an amine hydrochloride or an amine-$$\ce{CO2}$$ adduct either by forming the salt in situ in the process or by starting with the salt, to form the corresponding aromatic carbamyl chloride, which in turn is decomposed to the isocyanate with liberation of hydrogen chloride.
In addition, qualitative (and can develop for quantitative) analysis by GC/MS would work as well. However, Ref.1 has used UV-vis colorimetric quantitation described in Ref.2. The abstract states that:
A reagent consisting of 0.25% 4-p-nitrobenzylpyridine and 0.5% N-benzylaniline in diethylphthalate was tested and proved to be highly sensitive and relatively specific for the spectrophotometric determination of phosgene in air. Concentrations under $$\pu{0.1 ppm}$$ of phosgene in air may be determined with reliability. The color developed with the reagent was stable for several hours and only slightly affected by normal ambient humidity. None of the following substances interfered with the reaction: hydrogen chloride, chlorine, chlorine dioxide, carbon tetrachloride, chloroform, trichloroethylene, tetrachlorethylene, and dichlorodifluoromethane in concentrations likely to be encountered under industrial conditions.
It seems there is some development in teststrips containing a fluorophore for phosgene detection (See diagram below; Ref.3), which changes the color from blue to green with a hand-held UV light in the presence of phosgene.
References:
1. Eric Turk, “Phosgene from chloroform,” Chemical & Engineering News 1998, 76(9), p 6 (DOI: 10.1021/cen-v076n009.p006).
2. M. H. Noweir, E. A. Pfitzer, “An Improved Method for Determination of Phosgene in Air,” American Industrial Hygiene Association Journal 1971, 32(3), 163–169 (https://doi.org/10.1080/0002889718506431).
3. H. Xie, Y. Wu, F. Zeng, J. Chena, S. Wu, “An AIE-based fluorescent test strip for the portable detection of gaseous phosgene,” Chem. Comm. 2017, 53(70), 9813–9816 (DOI: 10.1039/c7cc05313d). | 2019-09-21 23:36:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5920066833496094, "perplexity": 4403.697228671251}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574710.66/warc/CC-MAIN-20190921231814-20190922013814-00068.warc.gz"} |
http://mathbook.ups.edu/doc/author-guide/html/overview-paragraphs.html | Once you have divisions, what do you put into them? Most likely, paragraphs. We use long, exact names for tags that are used infrequently, like <subsubsection>. But for frequently used elements, we use abbreviated tags, often identical to names used in HTML. So a paragraph is delimited by simply the <p> tag.
Lots of things can happen in paragraphs, some things can only happen in a paragraph, and some things are banned in paragraphs. Inside a paragraph, you can emphasize some text (<em>), you can quote some text (<q>), you can mark a phrase as being from another language (<foreign>), and much more. You can use special characters like an ampersand (&, <ampersand/>) or an octothorpe, aka “hash tag” (#, <hash/>). You must put a list inside a paragraph, and all mathematics (Section 3.5) will occur inside a paragraph. You cannot put a <table> or a <figure> in a paragraph, and many other structured components are prohibited in paragraphs.
Paragraphs are also used as part of the structure of other parts of your document. For example, a <remark> could be composed of several <p>. As you get started with PreTeXt, remember that much of your actual writing will occur inside of a <p> and you will have a collection of tags you can use there to express your meaning to your readers. | 2019-01-22 18:35:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5650750994682312, "perplexity": 1888.4611965353074}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583867214.54/warc/CC-MAIN-20190122182019-20190122204019-00481.warc.gz"} |
http://nrich.maths.org/7158/solution | ### Building Tetrahedra
Can you make a tetrahedron whose faces all have the same perimeter?
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
### Areas and Ratios
What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out.
# Square Product
##### Stage: 4 Short Challenge Level:
We make use of two key facts. First, we have the factorisation $n^2 -1 = (n-1)(n+1)$. Second, when a square number is factorised, each prime factor appears an even number of times.
Now $2^2 -1 = 1\times 3$.
We next get a prime factor $3$ with $4^2-1= 3\times 5$.
We next get a prime factor $5$ with $6^2-1= 5\times 7$.
We next get a prime factor $7$ with $8^2-1= 7\times 9$.
As $9=3^2$, it does not require any further factors. Hence we need $n\geq 8$. Checking the product $n=8$, we get\eqalign{ (2^2 -1)(3^2 -1)(4^2 -1)(5^2 -1)(6^2 -1)(7^2 -1)(8^2 -1) &= 1\times 3\times 2\times4\times3\times5\times4\times6\times5\times7\times6\times8\times7\times9\cr &= 2\times8 \times 3 \times 3\times4\times4\times5\times5\times6\times6\times7\times7\times3\times3\cr &=4\times4\times3\times3\times4\times4\times5\times5\times6\times6\times7\times7\times3\times3\cr &=(4\times3\times4\times5\times6\times7\times3)^2}
So in fact $n=8$ is sufficient, and is thus the minimum.
This problem is taken from the UKMT Mathematical Challenges. | 2017-06-24 18:59:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9944306015968323, "perplexity": 1127.2507748720602}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320323.17/warc/CC-MAIN-20170624184733-20170624204733-00360.warc.gz"} |
http://dshacker.blogspot.com/ | ## Thursday, August 25, 2016
### Calling Python from C++
The Python documentation has showed us how to call python scripts from C++. However, it has not shown us the performance penalty of calling Python from C++. Here I use the simple C++ code to benchmark the performance cost of calling Python from C++:
#include "python2.7/Python.h"
#include <iostream>
#include <chrono>
#define DURATION { \
auto dur = std::chrono::system_clock::now() - start; \
std::cout << "duration: " << std::chrono::duration_cast<std::chrono::milliseconds>(dur).count() << " milliseconds" << std::endl; \
start = std::chrono::system_clock::now(); \
}
int main(int argc, char *argv[])
{
std::cout << "start" << std::endl;
auto start = std::chrono::system_clock::now();
Py_SetProgramName(argv[0]); /* optional but recommended */
DURATION;
Py_Initialize();
DURATION;
PyRun_SimpleString("import numpy as np\n"
"\n");
DURATION;
int s = 0; for (std::size_t i = 0; i < 100; ++i) s += i;
DURATION;
Py_Finalize();
DURATION;
return 0;
}
The following command can be used to compile the code:
g++ test.c -std=c++11 -I/usr/include/python2.7 -lpython2.7 -o test
Running the code, we can see the time results as follows. Py_Initialized() cost 15 milliseconds, and just importing numpy cost 46 milliseconds. So It is quite expensive to call Python code from C++.
$./test start duration: 0 milliseconds duration: 15 milliseconds duration: 46 milliseconds duration: 0 milliseconds duration: 4 milliseconds ## Sunday, August 23, 2015 ### Installing zipline on Windows 7 First, install [Pythonxy](https://code.google.com/p/pythonxy/wiki/Downloads) on Windows, choose the latest version. Create a conda environment using the following command: conda create --name zipline python Then a folder named zipline will be created under C:\Python27\envs Type "activate [environment]" to make the environment active. After activating the environment, install any python packages using conda install [package]. For example, to install the python IDE Spyder, use conda install spyder Then install zipline using the following command: conda install -c Quantopian zipline For details regarding setting up environment in conda, please refer to: http://uoa-eresearch.github.io/eresearch-cookbook/recipe/2014/11/20/conda/ http://conda.pydata.org/docs/test-drive.html#managing-conda Zipline code structure: in TradingAlgorithm.run(), self.gen = self._create_generator(self.sim_params) is the main loop. _create_generator() calls AlgorithmSimulator.transform(self, stream_in), and for date, snapshot in stream_in: is the main loop that calls handle_data() for each bar ## Tuesday, August 4, 2015 ### Machine Learning and Statistics: a unified perspective | 4 Decision Trees Previously we have discussed using Generalized Linear Model(GLM) to estimate continuous variables (linear regressions) and categorical variables (logistical regression). GLM minimizes Loss function with respect to parameters $$\theta$$, and such $$\theta$$ is global in the sense that $$\theta$$ doesn’t change with the value of variable $$\mathbf{x}$$. GLM models are less prone to overfitting when effective degrees of freedom is not large, but it may not deal with local nonlinearities properly. Consider a k-nearest-neighbor (KNN) algorithm: Data point $$(y, \mathbf{x})$$’s target value y will be assigned to the mean of the nearest k data point to $$\mathbf{x}$$. This algorithm can adapt to high local nonlinearities, but it can also have high variance and significantly overfit the data. For models using adaptive local functions, we can write the model in the following form: $f(\mathbf{x}) = E[y|\mathbf{x}] = \sum_{m=1}^M w_mI(\mathbf{x} \in R_m) = \sum_m w_m\phi(\mathbf{x; v}_m)$ where $$R_m$$ is the m-th region, $$w_m$$ is the mean response weight, and more generally, $$\phi(\mathbf{x; v_m})$$ is a local basis function which is not necessarily $$I(\mathbf{x} \in R_m)$$. For decision trees, $$\mathbf{v}_m$$ encodes the choice of variable to split on, and the threshold value. This makes it clear that a decision tree based model is just a an adaptive basis-function model, where the basis functions define the regions, and the weights specify the response value in each region. ### Build a Tree Finding the optimal partitioning of the data is NP-complete (Hyafil and Rivest 1976), so it is common to use the greedy procedure locally optimal MLE. The split procedure is the following: $(j^*, t^*) = arg min_{j \in 1...K} min_{t \in T_j} [cost({\mathbf{x}, y : x_{i,j} \leq t}) + cost({\mathbf{x}, y : x_{i,j} \geq t})]$ For cost functions, we have the following considerations: • Regression: $$cost(\mathcal{D}) = \sum_{i \in \mathcal{D}}(y^{(i)} - \bar{y})^2$$ • Classification: if dependent variable y has C classes, we define the class-conditional probabilities as $$\pi_c = \frac{1}{|\mathcal{D}|}\sum_{i \in \mathcal{D}}I(y^{(i)} = c)$$, where $$|\mathcal{D}|$$ is the number of data points in data set $$\mathcal{D}$$. There are a few similar measures of cost: • Misclassification rate: define most probable class label as $$\hat{y}_c = argmax_c \hat{\pi}_c$$, then the corresponding error rate is $$1 - \pi_{\hat{y}}$$ • Entropy: $$H(\pi) = -\sum_{c=1}^C \pi_clog(\pi_c)$$. Note that minimizing entropy is the same as maximizing information gain. • Gini index, or the expected error rate: $$\sum_{c=1}^C \pi_c(1-\pi_c)$$ The 3 measures for classification error are similar. Let’s use two class case as an example. Assume the probability of class 1 is p, then: • misclassifcation rate = $$1-max(p, 1-p)$$ • Entropy = $$-plog(p) - (1-p)log(1-p)$$ • Gini index = $$2p(1-p)$$ All 3 measures is 0 when p = 0 or p = 1, and maximizes when p = 1/2. Let’s take an example of how to build a tree. Here we use the R package rpart and its example data Kyphosis. First we plot the data of kyphosis below. Data set Kyphosis has a dependent variable kyphosis of two categories: present and absent, and 3 varaibles Age, Number and Start. For more details, please refer to the (rpart documentation)[https://cran.r-project.org/web/packages/rpart/vignettes/longintro.pdf]. # Classification Tree with rpart library(rpart); library(ggplot2) plt <- ggplot(kyphosis, aes(x=Age, y=Start)) + ggtitle("kyphosis data") + geom_point(aes(fill=factor(Kyphosis)), size=3, pch=21) print(plt) We build a tree use all 3 parameters. To avoid overfitting, we need to set some limits on the complexity of the tree. In the rpart.control function below, minsplit is the minimum number of observations that must exist in a node in order for a split to be attempted. cp is the complexity parameter which means that any split that does not decrease the overall lack of fit by a factor of cp is not attempted. We can see a decision tree is built using the code below. fit <- rpart(Kyphosis ~ Age + Number + Start, method="class", data=kyphosis, control = rpart.control(minsplit = 20, cp = 0.01)) plot(fit, uniform=TRUE, main="Classification Tree for Kyphosis") text(fit, use.n=TRUE, all=TRUE, cex=.8) To illustrate how cp limits the depth of the tree, we print how the cp parameter changes as the split number goes up. As shown below, the last split, which corresponds to a tree of 4 splits, has a cp of 0.01, which is the limit we set in rpart.control. Thus the split stops at 4. Optimum value of cp can be determined by cross-validation. printcp(fit) # display the results ## ## Classification tree: ## rpart(formula = Kyphosis ~ Age + Number + Start, data = kyphosis, ## method = "class", control = rpart.control(minsplit = 20, ## cp = 0.01)) ## ## Variables actually used in tree construction: ## [1] Age Start ## ## Root node error: 17/81 = 0.20988 ## ## n= 81 ## ## CP nsplit rel error xerror xstd ## 1 0.176471 0 1.00000 1.00000 0.21559 ## 2 0.019608 1 0.82353 0.94118 0.21078 ## 3 0.010000 4 0.76471 1.00000 0.21559 Another criteria that can be used to prune is tree is to define a cost complexity criterion like the following. Let T be the number of nodes in a tree: $N_t = number\{x_i \in R_t\}$ $\hat{c}_t = 1/N_t\sum_{x_i \in R_t} y_i$ $Q_t = \sum_{x_i \in R_t} (y_i - \hat{c}_t)^2$ Then the cost complexity is: $C_{\alpha} = \sum_{t=1}^T Q_t + \alpha T$ where $$\alpha$$ is a coefficient which can be determined by cross-validation The advantanges of decisions trees include: • they are easy to interpret based on the split results • they can easily handle mixed discrete and continuous inputs • they are insensitive to monotone transformations of the inputs because the split points are based on ranking the data points • they perform automatic variable selection • they are relatively robust to outliers • they scale well to large data sets However, decision tree have an important disadvantage: overfitting due to the greedy nature of the tree construction algorithm, even with careful pruning. A related problem is that trees areunstable: small changes to the input data can have large effects on the structure of the tree. Solution to this problem will be discussed in the following section. ### Bagging and Random Forest An important solution to the problem of simple decision tree is to using boosting approach to create a large amount of sample data, create a decision tree for each of them, and average among all the trees. This technique is called bagging, which stands for “bootstrap aggregating”. Unfortunately, simply re-running the same learning algorithm on different subsets of the data can result in highly correlated predictors, which limits the amount of variance reduction that is possible. To decorrelate the base learners, a randomly chosen subset of input data sets as well as a randomly chosen subset of the features are combined to create decision trees. This technique is called Random Forest. Let’s take a look at an example of Random forest as shown below using the same data set of Kyphosis: library(randomForest) ## randomForest 4.6-10 ## Type rfNews() to see new features/changes/bug fixes. fit <- randomForest(Kyphosis ~ Age + Number + Start, data=kyphosis) print(fit) # view results ## ## Call: ## randomForest(formula = Kyphosis ~ Age + Number + Start, data = kyphosis) ## Type of random forest: classification ## Number of trees: 500 ## No. of variables tried at each split: 1 ## ## OOB estimate of error rate: 20.99% ## Confusion matrix: ## absent present class.error ## absent 60 4 0.0625000 ## present 13 4 0.7647059 Unfortunately, methods that use multiple trees lose their nice interpretability properties. Fortunately, various post-processing measures can be applied, and we can still see the relative importance of features from Random Forest. importance(fit) # importance of each predictor ## MeanDecreaseGini ## Age 8.837115 ## Number 5.305533 ## Start 9.797556 ### Boosting Boosting is a greedy algorithm for fitting adaptive basis-function by applying the weak learner sequentially to weighted versions of the data, where more weight is given to examples that were misclassified by earlier rounds. In math terms, boosting is a forward stepwise procedure: $\hat{\theta}_m = argmin_{\theta_m} \sum_{i=1}^N L(y^{(i)}, f_{m-1}(x^{(i)}) + T(x^{(i)}, \theta_m))$ If the Loss function is an exponential function $$L(y, f(x)) = exp(-yf(x))$$, the solution is particularly simple, and this stepwise method using exponential loss function is referred to as Adaboost. It can be relatively easily prooven that the solution for Adaboost is: $$T(x^{(i)}, \theta_m) = \alpha_m k_m(x^{(i)})$$ where $$k_m(x) = argmin_{k_m}\sum_i w_m^{(i)}exp(-y^{(i)}k_m(x^{(i)}))$$ is the decision tree that minimize the loss of data set $$(\mathbf{y, X})$$ with weight $$w_m^{(i)} = exp(-y^{(i)}f_{m-1}(x^{(i)}))$$. A realization of Adaboost algorithm is shown below. The in-sample error for data set kyphosis reduce to 0 when iteration is 10. adaboost <- function(y, X, ntree = 100) { library(rpart) y = sapply(y, function(x){if (x == levels(x)[1]) return(-1) else return(1)}) N = length(y) f = rep(0, N) in_sample_error = c() for (ii in 1:ntree) { w = exp(-y * f); w = w / sum(w) fit = rpart(y ~ ., X, w, method = "class") g = -1 + 2*(predict(fit, X)[,2] > 0.5) e = sum(w * ((y * g) < 0)) alpha = 0.5*log((1-e)/e) f = f + alpha*g in_sample_error = c(in_sample_error, sum((y*f) < 0)/N) } return(in_sample_error) } y = kyphosis$Kyphosis; X = kyphosis[, 2:4]
plot(1:20, in_sample_error, xlab = "Iteration", ylab = "In sample error")
Exponential loss function is easy to solve because it is analogous to the square loss in least square regression. However, it is prone to overfitting because it puts too much weight out the outlier. More robust loss functions such as Huber loss or Multinomial Deviance can be used, but numerical solutions are needed for more general loss functions. Gradient boost algrithm is developed to solve the problem: induce a tree $$T(\mathbf{x}, \theta)$$ at the m-th iteration whose predictions $$\mathbf{t}_m$$ are as close as possible to the negative gradient of each data point of the loss function $$g_{im} = [\frac{\partial L(y^{(i)}, f(x^{(i)}))}{\partial f(x^{(i)})}]$$: $\hat{\theta_m} = argmin_{\theta}\sum{i=1}^N(-g_{im}-T(x^{(i)}, \theta))^2$
Since the loss function in Gradient Boost can be of many different kinds, Gradient Boost can be used for either classification or regression. R package gbm implements Gradient Boost.
### Comparison with Linear Regression
Tree based methods have the following Pros and Cons:
Pros:
• handling of mixed data types
• hanlding of nonlinearity in response
• Insensitive to monotone transformations of inputs
• automatic feature selection / ranking
Cons:
• Can not generate linear combinations of features, and can not interpret the sign of correlation between variable and response
• Prone to overfitting and needs proper choice of tree parameters
Linear Regression has the following Pros and Cons:
Pros:
• Less prone to overfitting due to its simple linearity nature
• Easy interpretation on the sign of correlation between variable and response.
Cons:
• Relative importance of variables in multiple linear regression can only be achieved with normalized input data and same expected sign of correlation with response.
• Can not capture nonlinearity | 2017-01-20 13:51:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9606215953826904, "perplexity": 5551.182063552001}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280835.22/warc/CC-MAIN-20170116095120-00052-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://mathsmadeeasy.co.uk/gcse-maths-revision/proof-gcse-maths-revision-worksheets/ | ## What you need to know
Proof is at the heart of all the things you see in mathematics. That is, all the things that you use and take for granted, such as Pythagoras’ Theorem or the formula for the area of a circle, have, at some point over the last few thousand years, been proven to be true. You won’t have to delve into the proofs of everything you use (which is fortunate, because it would probably take a little while), instead this topic is intended as an introduction into how proofs work.
The idea of a proof is to make a universal statement – for example, you don’t just want to say that the angles in some triangles add up to 180, you want to say that the angles in all triangles add up to 180. This is a proof you actually do have to know, and you can see it here (https://mathsmadeeasy.co.uk/gcse-maths-revision/interior-and-exterior-angles-gcse-revision-and-worksheets/). The general structure of a proof is to begin with one statement, take a series of logical and mathematical steps, and end up at a desired conclusion. Of course, not everything we want can be proved true. In fact, all it takes to prove a statement false is to find a counterexample – a particular example for which the universal statement you’re trying to make doesn’t hold true.
Example: Hernan claims “if you square a number and add 1, the result is a prime number”. Find a counterexample to prove her statement wrong.
$1^2+1=1+1=2,\text{ which is prime.}$
$2^2+1=4+1=5,\text{ which is prime.}$
Now, at this point it might seem like her statement is true, but if we try the next square number:
$3^2+1=9+1=10,\text{ which is not prime.}$
This is a counterexample for her statement, so we have proved it to be false.
The proofs you see in this topic will involve a lot of algebraic expansion and collecting like terms together, so make sure you’re comfortable with that by going here (https://mathsmadeeasy.co.uk/gcse-maths-revision/multiplying-single-double-brackets/) before continuing.
Example: Prove that $(n+2)^2-(n-2)^2$ is divisible by 8 for any positive whole number $n$.
To do this, we need to show that $(n+2)^2-(n-2)^2$ can be written in some way that is clearly divisible by 8. To find a way to write an expression like this differently, we can try expanding it. So, the first bracket expands to
$(n+2)^2=n^2+2n+2n+4=n^2+4n+4$.
Then, the second bracket expands to
$(n-2)^2=n^2-2n-2n+4=n^2-4n+4$.
The expression in the question has the second bracket being subtracted from the first one. So, we will do this subtraction with the expansion of the two brackets:
$(n+2)^2-(n-2)^2=(n^2+4n+4)-(n^2-4n+4)$
We can see that the $n^2$ terms will cancel, as will the 4s, so all we’re left with is
$(n^2+4n+4)-(n^2-4n+4)=4n-(-4n)=8n$
So, the whole expression simplifies to $8n$. Now, if $n$ is a whole number, then $8n$ must be divisible by 8 (if we divide it by 8, we get the answer $n$). Since $8n$ is equivalent to the expression we started with, it must be the case that $(n+2)^2-(n-2)^2$ is divisible by 8 for any positive whole number $n$ – so the statement is now universal. Thus, we have completed the proof.
So, this proof gave us an expression to simplify and write in a way that satisfied the conclusion we wanted. Sometimes, the proof is phrased in words and we have to figure out how to turn those words into something we can do maths with.
Example: Prove that the square of an odd number is also odd.
Okay, so if we’re going to make a statement about odd numbers and we want to use algebra to prove the statement, we need some way to express an odd number with algebra. To do this, let’s consider even numbers: an even number is a multiple of 2, by definition, so if $n$ is any whole number, then $2n$ must be an even number. Indeed, any even number can be expressed by $2n$ if you pick the right $n$. Furthermore, any odd number is always the next one along from an even number. So, we can express an odd number like
$2n+1$.
So, now that we have our algebraic expression of an odd number, we can really get going. The question asks about the square of an odd number, so let’s square our expression for a general odd number:
$(2n+1)^2=4n^2+2n+2n+1=4n^2+4n+1$.
The question is now: how do we know this is odd? Well, we can’t take a factor of 2 out of the whole thing, but we can take a factor of 2 out of the first two terms (if you’re not sure where this is going that’s okay, just bear with me). Doing so, we get
$4n^2+4n+1=2(2n^2+2n)+1$
Now, $2n^2+2n$ might seem like a reasonably complicated expression but importantly, it must be a whole number. This is because $n$ is a whole number, and if you square a whole number/multiply it by other whole numbers, the result must still be a whole number. Already, I’ve said the phrase “whole number” far too many times here, but the whole point is that
$2(2n^2+2n)+1\,\,\,\text{ is just }\,\,\, 2\times(\text{some whole number})+1$.
And since $2\times(\text{some whole number})$ is how we defined an even number, it follows that $2\times(\text{some whole number})+1$ is how we define an odd number. Therefore, this expression – and thus, the square of any odd number – must be odd. This is our universal statement, and we have completed the proof.
Proof is quite different to a lot of the things you’ve seen in maths before so don’t worry if it’s tricky to get your head around. The more you see and practice following logical arguments like these ones, the less foreign they will seem. Don’t worry if you keep getting stuck at first – even the greatest mathematicians get stuck on proofs on a very regular basis.
## Example Questions
#### 1) Luke claims that if you square an even number and add 3, the answer is prime. Find a counterexample to show that Luke’s statement is false.
We will try the first few even numbers (squaring them and adding 3) until we find an example that isn’t prime. So, we get
$2^2+3=4+3=7,\text{ which is prime}.$
$4^2+3=16+3=19,\text{ which is prime}.$
$6^2+3=36+3=39,\text{ which is not prime}.$
Since 39 is divisible by 3, it must not be prime, so we have proved Luke’s statement to be false.
Note: there are many, many counter examples to Luke’s statement, and any one of them is an acceptable answer to this question.
#### 2) Prove that $(3n+1)^2+(n-1)^2$ is always even for any positive whole number $n$.
To answer this question, we will need to expand and simplify the expression given to us, so we can hopefully write it in a way that shows it is clearly divisible by 2 (since that’s the definition of even). So, expanding the first bracket, we get
$(3n+1)^2=9n^2+3n+3n+1=9n^2+6n+1$.
Then, expanding the second bracket, we get
$(n-1)^2=n^2-n-n+1=n^2-2n+1$.
Now, the expression in the question has the two brackets added together. So, adding the expansions together, we get
$(9n^2+6n+1)+(n^2-2n+1)=10n^2+4n+2$
Is this an even number? Well, if we take a factor of 2 out of the expression:
$2(5n^2+2n+1)$,
we see that since $5n^2+2n+1$ is a whole number because $n$ is a whole number, the expression in question is equal to $2\times(\text{some whole number})$ and so must be even. Thus, we have completed the proof.
#### 3) Prove that the sum of 3 consecutive odd numbers is an odd number. (HINT: if $2n+1$ is how we express an odd number, how can we express the next odd number along?)
Answering the question in the hint is the key to getting going with this question. So, we know that for any whole number $n$, $2n$ must be an even number and $2n+1$ must be an odd number. So, if we have one odd number and want to get to the next one, all we must do is add 2. Therefore, the next odd number after $2n+1$ is
$(2n+1)+2=2n+3$
Furthermore, for the next odd number after this one, we must add 2 again:
$(2n+3)+2=2n+5$.
So, now we have expressions for 3 consecutive odd numbers: $2n+1, 2n+3,$ and $2n+5$, all that remains is to add these up and show that the result is always odd. Adding them together gives us
$(2n+1)+(2n+3)+(2n+5)=6n+9$
Now, to show that this number is odd we have to be a little clever. If we write this expression instead like $6n+8+1$, then we get
$6n+8+1=2(3n+4)+1$
Since $n$ is a whole number, clearly $3n+4$ must also be a whole number, so we get that our expression is of the form $2\times\text{(some whole number)}+1$, which is precisely how we expressed an odd number in the first place, and so the result must be odd. Thus, we have completed the proof.
This is a tough question, so if you managed to get through any of the steps – good job. If you really had no idea, that’s okay too. Give the proof another read through to make sure you fully understood each step, and then next time a question similar comes up you’ll have a better idea.
## Proof Revision and Worksheets
Proofs
Level 8-9
Proofs 2
Level 8-9
Proofs Test
Level 8-9
Proof and algebraic proof are one of the most difficult topics to teach and yet there are very few great resources out there to help tutors and teachers to deliver this topic. Thankfully at Maths Made Easy we have taken the time to research the best resources out there as well as add in much of our own to help create a super resource for proof. | 2019-01-22 02:02:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 57, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8337868452072144, "perplexity": 148.62562790989628}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583823140.78/warc/CC-MAIN-20190122013923-20190122035923-00042.warc.gz"} |
http://www.math.tau.ac.il/~yekutiel/TAU%20Statistics%20Seminar%20Home%20Page%2015-6.htm | # Statistics Seminars
### Second Semester
###### Matan Gavish, HUJI
Optimal thresholding of singular values and eigenvalues
###### Tirza Routtenberg, BGU
Estimation after parameter selection: Estimation methods, performance analysis, and adaptive sampling
###### Saharon Rosset, TAU
Estimating Random-X Prediction Error of Regression Models
###### 17 May
Lucien Birge, Université Pierre et Marie Curie (Paris VI)
ρ-estimation – a robust alternative to maximum likelihood
###### Jeroen de Mast, University of Amsterdam
Estimation of the random error of binary tests using adaptive polynomials
###### 26 June
Asaf Cohen, University of Michigan
A Multiclass Queueing Model in the Moderate-Deviation Heavy-Traffic Regime
###### 5 July
Ofer Harel, University of Connecticut
### First Semester
###### 17 November
Omer Weissbrod, Tel Aviv University and Technion
Modern Applications of Linear Mixed Models for Large Genetic Studies
###### Shahar Mendelson, Technion
The small-ball method and its application in Learning Theory
###### Amichai Painsky
Generalized Independent Component Analysis Over Finite Alphabets
8 December
Armin Schwartzman, NC State University
###### Confidence Regions for Excursion Sets in Asymptotically Gaussian Random Fields
22 December
îòøëú úåîëú îåîçä ì÷áéòú äðãéøåú ùì ò÷áåú ðòìééí
5 January
Omer Bobrowski, Duke University
Random Topology and its Applications
Seminars are held on Tuesdays, 10.30 am, Schreiber Building, 309 (see the TAU map ). The seminar organizer is Daniel Yekutieli.
To join the seminar mailing list or any other inquiries - please call (03)-6409612 or email 12345yekutiel@post.tau.ac.il54321 (remove numbers unless you are a spammer…)
Seminars from previous years
ABSTRACTS
· Jelle Goeman, Radboud university medical center
Hommel's method for false discovery proportions
We study the combination of closed testing with local tests based on the Simes inequality that underlies Hochberg's and Hommel's methods for familywise error control. We show this combination can also be used to make simultaneous confidence statements for the false discovery proportion of arbitrary sets of hypotheses. These confidence statements are at least as powerful as those arising trivially from Hommel’s method, and can be much more powerful. Calculation time is linear in set size after an initial preparatory step that takes m log(m) time for m hypotheses. By their simultaneity, the coverage of these confidence statements is guaranteed even if sets of interest are chosen post hoc. With this method, the role of the user and the method can be reversed in multiple testing. Where normally the user chooses the error rate and the multiple testing method the rejected set, now the user can choose the rejected set and the multiple testing method calculates the error rate. This new method has some interesting connections to the Benjamini & Hochberg FDR method, which we hope to explore further.
· Abraham Wyner, University of Pennsylvania
Explaining the Success of AdaBoost and Random Forests as Interpolating Classifiers
There is a large literature explaining why AdaBoost is a successful classifier. The literature on AdaBoost focuses on classifier margins and boosting's interpretation as the optimization of an exponential likelihood function. These existing explanations, however, have been pointed out to be incomplete. A random forest is another popular ensemble method for which there is substantially less explanation in the literature. We introduce a novel perspective on AdaBoost and random forests that proposes that the two algorithms work for essentially similar reasons. While both classifiers achieve similar predictive accuracy, random forests cannot be conceived as a direct optimization procedure. Rather, random forests is a self-averaging, interpolating algorithm which creates a "spikey-smooth” classifier. We show that AdaBoost has the same property. We conjecture that both AdaBoost and random forests therefore succeed because of this mechanism. We provide a number of examples and some theoretical justification to support this explanation. In the process, we question the conventional wisdom that suggests that boosting algorithms for classification require regularization or early stopping and should be limited to low complexity classes of learners, such as decision stumps. We conclude that boosting should be used like random forests: with large decision trees and without direct regularization or early stopping.
· Omer Weissbrod, Tel Aviv University and Technion
Modern Applications of Linear Mixed Models for Large Genetic Studies
In recent years, linear mixed models (LMMs) have become an indispensable tool in large-scale genetic studies. At their core, LMMs are probabilistic models encoding the assumption that individuals sharing similar genotypes are more likely to share similar phenotypes. Although their origins date back at least to the 1950s, use of LMMs in genetic studies has recently skyrocketed, owing to increasing sample sizes and algorithmic improvements.
I will describe a model that improves on standard LMMs by assuming a complex parametric form for the covariance matrix, using multiple-kernel methods from the machine learning literature. The resulting model can automatically identify genetic interactions, and enables jointly inferring all kernel parameters efficiently. In an analysis of eight human genetic diseases and over a hundred mouse phenotypes, our method outperforms competing methods to attain state of the art phenotype prediction results.
If time allows, I will additionally describe modern methods for LMM hypothesis testing under binary phenotypes, with a special emphasis on proper handling of case-control ascertainment.
This work was done my PhD supervisors, Saharon Rosset and Dan Geiger.
· Shahar Mendelson, Technion
The small-ball method and its application in Learning Theory
The small-ball method leads to nontrivial lower bounds on the infimum of certain empirical processes, most notably, on $inf_{f \in F} \frac{1}{N}\sum_{i=1}^N f^2(X_i)$, where $F$ is a class of functions on a probability space $(\Omega,\mu)$ and $X_1,...,X_N$ are independent, distributed according to $\mu$. Such estimates play an essential role in many problems, for example, Sparse Recovery; the smallest singular value of random matrices with iid rows; (random) Gelfand widths of convex bodies, etc. The main feature of the small-ball method is that it requires rather minimal assumptions on the underlying class $F$. Thus, the lower bound may be established even in situations in which two-sided concentration is simply false.
I will present the highlights of the method and then focus on one application: obtaining a sharp (minimax) bound on the error rate of Empirical Risk Minimization performed in a convex class. Using the small-ball method one may obtain a rate that scales correctly with the "richness" of the class (the somewhat vague term will be clarified) and with the noise level of the problem; in particular, when the noise level tends to zero, the rate tends to "noise-free" (realizable) rate.
· Amichai Painsky, Tel Aviv University
Generalized Independent Component Analysis Over Finite Alphabets
Independent component analysis (ICA) is a statistical method for transforming an observable multidimensional random vector into components that are as statistically independent as possible from each other. Usually the ICA framework assumes a model according to which the observations are generated (such as a linear transformation with additive noise). ICA over finite fields is a special case of ICA in which both the observations and the independent components are over a finite alphabet.
In this work we consider a generalization of this framework in which an observation vector is decomposed to its independent components (as much as possible) with no prior assumption on the way it was generated. This generalization is also known as Barlow’s minimal redundancy representation problem and is considered an open problem. We propose several theorems and show that this NP hard problem can be accurately solved with a branch and bound search tree algorithm, or tightly approximated with a series of linear problems. Moreover, we show there exists a simple transformation (namely, order permutation) which provides a greedy yet very effective approximation of the optimal solution. We further show that while not every random vector can be efficiently decomposed into independent components, the vast majority of vectors do decompose very well (that is, with a small constant cost), as the dimension increases. Our contribution provides the first efficient set of solutions to Barlow’s problem.
The minimal redundancy representation has many applications, mainly in the fields of neural networks and deep learning. In this work we show that this formulation further applies to large alphabet source coding.
Joint work with Prof. Saharon Rosset and Prof. Meir Feder from the EE department.
· Armin Schwartzman, NC State University
Confidence Regions for Excursion Sets in Asymptotically Gaussian Random Fields, with an Application to Climate
The goal of this work is to give confidence regions for the excursion set of a spatial function above a given threshold from repeated noisy observations on a fine grid of fixed locations. Given an asymptotically Gaussian estimator of the target function, a pair of data-dependent nested excursion sets are constructed that are sub- and super-sets of the true excursion set, respectively, with a desired confidence. Asymptotic coverage probabilities are determined via a multiplier bootstrap method, not requiring Gaussianity of the original data nor stationarity or smoothness of the limiting Gaussian field. The method is used to determine geographical regions where the mean summer and winter temperatures are expected to increase by mid 21st century by more than 2 degrees Celsius.
· éåøí é÷åúéàìé, äîçì÷ä ìîãòé äîçùá, îëììä à÷ãîéú äãñä éøåùìéí.
îòøëú úåîëú îåîçä ì÷áéòú äðãéøåú ùì ò÷áåú ðòìééí
ò÷áåú ðòìééí îäååú àçú äøàéåú äçùåáåú áîùôè äôìéìé, àê äï ðúåðåú ìàçøåðä úçú äú÷ôä áãáø äúå÷ó äîãòé ùìäï. äôø÷èé÷ä ä÷ééîú îúáññú òì çéôåù ñéîðéí éçåãééí áò÷áåú äðîöàåú áæéøú äôùò åâí áäèáòåú ðñéåï ùì ðòìéé äçùåãéí äðáã÷åú áîòáãä. ñéîðéí éçåãééí àìä äí ôâîéí ÷èðéí áã"ë äðåöøéí áðòì áîäìê äùéîåù áä. ëîåú äôâîéí äúåàîéí ùîöìéçéí ìîöåà áò÷áú æéøú äôùò åáäèáòú äðñéåï ùáîòáãä åâí âåãìí åöåøúí îäååéí îãã ìîéãú ääúàîä áéï äò÷áåú. ëãé ìçæ÷ àú äúå÷ó äîãòé ùì òðó ò÷áåú äðòìééí ôéúçðå îòøëú ìäòøëä ëîåúéú ùì ðãéøåúí ùì ôâîéí áäèáòåú ðñéåï. àñôðå îàâø ðúåðéí âãåì ùì ë 13 àìôéí ôâîéí ùåðéí òì ôðé ëàøáò îàåú ðòìééí, åàôééðå àú äðãéøåú ùì äôâîéí. äàôééåï ëìì äòøëú äääúôìâåéåú ùì ùìåùä îàôééðéí ùì äôâîéí – îé÷åîí òì ôðé äðòì, ëéååðí, åöåøúí. ìöåøê æä ôéúçðå ùéèåú ìäòøëú îàôééðéí àìä, ëåìì ùâéàú äîãéãä òáåø ëì îàôééï. áäøöàä àöéâ àú äîòøëú, ùéèåú äàéñåó åäòøëú ääúôìâåéåú, äáòéåú ä÷ééîåú åëéååðéí ìäîùê äîç÷ø.
áùéúåó:
éøåï ùåø, ùøéðä åéæðø åöãå÷ öç, îòáãú ñéîðéí åçåîøéí, äîçì÷ä ìæéäåé ôìéìé, îùèøú éùøàì.
ôøåô îéëä îðãì åúìîéãú äãå÷èåøè ðòîé ÷ôìï-ãîøé, äîçì÷ä ìñèèéñèé÷ä, äàåðéáøñéèä äòáøéú.
· Omer Bobrowski, Duke University
Random Topology and its Applications
The study of random topology focuses on describing high-level qualitative properties of random spaces. This field has been rapidly developed in the past decade, and is driven both by deep theoretical questions and state-of-the-art data analysis applications. Topological Data Analysis (TDA) broadly refers to the use of concepts from mathematical topology to analyze data and networks. In the past decade a variety of powerful topological tools has been introduced, and were proven useful for applications in various fields (e.g. shape analysis, signal processing, neuroscience, and genomic research). The theory developed in random topology aims to provide a solid foundation for the statistical analysis of these methods, which to date is at a very preliminary stage.
This talk will be divided into three parts. First, I will provide an introduction and motivation to random topology and TDA. Next, we will discuss the theory of random geometric complexes. In particular, we will consider their Betti numbers (counting “cycles” in different dimensions), and phase transitions related to these. Finally, we will present a few examples of combining the theory of random complexes with statistical problems related to TDA. In particular, we are interested in analyzing topological noise” that appears in such problems, for the purposes of filtering and hypothesis testing.
· Sivan Sabato, Ben-Gurion University
Active linear regression
We propose a new active learning algorithm for parametric linear regression with random design. We provide finite sample convergence guarantees for general distributions in the misspecified model. This is the first active learner for this setting that provably can improve over passive learning.
Following the stratification technique advocated in Monte-Carlo function integration, our active learner approaches the optimal risk using piecewise constant approximations. Experiments demonstrate the algorithm's convergence to the oracle rates.
Joint work with Remi Munos, INRIA Lille.
· Yanyuan Ma, University of South Carolina
Robust mixed-effects model for clustered failure time data: application to Huntington's disease event measures
An important goal in clinical and statistical research is estimating the distribution for clustered failure times, which have a naturalintra-class dependency and are subject to censoring. We propose to handle these inherent challenges with a novel approach that does not impose restrictive modeling or distributional assumptions. Rather, using a logit transformation, we relate the distribution for clustered failure times to covariates and a random, subject-specific effect such that the covariates are modeled with unknown functional forms, and the random effect is distribution-free and potentially correlated with the covariates. Over a range of time points, the model is shown to be reminiscent of an additive logistic mixed effect model. Such a structure allows us to handle censoring via pseudo-value regression and to develop semiparametric techniques that completely factor out the unknown random effect. We show both theoretically and empirically that the resulting estimator is consistent for any choice of random effect distribution and for any dependency structure between the random effect and covariates. Lastly, we illustrate the method's utility in an application to the Cooperative Huntington's Observational Research Trial data, where our method provides new insights into differences between motor and cognitive impairment event times in subjects at risk for Huntington's disease.
· Matan Gavish, HUJI
Optimal thresholding of singular values and eigenvalues
It is common practice in multivariate and matrix-valued data analysis to reduce dimensionality by performing a Singular Value Decomposition or Principal Component Analysis, and keeping only $r$ singular values or principal components, the rest being presumably associated with noise. However, the literature does not propose a disciplined criterion to determine $r$; most practitioners still look for the elbow in the Scree Plot'', a 50-years-old heuristic performed by eye. I'll review a line of work which develops a systematic approach to eigenvalue and singular value thresholding. This approach assumes that the signal is low-rank and that the noise is rotationally invariant. Recent results derive optimal thresholds in the presence of quite general noise distributions.
Joint work with David Donoho, Iain Johnstone and Edgar Dobriban (Stanford).
· Tirza Routtenberg, BGU
Estimation after parameter selection: Estimation methods, performance analysis, and adaptive sampling
In many practical parameter estimation problems, such as medical experiments and cognitive radio communications, parameter selection is performed prior to estimation. The selection process has a major impact on subsequent estimation by introducing a selection bias and creating coupling between decoupled parameters. As a result, classical estimation theory may be inappropriate and inaccurate and a new methodology is needed. In this study, the problem of estimating a preselected unknown deterministic parameter, chosen from a parameter set based on a predetermined data-based selection rule, Ψ, is considered. In this talk, I present a general non-Bayesian estimation theory for estimation after parameter selection, includes estimation methods, performance analysis, and adaptive sampling strategies. First, I use the post-selection mean-square error (PSMSE) criterion as a performance measure instead of the commonly used mean-square-error (MSE). The corresponding Cramér-Rao-type bound on the PSMSE of any Ψ-unbiased estimator is derived, where the Ψ -unbiasedness is in the Lehmann unbiasedness sense. The post-selection maximum-likelihood (PSML) estimator is presented and its Ψ–efficiency properties are demonstrated. Practical implementations of the PSML estimator are proposed as well. Finally, I discuss the concept of adaptive sampling in a two-sampling stages scheme of selection and estimation.
· Lotem Kaplan, TAU
Empirical Approach for Sequential Bayesian Inference with Application to Active Learning
Active learning is a sequential learning framework, that holds two main components: (i) A classification model, and (ii) Querying strategy to choose the next data instance to observe. The querying strategy can have important consequences for the performance of the classifier.
The Active learning field is closely related to sequential experiment design, in the statistical literature. We approach this problem by applying sequential Bayesian inference to logistic regression. In this work we follow and extend the empirical representation of posterior distributions by Dror and Steinberg (2008) and show its application to binary data and active learning.
This seminar describes Lotem Kaplan's dissertation work under the supervision of Prof. David Steinberg and is presented as part of the PhD requirements of Tel Aviv University.
· Saharon Rosset, TAU
Estimating Random-X Prediction Error of Regression Models
The areas of model selection and model evaluation for predictive modeling have received extensive treatment in the statistics literature, leading to both theoretical results and practical methods based on covariance penalties and other approaches. However, the vast majority of this work is based on the Fixed-X assumption'', where covariate values are assumed to be non-random and known. By contrast, in most modern predictive modeling applications, it is more reasonable to take the Random-X'' view, where future prediction points are random and new. In this work we concentrate on examining the applicability of the covariance-penalty approaches to this problem. We propose a decomposition of the Random-X prediction error that clarifies the additional error due to Random-X, which is present in both the variance and bias components of the error. This decomposition is very general, but we focus on its application to the most fundamental case, of least squares regression. We show how to quantify the excess variance using simple random-matrix results, leading to a covariance penalty approach generalizing previous methods, like Tukey's MSPE and GCV. To account for excess bias, we propose to take only the bias component of the ordinary cross validation (OCV) estimate, resulting in a hybrid penalty term. This hybrid penalty approach is shown to be empirically superior to OCV in regression model evaluation simulations. This empirical observation is supported by a common representation of the two methods, which shows that the OCV has one more random component.
This talk is based on joint work with Ryan Tibshirani, and describes work in progress.
· Lucien Birge, Université Pierre et Marie Curie (Paris VI)
ρ-estimation – a robust alternative to maximum likelihood
It is well-known that the maximum likelihood, although widely used in Statistics, suffers from various defects. From a theoretical point of view, strong assumptions are needed to ensure that it performs in a satisfactory way and from a practical one, it is definitely not robust, i.e. quite sensitive to model misspecifications. These problems are mainly connected to the use of the log function, which is not bounded, to define the log-likelihood ratios. Replacing it by a specific bounded function leads to an alternative methods with better properties, including robustness (in the sense of Peter Huber). Moreover, in nice parametric models and when the true distribution does belong to the model, the new estimator is asymptotically equivalent to the maximum likelihood estimator.
The interested persons may look at http://arxiv.org/abs/1403.6057.1, this is joint work with Yannick Baraud and Mathieu Sart, this talk is a Sackler distinguished lecture.
· Jeroen de Mast, University of Amsterdam
Estimation of the random error of binary tests using adaptive polynomials
We propose an efficient and robust method for estimating the random error of binary tests and measurements in the situation that a gold standard or reference standard is not available. Recent studies have shown that in this situation, the widely used false acceptance probability (FAP) and false rejection probability (FRP), or their complements specificity and sensitivity, are difficult to estimate. We show by mathematical analysis that this problem is inherent to the unavailability of a gold standard. Instead, the proposed method determines the random components of the error probabilities: the inconsistent acceptance and rejection probabilities IAP and IRP . The estimation method is based on an extension of the beta distribution, and adapts the model complexity to the data. For estimating efficiency, the approach works from a sample taken from the stream of initially rejected parts and also incorporates a historical rejection rate (“baseline data”). A simulation study demonstrates the method’s robustness and is the basis for sample size recommendations. The usefulness of the method is shown from a real-life application, where it outperforms existing methods.
· Regev Schweiger, TAU
Fast and accurate construction of confidence intervals for heritability
Estimation of heritability is fundamental in genetic studies. Recently, heritability estimation using linear mixed models (LMMs) has gained popularity, because these estimates can be obtained from unrelated individuals collected in genome-wide association studies. Typically, heritability estimation under LMMs uses the restricted maximum likelihood (REML) approach. Existing methods for the construction of confidence intervals and standard errors for REML rely on asymptotic properties. However, these assumptions are often violated due to the bounded parameter space, statistical dependencies, and limited sample size, leading to biased estimates and inflated or deflated confidence intervals. In our work, we show that the construction of confidence intervals by current methods is inaccurate, especially when the true heritability is relatively low or relatively high. We propose a computationally efficient method for the estimation of the distribution of the heritability estimator, and for the construction of accurate confidence intervals.
Joint work with Eran Halperin, Saharon Rosset, Shachar Kaufman and Eleazar Eskin.
· Asaf Cohen, University of Michigan
A Multiclass Queueing Model in the Moderate-Deviation Heavy-Traffic Regime
Risk sensitive control with vanishing noise problems deals with a risk averse decision maker who observes a small noise controlled diffusion process. The model was presented by Jacobson in [4] and was broadly studied afterwards; see, e.g., [3]. By Girsanov's change of measure type of arguments one usually deduces that the limit of such problems is governed by a differential game, whose solution plays a role in finding an asymptotically optimal control in the original problem. Traditionally, queueing systems under heavy-traffic are approximated by diffusion processes. I will present a critically loaded system with small overload probability. Thus, the evolution of the system is approximated by a small noise diffusion. The controller is risk averse and uses a discounted version of the risk sensitive cost. Change of measure arguments do not apply here. We developed new tools to show that the model is governed by a differential (deterministic) game. We explicitly solved the game and showed that its solution eventually leads to an asymptotically optimal stationary feedback policy in the queueing problem.
This is joint work with Rami Atar. The talk is based on [1] and [2], with a focus on the new techniques that were recently developed in [1] to prove the convergence.
[1] R. Atar and A. Cohen. An asymptotically optimal control for a multiclass queueing model in the moderate-deviation heavy-traffic regime. submitted to Annals of Applied Probability, 2016.
[2] R. Atar and A. Cohen. A differential game for a multiclass queueing model in the moderate-deviation heavy-traffic regime. Mathematics of Operations Research, to appear, 2016.
[3] W. H. Fleming. Risk sensitive stochastic control and differential games. Communications in Information and Systems, 6(3):161-177, 2006.
[4] D. H. Jacobson. Optimal stochastic linear systems with exponential performance criteria and their relation to deterministic differential games. IEEE Trans. Automatic Control, AC-18(2):124-131, 1973.
· Ofer Harel, University of Connecticut
Clustering Incomplete Data via Normal Mixture Models
Model-based clustering using Normal mixture models provides a framework to describe how data groups together using Normal distributions. However, the existing methods for such analyses require complete data. One way to handle incomplete data is multiple imputation, a simulation-based approach which bypasses many of the disadvantages present in other methods for handling incomplete data. However, it is difficult to apply multiple imputation and cluster analysis in a straightforward manner. In this presentation, we develop MICA-N: Multiply Imputed Cluster Analysis for Normally distributed data. MICA-N adds clustering methods to particular steps in multiple imputation in order to create a way to cluster incomplete normal data. We illustrate how MICA-N outperforms existing methodology with a simulation, and then demonstrate the utility of MICA-N on yeast gene expression data. This is joint work with Chantal Larose and Dipak Dey. | 2019-05-21 11:50:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5877970457077026, "perplexity": 1981.7603501272945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256314.52/warc/CC-MAIN-20190521102417-20190521124417-00068.warc.gz"} |
https://testbook.com/question-answer/4th-term-of-a-g-p-is-8-and-10th-term-is-27-then--5c812007fdb8bb15c677b892 | # 4th term of a G. P is 8 and 10th term is 27. Then its 6th term is?
1. 12
2. 14
3. 16
4. 18
Option 1 : 12
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Electric charges and coulomb's law (Basic)
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## Detailed Solution
Concept:
Let us consider sequence a1, a2, a3 …. an is a G.P.
• Common ratio = r = $$\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}$$
• nth term of the G.P. is an = arn−1
• Sum of n terms = s = $$\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - \;1}}$$; where r >1
• Sum of n terms = s = $$\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - \;1}}$$; where r <1
Calculation:
Given:
4th term of a G. P is 8 and 10th term is 27
nth term of the G.P. is Tn = a rn-1
∴ T4 = a. r3 = 8 ----(1)
T10 = a r9 = 27 ----(2)
Equation (2) ÷ (1), we get
$${r^6} = \frac{{27}}{8}$$
$$\Rightarrow {\left( {{{\rm{r}}^2}} \right)^3} = {\rm{\;}}{\left( {\frac{3}{2}} \right)^3}$$
∴ $${r^2} = \frac{3}{2}$$
T6 = a r5
= a r3.r2
$$= 8 \times \frac{3}{2}$$
= 12 | 2023-03-31 18:35:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6438414454460144, "perplexity": 4482.2383943011}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00697.warc.gz"} |
https://stats.stackexchange.com/questions/102646/sampling-brownian-motion | # Sampling Brownian motion
I wish to sample standard linear Brownian motions on the interval $[0,1]$. I do this by dividing the interval into $n$ equal sub-intervals, deciding $B(0)=0$, and letting $B\left(\frac{k}{n}\right)=B\left(\frac{k-1}{n}\right)+\mathcal{N}\left(0,\frac{\sigma^2}{n}\right)$ for $k\ge 1$, after deciding for some $\sigma>0$. I do this $s$ times.
My question is as follows: how large should $n$ and $s$ be so I'll feel comfortable enough to say that the sampled Brownian motion represent, in some sense, the real distribution of Brownian motions?
In fact, there are two questions here:
• How large should $n$ be so that the random walks should resemble, in some sense, Brownian motions?
• How large should $s$ be so that the distribution of the sampled fractional Brownian motions would resemble, in some sense, the real distribution of such fractional Brownian motions?
• We cannot hope to judge your level of comfort. – Glen_b Jun 9 '14 at 8:38
• I've learned that many times in statistics there are bounds which are "empirically proven" to provide "satisfying results". Choice of optimal parameters is done often according to empirical results and intuition. I ask for experience here. – Bach Jun 9 '14 at 8:51
• Do you have any examples of such empirical proofs of "satisfaction"? – Glen_b Jun 9 '14 at 8:59
• Peters et al. (2007) say that a random forest is "empirically proven to be better than its individual members". Breiman has suggested on its random forest model that the mtry parameter will be $\sqrt{p}$ in classification forests and $p\over 3$ in regression forests, where $p$ is the number of variables. He did not claim that this should be "satisfying", but he did have his reasons to recommend this numbers. – Bach Jun 9 '14 at 10:07
• There's the rub; 'satisfying' was the criterion you brought up, and that's subjective. With a more objective criterion, some progress might be possible. – Glen_b Jun 9 '14 at 10:09
## 2 Answers
I illustrate a simulated Brownian Bridge on my blog, using the method described here. In your case, you would use covariance function $k(s,t)=\min(s,t)$
If you want it to look good, don't hold back. You might want to consider the number of pixels in your graph, since I'm not sure you will gain much by letting $n$ be larger than that.
• I don't think the question is either about the correctness of the algorithm nor about graphical representation. The word "resemble" most likely is intended in some vague sense of "is likely to have all the salient statistical characteristics of." This is the issue @Glen_b is pursuing in comments to the question: it needs clarification from the O.P. – whuber Jun 9 '14 at 14:19
• Well, in that case, you need $n=\infty$, since otherwise, how are you going to get a nowhere differentiable function? – Placidia Jun 9 '14 at 14:37
# You are worrying about nothing
One of the nice things about Brownian motion is that it has an explicit distributional form for its incremental change over any specified time period. In fact, this requirement is built into the definition. For a one-dimensional Brownian motion process (i.e., a Wiener process) $$W = \{ W_t | t \in \mathbb{R} \}$$ with variation rate $$\sigma^2$$, one of the stipulated requirements is that:
$$W_{r+t} - W_r \sim \text{N}(0, t \cdot \sigma^2) \quad \quad \quad \text{for all } r \in \mathbb{R} \text{ and } t \geqslant 0.$$
Thus, if we have a process with starting value $$W_0 = 0$$, and we want to "sample" from the process at time values $$0 < t_1 < \cdots < t_n$$, we can generate the sampled values $$W_{t_1}, ..., W_{t_n}$$ as follows:$$^\dagger$$
$$W_{t_k} = \sum_{i=1}^k Z_k \quad \quad \quad Z_k \sim \text{N}(0, (t_k-t_{k-1}) \cdot \sigma^2).$$
An important point is that there is no requirement for the time increments to be small. Regardless of whether the time increments are large or small (or a mixture of both), the displacement of the process over each time increment is independent of the other displacements, and has the distribution specified above. Thus, we can be "comfortable" that the generated values reflect a genuine Brownian motion process without worrying about making the increments small.
$$^\dagger$$ In this formula we take $$t_0 \equiv 0$$. | 2020-10-22 04:20:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7344337105751038, "perplexity": 494.2567637012714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107878879.33/warc/CC-MAIN-20201022024236-20201022054236-00021.warc.gz"} |
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https://www.groundai.com/project/decoherence-enhanced-measurements/ | Decoherence-enhanced measurements
# Decoherence-enhanced measurements
Daniel Braun and John Martin Université de Toulouse, UPS, Laboratoire de Physique Théorique (IRSAMC), F-31062 Toulouse, France CNRS, LPT (IRSAMC), F-31062 Toulouse, France Institut de Physique Nucléaire, Atomique et de Spectroscopie, Université de Liège, 4000 Liège, Belgium
Quantum-enhanced measurements use highly non-classical quantum states in order to enhance the sensitivity of the measurement of classical quantities, like the length of an optical cavity Giovannetti04 (). The major goal is to beat the standard quantum limit (SQL), i.e. a sensitivity of order , where is the number of quantum resources (e.g. the number of photons or atoms used), and to achieve a scaling , known as the Heisenberg limit. Doing so would have tremendous impact in many areas Huelga97 (); Goda08 (); Budker07 (), but so far very few experiments have demonstrated a slight improvement over the SQL Leibfried05 (); Nagata07 (); Higgins07 (). The required quantum states are generally difficult to produce, and very prone to decoherence. Here we show that decoherence itself may be used as an extremely sensitive probe of system properties. This should allow for a new measurement principle with the potential to achieve the Heisenberg limit without the need to produce highly entangled states.
Decoherence arises when a quantum system interacts with an environment with many uncontrolled degrees of freedom, such as the modes of the electromagnetic field, phonons in a solid, or simply a measurement instrument Zurek91 (). Decoherence destroys quantum mechanical interference, and plays an important role in the transition from quantum to classical mechanics Giulini96 (). It becomes extremely fast if the “distance” between the components of a “Schrödinger cat”-type superposition of quantum states reaches mesoscopic or even macroscopic proportions. Universal power laws rule the scaling of the decoherence rates in this regime Braun01 () and lead to time scales so small that in fact the founding fathers of quantum mechanics postulated an instantaneous collapse of the wave-function during measurement. Only recently could the collapse be time-resolved in experiments with relatively small “Schrödinger cat”–states Brune96 (); Guerlin07 (). However, different superpositions may decohere with very different rates. In particular, if the coupling of the quantum system to the environment enjoys a certain symmetry, entire decoherence-free subspaces (DFS) may exist, in which superpositions of states retain their coherence, regardless of the “distance” between the superposed states. In essence, the symmetry prevents the environment to distinguish the states, such that no information leaks out of the system and the quantum superpositions remain intact. DFS have found widespread use in quantum information theory after their formulation for Markovian master equations Zanardi97 (); Lidar98 (); Duan98 (), experimental demonstration Kwiat00 (); Kielpinski01 (); Viola01 (), and once it was realized that quantum computation might be performed inside a DFS Beige00 (). Given the reliance of the DFS on a symmetry in the coupling to the environment, it is clear that for a large “Schrödinger cat”-type superposition prepared in a DFS, the decoherence rate should be extremely sensitive to any changes that modify the symmetry of the coupling. This is the basic idea underlying the new measurement principle which we call “Decoherence-Enhanced Measurements” (DEM). Two fortunate circumstances make us believe that this idea may be turned into something of practical relevance. First, while it may seem that DEMs would again require the extremely difficult initial preparation of a highly entangled macroscopic state, surprisingly the Heisenberg limit can be reached with a much simpler to prepare product state of pairs of atoms. Second, the required initial “symmetry” of the coupling to the environment means nothing more but a degenerate eigenvalue of the Lindblad operators in the master equation, or, more generally, of the coupling Hamiltonian of the system to the environment Lidar98 (). Actual symmetries in the system (e.g. spatial symmetries) may lead to such degeneracy, but are by no means necessary Braun01B (). The scheme is therefore much more general than it may appear at first sight.
In order to illustrate the concept, we consider two–level atoms or ions ( even, ground and excited states , for atom , ) localized in a cavity with one semi-reflecting mirror, and resonantly coupled with coupling constants to a single e.m. mode of the cavity of frequency . The reduced density matrix of the atoms evolves according to the master equation in the interaction picture
ddtρ(t)=Ls[ρ(t)]+Lc[ρ(t)], (1)
where
Ls[ρ(t)]=Γ2N∑i=1([σ(i)−ρ(t),σ(i)+]+[σ(i)−,ρ(t)σ(i)+]) (2)
describes individual spontaneous emission with rate , while
Lc[ρ(t)]=γ([J−ρ(t),J+]+[J−,ρ(t)J+]) (3)
models collective decoherence, , , (), and where is the average coupling strength over all atoms coupled to the cavity field. The rate (with the single photon cavity decay rate) is independent of . Equation (1) is a well–known and experimentally verified Gross76 (); Skribanowitz73 () master equation which for and in the bad cavity limit describes superradiance Agarwal70 (); Bonifacio71a (); Glauber76 (); Gross82 (). Due to the spatial envelope of the e.m. mode in resonance with the atoms, the depend on the position of the atoms along the cavity axis and on the length of the cavity (the waist of the mode is taken to be much larger than the size of the atomic ensemble),
gi=√ℏωϵ0Vsin(kxxi)% \boldmath{ϵ}⋅d, (4)
where , denotes the dielectric constant of vacuum, the mode volume (with an effective cross–section ), the polarization vector of the mode, and the vector of electric dipole transition matrix elements between the states and , taken identical for all atoms. Decoherence in this system has been extensively studied, see Braun01B () for a review. The initial state belongs to the DFS with respect to collective emission, , if and only if Lidar98 (). If for , this DFS is well known Beige00b (); Braun01B (). It contains DF states, including a dimensional subspace in which the pair formed by the atoms and can be in a superposition of the triplet ground state and the singlet . For arbitrary , should be replaced by .
Consider now the situation where the atoms can be grouped into two sets with atoms each and coupling constants in the first set (), and in the second set (). One way of obtaining two coupling constants may be to trap the atoms in two two–dimensional lattices perpendicular to the cavity axis (see Fig.1).
Suppose that after preparing the atoms in a DFS state corresponding to the initial couplings (), the length of the cavity changes slightly. The coupling constants will evolve, , and so will the DFS. It is this collective change of the coupling constants which can be revealed very sensitively through the decoherence it induces as the original state becomes exposed to decoherence. The induced decoherence therefore provides for a very precise measurement of the change of the length of the cavity, as we shall show now.
In order to simplify notation we will assume in the following (i.e. the atoms are located for instance symmetrically with respect to an antinode of the cavity mode, or at a distance given by an integer multiple of the wavelength of the mode), but we emphasize that everything goes through for different initial couplings, unless otherwise mentioned. Assume that an initial pure product state of pairs of atoms is prepared in the initial DFS, , where
|ψ0⟩=N/2⨂l=1|φl⟩l (5)
with . The decoherence mechanism (3) is directly linked to photon loss from the cavity. The induced decoherence can be measured through the number of photons which escape through the cavity mirror during a small time interval . In the superradiant regime considered here (), any photon created leaves the cavity immediately, such that the quantum expectation value of is given by , where the collective pseudo-spin component measures total population inversion of the atoms. As long as resides in the DFS, we have . If the coupling constants undergo slight changes and get replaced by general values for atom , a straightforward calculation (see Methods) shows that
⟨˙Jz(0)⟩c = −γ(N/2∑i=1|~gi−~gi+N/2|2|bi|2 (6) +N/2∑\lx@stackreli,j=1i≠j(~g∗i−~g∗i+N/2)(~gj−~gj+N/2)b∗ibjaia∗j),
which is in general of order . In particular, if the undergo collective changes with and if all pairs of atoms were prepared in the same initial state , for , we have
⟨˙Jz(0)⟩c = −γ|δ~G|2f(N,b) (7) f(N,b) = 4|b|2[N2+N2(N2−1)(1−|b|2)], (8)
where . The term quadratic in is maximized for , i.e. an equal weight superposition of the two DF basis states and for each pair of atoms, and gives for a signal . As long as the two lattices are not situated at anti-nodes of the mode, the relation between and is linear to lowest order. If we choose with we have
δ~G=mπcot(nxπx1L)δLL, (9)
where we see that and are related by a factor independent of . Note that the measurement of allows the measurement of , and not just a detection of a change of : . The ultimate sensitivity achievable depends not only on the scaling of the signal with , but also of the noise, quantified through the standard deviation . The most fundamental noise associated with the measurement of is its fluctuation due to the quantum mechanical nature of the prepared state. Our approach of calculating initial time-derivatives of observables by tracing them over with the Lindbladian implies , as can change the number of excitations by at most 1, and this condition sets an upper bound on . In this regime, , and, therefore, . It follows that the signal-to-noise ratio is given by . The ultimate sensitivity achievable can be estimated from a fixed of order 1, independent of , which leads to a minimal . We have thus shown that a precision measurement based on the purely dissipative dynamics (3) and an initial product state can achieve the Heisenberg limit. This is in contrast to unitary dynamics of independent quantum resources, where the SQL cannot be surpassed when using an initial product state Giovannetti06 ().
In a real experiment there may be additional fundamental noise sources. One obvious concern is spontaneous emission. It is easily verified that leads to a contribution to which scales as and leads to a background signal against which, one might think, the collective decoherence signal has to be compared. However, note that spontaneous emission sends photons into the entire open space but not into the cavity, whereas the collective emission escapes exclusively through the leaky cavity mirror. Therefore, the two contributions can be well separated by observing only the photons escaping through the cavity mirror.
Another obvious concern are fluctuations of the coupling constants. In order to measure , the experiment has to be repeated times with . However, is independent of and only given by the desired signal/noise, or, equivalently, by itself. Increasing does therefore not influence the scaling with . It increases the sensitivity by a factor , but reduces the bandwidth by in the standard way. While we assume that the time scale of the mirror motion is sufficiently long compared to the time needed for averaging, the exact coupling constants might fluctuate about their slowly evolving mean values during the averaging, e.g. due to fluctuating traps caused by vibrations in the set up. But even for perfectly stable traps, the micro–motion of the atoms in their respective trapping potentials, thermal motion, or even quantum fluctuations in the traps will lead to fluctuating . The cost in sensitivity of these fluctuations depends on their correlations. To see this, let us consider fluctuations of the about their mean values , for , for . We introduce the correlation matrix , where the over-line denotes an average over the ensemble describing the fluctuations. Equation (6) then leads to a background in the photon counting rate and to fluctuations on top of , i.e. . The average background , given by
¯¯¯¯αbg =
can be determined independently at , and subtracted from the signal; it does therefore not influence the sensitivity of the measurement. The remaining noise fluctuates about zero,
δαf = 2γ{N/2∑i=1(δ~G∗Δ~gi+δ~GΔ~g∗i)|bi|2+N/2∑\lx@stackreli,j=1i≠j(δ~G∗Δ~gj+δ~GΔ~g∗i)b∗ibjaia∗j}, (10)
where . Assuming real coupling constants, we find the standard deviation , where , and is in general of order . This leads to fluctuations in the measured photon number with a standard deviation , and to a signal-to-noise ratio .
Several interesting cases can be considered:
1. Fully uncorrelated fluctuations, , where stands for the Kronecker-delta: Here we get , which is in general of order , and leads back to the SQL.
2. Pairwise identical fluctuations between the two sets: for . This can be the consequence of fully correlated fluctucations, . Alternatively, such a situation arises for example for atoms initially arranged symmetrically with respect to an anti-node such that , if the two atoms (or ions) in each pair () are locked into a common oscillation. This should be the case for two trapped ions repelling each other through a strong Coulomb interaction, and cooled below the temperature corresponding to the frequency of the breathing mode. Equation (10) then gives . Note, however, that for initial the more general DFS leads to a more complicated condition for the correlations, , which might be harder to achieve.
3. Correlated fluctuations within a set, but uncorrelated between the two sets, for or , but for and or vice versa. In this case both sums in (10) survive and lead to a noise of order , the worst case scenario. However, this comes to no surprise, as such correlations are indistinguishable from the signal: all the atoms in a given set move in a correlated fashion, but independently from the atoms of the other set. This leads to a collective difference in the couplings, just as if the length of the cavity was changed.
Case (2) above is clearly the most favorable situation. If there are no other background signals depending on , we keep the scaling of characteristic of the Heisenberg limit. In order to favor case (2) over cases (1),(3), it appears to be advantageous to work with ions and to try to bring the ions in a pair as closely together as possible, thus strongly correlating their fluctuations, while separating the ions in the same set as far as possible.
To summarize, we have shown for a particular example how the very sensitive dependence of collective decoherence on system parameters can be exploited to reach the Heisenberg limit in precision measurements while using an initial product state — something which is known to be impossible with unitary dynamics Giovannetti06 (). It should be clear that the principle of DEM is far more general than the example exposed here. Decoherence is itself a process in which interference effects play an important role. This is exemplified by the very existence of DFS, and can lead to exquisite sensitivity. One might therefore as well try to exploit these effects instead of trying to suppress decoherence at all costs.
Methods
Derivation of Eq. (6): The commutation relation , valid for any choice of couplings , allows to rewrite . A short calculation yields and leads immediately to Eq. (6). Preparation of initial state: In order to prepare the product state (5) it is helpful to use three–level atoms with a lambda structure. Let and the additional state be hyperfine (HF) states, and assume that their energies are split in a sufficiently strong magnetic field, such that only the transition resonates with the cavity mode. We assume further that the second optical lattice can be moved along the cavity axis, such that controlled pairwise collisions of corresponding atoms in the two lattices can be induced. Entangled pairs of atoms in their HF split ground states can thus be created (for atoms in the same lattice this has been demonstrated experimentally, see Bloch08 () for a review). After the creation of an entangled HF state , that differs from (5) by the replacement of states by states , the second lattice is moved back to its original position. Now one can selectively excite the states by a laser pulse in resonance with the transition, that replaces the singlets in the (very long lived) HF states by the desired singlets of the and states and thus produce the product state (5). However, as such, the method is not of much practical use yet, as it will be virtually impossible to park the second lattice at the exact position corresponding to coupling constants which render the state (5) decoherence free. The extreme sensitivity of the collective decoherence with respect to changes of the coupling constants plays against us here, and will lead to a superradiant flash of light from the cavity after the excitation , if the exact position corresponding to DFS is not achieved. But it is possible to position the second lattice at the required position with a precision of for the case (2) considered above, using a feed-back mechanism and a part of the quantum ressources. With the atoms in the state , do the following repeatedly in order to find the optimal position: Excite a part of the entangled HF pairs containing atoms with the laser, measure , and use the measurement results to bracket the minimum of as function of the lattice position. The minimum of indicates that the position corresponding to the DFS is achieved. Using golden section search, the minimum can be bracketed to precision in moves, as at each step the sensitivity of the measurement of the position of the lattice is of order . Once the minimum is found, excite the remaining unused pairs (there should be still a number of pairs of ) to the desired state . That state is now decoherence-free, and the system ready to detect small changes of the position of one of the mirrors. Note that for this method it is not necessary to know which exact state is produced in the controlled collisions and subsequent laser excitation.
Imperfections in preparation of : Suppose that instead of the state (5) a state
|~ψ0⟩=N/2⨂l=1|φ⟩l with |φ⟩l=a|t−⟩l+b|s⟩l+c|t0⟩l+d|t+⟩l (11)
was prepared (we consider the same state for all pairs for simplicity, but this is not essential). Repeating the calculation that leads to Eq. (8) and assuming real , we now find
⟨nph⟩ = γΔt[N2(((c−b)~G1+(c+b)~G2)2+d2(~G21+~G22)) (12) +N2(N2−1)((~G2−~G1)b(a−d)+(~G2+~G1)c(a+d))2].
The derivative of with respect to is of order , and thus still allows to find the minimum of as function of the position of the second lattice with a precision of order in case 2. At the minimum a component outside the DFS persists, such that photons will leak out of the cavity, but the average background is only of order , and can be measured separately and subtracted from the singal. Changes of away from the position of the minimum still lead to a signal that scales, for large , quadratically with , , and the previous analysis leading to the Heisenberg limit still applies.
Alternatively, one can get rid of the additional background by letting the system relax before measuring changes of . Indeed, any state with a component outside the DFS will relax to a DFS state (and thus a dark state) or mixtures of DFS states within a time of order or less. Components with large total pseudo-angular momentum relax in fact in much shorter time of order . The DFS states reached through relaxation starting from still allow a scaling of close to . We have shown this by simulating the relaxation process with the help of the stochastic Schrödinger equation (SSE) corresponding to Eq. (3). For real values of the coupling constants, the SSE reads
dψ(t) = D1(ψ(t))dt+D2(ψ(t))dW(t) (13) D1(ψ) = γ(2⟨J−⟩ψJ−−J+J−−⟨J−⟩2ψ)ψ (14) D2(ψ) = √2γ(J−−⟨J−⟩ψ)ψ, (15)
where is a Wiener process with average zero and variance , and Breuer06 (). Using an Euler scheme with a time step of , we followed the convergence of to DFS states for states with , until the norm of the difference dropped below . In these final dark states, randomly distributed over the DFS, we calculated , and averaged over a large number of realizations of the stochastic process (, , , , , , 250, 200, and 250 for , and ). Figure 2 shows the scaling of as function of for different values of for up to . Within this numerically accessible range of , follows a power law with an exponent that decays only gradually with for . Moreover, that decay might be a finite size effect: Note that, surprisingly, appears to be close to symmetric with respect to . This is corroborated by exact analytical calculations based on the diagonalization of , which lead to for , for , and for (in units ). The plot shows that all numerical data can be very well fitted by . From Eq. (8) we know that has to scale as for sufficiently large . Both and are negative for all for which we have data, and appears to be negligible. Fig. 2 shows that increases even more rapidly than (a fit in the range gives a power law ). But has to cross over to a power law with , unless other Fourier components start contributing significantly. Otherwise, would become negative for . This indicates that for large the scaling of is in fact for all .
In summary, our method still works, even if the product state (5) is not prepared perfectly. One has the choice to start measurement immediately after state preparation, which gives an additional background of order , or to wait a time of the order of a few after preparation of the initial state, until no more photons leave the cavity through the mirror, with no additional background. In both cases the scaling of the signal-to-noise ratio is still , and allows to reach the Heisenberg limit in the measurement of a subsequent small change of .
Another class of states in the DFS that allows quadratic scaling of with , are Schrödinger cat states of macroscopic pseudo-angular momentum in the two sublattices (i.e. states ),
⟨(ℓ,ℓ)j,−j|J1+J1−|(ℓ,ℓ)j,−j⟩= |~G1|2(2j+1)(2j+2)4(j+1)2−1[ℓ(ℓ+1)−j(j+2)4], (16)
with in agreement with the initial intuitive reasoning. If the total system is initially in a singlet state () and the angular momentum of each of the two sets of atoms has its maximal value , we have
⟨nph⟩=γΔt3|δ~G|2N(N+4). (17)
While these states are protected by the DFS and thus do not suffer the fate of rapid decoherence of the highly entangled states proposed for QEM, it appears to be still more challenging to produce them compared to the product states (5). Our numerical simulations also show that states chosen randomly inside the DFS lead on the average only to . Therefore, it is rather remarkable that the product states (5) and the above Schrödinger cat states share the property of scaling of .
The authors declare to have no competing financial interests.
Acknowledgments: DB thanks Eite Tiesinga and Peter Braun for useful discussions. This work was supported by the Agence National de la Recherche (ANR), project INFOSYSQQ. Numerical calculations were partly performed at CALMIP, Toulouse. J.M. thanks the Belgian F.R.S.-FNRS for financial support.
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The feedback must be of minimum 40 characters and the title a minimum of 5 characters | 2021-03-02 18:30:11 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8332181572914124, "perplexity": 904.8220330790518}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364027.59/warc/CC-MAIN-20210302160319-20210302190319-00503.warc.gz"} |
https://www.cheenta.com/a-problem-on-doppler-effect/ | Select Page
A train passes through a station with constant speed. A stationary observer at the station platform measures the tone of the train whistle as (484Hz) when it approaches the station and (442Hz) when it leaves the station. If the sound velocity is (330m/s), then the tone of the whistle and the speed of the train are
(a) (462hz, 54km/h)
(b) (463Hz, 52Km/h)
(c) (463Hz, 56Km/h)
(d) (464Hz, 52Knm/h)
Solution:
When train approaches the station, the frequency heard by the observer
$$n_1=n\frac{v}{v-v_s}=n(\frac{330}{330-v_s})$$
Here, $$v=330m/s$$
n is the actual frequency of the whistle
$$484 =n(330/330-v_s)$$….. (i)
When the train leaves the station $$n_2=n\frac{v}{v+v_s}=n(\frac{330}{330+v_s})$$
$$442=n(\frac{330}{330+v_s})$$…. (ii)
Divide Eqs (i) by (ii), we get
$$\frac{484}{442}=330+v_s/330-v_s$$
$$1.09=(330+v_s)/(330-v_s)$$
$$330+v_s=1.09(330-v_s)$$
$$v_s=\frac{31.35}{2.09}$$$$=15m/s$$
Substituting (v_s) in Eqn (i) gives $$484=n(330/330-15)$$ $$=n(330/315)$$ $$n=\frac{484*21}{22}$$
$$=462Hz$$ | 2020-03-29 09:29:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7075079679489136, "perplexity": 3786.6503765738057}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370494064.21/warc/CC-MAIN-20200329074745-20200329104745-00024.warc.gz"} |
https://www.groundai.com/project/deep-image-homography-estimation/ | Deep Image Homography Estimation
# Deep Image Homography Estimation
\authorblockNDaniel DeTone \authorblockAMagic Leap, Inc.
Mountain View, CA
ddetone@magicleap.com
\authorblockNTomasz Malisiewicz \authorblockAMagic Leap, Inc.
Mountain View, CA
tmalisiewicz@magicleap.com
\authorblockNAndrew Rabinovich \authorblockAMagic Leap, Inc.
Mountain View, CA
arabinovich@magicleap.com
###### Abstract
We present a deep convolutional neural network for estimating the relative homography between a pair of images. Our feed-forward network has 10 layers, takes two stacked grayscale images as input, and produces an 8 degree of freedom homography which can be used to map the pixels from the first image to the second. We present two convolutional neural network architectures for HomographyNet: a regression network which directly estimates the real-valued homography parameters, and a classification network which produces a distribution over quantized homographies. We use a 4-point homography parameterization which maps the four corners from one image into the second image. Our networks are trained in an end-to-end fashion using warped MS-COCO images. Our approach works without the need for separate local feature detection and transformation estimation stages. Our deep models are compared to a traditional homography estimator based on ORB features and we highlight the scenarios where HomographyNet outperforms the traditional technique. We also describe a variety of applications powered by deep homography estimation, thus showcasing the flexibility of a deep learning approach.
## I Introduction
Sparse 2D feature points are the basis of most modern Structure from Motion and SLAM techniques [9]. These sparse 2D features are typically known as corners, and in all geometric computer vision tasks one must balance the errors in corner detection methods with geometric estimation errors. Even the simplest geometric methods, like estimating the homography between two images, rely on the error-prone corner-detection method.
Estimating a 2D homography (or projective transformation) from a pair of images is a fundamental task in computer vision. The homography is an essential part of monocular SLAM systems in scenarios such as:
• Rotation only movements
• Planar scenes
• Scenes in which objects are very far from the viewer
It is well-known that the transformation relating two images undergoing a rotation about the camera center is a homography, and it is not surprising that homographies are essential for creating panoramas [3]. To deal with planar and mostly-planar scenes, the popular SLAM algorithm ORB-SLAM [14] uses a combination of homography estimation and fundamental matrix estimation. Augmented Reality applications based on planar structures and homographies have been well-studied [16]. Camera calibration techniques using planar structures [20] also rely on homographies.
The traditional homography estimation pipeline is composed of two stages: corner estimation and robust homography estimation. Robustness is introduced into the corner detection stage by returning a large and over-complete set of points, while robustness into the homography estimation step shows up as heavy use of RANSAC or robustification of the squared loss function. Since corners are not as reliable as man-made linear structures, the research community has put considerable effort into adding line features [18] and more complicated geometries [8] into the feature detection step. What we really want is a single robust algorithm that, given a pair of images, simply returns the homography relating the pair. Instead of manually engineering corner-ish features, line-ish features, etc, is it possible for the algorithm to learn its own set of primitives? We want to go even further, and add the transformation estimation step as the last part of a deep learning pipeline, thus giving us the ability to learn the entire homography estimation pipeline in an end-to-end fashion.
Recent research in âdenseâ or âdirectâ featureless SLAM algorithms such as LSD-SLAM [6] indicates promise in using a full image for geometric computer vision tasks. Concurrently, deep convolutional networks are setting state-of-the-art benchmarks in semantic tasks such as image classification, semantic segmentation and human pose estimation. Additionally, recent works such as FlowNet [7], Deep Semantic Matching [1] and Eigen et al.’s Multi-Scale Deep Network [5] present promising results for dense geometric computer vision tasks like optical flow and depth estimation. Even robotic tasks like visual odometry are being tackled with convolutional neural networks [4].
In this paper, we show that the entire homography estimation problem can be solved by a deep convolutional neural network (See Figure 1). Our contributions are as follows: we present a new VGG-style [17] network for the homography estimation task. We show how to use the 4-point parameterization [2] to get a well-behaved deep estimation problem. Because deep networks require a lot of data to be trained from scratch, we share our recipe for creating a seemingly infinite dataset of training triplets from an existing dataset of real images like the MS-COCO dataset. We present an additional formulation of the homography estimation problem as classification, which produces a distribution over homographies and can be used to determine the confidence of an estimated homography.
## Ii The 4-point Homography Parameterization
The simplest way to parameterize a homography is with a 3x3 matrix and a fixed scale. The homography maps , the pixels in the left image, to , the pixels in the right image, and is defined up to scale (see Equation 1).
⎛⎜⎝u′v′1⎞⎟⎠∼⎛⎜⎝H11H12H13H21H22H23H31H32H33⎞⎟⎠⎛⎜⎝uv1⎞⎟⎠ (1)
However, if we unroll the 8 (or 9) parameters of the homography into a single vector, we’ll quickly realize that we are mixing both rotational and translational terms. For example, the submatrix , represents the rotational terms in the homography, while the vector is the translational offset. Balancing the rotational and translational terms as part of an optimization problem is difficult.
We found that an alternate parameterization, one based on a single kind of âlocationâ variable, namely the corner location, is more suitable for our deep homography estimation task. The 4-point parameterization has been used in traditional homography estimation methods [2], and we use it in our modern deep manifestation of the homography estimation problem (See Figure 2). Letting be the u-offset for the first corner, the 4-point parameterization represents a homography as follows:
H4point=⎛⎜ ⎜ ⎜⎝Δu1Δv1Δu2Δv2Δu3Δv3Δu4Δv4⎞⎟ ⎟ ⎟⎠ (2)
Equivalently to the matrix formulation of the homography, the 4-point parameterization uses eight numbers. Once the displacement of the four corners is known, one can easily convert to . This can be accomplished in a number of ways, for example one can use the normalized Direct Linear Transform (DLT) algorithm [9], or the function
etPerspectiveTransform() in OpenCV.
\section{Data Generation for Homo
raphy Estimation Training deep convolutional networks from scratch requires a large amount of data. To meet this requirement, we generate a nearly unlimited number of labeled training examples by applying random projective transformations to a large dataset of natural images 111In our experiments, we used cropped MS-COCO [13] images, although any large-enough dataset could be used for training. The process is illustrated in Figure 3 and described below.
To generate a single training example, we first randomly crop a square patch from the larger image at position (we avoid the borders to prevent bordering artifacts later in the data generation pipeline). This random crop is . Then, the four corners of Patch A are randomly perturbed by values within the range [-, ]. The four correspondences define a homography . Then, the inverse of this homography is applied to the large image to produce image . A second patch is cropped from at position . The two grayscale patches, and are then stacked channel-wise to create the 2-channel image which is fed directly into our ConvNet. The 4-point parameterization of is then used as the associated ground-truth training label.
Managing the training image generation pipeline gives us full control over the kinds of visual effects we want to model. For example, to make our method more robust to motion blur, we can apply such blurs to the image in our training set. If we want the method to be robust to occlusions, we can insert random âoccludingâ shapes into our training images. We experimented with in-painting random occluding rectangles into our training images, as a simple mechanism to simulate real occlusions.
## Iii ConvNet Models
Our networks use 3x3 convolutional blocks with BatchNorm [10] and ReLUs, and are architecturally similar to Oxfordâs VGG Net [17] (see Figure 1). Both networks take as input a two-channel grayscale image sized 128x128x2. In other words, the two input images, which are related by a homography, are stacked channel-wise and fed into the network. We use 8 convolutional layers with a max pooling layer (2x2, stride 2) after every two convolutions. The 8 convolutional layers have the following number of filters per layer: 64, 64, 64, 64, 128, 128, 128, 128. The convolutional layers are followed by two fully connected layers. The first fully connected layer has 1024 units. Dropout with a probability of 0.5 is applied after the final convolutional layer and the first fully-connected layer. Our two networks share the same architecture up to the last layer, where the first network produces real-valued outputs and the second network produces discrete quantities (see Figure 4).
The regression network directly produces 8 real-valued numbers and uses the Euclidean (L2) loss as the final layer during training. The advantage of this formulation is the simplicity; however, without producing any kind of confidence value for the prediction, such a direct approach could be prohibitive in certain applications.
The classification network uses a quantization scheme, has a softmax at the last layer, and we use the cross entropy loss function during training. While quantization means that there is some inherent quantization error, the network is able to produce a confidence for each of the corners produced by the method. We chose to use 21 quantization bins for each of the 8 output dimensions, which results in a final layer with 168 output neurons. Figure 6 is a visualization of the corner confidences produced by our method — notice how the confidence is not equal for all corners.
## Iv Experiments
We train both of our networks for about 8 hours on a single Titan X GPU, using stochastic gradient descent (SGD) with momentum of 0.9. We use a base learning rate of 0.005 and decrease the learning rate by a factor of 10 after every 30,000 iterations. The networks are trained for for 90,000 total iterations using a batch size of 64. We use Caffe [11], a popular open-source deep learning package, for all experiments.
To create the training data, we use the MS-COCO Training Set. All images are resized to 320x240 and converted to grayscale. We then generate 500,000 pairs of image patches sized 128x128 related by a homography using the method described in Section II. We choose = 32, which means that each corner of the 128x128 grayscale image can be perturbed by a maximum of one quarter of the total image edge size. We avoid larger random perturbations to avoid extreme transformations. We did not use any form of pre-training; the weights of the networks were initialized to random values and trained from scratch. We use the MS-COCO validation set to monitor overfitting, of which we found very little.
To our knowledge there are no large, publicly available homography estimation test sets, thus we evaluate our homography estimation approach on our own Warped MS-COCO 14 Test Set. To create this test set, we randomly chose 5000 images from the test set and resized each image to grayscale 640x480, and generate a pairs of image patches sized 256x256 222We found that very few ORB features were detected when the patches were sized 128x128, while the HomographyNets had no issues working at the smaller scale. and corresponding ground truth homography, using the approach described in Figure 3 with = 64.
We compare the Classification and Regression variants of the HomographyNet with two baselines. The first baseline is a classical ORB [15] descriptor + RANSAC + etPerspectiveTransform() } OpenCV Homoraphy computation. We use the default OpenCV parameters in the traditional homography estimator. This estimates ORB features at multiple scales and uses the top 25 scoring matches as input to the RANSAC estimator. In scenarios where too few ORB features are computed, the ORB+RANSAC approach outputs an identity estimate. In scenarios where the ORB+RANSAC’s estimate is too extreme, the 4-point homography estimate is clipped at [-64,64]. The second baseline uses a 3x3 identity matrix for every pair of images in the test set.
Since the HomographyNets expect a fixed sized 128x128x2 input, the image pairs from the Warped MS-COCO 14 Test Set are resized from 256x256x2 to 128x128x2 before being passed through the network. The 4-point parameterized homography output by the network is then multiplied by a factor of two to account for this. When evaluating the Classification HomographyNet, the corner displacement with the highest confidence is chosen.
The results are reported in Figure 5. We report the Mean Average Corner Error for each approach. To measure this metric, one first computes the L2 distance between the ground truth corner position and the estimated corner position. The error is averaged over the four corners of the image, and the mean is computed over the entire test set. While the regression network performs the best, the classification network can produce confidences and thus a meaningful way to visually debug the results. In certain applications, it may be critical to have this measure of certainty.
We visualize homography estimations in Figure 7. The blue squares in column 1 are mapped to a blue quadrilateral in column 2 by a random homography generated from the process described in Section II. The green quadrilateral is the estimated homography. The more closely the blue and green quadrilateral align, the better. The red lines show the top scoring matches of ORB features across the image patches. A similar visualization is shown in columns 3 and 4, except the Deep Homography Estimator is used.
## V Applications
Our Deep Homography Estimation system enables a variety of interesting applications. Firstly, our system is fast. It runs at over 300fps with a batch size of one (i.e. real-time inference mode) on an NVIDIA Titan X GPU, which enables a host of applications that are simply not possible with a slower system. The recent emergence of specialized embedded hardware for deep networks will enable applications on many embedded systems or platforms with limited computational power which cannot afford an expensive and power-hungry desktop GPU. These embedded systems are capable of running much larger networks such as AlexNet [12] in real-time, and should have no problem running the relatively light-weight HomographyNets.
Secondly, by formulating homography estimation as a machine learning problem, one can build application-specific homography estimation engines. For example, a robot that navigates an indoor factory floor using planar SLAM via homography estimation could be trained solely with images captured from the robot’s image sensor of the indoor factory. While it is possible to optimize a feature detector such as ORB to work in specific environments, it is not straightforward. Environment and sensor-specific noise, motion blur, and occlusions which might restrict the ability of a homography estimation algorithm can be tackled in a similar fashion using a ConvNet. Other classical computer vision tasks such as image mosaicing (as in [19]) and markerless camera tracking systems for augmented reality (as in [16]) could also benefit from HomographyNets trained on image pair examples created from the target system’s sensors and environment.
## Vi Conclusion
In this paper we asked if one of the most essential computer vision estimation tasks, namely homography estimation, could be cast as a learning problem. We presented two Convolutional Neural Network architectures that are able to perform well on this task. Our end-to-end training pipeline contains two additional insights: using a 4-point âcorner parameterizationâ of homographies, which makes the parameterizationâs coordinates operate on the same scale, and using a large dataset of real image to synthetically create an seemingly unlimited-sized training set for homography estimation. We hope that more geometric problems in vision will be tackled using learning paradigms.
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https://www.mechamath.com/algebra/examples-of-polynomial-function-problems/ | Examples of Polynomial Function Problems
Polynomial functions are functions that only have non-negative integer exponents of the independent variable. Some examples of polynomial functions are the linear function, the quadratic function, and the cubic function. The graphs of these functions vary depending on the degree of the function.
Here, we will look at a summary of polynomial functions along with their most important characteristics. In addition, we will look at several examples with answers to learn about its characteristics and its applications.
ALGEBRA
Relevant for
Learning about polynomial functions with examples.
See examples
ALGEBRA
Relevant for
Learning about polynomial functions with examples.
See examples
Summary of polynomial functions
A polynomial function is a function such as a quadratic, cubic, quartic, among others, that only has non-negative integer powers of x. A polynomial of degree n is a function that has the general form:
$$f(x)=a_{n}{{x}^n}+a_{n-1}{{x}^{n-1}}+…+a_{2}{{x}^2}+a_{1}x+a_{0}$$
where the coefficients a are all real numbers. Although the general form looks very complicated, the particular examples are simpler. For example,
$latex f(x)=2{{x}^3}+4{{x}^2}+2x$
is a polynomial function of degree 3 since 3 is the largest power of the function. And the function:
$$f(x)=-4{{x}^5}+2{{x}^4}+3{{x}^2}+6$$
is a polynomial function of degree 5 since 5 is the greatest power of the function.
Roots of polynomial functions
When we have $latex (x-a)(x-b)=0$, we know that a and b are the roots of the function $latex f(x)=(x-a)(x-b)$. Therefore, we can find the polynomial roots by forming an equation by setting the polynomial part of the function equal to zero and factoring or solving for x.
It is also possible to use the opposite of this. For example, if we have that a and b are roots, we know that the polynomial function with these roots must be $latex f (x) = (x-a)(x-b)$, or a multiple of this.
If $latex x=3$ and $latex x=-4$ are the roots of the polynomial function, then this function must be $latex f(x)=(x-3)(x+4)$, or a constant multiple of this.
Examples with answers of polynomial function problems
The following polynomial functions examples can be used to learn about the most common applications and problems that can be encountered with these functions.
EXAMPLE 1
Is the function $latex f(x)=-2+3{{x}^2}+2{{x}^3}$ a polynomial function?
We can see that the function $latex f(x)=-2+3{{x}^2}+2{{x}^3}$ only has variables with positive integer exponents, therefore, this function is polynomial.
EXAMPLE 2
Is the function $latex f(x)=-2{{x}^3}+5{{x}^2}+\sqrt{x}$ a polynomial function?
Taking into account that radicals can be written as fractional powers, we can write the function $latex f(x)=-2{{x}^3}+5{{x}^2}+\sqrt{x}$ as:
$latex f(x)=-2{{x}^3}+5{{x}^2}+{{x}^{\frac{1}{2}}}$
We see that not all exponents of the variable are positive integers. Therefore, this function is not an exponential function.
EXAMPLE 3
Write a polynomial function that has roots 2, 3, 5, 7.
When we have a polynomial written in the form $latex (x-a)(x-b)=0$, we know that its roots area and b. Therefore, we can find a factored polynomial using those roots. In this case, the roots are 2, 3, 5, and 7, so we have:
$latex f(x)=(x-2)(x-3)(x-5)(x-7)$
This means that the polynomial function that has the given roots is $$f(x)=(x-2)(x-3)(x-5)(x-7)$$ or a multiple of this polynomial. For example, $$f(x)=2(x-2)(x-3)(x-5)(x-7)$$ and $$f(x)=-5(x-2)(x-3)(x-5)(x-7)$$ are also polynomial functions that have the given roots.
EXAMPLE 4
Find a polynomial function that has roots -4, -2, 2, 4.
Here, we have the roots -4, -2, 2, 4, so the polynomial function that contains these roots is:
$$f(x)=(x+4)(x+2)(x-2)(x-4)$$
Therefore, the polynomial function is $$f(x)=(x+4)(x+2)(x-2)(x-4)$$ or a multiple of this function.
We always have to remember to correctly assign the signs to the constants of each factor. For example, if the root is -4, the factor is $latex (x + 4)$ and if the root is 4, the factor is $latex (x-4)$.
EXAMPLE 5
Write a polynomial function that has roots -5, -2, 1, 5, 6.
Similar to the previous exercises, we simply form factors with the given roots. Therefore, we have:
$$f(x)=(x+5)(x+2)(x-1)(x-5)(x-6)$$
This means that the polynomial function that has the roots -5, -2, 1, 5, 6 is the function $$f(x)=(x+5)(x+2)(x-1)(x-5)(x-6)$$ or a multiple of this function like $$f(x)=4(x+5)(x+2)(x-1)(x-5)(x-6)$$ or $$f(x)=-3(x+5)(x+2)(x-1)(x-5)(x-6)$$.
Polynomial function problems – Practice
Use the following problems to practice what you have learned about polynomial functions. Select an answer and check it to see if you chose the correct one. | 2022-10-04 13:06:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7124513387680054, "perplexity": 205.84535640455184}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337504.21/warc/CC-MAIN-20221004121345-20221004151345-00783.warc.gz"} |
https://www.gfdl.noaa.gov/blog_held/30-extremes/ | # 30. Extremes
Posted on August 4th, 2012 in Isaac Held's Blog
Percentage change in the precipitation falling on days within which the daily precipitation is above the pth percentile (p is horizontal axis) as a function of latitude and averaged over longtitude, over the 21st century in a GCM projection for a business-as-usual scenario, from Pall et al 2007.
(I have added a paragraph under element (1) below in response to some off-line comments — Aug 15)
When I think about global warming enhancing “extremes”, I tend to distinguish in my own mind between different aspects of the problem as follows (there is nothing new here, but these distinctions are not always made very explicit):
1) increases in the frequency of extreme high temperatures that result from an increase in the mean of the temperature distribution without change in the shape of the distribution or in temporal correlations
The assumption that the distribution about the mean and correlations in time do not change certainly seems like an appropriately conservative starting point. But if you look far out on the distribution, the effects on the frequency of occurrence of days above a fixed high temperature, or of consecutive occurrences of very hot days (heat waves), can be surprisingly large. Just assuming a normal distribution, or playing with the shape of the tails of the distribution, and asking simple questions of this sort can be illuminating. I’m often struck by the statement that “we don’t care about the mean; we care about extremes” when these two things are so closely related (in the case of temperature). Uncertainty in the temperature response translates directly into uncertainty in changes in extreme temperatures in this fixed distribution limit. It would be nice if, in model projections, it was more commonplace to divide up the responses in extreme temperatures into a part due just to the increase in mean and a part due to everything else. It would make it easier to see if there was much that was robust across models in the “everything else” part. And it also emphasizes the importance of comparing the shape of the tails of the distributions in models and observations. Of course from this fixed-distribution perspective every statement about the increase in hot extremes is balanced by one about decreases in cold extremes.
(Added Aug 15) The discussion of this topic is often confused by the fact that people are asking different questions. Suppose we consider days that exceed some fixed temperature $T$ that is on the tail of the distribution of daily temperatures. If the mean temperature warms by $\delta T$, while holding the distribution about the mean fixed, this number could increase dramatically, depending on the shape of the distribution, even if $\delta T$ is much smaller than the width of the distribution. In this case, the mean warming is contributing a small fraction of the temperature anomaly in these extreme warm events even though the probability of these events has increased a lot (see Otto et al, 2012 for a discussion of the Russian heat wave along these lines). If we redefined our criterion for a very hot day by upping the criterion by the small amount $\delta T$ we would go from a description of what is going on as one in which the “number of very hot days increases dramatically” to one in which “the number of very hot days does not change but they are on average $\delta T$ warmer”: $P_{new}(T) >> P_{old}(T) = P_{new}(T + \delta T)$. My gut reactions to these two descriptions of the same physical situation are rather different. The goal has to be to relate these changes to impacts (things we care about) to decide what our level of concern should be, rather than relying on these emotional reactions to the way we phrase things.
2) increases in extreme precipitation that result from an increase in atmospheric moisture, this increase in turn resulting from the increase in saturation vapor pressure resulting from warming — without changes in the winds that are converging moisture into the region of interest during these extreme precipitation episodes;
There is an important sense in which the increase in high precipitation events is more basic, and more robust, than the changes in the mean precipitation. Some expectations for the latter are discussed in Post #13-14 and include regions of increasing and regions of decreasing mean precipitation. Changes in extremely high precipitation events seem to be simpler — we expect them to increase nearly everywhere. It is precisely when one is strongly converging water into some region, creating a lot of precipitation, that the upper bound on the water vapor in the atmosphere comes into play most strongly. irrespective of what the time mean humidity is doing. If you think of the dominant term that is trying to increase water vapor mixing ratios $q$ in regions of strong upward motion $w$ as $-w \partial q/\partial z$, and assume that the atmosphere is saturated $q = q_s$ over some depth, then the rain rate would be determined by integrating $w \partial q_s/\partial z = w (\partial q_s/\partial T)( \partial T/\partial z)$ over the layer within which condensation is preventing supersaturation. Since the saturation mixing ratio at a given pressure is just a function of temperature, and the temperature profile would be moist adiabatic, we have a straightforward null hypothesis connecting the warming and changes in these precipitation extremes, just as we do for temperature extremes.
The figure at the top of the page, from Pall et al (2007) illustrates this nicely. Take each grid point in a GCM and create a histogram of daily precipitation. Look at the change in total precip above the p-percentile of precip values, for a particular scenario by the end of the 21st century. To create a smooth zeroth-order picture, sum the p-percentile precip at each point over longitude and then compute the fractional change in the precip amount — as a function of p and of latitude. I like this plot because of the way it distinguishes between the subtropics (where mean precip is decreasing) and subpolar latitudes (where the mean is increasing) — but it does have the disadvantage, if I am interpreting it correctly, that these averaged results are dominated by the high precip regions at that latitude. In subpolar latitudes, precip is increasing in both heavy and light precip events. In the subtropics there is an increase in very heavy precip events (above the 90-95th percentile of daily values) but a decrease when the rainfall values are light. It is the latter that is evidently causing the reduction in the mean, along with an increase in the frequency of dry days not evident in this plot. SREX (Ch. 3) has a summary of observations of trends in extreme precipitation and a lot of references.
3) changes in the frequency or severity of storms or lower frequency climate anomalies, such as droughts, resulting from changes in atmospheric or oceanic circulations on large scales.
An example might be a poleward shift in the Atlantic storm track increasing the frequency of extreme wind and extreme surface wave events on the poleward flank, and decreasing these same extreme events on the equatorward flank of the storm track. These changes in extremes do not result from any subtle change in the underlying dynamics of the storms or waves — the robustness of the changes in extremes depends entirely on the robustness of the large-scale storm track shift.
Another example is the constructive superposition of la Nina and global warming-induced drought over the southern tier of the continental US. Radiative forcing due to increased well-mixed greenhouse gases expands the subtropics and shifts the midlatitude storm tracks polewards in a variety of models of different levels of complexity. El Nino has the opposite effect, especially over and downstream of the Pacific, where it shifts the jet and storm track equatorwards. So the opposite phase of the ENSO cycle, la Nina events, tends to reduce precipitation especially in the southern tier of the continental US (see here). See also this analysis by Bergman et al 2010 of the connections between Pacific ocean temperatures and medieval megadroughts. The la Nina response adds to the simulated effect of the greenhouse gases (here). See also Lau et al 2008 for discussion related to this superposition. By the same token, some of the effects of El Nino events on North America due to the changes in atmospheric circulation might be ameliorated. Even if the meteorology turns out to be basically a linear superposition, impacts of various kinds — forest fires, agricultural, etc, — will remain a source of strong nonlinearities. It is the existence of these nonlinearities in impacts that makes this constructive interference for US drought between la Nina and warming important, even if the effects of warming on the ENSO variability itself turn out to be modest.
Finally, we have–
4) Changes in the intensity of storms.
There is a tempting hand-waving argument that storms will intensify because there would be more heat of condensation released in rising air, creating more buoyancy and stronger upward motion, but there are a variety of reasons why this is not a convincing argument. In any case, you have to distinguish between extratropical storms and tropical cyclones — these have such different dynamics that they present us with two very different sets of problems. I’ll try to get back to some of these eventually. My point here is just to emphasize that, as outlined above, there are reasons to expect changes in extremes that do not depend on these changes in storm intensity.
[The views expressed on this blog are in no sense official positions of the Geophysical Fluid Dynamics Laboratory, the National Oceanic and Atmospheric Administration, or the Department of Commerce.] | 2022-12-06 14:10:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6165465116500854, "perplexity": 913.6556182421729}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00451.warc.gz"} |
https://publiclab.org/questions/warren/12-08-2017/where-can-microscope-lenses-objectives-be-bought | # Question: Where can microscope lenses (objectives) be bought?
warren asked on December 08, 2017 14:59
148 | 4 answers | #15334
The #microscopes project is moving along amazingly. I see that the raspberry pi microscope project uses a cheap lens (or objective) -- where can these be bought and for how much?
Do they need to be a certain type to fit the design posted by @partsandcrafts?
@bmela @kgradow1 @wmacfarl is this already listed in the activities you posted? I just wanted a place to list links and prices and options. Thanks!
Is this a question? Click here to post it to the Questions page.
We haven't tested a wide range of objectives -- that would be a great next step for this project
Here's one for $40 -- would it work? https://www.amazon.com/OMAX-Achromatic-Compound-Microscope-Objective/dp/B00FG8BC62/ Here's a 100x one for$25: https://www.aliexpress.com/item/100X-BIOLOGIC-MICROSCOPE-OBJECTIVE-RMS-thread-Achromatic-Objective-Brand-New-195-mm/32812997299.html
40x for $22: http://www.amscope.com/40x-achromatic-microscope-objective.html 100x for$31: http://www.amscope.com/100x-achromatic-microscope-oil-immersion-objective.html
Hi @kgradow1 wondering which lens you would recommend for a starter kit for imaging some of the PM materials we've been speaking about?
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Do you mean which magnification? Either 40x or 100x are good. I think the 40x is slightly easier to start with, since it's easier to find your material in the field of view. At 100x, any slight vibration changes the view dramatically, and it requires a significantly lighter touch to get the hang of being able to focus on anything specific
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hmm maybe we could get 1 or 2 10xes. Any idea why the cost on the 10x ones here vary so much? http://www.amscope.com/accessories/objective.html?i_magnifying_power=1186 Have you tried the $15 one? Is this a question? Click here to post it to the Questions page. The cost of all of them actually vary a lot. We took a 40x apart at the barnraising and realized it only has two lenses inside of it, which seemed strange, but it still works. We weren't able to get the high right on the 10x to actually see anything, which seems weird and is a thing worth troubleshooting. I don't think it will work as-is if we just hook up a 10x on the current setup @kgradow1 are you using one of the openflexure optics modules? If so, that might explain why the 10x lens didn't work as expected. Finite-conjugates objectives are designed to produce an image 150mm from the "shoulder" of the objective, with the sample a set distance (usually 35mm or 45mm) from the "shoulder", allowing you to easily swap objectives. If your camera sensor is closer than that (and you're not using a correcting lens) then different objectives will focus at different heights - specifically, the 10x will focus much further away than the 40x because it has a longer focal length. My apologies if that's obvious to you, just thought I'd mention it in case it helps debug what the issue was. Is this a question? Click here to post it to the Questions page. ohh interesting. I was about to buy the 10x. Any fixes for this? or any troubleshooting success @kgradow1 ? Is this a question? Click here to post it to the Questions page. Log in to comment We got ours from Amscope: http://www.amscope.com/4x-100x-four-achromatic-objective-lens-set.html#product_tabs_description_tabbed. 20mm RMS-thread mounting size. (I'm not sure what RMS-thread is, but I'd assume it's standard). And really, the thread mounting doesn't really matter, because we're screwing it into a 3D printed piece with no threads. Log in to comment Hey, for the Openflexure Microscope I tend to buy the lenses from Aliexpress, usually Sunlight Optical but there are many suppliers. The mass-produced objectives are generally all RMS-threaded and designed for "finite conjugates" (they produce an image 150mm from the back of the objectives), and the cost mostly reflects the level of correction (which is related to the number of lenses inside). The cheapest "achromatic" objectives may only have one doublet lens inside (corrected for colour imaging, but not for a wide/flat field of view). These typically cost around$20-$25 and are 35mm from the "shoulder" at the top of the thread to the sample. "plan" or "semi-plan" correction requires at least one more lens inside the objective, and usually is found in 45mm high objectives, costing$25-\$50. These will give a much bigger in-focus field of view. It's possible to add in more correction, or to image at "infinite conjugates", both of which are done in modern lab microscopes - and that's why you can pay thousands, rather than tens, of dollars for an objective if you want. However, I think for the PM work, a basic plan corrected lens will do. My experience so far is that semi-plan is probably acceptable at 40x, but for 100x you really want plan correction (and you need to use immersion oil too!) | 2018-08-18 06:18:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31083789467811584, "perplexity": 1716.847897314907}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213405.46/warc/CC-MAIN-20180818060150-20180818080150-00422.warc.gz"} |
https://indico.cern.ch/event/686555/contributions/2956914/ | ICHEP2018 SEOUL
Jul 4 – 11, 2018
COEX, SEOUL
Asia/Seoul timezone
The LHCf experiment: recent physics results
Jul 5, 2018, 5:00 PM
15m
104 (COEX, Seoul)
104
COEX, Seoul
Parallel Astro-particle Physics and Cosmology
Speaker
Eugenio Berti (Universita e INFN, Firenze (IT))
Description
The main aim of the LHC forward (LHCf) experiment [1] is to provide precise measurements of the particles production rate in the forward region. These high energy calibration data are very important for the tuning of hadronic interaction models used by ground-based cosmic rays experiments. LHC is the most suitable place where we can perform these measurements because a proton-proton collision at $\sqrt{s}$ = 14 TeV is equivalent to the interaction of a $10^{17}$ eV cosmic ray with the atmosphere. In order to do that, two small sampling calorimeters are installed at ±140 m from LHC IP1 (ATLAS Interaction Point), so that they can detect the neutral particles with $\eta$ > 8.4 produced in p-ion collisions [2].
In the past years, LHCf acquired data from p-ion collisions at different energies (p-p at $\sqrt{s}$ = 0.9, 2.76, 7 and 13 TeV; p-Pb at $\sqrt{s_{NN}}$ = 5.02 and 8.1 TeV). In this talk, we would like to present the analysis results relative to photons [3], neutrons [4] and $\pi^{0}$ [5] differential production cross sections, compared with models predictions. In particular, we will discuss the measurement of the energy distributions of secondary particles produced in $\sqrt{s}$ = 13 TeV p-p collisions (photons, already published [6], and neutrons) and in $\sqrt{s_{NN}}$ = 8.1 TeV p-Pb collisions (photons). In all these cases, no model resulted to be in good agreement with experimental observations in all the regions investigated by the analysis. We will also discuss about the ATLAS-LHCf joint analysis, based on the common data taking that the two experiments had in the last two operations at LHC. This activity is very important because the information of the ATLAS detector in the central region is an useful tag to distinguish between diffractive and non-diffractive events in the LHCf detector. Finally, we will present the measurement relative to the contribution of diffractive dissociation to the production of forward photons in $\sqrt{s}$ = 13 TeV p-p collisions [7], the first result from the ATLAS-LHCf joint analysis.
Primary authors
Eugenio Berti (Universita e INFN, Firenze (IT)) Oscar Adriani (Universita e INFN, Firenze (IT)) Lorenzo Bonechi (Universita e INFN, Firenze (IT)) Alessia Tricomi (Universita e INFN, Catania (IT)) Alessio Tiberio (Universita e INFN, Firenze (IT)) Sergio Bruno Ricciarini (Universita e INFN, Firenze (IT)) Lel D'Alessandro (Universita e INFN (IT)) Yoshitaka Ito (Nagoya University (JP)) Kimiaki Masuda (Nagoya University (JP)) Hiroaki Menjo (Nagoya University (JP)) Paolo Papini (INFN) Yasushi Muraki (STE-laboratory, Nagoya University (JP)) Takashi Sako (University of Tokyo (JP)) Nobuyuki Sakurai (Nagoya University (JP)) Tadahisa Tamura (Kanagawa University (JP)) Shoji Torii (Waseda University (JP)) William C Turner (Lawrence Berkeley Laboratory) Mana Ueno (Nagoya University (JP)) Qidong Zhou (Nagoya University (JP)) | 2022-08-09 01:34:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5608202815055847, "perplexity": 10178.214741059042}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00702.warc.gz"} |
http://spinningnumbers.org/v/lc-natural-response-derivation3.html | We use Euler’s Formula to change complex exponentials into a solution expressed with sines and cosines. Part 3 of 4.
Created by Willy McAllister. | 2018-01-24 07:32:45 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8684763312339783, "perplexity": 897.0264084439389}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084893530.89/warc/CC-MAIN-20180124070239-20180124090239-00638.warc.gz"} |
https://community.goactuary.com/t/economists-suggest-taxing-wfh-people-a-little-more/1189 | # Economists suggest taxing WFH people a little more
Of course, I object. But, mainly to the “one tax rate for all is fair” concept. I certainly don’t save the same percentage of my gross for working at home than, say, a claims processor at my company would. I don’t get three times as big of a lunch, nor drive three times as expensive (in maintenance… well, maybe I do, because it’s old) a car to get to work, nor do I drive three times the distance.
Also, I’d certainly be noting that I have a cubicle that has my name on it at an office, so that’s where I am supposed to work every day.
That’s bizarre. I can think of reasons you should pay less, starting with the cost of maintaining a home office. You’re supplying your own internet bandwidth and chair and heat and air conditioning and office supplies.
The cushion on my office chair is going to wear out faster given that I’m now sitting in it roughly 40 hours a week more than I was when I was working in the office.
My heat and A/C would go off during the work day if I was at the office and my employer would foot the bill to keep my workspace at an appropriate temperature. But when I’m home that cost falls on my shoulders. Office supplies are pretty small for me, but hey I do still take paper notes sometimes and print stuff out. There’s wear & tear on my printer and I’m using my own pens and paper. I go through toner so slowly that the incremental cost to me is probably $0, but it might not be$0 for others.
Many people are supplying their own computers/keyboards/mice/monitors although in most cases they probably already had those items and won’t necessarily go through them any faster due to additional use.
And even though most people are too stupid to understand this concept: en masse they’re paying for the extra internet bandwidth. We all might have “unlimited” plans, but the more data we use the more our providers will charge for those unlimited plans. So when large swathes of society start working at home, a price increase is likely to follow.
I always choose to live very close to work, so my commuting-to-in-person-office costs are extremely low, particularly considering that most of my errands were previously attached to commuting and now are separate trips.
So if anything, WFHers should get a tax BREAK to reflect their increased expenses.
3 Likes
Oh, and my employer used to provide lunches once a week and I’d pack the other days. So I’m also spending more on lunches now that I’m providing 100% of them rather than 80%. And my lunches aren’t as fancy as the ones my employer provided.
1 Like
I am saving on makeup, so there’s that.
But I’m spending more on hot water because I used to shower at work after working out in the company gym 2-3x a week.
And there’s all the toilet paper and water that I’m flushing down the toilet too.
Yeah, I don’t think the savings from makeup and a microscopic amount of mileage on my car trumps all of the other expenses I’m picking up. I think I’m worse off financially by working from home.
I mean, I make enough that it’s chump change, but the differential is definitely negative in my case.
1 Like
Oh, thought of another one. I never used them, but my employer actually supplied free tampons in the ladies room. So there’s that cost for some women.
I never drank the free coffee, but nearly every place I’ve ever worked has supplied free coffee that many employees drank. Gosh, Microsoft gave us free soda.
A previous employer had an on-site nurse. You could get free Tylenol and cold medicine and stuff from the nurse. One time I got a strep throat test from the nurse, saving myself the co-pay. Another employer didn’t have the free nurse but they did have free headache and cold medicine in the first aid kit along with the first aid supplies, which were also free.
Every place I’ve ever worked let you use the notaries in the legal department if you had a personal document you needed notarized and let you use the mail room and corporate shipping rates to send personal packages.
God, there was a ton of crap employers used to give us for free onsite that I’m no longer getting. Each expense piddly (which is why the employers provided them: cheaper to give free ibuprofen than pay for your time to run to the drugstore), but it all adds up!
1 Like
I can totally understand the economic argument. Right now, Mrs Waiting and I are spending nowhere near what we usually do, but both working our normal hours. As a household we could absorb higher taxes to help fund the support for those who can’t currently do their job.
But, logistically, I don’t know. I’m not sure how many households are in our position, it feels like most are operating on some level of reduced income - or aren’t seeing reduced outgoings - so taxing more isn’t going to help them.
But I find myself very much on the “keep everyone alive and we’ll worry about the economy later” side of the argument…
And that’s perfectly valid. If Biden and Harris come in and push a second round of stimulus and then later a tax bill to cover the expense, that’s fine. But an explicit tax on working from home seems like a really bad idea that is bound to cause both resentments and perverse incentives.
Income tax? Fine.
Wealth tax? Fine.
Lack of consumption tax? WTF.
1 Like
I make up for my transportation cost savings with an excessive cat food budget.
Also of note, I was surprised when the U.S. government gave me a stimulus cheque… (…check? How do you spell that in American?). I remember there being a note from Trump in the mail that came along with it that referred to me as a “fellow American citizen”.
Go stimulate the economy with your cheque.
I think I’ve stimulated Amazon enough already.
I suppose that it would be silly to point out that WFH alleviates strain on overloaded-in-many-locales transportation networks, and should slightly reduce the demand for taxpayer resources to support those networks and to offset costs arising from their congestion.
2 Likes
There is no need for you to eat cat food.
Some people think that hybrid and electric cars need to be taxed more (at registration, I guess), because they don’t consume gas, whose taxes maintain the roads, on which they are driven.
\textcolor{red}{\text{Wow, you should have been taxed for these benefits!!}}
Anywho, if I get taxed more, I’ll simply ask for a bigger raise. Or, go to work for an hour, then come home for the rest of it.
Definitely. As I say, I get the underlying idea - that folks who are financially better off due to Covid could be taxed more to fund supporting those who are struggling - but a WFH tax doesn’t quite do it, and would never work long term. If WFHing is genuinely cheaper/better then wages will adjust, a tax is just trying to accelerate it.
The Covid “solution” feels like it’s always going to be a case of building up debt now, and then finding a way to service it once we’re back to whatever normal looks like.
The WFH tax strikes me as too linked to the idea of maintaining the old normal.
1 Like
There’s more rationale for that at least. A hybrid Camry is just as hard on the road as a regular Camry and yet does pay less taxes that go towards maintaining the road. Historically gasoline consumption has been a reasonable proxy for how hard each vehicle is on the road and that connection is drying up. Now whether it is reasonable for the non-hybrid / non-electric vehicles to subsidize the hybrids and electrics due to their carbon footprint is a different question on which reasonable people could disagree.
But at least there’s a decent rationale behind it. A rationale that doesn’t hold for working from home, IMO.
It was considered de minimus. But so are any alleged cost savings in working from home. Cost “savings” that could well be negative. | 2022-08-09 16:50:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30723172426223755, "perplexity": 2254.7926607285567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00711.warc.gz"} |
https://qchu.wordpress.com/2009/11/16/groups-vs-abelian-groups/ | Feeds:
Posts
## Groups vs. abelian groups
A few weeks ago on MathOverflow Greg Muller asked, “why do groups and abelian groups feel so different?” The answers were very interesting and came from several different perspectives, but I still don’t feel as if the question was resolved. So I’ll try to synthesize and summarize some of the answers and hopefully something will be clearer in the end.
Morphisms
Groups naturally arise as automorphism groups of objects in categories. From this perspective abelian groups should just correspond to objects with particularly simple symmetries. But abelian groups have their own special type of symmetry: the pointwise sum of two homomorphisms is also a homomorphism, so $\text{Ab}$ is enriched over itself, something that doesn’t hold at all for $\text{Grp}$. What this implies is that the automorphisms of an abelian group have two composition structures: one coming from pointwise addition and another coming from composition. This is the “best” definition of a ring, at least in my opinion.
This observation also immediately tells us why $\mathbb{Z}$ is a special ring: it is the initial object in $\text{Ring}$, since every abelian group has a $\mathbb{Z}$-action given by $x \mapsto x^n$. From here one is readily led to the “best” definition of a module, exactly analogous to how one might go from sets to permutations to groups to group actions. Now we can think of abelian groups as $\mathbb{Z}$-modules and the action $x \mapsto x^n$ as a distinguished set of automorphisms of any abelian group. For some abelian groups these are the only automorphisms.
Non-abelian groups, on the other hand, have a totally different set of distinguished automorphisms: the inner automorphisms $x \mapsto gxg^{-1}$. Indeed, one way to define a non-abelian group is as a group with nontrivial inner automorphisms, and for some non-abelian groups these are the only automorphisms. In other words, while the structure of abelian groups is controlled in a strong way by the structure of $\mathbb{Z}$ (this is basically the content of the structure theorem), there is no such controlling object for general groups.
Perhaps precisely because their structure is so well-controlled, abelian groups lead to a lot of useful constructions in mathematics, such as abelian categories (the basic tool of homological algebra). Even to study groups one often passes to the category of modules over the group algebra!
What’s not clear to me is the following: should I think of abelianness as a powerful tool or as the easier version of a more difficult but more powerful theory of “non-abelian categories”?
A 2-categorical perspective
One particular complaint Greg had was that abelian groups rarely arise, in practice, as automorphism groups of objects. There is a way, however, one can get abelian groups “naturally” acting on things. The Eckmann-Hilton argument shows that two monoid structures on a set which are homomorphisms of each other must in fact be the same commutative monoid structure, and has the following consequences:
• The higher homotopy groups are abelian.
• The $2$-morphisms in a $2$-category with one object and one morphism form a commutative monoid. If the $2$-morphisms are invertible, they form an abelian group.
The latter is due to the fact that $2$-categories have two types of composition. This seems to have important philosophical implications; for example, awhile back at the n-Category cafe Bruce Bartlett asked a similar question about the difference between commutative and noncommutative algebras from an $n$-categorical point of view. It seems as if to understand commutativity one should really move up the $n$-categorical ladder; as Scott Carnahan puts it, if groups act on objects, then 2-groups act on categories, and the automorphisms of the identity functor automatically form an abelian group.
The big question, of course, is whether the abelian groups that naturally arise in mathematics can actually be fruitfully thought of in this way. Anyone have any thoughts?
### 8 Responses
1. I think part of what you’re getting at in the 2-categories section is that AbGroups is an abelian category but Groups is not(not even additive) and since if we consider the 2-category of an abelian category we get an abelian category(morphisms are Z-matrices since Hom sets are abelian groups). Thus moving up this ladder, the AbGroups look similar(the same?) but Groups does not. This may be similar to what your saying but at least this is something that _I_ find interesting about them.
B.t.w. nice blog, I enjoy it.
2. “should I think of abelianness as a powerful tool or as the easier version of a more difficult but more powerful theory of “non-abelian categories”?”
Well, obviously, the answer depends on context. For example, if you’re studying the automorphism group of some class of mathematical objects, and manage to prove that it’s always abelian, then that (may) turn into a powerful tool for telling you something about your objects.
You can ask similar questions in many areas of mathematics: e.g. replace abelianness w/ “field/vector space” and non-abelianness with “ring/module”. (Keeping in mind that modules over a group ring are morally equivalent to group representations) I don’t think of linear algebra as just an easier version of the representation theory of groups. I think of linear algebra as somehow more fundamental…but maybe I just haven’t thought about it too much.
• I think you’ve slightly misunderstood my question. What I mean is this: homological algebra is all about abelian categories, so when I say “abelianness” I’m including modules over rings. Is there a “non-abelian homological algebra” on “non-abelian categories” which generalizes homological algebra?
3. Well, here’s something I was thinking, although maybe it’s equivalent to what you said.
Groups are categories, right? Specifically, they’re groupoids that are monoids. So it makes sense to define functors between groups, which happen to coincide with homomorphisms. Now two functors (in the categorical setting)/homomorphisms (in the group-theoretic setting), which we’ll call $f, g: G \rightarrow H$, are naturally isomorphic iff there’s some element $a \in H$ with $f = aga^{-1}$. Now if H is abelian, then this happens iff $f = g$, so two parallel functors are naturally isomorphic iff they’re not just equivalent, but actually “the same.” But if H is nonabelian, there are plenty of natural transformations between different morphisms.
There’s gotta be a way of phrasing this categorically — I think we’re saying that the 2-category of abelian groups is a strict 2-groupoid, whereas the 2-category of groups is only a weak 2-groupoid? Again, this “feels” like it should tell us why the higher homotopy groups are abelian, although I don’t really see how.
• I interpret it as meaning that the Cartesian closure of $\text{Grp}$ is $\text{Gpd}$, but I don’t know if this is true or meaningful.
• Is $Gpd$ really cartesian closed? I think I see what you’re going for, though, although at best this seems like a 1-categorical shadow of a 2-categorical statement. | 2016-05-31 07:48:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9018148183822632, "perplexity": 345.6163837822643}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464051196108.86/warc/CC-MAIN-20160524005316-00146-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/621183/magnetic-field-induced-by-current-in-space | # Magnetic Field Induced by Current in Space
I'm learning about how a current in a wire can create a magnetic field, here is the diagram I see:
My question is about how the magnetic field effects things in space.
If there was a magnetic material that was a certain distance d away from the wire and then current goes through the wire, would the material orbit around the wire at distance d forever?
Update
After reading the responces, I understand that it would not rotate around the lines, but it would align to them as Claudio mentions.
My initial question actually arised after watching this video. So now I am going to change my question a little.
If you put a perfect circle of iron fillings in space around a wire, and allow current to flow through the wire, would they orbit around the wire?
The magnetic field generated by an electric current is highly non uniform. The magnetic field of a permanent small magnet is also non uniform.
The expected behavior in this case is the same of any 2 magnets: they rotate to align the fields and attract each other.
• I see. Do you think you could help answer my new question? Mar 15 '21 at 2:56
The magnetic field lines show the direction of the magnetic field. The separation of the field lines show how strong it is in that region - close field lines means a strong magnetic field and larger separation means a weaker field.
The field can then be used to find the force on a charged particle or on a wire carrying a current using $$F=BQv$$ where $$Q$$ is the charge and $$v$$ is the velocity of the charge. For the wire it's $$F=BIL$$ where $$I$$ is the current and $$L$$ is the length of wire in the field.
The direction of the force in either of the above situations is from 'Flemings Left Hand Rule'.
The magnetic material you mentioned would not necessarily be much influenced by the situation you described. The magnetic field created by the wire has most influence on moving charges, as described above. | 2022-01-27 21:36:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6730375289916992, "perplexity": 222.49355877476702}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00329.warc.gz"} |
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creating an .aux file. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the LaTeX Error: Missing p-arg in array arg ! It is a misinterpretation to conclude the chapter files should be treated by "separate TeX sessions", and Herb did not say in any way that they must have their own preambles.
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The first table or figure that is being hoarded may be a bottle neck. LaTeX Error: Something's wrong--perhaps a missing \item One thing that is obviously wrong is the supply of meaningful error messages. Nothing is absolute.) ! \script...font ... To get an actual %, you need \%. (That is a mistake even experienced latex users still make regularly.) For an & you need \&, for a # you need \#,
has an extra } ! Missing $inserted Usually, this happens to me if for some arcane reason I try to use a character like _ in the text. How does a Spatial Reference System like WGS84 have an elipsoid and a geoid? Missing { inserted LaTeX decided it really needed a { and inserted one. Parameters must be numbered consecutively ! Remember that 0 is not a valid length. Check for typos in the \begin{environment} line. For example, this error will be caused by doing: \documentclass{scrartcl} \begin{document}✘ \usepackage{graphicx} instead of \documentclass{scrartcl} \usepackage{graphicx}✔ \begin{document} ⇦ ⇧ ⇨ This book is also available as A4 PDF or 12.8cm x If stumped, try the general tricks. ! Doubt that it would happen with a modern latex version. If you can paste them directly from the log file, (so that the log file reader can recognize them), it would be even better. If so, enclose it between$ and \$.
LaTeX Error: Command \end{itemize} invalid in math mode ! Misplaced alignment tab character &. | 2019-03-18 13:28:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9018011093139648, "perplexity": 5472.277759044472}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201329.40/warc/CC-MAIN-20190318132220-20190318153644-00048.warc.gz"} |
https://math.stackexchange.com/questions/2515458/hamiltonian-path-on-cubic-graphs-and-whether-closed-triangle-meshes-are-triangl | # Hamiltonian Path on Cubic Graphs, and whether closed triangle meshes are triangle strips
The problem is this:
Can every closed triangle mesh (an approximation of a 3d object using triangles, eg. a tetrahedron) be 'peeled like an orange', that is, can we find a sequence of triangles such that consecutive triangles share edges, and all triangles in the mesh are included once (and only once) in our sequence.
This is my logic so far:
Model the mesh as its dual graph, where vertices are triangles, and edges join triangles that share edges, then the problem boils down to finding a Hamiltonian Path in the graph.
Each vertex will have degree three; this makes the graph by definition cubic.
I did some googling; I did find that Tait's conjecture, that every cubic polyhedral graph has a Hamiltonian cycle, is false. I guess then that every cubic polyhedral graph has a Hamiltonian path (otherwise a counterexample would disprove Tait's conjecture). Cubic polyhedral graphs are planar and 3-vertex-connected so if my guess is correct, we can 'peel' all convex polyhedra, but can we always 'peel' a torus?
I believe that if our 3d mesh is an object without holes, then the graph is always planar.
So my question is:
When are Hamiltonian paths guaranteed to exist in cubic graphs? Particularly those cubic graphs that correspond to physical triangle meshes.
In my quest for a counterexample I constructed a genus-2 surface out of triangles, but it was fairly easy for my computer to find a 'peeling', so I am now a little more convinced that everything can be peeled.
The first graph in this paper shows an 88 vertex planar cubic graph that doesn't have any Hamiltonian path:
The dual of this would be a 3d convex polyhedron made of 88 triangles (Garuanteed by Steinitz Theorem, as noted by Henning Makholm) that cannot be 'peeled'. It can be realised as a tetrahedron, with the base a single triangle, and the sides an arrangement of 29 triangles:
The example makes use of what it calls a 'Tutte Triangle': $abc$ - a graph with the property that any path joining $b$ and $c$ cannot pass through all vertices of $abc$.
The dual of the Tutte graph would be a counterexample to the existence of a Hamiltonian cycle. But the Tutte graph does have a Hamiltonian path.
On the other hand, we can use the Tutte fragment to construct a polyhedral 3-regular graph that doesn't even have a Hamiltonian path: Take the graph of a cube, allow two vertices opposite each other to remain vertices, and replace each of the other six vertices with a Tutte fragment "pointing towards" the selected cube vertex, producing this:
which (by Steinitz's theorem) you can draw on a sphere and dualize to get this unpeelable triangle mesh (imagine folding the figure into an octahedron):
I doubt you can find an easy necessary and sufficient criterion, since the NP-hardness of finding Hamiltonian cycles seems to be quite robust.
If your triangle mesh can be 2-colored (that is, the underlying cubic graph is bipartite), a Hamiltonian cycle may be guaranteed; this is the unsolved Barnette's conjecture.
[1]:
• Thanks, so the dual of this graph (replacing vertices with triangles) would be some triangle mesh that couldn't be peeled. Are we guaranteed that the dual of this can be realised a nice (eg. non-overlapping, finite area) 3d object? – Christian Fieldhouse Nov 11 '17 at 19:41
• @ChristianFieldhouse: Yes, that is fairly easy to check by Steinitz's theorem. – Henning Makholm Nov 11 '17 at 19:45
• How do you know that no Hamiltonian paths exist on this graph? I guess you could use any path to construct a Hamiltonian cycle (but I don't know how), a contradiction. Or is there a simpler way to see that no Hamiltonian Paths exist? – Christian Fieldhouse Nov 12 '17 at 10:32
• *a Hamiltonian cycle on the Tutte Graph – Christian Fieldhouse Nov 12 '17 at 10:41
• @ChristianFieldhouse: Hmm ... while trying to explain my argument I discovered a Hamiltonian path. Oops! The same idea will still work but we need three Tutte graphs rather than two. – Henning Makholm Nov 12 '17 at 16:42 | 2019-06-20 17:18:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6763367056846619, "perplexity": 582.1985046945055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999263.6/warc/CC-MAIN-20190620165805-20190620191805-00488.warc.gz"} |
https://zbmath.org/?q=ai%3Aqian.linyi+ai%3Acheng.gongpin | # zbMATH — the first resource for mathematics
An optimal investment strategy in a Markov-modulated risk model: maximizing the terminal utility. (English) Zbl 1289.91098
Summary: The surplus of an insurance company is governed by a jump-diffusion process, and it can be invested in a financial market with one risk-free asset and $$N$$ risky assets. The parameters of the surplus process and the asset price processes depend on the regime of the financial market, which is modeled by an observable finite-state continuous-time Markov chain. To maximize the terminal utility, we focus on finding an optimal investment strategy and solve it by using the HJB equation. An explicit expression for the optimal strategy and the corresponding objective function are presented when the company has an exponential utility function. Some interesting economic interpretations are involved.
##### MSC:
91B30 Risk theory, insurance (MSC2010) 91G10 Portfolio theory 62P05 Applications of statistics to actuarial sciences and financial mathematics 91B16 Utility theory | 2021-01-16 00:44:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3593592941761017, "perplexity": 569.0021440329886}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00270.warc.gz"} |
http://ncatlab.org/nlab/show/Kervaire+invariant | # nLab Kervaire invariant
### Context
#### Manifolds and cobordisms
manifolds and cobordisms
# Contents
## Idea
For $X$ a framed smooth manifold of dimension $4k+2$, $k\in ℕ$, the Kervaire invariant or Arf-Kervaire invariant
$\mathrm{Ker}\left(X\right)\in {ℤ}_{2}$Ker(X) \in \mathbb{Z}_2
with values in the group of order 2 is the Arf invariant? of the skew-quadratic form on the middle dimensional homology group.
## Properties
Manifolds with non-trivial Kervaire invariant, hence with Kervaire invariant 1, exist in dimension
• $d=2=4\cdot 0+2$
• $d=6=4\cdot 1+2$
• $d=14=4\cdot 3+2$
• $d=30=4\cdot 7+2$
• $d=62=4\cdot 15+2$
and in no other dimension, except possibly in $d=126$ (a case that is still open).
$4k$signature genusintersection pairingintegral Wu structure
$4k+2$Kervaire invariantframing | 2013-06-20 08:01:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9865298271179199, "perplexity": 5098.256299537075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368711005723/warc/CC-MAIN-20130516133005-00049-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/determinants-of-a-matrix.193941/ | Determinants of a matrix
1. Oct 25, 2007
JFonseka
1. The problem statement, all variables and given/known data
A 2 x 2 matrix B satisfies
B (3 1)$$^{T}$$ = B (5 2)$$^{T}$$
What is det (B) ? Give a reason
2. Relevant equations
None really
3. The attempt at a solution
I really have no idea how to start solving this. Does it involve inversing?
2. Oct 25, 2007
Dick
You are saying Bx=By. What's B(x-y)? Does that give you a hint?
3. Oct 25, 2007
JFonseka
So, It's B(-2 -1)$$^{T}$$
I don't get it though. That doesn't look like a determinant.
4. Oct 25, 2007
Dick
It's also equal to zero since Bx=By. It's not a determinant. But Bz=0 where z is a nonzero vector. What can that tell you about the determinant of B?
5. Oct 25, 2007
JFonseka
That the determinant is 0 ?
6. Oct 25, 2007
Dick
Yessssss.
7. Oct 25, 2007
JFonseka
How do we know B(x) = 0?
8. Oct 25, 2007
JFonseka
Ah nvm I see.
Thanks for the help | 2017-07-25 05:01:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4625386893749237, "perplexity": 2878.640788291715}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424961.59/warc/CC-MAIN-20170725042318-20170725062318-00058.warc.gz"} |
https://www.mathdoubts.com/derivative-of-sec-inverse-function/ | # Derivative Rule of Inverse Secant function
## Formula
$\dfrac{d}{dx}{\,\sec^{-1}{x}} \,=\, \dfrac{1}{|x|\sqrt{x^2-1}}$
### Introduction
Let $x$ represents a variable and also a real number, the inverse secant function is written as $\sec^{-1}{(x)}$ or $\operatorname{arcsec}{(x)}$ in inverse trigonometry. In differential calculus, the derivative or differentiation of the secant inverse function with respect to $x$ is written in two mathematical forms as follows.
$(1) \,\,\,$ $\dfrac{d}{dx}{\,\Big(\sec^{-1}{(x)}\Big)}$
$(2) \,\,\,$ $\dfrac{d}{dx}{\,\Big(\operatorname{arcsec}{(x)}\Big)}$
The derivative of the inverse secant function with respect to $x$ is equal to the reciprocal of product of modulus of $x$ and square root of the subtraction of one from $x$ squared.
$\implies$ $\dfrac{d}{dx}{\,\Big(\sec^{-1}{(x)}\Big)}$ $\,=\,$ $\dfrac{1}{|x|\sqrt{x^2-1}}$
##### Alternative forms
The derivative of secant inverse function can be written in terms of any variable. The following are some examples to learn how to write the derivative rule of inverse secant function in differential calculus.
$(1) \,\,\,$ $\dfrac{d}{dz}{\,\sec^{-1}{z}} \,=\, \dfrac{1}{|z|\sqrt{z^2-1}}$
$(2) \,\,\,$ $\dfrac{d}{dr}{\,\sec^{-1}{r}} \,=\, \dfrac{1}{|r|\sqrt{r^2-1}}$
$(3) \,\,\,$ $\dfrac{d}{dy}{\,\sec^{-1}{y}} \,=\, \dfrac{1}{|y|\sqrt{y^2-1}}$
#### Proof
Learn how to derive the differentiation formula for the inverse secant function by first principle.
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You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites. | 2022-05-20 15:10:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7143034934997559, "perplexity": 1203.1106464004285}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00568.warc.gz"} |
http://icpc.njust.edu.cn/Problem/Pku/3724/ | # Find the parameter
Time Limit: 10000MS
Memory Limit: 65536K
## Description
Panda is preparing for the mid-term examination by reading his notebook. However, some parts of his notebook have been ruined and some data are lost. For example, when he encounter a list of functions, he find out that all parameters are lost. The functions are similiar and can be written as y = exp(a0x) + exp(a1x) + ... + exp(a9x).Although all the ai are lost, the coordinates of some points on the curve of the function are known. Can you find back all the ai with these coordinates? Notice that all the ai are positive integers no more than 10 and aiai+1 (i=0,1,...,8). It is guaranteed to been only one correct answer.
## Input
The first line of input is N, the number of the coordinates. Each of the next N lines contains a pair of coordinates xi, yi separated by one space. 1 ≤ N ≤ 20 0 ≤ xi ≤ 5
## Output
The output contains 10 lines, each contains one ai in ascending order.
## Sample Input
20
0.200 12.214
0.400 14.918
0.600 18.221
0.800 22.255
1.000 27.183
1.200 33.201
1.400 40.552
1.600 49.530
1.800 60.496
2.000 73.891
2.200 90.250
2.400 110.232
2.600 134.637
2.800 164.446
3.000 200.855
3.200 245.325
3.400 299.641
3.600 365.982
3.800 447.012
4.000 545.982
## Sample Output
1
1
1
1
1
1
1
1
1
1
## Source
POJ Monthly Contest – 2009.04.05, Simon | 2019-01-23 07:33:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26093199849128723, "perplexity": 3968.4199991397513}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584203540.82/warc/CC-MAIN-20190123064911-20190123090911-00188.warc.gz"} |
https://codereview.stackexchange.com/questions/70536/protecting-webpage-text-from-being-selected | # Protecting webpage text from being selected
I've got a text I don't want to be copied by users.
To avoid that, I'm using this code:
index.php
<?php
session_start();
$characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';$randomString = '';
for ($i = 0;$i < 100; $i++) {$randomString .= $characters[rand(0, strlen($characters) - 1)];
}
$_SESSION['token'] =$randomString;
?>
<!DOCTYPE html>
<meta charset="utf-8">
<title>Test</title>
<script src="../jquery.js"></script>
<body>
<script>
$(document).ready(function(){$("#content").load("page.php?token=<?php echo $_SESSION['token']; ?>", function (responseText, textStatus, req) {if (textStatus == "error") {$("#content").html('An error occurred :(');
}
});
});
</script>
</body>
page.php
<?php
session_start();
if(!empty($_SESSION['token']) AND$_GET['token']==$_SESSION['token']){echo ' <script> function error(){ alert("This text is protected"); }$(document).ready(function(){
$(document).mousedown(function(e){ if( e.button == 2 ) { error(); return false; } });$(document).keydown(function(){
error();
return false;
});
});
</script>
Text...';
unset(\$_SESSION['token']);
}
else{echo 'Error.';
}
?>
Is this method reliable?
• why can't they just screenshot it and re-write it? – raptortech97 Nov 21 '14 at 19:30
• There's no reliable method to do this. They can just deactivate javascript. Even if you made images out of your text, there's plenty of tools that will extract text from images. – Michael Nov 21 '14 at 19:53
• watching the XHR requests you'll be able to copy the response as plain text, or are you concerned about normal copy/paste from the browser? – CᴴᵁᴮᴮʸNᴵᴺᴶᴬ Nov 21 '14 at 20:04
• @Michael I didn't think about OCR, thanks for pointing out that! – Stubborn Nov 21 '14 at 20:10
• @DannyHearnah No, I was trying to block every method (or at least the quickest ones) to copy the text, as the one you mentioned: I hadn't think about that, thanks for pointing it out! – Stubborn Nov 21 '14 at 20:12
That is an old skool question, the common wisdom is that you cannot prevent a user from getting to the text. There are a number of ways to get around your protection.
For example, in Chrome, open Developers Tools, open your site, check the web request, and then all the text will be perfectly copy pastable.
Other than that:
• alert() has been a bad practice for 20 years now, please use something else
• Indent your code, use jsbeautifier if you have trouble
• In the keydown, I would check whether the control key is pressed before throwing an error, unless the user never needs to press PageUp or PageDown
• I am not entire surely, but I think you will need event.preventDefault(); to really prevent right clicking etc. | 2019-10-22 20:32:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3580036759376526, "perplexity": 2681.471009379195}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00212.warc.gz"} |
https://physics.aps.org/articles/v11/s54 | # Synopsis: Revised Prediction for Mercury’s Orbit
Physics 11, s54
Mercury’s orbital ellipse is predicted to shift an additional ${1}^{\circ }$ every two billion years as a result of previously unaccounted for effects of general relativity.
Mercury’s orbit of the Sun isn’t fixed in space. Every 625 years, the ellipse shifts by ${1}^{\circ }$ because of its gravitational interactions with the planets and the Sun. Now Clifford Will from the University of Florida, Gainesville, has used the general theory of relativity to calculate the impact of indirect gravitational forces—such as the pull between the Sun and Jupiter—on Mercury’s orbit. He predicts that these additional forces add ${1}^{\circ }$of rotation to Mercury’s orbit every two billion years. Though tiny, and currently not measureable, this correction should be detectable by BepiColombo, a European and Japanese mission to Mercury scheduled to launch at the end of this year.
If Mercury were the only planet in the Solar System, its path around the Sun would stay fixed in space, according to Newtonian physics. But Mercury isn’t alone, and its Newtonian gravitational interactions with the other planets shift its orbit by 0.15 degrees per century (deg/cy). In addition, as Einstein famously predicted, general relativity affects the Sun-Mercury attraction and adds another 0.01 deg/cy to the planet’s orbital precession.
But the influence of the Sun doesn’t stop at Mercury. General relativistic effects of the Sun extend throughout the Solar System and alter the tug each planet exerts on Mercury. The theory also modifies the direct gravitational attraction between the planets and Mercury. Finally, the so-called gravitomagnetic force—a general relativistic effect of moving masses that is analogous to the magnetic force—also perturbs Mercury. Will predicts that these previously ignored effects should increase Mercury’s precession rate by about $6.2×{10}^{-8}$ deg/cy.
This research is published in Physical Review Letters.
–Katherine Wright
Katherine Wright is a Contributing Editor for Physics.
Gravitation
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https://zbmath.org/?q=an:06207353 | ## Real normalized differentials and Arbarello’s conjecture.(English. Russian original)Zbl 1278.30041
Funct. Anal. Appl. 46, No. 2, 110-120 (2012); translation from Funkts. Anal. Prilozh. 46, No. 2, 37-51 (2012).
Let $$M_g$$ denote the moduli space of compact Riemann surfaces of genus $$g\geq 2$$. About forty years ago, E. Arbarello considered in $$M_g$$ the natural flag $$W_2\subset W_3\subset\cdots\subset W_{g-1}\subset W_g= M_g$$ consisting of the subvarieties $$W_n=\{[C]\in M_g\mid\exists p\in C: h^0(C,{\mathcal O}_C(np)> 1\}$$ for $$2\leq n\leq g$$, the so-called Weierstrass flag in $$M_g$$. In the course of his study of the Weierstrass flag, he conjectured that any compact complex cycle in $$M_g$$ of dimension at least $$g-n$$ must intersect the $$(2g+ n-3)$$-dimensional, irreducible subvariety $$W_n$$, cf. [E. Arbarello, Compos. Math. 29, 325–342 (1974; Zbl 0355.14013)]. As $$W_2$$ is the affine subvariety of hyperelliptic curves in $$M_g$$, Arbarello’s conjecture implies that there are no compact complex cycles of dimension greater than $$g-2$$ in $$M_g$$. This statement was indeed proved by S. Diaz [Mem. Am. Math. Soc. 327, 69 p. (1985; Zbl 0581.14018)], while Arbarello’s general conjecture remained open until 2012. Actually, in the paper under review, the author finally presents a complete proof of Arbarello’s conjecture, the highly non-trivial nature of which is due to the fact that the higher Weierstrass loci $$W_n$$ are almost never affine, cf. [E. Arbarello and G. Mondello, Contemp. Math. 564, 137–144 (2012; Zbl 1255.14021)].
The author’s proof of Arbarello’s conjecture is based on applications of the so-called Whitham perturbation theory for integrable systems, which he himself had developed in the 1990s with a view toward topological quantum field theories and superymmetric gauge theories. Later on, the application of this framework to the study of moduli spaces of curves was initiated by I. M. Krichever and S. Grushevsky in their paper [in: Ji, Lizhen (ed.) et al., Geometry of Riemann surfaces and their moduli spaces. Somerville, MA: International Press. Surveys in Differential Geometry 14, 111–129 (2010; Zbl 1213.14055)]. Building upon this previous work in Whitham theory, especially on the central concept of real normalized meromorphic-differentials on Riemann surfaces, the author constructs certain foliations on moduli spaces of Riemann surfaces equipped with appropriate additional data, describes the corresponding leaves of these foliations defined by real normalized differentials as smooth complex varieties, and introduces then yet another new ingredient for the proof of Arbarello’s conjecture, namely the concept of cycles dual to critical points of real normalized meromorphic differentials on a Riemann surface. In this context, it is shown that the homology classes of these dual cycles generate the first homology group $$H_1(C,\mathbb{Z})$$ of the given surface $$C$$.
In the last section, the author gives the proof of Arbarello’s conjecture in its full generality, thereby making ingenious use of the new methods as developed in the previous sections. As an additional illustration of the power of these methods, the author has included a new proof of S. Diaz’s theorem (mentioned above) that is based on the properties of real normalized differentials with one pole of second-order. No doubt, this paper provides a decisive breakthrough in the study of the geometry of the Weierstrass flag in the moduli spaces of Riemann surfaces.
### MSC:
30F10 Compact Riemann surfaces and uniformization 30F30 Differentials on Riemann surfaces 14H10 Families, moduli of curves (algebraic) 14H55 Riemann surfaces; Weierstrass points; gap sequences 14H70 Relationships between algebraic curves and integrable systems 32G15 Moduli of Riemann surfaces, Teichmüller theory (complex-analytic aspects in several variables)
### Citations:
Zbl 0355.14013; Zbl 0581.14018; Zbl 1255.14021; Zbl 1213.14055
Full Text:
### References:
[1] E. Arbarello, ”Weierstrass points and moduli of curves,” Compositio Math., 29 (1974), 325–342. · Zbl 0355.14013 [2] E. Arbarello and G. Mondello, ”Two remarks on the Weierstrass flag,” in: Compact Moduli spaces and Vector Bundles, Contemp. Math. (Proceedings) series, vol. 564, Amer. Math. Soc., Providence, RI, 2012, 137–144. · Zbl 1255.14021 [3] Integrability: The Seiberg-Witten and Whitham Equations (eds. H. W. Braden and I. M. Krichever), Gordon and Breach Science Publishers, Amsterdam, 2000. [4] S. Diaz, ”Exceptional Weierstrass points and the divisor on moduli space that they define,” Mem. Amer. Math. Soc., 56:327 (1985). · Zbl 0581.14018 [5] A. Gorsky, I. Krichever, A. Marshakov, A. Mironov, and A. Morozov, ”Integrability and Seiberg-Witten exact solution,” Phys. Lett. B, 355 (1995), 466–474. · Zbl 0997.81567 [6] J. Harris and I. Morrison, Moduli of Curves, Graduate Texts in Math., vol. 187, Springer-Verlag, New York, 1998. · Zbl 0913.14005 [7] S. Grushevsky and I. Krichever, ”The universalWhitham hierarchy and geometry of the moduli space of pointed Riemann surfaces,” in: Surveys in Differ. Geom., vol. 14, Int. Press, Somerville, MA, 2010, 111–129. · Zbl 1213.14055 [8] S. Grushevsky and I. Krichever, Foliations on the Moduli Space of Curves, Vanishing in Cohomology, and Calogero-Moser Curves, http://arxiv.org/abs/1108.4211 . · Zbl 1213.14055 [9] I. M. Krichever, ”Spectral theory of finite-zone nonstationary Schrödinger operators. A nonstationary Peierls model,” Funkts. Anal. Prilozhen., 20:3 (1986), 42–54; English transl.: Funct. Anal. Appl., 20:3 (1986), 203–214. · Zbl 0637.35060 [10] I. M. Krichever, ”Method of an averaging for two-dimensional ”integrable” equations,” Funkts. Anal. Prilozhen., 22:3 (1988), 37–52; English transl.: Funct. Anal. Appl., 22:3 (1988), 200–213. · Zbl 0688.35088 [11] I. Krichever, ”The {$$\tau$$} -function of the universal Whitham hierarchy, matrix models, and topological field theories,” Comm. Pure Appl. Math., 47:4 (1994), 437–475. · Zbl 0811.58064 [12] I. Krichever and D. H. Phong, ”On the integrable geometry of soliton equations and N = 2 supersymmetric gauge theories,” J. Differential Geom., 45:2 (1997), 349–389. · Zbl 0889.58044 [13] I. Krichever and D. H. Phong, ”Symplectic forms in the theory of solitons,” in: Surveys in Differ. Geom., vol. 4, Int. Press, Boston, MA, 1998, 239–313. · Zbl 0931.35148 [14] E. Looijenga, ”On the tautological ring of Mg,” Invent. Math., 121:2 (1995), 411–419. · Zbl 0851.14017
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2023-02-03 10:52:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7048158645629883, "perplexity": 1241.5231422369916}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00137.warc.gz"} |
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# –1 < a < 0 < |a| < b < 1
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–1 < a < 0 < |a| < b < 1 [#permalink] 10 Aug 2018, 11:29
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$$–1 < a < 0 < |a| < b < 1$$
Quantity A Quantity B $$(\frac{a^2 \sqrt{b}}{\sqrt{a}})^2$$ $$\frac{ab^5}{(\sqrt{b})^4}$$
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 10 Sep 2018, 09:51
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obviously a<0 & b>0.
simplifying quantities A & B gives a^3*b and ab^3, respectively, both of which are negative.
taking a=-0.25, b=0.5, and plugging into the above simplified expressions, we get the answer A.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 16 Sep 2018, 13:05
saifee wrote:
obviously a<0 & b>0.
simplifying quantities A & B gives a^3*b and ab^3, respectively, both of which are negative.
taking a=-0.25, b=0.5, and plugging into the above simplified expressions, we get the answer A.
What if a= -(1/8) and b= (1/2) ??
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 24 Sep 2018, 08:16
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fixzion wrote:
I cant solve this, can someone help?
Both $$a$$ and $$b$$ are fractions. Now you have to select values of a and b such that $$b>a$$.
$$a= \frac{-1}{4}$$ and $$b= \frac{1}{2}$$.
Quantity A Quantity B $$(\frac{a^2 \sqrt{b}}{\sqrt{a}})^2$$ $$\frac{ab^5}{(\sqrt{b})^4}$$
or
Quantity A Quantity B $$(\frac{a^4 b}{a})$$ $$\frac{ab^5}{b^2}$$
or
Quantity A Quantity B $$a^3b$$ $$ab^3$$
$$a^3b=(\frac{-1}{4})^3\frac{1}{2}=\frac{-1}{128}$$
$$ab^3=\frac{-1}{4}(\frac{1}{2})^3=\frac{-1}{32}$$.
A is the larger number.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 16 Sep 2019, 18:09
sandy wrote:
fixzion wrote:
I cant solve this, can someone help?
Both $$a$$ and $$b$$ are fractions. Now you have to select values of a and b such that $$b>a$$.
$$a= \frac{-1}{4}$$ and $$b= \frac{1}{2}$$.
Quantity A Quantity B $$(\frac{a^2 \sqrt{b}}{\sqrt{a}})^2$$ $$\frac{ab^5}{(\sqrt{b})^4}$$
or
Quantity A Quantity B $$(\frac{a^4 b}{a})$$ $$\frac{ab^5}{b^2}$$
or
Quantity A Quantity B $$a^3b$$ $$ab^3$$
$$a^3b=(\frac{-1}{4})^3\frac{1}{2}=\frac{-1}{128}$$
$$ab^3=\frac{-1}{4}(\frac{1}{2})^3=\frac{-1}{32}$$.
A is the larger number.
Good one
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 23 Oct 2019, 14:40
sandy wrote:
fixzion wrote:
I cant solve this, can someone help?
Both $$a$$ and $$b$$ are fractions. Now you have to select values of a and b such that $$b>a$$.
$$a= \frac{-1}{4}$$ and $$b= \frac{1}{2}$$.
Quantity A Quantity B $$(\frac{a^2 \sqrt{b}}{\sqrt{a}})^2$$ $$\frac{ab^5}{(\sqrt{b})^4}$$
or
Quantity A Quantity B $$(\frac{a^4 b}{a})$$ $$\frac{ab^5}{b^2}$$
or
Quantity A Quantity B $$a^3b$$ $$ab^3$$
$$a^3b=(\frac{-1}{4})^3\frac{1}{2}=\frac{-1}{128}$$
$$ab^3=\frac{-1}{4}(\frac{1}{2})^3=\frac{-1}{32}$$.
A is the larger number.
Without putting in values to bring in the calculation, I think another way could be:
a^3.b a.b^3
so the comparison is between
a^2 & b^2
as we know |a| < b, and a,b both are fractions(positive or negative, doesn't really matter), 'b' is the larger fraction in magnitude hence would produce a smaller squared fraction
Is anything wrong with this approach?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 23 Oct 2019, 15:00
Expert's post
Not really
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 31 Oct 2019, 02:42
Can we plug in values for a and b with out simplifications?
we will have negative square root for a?
Last edited by mohmu123 on 31 Oct 2019, 02:59, edited 1 time in total.
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 31 Oct 2019, 02:48
mohmu123 wrote:
Can we plug in values for a and be with out simplifications?
we will have negative square root for a?
b, or be, which one u refer?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 31 Oct 2019, 03:00
I meant a and b
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 31 Oct 2019, 03:19
mohmu123 wrote:
Can we plug in values for a and b with out simplifications?
we will have negative square root for a?
Yes u can. For Negative value, just 1st put the only values without negative connotation then compare with ur result u got by putting values with the negative connotation which one u put aside before checking the values
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 01 Jan 2020, 02:27
sandy wrote:
fixzion wrote:
I cant solve this, can someone help?
Both $$a$$ and $$b$$ are fractions. Now you have to select values of a and b such that $$b>a$$.
$$a= \frac{-1}{4}$$ and $$b= \frac{1}{2}$$.
Quantity A Quantity B $$(\frac{a^2 \sqrt{b}}{\sqrt{a}})^2$$ $$\frac{ab^5}{(\sqrt{b})^4}$$
or
Quantity A Quantity B $$(\frac{a^4 b}{a})$$ $$\frac{ab^5}{b^2}$$
or
Quantity A Quantity B $$a^3b$$ $$ab^3$$
$$a^3b=(\frac{-1}{4})^3\frac{1}{2}=\frac{-1}{128}$$
$$ab^3=\frac{-1}{4}(\frac{1}{2})^3=\frac{-1}{32}$$.
A is the larger number.
why not more simplification? Like cancel a's from both sides and also b's ?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 24 Apr 2020, 04:44
1
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sandy wrote:
fixzion wrote:
I cant solve this, can someone help?
Both $$a$$ and $$b$$ are fractions. Now you have to select values of a and b such that $$b>a$$.
$$a= \frac{-1}{4}$$ and $$b= \frac{1}{2}$$.
Quantity A Quantity B $$(\frac{a^2 \sqrt{b}}{\sqrt{a}})^2$$ $$\frac{ab^5}{(\sqrt{b})^4}$$
or
Quantity A Quantity B $$(\frac{a^4 b}{a})$$ $$\frac{ab^5}{b^2}$$
or
Quantity A Quantity B $$a^3b$$ $$ab^3$$
$$a^3b=(\frac{-1}{4})^3\frac{1}{2}=\frac{-1}{128}$$
$$ab^3=\frac{-1}{4}(\frac{1}{2})^3=\frac{-1}{32}$$.
A is the larger number.
Why don't we simplify this further by dividing both sides by ab? In that case we will have to compare a^2 with b^2 and there the answer is B. What am I missing here? Can anyone explain?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 24 Apr 2020, 05:59
Expert's post
The fastest and simple approach is that above. reduce to a minimum term the two Qs
Then see which is greater.
regards
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 26 Apr 2020, 16:10
Ans choice A is right beacouse sum of A is greater then B?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 26 Apr 2020, 16:35
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Ans choice A is right beacouse sum of A is greater then B?
Posted from my mobile device
Here's what each answer choice means:
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Here's a video overview of this question type:
Cheers,
Brent
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 04 May 2020, 05:22
Carcass wrote:
The fastest and simple approach is that above. reduce to a minimum term the two Qs
The see which is greater.
regards
It comes down to a^2 & b^2
And here B is greater than A
What am I missing?
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 04 May 2020, 11:51
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Raj30 wrote:
Carcass wrote:
The fastest and simple approach is that above. reduce to a minimum term the two Qs
The see which is greater.
regards
It comes down to a^2 & b^2
And here B is greater than A
What am I missing?
$$((a^2 √b)/√a)^2$$ VS $$(ab^5)/(√b)^4$$
$$(a^4 b)/a$$ VS $$(ab^5)/b^2$$
$$a^3*b$$ VS $$a*b^3$$ … (i)
$$–1<a<0<|a|<b<1$$ => Thus, a is negative and bis positive
We have: $$|a|<b$$
Squaring: $$a^2<b^2$$
Multiply b: $$a^2 b<b^3$$
Multiply a (negative; hence reverse inequality): $$a^3 b>ab^3$$
Thus: A is greater
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 06 May 2020, 08:34
what about the a in the bar
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Re: –1 < a < 0 < |a| < b < 1 [#permalink] 12 May 2020, 11:56
After simplifying both sides quantity A is a^2 while quantity B is b^2.
if i take the value of a to be -1/2 and b to be 3/4, after calculation im getting quantity B to be greater. Can someone tell me where im going wrong?
Re: –1 < a < 0 < |a| < b < 1 [#permalink] 12 May 2020, 11:56
Display posts from previous: Sort by | 2020-09-29 16:39:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28298071026802063, "perplexity": 7132.507816503976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00601.warc.gz"} |
http://www.mathnet.ru/php/contents.phtml?wshow=issue&jrnid=mzm&series=0&year=1981&volume=29&volume_alt=&issue=2&issue_alt=&option_lang=eng | RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
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Connection between the 2-length and the derived length of a Sylow 2-subgroup of a finite solvable groupE. G. Bryukhanova 161 Semiperfect rings with nilpotent quasiregular groupV. A. Ratinov 171 Variation of the norms in the problem of best approximationV. I. Berdyshev 181 Computation of indefinite integrals involving certain special functionsYu. F. Filippov 197 Estimates and asymptotic formulas for generalized Faber polynomials in two variablesM. M. Tsvil' 201 Mean property for the heat-conduction equationL. P. Kuptsov 211 Description of the self-adjoint extensions of quasiregular operators generated by differential expressions with two termsG. A. Mirzoev 225 Necessary and sufficient conditions for quasipower basicity of certain systems of regular functionsV. A. Oskolkov 235 Manifolds of nonpositive curvature with small volumeS. V. Buyalo 243 Bonnet's theorem for geometric structuresP. Ya. Grushko 253 $K$-algebras and $K$-spaces of constant type with indefinite metricV. F. Kirichenko 265 Affine connections of second orderA. K. Rybnikov 279 Small deviations of Gaussian processA. A. Novikov 291 An extremal problem on partition of numbersV. I. Baranov 303 Multidimensional Jackson theorem in $L_2$V. A. Yudin 309 | 2021-04-13 04:37:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34418362379074097, "perplexity": 2822.4454427150336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00161.warc.gz"} |
http://www.dml.cz/handle/10338.dmlcz/128480 | # Article
Full entry | PDF (1.3 MB)
References:
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[A2] Albrecht, U.: Abelian groups $A$ such that the category of $A$-solvable groups is preabelian. Contemporary Mathematics 87 (1989), 117–131. MR 0995270 | Zbl 0691.20038
[A4] Albrecht, U.: Endomorphism rings and a generalization of torsion-freeness and purity. Communications in Algebra 17 (5) (1989), 1101–1135. MR 0993391 | Zbl 0691.20040
[ACH] Albrecht, U.: Extension functors on the category of $A$-solvable abelian groups. Czech. Math. J. 41 (116) (1991), 685–694. MR 1134957 | Zbl 0776.20018
[AWM] Albrecht, U.: Endomorphism rings and Fuchs’ Problem 47. (to appear).
[AH] Albrecht, U., and Hausen, J.: Modules with the quasi-summand intersection property. Bull. Austral. Math. Soc. 44 (1991), 189–201. MR 1126356
[AL] Arnold, D., and Lady, L.: Endomorphism rings and direct sums of torsion-free abelian groups. Trans. Amer. Math. Soc. 211 (1975), 225–237. MR 0417314
[AM] Arnold, D., and Murley, E.: Abelian groups, $A$, such that $\mathop {\mathrm Hom}\nolimits (A,-)$ preserves direct sums of copies of $A$. Pac. J. of Math. 56 (1975), 7–20. MR 0376901
[DG] Dugas, M., and Göbel, R.: Every cotorsion-free ring is an endomorphism ring. Proc. London Math. Soc. 45 (1982), 319–336. MR 0670040
[F] Faticoni, T.: Semi-local localization of rings and subdirect decomposition of modules. J. of Pure and Appl. Alg. 46 (1987), 137–163.
[FG] Faticoni, T., and Goeters, P.: Examples of torsion-free abelian groups flat as modules over their endomorphism rings. Comm. in Algebra 19 (1991), 1–27. MR 1092548
[FG1] Faticoni, T., and Goeters, P.: On torsion-free $\mathop {\mathrm Ext}\nolimits$. Comm. in Algebra 16 (9) (1988), 1853–1876. MR 0952214
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Partner of | 2014-11-01 13:11:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9676670432090759, "perplexity": 2453.7426573463867}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637906433.44/warc/CC-MAIN-20141030025826-00239-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-8-polynomials-and-factoring-chapter-review-8-5-and-8-6-factoring-quadratic-trinomials-page-525/44 | ## Algebra 1
$2(k-l)(3k+2l)$
Factor each expression. $6k^{2}-10kl+4l^{2}$ Factor out $2$ $2(3k^{2}-5kl+2l^{2}$ Rewrite as: $2(3k^{2}-2kl-3kl+2l^{2}$ Factor: $2(k(3k-2l)-l(3k+2l))$ Simplify: $2(k-l)(3k+2l)$ | 2018-09-22 19:00:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9259268045425415, "perplexity": 3911.428721948968}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158633.40/warc/CC-MAIN-20180922182020-20180922202420-00000.warc.gz"} |
https://articles.outlier.org/what-is-the-squeeze-theorem | Calculus
# Understanding the Squeeze Theorem
## Rachel McLean
Subject Matter Expert
Learn what the squeeze theorem is, how to prove it, and practice with some examples and tips.
## What Is the Squeeze Theorem?
The Squeeze Theorem is a method for evaluating the limit of a function. Also known as the Sandwich Theorem, the Squeeze Theorem traps one tricky function whose limit is hard to evaluate, between two different functions whose limits are easier to evaluate.
To introduce the logic behind this theorem, let’s recall a familiar algebraic property. This rule is similar in logic to the transitive property of equality in algebra. For example, if $a \leq b \leq c$ and $a = c$, then $b$ also equals $c$.
In the same vein of thought, if $g(x) \leq f(x) \leq h(x)$ for all $x$ on some open interval containing $c$, and $\lim_{x\to c}g(x) = \lim_{x\to c}h(x) = L$, then $\lim_{x\to c}f(x) = L$. This is the Squeeze Theorem definition.
The following figure provides a graphical proof of the Squeeze Theorem. Do you see how “squeezing” or “sandwiching” a hard function between two easier functions allows us to find its limit?
## Squeeze Theorem Proof
If graphical proofs aren’t your jam, we can also prove the Squeeze Theorem using the epsilon-delta ($\epsilon - \delta$) definition of limits. The $\epsilon - \delta$ definition goes like this:
Let the function f(x) be defined on an open interval, and let c be on this interval. Then \lim_{x\to c}f(x) = L if for all \epsilon > 0, there exists some \delta > 0 such that when 0 < |x - c | < \delta, then this immediately implies that |f(x) - L| < \epsilon.
To begin our proof, let’s start with our assumptions: $g(x) \leq f(x) \leq h(x)$ for all $x$ on some open interval containing $c$, and $\lim_{x\to c}g(x) = \lim_{x\to c}h(x) = L$. We want to prove that $\lim_{x\to c}f(x) = L$ using the above definition.
Using this definition, our second assumption tells us that for all \epsilon > 0, there exists \delta_1 > 0 such that when 0 < |x - c | < \delta_1, then |g(x) - L| < \epsilon. Similarly, for all \epsilon > 0, there exists \delta_2 > 0 such that when 0 < |x - c | < \delta_2, then |h(x) - L| < \epsilon.
We’ll let \delta = \min \{\delta_1, \delta_2\}. This implies that when 0 < |x - c | < \delta, then |g(x) - L| < \epsilon and |h(x) - L| < \epsilon. We can rewrite this to say -\epsilon < g(x) - L< \epsilon and -\epsilon < h(x) - L < \epsilon. Adding L to each side, if 0 < |x - c | < \delta, then L -\epsilon < g(x) < L + \epsilon and L -\epsilon < h(x) < L + \epsilon.
Applying these new inequalities to our first assumption, we get L - \epsilon < g(x) \leq f(x) \leq h(x) < L + \epsilon, which implies that L - \epsilon < f(x) < L + \epsilon. Subtracting L from each side, we get - \epsilon < f(x) - L < \epsilon. We can rewrite this as |f(x) - L| < \epsilon.
This concludes our proof. Thus, $\lim_{x\to c}f(x) = L$.
## Step-By-Step Guide to the Squeeze Theorem
How do you use the squeeze theorem? Well, if $f(x)$ is sandwiched between two functions $g(x)$ and $h(x)$ near $c$, and if $g(x)$ and $h(x)$ have the same limit $L$ at $c$, then $f(x)$ is trapped between them and must also have the same limit $L$ at $c$!
Here are 4 simple steps to using the Squeeze Theorem:
1. Use equalities to find two functions $g(x)$ and $h(x)$.
2. Show that $f(x)$ lies between $g(x)$ and $h(x)$ near $c$.
3. Show that $\lim_{x\to c}g(x) = L$.
4. Show that $\lim_{x\to c}h(x) = L$.
## Squeeze Theorem Examples
We’ll walk through a few examples of the squeeze theorem together.
### 1. Use the Squeeze Theorem To Evaluate
Use the Squeeze Theorem to evaluate $\lim_{x\to0}f(x)$ where $f(x) = x^2\cos(\frac{1}{x^2})$.
How can we sandwich $f(x)$ between two other functions? Let’s use equalities. For now, we’ll just focus on $\cos{(\frac{1}{x^2})}$. The fraction might look daunting, but since this is an oscillating cosine function, we know the following statement is true:
$-1 \leq \cos (\frac{1}{x^2}) \leq 1$
Multiply all sides by $x^2$ to get our original function in the middle.
This gives us:
$-x^2 \leq x^2\cos (\frac{1}{x^2}) \leq x^2$
This equality gives us two functions we can use, $g(x) = -x^2$ and $h(x) = x^2$. Our next step is to take the limit of $g(x)$ and $h(x)$ as $x$ approaches 0, and see if their limits match:
$\lim_{x\to 0}(-x^2) = \text{0 and} \lim_{x\to 0}(x^2) = 0$
Their limits match! We’ve now shown $f(x)$ is squeezed between $g(x)$ and $h(x)$, and $\lim_{x\to 0 }g(x) = \lim_{x\to 0 }h(x) = 0$.
By the Squeeze Theorem, we can conclude:
$0 \leq \lim_{x\to 0 }x^2 \cos{(\frac{1}{x^2})} \leq 0$
Our answer is $\lim_{x\to 0 }f(x) = 0$.
### 2. Use the Squeeze Theorem To Compute
Use the Squeeze Theorem to compute $lim_{x\to0}f(x)$ where $f(x) = x \sin{(\frac{1}{x})}$.
We want to pinch f(x) between two functions whose limits are more easily evaluated at x = 0. Let’s break this down. We’ll ignore x for now, and just focus on \sin{(\frac{1}{x})}.
Since this is a sine function, we know that:
$-1 \leq \sin{(\frac{1}{x})} \leq 1$
We have to be careful here. We can’t multiply both sides by $x$, since $x$ can be negative and our inequality will be thrown off.
We’ll rewrite our expression using absolute value instead:
$0 \leq |\sin{(\frac{1}{x})}| \leq 1$
Now, we’ll multiply both sides of our equation by $|x|$:
$0 \leq |x| \cdot |\sin{(\frac{1}{x})}| \leq |x|$
$0 \leq |x \sin{(\frac{1}{x})}| \leq |x|$
We now have two functions we can use, $g(x) = 0$ and $h(x) = |x|$. Let’s take the limit of $g(x)$ and $h(x)$ as $x$ approaches 0, and see if these limits match:
$\lim_{x\to 0 }0 = \text{0 and} \lim_{x\to 0 }|x| = 0$
The limits match! We’ve shown that g(x) < |f(x)| < h(x) and \lim_{x\to 0 }g(x) = \lim_{x\to 0 }h(x) = 0.
By the Squeeze Theorem, we can conclude:
$0 \leq \lim_{x\to 0 }|x \sin{(\frac{1}{x})}| \leq 0$
Therefore, $\lim_{x\to 0 }|x \sin{(\frac{1}{x})}| = 0$. Since this answer is 0, this immediately implies that $\lim_{x\to 0 }x \sin{(\frac{1}{x})} = 0$.
## 5 Tips When Using the Squeeze Theorem
Here are a few additional tips to consider when using the Squeeze Theorem.
1. What functions is this pinching theorem most used for? This method is especially useful for oscillating sine and cosine functions, as well as other trigonometric functions.
2. We use the Squeeze Theorem when other methods don’t work, such as factoring, trigonometry substitutions, rationalization, or other algebraic manipulations.
3. It is a good idea to be familiar with radians and the unit circle. Many Squeeze Theorem problems will use trig functions.
4. When in doubt, graph it! Is $f(x)$ sandwiched between $g(x)$ and $h(x)$ on the necessary interval? Do their limits look the same at the right particular point? Graphing can help validate your answer.
5. Remember, the limits of your lower and upper bounds must match. Otherwise the theorem can’t be used.
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Explore course | 2023-03-27 09:35:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 76, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9575804471969604, "perplexity": 414.9263812065376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00793.warc.gz"} |
http://nuit-blanche.blogspot.com/2011/02/on-local-correctness-of-l1-minimization.html | ## Thursday, February 03, 2011
### On the Local Correctness of L^1 Minimization for Dictionary Learning, A Panorama on Multiscale Geometric Representations, Intertwining Spatial, Directional and Frequency Selectivity
At the SMALL meeting, there was a presentation by John Wright, ( here is the video and the attendant presentation ), today we have the attendant paper: On the Local Correctness of L^1 Minimization for Dictionary Learning by Quan Geng, Huan Wang, John Wright. The abstract reads:
The idea that many important classes of signals can be well-represented by linear combinations of a small set of atoms selected from a given dictionary has had dramatic impact on the theory and practice of signal processing. For practical problems in which an appropriate sparsifying dictionary is not known ahead of time, a very popular and successful heuristic is to search for a dictionary that minimizes an appropriate sparsity surrogate over a given set of sample data. While this idea is appealing, the behavior of these algorithms is largely a mystery; although there is a body of empirical evidence suggesting they do learn very effective representations, there is little theory to guarantee when they will behave correctly, or when the learned dictionary can be expected to generalize. In this paper, we take a step towards such a theory. We show that under mild hypotheses, the dictionary learning problem is locally well-posed: the desired solution is indeed a local minimum of the $\ell^1$ norm. Namely, if $\mb A \in \Re^{m \times n}$ is an incoherent (and possibly overcomplete) dictionary, and the coefficients $\mb X \in \Re^{n \times p}$ follow a random sparse model, then with high probability $(\mb A,\mb X)$ is a local minimum of the $\ell^1$ norm over the manifold of factorizations $(\mb A',\mb X')$ satisfying $\mb A' \mb X' = \mb Y$, provided the number of samples $p = \Omega(n^3 k)$. For overcomplete $\mb A$, this is the first result showing that the dictionary learning problem is locally solvable. Our analysis draws on tools developed for the problem of completing a low-rank matrix from a small subset of its entries, which allow us to overcome a number of technical obstacles; in particular, the absence of the restricted isometry property.
This is a very interesting paper. An eloquent introduction states:
This idea has several appeals: Given the recent proliferation of new and exotic types of data (images, videos, web and bioinformatic data, ect.), it may not be possible to invest the intellectual e ort required to develop optimal representations for each new class of signal we encounter. At the same time, data are becoming increasingly high-dimensional, a fact which stretches the limitations of our human intuition, potentially limiting our ability to develop e ective data representations. It may be possible for an automatic procedure to discover useful structure in the data that is not readily apparent to us.
so we are getting closer to Skynet :). Anyway, I note that once more the work of David Gross has helped a lot once more and that this work has a direct impact on calibration type of exercises.
On a different note, here is a paper on dictionaries we have built in the past few years based on our intuition:; A Panorama on Multiscale Geometric Representations, Intertwining Spatial, Directional and Frequency Selectivity by Laurent Jacques, Laurent Duval, Caroline Chaux, Gabriel Peyré. The abstract reads:
The richness of natural images makes the quest for optimal representations in image processing and computer vision challenging. This fact has not prevented the design of candidates, convenient for rendering smooth regions, contours and textures at the same time, with compromises between efficiency and complexity. The most recent ones, proposed in the past decade, share an hybrid heritage highlighting the multiscale and oriented nature of edges and patterns in images. This paper endeavors a panorama of the aforementioned literature on decompositions in multiscale, multi-orientation bases or dictionaries. They typically exhibit redundancy to improve both the sparsity of the representation, and sometimes its invariance to various geometric deformations. Oriented multiscale dictionaries extend traditional wavelet processing and may offer rotation invariance. Highly redundant dictionaries require specific algorithms to simplify the search for an efficient (sparse) representation. We also discuss the extension of multiscale geometric decompositions to non-Euclidean domains such as the sphere $S^2$ and arbitrary meshed surfaces in $\Rbb^3$. The apt and patented etymology of panorama suggests an overview based on a choice of overlapping "pictures", i.e., selected from a broad set of computationally efficient mathematical tools. We hope this work help enlighten a substantial fraction of the present exciting research in image understanding, targeted to better capture data diversity. | 2017-09-23 07:34:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4768373966217041, "perplexity": 727.4989881760275}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689572.75/warc/CC-MAIN-20170923070853-20170923090853-00418.warc.gz"} |
https://socratic.org/questions/euclid-showed-geometrically-the-distributed-law-of-multiplication-let-bar-ab-and | # Euclid showed geometrically the distributed law of multiplication. Let bar(AB) and bar(BC) be two straight lines, tracing a rectangle and let bar(BC)be cut at random at the points D & E. Use this fact to show the distributed law?
Sep 13, 2016
#### Explanation:
Let $B D = p$, $D E = q$ and $E C = r$
then $B C = p + q + r$. Also let $A B = a$ and hence $A B = D H = E J = a$.
Therefore area of rectangle $A B C N = A B \times B C = a \times \left(p + q + r\right)$ as $B C = B D + D E + E C = p + q + r$
Also area of rectangles $A B D H$, $D E J H$ and $E C N J$ are
$A B D H = A B \times B D = a \times p$
$D E J H = D H \times D E = A B \times D E = a \times q$
$E C N J = E J \times E C = A B \times E C = a \times r$
it is evident from the image that area of rectangle $A B C N$ is sum of areas of rectangles $A B D H$, $D E J H$ and $E C N J$.
Hence, $a \times \left(p + q + r\right) = a \times p + a \times q + a \times r$
which is nothing but the distributive law. | 2019-12-08 03:17:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 19, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9487910866737366, "perplexity": 629.8070339073532}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00200.warc.gz"} |
https://fr.mathworks.com/help/fusion/ref/imufilter-system-object.html | # imufilter
Orientation from accelerometer and gyroscope readings
## Description
The `imufilter` System object™ fuses accelerometer and gyroscope sensor data to estimate device orientation.
To estimate device orientation:
1. Create the `imufilter` object and set its properties.
2. Call the object with arguments, as if it were a function.
## Creation
### Syntax
``FUSE = imufilter``
``FUSE = imufilter('ReferenceFrame',RF)``
``FUSE = imufilter(___,Name,Value)``
### Description
example
````FUSE = imufilter` returns an indirect Kalman filter System object, `FUSE`, for fusion of accelerometer and gyroscope data to estimate device orientation. The filter uses a nine-element state vector to track error in the orientation estimate, the gyroscope bias estimate, and the linear acceleration estimate.```
````FUSE = imufilter('ReferenceFrame',RF)` returns an `imufilter` filter System object that fuses accelerometer and gyroscope data to estimate device orientation relative to the reference frame `RF`. Specify `RF` as `'NED'` (North-East-Down) or `'ENU'` (East-North-Up). The default value is `'NED'`.```
example
````FUSE = imufilter(___,Name,Value)` sets each property `Name` to the specified `Value`. Unspecified properties have default values.Example: `FUSE = imufilter('SampleRate',200,'GyroscopeNoise',1e-6)` creates a System object, `FUSE`, with a 200 Hz sample rate and gyroscope noise set to 1e-6 radians per second squared.```
## Properties
expand all
Unless otherwise indicated, properties are nontunable, which means you cannot change their values after calling the object. Objects lock when you call them, and the `release` function unlocks them.
If a property is tunable, you can change its value at any time.
Sample rate of the input sensor data in Hz, specified as a positive finite scalar.
Tunable: No
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Decimation factor by which to reduce the sample rate of the input sensor data, specified as a positive integer scalar.
The number of rows of the inputs, `accelReadings` and `gyroReadings`, must be a multiple of the decimation factor.
Tunable: No
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Variance of accelerometer signal noise in (m/s2)2, specified as a positive real scalar.
Tunable: Yes
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Variance of gyroscope signal noise in (rad/s)2, specified as a positive real scalar.
Tunable: Yes
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Variance of gyroscope offset drift in (rad/s)2, specified as a positive real scalar.
Tunable: Yes
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Variance of linear acceleration noise in (m/s2)2, specified as a positive real scalar. Linear acceleration is modeled as a lowpass filtered white noise process.
Tunable: Yes
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Decay factor for linear acceleration drift, specified as a scalar in the range [0,1]. If linear acceleration is changing quickly, set `LinearAccelerationDecayFactor` to a lower value. If linear acceleration changes slowly, set `LinearAccelerationDecayFactor` to a higher value. Linear acceleration drift is modeled as a lowpass-filtered white noise process.
Tunable: Yes
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Covariance matrix for process noise, specified as a 9-by-9 matrix. The default is:
``` Columns 1 through 6 0.000006092348396 0 0 0 0 0 0 0.000006092348396 0 0 0 0 0 0 0.000006092348396 0 0 0 0 0 0 0.000076154354947 0 0 0 0 0 0 0.000076154354947 0 0 0 0 0 0 0.000076154354947 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 7 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.009623610000000 0 0 0 0.009623610000000 0 0 0 0.009623610000000```
The initial process covariance matrix accounts for the error in the process model.
Data Types: `single` | `double` | `uint8` | `uint16` | `uint32` | `uint64` | `int8` | `int16` | `int32` | `int64`
Output orientation format, specified as `'quaternion'` or `'Rotation matrix'`. The size of the output depends on the input size, N, and the output orientation format:
• `'quaternion'` –– Output is an N-by-1 `quaternion`.
• `'Rotation matrix'` –– Output is a 3-by-3-by-N rotation matrix.
Data Types: `char` | `string`
## Usage
### Syntax
``[orientation,angularVelocity] = FUSE(accelReadings,gyroReadings)``
### Description
example
````[orientation,angularVelocity] = FUSE(accelReadings,gyroReadings)` fuses accelerometer and gyroscope readings to compute orientation and angular velocity measurements. The algorithm assumes that the device is stationary before the first call.```
### Input Arguments
expand all
Accelerometer readings in the sensor body coordinate system in m/s2, specified as an N-by-3 matrix. N is the number of samples, and the three columns of `accelReadings` represent the [x y z] measurements. Accelerometer readings are assumed to correspond to the sample rate specified by the SampleRate property.
Data Types: `single` | `double`
Gyroscope readings in the sensor body coordinate system in rad/s, specified as an N-by-3 matrix. N is the number of samples, and the three columns of `gyroReadings` represent the [x y z] measurements. Gyroscope readings are assumed to correspond to the sample rate specified by the SampleRate property.
Data Types: `single` | `double`
### Output Arguments
expand all
Orientation that can rotate quantities from a global coordinate system to a body coordinate system, returned as quaternions or an array. The size and type of `orientation` depends on whether the OrientationFormat property is set to `'quaternion'` or ```'Rotation matrix'```:
• `'quaternion'` –– The output is an M-by-1 vector of quaternions, with the same underlying data type as the inputs.
• `'Rotation matrix'` –– The output is a 3-by-3-by-M array of rotation matrices the same data type as the inputs.
The number of input samples, N, and the DecimationFactor property determine M.
You can use `orientation` in a `rotateframe` function to rotate quantities from a global coordinate system to a sensor body coordinate system.
Data Types: `quaternion` | `single` | `double`
Angular velocity with gyroscope bias removed in the sensor body coordinate system in rad/s, returned as an M-by-3 array. The number of input samples, N, and the DecimationFactor property determine M.
Data Types: `single` | `double`
## Object Functions
To use an object function, specify the System object as the first input argument. For example, to release system resources of a System object named `obj`, use this syntax:
`release(obj)`
expand all
`tune` Tune `imufilter` parameters to reduce estimation error
`step` Run System object algorithm `release` Release resources and allow changes to System object property values and input characteristics `reset` Reset internal states of System object
## Examples
collapse all
Load the `rpy_9axis` file, which contains recorded accelerometer, gyroscope, and magnetometer sensor data from a device oscillating in pitch (around y-axis), then yaw (around z-axis), and then roll (around x-axis). The file also contains the sample rate of the recording.
```load 'rpy_9axis.mat' sensorData Fs accelerometerReadings = sensorData.Acceleration; gyroscopeReadings = sensorData.AngularVelocity;```
Create an `imufilter` System object™ with sample rate set to the sample rate of the sensor data. Specify a decimation factor of two to reduce the computational cost of the algorithm.
```decim = 2; fuse = imufilter('SampleRate',Fs,'DecimationFactor',decim);```
Pass the accelerometer readings and gyroscope readings to the `imufilter` object, `fuse`, to output an estimate of the sensor body orientation over time. By default, the orientation is output as a vector of quaternions.
`q = fuse(accelerometerReadings,gyroscopeReadings);`
Orientation is defined by the angular displacement required to rotate a parent coordinate system to a child coordinate system. Plot the orientation in Euler angles in degrees over time.
`imufilter` fusion correctly estimates the change in orientation from an assumed north-facing initial orientation. However, the device's x-axis was pointing southward when recorded. To correctly estimate the orientation relative to the true initial orientation or relative to NED, use `ahrsfilter`.
```time = (0:decim:size(accelerometerReadings,1)-1)/Fs; plot(time,eulerd(q,'ZYX','frame')) title('Orientation Estimate') legend('Z-axis', 'Y-axis', 'X-axis') xlabel('Time (s)') ylabel('Rotation (degrees)')```
Model a tilting IMU that contains an accelerometer and gyroscope using the `imuSensor` System object™. Use ideal and realistic models to compare the results of orientation tracking using the `imufilter` System object.
Load a struct describing ground-truth motion and a sample rate. The motion struct describes sequential rotations:
1. yaw: 120 degrees over two seconds
2. pitch: 60 degrees over one second
3. roll: 30 degrees over one-half second
4. roll: -30 degrees over one-half second
5. pitch: -60 degrees over one second
6. yaw: -120 degrees over two seconds
In the last stage, the motion struct combines the 1st, 2nd, and 3rd rotations into a single-axis rotation. The acceleration, angular velocity, and orientation are defined in the local NED coordinate system.
```load y120p60r30.mat motion fs accNED = motion.Acceleration; angVelNED = motion.AngularVelocity; orientationNED = motion.Orientation; numSamples = size(motion.Orientation,1); t = (0:(numSamples-1)).'/fs; ```
Create an ideal IMU sensor object and a default IMU filter object.
```IMU = imuSensor('accel-gyro','SampleRate',fs); aFilter = imufilter('SampleRate',fs); ```
In a loop:
1. Simulate IMU output by feeding the ground-truth motion to the IMU sensor object.
2. Filter the IMU output using the default IMU filter object.
```orientation = zeros(numSamples,1,'quaternion'); for i = 1:numSamples [accelBody,gyroBody] = IMU(accNED(i,:),angVelNED(i,:),orientationNED(i,:)); orientation(i) = aFilter(accelBody,gyroBody); end release(aFilter) ```
Plot the orientation over time.
```figure(1) plot(t,eulerd(orientation,'ZYX','frame')) xlabel('Time (s)') ylabel('Rotation (degrees)') title('Orientation Estimation -- Ideal IMU Data, Default IMU Filter') legend('Z-axis','Y-axis','X-axis') ```
Modify properties of your `imuSensor` to model real-world sensors. Run the loop again and plot the orientation estimate over time.
```IMU.Accelerometer = accelparams( ... 'MeasurementRange',19.62, ... 'Resolution',0.00059875, ... 'ConstantBias',0.4905, ... 'AxesMisalignment',2, ... 'NoiseDensity',0.003924, ... 'BiasInstability',0, ... 'TemperatureBias', [0.34335 0.34335 0.5886], ... 'TemperatureScaleFactor',0.02); IMU.Gyroscope = gyroparams( ... 'MeasurementRange',4.3633, ... 'Resolution',0.00013323, ... 'AxesMisalignment',2, ... 'NoiseDensity',8.7266e-05, ... 'TemperatureBias',0.34907, ... 'TemperatureScaleFactor',0.02, ... 'AccelerationBias',0.00017809, ... 'ConstantBias',[0.3491,0.5,0]); orientationDefault = zeros(numSamples,1,'quaternion'); for i = 1:numSamples [accelBody,gyroBody] = IMU(accNED(i,:),angVelNED(i,:),orientationNED(i,:)); orientationDefault(i) = aFilter(accelBody,gyroBody); end release(aFilter) figure(2) plot(t,eulerd(orientationDefault,'ZYX','frame')) xlabel('Time (s)') ylabel('Rotation (degrees)') title('Orientation Estimation -- Realistic IMU Data, Default IMU Filter') legend('Z-axis','Y-axis','X-axis') ```
The ability of the `imufilter` to track the ground-truth data is significantly reduced when modeling a realistic IMU. To improve performance, modify properties of your `imufilter` object. These values were determined empirically. Run the loop again and plot the orientation estimate over time.
```aFilter.GyroscopeNoise = 7.6154e-7; aFilter.AccelerometerNoise = 0.0015398; aFilter.GyroscopeDriftNoise = 3.0462e-12; aFilter.LinearAccelerationNoise = 0.00096236; aFilter.InitialProcessNoise = aFilter.InitialProcessNoise*10; orientationNondefault = zeros(numSamples,1,'quaternion'); for i = 1:numSamples [accelBody,gyroBody] = IMU(accNED(i,:),angVelNED(i,:),orientationNED(i,:)); orientationNondefault(i) = aFilter(accelBody,gyroBody); end release(aFilter) figure(3) plot(t,eulerd(orientationNondefault,'ZYX','frame')) xlabel('Time (s)') ylabel('Rotation (degrees)') title('Orientation Estimation -- Realistic IMU Data, Nondefault IMU Filter') legend('Z-axis','Y-axis','X-axis') ```
To quantify the improved performance of the modified `imufilter`, plot the quaternion distance between the ground-truth motion and the orientation as returned by the `imufilter` with default and nondefault properties.
```qDistDefault = rad2deg(dist(orientationNED,orientationDefault)); qDistNondefault = rad2deg(dist(orientationNED,orientationNondefault)); figure(4) plot(t,[qDistDefault,qDistNondefault]) title('Quaternion Distance from True Orientation') legend('Realistic IMU Data, Default IMU Filter', ... 'Realistic IMU Data, Nondefault IMU Filter') xlabel('Time (s)') ylabel('Quaternion Distance (degrees)') ```
This example shows how to remove gyroscope bias from an IMU using `imufilter`.
Use `kinematicTrajectory` to create a trajectory with two parts. The first part has a constant angular velocity about the y- and z-axes. The second part has a varying angular velocity in all three axes.
```duration = 60*8; fs = 20; numSamples = duration * fs; rng('default') % Seed the RNG to reproduce noisy sensor measurements. initialAngVel = [0,0.5,0.25]; finalAngVel = [-0.2,0.6,0.5]; constantAngVel = repmat(initialAngVel,floor(numSamples/2),1); varyingAngVel = [linspace(initialAngVel(1), finalAngVel(1), ceil(numSamples/2)).', ... linspace(initialAngVel(2), finalAngVel(2), ceil(numSamples/2)).', ... linspace(initialAngVel(3), finalAngVel(3), ceil(numSamples/2)).']; angVelBody = [constantAngVel; varyingAngVel]; accBody = zeros(numSamples,3); traj = kinematicTrajectory('SampleRate',fs); [~,qNED,~,accNED,angVelNED] = traj(accBody,angVelBody);```
Create an `imuSensor` System object™, `IMU`, with a nonideal gyroscope. Call `IMU` with the ground-truth acceleration, angular velocity, and orientation.
```IMU = imuSensor('accel-gyro', ... 'Gyroscope',gyroparams('RandomWalk',0.003,'ConstantBias',0.3), ... 'SampleRate',fs); [accelReadings, gyroReadingsBody] = IMU(accNED,angVelNED,qNED);```
Create an `imufilter` System object, `fuse`. Call `fuse` with the modeled accelerometer readings and gyroscope readings.
```fuse = imufilter('SampleRate',fs, 'GyroscopeDriftNoise', 1e-6); [~,angVelBodyRecovered] = fuse(accelReadings,gyroReadingsBody);```
Plot the ground-truth angular velocity, the gyroscope readings, and the recovered angular velocity for each axis.
The angular velocity returned from the `imufilter` compensates for the effect of the gyroscope bias over time and converges to the true angular velocity.
```time = (0:numSamples-1)'/fs; figure(1) plot(time,angVelBody(:,1), ... time,gyroReadingsBody(:,1), ... time,angVelBodyRecovered(:,1)) title('X-axis') legend('True Angular Velocity', ... 'Gyroscope Readings', ... 'Recovered Angular Velocity') ylabel('Angular Velocity (rad/s)')```
```figure(2) plot(time,angVelBody(:,2), ... time,gyroReadingsBody(:,2), ... time,angVelBodyRecovered(:,2)) title('Y-axis') ylabel('Angular Velocity (rad/s)')```
```figure(3) plot(time,angVelBody(:,3), ... time,gyroReadingsBody(:,3), ... time,angVelBodyRecovered(:,3)) title('Z-axis') ylabel('Angular Velocity (rad/s)') xlabel('Time (s)')```
## Algorithms
expand all
Note: The following algorithm only applies to an NED reference frame.
The `imufilter` uses the six-axis Kalman filter structure described in [1]. The algorithm attempts to track the errors in orientation, gyroscope offset, and linear acceleration to output the final orientation and angular velocity. Instead of tracking the orientation directly, the indirect Kalman filter models the error process, x, with a recursive update:
`${x}_{k}=\left[\begin{array}{c}{\theta }_{k}\\ {b}_{k}\\ {a}_{k}\end{array}\right]={F}_{k}\left[\begin{array}{c}{\theta }_{k-1}\\ {b}_{k-1}\\ {a}_{k-1}\end{array}\right]+{w}_{k}$`
where xk is a 9-by-1 vector consisting of:
• θk –– 3-by-1 orientation error vector, in degrees, at time k
• bk –– 3-by-1 gyroscope zero angular rate bias vector, in deg/s, at time k
• ak –– 3-by-1 acceleration error vector measured in the sensor frame, in g, at time k
• wk –– 9-by-1 additive noise vector
• Fk –– state transition model
Because xk is defined as the error process, the a priori estimate is always zero, and therefore the state transition model, Fk, is zero. This insight results in the following reduction of the standard Kalman equations:
Standard Kalman equations:
`$\begin{array}{l}{x}_{k}^{-}={F}_{k}{x}_{k-1}^{+}\\ {P}_{k}{}^{-}={F}_{k}{P}_{k-1}^{+}{F}_{k}^{T}+{Q}_{k}\\ \\ {y}_{k}={z}_{k}-{H}_{k}{x}_{k}^{-}\\ {S}_{k}\text{\hspace{0.17em}}={R}_{k}+{H}_{k}{P}_{k}^{-}{H}_{k}{}^{T}\\ {K}_{k}={P}_{k}^{-}{H}_{k}^{T}{\left({S}_{k}\right)}^{-1}\\ {x}_{k}^{+}={x}_{k}^{-}+{K}_{k}{y}_{k}\\ {P}_{k}^{+}={P}_{k}{}^{-}-{K}_{k}{H}_{k}{P}_{k}{}^{-}\end{array}$`
Kalman equations used in this algorithm:
`$\begin{array}{l}{x}_{k}^{-}=0\\ {P}_{k}{}^{-}={Q}_{k}\\ \\ {y}_{k}={z}_{k}\\ {S}_{k}\text{\hspace{0.17em}}={R}_{k}+{H}_{k}{P}_{k}^{-}{H}_{k}{}^{T}\\ {K}_{k}={P}_{k}^{-}{H}_{k}^{T}{\left({S}_{k}\right)}^{-1}\\ {x}_{k}^{+}={K}_{k}{y}_{k}\\ {P}_{k}^{+}={P}_{k}{}^{-}-{K}_{k}{H}_{k}{P}_{k}{}^{-}\end{array}$`
where
• xk –– predicted (a priori) state estimate; the error process
• Pk –– predicted (a priori) estimate covariance
• yk –– innovation
• Sk –– innovation covariance
• Kk –– Kalman gain
• xk+ –– updated (a posteriori) state estimate
• Pk+ –– updated (a posteriori) estimate covariance
k represents the iteration, the superscript + represents an a posteriori estimate, and the superscript represents an a priori estimate.
The graphic and following steps describe a single frame-based iteration through the algorithm.
Before the first iteration, the `accelReadings` and `gyroReadings` inputs are chunked into 1-by-3 frames and `DecimationFactor`-by-3 frames, respectively. The algorithm uses the most current accelerometer readings corresponding to the chunk of gyroscope readings.
## References
[2] Roetenberg, D., H.J. Luinge, C.T.M. Baten, and P.H. Veltink. "Compensation of Magnetic Disturbances Improves Inertial and Magnetic Sensing of Human Body Segment Orientation." IEEE Transactions on Neural Systems and Rehabilitation Engineering. Vol. 13. Issue 3, 2005, pp. 395-405. | 2020-09-22 02:54:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6556589603424072, "perplexity": 4170.987493892895}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00525.warc.gz"} |
https://physics.uchicago.edu/news/year/2021/ | # News: 2021
## April
### Graduate Student Ihar Lobach: Using Fluctuations to Measure Beam Properties
April 1, 2021
Scientists planning future particle accelerators and synchrotron light sources strive for tighter, more powerful electron beams. But as beams get narrower, it becomes harder to measure important properties. Now, Ihar Lobach of the University of Chicago and colleagues have demonstrated a new way to measure a beam’s vertical emittance more precisely than existing methods.
## January
### Roger Hildebrand, Manhattan Project veteran and ‘giant’ of physics and astrophysics, 1922-2021
January 31, 2021
Prof. Emeritus Roger Hildebrand, who worked on the Manhattan Project as an undergraduate before conducting pioneering work in particle physics and astrophysics at the University of Chicago, died Jan. 21. He was 98.
### PUEO Selected by NASA for Pioneers Program
January 8, 2021
NASA has chosen four small-scale astrophysics missions for further concept development in a new program called Pioneers.
PUEO is a balloon mission designed to launch from Antarctica that will detect signals from ultra-high energy neutrinos. The principal investigator is Abigail Vieregg of the University of Chicago. | 2021-04-14 07:58:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2914033830165863, "perplexity": 6359.924931294607}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00353.warc.gz"} |
https://tex.stackexchange.com/questions/453398/keeping-a-diagram-question-on-the-same-page-column-with-the-exam-class | Keeping a diagram + question on the same page/column with the exam class
I use the exam class to write multiple choice questions, and I frequently use tikz to make a diagram that precedes a question. How can I make sure that each diagram+question doesn't break across a column or page break?
Wrapping the tikzpicture and question in a minipage works for the 2nd question onward, but not for the very first question after \begin{questions}. Is there a way I can do something similar for the first question?
This works:
\documentclass[twocolumn]{exam}
\usepackage{tikz}
\begin{document}
\twocolumn
\begin{questions}
\question What's the length of the line above?
\begin{choices}
\choice 1
\choice 8
\end{choices}
\begin{minipage}{\linewidth}
\begin{tikzpicture}
\draw (0,0) -- (2,0);
\end{tikzpicture}
\question What's the length of the line above?
\begin{choices}
\choice 2
\choice 1
\end{choices}
\end{minipage}
\end{questions}
\end{document}
But this doesn't compile (presumably since \begin{questions} expects \question right away):
\documentclass[twocolumn]{exam}
\usepackage{tikz}
\begin{document}
\twocolumn
\begin{questions}
\begin{minipage}{\linewidth}
\begin{tikzpicture}
\draw (0,0) -- (2,0);
\end{tikzpicture}
\question What's the length of the line above?
\begin{choices}
\choice 2
\choice 1
\end{choices}
\end{minipage}
\question What's the length of the line above?
\begin{choices}
\choice 1
\choice 8
\end{choices}
\end{questions}
\end{document}
Welcome to TeX.SE! I am not sure I understand the purpose of the minipage here. Regardless of whether or not you want to use it, here is a simple cheat that allows you to add a picture above a question. It adds an item, but does not print the number and makes sure the counter does not get increased.
\documentclass[twocolumn]{exam}
\usepackage{tikz}
\begin{document}
\twocolumn
\begin{questions}
\begin{pictureabovequestion}
\begin{tikzpicture}
\draw (0,0) -- (2,0);
\end{tikzpicture}
\end{pictureabovequestion}
\question What's the length of the line above?
\begin{choices}
\choice 1
\choice 8
\end{choices}
\begin{pictureabovequestion}
\begin{tikzpicture}
\draw[blue] (0,0) -- (1,0);
\end{tikzpicture}
\end{pictureabovequestion}
\question What's the length of the line above?
\begin{choices}
\choice 2
\choice 1
\end{choices}
\end{questions}
\end{document}
• Thanks! If I wasn't clear, I was trying to use minipage to prevent page/column breaks between either the picture and the question or the question and its answers. minipage works around your "pictureabovequestion" environment, so your answer helps. – lwad Oct 2 '18 at 3:25
• @creakyshrimp You're welcome! (A single tikzpicture won't get broken up, but yes, if you have a whole bunch of things, a minipage does make sense.) – user121799 Oct 2 '18 at 3:28 | 2020-04-02 23:16:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7798555493354797, "perplexity": 1418.827547904378}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370508367.57/warc/CC-MAIN-20200402204908-20200402234908-00555.warc.gz"} |
http://clear-lines.com/blog/2009/07/default.aspx | Mathias Brandewinder on .NET, F#, VSTO and Excel development, and quantitative analysis / machine learning.
31. July 2009 18:00
I have been slammed with work lately, and couldn’t find the time to do this, even though I have wanted to for a while. Well, it’s done – this blog now runs on BlogEngine.Net 1.5.1.14! One of the reasons I wanted to give it a try is the multiple-widget zone feature, which is unfortunately not yet in the 1.5 “official” release, so I went the adventurous route, and here we are! I still need to do some minor polishing on the styling in the next few days, but so far, everything looks like it’s running smoothly.
29. July 2009 11:24
Today I came across a post which demonstrates how to use Goal Seek to determine how to save for your retirement. Goal Seek is essentially a simplified solver: point Goal Seek at a cell, tell it how much you want it to be and what cells it can tinker with, and Goal Seek will try to find values that reach that goal. The post is an excellent illustration of what’s great about it: it’s super easy to use, and very practical.
However, there is no such thing as a perfect tool, and Goal Seek can fail miserably at finding the optimal answer to very simple problems. After reading this, I thought it would be a good public service to illustrate what its shortcomings are, especially if you are going to trust it for questions as important as your retirement!
For our illustration, we will use the following setup.
Now let’s say we want to find a value such that B2 = 250. Following Pointy Haired Dilbert, let’s use Goal Seek:
Put in a value of 1 in cell B1, and run Goal Seek - here is what happens:
Goal Seek fails to find a value in B1 such that B2 = 250. Complete failure.
More...
24. July 2009 15:54
I really like the addition of [TestCase] in NUnit 2.5. A significant part of the code I write is math or finance oriented, and I find Data-Driven tests more convenient that “classic” unit tests to validate numeric procedures.
However, I got a bit frustrated today, because of the lack of tolerance mechanism in data-driven tests. Tolerance allows you to specify a margin of error (delta) on your test, and is supported in classic asserts:
[Test]
public void ClassicToleranceAssert()
{
double numerator = 10d;
double denominator = 3d;
Assert.AreEqual(3.33d, numerator / denominator, 0.01);
Assert.That(3.33d, Is.EqualTo(numerator / denominator).Within(0.01));
}
You can specify how close the result should be from the expected test result, here +/- 0.01.
I came into some rounding problems with data driven tests today, and hoped I would be able to resolve them with the same mechanism. Here is roughly the situation:
[TestCase(10d, 2d, Result = 5d)]
[TestCase(10d, 3d, Result = 3.33d)]
public double Divide(double numerator, double denominator)
{
return numerator / denominator;
}
Not surprisingly, the second test case fails – and when I looked for a similar tolerance mechanism, I found zilch.
The best solution I got was to do something like this:
[TestCase(10d, 2d, Result = 5d)]
[TestCase(10d, 3d, Result = 3.33d)]
public double Divide(double numerator, double denominator)
{
return Math.Round(numerator / denominator, 2);
}
Of course, this works – but this is clumsy. I was really hoping that TestCase would support the same functionality as Assert, with a built-in delta tolerance. It seems particularly relevant: rounding error issues are typical in numerical procedures, a field where data-driven tests are especially adapted.
Maybe the feature exists, but is undocumented. If you know how to do this, sharing your wisdom will earn you a large serving of gratitude, and if it the feature doesn’t exist yet… maybe in NUnit 2.5.1?
14. July 2009 06:58
While shopping at PetCo the other day, I saw this scene, and couldn’t get enough of it. It’s a great illustration of what happens when teamwork goes wrong. If I saw this in real life, I would feel sorry for the team, but the mice don’t seem to be suffering (besides some confusion), so sit back and enjoy the show, guilt-free.
7. July 2009 11:25
I was going through my digital camera today when I came across this:
I took this picture in the Seattle public library elevator. It’s a great building, with lots of unusual features.
Why did I take it? Because I thought the design was interesting. The typical elevator button panels has a series of buttons, each of them the same size, ordered with the corresponding floor elevation. This panel has buttons of various sizes, and given what I saw in the building, I guess the size corresponds to the frequentation of each floor, or, if you prefer, the relative likelihood that you want to go to each floor.
In my opinion, there are some flaws in the design (for instance, the buttons are very far from the labels saying what is on the floor, which slows down figuring out which button you really want to push), but I thought the idea of various button sizes was smart, and bizarrely under-utilized in computer user interfaces. If you are going to expose different functionalities to your user, chances are, some will be used more often than others, and giving them a larger button should make the user’s navigation easier.
I found an example of this on Netflix’s website, for instance:
Note how the 2 right-most elements of the tab are smaller than the rest?
So why is this design style rare? I guess it has to do with the fact that regularity looks good, without much effort. Give your buttons the same size and font, and everything looks organized, and easy on the eye. Once you start playing with elements of various sizes, it is much harder to achieve balance and cohesion. Usability might be improved, but chances are, you will need someone with a sense of visual design to help you make it look good – whereas even the most lacking in aesthetic sense can align buttons.
#### Need help with F#?
The premier team for
F# training & consulting. | 2019-03-19 11:04:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36842918395996094, "perplexity": 1227.0066764246349}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201953.19/warc/CC-MAIN-20190319093341-20190319115341-00198.warc.gz"} |
https://tex.stackexchange.com/questions/159429/a-math-font-error-i-fixed-but-dont-know-what-was-wrong | # A “math font error” I fixed but, don't know what was wrong
I was typing my homework and got this error:
! LaTeX Error: \mathfrak allowed only in math mode.
Recently I added some macros to my macro file that gives mathfrak text when used so, I figured I must have used that command without using math mode. I looked around and saw that since I last compiled I had not used that command, so I start commenting out stuff till my document would compile. Anyway long story short it looked like when my document got over a page it would output the error and not compile. I confirmed this by opening a new document (a template I use) that loads my macros file and typed a bit, then made a new page then typed a bit, no math at all, and it would output the same error.
I would like to know why it would only output an error after a page. Also it would be cool to know how I could get the \sl command I defined without that undesirable behavior.
Edit: Here is an example
\documentclass[a4paper,10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{amssymb}
\usepackage{fancyhdr}
\pagestyle{fancy}
\renewcommand{\sl}{\mathfrak{sl}}
\begin{document}
P
\end{document}
For some reason this example doesn't need a second page. Also it looks like the error has something to do with fancy.
• \sl gives slanted text and it is already defined. You may use \newcommand{\Sl}{\mathfrak{sl}} instead. – user11232 Feb 9 '14 at 7:59
• Welcome to TeX.SE! The command \sl, while OK in Plain-TeX documents, is considered deprecated for LaTeX documents; use either \textsl{…} or {\slshape …} instead of \sl. However, since you're not actually looking to use slanted glyphs, it's best to stay away from this command; in particular, do not try to redefine the command \sl, as doing so may break other things in your document. Instead, use a name for your macro; \sla will work, though I'd like to encourage you to find a more mnemonic name that's not already been taken. – Mico Feb 9 '14 at 8:03
• I think this has nothing to do with the fact you changed \sl to \sla. In the first case you were using \sl outside math mode. This is the error you got. And anyway, as said in other comments, don't redefine \sl – karlkoeller Feb 9 '14 at 8:04
• Welcome to TeX.SE! Please try to give the minimal working example (MWE) in which your problem occurs. Creating it is the first step in debugging and helps us tremendously in pin-pointing where the issue may lie. In particular it should start with \documentclass and end with \end{document}. – bodo Feb 9 '14 at 8:17
• @PaulPlummer - Because \sl is a fairly low level (as well as well-established, though deprecated) command, it may still be in use in various LaTeX packages, whether loaded explicitly by you or by other packages that are in use. Redefining the \sl macro may therefore cause unexpected and unforeseeable problems. It sure looks like you've encountered one such problem, and more might be lurking. That's why I recommend you not redefine the \sl command at all and, rather, come up with an entirely new name for your shortcut macro. – Mico Feb 9 '14 at 8:36
Thanks for posting an MWE that generates the problem behavior you're looking to fix.
The problem -- i.e., the error message about \mathfrak being used outside of math mode -- seems to arise because you're also loading the fancyhdr package: Comment out the instructions \usepackage{fancyhdr} and \pagestyle{fancy}, and no more error message is produced.
Sure enough, inside the file fancyhdr.sty, one finds the following instructions:
\if@twoside
\else
\fi
Given that \sl really shouldn't be used in LaTeX documents anymore, the best solution would be for the package to be amended to use \slshape instead of \sl in the code shown above. The next best solution -- and the one that's available immediately -- is for the user (you!) not to try to redefine \sl in the first place.
Update March 2020: At some point since I originally wrote this answer, the fancyhdr package was updated to use \slshape rather than \sl in all four instances shown in the code chunk above. The following paragraphs are thus mainly (hopefully...) of general interest; however, they no longer describe the current properties of the fancyhdr package.
What is it about the command \sl -- and the related font-changing commands \rm, \sf, \tt, \bf, \it, and \sc -- that makes them "deprecated" for use in LaTeX? For one, these commands are no longer defined in the LaTeX "kernel" (i.e., you won't find them in the file latex.ltx in your TeX distribution) but only in "class" files, e.g., in article.cls. Second, to the extent the command \sl is defined at all, it's generally set up as follows:
\DeclareOldFontCommand{\sl}{\normalfont\slshape}{\@nomath\sl}
(The macros \normalfont and \slshape, naturally, are defined in the LaTeX kernel.)
This definition of \sl, which invokes \normalfont before \slshape, is deliberately crafted to mimic the action of the original Plain-TeX \sl macro. Because of the presence of the \normalfont instruction -- which executes the four instructions \encodingdefault, \familydefault, \seriesdefault, and \shapedefault -- one cannot "combine" several old-style font declaration commands. Thus, an instruction such as \bf\it will not create bold-italic output; it will only create italic output. (Similarly, \it\bf will only create bold output.)
One of the (many) advantages of using \bfseries\itshape instead of \bf\it is precisely that it lets you "combine" the two font-changing commands. (Of course, if the font family in use doesn't provide bold-italic glyphs, TeX/LaTeX can't employ them. However, the availability of specific combinations of font weights and shapes is an entirely separate issue.)
• The code in fancyhdr.sty should be changed. Another possibility could be to define the header in the document, but the possibility that another non updated package calls \sl would still be there. :-( – egreg Feb 9 '14 at 10:10
• @egreg - I agree with you that the code in the fancyhdr should be changed. But, as you say, there are lots and lots of LaTeX packages out there that still use the deprecated font-changing commands. I suspect that tracking down all deprecated commands used in files stored on the CTAN, contacting the packages' maintainers, and asking them to update the deprecated code is a task that even Hercules would take a rather long time to achieve. (No diversion of rivers allowed -- or possible...) – Mico Feb 9 '14 at 10:28 | 2020-07-02 19:44:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8348917961120605, "perplexity": 1912.6002658229881}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655879738.16/warc/CC-MAIN-20200702174127-20200702204127-00048.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-9-polar-coordinates-vectors-9-4-vectors-9-4-assess-your-understanding-page-606/30 | Precalculus (10th Edition)
$-3i-5j.$ This is also the position vector.
If a vector $v$ initiates at point $P(x_1,y_1)$ and terminates at $Q(x_2,y_2)$ then $v=(x_2-x_1,y_2-y_1)=(x_2-x_1)i+(y_2-y_1)j.$ Hence here $v=(-3-0)i+(-5-0)j=-3i-5j.$ This is also the position vector. | 2021-10-25 01:35:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9167278409004211, "perplexity": 504.5466029443694}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587608.86/warc/CC-MAIN-20211024235512-20211025025512-00192.warc.gz"} |
http://blog.stata.com/category/performance/ | ### Archive
Archive for the ‘Performance’ Category
## Multiprocessor (core) software (think Stata/MP) and percent parallelization
When most people first think about software designed to run on multiple cores such as Stata/MP, they think to themselves, two cores, twice as fast; four cores, four times as fast. They appreciate that reality will somehow intrude so that two cores won’t really be twice as fast as one, but they imagine the intrusion is something like friction and nothing that an intelligently placed drop of oil can’t improve.
In fact, something inherent intrudes. In any process to accomplish something — even physical processes — some parts may be able to to be performed in parallel, but there are invariably parts that just have to be performed one after the other. Anyone who cooks knows that you sometimes add some ingredients, cook a bit, and then add others, and cook some more. So it is, too, with calculating xt = f(xt-1) for t=1 to 100 and t0=1. Depending on the form of f(), sometimes there’s no alternative to calculating x1 = f(x0), then calculating x2 = f(x1), and so on.
In any calculation, some proportion p of the calculation can be parallelized and the remainder, 1-p, cannot. Consider a calculation that takes T hours if it were performed sequentially on a single core. If we had an infinite number of cores and the best possible implementation of the code in parallelized form, the execution time would fall to (1-p)T hours. The part that could be parallelized, which ordinarily would run in pT hours, would run in literally no time at all once split across an infinite number of cores, and that would still leave (1-p)T hours to go. This is known as Amdahl’s Law.
We can generalize this formula to computers with a finite number of cores, say n of them. The parallelizable part of the calculation, the part that would ordinarily run in pT hours, will run in pT/n. The unparallelizable part will still take (1-p)T hours, so we have
Tn = pT/n + (1-p)T
As n goes to infinity, Tn goes to (1-pT).
Stata/MP is pretty impressively parallelized. We achieve p of 0.8 or 0.9 in many cases. We do not claim to have hit the limits of what is possible, but in most cases, we believe we are very close to those limits. Most estimation commands have p above 0.9, and linear regression is actually above 0.99! This is explained in more detail along with percentage parallelization details for all Stata commands in the Stata/MP Performance Report.
Let’s figure out the value of having more cores. Consider a calculation that would ordinarily require T = 1 hour. With p=0.8 and 2 cores, run times would fall to 0.6 hours; With p=0.9, 0.55 hours. That is very close to what would be achieved even with p=1, which is not possible. For 4 cores, run times would fall to 0.4 (p=0.8) and 0.325 (p=0.9). That’s good, but no where near the hoped for 0.25 that we would observe if p were 1.
In fact, to get to 0.25, we need about 16 cores. With 16 cores, run times fall to 0.25 (p=0.8) and 0.15625 (p=0.9). Going to 32 cores improves run times just a little, to 0.225 (p=0.8) and 0.128125 (p=0.9). Going to 64 cores, we would get 0.2125 (p=0.8) and 0.11384615 (p=0.9). There’s little gain at all because all the cores in the world combined, and more, cannot reduce run times to below 0.2 (p=0.8) and 0.1 (p=0.9).
Stata/MP supports up to 64 cores. We could make a version that supports 128 cores, but it would be a lot of work even though we would not have to write even one line of code. The work would be in running the experiments to set the tuning parameters.
It turns out there are yet other ways in which reality intrudes. In addition to some calculations such as xt = f(xt-1) not being parallelizable at all, it’s an oversimplification to say any calculation is parallelizable because there are issues of granularity and of diseconomies of scale, two related, but different, problems.
Let’s start with granularity. Consider making the calculation xt = f(zt) for t = 1 to 100, and let’s do that by splitting on the subscript t. If we have n=2 cores, we’ll assign the calculation for t = 1 to 50 to one core, and for t=51 to 100 to another. If we have four cores, we’ll split t into four parts. Granularity concerns what happens when we move from n=100 to n=101 cores. This problem can be split into only 100 parallelizable parts and the minimum run time is therefore max(T/n, T/100) and not T/n, as we previously assumed.
All problems suffer from granularity. Diseconomies of scale is a related issue, and it strikes sooner than granularity. Many, but not all problems suffer from diseconomies of scale. Rather than calculating f(zt) for t = 1 to 100, let’s consider calculating the sum of f(zt) for t = 1 to 100. We’ll make this calculation in parallel in the same way as we made the previous calculation, by splitting on t. This time, however, each subprocess will report back to us the sum over the subrange. To obtain the overall sum, we will have to add sub-sums. So if we have n=2 cores, core 1 will calculate the sum over t = 1 to 50, core 2 will calculate the sum for t = 51 to 100, and then, the calculation having come back together, the master core will have to calculate the sum of two numbers. Adding two numbers can be done in a blink of an eye.
But what if we split the problem across 100 cores? We would get back 100 numbers which we would then have to sum. Moreover, what if the calculation of f(zt) is trivial? In that case, splitting the calculation among all 100 cores might result in run times that are nearly equal to what we would observe performing the calculation on just one core, even though splitting the calculation between two cores would nearly halve the execution time, and splitting among four would nearly quarter it!
So what’s the maximum number of cores over which we should split this problem? It depends on the relative execution times of f(zt) and the the combination operator to be performed on those results (addition in this case).
It is the diseconomies of scale problem that bit us in the early versions of Stata/MP, at least in beta testing. We did not adequately deal with the problem of splitting calculations among fewer cores than were available. Fixing that problem was a lot of work and, for your information, we are still working on it as hardware becomes available with more and more cores. The right way to address the issue is to have calculation-by-calculation tuning parameters, which we do. But it takes a lot of experimental work to determine the values of those tuning parameters, and the greater the number of cores, the more accurately the values need to be measured. We have the tuning parameters determined accurately enough for up to 64 cores, although there are one or two which we suspect we could improve even more. We would need to do a lot of experimentation, however, to ensure we have values adequate for 128 cores. The irony is that we would be doing that to make sure we don’t use them all except when problems are large enough!
In any case, I have seen articles predicting and in some cases, announcing, computers with hundreds of cores. For applications with p approaching 1, those are exciting announcements. In the world of statistical software, however, these announcements are exciting only for those running with immense datasets.
Categories: Multiprocessing Tags:
## Stata/MP — having fun with millions
I was reviewing some timings from the Stata/MP Performance Report this morning. (For those who don’t know, Stata/MP is the version of Stata that has been programmed to take advantage of multiprocessor and multicore computers. It is functionally equivalent to the largest version of Stata, Stata/SE, and it is faster on multicore computers.)
What was unusual this morning is that I was running Stata/MP interactively. We usually run MP for large batch jobs that run thousands of timings on large datasets — either to tune performance or to produce reports like the Performance Report. That is the type of work Stata/MP was designed for — big jobs on big datasets.
I will admit right now that I mostly run Stata interactively using the auto dataset, which has 74 observations. I run Stata/MP using all 4 cores of my quad-core computer, but I am mostly wasting 3 of them — there is no speeding up the computations on 74 observations. This morning I was running Stata/MP interactively on a 24-core computer using a somewhat larger dataset.
After a while, I was struck by the fact that I wasn’t noticing any annoying delays waiting for commands to run. It felt almost as though I were running on the auto dataset. But I wasn’t. I was running commands using 50 covariates on 1 million observations! Regressions, summary statistics, etc.; this was fun. I had never played interactively with a million-observation dataset before.
Out of curiousity, I turned off multicore support. The change was dramatic. Commands that were taking less than a second were now taking longer, too long. My coffee cup was full, but I contemplated fetching a snack. Running on only one processor was not so much fun.
For your information, I set rmsg on and ran a few timings:
Timing (seconds) Analysis 24 cores 1 core generate a new variable .03 .33 summarize 50 variables .88 19.55 twoway tabulation .45 .45 linear regression .65 11.48 logistic regression 7.19 59.27 All timings are on a 1 million observation dataset. The two regressions included 50 covariates.
OK, the timings with 24 cores are not quite the same as with the auto dataset, but well within comfortable interactive use.
Careful readers will have noticed that the 24-core and 1-core timings for twoway tabulation are the same. We have not rewritten the code for tabulate to support multiple cores, partly because tabulate is already very fast, and partly because the code for tabulate is isolated, so changing it will not improve the performance of other commands. Thus, parallelizing tabulate is on our long-run, not short-run, list of additions to Stata/MP. We have rewritten about 250 sections of Stata’s internal code to support Symmetric Multi Processing (SMP). Each rewritten section typically improves the performance of many commands.
I switched back to using all 24 cores and returned to my original work — stress testing changes in the number of covariates and observations. My fun was quelled when I started running some timings of Cox proportional hazards regressions. With my 50 covariates and 1 million observations, a Cox regression took just over two minutes. Most estimators in Stata are parallelized, including the estimators for parametric survival models. The Cox proportional hazards estimator is not. It is not parallelized because it uses a clever algorithm that requires sequential computations. When I say sequential I mean that some computations are wholly dependent on previous computations so that they simply cannot be performed simultaneously, in parallel. There are other algorithms for fitting the Cox model, but they are orders of magnitude slower. Even parallelized, they would not be faster than our current sequential algorithm unless run on 20 or more processors. When more computers start shipping with dozens of cores, we will evaluate adding a parallelized algorithm for the Cox estimator.
The computer I was running on is about a year old. There have been a spate of new and faster server-grade processors from Intel and AMD in the past year. You can get reasonably close to the performance of my 24-core computer using just 8-cores and the newer chips. That means that with a newer 32-core computer, I could increase my threshold for interactive analysis to about 4 million observations.
There are four speed comparisons above. To see 450 more, including graphs and a discussion of SMP and its implementation in Stata, see the Stata/MP white paper, a.k.a. the Stata/MP Performance Report.
Categories: Multiprocessing Tags:
## Big computers
We here at Stata are often asked to make recommendations on the “best” computer on which to run Stata, and such discussions sometimes pop up on Statalist. Of course, there is no simple answer, as it depends on the analyses a given user wishes to run, the size of their datasets, and their budget. And, we do not recommend particular computer or operating system vendors. Many manufacturers use similar components in their computers, and the choice of operating system comes down to personal preference of the user. We take pride in making sure Stata works well regardless of operating system and hardware configuration.
For some users, the analyses they wish to run are demanding, the datasets they have are huge, and their budgets are large. For these users, it is useful to know what kind of off-the-shelf hardware they can easily get their hands on. To give you an idea of what is available, HP makes a server with up to 1 TB of memory. Yes, 1 terabyte! This computer can be configured and ordered online at hp.com.
It can have up to 4 processors, each with 8 cores, for a total of 32 cores of processing power. A sample rack-mount configuration with the fastest 8-core Intel Xeon processors available for this computer and a full 1 TB of memory totals roughly $100,000. We mention HP because they were one of the first to allow such large memory configurations without going to a much more expensive completely custom-built solution. Wouldn’t you love to have one of these running Stata/MP (or Halo)? You can run Windows or Linux on a computer like the above. If you prefer Mac OS X, the largest current configuration from Apple allows a total of 12 cores and 32 GB of memory. This is a tower case unit and costs around$10,000. Visit store.apple.com to configure such a computer.
The largest fastest laptops easily purchased these days allow up to 4 cores and 16 GB of RAM. That much power in a small package will cost you though, with such a configuration costing over \$7,000. Here is one such example you can configure from Dell: dell.com.
We’ll keep you updated periodically with the state of the high end of the computer market as memory capacities and number of cores increase.
Categories: Hardware Tags: | 2014-10-01 00:14:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40680280327796936, "perplexity": 1209.904832659858}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663218.28/warc/CC-MAIN-20140930004103-00208-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://mathematics-diary.blogspot.nl/2011/04/magic-squares-of-type-5-by-5.html | As of May 4 2007 the scripts will autodetect your timezone settings. Nothing here has to be changed, but there are a few things
## Tuesday, April 19, 2011
### Magic squares of type 5-by-5
Still in 're-discovery mode' generating magic squares has become trivial, although the computational resources increase fast depending on the size of the square.
$$\left( \begin{array}{ccccc} 14 & 32 & 6 & 8 & 22 \\ 14 & 7 & 30 & 17 & 14 \\ 0 & 26 & 24 & 22 & 10 \\ 44 & 9 & 16 & 7 & 6 \\ 10 & 8 & 6 & 28 & 30 \end{array} \right)$$
$$\left( \begin{array}{ccccc} 4 & 22 & 6 & 28 & 42 \\ 4 & 37 & 30 & 17 & 14 \\ 20 & 26 & 24 & 22 & 10 \\ 64 & 9 & 16 & 7 & 6 \\ 10 & 8 & 26 & 28 & 30 \end{array} \right)$$
$$\left( \begin{array}{ccccc} 4 & 6 & 22 & 26 & 64 \\ 6 & 55 & 30 & 17 & 14 \\ 16 & 34 & 24 & 42 & 6 \\ 86 & 7 & 16 & 7 & 6 \\ 10 & 20 & 30 & 30 & 32 \end{array} \right)$$
##### Magic square
A magic square is a square matrix with elements in $\mathbf{Z}$ such that the totals of rows, columns and both diagonals are equal.
##### Normal magic square
A normal magic square is a magic square with elements $1,2, \cdots, n^2$ where $n$ is the size of the matrix.
##### Latin square
A latin square is a $n$ by $n$ square matrix containing $n$ times the first $n$ elements of the alphabet such that each row and each column contains each letter only once. (i.e. Cayley Table)
#### Further developments
Clearly, normal magic squares are the most desirable objects in the realm of matrices. I am making detailed notes about this work in the ( still? ) unpublished personal mathematics wiki I am setting up.
## Welcome to The Bridge
Mathematics: is it the fabric of MEST?
This is my voyage
My continuous mission
To uncover hidden structures
To create new theorems and proofs
To boldly go where no man has gone before
(Raumpatrouille – Die phantastischen Abenteuer des Raumschiffes Orion, colloquially aka Raumpatrouille Orion was the first German science fiction television series. Its seven episodes were broadcast by ARD beginning September 17, 1966. The series has since acquired cult status in Germany. Broadcast six years before Star Trek first aired in West Germany (in 1972), it became a huge success.) | 2017-08-17 01:37:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39042070508003235, "perplexity": 2157.5291615514025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00330.warc.gz"} |
https://www.physicsforums.com/threads/almost-sure-convergance-of-sum-of-rv.628170/ | # Almost sure convergance of sum of rv
1. Aug 14, 2012
### logarithmic
I'm trying to prove that if $\{X_n\}$ is independent and $E(X_n)=0$ for all n, and $\sum_{n}E(X_n^2) <\infty$, then $\sum_{n}X_n$ converges almost surely.
What I've got so far is the following: Denote the partial sums by $\{S_n\}$, then proving almost sure convergence is equivalent to showing that there exists a random variable, X, such that for all $\epsilon > 0$, $\sum_{n}P(|S_n-X|>\epsilon)<\infty$. Using the Markov inequality gives $\sum_{n}P(|S_n-X|>\epsilon)\leq\sum_{n}E(|S_n-X|)/\epsilon$. Then we just need to show that this sum converges. But I can't find anyway to do this.
It's easy to show that $\sum_{n}X_n$ converges in the $L^2$ sense, so it also converges in the $L^1$ sense, which means that $\lim_{n\to\infty}E(|S_n-X|)=0$, but this isn't strong enough to imply that $\sum_{n}E(|S_n-X|)<\infty$, which is what I need.
Can anyone help me with this?
2. Aug 14, 2012
### chiro
Hey logarithmic.
Just for clarification is the thing you are converging to a random variable? (I'm guessing it is).
If this is the case, then if you show that the moments converge to some particular value, then the distribution itself that is represented by those moments is the analogous to the resulting random variable.
If they are independent, then the MGF of the final random variable is the sum of all the MGF's and if this defines a valid unique random variable, then you're done. As long as the random variable is a valid random variable, then this is what matters.
You will not in general get convergence for a sum of random variables like that to converge to some single value (like you would for say a consistent estimator for say a parameter or a mean).
The other thing is that using the expectation is not an indicator of whether the value converges. For example a Cauchy distribution doesn't have a finite mean but it definitely has values at which it can converge with a non-zero probability (intervals I mean).
At a more abstract level, you could even through an analysis argument, that if the MGF is bounded by some function with the exponent of the nth power (n random variables) that shows that a unique PDF exists with only finite values that have non-zero probability, then you have shown that the random variable converges for all possible probabilities.
This kind of converges shows convergence to probability and there are many ways to show this, but effectively they are all about showing that a particular distribution exists.
It might also be useful to consider that the variance of the final random variable is finite and how this affects the nature of the final distribution.
3. Aug 14, 2012
### logarithmic
Yes, X is a random variable, not a number. I'm not sure if looking at the MGF will help though, as the only results I know about MGF and convergence (e.g. Levy's continuity theorem, and the theorem that if $E(X_n^p)\to E(X^p)$ for all p, then we have convergence in distribution) are about convergence in distribution, not almost sure convergence.
Is there some result that relates MGFs or moments to almost sure convergence?
4. Aug 14, 2012
### chiro
If you show the moments converge you show the PDF converges: the MGF is unique for a distribution just like a Laplace or Fourier transform of a valid function is unique for that function.
5. Aug 14, 2012
### logarithmic
If the PDF converges, that's convergence in distribution, which is weaker than almost sure convergence.
6. Aug 14, 2012
### chiro
Ok sorry: what's the definition of almost sure convergence? Kinda sounds made up (i.e. converges "almost' surely is not what I'd expect a mathematician to say) and I know it's not your definition either ;).
7. Aug 14, 2012
### logarithmic
8. Aug 14, 2012
### chiro
I don't know how that can be weaker: proving the case of equality is even stronger than proving the limiting case.
Besides: With Levy's theorem that you stated, if you can use that then you're done since all the moments of a distribution uniquely define the distribution. Proving that all moments converge to a particular value means you have proven that the distribution is represented by the definition and values of those moments.
If you want to prove the above, you use the definition of the MGF and show that if the moments exist and are finite, then the MGF exists and is valid and if that's valid then you can get a unique PDF from an Fourier transform that represents the distribution representing by those moments that Levy's theorem proved. | 2017-10-20 06:13:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9096938967704773, "perplexity": 249.60409045835854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823731.36/warc/CC-MAIN-20171020044747-20171020064747-00730.warc.gz"} |
https://neuralprophet.com/html/model-overview.html | # Overview of the NeuralProphet Model¶
NeuralProphet is a Neural Network based PyTorch implementation of a user-friendly time series forecasting tool for practitioners. This is heavily inspired by Prophet, which is the popular forecasting tool developed by Facebook. NeuralProphet is developed in a fully modular architecture which makes it scalable to add any additional components in the future. Our vision is to develop a simple to use forecasting tool for users while retaining the original objectives of Prophet such as interpretability, configurability and providing much more such as the automatic differencing capabilities by using PyTorch as the backend.
## Time Series Components¶
NeuralProphet is a decomposable time series model with the components, trend, seasonality, auto-regression, special events, future regressors and lagged regressors. Future regressors are external variables which have known future values for the forecast period whereas the lagged regressors are those external variables which only have values for the observed period. Trend can be modelled either as a linear or a piece-wise linear trend by using changepoints. Seasonality is modelled using fourier terms and thus can handle multiple seasonalities for high-frequency data. Auto-regression is handled using an implementation of AR-Net, an Auto-Regressive Feed-Forward Neural Network for time series. Lagged regressors are also modelled using separate Feed-Forward Neural Networks. Future regressors and special events are both modelled as covariates of the model with dedicated coefficients. For more details, refer to the documentation of the individual components.
## Data Preprocessing¶
We perform a few data pre-processing steps in the model. For the observed values of the time series, users can specify whether they would like the values to be normalized. By default, the y values would be min-max normalized. If the user specifically, sets the normalize_y argument to true, the data is z-score normalized. Normalization can be performed for covariates as well. The default mode for normalization of covariates is auto. In this mode, apart from binary features such as events, all others are z-score normalized.
We also perform an imputation in-case there are missing values in the data. However, imputation is only done if auto-regression is enabled in the model. Otherwise, the missing values do not really matter for the regression model. No special imputation is done for binary data. They are simply taken as 0 for the missing dates. For the numeric data, including the y values, normalization is a two-step process. First, small gaps are filled with a linear imputation and then the more larger gaps are filled with rolling averages. When auto-regression is enabled, the observed y values are preprocessed in a moving window format to learn from lagged values. This is done for lagged regressors as well.
## When to Use NeuralProphet¶
NeuralProphet can produce both single step and multi step-ahead forecasts. At the moment, NeuralProphet builds models univariately. This means that if you have many series that you expect to produce forecasts for, you need to do this one at a time. However, in future we hope to integrate the capability of global forecasting models into NeuralProphet.
NeuralProphet helps build forecasting models for scenarios where there are other external factors which can drive the behaviour of the target series over time. Using such external information can heavily improve forecasting models rather than relying only on the autocorrelation of the series. NeuralProphet tool is suitable for forecasting practitioners that wish to gain insights into the overall modelling process by visualizing the forecasts, the individual components as well as the underlying coefficients of the model. Through our descriptive plots, users can visualize the interaction of the individual components. They also have the power to control these coefficients as required by introducing sparsity through regularization. They can combine the components additively or multiplicatively as per their domain knowledge.
This is an ongoing effort. Therefore, NeuralProphet will be equipped with even much more features in the upcoming releases. | 2022-05-22 11:50:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5090339183807373, "perplexity": 1140.4048112592086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00271.warc.gz"} |
http://en.wikipedia.org/wiki/Matter_waves | Matter wave
(Redirected from Matter waves)
In quantum mechanics, the concept of matter waves or de Broglie waves reflects the wave–particle duality of matter. The theory was proposed by Louis de Broglie in 1924 in his PhD thesis.[1] The de Broglie relations show that the wavelength is inversely proportional to the momentum of a particle and is also called de Broglie wavelength. Also the frequency of matter waves, as deduced by de Broglie, is directly proportional to the total energy E (sum of its rest energy and the kinetic energy) of a particle.[2]
Historical context
Propagation of de Broglie waves in 1d – real part of the complex amplitude is blue, imaginary part is green. The probability (shown as the colour opacity) of finding the particle at a given point x is spread out like a waveform, there is no definite position of the particle. As the amplitude increases above zero the curvature decreases, so the amplitude decreases again, and vice versa – the result is an alternating amplitude: a wave. Top: plane wave. Bottom: wave packet.
At the end of the 19th century, light was thought to consist of waves of electromagnetic fields which propagated according to Maxwell’s equations, while matter was thought to consist of localized particles (See history of wave and particle viewpoints). This division was challenged when, in his 1905 paper on the photoelectric effect, Albert Einstein postulated that light was emitted and absorbed as localized packets, or "quanta" (now called photons). These quanta would have an energy
$E=h\nu$
where $\scriptstyle \nu$ is the frequency of the light and h is Planck’s constant. In the modern convention, frequency is symbolized by f as is done in the rest of this article. Einstein’s postulate was confirmed experimentally by Robert Millikan and Arthur Compton over the next two decades. Thus it became apparent that light has both wave-like and particle-like properties. De Broglie, in his 1924 PhD thesis, sought to expand this wave-particle duality to all particles:
“ When I conceived the first basic ideas of wave mechanics in 1923–24, I was guided by the aim to perform a real physical synthesis, valid for all particles, of the coexistence of the wave and of the corpuscular aspects that Einstein had introduced for photons in his theory of light quanta in 1905. ” —De Broglie[3]
In 1926, Erwin Schrödinger published an equation describing how this matter wave should evolve—the matter wave equivalent of Maxwell’s equations—and used it to derive the energy spectrum of hydrogen. That same year Max Born published his now-standard interpretation that the square of the amplitude of the matter wave gives the probability of finding the particle at a given place. This interpretation was in contrast to De Broglie’s own interpretation, in which the wave corresponds to the physical motion of a localized particle.
de Broglie relations
Quantum mechanics
The de Broglie equations relate the wavelength λ to the momentum p, and frequency f to the total energy E of a particle:[2]
\begin{align} & \lambda = h/p\\ & f = E/h \end{align}
where h is Planck's constant. The equation can be equivalently written as
\begin{align} & p = \hbar k\\ & E = \hbar \omega\\ \end{align}
using the definitions
• $\hbar=h/2\pi$ is the reduced Planck's constant (also known as Dirac's constant, pronounced "h-bar"),
• $k= 2\pi/\lambda$ is the angular wavenumber,
• $\omega=2\pi f$ is the angular frequency.
In each pair, the second is also referred to as the Planck-Einstein relation, since it was also proposed by Planck and Einstein.
Special relativity
Using the relativistic momentum formula from special relativity
$p = \gamma m_0v$
allows the equations to be written as[4]
\begin{align}&\lambda = \frac {h}{\gamma m_0v} = \frac {h}{m_0v} \sqrt{1 - \frac{v^2}{c^2}}\\ & f = \frac{\gamma\,m_0c^2}{h} = \frac {m_0c^2}{h\sqrt{1 - \frac{v^2}{c^2}}} \end{align}
where m0 is the particle's rest mass, v is the particle's velocity, γ is the Lorentz factor, and c is the speed of light in a vacuum. See below for details of the derivation of the de Broglie relations. Group velocity (equal to the particle's speed) should not be confused with phase velocity (equal to the product of the particle's frequency and its wavelength). In the case of a non-dispersive medium, they happen to be equal, but otherwise they are not.
Group velocity
Albert Einstein first explained the wave–particle duality of light in 1905. Louis de Broglie hypothesized that any particle should also exhibit such a duality. The velocity of a particle, he concluded then (but may be questioned today, see above), should always equal the group velocity of the corresponding wave. De Broglie deduced that if the duality equations already known for light were the same for any particle, then his hypothesis would hold. This means that
$v_g = \frac{\partial \omega}{\partial k} = \frac{\partial (E/\hbar)}{\partial (p/\hbar)} = \frac{\partial E}{\partial p}$
where E is the total energy of the particle, p is its momentum, ħ is the reduced Planck constant. For a free non-relativistic particle it follows that
\begin{align} v_g &= \frac{\partial E}{\partial p} = \frac{\partial}{\partial p} \left( \frac{1}{2}\frac{p^2}{m} \right),\\ &= \frac{p}{m},\\ &= v. \end{align}
where m is the mass of the particle and v its velocity.
Also in special relativity we find that
\begin{align} v_g &= \frac{\partial E}{\partial p} = \frac{\partial}{\partial p} \left( \sqrt{p^2c^2+m^2c^4} \right),\\ &= \frac{pc^2}{\sqrt{p^2c^2 + m^2c^4}},\\ &= \frac{p}{m\sqrt{\left(\frac{p}{mc}\right)^2+1}},\\ &= \frac{p}{m\gamma},\\ &= \frac{mv\gamma}{m\gamma},\\ &= v. \end{align}
where m is the mass of the particle, c is the speed of light in a vacuum, $\gamma$ is the Lorentz factor, and v is the velocity of the particle regardless of wave behavior.
Group velocity (equal to an electron's speed) should not be confused with phase velocity (equal to the product of the electron's frequency multiplied by its wavelength).
Both in relativistic and non-relativistic quantum physics, we can identify the group velocity of a particle's wave function with the particle velocity. Quantum mechanics has very accurately demonstrated this hypothesis, and the relation has been shown explicitly for particles as large as molecules.[citation needed]
Phase velocity
In quantum mechanics, particles also behave as waves with complex phases. By the de Broglie hypothesis, we see that
$v_\mathrm{p} = \frac{\omega}{k} = \frac{E/\hbar}{p/\hbar} = \frac{E}{p}.$
Using relativistic relations for energy and momentum, we have
$v_\mathrm{p} = \frac{E}{p} = \frac{\gamma m c^2}{\gamma m v} = \frac{c^2}{v} = \frac{c}{\beta}$
where E is the total energy of the particle (i.e. rest energy plus kinetic energy in kinematic sense), p the momentum, $\gamma$ the Lorentz factor, c the speed of light, and β the speed as a fraction of c. The variable v can either be taken to be the speed of the particle or the group velocity of the corresponding matter wave. Since the particle speed $v < c$ for any particle that has mass (according to special relativity), the phase velocity of matter waves always exceeds c, i.e.
$v_\mathrm{p} > c, \,$
and as we can see, it approaches c when the particle speed is in the relativistic range. The superluminal phase velocity does not violate special relativity, as it carries no information. See the article on signal velocity for details.
Four-vectors
Using the four-momentum P = (E/c, p) and the four-wavevector K = (ω/c, k), the De Broglie relations form a single equation:
$\mathbf{P}= \hbar\mathbf{K}$
which is frame-independent.
Experimental confirmation
Matter waves were first experimentally confirmed to occur in the Davisson-Germer experiment for electrons, and the de Broglie hypothesis has been confirmed for other elementary particles. Furthermore, neutral atoms and even molecules have been shown to be wave-like.
Electrons
In 1927 at Bell Labs, Clinton Davisson and Lester Germer fired slow-moving electrons at a crystalline nickel target. The angular dependence of the reflected electron intensity was measured, and was determined to have the same diffraction pattern as those predicted by Bragg for x-rays. Before the acceptance of the de Broglie hypothesis, diffraction was a property that was thought to be only exhibited by waves. Therefore, the presence of any diffraction effects by matter demonstrated the wave-like nature of matter. When the de Broglie wavelength was inserted into the Bragg condition, the observed diffraction pattern was predicted, thereby experimentally confirming the de Broglie hypothesis for electrons.[5]
This was a pivotal result in the development of quantum mechanics. Just as the photoelectric effect demonstrated the particle nature of light, the Davisson–Germer experiment showed the wave-nature of matter, and completed the theory of wave-particle duality. For physicists this idea was important because it means that not only can any particle exhibit wave characteristics, but that one can use wave equations to describe phenomena in matter if one uses the de Broglie wavelength.
Neutral atoms
Experiments with Fresnel diffraction[6] and specular reflection[7][8] of neutral atoms confirm the application of the de Broglie hypothesis to atoms, i.e. the existence of atomic waves which undergo diffraction, interference and allow quantum reflection by the tails of the attractive potential.[9] Advances in laser cooling have allowed cooling of neutral atoms down to nanokelvin temperatures. At these temperatures, the thermal de Broglie wavelengths come into the micrometre range. Using Bragg diffraction of atoms and a Ramsey interferometry technique, the de Broglie wavelength of cold sodium atoms was explicitly measured and found to be consistent with the temperature measured by a different method.[10]
This effect has been used to demonstrate atomic holography, and it may allow the construction of an atom probe imaging system with nanometer resolution.[11][12] The description of these phenomena is based on the wave properties of neutral atoms, confirming the de Broglie hypothesis.
Molecules
Recent experiments even confirm the relations for molecules and even macromolecules, which are normally considered too large to undergo quantum mechanical effects. In 1999, a research team in Vienna demonstrated diffraction for molecules as large as fullerenes.[13] The researchers calculated a De Broglie wavelength of the most probable C60 velocity as 2.5 pm. More recent experiments prove the quantum nature of molecules with a mass up to 6910 amu.[14] In general, the De Broglie hypothesis is expected to apply to any well isolated object.
Spatial Zeno effect
The matter wave leads to the spatial version of the quantum Zeno effect. If an object (particle) is observed with frequency Ω >> ω in a half-space (say, y < 0), then this observation prevents the particle, which stays in the half-space y > 0 from entry into this half-space y < 0. Such an "observation" can be realized with a set of rapidly moving absorbing ridges, filling one half-space. In the system of coordinates related to the ridges, this phenomenon appears as a specular reflection of a particle from a ridged mirror, assuming the grazing incidence (small values of the grazing angle). Such a ridged mirror is universal; while we consider the idealised "absorption" of the de Broglie wave at the ridges, the reflectivity is determined by wavenumber k and does not depend on other properties of a particle.[8]
References
1. ^ L. de Broglie, Recherches sur la théorie des quanta (Researches on the quantum theory), Thesis (Paris), 1924; L. de Broglie, Ann. Phys. (Paris) 3, 22 (1925).
2. ^ a b Resnick, R.; Eisberg, R. (1985). Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles (2nd ed.). New York: John Wiley & Sons. ISBN 0-471-87373-X.
3. ^
4. ^ Holden, Alan (1971). Stationary states. New York: Oxford University Press. ISBN 0-19-501497-9.
5. ^ Mauro Dardo, Nobel Laureates and Twentieth-Century Physics, Cambridge University Press 2004, pp. 156–157
6. ^ R.B.Doak; R.E.Grisenti, S.Rehbein, G.Schmahl, J.P.Toennies2, and Ch. Wöll (1999). "Towards Realization of an Atomic de Broglie Microscope: Helium Atom Focusing Using Fresnel Zone Plates". Physical Review Letters 83 (21): 4229–4232. Bibcode:1999PhRvL..83.4229D. doi:10.1103/PhysRevLett.83.4229.
7. ^ F. Shimizu (2000). "Specular Reflection of Very Slow Metastable Neon Atoms from a Solid Surface". Physical Review Letters 86 (6): 987–990. Bibcode:2001PhRvL..86..987S. doi:10.1103/PhysRevLett.86.987. PMID 11177991.
8. ^ a b D. Kouznetsov; H. Oberst (2005). "Reflection of Waves from a Ridged Surface and the Zeno Effect". Optical Review 12 (5): 1605–1623. Bibcode:2005OptRv..12..363K. doi:10.1007/s10043-005-0363-9.
9. ^ H.Friedrich; G.Jacoby, C.G.Meister (2002). "quantum reflection by Casimir–van der Waals potential tails". Physical Review A 65 (3): 032902. Bibcode:2002PhRvA..65c2902F. doi:10.1103/PhysRevA.65.032902.
10. ^ Pierre Cladé; Changhyun Ryu, Anand Ramanathan, Kristian Helmerson, William D. Phillips (2008). "Observation of a 2D Bose Gas: From thermal to quasi-condensate to superfluid". arXiv:0805.3519.
11. ^ Shimizu; J.Fujita (2002). "Reflection-Type Hologram for Atoms". Physical Review Letters 88 (12): 123201. Bibcode:2002PhRvL..88l3201S. doi:10.1103/PhysRevLett.88.123201. PMID 11909457.
12. ^ D. Kouznetsov; H. Oberst, K. Shimizu, A. Neumann, Y. Kuznetsova, J.-F. Bisson, K. Ueda, S. R. J. Brueck (2006). "Ridged atomic mirrors and atomic nanoscope". Journal of Physics B 39 (7): 1605–1623. Bibcode:2006JPhB...39.1605K. doi:10.1088/0953-4075/39/7/005.
13. ^ Arndt, M.; O. Nairz, J. Voss-Andreae, C. Keller, G. van der Zouw, A. Zeilinger (14 October 1999). "Wave-particle duality of C60". Nature 401 (6754): 680–682. Bibcode:1999Natur.401..680A. doi:10.1038/44348. PMID 18494170.
14. ^ Gerlich, S.; S. Eibenberger, M. Tomandl, S. Nimmrichter, K. Hornberger, P. J. Fagan, J. Tüxen, M. Mayor & M. Arndt (5 April 2011). "Quantum interference of large organic molecules". Nature Communications 2 (263): 263–. Bibcode:2011NatCo...2E.263G. doi:10.1038/ncomms1263. PMC 3104521. PMID 21468015. | 2014-03-13 21:41:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9171647429466248, "perplexity": 1095.981107572352}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678682243/warc/CC-MAIN-20140313024442-00013-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://community.wolfram.com/groups/-/m/t/2381118 | [GiF] Ceiling tower equal to a positive integer
Posted 16 days ago
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Grid lines are solutions for valid pair of $r$ and positive integer $n$ in the ceiling tower equation with 6 levels | 2021-10-22 22:30:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23013906180858612, "perplexity": 3406.930213089796}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585522.78/warc/CC-MAIN-20211022212051-20211023002051-00121.warc.gz"} |
https://codegolf.stackexchange.com/questions/179626/heart-shaped-quine | # Heart-shaped quine [closed]
As you may or may not know, Valentine's day is getting close. So today, my task for you is:
# Write a heart-shaped quine
Rules:
• Code should look like an identifyable heart shape, i.e. Has two round parts on top and a pointy bottom. It does NOT have to be filled in, but this is not disallowed.
• Standard Quine rules apply (no opening/reading files, etc)
• This is a , so try to be as creative as possible in your solutions!
The closing date will be the day after valentine's day (15th of February), 12:00 UTC
• A popularity contest must always include an objective validity criterion. The validity criterion identifyable heart shape. It doesn't have to be perfect, but make sure it resembles a heart is clearly not objective – Luis Mendo Feb 7 '19 at 12:32
• @Martmists I do not think round and pointy are really objective with regards to text ... – Jonathan Frech Feb 7 '19 at 12:57
• popularity-contests are (ironically) not very popular these days, simply because they're very hard to define in a way that is objective while also allowing for variation in answers. – Jo King Feb 7 '19 at 13:22
• Title suggestion: A program that only loves itself – Jo King Feb 7 '19 at 13:23
• from a perl quine – Nahuel Fouilleul Feb 8 '19 at 17:02 | 2020-01-23 07:01:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27341052889823914, "perplexity": 2448.2179390576043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00495.warc.gz"} |
https://www.hackmath.net/en/math-problem/7546 | # Pupils
There are 32 pupils in the classroom, and girls are two-thirds more than boys.
a) How many percents are more girls than boys? Round the result to a whole percentage.
b) How many are boys in the class?
c) Find the ratio of boys and girls in the class. Write in the simplest form.
Correct result:
c = 12
d = 20
p = 67 %
f = 3:5
#### Solution:
c+d=32
d = c+ 2/3c
c+d=32
d = c+ 2/3•c
c+d = 32
5c-3d = 0
c = 12
d = 20
Our linear equations calculator calculates it.
$f=\frac{c}{d}=\frac{12}{20}=\frac{3}{5}=0.6=3:5$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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In the bag there are 3 red, 12 blue and 8 green balls. How many red balls we must be attached to the bag if we want the probability of pulling out the red balls was 20%? | 2020-10-27 06:19:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42415380477905273, "perplexity": 2140.438028863375}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107893402.83/warc/CC-MAIN-20201027052750-20201027082750-00543.warc.gz"} |
http://nrich.maths.org/public/leg.php?code=71&cl=4&cldcmpid=7440 | Search by Topic
Resources tagged with Mathematical reasoning & proof similar to Interactive Workout - Further:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
Square Mean
Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
Stage: 4 and 5 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
Target Six
Stage: 5 Challenge Level:
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
Fractional Calculus III
Stage: 5
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
Direct Logic
Stage: 5 Challenge Level:
Can you work through these direct proofs, using our interactive proof sorters?
Thousand Words
Stage: 5 Challenge Level:
Here the diagram says it all. Can you find the diagram?
To Prove or Not to Prove
Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
Stage: 5 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
Problem Solving, Using and Applying and Functional Mathematics
Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
Sums of Squares and Sums of Cubes
Stage: 5
An account of methods for finding whether or not a number can be written as the sum of two or more squares or as the sum of two or more cubes.
Modulus Arithmetic and a Solution to Differences
Stage: 5
Peter Zimmerman, a Year 13 student at Mill Hill County High School in Barnet, London wrote this account of modulus arithmetic.
Stage: 5 Challenge Level:
Find all positive integers a and b for which the two equations: x^2-ax+b = 0 and x^2-bx+a = 0 both have positive integer solutions.
Sprouts Explained
Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
Mechanical Integration
Stage: 5 Challenge Level:
To find the integral of a polynomial, evaluate it at some special points and add multiples of these values.
Proof Sorter - Geometric Series
Stage: 5 Challenge Level:
This is an interactivity in which you have to sort into the correct order the steps in the proof of the formula for the sum of a geometric series.
Logic, Truth Tables and Switching Circuits Challenge
Stage: 3, 4 and 5
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and fill in the blanks in truth tables to record. . . .
Logic, Truth Tables and Switching Circuits
Stage: 3, 4 and 5
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and record your findings in truth tables.
Plus or Minus
Stage: 5 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
Integral Inequality
Stage: 5 Challenge Level:
An inequality involving integrals of squares of functions.
Recent Developments on S.P. Numbers
Stage: 5
Take a number, add its digits then multiply the digits together, then multiply these two results. If you get the same number it is an SP number.
Sperner's Lemma
Stage: 5
An article about the strategy for playing The Triangle Game which appears on the NRICH site. It contains a simple lemma about labelling a grid of equilateral triangles within a triangular frame.
Rational Roots
Stage: 5 Challenge Level:
Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.
For What?
Stage: 4 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
Water Pistols
Stage: 5 Challenge Level:
With n people anywhere in a field each shoots a water pistol at the nearest person. In general who gets wet? What difference does it make if n is odd or even?
Binomial
Stage: 5 Challenge Level:
By considering powers of (1+x), show that the sum of the squares of the binomial coefficients from 0 to n is 2nCn
Big, Bigger, Biggest
Stage: 5 Challenge Level:
Which is the biggest and which the smallest of $2000^{2002}, 2001^{2001} \text{and } 2002^{2000}$?
The Triangle Game
Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
Proofs with Pictures
Stage: 5
Some diagrammatic 'proofs' of algebraic identities and inequalities.
DOTS Division
Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
Long Short
Stage: 4 Challenge Level:
A quadrilateral inscribed in a unit circle has sides of lengths s1, s2, s3 and s4 where s1 ≤ s2 ≤ s3 ≤ s4. Find a quadrilateral of this type for which s1= sqrt2 and show s1 cannot. . . .
Stonehenge
Stage: 5 Challenge Level:
Explain why, when moving heavy objects on rollers, the object moves twice as fast as the rollers. Try a similar experiment yourself.
Impossible Sandwiches
Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
Continued Fractions II
Stage: 5
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
Pair Squares
Stage: 5 Challenge Level:
The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.
Unit Interval
Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
Euclid's Algorithm II
Stage: 5
We continue the discussion given in Euclid's Algorithm I, and here we shall discover when an equation of the form ax+by=c has no solutions, and when it has infinitely many solutions.
Stage: 4 Challenge Level:
Four jewellers possessing respectively eight rubies, ten saphires, a hundred pearls and five diamonds, presented, each from his own stock, one apiece to the rest in token of regard; and they. . . .
Magic Squares II
Stage: 4 and 5
An article which gives an account of some properties of magic squares.
The Great Weights Puzzle
Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
Dodgy Proofs
Stage: 5 Challenge Level:
These proofs are wrong. Can you see why?
Whole Number Dynamics I
Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
Tree Graphs
Stage: 4 Challenge Level:
A connected graph is a graph in which we can get from any vertex to any other by travelling along the edges. A tree is a connected graph with no closed circuits (or loops. Prove that every tree. . . .
Proof Sorter - Sum of an AP
Stage: 5 Challenge Level:
Use this interactivity to sort out the steps of the proof of the formula for the sum of an arithmetic series. The 'thermometer' will tell you how you are doing
Whole Number Dynamics IV
Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
Stage: 3, 4 and 5 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
Diverging
Stage: 5 Challenge Level:
Show that for natural numbers x and y if x/y > 1 then x/y>(x+1)/(y+1}>1. Hence prove that the product for i=1 to n of [(2i)/(2i-1)] tends to infinity as n tends to infinity.
Basic Rhythms
Stage: 5 Challenge Level:
Explore a number pattern which has the same symmetries in different bases.
Euler's Squares
Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square. Three of the numbers that he found are a = 18530, b=65570, c=45986. Find the fourth number, x. You. . . .
Whole Number Dynamics III
Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
Whole Number Dynamics II
Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point. | 2016-07-30 16:58:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39563503861427307, "perplexity": 1257.3725821083115}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469258936356.77/warc/CC-MAIN-20160723072856-00306-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/535878/field-extensions-and-irreducible-polynomial-question/535912 | # Field extensions and irreducible polynomial question
I'm having some trouble solving this homework problem and understanding what the hint is trying to tell me.
Suppose $K$ is a field extension of $F$ of finite degree. Prove that if $\alpha$ is in $K$, then there is an irreducible polynomial $f(x)$ in $F[x]$ having a as a root. (Hint: If $[K : F]=n$, consider $1,a,a^2,...,a^n$)
-
Here's a little hint: the family of $1,a,a^2,\dots$ generates a subring of $K$ which is also a $F$-vector space, so subspace of $K$ as $F$-vector space. | 2014-12-21 00:01:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212873578071594, "perplexity": 86.46473706782238}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802770432.4/warc/CC-MAIN-20141217075250-00015-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://collegephysicsanswers.com/openstax-solutions/cruise-ship-mass-100-times-107-textrm-kg-strikes-pier-speed-0750-ms-it-comes | Question
A cruise ship with a mass of $1.00 \times 10^7 \textrm{ kg}$ strikes a pier at a speed of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s finances. Calculate the average force exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest.)
$4.69 \times 10^5 \textrm{ N}$
Solution Video | 2019-04-21 15:01:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31263232231140137, "perplexity": 1293.359031811716}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578531984.10/warc/CC-MAIN-20190421140100-20190421162100-00351.warc.gz"} |
http://www.imar.ro/~apopa/NTday5.html | # Fifth Bucharest Number Theory Day July 10-11, 2017
Organizers: Alina Cojocaru, Vicențiu Pașol, Alexandru Popa
The talks will take place at IMAR in Amfiteatrul "Miron Nicolescu" (parter). No registration is necessary.
The workshop is partly supported by BitDefender and by CNCS-UEFISCDI grant TE-2014-4-2077.
## Speakers
• Nicu Beli, IMAR
• Nicolae Bonciocat, IMAR
• Cristian Cobeli, IMAR
• Alina Cojocaru, University of Illinois at Chicago and IMAR
• Adrian Diaconu, University of Minnesotta and IMAR
• Alexandra Florea, Stanford University
• Nathan Jones, University of Illinois at Chicago
• Florin Nicolae, IMAR
• Alexandru Popa, IMAR
## Schedule
Monday, July 10th Tuesday, July 11th Florin Nicolae On the class group of imaginary quadratic fields Radu Gaba A generalization of a congruence of Ramanujan Coffee break Nathan Jones Elliptic curves with non-abelian entanglements Alina Cojocaru Constants in Titchmarsh divisor problems for elliptic curves Adrian Diaconu On averages of families of L-functions associated to moduli of curves over finite fields Nicolae Bonciocat Using prime numbers in attempts to understand polynomials Lunch break Alexandra Florea The fourth moment of quadratic Dirichlet L-functions over function fields Nicu Beli Analogues of the pn-th Hilbert symbol in characteristic p Coffee break Alexandru Popa On the trace formula for Hecke operators Cristian Cobeli On the numbers behind some beautiful pictures
## Abstracts
Nicu Beli: Analogues of the pn-th Hilbert symbol in characteristic p
The p-th Hilbert symbol (_,_) from characteristic different from p has two analogues in characteristic p: [_,_) and the lesser known ((_,_)). The symbol [_,_) generalizes via Witt vectors to an analogue of the pn-th Hilbert symbol, which takes values in the pn torsion of the Brauer group. We prove that ((_,_)) admits similar generalizations. The symbols we introduce are in terms of central simple algebras. Our construction involves Weyl algebras and Witt vectors. With the help of the new symbols we give a representation theorem for the pn torsion of the Brauer group, that extends a known result when n=1.
Nicolae Bonciocat: Using prime numbers in attempts to understand polynomials
We describe several ways in which prime numbers help us understand polynomials, especially factorization properties such as irreducibility and separability. We will analyse this from a number theoretical point of view, then from a geometrical point of view by Newton polygon methods, and finally from an algebraic point of view, in a non-archimedean setting.
Cristian Cobeli: On the numbers behind some beautiful pictures
We discuss a few problems related to the distribution of some nice sequences.
Alina Cojocaru: Constants in Titchmarsh divisor problems for elliptic curves
Inspired by the analogy between the group of units of the finite field $$\mathbb{F}_p$$ and the group of points of an elliptic curve $$E/\mathbb{F}_p$$, we consider elliptic curve analogues of the classical Titchmarsh divisor problem, such as analogues investigated by E. Kowalski, A. Akbary & D. Ghioca, and T. Freiberg & P. Kurlberg in the context of reductions modulo primes of an elliptic curve $$E/\mathbb{Q}$$. We pursue a comprehensive study of the constants emerging in these problems and prove upper bounds, explicit formulae, and averages for these constants. This is joint work with Renee Bell (MIT), Clifford Blakestad (Univ. Colorado), Alexander Cowan (Columbia Univ.), Nathan Jones (Univ. Illinois, Chicago), Vlad Matei (Univ. Wisconsin, Madison), Geoffrey Smith (Harvard Univ.), and Isabel Vogt (MIT), based on a research project started at the 2016 Arizona Winter School.
Adrian Diaconu: On averages of families of L-functions associated to moduli of curves over finite fields
In this talk I will discuss some aspects of equivariant Euler characteristics of the moduli space of stable curves over finite fields. These Euler characteristics occur naturally in the study of mean values of the corresponding families of L-functions.
Alexandra Florea: The fourth moment of quadratic Dirichlet L-functions over function fields
I will discuss moments of L-functions over function fields, and I will focus on the fourth moment in the family of quadratic Dirichlet L-functions, obtaining part of an asymptotic formula conjectured by Andrade and Keating.
Radu Gaba: A generalization of a congruence of Ramanujan
In this talk, we present a generalization to modular forms of prime level of Ramanujan's congruence between the Fourier coefficients of a cusp form and those of an Eisenstein series of weight 12 for the modular group. The proof is based on methods introduced by Ihara and used by Ribet and Serre in order to prove congruences between cusp forms of different level.
Nathan Jones: Elliptic curves with non-abelian entanglements
Let $$K$$ be a number field. An elliptic curve $$E/K$$ is said to have a non-abelian entanglement if there are relatively prime positive integers, $$m_1$$ and $$m_2$$, such that $$K(E[m_1])\cap K(E[m_2])$$ is a non-abelian Galois extension of $$K$$. In this talk, we will discuss our ongoing efforts to classify, using explicit methods, all infinite families of elliptic curves $$E/K$$, for a fixed $$K$$, with non-abelian entanglements. This problem is closely related to that of determining when the image of $$\rho_E$$ in $$GL_2(\hat{Z})$$ is maximal, and to the study of correction factors for various conjectural constants for elliptic curves over $$\mathbb{Q}$$. This is based on joint work with Ken McMurdy.
Florin Nicolae: On the class group of imaginary quadratic fields
Which are the imaginary quadratic fields with class group of exponent 4?
Alexandru Popa: On the trace formula for Hecke operators
In joint work with Don Zagier, we gave a new, algebraic proof of the trace formula for Hecke operators on modular forms for the modular group. I will present a generalization for congruence subgroups, which can be stated as a very simple cohomological trace formula. | 2017-11-18 19:42:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5723104476928711, "perplexity": 1192.1033281002944}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00639.warc.gz"} |
https://peoplepill.com/people/zacharias-dase | peoplepill id: zacharias-dase
ZD
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# Zacharias Dase
German mental calculator
Zacharias Dase
The basics
## Quick Facts
Intro German mental calculator A.K.A. Johann Martin Zacharias Dase Was Mathematician From Germany Type Mathematics Gender male Birth 1824, Hamburg, Germany Death 1861, Hamburg, Germany (aged 37 years)
The details (from wikipedia)
## Biography
Johann Martin Zacharias Dase (June 23, 1824, Hamburg – September 11, 1861, Hamburg) was a German mental calculator.
He attended schools in Hamburg from a very early age, but later admitted that his instruction had little influence on him. He used to spend a lot of time playing dominoes, and suggested that this played a significant role in developing his calculating skills. Dase suffered from epilepsy from early childhood throughout his life.
At age 15 he began to travel extensively, giving exhibitions in Germany, Austria and England. Among his most impressive feats, he multiplied 79532853 × 93758479 in 54 seconds. He multiplied two 20-digit numbers in 6 minutes; two 40-digit numbers in 40 minutes; and two 100-digit numbers in 8 hours 45 minutes. The famous mathematician Carl Friedrich Gauss commented that someone skilled in calculation could have done the 100-digit calculation in about half that time with pencil and paper.
These exhibitions however did not earn him enough money, so he tried to find other employments. In 1844 he obtained a position in the Railway Department of Vienna, but this didn't last long since in 1845 he was reported in Mannheim and in 1846 in Berlin.
In 1844, Dase calculated π to 200 decimal places over the course of approximately two months, a record for the time, from the Machin-like formula:
${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{8}}.}$
He also calculated a 7-digit logarithm table and extended a table of integer factorizations from 6,000,000 to 9,000,000.
Dase had very little knowledge of mathematical theory. The mathematician Julius Petersen tried to teach him some of Euclid's theorems, but gave up the task once he realized that their comprehension was beyond Dase's capabilities. Gauss however was very impressed with his calculating skill, and he recommended that the Hamburg Academy of Sciences should allow Dase to do mathematical work on a full-time basis, but Dase died shortly thereafter.
The book "Gödel, Escher, Bach" by Douglas Hofstadter mentions his calculating abilities. "... he also had an uncanny sense of quantity. That is, he could just 'tell', without counting, how many sheep were in a field, or words in a sentence, and so forth, up to about 30."
The contents of this page are sourced from Wikipedia article on 09 Mar 2020. The contents are available under the CC BY-SA 4.0 license. | 2021-12-06 20:10:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6183615922927856, "perplexity": 3798.9421313099506}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00090.warc.gz"} |
https://codegolf.stackexchange.com/questions/178344/oreoorererereoo/178494 | # Oreoorererereoo
Given an input string that is similar to the word "oreo", give an ASCII representation of the cookie that is as wide as the input string (to ensure cookie stability).
## Rules
• The input is lowercase, a non-empty string with no whitespace containing any combination of the strings "o" and "re", and containing only those strings.
• The string "o" represents the solid cookie, while the string "re" represents the filling.
• The output must be a stacked cookie that is as wide as the input string.
• The output may not be an array of strings
• The cookie must overlap the filling by one character on each side
• The characters used for the output don't have to match the output below (█ and ░), they just have to be different non-whitespace characters for the two parts of the cookie
• The whitespace padding on the left side of the filling is required, and any trailing whitespace is optional
## Examples
Input: oreo
Output:
████
░░
████
Input: o
Output:
█
Input: re
Output: (two spaces)
Input: rere
Output:
░░
░░
Input: oreoorererereoo
Output:
███████████████
░░░░░░░░░░░░░
███████████████
███████████████
░░░░░░░░░░░░░
░░░░░░░░░░░░░
░░░░░░░░░░░░░
░░░░░░░░░░░░░
███████████████
███████████████
Since this is code golf the shortest answer wins, good luck :)
• "The whitespace padding on each side of the filling is required". Does this actually mean that there must be a space character at the end of each line of filling? If so why? As long as it works visually then what does this requirement add to the challenge? – ElPedro Jan 6 at 0:23
• @ ElPedro Good point, I modified the rules and @Dennis I edited the rules so the comments should be okay to clean up – GammaGames Jan 7 at 3:22
• @JonathanAllan Since it's printing "ascii-art" I removed that rule, it looks like I forgot to update the question. Should be updated now. – GammaGames Jan 7 at 16:05
• Awesome, thanks! – Jonathan Allan Jan 7 at 16:18
• @GammaGames, if whitespace on the right is not required anymore, I assume the output for test case re should be now acceptable as 1 or 2 spaces, not necessarily 2? – Kirill L. Jan 10 at 8:24
# Jelly, 16 14 13 bytes
-1 Thanks to Erik the Outgolfer
OḂƇẒṁ€aØ.¦€⁶Y
Uses 1 for the cream and 0 for the cookie.
Try it online!
### How?
OḂƇẒṁ€aØ.¦€⁶Y - Main Link: list of characters, V e.g. 'orereo'
O - ordinal (vectorises) [111,114,101,114,101,111]
Ƈ - filter keep those for which:
Ḃ - modulo 2 [111, 101, 101,111]
Ẓ - is prime? (vectorises) [ 0, 1, 1, 0]
ṁ€ - mould each like V [[0,0,0,0,0,0],[1,1,1,1,1,1],[1,1,1,1,1,1],[0,0,0,0,0,0]]
€ - for each:
¦ - sparse application...
Ø. - ...to indices: literal [0,1] (0 is the rightmost index, 1 is the leftmost)
a - ...apply: logical AND with:
⁶ - space character [[0,0,0,0,0,0],[' ',1,1,1,1,' '],[' ',1,1,1,1,' '],[0,0,0,0,0,0]]
Y - join with newline characters [0,0,0,0,0,0,'\n',' ',1,1,1,1,' ','\n',' ',1,1,1,1,' ','\n',0,0,0,0,0,0]
- implicit print ...smashes everything together:
- 000000
- 1111
- 1111
- 000000
Previous 16 byter:
ḟ”eẋ€Ly@Ø.¦€⁾r Y
Uses r for the cream and o for the cookie.
Try it online!
• I was hoping for a Jelly entry, such an interesting language! – GammaGames Jan 4 at 17:39
# Pepe, 364 bytes
Unfortunately the online interpreter does not take care of compressing comments, hence all o characters will be replaced by a space.. Neither the spaces nor the o are necessary, so this could be 295 bytes, but I like it more this way:
rEeEEeeEeEororEEoreoreeeEeeeeeorEEEEeoREeoreorEeEEeEEEEororEEoreorEEEEEoREeoreorEeEEEeeEeororEEoreoReoREoREEEeoREEEEEoreorEorEEEeorEEEEEoreEoREeoreoREEeoREEEEeEeeoREEEeoREeeEoREEEeoREEEEEEEorEEEeEorEEEeoREoREEEeoREEEEEoREEoReoreorEEEeEoREEEEEEeorEEEeoReEoREoREEEeoREEoReoroReEeoREoREEEeorEEEEeoReeoREEEeoREeeEoREEEeoREEEEEEEoreoReoReoREoREEEeoREEEEEoreeeeeEeEeoRee
Try it online!
## Ungolfed
There might be some golfing oppurtunities with flags which I missed, but I'm done for now:
# "function" for 'e'
rEeEEeeEeE rrEE
re # remove duplicated argument
reeeEeeeee # print space
rEEEEe # decrement counter twice
REe re
# "function" for 'o'
rEeEEeEEEE rrEE
re # remove duplicated argument
rEEEEE # increment counter
REe re
# "function for 'r'
rEeEEEeeEe rrEE
re Re # remove duplicated argument & char
RE REEEe REEEEE # push 1
re rE rEEEe rEEEEE # replace 1
reE # goto 1
REe re
# Main
REEe REEEEeEee # read input & reverse
REEEe REeeE REEEe REEEEEEE # push length-1 & move to r
rEEEeE rEEEe # dummy loop-var (fucking do-whiles...)
RE REEEe REEEEE REE # while [label-1]
# Call the right procedure depending on current character,
# sets stacks up as follows:
# R [ .... *currentChar ]
# r [ (N-1) *count ]
Re re # pop 1 & loop-counter
rEEEeE # duplicate counter
REEEEEEe rEEEe # copy current char to other stack
ReE # jeq to 'o'-label or 'e'-label
# Output currentChar count times:
RE REEEe REE # while [label-0]:
Re # pop 0
rReEe # print character
RE REEEe # push 0
rEEEEe # decrement counter
Ree
REEEe REeeE REEEe REEEEEEE # push length-1 & move to r
re Re Re # pop 0, counter and 9((((currentChar
RE REEEe REEEEE # push 1
reeeeeEeEe # print new-line
Ree
# Canvas, 1918 17 bytes
e ∙╋
：r≠＊┤］；Ｌ×⁸↔⁸
Try it here!
Uses the annoyingly long code of ：r≠＊┤］ to remove rs from the input..
• That's a handy feature, and cool language! – GammaGames Jan 4 at 1:48
# Japt-R, 16 15 bytes
re ¬£çX sX²èrÃû
Try it
:Implicit input of string U
re :Remove all "e"s
¬ :Split to array of characters
£ :Map each X
çX : Repeat X to the length of U
s : Slice from index
X² : Duplicate X
èr : Count the occurrences of "r"
à :End map
û :Centre pad each element with spaces to the length of the longest
:Implicitly join with newlines and output
re ¬£îX rr²i^Ãû
Try it online!
# Alternative using Aggregate, 108 bytes
Try it online!
• now it trims trailing spaces.. – dzaima Jan 4 at 1:50
• There was enough feedback that I removed the trailing newline rule. Feel free to update your entry. – GammaGames Jan 4 at 4:40
• Your replace doesn't work when the input is o, since the n.Length-2 will result in -1. – Kevin Cruijssen Jan 4 at 10:18
• The n.Length-2 is is for when the input has re. – Embodiment of Ignorance Jan 4 at 16:27
# R, 106 bytes
function(s,N=nchar(s)){m=rep(el(strsplit(gsub('re',0,s),'')),e=N)
m[m<1&seq(m)%%N<2]=' '
write(m,1,N,,"")}
Try it online!
• -12 bytes thanks to @Giuseppe
Previous version with explanation :
# R, 118 bytes
function(s,N=nchar(s)){m=t(replicate(N,el(strsplit(gsub('re',0,s),''))))
m[m<1&row(m)%in%c(1,N)]=' '
write(m,1,N,,'')}
Try it online!
• -1 byte thanks to @Giuseppe
Unrolled code and explanation :
function(s){ # s is the input string, e.g. 'oreo'
N = nchar(s) # store the length of s into N, e.g. 4
s1 = gsub('re',0,s) # replace 're' with '0' and store in s1, e.g. 'o0o'
v = el(strsplit(s1,'')) # split s1 into a vector v of single characters
# e.g. 'o','0','o'
m = replicate(N,v) # evaluate N times the vector v and arrange
# the result into a matrix m (nchar(s1) x N)
# e.g.
# 'o' 'o' 'o' 'o'
# '0' '0' '0' '0'
# 'o' 'o' 'o' 'o'
m = t(m) # transpose the matrix
m[m<1 & row(m)%in%c(1,N)] = ' ' # substitute the zeros (i.e. where < 1)
# on the 1st and last row of the matrix with ' ' (space)
# e.g.
# 'o' ' ' 'o'
# 'o' '0' 'o'
# 'o' '0' 'o'
# 'o' ' ' 'o'
write(m,1,N,,'') # write the matrix to stdout (write function transposes it)
# e.g.
# oooo
# 00
# oooo
}
• 106 bytes – Giuseppe Jan 4 at 22:40
• aaand 104 bytes returning a list of lines, which isn't acceptable here, but it's an interesting idea (essentially my SNOBOL submission translated to R) – Giuseppe Jan 4 at 23:11
# 05AB1E, 1817 16 bytes
'eKεD'rQ2*Igα×}.c
-1 byte thanks to @Emigna
Uses o for the cookie and r for the filling.
Explanation:
'eK '# Remove all "e" from the (implicit) input
# i.e. "orereo" → "orro"
ε } # Map all characters to:
D # Duplicate the current character
'rQ '# Check if it's an "r" (1 if truthy; 0 if falsey)
# i.e. "r" → 1
# i.e. "o" → 0
· # Double that
# i.e. 1 → 2
# i.e. 0 → 0
Ig # Take the length of the input
# i.e. "orereo" → 6
α # Take the absolute difference between the two
# i.e. 2 and 6 → 4
# i.e. 0 and 6 → 6
× # Repeat the duplicated character that many times
# i.e. "r" and 4 → "rrrr"
# i.e. "o" and 6 → "oooooo"
.c # Then centralize it, which also imlicitly joins by newlines
# (and the result is output implicitly)
# i.e. ["oooooo","rrrr","rrrr","oooooo"]
# → "oooooo\n rrrr\n rrrr\noooooo"
• Creative solution, But it does not solve the problem entirely: oro would give a wrong answer – Mark Smit Jan 6 at 9:15
• @MarkSmit oro isn't a possible input, since the input will only contain os and res. Regardless, oro still seems to output correctly following the spec, since it outputs ooo\n r\nooo. What is wrong about it? – Kevin Cruijssen Jan 6 at 10:44
• This is invalid: "The whitespace padding on each side of the filling is required" – NieDzejkob Jan 6 at 15:14
• 2* can be · and the missing whitespace can be fixed by changing ».c to .c.B» – Emigna Jan 7 at 9:29
• @Emigna Ah, can't believe I haven't thought about ·, thanks! :) And always nice to have changing specs during the challenge, sigh.. – Kevin Cruijssen Jan 7 at 9:40
# Retina, 74 73 bytes
I feel like I haven't posted an answer in a very long time. Well, here I am. Also, Retina has changed a lot, and I feel like I suck at it now.
.+
$0$.0
(\d+)
*
e
o|r
$&¶ _$
+(/_/&o¶
oo¶
_$)/_/&r¶ rr¶ ¶$
m^r
Try it online!
• Whoa, what a crazy looking language. I like it! – GammaGames Jan 4 at 1:58
• doesn't include trailing whitespaces.. – dzaima Jan 4 at 2:00
• I like how [or] means o or r instead of [ or ]. Makes my head hurt. – nedla2004 Jan 4 at 2:43
• @dzaima The question does not specify that trailing whitespaces are required. A comment asked, but no reply was given. – mbomb007 Jan 4 at 19:44
• @nedla2004 That actually helped me notice a way to save a byte. Thanks. – mbomb007 Jan 4 at 19:44
# Retina, 21 bytes
r
L$.$.+*$& \bee Try it online! Explanation: r Delete the rs. L$.
$.+*$&
List each letter on its own line repeated to the length of the original input.
\bee
Replace the first two ees on each line with a space.
• This breaks the rules: "The whitespace padding on each side of the filling is required" – NieDzejkob Jan 6 at 15:16
• @NieDzejkob Sorry for overlooking that, should be fixed now. – Neil Jan 6 at 16:14
• FYI trailing whitespace requirement lifted. – Jacktose Jan 8 at 0:26
• @Neil You should fix that & :P – ASCII-only Jan 8 at 0:35
# C (gcc), 135113109 104 bytes
#define $putchar(33 O(char*r){for(char*e,*o=r,x;*r;$-23))for(x=*r++>111,e=x?$-1),r++,o+2:o;*e++;$+x));}
Try it online!
• Shave off a few bytes with -D$=putchar – user77406 Jan 4 at 13:55 • 131 bytes if you add a trailing newline as allowed by the rules. – NieDzejkob Jan 6 at 13:54 • 127 bytes if you move the e=o to the condition of the first for loop and then remove the else. – NieDzejkob Jan 6 at 14:20 • 118 bytes if you choose the cookie and filling characters carefully. – NieDzejkob Jan 6 at 14:58 • 113 bytes – NieDzejkob Jan 6 at 15:02 # JavaScript ES6, 103 bytes ### Using replace 103 bytes: x=>x.replace(/o/g,"-".repeat(s=x.length)+ ).replace(/re/g," "+"|".repeat(s>1?s-2:0)+ ).slice(0,-1) Try it online! ### Using split and map 116 bytes: x=>x.split("re").map(y=>("-"[h='repeat'](r=x.length)+ )[h](y.length)).join(" "+"|"[h](r>1?r-2:0)+ ).slice(0,-1) Try it online! • JS, nice! You reminded me that I was going to add a rule about not having line returns at the end of the output, I've added it. Sorry about that! – GammaGames Jan 4 at 1:45 • just removing the final newline is 12 bytes – fəˈnɛtɪk Jan 4 at 1:51 • There was enough feedback that I removed the trailing newline rule. Feel free to update your entry. – GammaGames Jan 4 at 4:44 • You can save a byte by using a template string with${"|".repeat(s>1?s-2:0)} and its whitespaces, instead of using " "+"|".repeat(s>1?s-2:0). – Ismael Miguel Jan 4 at 10:26
• If you use backticks for the string in the first split, you can remove the parentheses around it. – skiilaa Jan 11 at 19:56
# Perl 5-p, 47 bytes
s|o|X x($i=y///c).$/|ge;s|re|$".O x($i-2).$/|ge Try it online! • This breaks the rules: "The whitespace padding on each side of the filling is required" – NieDzejkob Jan 6 at 15:16 • with some variations tio.run/##K0gtyjH9/… – Nahuel Fouilleul Jan 7 at 15:16 # Python 3, 77 bytes lambda x:x.replace("o","-"*len(x)+"\n").replace("re"," "+'.'*(len(x)-2)+"\n") Try it online! • Clever! I did intend the output to not be printing whitespace for the filling (it's pretty much oreo ascii), so I have edited the rules accordingly. Sorry about that! And I always love a python answer :) – GammaGames Jan 4 at 4:50 • @JonathanFrech migth as well delete the comments, that approach was invalidated. I'll work on golfing more tomorrow. – Rɪᴋᴇʀ Jan 4 at 4:55 • You can remove the space at +" \n" to save a byte. – Kevin Cruijssen Jan 4 at 10:37 • @KevinCruijssen can I? The input program says the whole cookie must be as wide as the input. – Rɪᴋᴇʀ Jan 4 at 16:18 • I interpreted that as meaning that a trailing space is the same (visually) as no space. That's the beauty of answers to ascii art challenges. If they look right they are right :-) – ElPedro Jan 4 at 21:33 # Mathematica, 111 91 bytes #~StringReplace~{"o"->"O"~Table~(n=StringLength@#)<>"\n","re"->" "<>Table["R",n-2]<>" \n"}& Try It Online! This was majorly shortened thanks to Misha's edits. My original code: (z=StringRepeat;n=StringLength@#;#~StringReplace~{"o"->"O"~z~n<>"\n","re"->" "<>If[n>2,z["R",n-2],""]<>" \n"})& This code is not very fancy but it seems too expensive to convert away from strings and then back or to do anything else clever. In particular, with only 3-4 commands that have the name String, my original approach couldn't save bytes at all by trying to abstract that away. For example, the following is 129 bytes: (w=Symbol["String"<>#]&;z=w@"Repeat";n=w["Length"]@#;#~w@"Replace"~{"o"->"O"~z~n<>"\n","re"->" "<>If[n>2,z["R",n-2],""]<>" \n"})& • A few improvements: StringRepeat can be Table since <> will convert the list into a string later; the If is unnecessary since we take the re branch only when n is at least 2; we can save on parentheses by defining n only when we use it. Try it online! – Misha Lavrov Jan 5 at 20:32 • @MishaLavrov The If was added because StringRepeat would throw an error on the case of "re"; it doesn't allow you to repeat a string 0 times. Table has no such limitation, so that's a big save! – Mark S. Jan 5 at 21:33 # Perl 6, 37 bytes {m:g/o|r/>>.&({S/rr/ /.say}o*x.comb)} Try it online! Anonymous code block that takes a string and prints the oreo, with o as the cookie and r as the cream. ### Explanation: { } # Anonymous code block m:g/o|r/ # Select all o s and r s >>.&( ) # Map each letter to *x.comb # The letter padded to the width S/rr/ / # Substitute a leading rr with a space .say # And print with a newline • I didn't realize o could be used in place of . Very nicely golfed. – primo Jan 16 at 13:22 # Java 11, 110 bytes s->{int l=s.length();return s.replace("re"," "+"~".repeat(l-(l<2?1:2))+"\n").replace("o","=".repeat(l)+"\n");} Uses = for the cookie and ~ for the filling. Try it online. Explanation: s->{ // Method with String as both parameter and return-type int l=s.length(); // Get the length of the input return s // Return the input .replace("re", // After we've replaced all "re" with: " " // A space +"~".repeat(l-(l<2?1:2)) // Appended with length-2 amount of "~" // (or length-1 if the input-length was 1) +"\n") // Appended with a newline .replace("o", // And we've also replaced all "o" with: "=".repeat(l) // Length amount of "=" +"\n");} // Appended with a newline The above solution uses a replace. The following maps over the characters of the input instead: # Java 11, 113 112 bytes s->s.chars().forEach(c->{if(c>101)System.out.println((c>111?" ":"")+(""+(char)c).repeat(s.length()-2*(~c&1)));}) -1 byte thanks to @Neil. Try it online. Explanation: s-> // Method with String parameter and no return-type s.chars().forEach(c->{ // Loop over the characters as codepoint-integers if(c>101) // If it's not an 'e': System.out.println( // Print with trailing newline: (c>111? // If it's an 'r' " " // Start with a space : // Else (it's an 'o' instead) "") // Start with an empty string +(""+(char)c).repeat( // And append the character itself .repeat( // Repeated the following amount of times: s.length() // The input-length -2*(~c&1)));}) // Minus 2 if it's an "r", or 0 if it's an "o" • Can you use ~c&1? – Neil Jan 4 at 11:34 • @Neil I indeed can, thanks. – Kevin Cruijssen Jan 4 at 11:57 • This is invalid: "The whitespace padding on each side of the filling is required" – NieDzejkob Jan 6 at 15:15 • @NieDzejkob Fixed.. Always nice to have changing specs during the challenge, sigh.. – Kevin Cruijssen Jan 7 at 9:43 • @KevinCruijssen not anymore :P – ASCII-only Jan 7 at 23:32 # JavaScript, 7265 64 bytes s=>s.replace(/.e?/g,([x,y])=>(y? : ).padEnd(s.length+!y,x)) Try it online # Powershell, 7169 66 bytes -2 bytes thanks @Veskah -3 bytes thanks @AdmBorkBork$l=$args|% le* switch($args|% t*y){'o'{'#'*$l}'r'{" "+'%'*($l-2)}}
Less golfed test script:
$f = {$l=$args|% length switch($args|% t*y){
'o'{'#'*$l} 'r'{" "+'%'*($l-2)}
}
}
@(
,(
'oreo',
'####',
' %%',
'####'
)
,(
'o',
'#'
)
,(
're',
' '
)
,(
'rere',
' %%',
' %%'
)
,(
'oreoorererereoo',
'###############',
' %%%%%%%%%%%%%',
'###############',
'###############',
' %%%%%%%%%%%%%',
' %%%%%%%%%%%%%',
' %%%%%%%%%%%%%',
' %%%%%%%%%%%%%',
'###############',
'###############'
)
) | % {
$s,$expected = $_$result = &$f$s
"$result"-eq"$expected"
# $result # uncomment this line to display a result } Output: True True True True True • Looks like you don't need parens around the$args 69 bytes – Veskah Jan 4 at 23:30
• The length of [string[]] is an [int[]]... The [int[]] is [int] if the array contains one element only. Great! Thanks! – mazzy Jan 5 at 8:55
• The OP updated the challenge so you don't need trailing spaces anymore. This means your r can be " "+'%'*($l-2) instead for -3 bytes. – AdmBorkBork Jan 18 at 20:19 # Charcoal, 19 bytes Ｆθ≡ιo⟦⭆θ#⟧e«→Ｐ⁻Ｌθ²↙ Try it online! Link is to verbose version of code. Explanation: Ｆθ Loop through the characters of the input string. ≡ι Switch on each character. o⟦⭆θ#⟧ If it's an o then print the input string replaced with #s on its own line. e«→Ｐ⁻Ｌθ²↙ If it's an e then move right, print a line of -s that's two less than the length of the input string, then move down and left. # Bash, 87 bytes Without sed: f(){ printf %$1s|tr \ $2;} c=${1//o/f ${#1} B } echo "${c//re/ f $[${#1}-2] F
}"
Thanks to @manatwork.
With sed (90 bytes):
f(){ printf %$1s|tr \$2;}
echo $1|sed "s/o/f${#1} B\n/g;s/re/ f $[${#1}-2] F` \n/g"
• Could you show us some sample usage? I'm a bit confused by your function expecting 2 parameters. – manatwork Jan 6 at 18:34
• You write that into a script called test.sh. Then, you call test.sh from the command line as follows: bash test.sh oreoorererereoo. f is needed to repeat the character $2$1 number of times – Green Jan 6 at 18:41
• Oops. I completely misunderstood function f. Some further minor changes could be made there: Try it online! – manatwork Jan 6 at 18:55
• 75 bytes. – Dennis Jan 7 at 4:15
• 60 bytes – Nahuel Fouilleul Jan 8 at 8:27
# C# (Visual C# Interactive Compiler), 71 bytes
Try it online!
Borrowed some ideas from on Embodiment of Ignorance's answer for sure.
-6 bytes thanks to @ASCIIOnly!
The overall concept is to compute a string aggregate over the input characters following these rules:
• If an r is encountered, append a single space character for indentation. We know the next character will be an e.
• If an o or an e is encountered, generate a string by repeating the current character a specific number of times and prepending it to a newline or some padding and a newline.
• The number of times to repeat is determined by length of input string and whether the current line is indented.
• The PadLeft function is used to generate the repeating character string.
The result is the concatenation of all of these strings.
• 71 – ASCII-only Jan 5 at 1:09
• @ASCIIOnly - Thanks :) – dana Jan 5 at 1:57
• > The whitespace padding on each side of the filling is required – ASCII-only Jan 5 at 4:57
• 85? – ASCII-only Jan 5 at 5:01
• I didn't notice that :) Although, in reviewing the posted answers about 1/2 have done this incorrectly as well. Good catch though! – dana Jan 5 at 5:08
# PHP, 10099 93 bytes
$l=strlen($i=$argv[1]);$r=str_repeat;echo strtr($i,[o=>$r(X,$l)." ",re=>' '.$r(o,$l-2)." "]); Try it online! OUCH. PHP's waaaay_too_long function names strike again! Output:$php oreo.php oreo
XXXX
oo
XXXX
$php oreo.php o X$php oreo.php rere
oo
oo
$php oreo.php oreoorererereoo XXXXXXXXXXXXXXX ooooooooooooo XXXXXXXXXXXXXXX XXXXXXXXXXXXXXX ooooooooooooo ooooooooooooo ooooooooooooo ooooooooooooo XXXXXXXXXXXXXXX XXXXXXXXXXXXXXX • Invalid, cream lines need a trailing space – ASCII-only Jan 5 at 5:03 • Fixed the trailing space. Thanks! – 640KB Jan 5 at 12:09 • Oh boy, PHP! Also any trailing whitespace is now optional, there were enough people that pointed out that since it's printing out ascii it shouldn't really be required. – GammaGames Jan 7 at 3:26 # Pyth, 28 bytes FNzIqN"o"*lzN)IqN"r"+d*-lz2N FNz For each value, N, in input IqN"o" if the character is "o" *lzN return the character times the length of the input ) end if IqN"r" if the character is "r" FNzIqN"o"*lzN)IqN"r"+d*-lz2N *-lz2N return the character times length - 2 +d padded on the left with " " Try it here! This one uses a loop. # Pyth, 30 bytes (As string replace) ::z"o"+*lz"="b"re"++d*-lz2"~"b :z"o" With the input, replace "o" with *lz"=" "=" times the length of the input + b and a newline added to the end : "re" With the input, replace "re" with * "~" "~" times -lz2 the length of the input minus 2 +d padded on the left with " " + b and a newline added to the end Try it here! This one uses string replacement. I really like python (it's what I wrote my original test scripts in), so I thought I'd do a pyth entry for fun :) • Isn't this 37 bytes? I thought Pyth uses default ASCII as its codepage just like Python, if I remember correctly. So even though your code is 33 characters, both and are three bytes each. Or am I missing something here? – Kevin Cruijssen Jan 4 at 7:05 • Good call, I didn't realize that (I couldn't get pyth to work on tio.run, so I used the length counter on the herokuapp page). In the for loop I could just replace the character with N, even saving a few bytes! – GammaGames Jan 4 at 14:38 • Thought something like that happened. :) I once had the same issue with a 05AB1E answer of mine that was using characters outside its code page. Unfortunately TIO displays chars and bytes the same for most golfing languages. For Java or Python TIO will correctly state 33 chars, 37 bytes, but not in golfing languages on TIO. But in your solutions just changing those characters indeed fixes the issue, so it's not that big of a deal here. – Kevin Cruijssen Jan 4 at 14:41 • @KevinCruijssen Wait, 05AB1E doesn't use an actual SBCS? – ASCII-only Jan 6 at 1:08 • If you're interested, it seems to work effortlessly on TIO for me. – NieDzejkob Jan 9 at 16:07 # PHP, 9687 85 bytes Thanks to @gwaugh -9 Bytes Thanks to @manatwork -2 Bytes <?=strtr($i=$argv[1],[o=>($r=str_repeat)(X,$l=strlen($i))."
",re=>" {$r(o,$l-2)}
"]);
Try it online!
Try it online! (87 Bytes)
And a recursive function
# PHP, 135 bytes
function f($w,$x=0){$f=str_repeat;echo($x<($l=strlen($w)))?($w[$x]=='o')?$f(█,$l)."
".f($w,$x+1):" ".$f(░,$l-2)."
".f($w,$x+2):"";}
Try it online! (recursive)
• by combining the best of our two submissions I was able to get it down to 87 bytes TIO. Would you be game to go in with this as a collaborative submission? :) – 640KB Jan 9 at 16:55
• I think we can remove 1 more byte by using the command short_tag_open, and instead of <?= we can use <?, or am i mistaken? – Francisco Hahn Jan 9 at 18:46
• 2 characters shorter with string interpolation: ' '.$r(o,$l-2).""" {$r(o,$l-2)}". – manatwork Jan 9 at 18:48
• Thanks @manatwork sometimes i forgot php vars are evaluated in a string if the entire string is declared with "" istead of '' – Francisco Hahn Jan 9 at 18:58
• This can be 3 bytes shorter using $argn: Try it online! – Night2 Sep 2 at 12:00 # Ruby, 62 60 bytes ->s{s.gsub /./,?r=>" #{(?**z=s.size)[0..-3]} ",?o=>?O*z+?\n} Try it online! Uses O for the cookie, * for the filling. -1 thanks to @manatwork pointing out a silly mistake and another -1 due to relaxation of the rules about whitespaces. • No need for parenthesis around .gsub's parameters. – manatwork Jan 9 at 20:54 # C# (.NET Core), 143 bytes Without LINQ. p=>{var q="";foreach(char c in p){if(c!='e'){for(var j=0;j<p.Length;j++)q+=(j<1|j>p.Length-2)&c>'q'?" ":c<'p'?"█":"░";q+="\n";}}return q;}; Try it online! # Clojure, 137 bytes (fn[f](let[w(count f)r #(apply str(repeat % %2))](clojure.string/join"\n"(replace{\o(r w \#)\e(str \ (r(- w 2)\-) \ )}(remove #{\r}f))))) I'm not using the nice characters in the printout in the golfed version since those are expensive. Returns a string to be printed. Try it online! See below for explanation. Pre-golfed: ; Backslashes indicate a character literal (defn oreo [format-str] (let [width (count format-str) ; A helper function since Clojure doesn't have built-in string multiplication str-repeat #(apply str (repeat % %2)) ; Define the layers cookie (str-repeat width \█) cream (str \ (str-repeat (- width 2) \░) \ )] (->> format-str ; Take the input string, (remove #{\r}) ; remove r for simplcity, (replace {\o cookie, \e cream}) ; replace the remaining letters with the layers, (clojure.string/join "\n")))) ; and join the layers together with newlines # Dart, 120106 107 bytes f(s)=>s.replaceAll('o',''.padRight(s.length,'#')+'\n').replaceAll('re',' '.padRight(s.length-1,'-')+' \n'); Try it online! • +1 byte : Added trailing whitespace • This is invalid: "The whitespace padding on each side of the filling is required" – NieDzejkob Jan 6 at 15:15 • Oh, never mind then, I'll correct it soon. Thanks for the info, I missed it – Elcan Jan 6 at 22:14 # Python 2, 7776 72 bytes lambda i:'\n'.join((x*len(i),' '+x*(len(i)-2))[x>'o']for x in i if'e'<x) Try it online! The outer part of the cookie is 'o' and the filling is 'r'. • 68 bytes. Although I doubt if you can really omit the trailing spaces, the spec does say "The whitespace padding on each side of the filling is required"... – Erik the Outgolfer Jan 5 at 22:35 • Thanks @EriktheOutgolfer. Thought a lambda would be shorter! Guess in this case not. Had missed the requirement about the mandatory trailing space on the filling. Really can't see the point with an ascii art challenge but if that's what OP requires then I guess my answer is invalid anyway. – ElPedro Jan 6 at 0:13 • Now corrected... – ElPedro Jan 6 at 0:18 • Why bring it back to 76? Just put +' ' after (l-2). Also, you have a typo, *' ' must be +' '. – Erik the Outgolfer Jan 6 at 0:23 • That's what I did with my current solution. Will take a closer look at your hints tomorrow (later today). It's late here and I have been shoveling snow all day so too tired for golf. Thanks for the tips though :) – ElPedro Jan 6 at 0:27 # x86-64 machine code (Linux), 97 bytes 0000000000000000 <oreo_asm>: 0: 56 push %rsi 1: 57 push %rdi 0000000000000002 <len>: 2: 48 ff c7 inc %rdi 5: 80 3f 00 cmpb$0x0,(%rdi)
8: 75 f8 jne 2 <len>
a: 49 89 fc mov %rdi,%r12
d: 5f pop %rdi
e: 49 29 fc sub %rdi,%r12
11: 4d 31 f6 xor %r14,%r14
14: eb 18 jmp 2e <outer_loop.skip>
0000000000000016 <extra>:
16: 41 c6 01 20 movb $0x20,(%r9) 1a: c6 03 20 movb$0x20,(%rbx)
1d: 49 ff ce dec %r14
20: eb 06 jmp 28 <outer_loop>
0000000000000022 <newline>:
22: c6 06 0a movb $0xa,(%rsi) 25: 48 ff c6 inc %rsi 0000000000000028 <outer_loop>: 28: 49 ff c6 inc %r14 2b: 48 ff c7 inc %rdi 000000000000002e <outer_loop.skip>: 2e: 44 8a 07 mov (%rdi),%r8b 31: 41 80 f8 65 cmp$0x65,%r8b
35: 74 df je 16 <extra>
37: 45 84 c0 test %r8b,%r8b
3a: 74 23 je 5f <done>
3c: 48 89 f3 mov %rsi,%rbx
000000000000003f <inner_loop>:
3f: 44 88 06 mov %r8b,(%rsi)
42: 49 89 f1 mov %rsi,%r9
45: 48 ff c6 inc %rsi
48: 48 31 d2 xor %rdx,%rdx
4b: 48 89 f0 mov %rsi,%rax
4e: 48 2b 04 24 sub (%rsp),%rax
52: 4c 29 f0 sub %r14,%rax
55: 49 f7 f4 div %r12
58: 48 85 d2 test %rdx,%rdx
5b: 74 c5 je 22 <newline>
5d: eb e0 jmp 3f <inner_loop>
000000000000005f <done>:
5f: 5e pop %rsi
60: c3 retq
This x86-64 function takes in the pointer to the input string in rsi and builds the output starting at the pointer in rdi (these are the registers used to pass the first two arguments from a C function on Linux). For convenience, I've written a C++ wrapper for this which also does nice input sanitization and prints the output. That code can be located here. This also shows the original nasm syntax assembly I wrote for this function (as well as the non-golfed version I got working first).
A few things to note is that this code doesn't respect any callee saved registers, which means that the C++ code likely will crash if run for a while after calling this function. On my machine it doesn't, but that's rather surprising. I also don't add a null byte to delimit the output string, and instead the space allocated for the output string is pre-filled with bytes. (If this isn't allowed I can add the null terminator at a cost of 3 bytes).
The logic for this code is essentially counting the length of the string, then building a line of this length for each 'o' and 'r' characters seen in the input string, and then for any 'e' character seen, replacing the first and last characters on the previous line with space characters.
I can't find anywhere online to compile and run a mix of C++ and nasm source code, so I might write some small wrapper code for this to prove it works. Otherwise you should be able to compile and run this with the makefile in the link I gave with the command:
$make oreo ASM_FILE=oreo_golf.nasm$ ./oreo oreoorererereoo --use_asm
I was able to format the assembly to something acceptable by gcc, so try it online!
• Oh my, now this is an entry! – GammaGames Jan 8 at 15:09 | 2019-10-15 18:29:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2572019696235657, "perplexity": 8596.631215187082}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00490.warc.gz"} |
https://forum.azimuthproject.org/plugin/ViewComment/17029 | [Artur Grzesiak wrote:](https://forum.azimuthproject.org/discussion/comment/16937/#Comment_16937)
> There is a difference in defining symmetry.
> In the lecture: for all \$$x,y \in X\$$, \$$x \sim y\$$ implies \$$y \sim x.\$$
> In the book (definition 1.8): \$$a \sim b\$$ iff \$$b \sim a\$$, for all \$$a,b \in X\$$
> Implication is weaker than *iff*, but to be honest I am not sure if this is the root cause of the problem, as for me they *feel*... *equivalent* in this particular case.
As you note, the book's definition implies mine, because "P iff Q" implies "P implies Q".
**Puzzle.** Prove that my definition implies the book's definition.
(I have changed the set \$$A\$$ in your comment to \$$X\$$, so that the two definitions are talking about the same set.) | 2019-08-21 18:03:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9699987173080444, "perplexity": 1635.432212474564}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00063.warc.gz"} |
https://math.stackexchange.com/questions/1586725/are-the-two-matrices-similar | # Are the two matrices similar?
Consider the matrices $$A= \begin{bmatrix} 2 & 2 & 1 \\0 & 2 & -1\\ 0 & 0 & 3\end{bmatrix}$$ and $$B= \begin{bmatrix} 2 & 1 & 0 \\0 & 2 & 0\\ 0 & 0 & 3\end{bmatrix}.$$
Which are true:
1. $A$ and $B$ are similar over $\Bbb Q$.
2. $A$ is diagonalizable over $\Bbb Q$.
3. $B$ is Jordan Canonical Form of $A$.
4. The minimal and characteristic polynomial of $A$ are same.
The characteristic polynomial of $A$ is coming as $(x-2)^2(x-3)$. Since minimal polynomial and characteristic polynomial have the same roots so minimal polynomial must be $(x-2)(x-3)$ or $(x-2)^2(x-3)$.$A$ does not satisfy $(x-2)(x-3)$ so it's minimal polynomial must be $(x-2)^2(x-3)$.Obviously $B$ is JCF of $A$.
And since roots of minimal polynomial are not distinct so $A$ is not diagonalisable.Hence $3,4$ are true; $1$ is false.
How to prove/disprove $1$
The characteristic polynomials of $A$ and $B$ are obviously the same and factors over $\mathbb Q$.
Since it it also easy to see that the eigenspace corresponding to the eigenvalue $2$ for $A$ is one-dimensional, we can conclude that $B$ is in fact the Jordan canonical form of $A$.
This implies the answers to (1) and (2), which again gives information about the minimal polynomials.
• How can we say that $A$ and $B$ are similar? – Learnmore Dec 23 '15 at 14:42
• @Amartya: By definition a matrix and its Jordan canonical form are always similar. Otherwise it wouldn't be its Jordan form. – Henning Makholm Dec 23 '15 at 14:43
You have the following result:
If $\mathbb{F} \subseteq \mathbb{K}$ are two fields and $A,B \in M_n(\mathbb{F})$ are similar as matrices in $M_n(\mathbb{K})$ (so there exists an invertible $P \in M_n(\mathbb{K})$ such that $P^{-1}AP = B$) then they are also similar as matrices in $M_n(\mathbb{F})$.
In your case, $A$ and $B$ are clearly similar over $\mathbb{C}$ and have rational entries and so they are also similar over $\mathbb{Q}$. The matrix $A$ is not diagonalizable over $\mathbb{Q}$ so (1) is true and (2) is false. | 2019-09-22 18:41:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9722416400909424, "perplexity": 123.9420856555215}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575627.91/warc/CC-MAIN-20190922180536-20190922202536-00510.warc.gz"} |
https://cs.stackexchange.com/questions/80896/is-there-a-name-for-the-tree-aproach-to-listing-things | # Is there a name for the "tree" aproach to listing things?
I've only been studying computer science for a few weeks, so I apologize if this is a silly or naive question.
Suppose we're trying to list all elements of the set $$C(r) = \{(x,y,z) \in \mathbb{N}^3 : \sqrt{x^2+y^2+z^2} \leq r\}.$$
An obvious approach would be this: first, list all integers between $0$ and $r$; these become our $x$-values. Then, for each $x$-value, list all integers between $0$ and $\sqrt{r^2-x^2}.$ These become our $y$ values. Then, for each $(x,y)$ pair, list all integers between $0$ and $\sqrt{r^2-x^2-y^2}$. These become our $z$ values. We can visualize this process as a tree; the branches of the tree are the elements of $C(r)$.
A similar approach can be applied to solve the Knapsack problem a little quicker than brute force. Begin by forgetting the value of each item and just focusing on its weight. Now go through the items from left to right, bifurcating each node into two as you go, so that one edge corresponds to "take the item" and the other corresponds to "don't take the item." If taking the item would send you overweight, you refrain from adjoining the edge that corresponds to taking the item, but you still adjoin the one corresponding to not taking the item. In the end, we get a binary tree whose branches can be viewed as subsets of the set of items, which ultimately means fewer numbers that need comparing.
Well, I can see my computer running out of memory pretty quickly trying to do this - I'm sure there's better approaches out there - but nonetheless, if there's a name for this kind of thing, I'd like to know it.
• Surely this is more efficient than brute-forcing it? We did brute-force in class; for Knapsack, it was described as checking all $2^n$ possibilities, where $n$ is the number of items. The tree method would be quicker than this, right? Sep 6, 2017 at 1:40 | 2022-08-18 14:58:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5909192562103271, "perplexity": 256.3408837373554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00123.warc.gz"} |
https://www.math.princeton.edu/events/trace-reconstruction-deletion-channel-2017-10-02t200010 | # Trace reconstruction for the deletion channel
-
Yuval Peres , Microsoft Research
Fine Hall 214
In the trace reconstruction problem, an unknown string $x$ of $n$ bits is observed through the deletion channel, which deletes each bit with some constant probability $q$, yielding a contracted string. How many independent outputs (traces) of the deletion channel are needed to reconstruct $x$ with high probability?The best lower bound known is linear in $n$. Until recently, the best upper bound was exponential in the square root of $n$. We improve the square root to a cube root using statistics of individual output bits and some complex analysis; this bound is sharp for reconstruction algorithms that only use this statistical information. (Similar results were obtained independently and concurrently by De, O’Donnell and Servedio). If the string $x$ is random and \$q | 2018-02-20 23:07:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7646604180335999, "perplexity": 993.8619593057664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813109.8/warc/CC-MAIN-20180220224819-20180221004819-00296.warc.gz"} |
https://www.gktoday.in/aptitude/a-can-do-a-piece-of-work-in-12-days-and-b-in-15-days-they-work-together-for-5-days-and-then-b-left-the-days-taken-a-to-finish-the-remaining-work-is/ | A can do a piece of work in 12 days, and B in 15 days. They work together for 5 days and then B left. The days taken A to finish the remaining work is :
[A]3 days
[B]5 days
[C]10 days
[D]121 days
3 days
Work done by A and B in 5 days
$= 5\left ( \frac{1}{12}+\frac{1}{15} \right ) = 5\left ( \frac{5+4}{60} \right )$
$= 5\times \frac{9}{60} = \frac{9}{12} = \frac{3}{4}$
Remaining work $= 1-\frac{3}{4} = \frac{1}{4}$
∴ Time taken by A $= \frac{1}{4}\times 12 = 3\ days$
Hence option [A] is correct answer. | 2019-04-23 18:14:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8625536561012268, "perplexity": 1038.861824328475}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578610036.72/warc/CC-MAIN-20190423174820-20190423200820-00300.warc.gz"} |