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https://bioinformatics.stackexchange.com/questions/14891/moving-umi-from-tag-into-read-name	# Moving UMI from tag into read name I have a bam file with Unique Molecular Identifiers (UMIs) for each read present on the RX tag ( such as RX:Z:TGAGAAGGG), as expected by picard and fgbio tools. However, several other tools (such as UMI-tools) expect the UMIs to be present at the end of the read name, such as (NB501943:73:HK333BGXC:3:12402:20313:16437_TGAGAAGGG). Does anyone know of a tool to do this? I have an awk solution which I'll self-answer with, but I don't know whether that is robust. • Did you check UMI-tools, this is afaik exactly what it is doing? Nov 22 '20 at 11:11 • @ATpoint, unless I'm incorrect, UMI-tools can't do this directly. umi_tools extract moves sequences from a position in the FASTA sequence onto the read name, but in this case the reads have been added to a tag via fgbio. Nov 22 '20 at 11:38 My current solution is (awk is not a tool I'm very familiar with, so this may not be good awk code, there is presumably a way to get it to print the header lines unmodified.): samtools view -H temp.sam > temp2.sam awk '{OFS = "\t"} {for(i=1;i<=NF;i++) if ($$i ~ /^RX:Z:/) {1=1"_"i; gsub("RX:Z:","",$$1); print}}' >> temp2.sam	2022-01-20 17:57:01	{"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2025497406721115, "perplexity": 3407.269395693766}, "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/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00286.warc.gz"}
https://the-learning-machine.com/article/optimization/gradient-descent	##### Optimization Gradient descent is an approach for unconstrained optimization. In this article, we will motivate the formulation for gradient descent and provide interactive demos over multiple univariate and multivariate functions to show it in action. ## Prerequisites Follow the above links to first get acquainted with the corresponding concepts. ## Problem statement In this article, we will consider an objective function $f : \real^\ndim \to \real$ that we wish to minimize. We express the unconstrained optimization problem as, \begin{aligned} \min_{\vx} \text{ }& f(\vx) \\ \end{aligned} Gradient descent is an iterative approach. Starting from an arbitrary point, we will progressively discover points with lower function values till we converge to a point with minimum function value. The next few sections will outline the strategy used by gradient descent to discover the next point $\va + \vp$, from the current point $\va$, on its journey to the optimal solution. ## Taylor's theorem and optimization: A recap Owing to the Taylor's Theorem, for a twice differentiable function $f: \real^\ndim \to \ndim$, the value of the function at a point $\va+\vp$ can be written in terms of the function value at a point $\va$ as follows. $$f(\va+\vp) = f(\va) + \nabla_{\va}^T \vp + \frac{1}{2} \vp^T \mH_{\va} \vp \label{eqn:second-order-taylor-approx}$$ Here, $\nabla_{\va}$ and $\mH_{\va}$ denote the gradient and Hessian of the function, evaluated at the point $\va$. To choose the next point $\va + \vp$, we need to discover a suitable value of $\vp$. Since $\va$ is already known (the current point), this is a function in $\vp$. At its local minimizer, $\star{\vp}$, the gradient with respect to $\vp$ will be zero. \begin{aligned} \doh{f(\va + \star{\vp})}{\vp} = 0 \\\\ \label{eqn:first-order-condition-on-p} \end{aligned} Thus, we now have a system of equations to solve for the value of $\vp$ to choose our next point $\va + \vp$. ## Ignoring the Hessian term We studied that Newton's method solves the equations above in its entirey. When the Hessian is difficult to compute, an alternative is a quasi-Newton method such as BFGS that locally approximate the Hessian. Another alternative is to completely ignore the term with the Hessian! Re-writing the function approximation from the previous section using Taylor's Theorem by ignoring the Hessian term, we get. $$f(\va+\vp) = f(\va) + \nabla_{\va}^T \vp \label{eqn:first-order-taylor-approx}$$ This formulation only involves the gradient term. And the approach that utilizes such first-order approximation to discover the next point is known as gradient descent. For minimization, we wish to move from an iterate $\va$ to $\va + \vp$ such that $f(\va + \vp) < f(\va)$. From the first order Taylor approximation, this requires that \begin{aligned} & f(\va + \vp) < f(\va) \\ \implies& f(\va) + \nabla_{\va}^T \vp < f(\va) \\ \implies& \nabla_{\va}^T \vp < 0 \label{eqn:first-order-decrease-criteria} \end{aligned} The easiest way of ensuring this is to choose $\vp = -\nabla_{\va}$. Easy to verify that this might work. \begin{aligned} \nabla_{\va}^T \vp = - \nabla_{\va}^T \nabla_{\va} < 0 \end{aligned} because, $\nabla_{\va}^T \nabla_{\va}$ is a square and hence always positive. Great! By setting $\vp = -\nabla_{\va}$, we are moving in the direction opposite (negative) of the gradient. Therefore, this method is called gradient descent. ## Learning rate We saw that the update in the Newton's method was $\vp = -\inv{\mH_{\va}} \nabla_{\va}$. There was a scaling factor $\inv{\mH_\va}$ — the inverse of the Hessian. We can introduce a similar scaling factor for gradient descent as well — a positive scalar $\alpha$. The new update, with the scaling factor, in the case of gradient descent is $$\vp = -\alpha \nabla_{\va}$$ Because $\alpha$ controls the size of the step from $\va$ to $\va + \vp$, it is known as the step length or learning rate. We will see the effect of the learning rate on the performance of gradient descent in the upcoming interactive demos. Thus, the gradient descent proceeds as follows: 1. Start from a suitable point $\vx$ 2. Apply the following update to $\vx$ till convergence in the function value or until a maximum number of iterations have been completed: $\vx \leftarrow \vx - \alpha \nabla_{\vx}$. See the similarity to the Newton's method? Indeed, the Newton's method is a special form of this structure with a step length equal to the inverse of the Hessian $\alpha_{\text{Newton}} = \inv{\mH_{\vx}}$. ## Gradient descent demo: $\min x^2$ Let's see gradient descent in action with a simple univariate function $f(x) = x^2$, where $x \in \real$. Note that the function has a global minimum at $x = 0$. The goal of the gradient descent method is to discover this point of least function value, starting at any arbitrary point. In this accompanying demo, change the starting point by dragging the orange circle and see the trajectory of gradient descent method towards the global minimum. Also study the effect of the learning rate on the convergence rate. ## Gradient descent analysis: $\min x^2$ Remember how quickly Newton's method minimized the convex quadratic objective $x^2$? Well, note that gradient descent is not that fast, in terms of the number of steps. However, computationally, it may still be faster for some problems, as it does not require the computation of the Hessian inverse, a major disadvantage of the Newton's method. Also, note the impact of the step size or learning rate $\alpha$ on the performance. For this particular problem, the Newton's method suggests a step size of $0.5$. So, note that a step size of 0.5 with gradient descent will reach the minimum in one step. Any learning rate lower than 0.5 leads to slower convergence. Any learning rate higher than 0.5 leads to oscillations around the minimum till it finally lands at the minimum. In fact, try the learning rate $\alpha = 1$ for this function. The iterations just keep oscillating. Clearly, the learning rate is a crucial parameter of the gradient descent approach. We will look at practical strategies to choose a good value of the learning rate in the upcoming sections. ## Gradient descent demo: $\min x^4 - 8x^2$ Let's see gradient descent in action on a nonconvex function with multiple minima and a single maximum. $$f(x) = x^4 - 8x^2$$ Note that the learning rate plays a huge role in where the gradient descent solution ends up. Also, note that due to the bumpy nature of the function, we need a much slower learning rate than what we used in the previous demo. If not for a low learning rate, the iterations oscillate across the minima. ## Gradient descent demo: Rosenbrock's function We introduced Rosenbrock's function in our article on multivariate functions in calculus. Rosenbrock is a very hard problem for many optimization algorithms. Notice in this next demo that gradient descent can quickly find the valley in the center, but the convergence towards the optimal (green point) is extremely slow, as the iterates oscillate on the two sides of the valley. ## Gradient descent demo: Himmelblau's function We introduced Himmelblau's function in our article on multivariate functions in calculus. It has 4 local minima (highlighted in green) and 1 maximum (highlighted in blue). Turns out, Himmelblaus, in spite of its bump, is actually quite easy for gradient descent. Unlike Newton's method, the trajectory does not go over the bump, but rather quickly seeks out the nearest minimum. ## A good learning rate In all the previous demos, we observed that gradient descent is heavily dependent on the learning rate parameter. The functions we studied were easy to visualize, but the ones that we encounter in practice are much harder, with many more bumps and local minima. How do we choose a good learning rate? Check out our comprehensive interactive article on choosing a good learning rate where we talk about strategies such as dampening and Wolfe conditions. ## Where to next? Expand your knowledge of optimization approaches with our detailed interactive articles on this subject. To brush up on calculus, explore our articles on topics in calculus and multivariate calculus. Or start with strong mathematical foundations. Already an optimization expert? Check out comprehensive courses on machine learning or deep learning.	2022-01-21 14:26:31	{"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": 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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9066123962402344, "perplexity": 471.36785927204886}, "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/1642320303385.49/warc/CC-MAIN-20220121131830-20220121161830-00142.warc.gz"}
https://juliahub.com/docs/CalculusWithJulia/AZHbv/0.0.5/integral_vector_calculus/stokes_theorem.html	# Green's Theorem, Stokes' Theorem, and the Divergence Theorem The fundamental theorem of calculus is a fan favorite, as it reduces a definite integral, $\int_a^b f(x) dx$, into the evaluation of a related function at two points: $F(b)-F(a)$, where the relation is $F$ is an antiderivative of $f$. It is a favorite as it makes life much easier than the alternative of computing a limit of a Riemann sum. This relationship can be generalized. The key is to realize that the interval $[a,b]$ has boundary $\{a, b\}$ (a set) and then expressing the theorem as: the integral around some region of $f$ is the integral, suitably defined, around the boundary of the region for a function related to $f$. In an abstract setting, Stokes' theorem says exactly this with the relationship being the exterior derivative. Here we are not as abstract, we discuss below: • Green's theorem, a $2$-dimensional theorem, where the region is a planar region, $D$, and the boundary a simple curve $C$; • Stokes' theorem in $3$ dimensions, where the region is an open surface, $S$, in $R^3$ with boundary, $C$; • The Divergence theorem in $3$ dimensions, where the region is a volume in three dimensions and the boundary its $2$-dimensional closed surface. The related functions will involve the divergence and the curl, previously discussed. Many of the the examples in this section come from either Strang or Schey. Before beginning, we load our usual packages. using CalculusWithJulia using Plots To make the abstract concrete, consider the one dimensional case of finding the definite integral $\int_a^b F'(x) dx$. The Riemann sum picture at the microscopic level considers a figure like: The total area under the blue curve from $a$ to $b$, is found by adding the area of each segment of the figure. Let's consider now what an integral over the boundary would mean. The region, or interval, $[x_{i-1}, x_i]$ has a boundary that clearly consists of the two points $x_{i-1}$ and $x_i$. If we orient the boundary, as we need to for higher dimensional boundaries, using the outward facing direction, then the oriented boundary at the right-hand end point, $x_i$, would point towards $+\infty$ and the left-hand end point, $x_{i-1}$, would be oriented to point to $-\infty$. An "integral" on the boundary of $F$ would naturally be $F(b) \times 1$ plus $F(a) \times -1$, or $F(b)-F(a)$. With this choice of integral over the boundary, we can see much cancellation arises were we to compute this integral for each piece, as we would have with $a=x_0 < x_1 < \cdots x_{n-1} < x_n=b$: $~ (F(x_1) - F(x_0)) + (F(x_2)-F(x_1)) + \cdots + (F(x_n) - F(x_{n-1})) = F(x_n) - F(x_0) = F(b) - F(a). ~$ That is, with this definition for a boundary integral, the interior pieces of the microscopic approximation cancel and the total is just the integral over the oriented macroscopic boundary $\{a, b\}$. But each microscopic piece can be reimagined, as $~ F(x_{i}) - F(x_{i-1}) = \left(\frac{F(x_{i}) - F(x_{i-1})}{\Delta{x}}\right)\Delta{x} \approx F'(x_i)\Delta{x}. ~$ The approximation could be exact were the mean value theorem used to identify a point in the interval, but we don't pursue that, as the key point is the right hand side is a Riemann sum approximation for a different integral, in this case the integral $\int_a^b F'(x) dx$. Passing from the microscopic view to an infinitesimal view, the picture gives two interpretations, leading to the Fundamental Theorem of Calculus: $~ \int_a^b F'(x) dx = F(b) - F(a). ~$ The three theorems of this section, Green's theorem, Stokes' theorem, and the divergence theorem, can all be seen in this manner: the sum of microscopic boundary integrals leads to a macroscopic boundary integral of the entire region; whereas, by reinterpretation, the microscopic boundary integrals are viewed as Riemann sums, which in the limit become integrals of a related function over the region. ## Green's theorem To continue the above analysis for a higher dimension, we consider the following figure hinting at a decomposition of a macroscopic square into subsequent microscopic sub-squares. The boundary of each square is oriented so that the right hand rule comes out of the picture. Consider the boundary integral $\oint_c F\cdot\vec{T} ds$ around the smallest (green) squares. We have seen that the curl at a point in a direction is given in terms of the limit. Let the plane be the $x-y$ plane, and the $\hat{k}$ direction be the one coming out of the figure. In the derivation of the curl, we saw that the line integral for circulation around the square satisfies: $~ \lim \frac{1}{\Delta{x}\Delta{y}} \oint_C F \cdot\hat{T}ds = \frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}}. ~$ If the green squares are small enough, then the line integrals satisfy: $~ \oint_C F \cdot\hat{T}ds \approx \left( \frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}} \right) \Delta{x}\Delta{y} . ~$ We interpret the right hand side as a Riemann sum approximation for the $2$ dimensional integral of the function $f(x,y) = \frac{\partial{F_x}}{\partial{y}} - \frac{\partial{F_y}}{\partial{x}}=\text{curl}(F)$, the two-dimensional curl. Were the green squares continued to fill out the large blue square, then the sum of these terms would approximate the integral $~ \iint_S f(x,y) dA = \iint_S \left(\frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}}\right) dA = \iint_S \text{curl}(F) dA. ~$ However, the microscopic boundary integrals have cancellations that lead to a macroscopic boundary integral. The sum of $\oint_C F \cdot\hat{T}ds$ over the $4$ green squares will be equal to $\oint_{C_r} F\cdot\hat{T}ds$, where $C_r$ is the red square, as the interior line integral pieces will all cancel off. The sum of $\oint_{C_r} F \cdot\hat{T}ds$ over the $4$ red squares will equal $\oint_{C_b} F \cdot\hat{T}ds$, where $C_b$ is the oriented path around the blue square, as again the interior line pieces will cancel off. Etc. This all suggests that the flow integral around the surface of the larger region (the blue square) is equivalent to the integral of the curl component over the region. This is Green's theorem, as stated by Wikipedia: Green's theorem: Let $C$ be a positively oriented, piecewise smooth, simple closed curve in the plane, and let $D$ be the region bounded by $C$. If $F=\langle F_x, F_y\rangle$, is a vector field on an open region containing $D$ having continuous partial derivatives then: $~ \oint_C F\cdot\hat{T}ds = \iint_D \left( \frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}} \right) dA= \iint_D \text{curl}(F)dA. ~$ The statement of the theorem applies only to regions whose boundaries are simple closed curves. Not all simple regions have such boundaries. An annulus for example. This is a restriction that will be generalized. ### Examples Some examples, following Strang, are: #### Computing area Let $F(x,y) = \langle -y, x\rangle$. Then $\frac{\partial{F_y}}{\partial{x}} - \frac{\partial{F_x}}{\partial{y}}=2$, so $~ \frac{1}{2}\oint_C F\cdot\hat{T}ds = \frac{1}{2}\oint_C (xdy - ydx) = \iint_D dA = A(D). ~$ This gives a means to compute the area of a region by integrating around its boundary. To compute the area of an ellipse, we have: F(x,y) = [-y,x] F(v) = F(v...) r(t) = [acos(t),bsin(t)] @vars a b t positive=true (1//2) * integrate( F(r(t)) ⋅ diff.(r(t),t), (t, 0, 2PI)) \begin{equation}\pi a b\end{equation} To compute the area of the triangle with vertices $(0,0)$, $(a,0)$ and $(0,b)$ we can orient the boundary counter clockwise. Let $A$ be the line segment from $(0,b)$ to $(0,0)$, $B$ be the line segment from $(0,0)$ to $(a,0)$, and $C$ be the other. Then ~ \begin{align} \frac{1}{2} \int_A F\cdot\hat{T} ds &=\frac{1}{2} \int_A -ydx = 0\\ \frac{1}{2} \int_B F\cdot\hat{T} ds &=\frac{1}{2} \int_B xdy = 0, \end{align} ~ as on $A$, $y=0$ and $dy=0$ and on $B$, $x=0$ and $dx=0$. On $C$ we have $\vec{r}(t) = (0, b) + t\cdot(1,-b/a) =\langle t, b-(bt)/a\rangle$ from $t=a$ to $0$ $~ \int_C F\cdot \frac{d\vec{r}}{dt} dt = \int_a^0 \langle -b + (bt)/a), t\rangle\cdot\langle 1, -b/a\rangle dt = \int_a^0 -b dt = -bt\mid_{a}^0 = ba. ~$ Dividing by $1/2$ give the familiar answer $A=(1/2) a b$. #### Conservative fields A vector field is conservative if path integrals for work are independent of the path. We have seen that a vector field that is the gradient of a scalar field will be conservative and vice versa. This led to the vanishing identify $\nabla\times\nabla(f) = 0$ for a scalar field $f$. Is the converse true? Namely, if for some vector field $F$, $\nabla\times{F}$ is identically $0$ is the field conservative? The answer is yes if vector field has continuous partial derivatives and the curl is $0$ in a simply connected domain. For the two dimensional case the curl is a scalar. If $F = \langle F_x, F_y\rangle = \nabla{f}$ is conservative, then $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y} = 0$. Now assume $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y} = 0$. Let $P$ and $Q$ be two points in the plane. Take any path, $C_1$ from $P$ to $Q$ and any return path, $C_2$, from $Q$ to $P$ that do not cross and such that $C$, the concatenation of the two paths, satisfies Green's theorem. Then, as $F$ is continuous on an open interval containing $D$, we have: $~ 0 = \iint_D 0 dA = \iint_D \left(\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y}\right)dA = \oint_C F \cdot \hat{T} ds = \int_{C_1} F \cdot \hat{T} ds + \int_{C_2}F \cdot \hat{T} ds. ~$ Reversing $C_2$ to go from $P$ to $Q$, we see the two work integrals are identical, that is the field is conservative. Summarizing: • If $F=\nabla{f}$ then $F$ is conservative. • If $F=\langle F_x, F_y\rangle$ has continuous partial derivatives in a simply connected open region with $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y}=0$, then in that region $F$ is conservative and can be represented as the gradient of a scalar function. For example, let $F(x,y) = \langle \sin(xy), \cos(xy) \rangle$. Is this a conservative vector field? We can check by taking partial derivatives. Those of interest are: ~ \begin{align} \frac{\partial{F_y}}{\partial{x}} &= \frac{\partial{(\cos(xy))}}{\partial{x}} = -\sin(xy) y,\\ \frac{\partial{F_x}}{\partial{y}} &= \frac{\partial{(\sin(xy))}}{\partial{y}} = \cos(xy)x. \end{align} ~ It is not the case that $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y}=0$, so this vector field is not conservative. The conditions of Green's theorem are important, as this next example shows. Let $D$ be the unit disc, $C$ the unit circle parameterized counter clockwise. Let $R(x,y) = \langle -y, x\rangle$ be a rotation field and $F(x,y) = R(x,y)/(R(x,y)\cdot R(x,y))$. Then: R(x,y) = [-y,x] F(x,y) = R(x,y)/(R(x,y)⋅R(x,y)) @vars x y real=true Fx, Fy = F(x,y) diff(Fy, x) - diff(Fx, y) \|> simplify \begin{equation}0\end{equation} Then, $\iint_D \left( \partial{F_y}/{\partial{x}}-\partial{F_xy}/{\partial{y}}\right)dA = 0$. But, $~ F\cdot\hat{T} = \frac{R}{R\cdot{R}} \cdot \frac{R}{R\cdot{R}} = \frac{R\cdot{R}}{(R\cdot{R})^2} = \frac{1}{R\cdot{R}}, ~$ so $\oint_C F\cdot\hat{T}ds = 2\pi$, $C$ being the unit circle so $R\cdot{R}=1$. That is, for this example, Green's theorem does not apply, as the two integrals are not the same. What isn't satisfied in the theorem? $F$ is not continuous at the origin and our curve $C$ defining $D$ encircles the origin. So, $F$ does not have continuous partial derivatives, as is required for the theorem. #### More complicated boundary curves A simple closed curve is one that does not cross itself. Green's theorem applies to regions bounded by curves which have finitely many crosses provided the orientation used is consistent throughout. Consider the curve $y = f(x)$, $a \leq x \leq b$, assuming $f$ is continuous, $f(a) > 0$, and $f(b) < 0$. We can use Green's theorem to compute the signed "area" under under $f$ if we consider the curve in $R^2$ from $(b,0)$ to $(a,0)$ to $(a, f(a))$, to $(b, f(b))$ and back to $(b,0)$ in that orientation. This will cross at each zero of $f$. Let $A$ label the red line, $B$ the green curve, $C$ the blue line, and $D$ the black line. Then the area is given from Green's theorem by considering half of the the line integral of $F(x,y) = \langle -y, x\rangle$ or $\oint_C (xdy - ydx)$. To that matter we have: ~ \begin{align} \int_A (xdy - ydx) &= a f(a)\\ \int_C (xdy - ydx) &= b(-f(b))\\ \int_D (xdy - ydx) &= 0\\ \end{align} ~ Finally the integral over $B$, using integration by parts: ~ \begin{align} \int_B F(\vec{r}(t))\cdot \frac{d\vec{r}(t)}{dt} dt &= \int_b^a \langle -f(t),t)\rangle\cdot\langle 1, f'(t)\rangle dt\\ &= \int_a^b f(t)dt - \int_a^b tf'(t)dt\\ &= \int_a^b f(t)dt - \left(tf(t)\mid_a^b - \int_a^b f(t) dt\right). \end{align} ~ Combining, we have after cancellation $\oint (xdy - ydx) = 2\int_a^b f(t) dt$, or after dividing by $2$ the signed area under the curve. The region may not be simply connected. A simple case might be the disc: $1 \leq x^2 + y^2 \leq 4$. In this figure we introduce a cut to make a simply connected region. The cut leads to a counter-clockwise orientation on the outer ring and a clockwise orientation on the inner ring. If this cut becomes so thin as to vanish, then the line integrals along the lines introducing the cut will cancel off and we have a boundary consisting of two curves with opposite orientations. (If we follow either orientation the closed figure is on the left.) To see that the area integral of $F(x,y) = (1/2)\langle -y, x\rangle$ produces the area for this orientation we have, using $C_1$ as the outer ring, and $C_2$ as the inner ring: ~ \begin{align} \oint_{C_1} F \cdot \hat{T} ds &= \int_0^{2\pi} (1/2)(2)\langle -\sin(t), \cos(t)\rangle \cdot (2)\langle-\sin(t), \cos(t)\rangle dt = (1/2) (2\pi) 4 = 4\pi\\ \oint_{C_2} F \cdot \hat{T} ds &= \int_{0}^{2\pi} (1/2) \langle \sin(t), \cos(t)\rangle \cdot \langle-\sin(t), -\cos(t)\rangle dt\\ &= -(1/2)(2\pi) = -\pi. \end{align} ~ (Using $\vec{r}(t) = 2\langle \cos(t), \sin(t)\rangle$ for the outer ring and $\vec{r}(t) = 1\langle \cos(t), -\sin(t)\rangle$ for the inner ring.) Adding the two gives $4\pi - \pi = \pi \cdot(b^2 - a^2)$, with $b=2$ and $a=1$. #### Flow not flux Green's theorem has a complement in terms of flow across $C$. As $C$ is positively oriented (so the bounded interior piece is on the left of $\hat{T}$ as the curve is traced), a normal comes by rotating $90^\circ$ counterclockwise. That is if $\hat{T} = \langle a, b\rangle$, then $\hat{N} = \langle b, -a\rangle$. Let $F = \langle F_x, F_y \rangle$ and $G = \langle F_y, -F_x \rangle$, then $G\cdot\hat{T} = -F\cdot\hat{N}$. The curl formula applied to $G$ becomes $~ \frac{\partial{G_y}}{\partial{x}} - \frac{\partial{G_x}}{\partial{y}} = \frac{\partial{-F_x}}{\partial{x}}-\frac{\partial{(F_y)}}{\partial{y}} = -\left(\frac{\partial{F_x}}{\partial{x}} + \frac{\partial{F_y}}{\partial{y}}\right)= -\nabla\cdot{F}. ~$ Green's theorem applied to $G$ then gives this formula for $F$: $~ \oint_C F\cdot\hat{N} ds = -\oint_C G\cdot\hat{T} ds = -\iint_D (-\nabla\cdot{F})dA = \iint_D \nabla\cdot{F}dA. ~$ The right hand side integral is the $2$-dimensional divergence, so this has the interpretation that the flux through $C$ ($\oint_C F\cdot\hat{N} ds$) is the integral of the divergence. (The divergence is defined in terms of a limit of this picture, so this theorem extends the microscopic view to a bigger view.) Rather than leave this as an algebraic consequence, we sketch out how this could be intuitively argued from a microscopic picture, the reason being similar to that for the curl, where we considered the small green boxes. In the generalization to dimension $3$ both arguments are needed for our discussion: Consider now a $2$-dimensional region split into microscopic boxes; we focus now on two adjacent boxes, $A$ and $B$: The integrand $F\cdot\hat{N}$ for $A$ will differ from that for $B$ by a minus sign, as the field is the same, but the normal carries an opposite sign. Hence the contribution to the line integral around $A$ along this part of the box partition will cancel out with that around $B$. The only part of the line integral that will not cancel out for such a partition will be the boundary pieces of the overall shape. This figure shows in red the parts of the line integrals that will cancel for a more refined grid. Again, the microscopic boundary integrals when added will give a macroscopic boundary integral due to cancellations. But, as seen in the derivation of the divergence, only modified for $2$ dimensions, we have $\nabla\cdot{F} = \lim \frac{1}{\Delta S} \oint_C F\cdot\hat{N}$, so for each cell $~ \oint_{C_i} F\cdot\hat{N} \approx \left(\nabla\cdot{F}\right)\Delta{x}\Delta{y}, ~$ an approximating Riemann sum for $\iint_D \nabla\cdot{F} dA$. This yields: $~ \oint_C (F \cdot\hat{N}) dA = \sum_i \oint_{C_i} (F \cdot\hat{N}) dA \approx \sum \left(\nabla\cdot{F}\right)\Delta{x}\Delta{y} \approx \iint_S \nabla\cdot{F}dA, ~$ the approximations becoming equals signs in the limit. ##### Example Let $F(x,y) = \langle ax , by\rangle$, and $D$ be the square with side length $2$ centered at the origin. Verify that the flow form of Green's theorem holds. We have the divergence is simply $a + b$ so $\iint_D (a+b)dA = (a+b)A(D) = 4(a+b)$. The integral of the flow across $C$ consists of $4$ parts. By symmetry, they all should be similar. We consider the line segment connecting $(1,-1)$ to $(1,1)$ (which has the proper counterclockwise orientation): $~ \int_C F \cdot \hat{N} ds= \int_{-1}^1 \langle F_x, F_y\rangle\cdot\langle 0, 1\rangle ds = \int_{-1}^1 b dy = 2b. ~$ Integrating across the top will give $2a$, along the bottom $2a$, and along the left side $2b$ totaling $4(a+b)$. Next, let $F(x,y) = \langle -y, x\rangle$. This field rotates, and we see has no divergence, as $\partial{F_x}/\partial{x} = \partial{(-y)}/\partial{x} = 0$ and $\partial{F_y}/\partial{y} = \partial{x}/\partial{y} = 0$. As such, the area integral in Green's theorem is $0$. As well, $F$ is parallel to $\hat{T}$ so orthogonal to $\hat{N}$, hence $\oint F\cdot\hat{N}ds = \oint 0ds = 0$. For any region $S$ there is no net flow across the boundary and no source or sink of flow inside. ##### Example: stream functions Strang compiles the following equivalencies (one implies the others) for when the total flux is $0$ for a vector field with continuous partial derivatives: • $\oint F\cdot\hat{N} ds = 0$ • for all curves connecting $P$ to $Q$, $\int_C F\cdot\hat{N}$ has the same value • There is a stream function $g(x,y)$ for which $F_x = \partial{g}/\partial{y}$ and $F_y = -\partial{g}/\partial{x}$. (This says $\nabla{g}$ is orthogonal to $F$.) • the components have zero divergence: $\partial{F_x}/\partial{x} + \partial{F_y}/\partial{y} = 0$. Strang calls these fields source free as the divergence is $0$. A stream function plays the role of a scalar potential, but note the minus sign and order of partial derivatives. These are accounted for by saying $\langle F_x, F_y, 0\rangle = \nabla\times\langle 0, 0, g\rangle$, in Cartesian coordinates. Streamlines are tangent to the flow of the velocity vector of the flow and in two dimensions are perpendicular to field lines formed by the gradient of a scalar function. Potential flow uses a scalar potential function to describe the velocity field through $\vec{v} = \nabla{f}$. As such, potential flow is irrotational due to the curl of a conservative field being the zero vector. Restricting to two dimensions, this says the partials satisfy $\partial{v_y}/\partial{x} - \partial{v_x}/\partial{y} = 0$. For an incompressible flow (like water) the velocity will have $0$ divergence too. That is $\nabla\cdot\nabla{f} = 0$ - $f$ satisfies Laplace's equation. By the equivalencies above, an incompressible potential flow means in addition to a potential function, $f$, there is a stream function $g$ satisfying $v_x = \partial{g}/\partial{y}$ and $v_y=-\partial{g}/\partial{x}$. The gradient of $f=\langle v_x, v_y\rangle$ is orthogonal to the contour lines of $f$. The gradient of $g=\langle -v_y, v_x\rangle$ is orthogonal to the gradient of $f$, so are tangents to the contour lines of $f$. Reversing, the gradient of $f$ is tangent to the contour lines of $g$. If the flow follows the velocity field, then the contour lines of $g$ indicate the flow of the fluid. As an example consider the following in polar coordinates: $~ f(r, \theta) = A r^n \cos(n\theta),\quad g(r, \theta) = A r^n \sin(n\theta). ~$ The constant $A$ just sets the scale, the parameter $n$ has a qualitative effect on the contour lines. Consider $n=2$ visualized below: n = 2 f(r,theta) = r^n * cos(ntheta) g(r, theta) = r^n sin(ntheta) f(v) = f(v...); g(v)= g(v...) Φ(x,y) = [sqrt(x^2 + y^2), atan(y,x)] Φ(v) = Φ(v...) xs = ys = range(-2,2, length=50) contour(xs, ys, f∘Φ, color=:red, legend=false, aspect_ratio=:equal) contour!(xs, ys, g∘Φ, color=:blue, linewidth=3) The fluid would flow along the blue (stream) lines. The red lines have equal potential along the line. ## Stokes' theorem Were the figure of Jiffy Pop popcorn animated, the surface of foil would slowly expand due to pressure of popping popcorn until the popcorn was ready. However, the boundary would remain the same. Many different surfaces can have the same boundary. Take for instance the upper half unit sphere in $R^3$ it having the curve $x^2 + y^2 = 1$ as a boundary curve. This is the same curve as the surface of the cone $z = 1 - (x^2 + y^2)$ that lies above the $x-y$ plane. This would also be the same curve as the surface formed by a Mickey Mouse glove if the collar were scaled and positioned onto the unit circle. Imagine if instead of the retro labeling, a rectangular grid were drawn on the surface of the Jiffy Pop popcorn before popping. By Green's theorem, the integral of the curl of a vector field $F$ over this surface reduces to just an accompanying line integral over the boundary, $C$, where the orientation of $C$ is in the $\hat{k}$ direction. The intuitive derivation being that the curl integral over the grid will have cancellations due to adjacent cells having shared paths being traversed in both directions. Now imagine the popcorn expanding, but rather than worry about burning, focusing instead on what happens to the integral of the curl in the direction of the normal, we have $~ \nabla\times{F} \cdot\hat{N} = \lim \frac{1}{\Delta{S}} \oint_C F\cdot\hat{T} ds \approx \frac{1}{\Delta{S}} F\cdot\hat{T} \Delta{s}. ~$ This gives the series of approximations: $~ \oint_C F\cdot\hat{T} ds = \sum \oint_{C_i} F\cdot\hat{T} ds \approx \sum F\cdot\hat{T} \Delta s \approx \sum \nabla\times{F}\cdot\hat{N} \Delta{S} \approx \iint_S \nabla\times{F}\cdot\hat{N} dS. ~$ In terms of our expanding popcorn, the boundary integral - after accounting for cancellations, as in Green's theorem - can be seen as a microscopic sum of boundary integrals each of which is approximated by a term $\nabla\times{F}\cdot\hat{N} \Delta{S}$ which is viewed as a Riemann sum approximation for the the integral of the curl over the surface. The cancellation depends on a proper choice of orientation, but with that we have: Stokes' theorem. Let $S$ be an orientable smooth surface in $R^3$ with boundary $C$, $C$ oriented so that the chosen normal for $S$ agrees with the right-hand rule for $C$'s orientation. Then if $F$ has continuous partial derivatives $~ \oint_C F \cdot\hat{T} ds = \iint_S (\nabla\times{F})\cdot\hat{N} dA. ~$ Green's theorem is an immediate consequence upon viewing the region in $R^2$ as a surface in $R^3$ with normal $\hat{k}$. ### Examples ##### Example Our first example involves just an observation. For any simply connected surface $S$ without boundary (such as a sphere) the integral $\oint_S \nabla\times{F}dS=0$, as the line integral around the boundary must be $0$, as there is no boundary. ##### Example Let $F(x,y,z) = \langle x^2, 0, y^2\rangle$ and $C$ be the circle $x^2 + z^2 = 1$ with $y=0$. Find $\oint_C F\cdot\hat{T}ds$. We can use Stoke's theorem with the surface being just the disc, so that $\hat{N} = \hat{j}$. This makes the computation easy: @vars x y z real=true F(x,y,z) = [x^2, 0, y^2] CurlF = curl(F(x,y,z), [x,y,z]) $\left[ \begin{array}{r}2 y\\0\\0\end{array} \right]$ We have $\nabla\times{F}\cdot\hat{N} = 0$, so the answer is $0$. We could have directly computed this. Let $r(t) = \langle \cos(t), 0, \sin(t)\rangle$. Then we have: @vars t real=true r(t) = [cos(t), 0, sin(t)] rp = diff.(r(t), t) integrand = F(r(t)...) ⋅ rp \begin{equation}- \sin{\left(t \right)} \cos^{2}{\left(t \right)}\end{equation} The integrand isn't obviously going to yield $0$ for the integral, but through symmetry: integrate(integrand, (t, 0, 2PI)) \begin{equation}0\end{equation} ##### Example: Ampere's circuital law (Schey) Suppose a current $I$ flows along a line and $C$ is a path encircling the current with orientation such that the right hand rule points in the direction of the current flow. Ampere's circuital law relates the line integral of the magnetic field to the induced current through: $~ \oint_C B\cdot\hat{T} ds = \mu_0 I. ~$ The goal here is to re-express this integral law to produce a law at each point of the field. Let $S$ be a surface with boundary $C$, Let $J$ be the current density - $J=\rho v$, with $\rho$ the density of the current (not time-varying) and $v$ the velocity. The current can be re-expressed as $I = \iint_S J\cdot\hat{n}dA$. (If the current flows through a wire and $S$ is much bigger than the wire, this is still valid as $\rho=0$ outside of the wird.) We then have: $~ \mu_0 \iint_S J\cdot\hat{N}dA = \mu_0 I = \oint_C B\cdot\hat{T} ds = \iint_S (\nabla\times{B})\cdot\hat{N}dA. ~$ As $S$ and $C$ are arbitrary, this implies the integrands of the surface integrals are equal, or: $~ \nabla\times{B} = \mu_0 J. ~$ (Strang) Suppose $C$ is a wire and there is a time-varying magnetic field $B(t)$. Then Faraday's law says the flux passing within $C$ through a surface $S$ with boundary $C$ of the magnetic field, $\phi = \iint B\cdot\hat{N}dS$, induces an electric field $E$ that does work: $~ \oint_C E\cdot\hat{T}ds = -\frac{\partial{\phi}}{\partial{t}}. ~$ Faraday's law is an empirical statement. Stokes' theorem can be used to produce one of Maxwell's equations. For any surface $S$, as above with its boundary being $C$, we have both: $~ -\iint_S \left(\frac{\partial{B}}{\partial{t}}\cdot\hat{N}\right)dS = -\frac{\partial{\phi}}{\partial{t}} = \oint_C E\cdot\hat{T}ds = \iint_S (\nabla\times{E}) dS. ~$ This is true for any capping surface for $C$. Shrinking $C$ to a point means it will hold for each point in $R^3$. That is: $~ \nabla\times{E} = -\frac{\partial{B}}{\partial{t}}. ~$ ##### Example: Conservative fields Green's theorem gave a characterization of $2$-dimensional conservative fields, Stokes' theorem provides a characterization for $3$ dimensional conservative fields (with continuous derivatives): • The work $\oint_C F\cdot\hat{T} ds = 0$ for every closed path • The work $\int_P^Q F\cdot\hat{T} ds$ is independent of the path between $P$ and $Q$ • for a scalar potential function $\phi$, $F = \nabla{\phi}$ • The curl satisfies: $\nabla\times{F} = \vec{0}$ (and the domain is simply connected). Stokes's theorem can be used to show the first and fourth are equivalent. First, $0 = \oint_C F\cdot\hat{T} ds$, then by Stokes' theorem $0 = \int_S \nabla\times{F} dS$ for any orientable surface $S$ with boundary $C$. For a given point, letting $C$ shrink to that point can be used to see that the cross product must be $0$ at that point. Conversely, if the cross product is zero in a simply connected region, then tke any simple closed curve, $C$ in the region. If the region is simply connected then there exists an orientable surface, $S$ in the region with boundary $C$ for which: $\oint_C F\cdot{N} ds = \iint_S (\nabla\times{F})\cdot\hat{N}dS= \iint_S \vec{0}\cdot\hat{N}dS = 0$. The construction of a scalar potential function from the field can be done as illustrated in this next example. Take $F = \langle yz^2, xz^2, 2xyz \rangle$. Verify $F$ is conservative and find a scalar potential $\phi$. To verify that $F$ is conservative, we find its curl to see that it is $\vec{0}$: F(x,y,z) = [yz^2, xz^2, 2xyz] @vars x y z real=true curl(F(x,y,z), [x,y,z]) $\left[ \begin{array}{r}0\\0\\0\end{array} \right]$ We need $\phi$ with $\partial{\phi}/\partial{x} = F_x = yz^2$. To that end, we integrate in $x$: $~ \phi(x,y,z) = \int yz^2 dx = xyz^2 + g(y,z), ~$ the function $g(y,z)$ is a "constant" of integration (it doesn't depend on $x$). That $\partial{\phi}/\partial{x} = F_x$ is true is easy to verify. Now, consider the partial in $y$: $~ \frac{\partial{\phi}}{\partial{y}} = xz^2 + \frac{\partial{g}}{\partial{y}} = F_y = xz^2. ~$ So we have $\frac{\partial{g}}{\partial{y}}=0$ or $g(y,z) = h(z)$, some constant in $y$. Finally, we must have $\partial{\phi}/\partial{z} = F_z$, or $~ \frac{\partial{\phi}}{\partial{z}} = 2xyz + h'(z) = F_z = 2xyz, ~$ So $h'(z) = 0$. This value can be any constant, even $0$ which we take, so that $g(y,z) = 0$ and $\phi(x,y,z) = xyz^2$ is a scalar potential for $F$. ##### Example Let $F(x,y,z) = \nabla(xy^2z^3) = \langle y^2z^3, 2xyz^3, 3xy^2z^2\rangle$. Show that the line integrals around the unit circle in the $x-y$ plane and the $y-z$ planes are $0$, as $F$ is conservative. @vars x y z t Fxyz = ∇(xy^2z^3) r(t) = [cos(t), sin(t), 0] rp = diff.(r(t), t) Ft = subs.(Fxyz, x .=> r(t)[1], y.=> r(t)[2], z .=> r(t)[3]) integrate(Ft ⋅ rp, (t, 0, 2PI)) \begin{equation}0\end{equation} (This is trivial, as Ft is $0$, as each term as a $z$ factor of $0$.) In the $y-z$ plane we have: r(t) = [0, cos(t), sin(t)] rp = diff.(r(t), t) Ft = subs.(Fxyz, x .=> r(t)[1], y.=> r(t)[2], z .=> r(t)[3]) integrate(Ft ⋅ rp, (t, 0, 2PI)) \begin{equation}0\end{equation} This is also easy, as Ft has only an x component and rp has only y and z components, so the two are orthogonal. ##### Example In two dimensions the vector field $F(x,y) = \langle -y, x\rangle/(x^2+y^2) = S(x,y)/\\|R\\|^2$ is irrotational ($0$ curl) and has $0$ divergence, but is not conservative in $R^2$, as with $C$ being the unit disk we have $\oint_C F\cdot\hat{T}ds = \int_0^{2\pi} \langle -\sin(\theta),\cos(\theta)\rangle \cdot \langle-\sin(\theta), \cos(\theta)\rangle/1 d\theta = 2\pi$. This is because $F$ is not continuously differentiable at the origin, so the path $C$ is not in a simply connected domain where $F$ is continuously differentiable. (Were $C$ to avoid the origin, the integral would be $0$.) In three dimensions, removing a single point in a domain does change simple connectedness, but removing an entire line will. So the function $F(x,y,z) =\langle -y,x,0\rangle/(x^2+y^2)\rangle$ will have $0$ curl, $0$ divergence, but won't be conservative in a domain that includes the $z$ axis. However, the function $F(x,y,z) = \langle x, y,z\rangle/\sqrt{x^2+y^2+z^2}$ has curl $0$, except at the origin. However, $R^3$ less the origin, as a domain, is simply connected, so $F$ will be conservative. ## Divergence theorem The divergence theorem is a consequence of a simple observation. Consider two adjacent cubic regions that share a common face. The boundary integral, $\oint_S F\cdot\hat{N} dA$, can be computed for each cube. The surface integral requires a choice of normal, and the convention is to use the outward pointing normal. The common face of the two cubes has different outward pointing normals, the difference being a minus sign. As such, the contribution of the surface integral over this face for one cube is cancelled out by the contribution of the surface integral over this face for the adjacent cube. As with Green's theorem, this means for a cubic partition, that only the contribution over the boundary is needed to compute the boundary integral. In formulas, if $V$ is a $3$ dimensional cubic region with boundary $S$ and it is partitioned into smaller cubic subregions, $V_i$ with surfaces $S_i$, we have: $~ \oint_S F\cdot{N} dA = \sum \oint_{S_i} F\cdot{N} dA. ~$ If the partition provides a microscopic perspective, then the divergence approximation $\nabla\cdot{F} \approx (1/\Delta{V_i}) \oint_{S_i} F\cdot{N} dA$ can be used to say: $~ \oint_S F\cdot{N} dA = \sum \oint_{S_i} F\cdot{N} dA \approx \sum (\nabla\cdot{F})\Delta{V_i} \approx \iiint_V \nabla\cdot{F} dV, ~$ the last approximation through a Riemann sum approximation. This heuristic leads to: The divergence theorem. Suppose $V$ is a $3$-dimensional volume which is bounded (compact) and has a boundary, $S$, that is piecewise smooth. If $F$ is a continuously differentiable vector field defined on an open set containing $V$, then: $~ \iiint_V (\nabla\cdot{F}) dV = \oint_S (F\cdot\hat{N})dS. ~$ That is, the volume integral of the divergence can be computed from the flux integral over the boundary of $V$. ### Examples ##### Example Verify the divergence theorem for the vector field $F(x,y,z) = \langle xy, yz, zx\rangle$ for the cubic box centered at the origin with side lengths $2$. We need to compute two terms and show they are equal. We begin with the volume integral: F(x,y,z) = [xy, yz, zx] @vars x y z real=true DivF = divergence(F(x,y,z), [x,y,z]) integrate(DivF, (x, -1,1), (y,-1,1), (z, -1,1)) \begin{equation}0\end{equation} The total integral is $0$ by symmetry, not due to the divergence being $0$, as it is $x+y+z$. As for the surface integral, we have $6$ sides to consider. We take the sides with $\hat{N}$ being $\pm\hat{i}$: Nhat = [1,0,0] integrate((F(x,y,z) ⋅ Nhat), (y, -1, 1), (z, -1,1)) # at x=1 \begin{equation}0\end{equation} In fact, all $6$ sides will be $0$, as in this case $F \cdot \hat{i} = xy$ and at $x=1$ the surface integral is just $\int_{-1}^1\int_{-1}^1 y dy dz = 0$, as $y$ is an odd function. As such, the two sides of the Divergence theorem are both $0$, so the theorem is verified. ###### Example (From Strang) If the temperature inside the sun is $T = \log(1/\rho)$ find the heat flow $F=-\nabla{T}$; the source, $\nabla\cdot{F}$; and the flux, $\iint F\cdot\hat{N}dS$. Model the sun as a ball of radius $\rho_0$. We have the heat flow is simply: @vars x y z real=true R(x,y,z) = norm([x,y,z]) T(x,y,z) = log(1/R(x,y,z)) HeatFlow = -diff.(T(x,y,z), [x,y,z]) $\left[ \begin{array}{r}\frac{x}{x^{2} + y^{2} + z^{2}}\\\frac{y}{x^{2} + y^{2} + z^{2}}\\\frac{z}{x^{2} + y^{2} + z^{2}}\end{array} \right]$ We may recognize this as $\rho/\\|\rho\\|^2 = \hat{\rho}/\\|\rho\\|$. The source is DivF = divergence(HeatFlow, [x,y,z]) \|> simplify \begin{equation}\frac{1}{x^{2} + y^{2} + z^{2}}\end{equation} Which would simplify to $1/\rho^2$. Finally, the surface integral over the surface of the sun is an integral over a sphere of radius $\rho_0$. We could use spherical coordinates to compute this, but note instead that the normal is $\hat{\rho}$ so, $F \cdot \hat{N} = 1/\rho = 1/\rho_0$ over this surface. So the surface integral is simple the surface area times $1/\rho_0$: $4\pi\rho_0^2/\rho_0 = 4\pi\rho_0$. Finally, though $F$ is not continuous at the origin, the divergence theorem's result holds. Using spherical coordinates we have: @vars rho rho_0 phi theta real=true Jac = rho^2 sin(phi) integrate(1/rho^2 * Jac, (rho, 0, rho_0), (theta, 0, 2PI), (phi, 0, PI)) \begin{equation}4 \pi \rho_{0}\end{equation} ##### Example: Continuity equation (Schey) Imagine a venue with a strict cap on the number of persons at one time. Two ways to monitor this are: at given times, a count, or census, of all the people in the venue can be made. Or, when possible, a count of people coming in can be compared to a count of people coming out and the difference should yield the number within. Either works well when access is limited and the venue small, but the latter can also work well on a larger scale. For example, for the subway system of New York it would be impractical to attempt to count all the people at a given time using a census, but from turnstile data an accurate count can be had, as turnstiles can be used to track people coming in and going out. But turnstiles can be restricting and cause long(ish) lines. At some stores, new technology is allowing checkout-free shopping. Imagine if each customer had an app on their phone that can be used to track location. As they enter a store, they can be recorded, as they exit they can be recorded and if RFID tags are on each item in the store, their "purchases" can be tallied up and billed through the app. (As an added bonus to paying fewer cashiers, stores can also track on a step-by-step basis how a customer interacts with the store.) In any of these three scenarios, a simple thing applies: the total number of people in a confined region can be counted by counting how many crossed the boundary (and in which direction) and the change in time of the count can be related to the change in time of the people crossing. For a more real world example, the New York Times ran an article about estimating the size of a large protest in Hong Kong: Crowd estimates for Hong Kong’s large pro-democracy protests have been a point of contention for years. The organizers and the police often release vastly divergent estimates. This year’s annual pro-democracy protest on Monday, July 1, was no different. Organizers announced 550,000 people attended; the police said 190,000 people were there at the peak. But for the first time in the march’s history, a group of researchers combined artificial intelligence and manual counting techniques to estimate the size of the crowd, concluding that 265,000 people marched. On Monday, the A.I. team attached seven iPads to two major footbridges along the march route. Volunteers doing manual counts were also stationed next to the cameras, to help verify the computer count. The article describes some issues in counting such a large group: The high density of the crowd and the moving nature of these protests make estimating the turnout very challenging. For more than a decade, groups have stationed teams along the route and manually counted the rate of people passing through to derive the total number of participants. As there are no turnstiles to do an accurate count and too many points to come and go, this technique can be too approximate. The article describes how artificial intelligence was used to count the participants. The Times tried their own hand: Analyzing a short video clip recorded on Monday, The Times’s model tried to detect people based on color and shape, and then tracked the figures as they moved across the screen. This method helps avoid double counting because the crowd generally flowed in one direction. The divergence theorem provides two means to compute a value, the point here is to illustrate that there are (at least) two possible ways to compute crowd size. Which is better depends on the situation. Following Schey, we now consider a continuous analog to the crowd counting problem through a flow with a non-uniform density that may vary in time. Let $\rho(x,y,z;t)$ be the time-varying density and $v(x,y,z;t)$ be a vector field indicating the direction of flow. Consider some three-dimensional volume, $V$, with boundary $S$ (though two-dimensional would also be applicable). Then these integrals have interpretations: ~ \begin{align} \iiint_V \rho dV &&\quad\text{Amount contained within }V\\ \frac{\partial}{\partial{t}} \iiint_V \rho dV &= \iiint_V \frac{\partial{\rho}}{\partial{t}} dV &\quad\text{Change in time of amount contained within }V \end{align} ~ Moving the derivative inside the integral requires an assumption of continuity. Assume the material is conserved, meaning that if the amount in the volume $V$ changes it must flow in and out through the boundary. The flow out through $S$, the boundary of $V$, is $~ \oint_S (\rho v)\cdot\hat{N} dS, ~$ using the customary outward pointing normal for the orientation of $S$. So we have: $~ \iiint_V \frac{\partial{\rho}}{\partial{t}} dV = -\oint_S (\rho v)\cdot\hat{N} dS = - \iiint_V \nabla\cdot\left(\rho v\right)dV. ~$ The last equality by the divergence theorem, the minus sign as a positive change in amount within $V$ means flow opposite the outward pointing normal for $S$. The volume $V$ was arbitrary. While it isn't the case that two integrals being equal implies the integrands are equal, it is the case that if the two integrals are equal for all volumes and the two integrands are continuous, then they are equal. That is, under the assumptions that material is conserved and density is continuous a continuity equation can be derived from the divergence theorem: $~ \nabla\cdot(\rho v) = - \frac{\partial{\rho}}{dt}. ~$ ##### Example: The divergence theorem can fail to apply The assumption of the divergence theorem that the vector field by continuously differentiable is important, as otherwise it may not hold. With $R(x,y,z) = \langle x,y,z\rangle$ take for example $F = (R/\\|R\\|) / \\|R\\|^2)$. This has divergence R(x,y,z) = [x,y,z] F(x,y,z) = R(x,y,z) / norm(R(x,y,z))^3 @vars x y z real=true divergence(F(x,y,z), [x,y,z]) \|> simplify \begin{equation}0\end{equation} The simplification done by SymPy masks the presence of $R^{-5/2}$ when taking the partial derivatives, which means the field is not continuously differentiable at the origin. Were the divergence theorem applicable, then the integral of $F$ over the unit sphere would mean: $~ 0 = \iiint_V \nabla\cdot{F} dV = \oint_S F\cdot{N}dS = \oint_S \frac{R}{\\|R\\|^3} \cdot{R} dS = \oint_S 1 dS = 4\pi. ~$ Clearly, as $0$ is not equal to $4\pi$, the divergence theorem can not apply. However, it does apply to any volume not enclosing the origin. So without any calculation, if $V$ were shifted over by $2$ units the volume integral over $V$ would be $0$ and the surface integral over $S$ would be also. As already seen, the inverse square law here arises in the electrostatic force formula, and this same observation was made in the context of Gauss's law. ## Questions ###### Question (Schey) What conditions on $F: R^2 \rightarrow R^2$ imply $\oint_C F\cdot d\vec{r} = A$? ($A$ is the area bounded by the simple, closed curve $C$) ###### Question For $C$, a simple, closed curve parameterized by $\vec{r}(t) = \langle x(t), y(t) \rangle$, $a \leq t \leq b$. The area contained can be computed by $\int_a^b x(t) y'(t) dt$. Let $\vec{r}(t) = \sin(t) \cdot \langle \cos(t), \sin(t)\rangle$. Find the area inside $C$ ###### Question Let $\hat{N} = \langle \cos(t), \sin(t) \rangle$ and $\hat{T} = \langle -\sin(t), \cos(t)\rangle$. Then polar coordinates can be viewed as the parametric curve $\vec{r}(t) = r(t) \hat{N}$. Applying Green's theorem to the vector field $F = \langle -y, x\rangle$ which along the curve is $r(t) \hat{T}$ we know the area formula $(1/2) (\int xdy - \int y dx)$. What is this in polar coordinates (using $\theta=t$?) (Using $(r\hat{N}' = r'\hat{N} + r \hat{N}' = r'\hat{N} +r\hat{T}$ is useful.) ###### Question Let $\vec{r}(t) = \langle \cos^3(t), \sin^3(t)\rangle$, $0\leq t \leq 2\pi$. (This describes hypocycloid.) Compute the area enclosed by the curve $C$ using Green's theorem. ###### Question Let $F(x,y) = \langle y, x\rangle$. We verify Green's theorem holds when $S$ is the unit square, $[0,1]\times[0,1]$. The curl of $F$ is As the curl is a constant, say $c$, we have $\iint_S (\nabla\times{F}) dS = c \cdot 1$. This is? To integrate around the boundary we have 4 terms: the path $A$ connecting $(0,0)$ to $(1,0)$ (on the $x$ axis), the path $B$ connecting $(1,0)$ to $(1,1)$, the path $C$ connecting $(1,1)$ to $(0,1)$, and the path $D$ connecting $(0,1)$ to $(0,0)$ (along the $y$ axis). Which path has tangent $\hat{j}$? Along path $C$, $F(x,y) = [1,x]$ and $\hat{T}=-\hat{i}$ so $F\cdot\hat{T} = -1$. The path integral $\int_C (F\cdot\hat{T})ds = -1$. What is the value of the path integral over $A$? What is the integral over the oriented boundary of $S$? ###### Question Suppose $F: R^2 \rightarrow R^2$ is a vector field such that $\nabla\cdot{F}=0$ except at the origin. Let $C_1$ and $C_2$ be the unit circle and circle with radius $2$ centered at the origin, both parameterized counterclockwise. What is the relationship between $\oint_{C_2} F\cdot\hat{N}ds$ and $\oint_{C_1} F\cdot\hat{N}ds$? ###### Question Let $F(x,y) = \langle x, y\rangle/(x^2+y^2)$. Though this has divergence $0$ away from the origin, the flow integral around the unit circle, $\oint_C (F\cdot\hat{N})ds$, is $2\pi$, as Green's theorem in divergence form does not apply. Consider the integral around the square centered at the origin, with side lengths $2$. What is the flow integral around this closed curve? ###### Question Using the divergence theorem, compute $\iint F\cdot\hat{N} dS$ where $F(x,y,z) = \langle x, x, y \rangle$ and $V$ is the unit sphere. ###### Question Using the divergence theorem, compute $\iint F\cdot\hat{N} dS$ where $F(x,y,z) = \langle y, y,x \rangle$ and $V$ is the unit cube $[0,1]\times[0,1]\times[0,1]$. ###### Question Let $R(x,y,z) = \langle x, y, z\rangle$ and $\rho = \\|R\\|^2$. If $F = 2R/\rho^2$ then $F$ is the gradient of a potential. Which one?	2021-06-14 05:09:46	{"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": 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": 6, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.960620641708374, "perplexity": 374.7657935679705}, "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-2021-25/segments/1623487611445.13/warc/CC-MAIN-20210614043833-20210614073833-00129.warc.gz"}
https://www.jlab.org/indico/event/282/session/19/	# 8th Workshop of the APS Topical Group on Hadronic Physics 10-12 April 2019 Denver, CO US/Mountain timezone Home > Timetable > Session details PDF \| iCal ## Place Location: Denver, CO Address: Sheraton Denver Downtown Hotel, 1550 Court Pl. lobby level of the Plaza building Date: 10 Apr 16:00 - 17:35 ## Conveners • Prof. Perdrisat, Charles (The College of William and Mary) ## Timetable \| Contribution List Displaying 4 contributions out of 4 Type: invited talk Session: Hadron Form Factors In 2008 the Jefferson Laboratory F_pi-2 Collaboration released results for the pion electromagnetic form factor, which they had extracted from pion electro-production data. The measured values for the pion form factor were extracted using the Vanderhagen, Guidal and Laget Regge Model. While agreement betweenthis model and data is impressive, the theoretical implementation of gauge invariance is le ... More Presented by Prof. Anthony THOMAS on 10/4/2019 at 22:50 Type: contributed talk Session: Hadron Form Factors We adopt a chiral nucleon-pion Lagrangian and formulate a Basis Light-Front Quantization approach to solve the resulting light-front Hamiltonian of a nucleon as an eigenvalue problem. We obtain the nucleon mass spectrum and the corresponding light-front wave functions. Based on the light-front wave functions, we calculate the parton distribution functions as well as the elastic electromagnetic for ... More Presented by Mr. Weijie DU on 10/4/2019 at 23:15 Type: invited talk Session: Hadron Form Factors We present a calculation of the pion form factor using overlap fermions on 2+1-flavor domain-wall configurations on a $24^3\times 64$ lattice with $a=0.11 \, {\rm{fm}}$ and on a $32^3 \times 64$ lattice with $a=0.143 \, {\rm{fm}}$ generated by the RBC/UKQCD collaboration. Using the multi-mass algorithm, a simulation has been done with various valence quark masses with a range of space-like $Q^2$ f ... More Presented by Gen WANG on 10/4/2019 at 22:25 Type: invited talk Session: Hadron Form Factors The GMp Experiment which ran in Experimental Hall A at Jefferson Lab as one of the 12 GeV commissioning experiments measured the elastic cross section from electron-proton scattering with 2-3% accuarcy in the Q^2 range from 7 to 16 (GeV/c)^2. Such measurements test our understanding of nucleon structure in terms of Quantum Chromodynamics (QCD), with high-Q^2 , high-precision data providing strong ... More Presented by Prof. M. Eric Christy CHRISTY on 10/4/2019 at 22:00 Building timetable...	2020-08-15 20:44:58	{"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": 0.730151355266571, "perplexity": 9056.848675261059}, "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-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00068.warc.gz"}
https://answers.opencv.org/question/3018/how-to-verify-the-correctness-of-calibration-of-a-webcam/	# How to verify the correctness of calibration of a webcam i am totaly new to camera calibration techniques...i am using OpenCV chessboard technique...i am using a webcam from Quantum...here are my observations and steps... 1. i have kept each chess square side = 3.5 cm. It is a 7 x 5 chessboard with 6 x 4 internal corners. i am taking total of 10 images in differnet views/poses at a distance of 1 to 1.5 mtr from the webcam 2. i am following the C code in Learning OpenCV by Bradski for the calibration. my code for calibration is cvCalibrateCamera2(object_points,image_points,point_counts,cvSize(640,480),intrinsic_matrix,distortion_coeffs,NULL,NULL,CV_CALIB_FIX_ASPECT_RATIO); 3. before calling this function i am making the first and 2nd element along the diagonal of the intrinsic matrix as one to keep the ratio of focal lengths constant and using CV_CALIB_FIX_ASPECT_RATIO 4. with the change in distance of the chess board from the webcam the fx and fy are changing with fx:fy almost equal to 1. there are cx and cy values in order of 200 to 400. the fx and fy are in the order of 300 - 700 when i change the distance. 5. presently i have put all the distortion coeficients to zero because i didnot get good result including distortion coeffs. my original image looked handsome than the undistorted one!! am i doing the calibration correctly? will i use any other option than CV_CALIB_FIX_ASPECT_RATIO? if yes which one? edit retag close merge delete Sort by » oldest newest most voted Well,well,well. I am also a totally new to camera calibration techniques.But I still hope my words can help you. 1. Learning OpenCV is an older edition ,based on OpenCV 1.0. If you want to use the new OpenCV functions, I think you should follow the online documentions. And you also can find it in ../build/doc/.(but I dont know why, I cant find it in the OpenCV 3.0); 2. The function calirateCamera() returns a value , called rms(reprojection error) which can tell you your calibration precision. The rms should be between 0.1~1,with best something <0.5. 3. And there are some calibration samples in ../opencv/sources/samples/cpp,you can try it first when you write your own code. 4. There are some tips @StevenPuttemans gave me, and I think it may be useful to you. • Make sure the dimensions of your calibration pattern are uneven. • Make sure you have enough calibration images, using 2 images will never yield a good camera calibration. • Be sure to calibrate every single region of your camera range, if not, you will have large deformations towards the non calibrated areas. more Hi, There are some methods reported in literature for example try googling: A Comparative Review Of Camera Calibrating Methods with Accuracy Evaluation and you can also refer Accuracy assesment section of Tsai's seminal paper were he proposes such tests: A versatile camera calibration technique for high accuracy 3d machine vision metrology using off-the-shelf tv cameras and lenses (1987). Further, you can refer section-2 in paper : Accuracy Improvement in Camera Calibration by FaJie Li, Qi Zang and Reinhard Klette, which also mentions some approaches for evaluating camera calibration accuracy. more Official site GitHub Wiki Documentation	2020-02-18 22:32:28	{"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": 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": 0.22205178439617157, "perplexity": 1685.0454520351504}, "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-2020-10/segments/1581875143815.23/warc/CC-MAIN-20200218210853-20200219000853-00431.warc.gz"}
https://www.mathlearnit.com/what-is-51-288-as-a-decimal	# What is 51/288 as a decimal? ## Solution and how to convert 51 / 288 into a decimal 51 / 288 = 0.177 Convert 51/288 to 0.177 decimal form by understanding when to use each form of the number. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. So let’s dive into how and why you can convert 51/288 into a decimal. ## 51/288 is 51 divided by 288 Converting fractions to decimals is as simple as long division. 51 is being divided by 288. For some, this could be mental math. For others, we should set the equation. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. To solve the equation, we must divide the numerator (51) by the denominator (288). Here's how our equation is set up: ### Numerator: 51 • Numerators sit at the top of the fraction, representing the parts of the whole. 51 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Large two-digit conversions are tough. Especially without a calculator. Now let's explore X, the denominator. ### Denominator: 288 • Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 288 is one of the largest two-digit numbers to deal with. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Ultimately, don't be afraid of double-digit denominators. So without a calculator, let's convert 51/288 from a fraction to a decimal. ## Converting 51/288 to 0.177 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 288 \enclose{longdiv}{ 51 }$$ To solve, we will use left-to-right long division. Yep, same left-to-right method of division we learned in school. This gives us our first clue. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 288 \enclose{longdiv}{ 51.0 }$$ We've hit our first challenge. 51 cannot be divided into 288! So that means we must add a decimal point and extend our equation with a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 288 into 51 + 0 or 510. ### Step 3: Solve for how many whole groups you can divide 288 into 510 $$\require{enclose} 00.1 \\ 288 \enclose{longdiv}{ 51.0 }$$ Since we've extended our equation we can now divide our numbers, 288 into 510 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply by the left of our equation (288) to get the first number in our solution. ### Step 4: Subtract the remainder $$\require{enclose} 00.1 \\ 288 \enclose{longdiv}{ 51.0 } \\ \underline{ 288 \phantom{00} } \\ 222 \phantom{0}$$ If you don't have a remainder, congrats! You've solved the problem and converted 51/288 into 0.177 If you still have a remainder, continue to the next step. ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable. ### Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals is a necessity. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples: ### When you should convert 51/288 into a decimal Tax - Taxes are also in decimal form. Example. \$0.06 on every dollar spent. This is also represented in percentages. ### When to convert 0.177 to 51/288 as a fraction Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves. ### Practice Decimal Conversion with your Classroom • If 51/288 = 0.177 what would it be as a percentage? • What is 1 + 51/288 in decimal form? • What is 1 - 51/288 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.177 + 1/2?	2023-03-28 02:56:47	{"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": 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": 0.630963146686554, "perplexity": 1116.2337931698496}, "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/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00032.warc.gz"}
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Like amplitude in wavelength of water waves signify the displacement of water particles about their mean position.	2014-12-19 09:40:31	{"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": 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": 0.8679495453834534, "perplexity": 920.5919126809606}, "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/1418802768378.98/warc/CC-MAIN-20141217075248-00175-ip-10-231-17-201.ec2.internal.warc.gz"}
http://slideplayer.com/slide/7080493/	# 2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt Matter Properties of matter Mass. ## Presentation on theme: "2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt Matter Properties of matter Mass."— Presentation transcript: 2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt Matter Properties of matter Mass Volume Density These are the properties that all matter have. What are physical and chemical properties? These describe a substances’ ability to form a new substance What is a Chemical property? If it can be measured then it is this type of property. What is a physical property? The melting point of a substance is this type of property. What is a physical property? This property can be observed without forming a new substance. What is a physical property? This is the amount of matter in an object. What is the mass of an object? This has mass and takes up space. What is matter? This is the amount of space an object occupies. What is volume? This is the force of gravity on an object. What is weight? True or False: You will weigh more on the moon than on Earth. What is false? This is the formula for density. What is Density = _Mass_ volume? True or False: The density of an object decreases when it is broken up into smaller pieces. What is False? The boiling point of water is 100 degrees C. This is what type of property. What is a physical property? Tarnishing is this type of property. What is a chemical property? This what you need to measure the density of an object. What is mass and volume? True or False: If an object floats in water then we can say it is more dense than water What is false? True or False: If 2 objects have the same mass, they still may not have the same density. What is true? Block 1: 0.98g/cm3 Block 2: 0.83g/cm3 This is the least dense item. What is Block 2? This is the formula for volume. What is Width x Length x Height? This is how we find the volume of an irregular shaped object. What is displacement? Complete the following sentence: Volume is measured in _______ and cm3. What is ml? True or False: Weight is measured on a balance. What is false? This is what we use to measure a liquid accurately. The freezing point of a substance is this kind of property. What is physical property? Oil not reacting to iron is this kind of property. What is a chemical property? Download ppt "2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt Matter Properties of matter Mass." Similar presentations	2022-01-17 10:40:08	{"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": 0.8566555380821228, "perplexity": 2077.790484364206}, "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/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00139.warc.gz"}
https://d2mvzyuse3lwjc.cloudfront.net/doc/en/python/PyOrigin/Global-Functions/ActiveNotePage	8.2.3 ActiveNotePage Description Get the active Notes in the project. Syntax ActiveNotePage() Return The active Notes in project. Examples EX1 # Open a Notes in project import PyOrigin Notes=PyOrigin.ActiveNotePage() Notes.SetText('This example shows how to set text for Notes by using PyOrigin') NoteText=Notes.GetText() print(NoteText)	2022-08-16 10:53:58	{"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": 0.353128045797348, "perplexity": 13027.712446681708}, "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-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00573.warc.gz"}
https://bt.gateoverflow.in/827/gate-bt-2022-question-3	For a double-pipe heat exchanger, the inside and outside heat transfer coefficients are $100$ and $200 \; \text{W m}^{-2} \text{K}^{-1},$ respectively. The thickness and thermal conductivity of the thin-walled inner pipe are $1 \; \text{cm}$ and $10 \; \text{W m}^{-1} \text{K}^{-1},$ respectively. The value of the overall heat transfer coefficient is __________ $\text{W m}^{-2} \text{K}^{-1}.$ 1. $0.016$ 2. $42.5$ 3. $62.5$ 4. $310$	2022-06-25 14:24:15	{"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": 0.8585749864578247, "perplexity": 217.7084089341441}, "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-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00123.warc.gz"}
https://acpc18.kattis.com/problems/earlywinter	Alberta Collegiate Programming Contest 2018 #### Start 2018-10-27 18:00 UTC ## Alberta Collegiate Programming Contest 2018 #### End 2018-10-27 23:00 UTC The end is near! Contest is over. Not yet started. Contest is starting in -485 days 7:53:00 5:00:00 0:00:00 # Problem AEarly Winter Yraglac really hates early winter. So he was greatly annoyed when it snowed early this year. His friend however claimed that summer is always too short in Galcary so there’s nothing to complain about. Yraglac doesn’t believe his friend. So he decides to look up the historical weather report for the past $n$ years to figure out when it usually starts snowing. For each year in the past $n$ years, the historical weather report records $d_ i$, the number of days between the end of summer and the first day of snow on the $i^\textrm {th}$ year. You may assume it never snows during summer in Galcary, as that would be absurd. Given the historical data, Yraglac would like to know the number of consecutive years right before the current year with a larger gap between the end of summer and the first day of snow. More formally, suppose the current year is $m$. Then he’d like to determine the largest integer $k$ for which $d_{m-1}, d_{m-2},\ldots ,d_{m-k}> d_ m$, or determine that it had never snowed this early in the last $n$ years. ## Input The first line of the input contains two integers $n$ and $d_ m$. It is guaranteed that $1\leq n\leq 100$ and $0\leq d_ m\leq 100$. The next line of the input contains $n$ integers. The $i^\textrm {th}$ integer denotes $d_{m-i}$. It is guaranteed that $0\leq d_{m-i}\leq 100$. ## Output If there exists an integer $k$ for which $d_{m-k}\leq d_ m$, print “It hadn’t snowed this early in $k$ years!” (without quotes). Otherwise, print “It had never snowed this early!” (without quotes). Sample Input 1 Sample Output 1 4 2 3 3 3 2 It hadn't snowed this early in 3 years! Sample Input 2 Sample Output 2 2 10 0 100 It hadn't snowed this early in 0 years! Sample Input 3 Sample Output 3 3 1 42 43 44 It had never snowed this early!	2020-02-25 01:53:00	{"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": 0.42985999584198, "perplexity": 2291.8651394958624}, "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-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00023.warc.gz"}
https://tex.stackexchange.com/questions/291524/using-standard-circuits-and-tikzcircuits-in-the-same-document	# using “standard” circuits and tikzcircuits in the same document? Are the "standard" tikz circuit library and circuitikz incompatible? Consider the following MWE: \documentclass{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{tikz} \usepackage{pgfplots}\pgfplotsset{compat=1.9} \usepackage[siunitx]{circuitikz} % circuits, circuits.ee.IEC } \begin{document} %Standard libraries battery: % %\begin{tikzpicture}[baseline, huge circuit symbols, circuit ee IEC, % elec/.style={circle, inner sep=2pt, draw=blue, fill=blue}, % ] % \node (E) [battery, info=135:$E$] at (0,3.4) {}; % \end{tikzpicture} tikz circuit: \begin{circuitikz}[ american, ] \draw (0,0) to[R] (0,2); \end{circuitikz} \end{document} The output is as expected: but if I uncomment the commented parts, I have: ...and if I add a labels for example I often have errors in one of the two environments. Is there a way to use both of the circuit facilities in the same document of I have simply to choose one? Thanks! • Circuitiz is TikZ. The problem is that the circuits.ee.IEC library is incompatible with the circuitikz definitions. Originally circuitz components were spelled differently (R vs. resistor), but now both names do the same thing. – John Kormylo Feb 6 '16 at 18:33 Actually, one could use this trick to load any tikzlibrary temporarily. \documentclass{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{pgfplots}\pgfplotsset{compat=1.9} \usepackage[siunitx]{circuitikz} \begin{document} circuit ee IEC \bgroup% local definitions \usetikzlibrary{circuits.ee.IEC} \begin{circuitikz}[circuit ee IEC] \node (E) [battery, info=135:$E$] at (0,3.4) {}; \end{circuitikz} \egroup circuitikz components \begin{tikzpicture}[american] \draw (0,0) to[R] (0,2); \end{tikzpicture} \end{document} • For some reason it is not working in beamer... grrrr. – Rmano Feb 7 '16 at 11:27 • It worked fine for me. Just be careful not to load the circuit tikzlibraries globally. – John Kormylo Feb 7 '16 at 16:45	2021-03-08 22:02:36	{"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": 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": 0.7832552194595337, "perplexity": 8376.240066420887}, "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/1614178385529.97/warc/CC-MAIN-20210308205020-20210308235020-00115.warc.gz"}
https://math.stackexchange.com/questions/2027525/if-m-is-an-odd-natural-number-show-that-m-mid-2-phim-1-where-phi-i	# If $m$ is an odd natural number, show that $m\mid 2^{\phi(m)}-1,$ where $\phi$ is the Euler totient function. If $m$ is an odd natural number, show that $$m\mid 2^{\phi(m)}-1,$$ where $\phi$ is the Euler totient function. Can someone provide me some hints. ## 1 Answer The result you want to prove is a straight consequence of Euler's totient theorem. On a quick look at the theorem, you will find that: If n and a are coprime positive integers, then $${\displaystyle a^{\varphi (n)}\equiv 1{\pmod {n}}}$$ Since m is odd and 2 is even so m and 2 are coprime and hence you can apply Euler's totient theorem here, without any hesitation.	2019-10-17 10:14:12	{"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": 0.9407148361206055, "perplexity": 62.89160870356944}, "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-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00184.warc.gz"}
https://www.ncatlab.org/nlab/show/John+David+Stuart+Jones	# nLab John David Stuart Jones John David Stuart Jones (J.D.S. Jones) is a mathematician at the University of Warwick. ## selected writings category: people Last revised on January 11, 2019 at 10:24:48. See the history of this page for a list of all contributions to it.	2019-10-17 10:26:12	{"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": 0.8275418281555176, "perplexity": 4030.577361907203}, "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-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00421.warc.gz"}
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https://jeremykun.com/tag/cryptography-2/	# Sample Extraction from RLWE to LWE In this article I’ll derive a trick used in FHE called sample extraction. In brief, it allows one to partially convert a ciphertext in the Ring Learning With Errors (RLWE) scheme to the Learning With Errors (LWE) scheme. Here are some other articles I’ve written about other FHE building blocks, though they are not prerequisites for this article. ## LWE and RLWE The first two articles in the list above define the Learning With Errors problem (LWE). I will repeat the definition here: LWE: The LWE encryption scheme has the following parameters: • A plaintext space $\mathbb{Z}/q\mathbb{Z}$, where $q \geq 2$ is a positive integer. This is the space that the underlying message $m$ comes from. • An LWE dimension $n \in \mathbb{N}$. • A discrete Gaussian error distribution $D$ with a mean of zero and a fixed standard deviation. An LWE secret key is defined as a vector $s \in \{0, 1\}^n$ (uniformly sampled). An LWE ciphertext is defined as a vector $a = (a_1, \dots, a_n)$, sampled uniformly over $(\mathbb{Z} / q\mathbb{Z})^n$, and a scalar $b = \langle a, s \rangle + m + e$, where $m$ is the message, $e$ is drawn from $D$ and all arithmetic is done modulo $q$. Note: the message $m$ usually is represented by placing an even smaller message (say, a 4-bit message) in the highest-order bits of a 32-bit unsigned integer. So then decryption corresponds to computing $b – \langle a, s \rangle = m + e$ and rounding the result to recover $m$ while discarding $e$. Without the error term, an attacker could determine the secret key from a polynomial-sized collection of LWE ciphertexts with something like Gaussian elimination. The set of samples looks like a linear (or affine) system, where the secret key entries are the unknown variables. With an error term, the problem of solving the system is believed to be hard, and only exponential time/space algorithms are known. RLWE: The Ring Learning With Errors (RLWE) problem is the natural analogue of LWE, where all scalars involved are replaced with polynomials over a (carefully) chosen ring. Formally, the RLWE encryption scheme has the following parameters: • A ring $R = \mathbb{Z}/q\mathbb{Z}$, where $q \geq 2$ is a positive integer. This is the space of coefficients of all polynomials in the scheme. I usually think of $q$ as $2^{32}$, i.e., unsigned 32-bit integers. • A plaintext space $R[x] / (x^N + 1)$, where $N$ is a power of 2. This is the space that the underlying message $m(x)$ comes from, and it is encoded as a list of $N$ integers forming the coefficients of the polynomial. • An RLWE dimension $n \in \mathbb{N}$. • A discrete Gaussian error distribution $D$ with a mean of zero and a fixed standard deviation. An RLWE secret key $s$ is defined as a list of $n$ polynomials with binary coefficients in $\mathbb{B}[x] / (x^N+1)$, where $\mathbb{B} = \{0, 1\}$. The coefficients are uniformly sampled, like in LWE. An RLWE ciphertext is defined as a vector of $n$ polynomials $a = (a_1(x), \dots, a_n(x))$, sampled uniformly over $(R[x] / (x^N+1))^n$, and a polynomial $b(x) = \langle a, s \rangle + m(x) + e(x)$, where $m(x)$ is the message (with a similar “store it in the top bits” trick as LWE), $e(x)$ is a polynomial with coefficients drawn from $D$ and all the products of the inner product are done in $R[x] / (x^N+1)$. Decryption in RLWE involves computing $b(x) – \langle a, s \rangle$ and rounding appropriately to recover $m(x)$. Just like with RLWE, the message is “hidden” in the noise added to an equation corresponding to the polynomial products (i.e., without the noise and with enough sample encryptions of the same message/secret key, you can solve the system and recover the message). For more notes on how polynomial multiplication ends up being tricker in this ring, see my negacyclic polynomial multiplication article. The most common version of RLWE you will see in the literature sets the vector dimension $n=1$, and so the secret key $s$ is a single polynomial, the ciphertext is a single polynomial, and RLWE can be viewed as directly replacing the vector dot product in LWE with a polynomial product. However, making $n$ larger is believed to provide more security, and it can be traded off against making the polynomial degree smaller, which can be useful when tweaking parameters for performance (keeping the security level constant). ## Sample Extraction Sample extraction is the trick of taking an RLWE encryption of $m(x) = m_0 + m_1(x) + \dots + m_{N-1}x^{N-1}$, and outputting an LWE encryption of $m_0$. In our case, the degree $N$ and the dimension $n_{\textup{RLWE}}$ of the input RLWE ciphertext scheme is fixed, but we may pick the dimension $n_{\textup{LWE}}$ of the LWE scheme as we choose to make this trick work. This is one of those times in math when it is best to “just work it out with a pencil.” It turns out there are no serious obstacles to our goal. We start with polynomials $a = (a_1(x), \dots, a_n(x))$ and $b(x) = \langle a, s \rangle + m(x) + e(x)$, and we want to produce a vector of scalars $(x_1, \dots, x_D)$ of some dimension $D$, a corresponding secret key $s$, and a $b = \langle a, s \rangle + m_0 + e’$, where $e’$ may be different from the input error $e(x)$, but is hopefully not too much larger. As with many of the articles in this series, we employ the so-called “phase function” to help with the analysis, which is just the partial decryption of an RLWE ciphertext without the rounding step: $\varphi(x) = b(x) – \langle a, s \rangle = m(x) + e(x)$. The idea is as follows: inspect the structure of the constant term of $\varphi(x)$, oh look, it’s an LWE encryption. So let’s expand the constant term of $b(x) – \langle a, s \rangle$. Given a polynomial expression, I will use the notation $(-)[0]$ to denote the constant coefficient, and $(-)[k]$ for the $k$-th coefficient. \begin{aligned}(b(x) – \langle a, s \rangle)[0] &= b[0] – \left ( (a_1s_1)[0] + \dots + (a_n s_n)[0] \right ) \end{aligned} Each entry in the dot product is a negacyclic polynomial product, so its constant term requires summing all the pairs of coefficients of $a_i$ and $s_i$ whose degrees sum to zero mod $N$, and flipping signs when there’s wraparound. In particular, a single product above for $a_i s_i$ has the form: $$(a_is_i) [0] = s_i[0]a_i[0] – s_i[1]a_i[N-1] – s_i[2]a_i[N-2] – \dots – s_i[N-1]a_i[1]$$ Notice that I wrote the coefficients of $s_i$ in increasing order. This was on purpose, because if we re-write this expression $(a_is_i)[0]$ as a dot product, we get $$(a_is_i[0]) = \left \langle (s_i[0], s_i[1], \dots, s_i[N-1]), (a_i[0], -a_i[N-1], \dots, -a_i[1])\right \rangle$$ In particular, the $a_i[k]$ are public, so we can sign-flip and reorder them easily in our conversion trick. But $s_i$ is unknown at the time the sample extraction needs to occur, so it helps if we can leave the secret key untouched. And indeed, when we apply the above expansion to all of the terms in the computation of $\varphi(x)[0]$, we end up manipulating the $a_i$’s a lot, but merely “flattening” the coefficients of $s = (s_1(x), \dots, s_n(x))$ into a single long vector. So combining all of the above products, we see that $(b(x) – \langle a, s \rangle)[0]$ is already an LWE encryption with $(x, y) = ((x_1, \dots, x_D), b[0])$, and $x$ being the very long ($D = nN$) vector \begin{aligned} x = (& a_0[0], -a_0[N-1], \dots, -a_0[1], \\ &a_1[0], -a_1[N-1], \dots, -a_1[1], \\ &\dots , \\ &a_n[0], -a_n[N-1], \dots, -a_n[1] ) \end{aligned} And the corresponding secret key is \begin{aligned} s_{\textup{LWE}} = (& (s_0[0], s_0[1], \dots, s_0[N-1] \\ &(s_1[0], s_1[1], \dots, s_1[N-1], \\ &\dots , \\ &s_n[0], s_n[1], \dots, s_n[N-1] ) \end{aligned} And the error in this ciphertext is exactly the constant coefficient of the error polynomial $e(x)$ from the RLWE encryption, which is independent of the error of all the other coefficients. ## Commentary This trick is a best case scenario. Unlike with key switching, we don’t need to encrypt the output LWE secret key to perform the conversion. And unlike modulus switching, there is no impact on the error growth in the conversion from RLWE to LWE. So in a sense, this trick is “perfect,” though it loses information about the other coefficients of $m(x)$ in the process. As it happens, the CGGI FHE scheme that these articles are building toward only uses the constant coefficient. The only twist to think about is that the output LWE ciphertext is dependent on the RLWE scheme parameters. What if you wanted to get a smaller-dimensional LWE ciphertext as output? This is a realistic concern, as in the CGGI FHE scheme one starts from an LWE ciphertext of one dimension, goes to RLWE of another (larger) dimension, and needs to get back to LWE of the original dimension by the end. To do this, you have two options: one is to pick the RLWE ciphertext parameters $n, N$, so that their product is the value you need. A second is to allow the RLWE parameters to be whatever you need for performance/security, and then employ a key switching operation after the sample extraction to get back to the LWE parameters you need. It is worth mentioning—though I am far from fully understanding the methods—there other ways to convert between LWE and RLWE. One can go from LWE to RLWE, or from a collection of LWEs to RLWE. Some methods can be found in this paper and its references. Until next time! # Estimating the Security of Ring Learning with Errors (RLWE) This article was written by my colleague, Cathie Yun. Cathie is an applied cryptographer and security engineer, currently working with me to make fully homomorphic encryption a reality at Google. She’s also done a lot of cool stuff with zero knowledge proofs. In previous articles, we’ve discussed techniques used in Fully Homomorphic Encryption (FHE) schemes. The basis for many FHE schemes, as well as other privacy-preserving protocols, is the Learning With Errors (LWE) problem. In this article, we’ll talk about how to estimate the security of lattice-based schemes that rely on the hardness of LWE, as well as its widely used variant, Ring LWE (RLWE). A previous article on modulus switching introduced LWE encryption, but as a refresher: ## Reminder of LWE A literal repetition from the modulus switching article. The LWE encryption scheme I’ll use has the following parameters: • A plaintext space $\mathbb{Z}/q\mathbb{Z}$, where $q \geq 2$ is a positive integer. This is the space that the underlying message comes from. • An LWE dimension $n \in \mathbb{N}$. • A discrete Gaussian error distribution $D$ with a mean of zero and a fixed standard deviation. An LWE secret key is defined as a vector in $\{0, 1\}^n$ (uniformly sampled). An LWE ciphertext is defined as a vector $a = (a_1, \dots, a_n)$, sampled uniformly over $(\mathbb{Z} / q\mathbb{Z})^n$, and a scalar $b = \langle a, s \rangle + m + e$, where $e$ is drawn from $D$ and all arithmetic is done modulo $q$. Note that $e$ must be small for the encryption to be valid. ## Learning With Errors (LWE) security Choosing appropriate LWE parameters is a nontrivial challenge when designing and implementing LWE based schemes, because there are conflicting requirements of security, correctness, and performance. Some of the parameters that can be manipulated are the LWE dimension $n$, error distribution $D$ (referred to in the next few sections as $X_e$), secret distribution $X_s$, and plaintext modulus $q$. ## Lattice Estimator Here is where the Lattice Estimator tool comes to our assistance! The lattice estimator is a Sage module written by a group of lattice cryptography researchers which estimates the concrete security of Learning with Errors (LWE) instances. For a given set of LWE parameters, the Lattice Estimator calculates the cost of all known efficient lattice attacks – for example, the Primal, Dual, and Coded-BKW attacks. It returns the estimated number of “rops” or “ring operations” required to carry out each attack; the attack that is the most efficient is the one that determines the security parameter. The bits of security for the parameter set can be calculated as $\log_2(\text{rops})$ for the most efficient attack. ## Running the Lattice Estimator For example, let’s estimate the security of the security parameters originally published for the popular TFHE scheme: n = 630 q = 2^32 Xs = UniformMod(2) Xe = DiscreteGaussian(stddev=2^17) After installing the Lattice Estimator and sage, we run the following commands in sage: > from estimator import > schemes.TFHE630 LWEParameters(n=630, q=4294967296, Xs=D(σ=0.50, μ=-0.50), Xe=D(σ=131072.00), m=+Infinity, tag='TFHE630') > _ = LWE.estimate(schemes.TFHE630) bkw :: rop: ≈2^153.1, m: ≈2^139.4, mem: ≈2^132.6, b: 4, t1: 0, t2: 24, ℓ: 3, #cod: 552, #top: 0, #test: 78, tag: coded-bkw usvp :: rop: ≈2^124.5, red: ≈2^124.5, δ: 1.004497, β: 335, d: 1123, tag: usvp bdd :: rop: ≈2^131.0, red: ≈2^115.1, svp: ≈2^131.0, β: 301, η: 393, d: 1095, tag: bdd bdd_hybrid :: rop: ≈2^185.3, red: ≈2^115.9, svp: ≈2^185.3, β: 301, η: 588, ζ: 0, \|S\|: 1, d: 1704, prob: 1, ↻: 1, tag: hybrid bdd_mitm_hybrid :: rop: ≈2^265.5, red: ≈2^264.5, svp: ≈2^264.5, β: 301, η: 2, ζ: 215, \|S\|: ≈2^189.2, d: 1489, prob: ≈2^-146.6, ↻: ≈2^148.8, tag: hybrid dual :: rop: ≈2^128.7, mem: ≈2^72.0, m: 551, β: 346, d: 1181, ↻: 1, tag: dual dual_hybrid :: rop: ≈2^119.8, mem: ≈2^115.5, m: 516, β: 314, d: 1096, ↻: 1, ζ: 50, tag: dual_hybrid In this example, the most efficient attack is the dual_hybrid attack. It uses 2^119.8 ring operations, and so these parameters provide 119.8 bits of security. The reader may notice that the TFHE website claims those parameters give 128 bits of security. This discrepancy is due to the fact that they used an older library (the LWE estimator, which is no longer maintained), which doesn’t take into account the most up-to-date lattice attacks. For further reading, Benjamin Curtis wrote an article about parameter selection for the CONCRETE implementation of the TFHE scheme. Benjamin Curtis, Martin Albrecht, and other researchers also used the Lattice Estimator to estimate all the LWE and NTRU schemes. ## Ring Learning with Errors (RLWE) security It is often desirable to use Ring LWE instead of LWE, for greater efficiency and smaller key sizes (as Chris Peikert illustrates via meme). We’d like to estimate the security of a Ring LWE scheme, but it wasn’t immediately obvious to us how to do this, since the Lattice Estimator only operates over LWE instances. In order to use the Lattice Estimator for this security estimate, we first needed to do a reduction from the RLWE instance to an LWE instance. ## Attempted RLWE to LWE reduction Given an RLWE instance with $\text{RLWE_dimension} = k$ and $\text{poly_log_degree} = N$, we can create a relation that looks like an LWE instance of $\text{LWE_dimension} = N * k$ with the same security, as long as $N$ is a power of 2 and there are no known attacks that target the ring structure of RLWE that are more efficient than the best LWE attacks. Note: $N$ must be a power of 2 so that $x^N+1$ is a cyclotomic polynomial. An RLWE encryption has the following form: $(a_0(x), a_1(x), … a_{k-1}(x), b(x))$ • Public polynomials: $a_0(x), a_1(x), \dots a_{k-1}(x) \overset{{\scriptscriptstyle\$}}{\leftarrow} (\mathbb{Z}/{q \mathbb{Z}[x]} ) / (x^N + 1)^k$• Secret (binary) polynomials:$ s_0(x), s_1(x), \dots s_{k-1}(x) \overset{{\scriptscriptstyle\$}}{\leftarrow} (\mathbb{B}_N[x])^k$ • Error: $e(x) \overset{{\scriptscriptstyle\$}}{\leftarrow} \chi_e$• RLWE instance:$ b(x) = \sum_{i=0}^{k-1} a_i(x) \cdot s_i(x) + e(x) \in (\mathbb{Z}/{q \mathbb{Z}[x]} ) / (x^N + 1)$We would like to express this in the form of an LWE encryption. We can make start with the simple case, where$ k=1 $. Therefore, we will only be working with the zero-entry polynomials,$a_0(x)$and$s_0(x)$. (For simplicity, in the next example you can ignore the zero-subscript and think of them as$a(x)$and$s(x)$). ## Naive reduction for$k=1$(wrong!) Naively, if we simply defined the LWE$A$matrix to be a concatenation of the coefficients of the RLWE polynomial$a(x)$, we get: $$A_{\text{LWE}} = ( a_{0, 0}, a_{0, 1}, \dots a_{0, N-1} )$$ We can do the same for the LWE$s$vector: $$s_{\text{LWE}} = ( s_{0, 0}, s_{0, 1}, \dots s_{0, N-1} )$$ But this doesn’t give us the value of$b_{LWE}$for the LWE encryption that we want. In particular, the first entry of$b_{LWE}$, which we can call$b_{\text{LWE}, 0}$, is simply a product of the first entries of$a_0(x)$and$s_0(x)$: $$b_{\text{LWE}, 0} = a_{0, 0} \cdot s_{0, 0} + e_0$$ However, we want$b_{\text{LWE}, 0}$to be a sum of the products of all the coefficients of$a_0(x)$and$s_0(x)$that give us a zero-degree coefficient mod$x^N + 1$. This modulus is important because it causes the product of high-degree monomials to “wrap around” to smaller degree monomials because of the negacyclic property, such that$x^N \equiv -1 \mod x^N + 1$. So the constant term$b_{\text{LWE}, 0}should include all of the following terms: \begin{aligned} b_{\text{LWE}, 0} = & a_{0, 0} \cdot s_{0, 0} \\ – & a_{0, 1} \cdot s_{0, N-1} \\ – & a_{0, 2} \cdot s_{0, N-2} \\ – & \dots \\ – & a_{0, N-1} \cdot s_{0, 1}\\ + & e_0\\ \end{aligned} ## Improved reduction fork=1$We can achieve the desired value of$b_{\text{LWE}}$by more strategically forming a matrix$A_{\text{LWE}}$, to reflect the negacyclic property of our polynomials in the RLWE space. We can keep the naive construction for$s_\text{LWE}$. $$A_{\text{LWE}} = \begin{pmatrix} a_{0, 0} & -a_{0, N-1} & -a_{0, N-2} & \dots & -a_{0, 1}\\ a_{0, 1} & a_{0, 0} & -a_{0, N-1} & \dots & -a_{0, 2}\\ \vdots & \ddots & & & \vdots \\ a_{0, N-1} & \dots & & & a_{0, 0} \\ \end{pmatrix}$$ This definition of$A_\text{LWE}$gives us the desired value for$b_\text{LWE}$, when$b_{\text{LWE}}$is interpreted as the coefficients of a polynomial. As an example, we can write out the elements of the first row of$b_\text{LWE}: \begin{aligned} b_{\text{LWE}, 0} = & \sum_{i=0}^{N-1} A_{\text{LWE}, 0, i} \cdot s_{0, i} + e_0 \\ b_{\text{LWE}, 0} = & a_{0, 0} \cdot s_{0, 0} \\ – & a_{0, 1} \cdot s_{0, N-1} \\ – & a_{0, 2} \cdot s_{0, N-2} \\ – & \dots \\ – & a_{0, N-1} \cdot s_{0, 1}\\ + & e_0 \\ \end{aligned} ## Generalizing for allk$In the generalized$k$case, we have the RLWE equation: $$b(x) = a_0(x) \cdot s_0(x) + a_1(x) \cdot s_1(x) \cdot a_{k-1}(x) \cdot s_{k-1}(x) + e(x)$$ We can construct the LWE elements as follows: $$A_{\text{LWE}} = \left ( \begin{array}{c\|c\|c\|c} A_{0, \text{LWE}} & A_{1, \text{LWE}} & \dots & A_{k-1, \text{LWE}} \end{array} \right )$$ where each sub-matrix is the construction from the previous section: $$A_{\text{LWE}} = \begin{pmatrix} a_{i, 0} & -a_{i, N-1} & -a_{i, N-2} & \dots & -a_{i, 1}\\ a_{i, 1} & a_{i, 0} & -a_{i, N-1} & \dots & -a_{i, 2}\\ \vdots & \ddots & & & \vdots \\ a_{i, N-1} & \dots & & & a_{i, 0} \\ \end{pmatrix}$$ And the secret keys are stacked similarly: $$s_{\text{LWE}} = ( s_{0, 0}, s_{0, 1}, \dots s_{0, N-1} \mid s_{1, 0}, s_{1, 1}, \dots s_{1, N-1} \mid \dots )$$ This is how we can reduce an RLWE instance with RLWE dimension$k$and polynomial modulus degree$N$, to a relation that looks like an LWE instance of LWE dimension$N * k$. ## Caveats and open research This reduction does not result in a correctly formed LWE instance, since an LWE instance would have a matrix$A$that is randomly sampled, whereas the reduction results in an matrix$A$that has cyclic structure, due to the cyclic property of the RLWE instance. This is why I’ve been emphasizing that the reduction produces an instance that looks like LWE. All currently known attacks on RLWE do not take advantage of the structure, but rather directly attack this transformed LWE instance. Whether the additional ring structure can be exploited in the design of more efficient attacks remains an open question in the lattice cryptography research community. In her PhD thesis, Rachel Player mentions the RLWE to LWE security reduction: In order to try to pick parameters in Ring-LWE-based schemes (FHE or otherwise) that we hope are sufficiently secure, we can choose parameters such that the underlying Ring-LWE instance should be hard to solve according to known attacks. Each Ring-LWE sample can be used to extract$n$LWE samples. To the best of our knowledge, the most powerful attacks against$d$-sample Ring-LWE all work by instead attacking the$nd$-sample LWE problem. When estimating the security of a particular set of Ring-LWE parameters we therefore estimate the security of the induced set of LWE parameters. This indicates that we can do this reduction for certain RLWE instances. However, we must be careful to ensure that the polynomial modulus degree$N$is a power of two, because otherwise the error distribution “breaks”, as my colleague Baiyu Li explained to me in conversation: The RLWE problem is typically defined in using the ring of integers of the cyclotomic field$\mathbb{Q}[X]/(f(X))$, where$f(X)$is a cyclotomic polynomial of degree$k=\phi(N)$(where$\phi$is Euler’s totient function), and the error is a spherical Gaussian over the image of the canonical embedding into the complex numbers$\mathbb{C}^k$(basically the images of primitive roots of unity under$f$). In many cases we set$N$to be a power of 2, thus$f(X)=X^{N/2}+1$, since the canonical embedding for such$N$has a nice property that the preimage of the spherical Gaussian error is also a spherical Gaussian over the coefficients of polynomials in$\mathbb{Q}[X]/(f(X))$. So in this case we can sample$k=N/2$independent Gaussian numbers and use them as the coefficients of the error polynomial$e(x)$. For$N$not a power of 2,$f(X)$may have some low degree terms, and in order to get the spherical Gaussian with the same variance$s^2$in the canonical embedding, we probably need to use a larger variance when sampling the error polynomial coefficients. The RLWE we frequently use in practice is actually a specialized version called “polynomial LWE”, and instantiated with$N$= power of 2 and so$f(X)=X^{N/2}+1$. For other parameters the two are not exactly the same. This paper has some explanations: https://eprint.iacr.org/2018/170.pdf The error distribution “breaks” if$N$is not a power of 2 due to the fact that the precise form of RLWE is not defined on integer polynomial rings$R = \mathbb{Z}[X]/(f(X))$, but is defined on its dual (or the dual in the underlying number field, which is a fractional ideal of$\mathbb{Q}[X]/(f(x))$), and the noise distribution is on the Minkowski embedding of this dual ring. For non-power of 2$N$, the product mod$f$of two small polynomials in$\mathbb{Q}[X]/(f(x))$may be large, where small/large means their L2 norm on the coefficient vector. This means that in order to sample the required noise distribution, you may need a skewed coefficient distribution. Only when$N$is a power of 2, the dual of$R$is a scaling of$R$, and distance in the embedding of$R^{\text{dual}}$is preserved in$R$, and so we can just sample iid gaussian coefficient to get the required noise. Because working with a power-of-two RLWE polynomial modulus gives “nice” error behavior, this parameter choice is often recommended and chosen for concrete instantiations of RLWE. For example, the Homomorphic Encryption Standard recommends and only analyzes the security of parameters for power-of-two cyclotomic fields for use in homomorphic encryption (though future versions of the standard aim to extend the security analysis to generic cyclotomic rings): We stress that when the error is chosen from sufficiently wide and “well spread” distributions that match the ring at hand, we do not have meaningful attacks on RLWE that are better than LWE attacks, regardless of the ring. For power-of-two cyclotomics, it is sufficient to sample the noise in the polynomial basis, namely choosing the coefficients of the error polynomial$e \in \mathbb{Z}[x] / \phi_k(x)$independently at random from a very “narrow” distribution. Existing works analyzing and targeting the ring structure of RLWE include: It would of course be great to have a definitive answer on whether we can be confident using this RLWE to LWE reduction to estimate the security of RLWE based schemes. In the meantime, we have seen many Fully Homomorphic Encryption (FHE) schemes using this RLWE to LWE reduction, and we hope that this article helps explain how that reduction works and the existing open questions around this approach. # Negacyclic Polynomial Multiplication In this article I’ll cover three techniques to compute special types of polynomial products that show up in lattice cryptography and fully homomorphic encryption. Namely, the negacyclic polynomial product, which is the product of two polynomials in the quotient ring$\mathbb{Z}[x] / (x^N + 1)$. As a precursor to the negacyclic product, we’ll cover the simpler cyclic product. All of the Python code written for this article is on GitHub. ## The DFT and Cyclic Polynomial Multiplication A recent program gallery piece showed how single-variable polynomial multiplication could be implemented using the Discrete Fourier Transform (DFT). This boils down to two observations: 1. The product of two polynomials$f, g$can be computed via the convolution of the coefficients of$f$and$g$. 2. The Convolution Theorem, which says that the Fourier transform of a convolution of two signals$f, g$is the point-wise product of the Fourier transforms of the two signals. (The same holds for the DFT) This provides a much faster polynomial product operation than one could implement using the naïve polynomial multiplication algorithm (though see the last section for an implementation anyway). The DFT can be used to speed up large integer multiplication as well. A caveat with normal polynomial multiplication is that one needs to pad the input coefficient lists with enough zeros so that the convolution doesn’t “wrap around.” That padding results in the output having length at least as large as the sum of the degrees of$f$and$g$(see the program gallery piece for more details). If you don’t pad the polynomials, instead you get what’s called a cyclic polynomial product. More concretely, if the two input polynomials$f, g$are represented by coefficient lists$(f_0, f_1, \dots, f_{N-1}), (g_0, g_1, \dots, g_{N-1})$of length$N$(implying the inputs are degree at most$N-1$, i.e., the lists may end in a tail of zeros), then the Fourier Transform technique computes $f(x) \cdot g(x) \mod (x^N – 1)$ This modulus is in the sense of a quotient ring$\mathbb{Z}[x] / (x^N – 1)$, where$(x^N – 1)$denotes the ring ideal generated by$x^N-1$, i.e., all polynomials that are evenly divisible by$x^N – 1$. A particularly important interpretation of this quotient ring is achieved by interpreting the ideal generator$x^N – 1$as an equation$x^N – 1 = 0$, also known as$x^N = 1$. To get the canonical ring element corresponding to any polynomial$h(x) \in \mathbb{Z}[x]$, you “set”$x^N = 1$and reduce the polynomial until there are no more terms with degree bigger than$N-1$. For example, if$N=5$then$x^{10} + x^6 – x^4 + x + 2 = -x^4 + 2x + 3$(the$x^{10}$becomes 1, and$x^6 = x$). To prove the DFT product computes a product in this particular ring, note how the convolution theorem produces the following formula, where$\textup{fprod}(f, g)$denotes the process of taking the Fourier transform of the two coefficient lists, multiplying them entrywise, and taking a (properly normalized) inverse FFT, and$\textup{fprod}(f, g)(j)$is the$j$-th coefficient of the output polynomial: $\textup{fprod}(f, g)(j) = \sum_{k=0}^{N-1} f_k g_{j-k \textup{ mod } N}$ In words, the output polynomial coefficient$j$equals the sum of all products of pairs of coefficients whose indices sum to$j$when considered “wrapping around”$N$. Fixing$j=1$as an example,$\textup{fprod}(f, g)(1) = f_0 g_1 + f_1g_0 + f_2 g_{N-1} + f_3 g_{N-2} + \dots$. This demonstrates the “set$x^N = 1$” interpretation above: the term$f_2 g_{N-1}$corresponds to the product$f_2x^2 \cdot g_{N-1}x^{N-1}$, which contributes to the$x^1$term of the polynomial product if and only if$x^{2 + N-1} = x$, if and only if$x^N = 1$. To achieve this in code, we simply use the version of the code from the program gallery piece, but fix the size of the arrays given to numpy.fft.fft in advance. We will also, for simplicity, assume the$N$one wishes to use is a power of 2. The resulting code is significantly simpler than the original program gallery code (we omit zero-padding to length$N$for brevity). import numpy from numpy.fft import fft, ifft def cyclic_polymul(p1, p2, N): """Multiply two integer polynomials modulo (x^N - 1). p1 and p2 are arrays of coefficients in degree-increasing order. """ assert len(p1) == len(p2) == N product = fft(p1) * fft(p2) inverted = ifft(product) return numpy.round(numpy.real(inverted)).astype(numpy.int32) As a side note, there’s nothing that stops this from working with polynomials that have real or complex coefficients, but so long as we use small magnitude integer coefficients and round at the end, I don’t have to worry about precision issues (hat tip to Brad Lucier for suggesting an excellent paper by Colin Percival, “Rapid multiplication modulo the sum and difference of highly composite numbers“, which covers these precision issues in detail). ## Negacyclic polynomials, DFT with duplication Now the kind of polynomial quotient ring that shows up in cryptography is critically not$\mathbb{Z}[x]/(x^N-1)$, because that ring has enough easy-to-reason-about structure that it can’t hide secrets. Instead, cryptographers use the ring$\mathbb{Z}[x]/(x^N+1)$(the minus becomes a plus), which is believed to be more secure for cryptography—although I don’t have a great intuitive grasp on why. The interpretation is similar here as before, except we “set”$x^N = -1$instead of$x^N = 1$in our reductions. Repeating the above example, if$N=5$then$x^{10} + x^6 – x^4 + x + 2 = -x^4 + 3$(the$x^{10}$becomes$(-1)^2 = 1$, and$x^6 = -x$). It’s called negacyclic because as a term$x^k$passes$k \geq N$, it cycles back to$x^0 = 1$, but with a sign flip. The negacyclic polynomial multiplication can’t use the DFT without some special hacks. The first and simplest hack is to double the input lists with a negation. That is, starting from$f(x) \in \mathbb{Z}[x]/(x^N+1)$, we can define$f^(x) = f(x) – x^Nf(x)$in a different ring$\mathbb{Z}[x]/(x^{2N} – 1)$(and similarly for$g^$and$g$). Before seeing how this causes the DFT to (almost) compute a negacyclic polynomial product, some math wizardry. The ring$\mathbb{Z}[x]/(x^{2N} – 1)$is special because it contains our negacyclic ring as a subring. Indeed, because the polynomial$x^{2N} – 1$factors as$(x^N-1)(x^N+1)$, and because these two factors are coprime in$\mathbb{Z}[x]/(x^{2N} – 1)$, the Chinese remainder theorem (aka Sun-tzu’s theorem) generalizes to polynomial rings and says that any polynomial in$\mathbb{Z}[x]/(x^{2N} – 1)$is uniquely determined by its remainders when divided by$(x^N-1)$and$(x^N+1)$. Another way to say it is that the ring$\mathbb{Z}[x]/(x^{2N} – 1)$factors as a direct product of the two rings$\mathbb{Z}[x]/(x^{N} – 1)$and$\mathbb{Z}[x]/(x^{N} + 1)$. Now mapping a polynomial$f(x)$from the bigger ring$(x^{2N} – 1)$to the smaller ring$(x^{N}+1)$involves taking a remainder of$f(x)$when dividing by$x^{N}+1$(“setting”$x^N = -1$and reducing). There are many possible preimage mappings, depending on what your goal is. In this case, we actually intentionally choose a non preimage mapping, because in general to compute a preimage requires solving a system of congruences in the larger polynomial ring. So instead we choose$f(x) \mapsto f^(x) = f(x) – x^Nf(x) = -f(x)(x^N – 1)$, which maps back down to$2f(x)$in$\mathbb{Z}[x]/(x^{N} + 1)$. This preimage mapping has a particularly nice structure, in that you build it by repeating the polynomial’s coefficients twice and flipping the sign of the second half. It’s easy to see that the product$f^(x) g^(x)$maps down to$4f(x)g(x)$. So if we properly account for these extra constant factors floating around, our strategy to perform negacyclic polynomial multiplication is to map$f$and$g$up to the larger ring as described, compute their cyclic product (modulo$x^{2N} – 1$) using the FFT, and then the result should be a degree$2N-1$polynomial which can be reduced with one more modular reduction step to the right degree$N-1$negacyclic product, i.e., setting$x^N = -1$, which materializes as taking the second half of the coefficients, flipping their signs, and adding them to the corresponding coefficients in the first half. The code for this is: def negacyclic_polymul_preimage_and_map_back(p1, p2): p1_preprocessed = numpy.concatenate([p1, -p1]) p2_preprocessed = numpy.concatenate([p2, -p2]) product = fft(p1_preprocessed) fft(p2_preprocessed) inverted = ifft(product) rounded = numpy.round(numpy.real(inverted)).astype(p1.dtype) return (rounded[: p1.shape[0]] - rounded[p1.shape[0] :]) // 4 However, this chosen mapping hides another clever trick. The product of the two preimages has enough structure that we can “read” the result off without doing the full “set$x^N = -1$” reduction step. Mapping$f$and$g$up to$f^, g^$and taking their product modulo$(x^{2N} – 1)gives \begin{aligned} f^g^ &= -f(x^N-1) \cdot -g(x^N – 1) \\ &= fg (x^N-1)^2 \\ &= fg(x^{2N} – 2x^N + 1) \\ &= fg(2 – 2x^N) \\ &= 2(fg – x^Nfg) \end{aligned} This has the same syntactical format as the original mappingf \mapsto f – x^Nf$, with an extra factor of 2, and so its coefficients also have the form “repeat the coefficients and flip the sign of the second half” (times two). We can then do the “inverse mapping” by reading only the first half of the coefficients and dividing by 2. def negacyclic_polymul_use_special_preimage(p1, p2): p1_preprocessed = numpy.concatenate([p1, -p1]) p2_preprocessed = numpy.concatenate([p2, -p2]) product = fft(p1_preprocessed) * fft(p2_preprocessed) inverted = ifft(product) rounded = numpy.round(0.5 * numpy.real(inverted)).astype(p1.dtype) return rounded[: p1.shape[0]] Our chosen mapping$f \mapsto f-x^Nf$is not particularly special, except that it uses a small number of pre and post-processing operations. For example, if you instead used the mapping$f \mapsto 2f + x^Nf$(which would map back to$f$exactly), then the FFT product would result in$5fg + 4x^Nfg$in the larger ring. You can still read off the coefficients as before, but you’d have to divide by 5 instead of 2 (which, the superstitious would say, is harder). It seems that “double and negate” followed by “halve and take first half” is the least amount of pre/post processing possible. ## Negacyclic polynomials with a “twist” The previous section identified a nice mapping (or embedding) of the input polynomials into a larger ring. But studying that shows some symmetric structure in the FFT output. I.e., the coefficients of$f$and$g$are repeated twice, with some scaling factors. It also involves taking an FFT of two$2N$-dimensional vectors when we start from two$N$-dimensional vectors. This sort of situation should make you think that we can do this more efficiently, either by using a smaller size FFT or by packing some data into the complex part of the input, and indeed we can do both. [Aside: it’s well known that if all the entries of an FFT input are real, then the result also has symmetry that can be exploted for efficiency by reframing the problem as a size-N/2 FFT in some cases, and just removing half the FFT algorithm’s steps in other cases, see Wikipedia for more] This technique was explained in Fast multiplication and its applications (pdf link) by Daniel Bernstein, a prominent cryptographer who specializes in cryptography performance, and whose work appears in widely-used standards like TLS, OpenSSH, and he designed a commonly used elliptic curve for cryptography. [Aside: Bernstein cites this technique as using something called the “Tangent FFT (pdf link).” This is a drop-in FFT replacement he invented that is faster than previous best (split-radix FFT), and Bernstein uses it mainly to give a precise expression for the number of operations required to do the multiplication end to end. We will continue to use the numpy FFT implementation, since in this article I’m just focusing on how to express negacyclic multiplication in terms of the FFT. Also worth noting both the Tangent FFT and “Fast multiplication” papers frame their techniques—including FFT algorithm implementations!—in terms of polynomial ring factorizations and mappings. Be still, my beating cardioid.] In terms of polynomial mappings, we start from the ring$\mathbb{R}[x] / (x^N + 1)$, where$N$is a power of 2. We then pick a reversible mapping from$\mathbb{R}[x]/(x^N + 1) \to \mathbb{C}[x]/(x^{N/2} – 1)$(note the field change from real to complex), apply the FFT to the image of the mapping, and reverse appropriately it at the end. One such mapping takes two steps, first mapping$\mathbb{R}[x]/(x^N + 1) \to \mathbb{C}[x]/(x^{N/2} – i)$and then from$\mathbb{C}[x]/(x^{N/2} – i) \to \mathbb{C}[x]/(x^{N/2} – 1)$. The first mapping is as easy as the last section, because$(x^N + 1) = (x^{N/2} + i) (x^{N/2} – i)$, and so we can just set$x^{N/2} = i$and reduce the polynomial. This as the effect of making the second half of the polynomial’s coefficients become the complex part of the first half of the coefficients. The second mapping is more nuanced, because we’re not just reducing via factorization. And we can’t just map$i \mapsto 1$generically, because that would reduce complex numbers down to real values. Instead, we observe that (momentarily using an arbitrary degree$k$instead of$N/2$), for any polynomial$f \in \mathbb{C}[x]$, the remainder of$f \mod x^k-i$uniquely determines the remainder of$f \mod x^k – 1$via the change of variables$x \mapsto \omega_{4k} x$, where$\omega_{4k}$is a$4k$-th primitive root of unity$\omega_{4k} = e^{\frac{2 \pi i}{4k}}$. Spelling this out in more detail: if$f(x) \in \mathbb{C}[x]$has remainder$f(x) = g(x) + h(x)(x^k – i)$for some polynomial$h(x), then \begin{aligned} f(\omega_{4k}x) &= g(\omega_{4k}x) + h(\omega_{4k}x)((\omega_{4k}x)^{k} – i) \\ &= g(\omega_{4k}x) + h(\omega_{4k}x)(e^{\frac{\pi i}{2}} x^k – i) \\ &= g(\omega_{4k}x) + i h(\omega_{4k}x)(x^k – 1) \\ &= g(\omega_{4k}x) \mod (x^k – 1) \end{aligned} Translating this back tok=N/2$, the mapping from$\mathbb{C}[x]/(x^{N/2} – i) \to \mathbb{C}[x]/(x^{N/2} – 1)$is$f(x) \mapsto f(\omega_{2N}x)$. And if$f = f_0 + f_1x + \dots + f_{N/2 – 1}x^{N/2 – 1}$, then the mapping involves multiplying each coefficient$f_k$by$\omega_{2N}^k$. When you view polynomials as if they were a simple vector of their coefficients, then this operation$f(x) \mapsto f(\omega_{k}x)$looks like$(a_0, a_1, \dots, a_n) \mapsto (a_0, \omega_{k} a_1, \dots, \omega_k^n a_n)$. Bernstein calls the operation a twist of$\mathbb{C}^n$, which I mused about in this Mathstodon thread. What’s most important here is that each of these transformations are invertible. The first because the top half coefficients end up in the complex parts of the polynomial, and the second because the mapping$f(x) \mapsto f(\omega_{2N}^{-1}x)$is an inverse. Together, this makes the preprocessing and postprocessing exact inverses of each other. The code is then def negacyclic_polymul_complex_twist(p1, p2): n = p2.shape[0] primitive_root = primitive_nth_root(2 * n) root_powers = primitive_root ** numpy.arange(n // 2) p1_preprocessed = (p1[: n // 2] + 1j * p1[n // 2 :]) * root_powers p2_preprocessed = (p2[: n // 2] + 1j * p2[n // 2 :]) * root_powers p1_ft = fft(p1_preprocessed) p2_ft = fft(p2_preprocessed) prod = p1_ft * p2_ft ifft_prod = ifft(prod) ifft_rotated = ifft_prod * primitive_root ** numpy.arange(0, -n // 2, -1) return numpy.round( numpy.concatenate([numpy.real(ifft_rotated), numpy.imag(ifft_rotated)]) ).astype(p1.dtype) And so, at the cost of a bit more pre- and postprocessing, we can negacyclically multiply two degree$N-1$polynomials using an FFT of length$N/2$. In theory, no information is wasted and this is optimal. ## And finally, a simple matrix multiplication The last technique I wanted to share is not based on the FFT, but it’s another method for doing negacyclic polynomial multiplication that has come in handy in situations where I am unable to use FFTs. I call it the Toeplitz method, because one of the polynomials is converted to a Toeplitz matrix. Sometimes I hear it referred to as a circulant matrix technique, but due to the negacyclic sign flip, I don’t think it’s a fully accurate term. The idea is to put the coefficients of one polynomial$f(x) = f_0 + f_1x + \dots + f_{N-1}x^{N-1}$into a matrix as follows: $\begin{pmatrix} f_0 & -f_{N-1} & \dots & -f_1 \\ f_1 & f_0 & \dots & -f_2 \\ \vdots & \vdots & \ddots & \vdots \\ f_{N-1} & f_{N-2} & \dots & f_0 \end{pmatrix}$ The polynomial coefficients are written down in the first column unchanged, then in each subsequent column, the coefficients are cyclically shifted down one, and the term that wraps around the top has its sign flipped. When the second polynomial is treated as a vector of its coefficients, say,$g(x) = g_0 + g_1x + \dots + g_{N-1}x^{N-1}$, then the matrix-vector product computes their negacyclic product (as a vector of coefficients): $\begin{pmatrix} f_0 & -f_{N-1} & \dots & -f_1 \\ f_1 & f_0 & \dots & -f_2 \\ \vdots & \vdots & \ddots & \vdots \\ f_{N-1} & f_{N-2} & \dots & f_0 \end{pmatrix} \begin{pmatrix} g_0 \\ g_1 \\ \vdots \\ g_{N-1} \end{pmatrix}$ This works because each row$j$corresponds to one output term$x^j$, and the cyclic shift for that row accounts for the degree-wrapping, with the sign flip accounting for the negacyclic part. (If there were no sign attached, this method could be used to compute a cyclic polynomial product). The Python code for this is def cylic_matrix(c: numpy.array) -> numpy.ndarray: """Generates a cyclic matrix with each row of the input shifted. For input: [1, 2, 3], generates the following matrix: [[1 2 3] [2 3 1] [3 1 2]] """ c = numpy.asarray(c).ravel() a, b = numpy.ogrid[0 : len(c), 0 : -len(c) : -1] indx = a + b return c[indx] def negacyclic_polymul_toeplitz(p1, p2): n = len(p1) # Generates a sign matrix with 1s below the diagonal and -1 above. up_tri = numpy.tril(numpy.ones((n, n), dtype=int), 0) low_tri = numpy.triu(numpy.ones((n, n), dtype=int), 1) * -1 sign_matrix = up_tri + low_tri cyclic_matrix = cylic_matrix(p1) toeplitz_p1 = sign_matrix * cyclic_matrix return numpy.matmul(toeplitz_p1, p2) Obviously on most hardware this would be less efficient than an FFT-based method (and there is some relationship between circulant matrices and Fourier Transforms, see Wikipedia). But in some cases—when the polynomials are small, or one of the two polynomials is static, or a particular hardware choice doesn’t handle FFTs with high-precision floats very well, or you want to take advantage of natural parallelism in the matrix-vector product—this method can be useful. It’s also simpler to reason about. Until next time! # Key Switching in LWE Last time we covered an operation in the LWE encryption scheme called modulus switching, which allows one to switch from one modulus to another, at the cost of introducing a small amount of extra noise, roughly$\sqrt{n}$, where$n$is the dimension of the LWE ciphertext. This time we’ll cover a more sophisticated operation called key switching, which allows one to switch an LWE ciphertext from being encrypted under one secret key to another, without ever knowing either secret key. ## Reminder of LWE A literal repetition of the last article. The LWE encryption scheme I’ll use has the following parameters: • A plaintext space$\mathbb{Z}/q\mathbb{Z}$, where$q \geq 2$is a positive integer. This is the space that the underlying message comes from. • An LWE dimension$n \in \mathbb{N}$. • A discrete Gaussian error distribution$D$with a mean of zero and a fixed standard deviation. An LWE secret key is defined as a vector in$\{0, 1\}^n$(uniformly sampled). An LWE ciphertext is defined as a vector$a = (a_1, \dots, a_n)$, sampled uniformly over$(\mathbb{Z} / q\mathbb{Z})^n$, and a scalar$b = \langle a, s \rangle + m + e$, where$e$is drawn from$D$and all arithmetic is done modulo$q$. Note that$e$must be small for the encryption to be valid. Sometimes I will denote by$\textup{LWE}_s(x)$the LWE encryption of plaintext$x$under the secret key$s$, and it should be understood that this is a fixed (but arbitrary) draw from the distribution of LWE ciphertexts described above. ## Main idea: homomorphically almost-decrypt The main idea is to encrypt each entry of the original secret key using the new secret key (this collection of encryptions is jointly called a key-switching key), and then use this to homomorphically evaluate the first step of the decryption function (i.e., compute$b – \langle a, s \rangle$). The result is an encryption of the (noisy) message under the new key. First we’ll show how this works in a naïve sense. In particular, doing what I said in the last paragraph verbatim won’t work because the error will grow too large. But we’ll do it anyway, measure the error, and the remainder of the article will show how the gadget decomposition can be used to reduce the error. ## Key switching, without gadget decompositions Start with an LWE ciphertext for the plaintext$m$. Call it$\displaystyle c = (a_1, \dots, a_n, b) \in (\mathbb{Z}/q\mathbb{Z})^{n+1}$where$\displaystyle b = \left ( \sum_{i=1}^n a_i s_i \right ) + m + e_{\textup{original}}$and$s = (s_1, \dots, s_n) \in \{ 0,1\}^n$is the secret key. Now say we have another secret key, possibly of a different dimension$t = (t_1, \dots, t_m) \in \{ 0, 1\}^m$, and we would like to switch the ciphertext$c$to a ciphertext$c’$which encrypts the same underlying message$m$, but under the new secret key$t$. That is, we would like to write$\displaystyle c’ = (a’_1, \dots, a’_m, b’) \in (\mathbb{Z}/q\mathbb{Z})^{m+1}$where$\displaystyle b’ = \left ( \sum_{i=1}^n a’_i t_i \right ) + m + e_{\textup{original}} + e_{\textup{new}}$implying that there is possibly some additional error introduced as a result. As usual, so long as the total error in the ciphertext remains small enough (and$m$is stored in the significant bits of the underlying integer space), the result will still be a valid LWE ciphertext. Define the key switching key$\textup{KSK}(s, t)$as follows (I will omit the$s, t$and just call it KSK from now on):$\displaystyle \textup{KSK} = \{ \textup{KSK}_i = \textup{LWE}_t(s_i) = (x_{i, 1}, \dots, x_{i, m}, y_i) \mid i=1, \dots, n\}$In other words,$\textup{KSK}_i$encrypts bit$s_i$, and$y_i = \langle x_i, t \rangle + s_i + e_i$makes it a valid LWE encryption. Now the algorithm to switch keys is merely as follows (where the first vector has$mleading zeros to ensure the dimensions align):\displaystyle c’ = (0, \dots, 0, b) – \sum_{i=1}^n a_i \textup{KSK}_i$This is computing a linear combination of the$\textup{KSK}_i$. The specific linear combination is the first step of LWE decryption ($b – \langle a, s \rangle$), but performed on ciphertexts of$b$and the$s_i$. Note,$(0, \dots, 0, b)$is a valid (but insecure) LWE ciphertext of$b$under any secret key, in part because we’re pretending the LWE samples and error were all sampled as zero; an unlikely but coherent outcome used to jumpstart a homomorphic computation in more places than key switching. So if you wanted to, you could write$c’$as follows, to highlight how we’re computing additions and linear scalings of LWE ciphertexts.$\displaystyle c’ = \textup{LWE}_{\textup{t}}(b) – \sum_{i=1}^n a_i \textup{LWE}_t(s_i)$This should be enough to show that$c’$is a valid LWE encryption (if we accept that adding and scaling preserves LWE validity). But to warm up for the rest of the article we’ll reprove it with a slightly different technique. This will also help us understand the error growth. Because LWE naturally admits sums and scalar products with corresponding added error, we expect the error to grow proportionally to the number of additions and the magnitudes of the$a_i$’s. And you may already be able to tell that because the$a_i$’s are uniform$\mathbb{Z}/q\mathbb{Z}$elements, this part will be far too large to be useful. Let’s make this explicit now. To show it’s a valid LWE encryption, we define the function$\varphi_s$, defined on any LWE ciphertext$c = (a_1, \dots, a_n, b)$as$\varphi_s(c) = b – \langle a, s \rangle$. Some authors call$\varphi_s$the “phase” function, but I think of it as a close friend: the first step of the decryption function for LWE (the second step would be rounding off the error). Critically, an LWE encryption is valid if and only if$\varphi_s(c) = m + e$(provided$e$is sufficiently small). Because$\varphi_s$is a linear function, it factors through the definition of$c’$nicely, and we get$\displaystyle \begin{aligned} \varphi_t(c’) &= \varphi_t((0, \dots, 0, b)) – \sum_{i=1}^n a_i \varphi_t(\textup{KSK}_i) \\ &= b – \sum_{i=1}^n a_i (y_i – \langle x_i, t \rangle) \\ &= b – \sum_{i=1}^n a_i (s_i + e_i) \end{aligned}$where (reminder)$e_i$is the error sample from$\textup{KSK}_i$’s definition. Distributing$a_i$across the$(s_i + e_i)$simplifies everything nicely$\displaystyle \begin{aligned} &= b – \sum_{i=1}^n a_i s_i – \sum_{i=1}^n a_i e_i \\ &= m + e_{\textup{original}} – \sum_{i=1}^n a_i e_i \end{aligned}$Now as we foreshadowed,$e_{\textup{new}} = -\sum_{i=1}^n a_i e_i$is simply too large. A typical LWE ciphertext will have error at least 1 (or it would be useless), and if$q = 2^{32}$, the$a_i$’s would also be of magnitude roughly$2^{31}$, so summing even two of those would corrupt even a 1-bit message stored in the most significant bit of the plaintext. The way to deal with this is to use a bit decomposition. ## Key switching, with gadget decompositions Recall from the gadget decomposition article that the core function of a gadget decomposition is to preserve the ultimate value of a dot product while making the vectors multiplicands larger (spending space/time) but also making the size of the coefficients of one of the vectors smaller (reducing the accumulation of error due to that dot product). This is exactly the approach we’ll take here. The “dot product” in question is$(a_1, \dots, a_n) \cdot \textup{KSK}$(where KSK is viewed as a matrix), and we’ll expand the values$a_i$into a vector of its digits in a base-$B$number system, while modifying the key switching key so that those missing powers of$B$are part of the LWE encryption. This will result in replacing the error term that looked like$\sum_{i=1}^n a_i e_i$with an error term like$\sum_{i=1}^n c B e_i$for some small constant$c$(expect it to be even less than$B$). More specifically, define decomposition parameters as a triple of numbers$(B, k, L)$. The number$B$is a power of 2 no bigger than$q/2$, and$L$, or the number of levels of the decomposition, is the positive integer such that$B^L = q$(this is forced by the choice of$B$). Then finally,$k$is a number between$0$and$L-1$describing the “lowest level” (or least-significant digit) included in the decomposition. An error-free decomposition sets the parameter$k=0$, and this is defined simply as a base-$B$representation of a number. For example, suppose$q = 2^{32}$, and$(B, k, L) = (256, 0, 4)$, and we’re decomposing$x=2^{32} – 2$. Then$\textup{Decomp}_{256, 0, 4}(x) = (254, 255, 255, 255)$. I subtracted 2 to emphasize that the digits are little-Endian (the right-most entry is the most significant, representing the$256^3$place). An approximate decomposition is one with$k > 0$. For example, suppose$(B, k, L) = (256, 2, 4)$and again$x=2^{32} – 2$. Setting$k=2$means that we represent this number as if it were$(0, 0, 255, 255)$, wiping out the two least significant digits. The error of this approximation is$65534 = 254 + 255 \cdot 256^1$. As we will see, an approximate decomposition may help reduce overall error by splitting the newly introduced error into a sum of two terms, where$k$scales the error differently in each term. Let’s go through the key-switching key derivation again, using an error-free decomposition$(B, 0, L)$. First, re-define the key switching key as follows.$\displaystyle \textup{KSK} = \{ \textup{KSK}_{i, j} = \textup{LWE}_t(s_i B^j) \mid i=1, \dots, n ; j = 0, \dots, L-1\}$Note that this increases the dimension of the key-switching key by 1. Previously the key-switching key was a list of LWE ciphertexts (2-dimensional array of numbers), and now it’s a 3-dimensional array, with the new dimension corresponding to the decomposition digit$j$. Because the powers of$B$are attached to the message, they will factor out and allow us to reconstruct the original$a_i$’s, but they will not be included in the error part because error is added to the message during encryption. Next, to perform the key switch, define$\textup{Decomp}(a_i) = (a_{i,0}, \dots, a_{i,L-1})$and compute$\displaystyle c’ = (0, \dots, 0, b) – \sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} \textup{KSK}_{i,j}$This is the same as the original key switch, but the extra summation accounts for the extra dimension introduced by the gadget decomposition. Then we can repeat the same$\varphi_t$trick and see how the original$a_i$’s are reconstructed.$\displaystyle \begin{aligned} \varphi_t(c’) &= b – \sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} \varphi_t(\textup{KSK}_{i,j}) \\ &= b -\sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} (s_i B^j + e_i) \\ &= b -\sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} s_i B^j – \sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} e_i \\ &= b -\sum_{i=1}^n a_i s_i – \sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} e_i \\ &= m + e_{\textup{original}} – \sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} e_i \end{aligned}$One key ingredient above is noticing that in$\sum_{i=1}^n \sum_{j=0}^{L-1} a_{i,j} s_i B^j$, the$s_i$factors out of the innermost sum, and what you have left is$\sum_{j=0}^{L-1} a_{i,j} B^j$, which is exactly how to reconstruct$a_i$from its base-$B$digits. The second key ingredient is that the innermost term on the second line is$a_{i,j} (s_i B^j + e_i)$, which means that only the digits$a_{i,j}$are multiplied by the error terms, not including the powers of$B$, and so the final error can be bounded by the largest allowable value of a single digit$B-1$, resulting in the new error being$L (B-1) \sum_{i=1}^n e_i$. For a Gaussian centered at zero, the expectation of these errors is zero, and using standard bounding arguments like Chernoff bounds, you can prove that with high probability this new error is at most$L(B-1) \sigma \sqrt{2n \log n}$, where$\sigma$is the standard deviation of the error distribution. Now, finally, we can run through this argument one more time, but using an approximate decomposition. This merely changes the sum’s lower bound from$j=0$to$j=k$. Start by calling$\tilde{a}_i = \sum_{j=k}^{L-1} a_{i,j} B^j$, the approximation of$a_i$from its most significant bits. Then the error of this approximation is$a_i – \tilde{a}_i = \sum_{j=0}^{k-1} a_{i,j} B^j$, a relatively small quantity at most$(B^k – 1) / (B-1)$(if each$a_{i,j} = B-1$is as large as possible).$\displaystyle \begin{aligned} \varphi_t(c’) &= b – \sum_{i=1}^n \sum_{j=k}^{L-1} a_{i,j} \varphi_t(\textup{KSK}_{i,j}) \\ &= b -\sum_{i=1}^n \sum_{j=k}^{L-1} a_{i,j} (s_i B^j + e_i) \\ &= b -\sum_{i=1}^n s_i \sum_{j=k}^{L-1} a_{i,j} B^j – \sum_{i=1}^n \sum_{j=k}^{L-1} a_{i,j} e_i \\ &= b -\sum_{i=1}^n s_i \tilde{a}_i – \sum_{i=1}^n \sum_{j=k}^{L-1} a_{i,j} e_i \end{aligned}$Mentally zoom in on the first sum$\sum_{i=1}^n s_i \tilde{a}_i$. Use the trick of adding zero to get$\displaystyle \sum_{i=1}^n s_i \tilde{a}_i = \sum_{i=1}^n s_i (a_i + \tilde{a}_i – a_i) = \sum_{i=1}^n s_i a_i – \sum_{i=1}^n s_i(a_i – \tilde{a}_i)$The term$\sum_{i=1}^n s_i(a_i – \tilde{a}_i)$is part of our new error term, and recalling that the secret key bits are binary, you should think of this in expectation as roughly$\frac{n}{2} B^{k-1}$(more precisely,$\frac{n}{2} (B^{k}-1)/(B-1)$). Continuing, we arrive at$\displaystyle \begin{aligned} \varphi_t(c’) &= b -\sum_{i=1}^n a_i s_i – \sum_{i=1}^n s_i(a_i – \tilde{a}_i) – \sum_{i=1}^n \sum_{j=k}^{L-1} a_{i,j} e_i \\ &= m + e_{\textup{original}} – \sum_{i=1}^n s_i(a_i – \tilde{a}_i) – \sum_{i=1}^n \sum_{j=k}^{L-1} a_{i,j} e_i \end{aligned}$## Rough error analysis Now the choice of$k$admits a tradeoff that one can optimize for to minimize the total newly introduced error. I’m going to switch to a sloppy mode of math to heuristically navigate this tradeoff. The triangle inequality lets us bound the magnitude of the error by the sum of the magnitudes of the parts, i.e., the error is bounded from above by$\displaystyle \left \| \sum_{i=1}^n s_i(a_i – \tilde{a}_i) \right \| + \left \| \sum_{i=1}^n \sum_{j=k}^{L-1} a_{i,j} e_i \right \|$The left term is like$\frac{n}{2} B^{k-1}$as we stated earlier, and with high probability it’s at most$(n/2 + \sqrt{n \log n}) B^{k-1}$. The right term is at most$(L-k)B \sum_{i=1}^n e_i$, (worst case size of$a_{i,j}$, increasing$B-1$to$B$because why not), and with high probability the sum of the$e_i$is like$\sigma \sqrt{2n \log n}$, making the whole term bounded by$(L-k)B \sigma \sqrt{2n \log n}$. So we want to minimize the sum$\displaystyle (n/2 + \sqrt{n \log n}) B^{k-1} + (L-k)B \sigma \sqrt{2n \log n}$We could try to explicitly optimize this for$k$, treating the other terms as constant, but it won’t be nice because$k$is present in both a linear term and an exponent. We could also just stare at it and think. The approximation error (the term on the left) is going to get exponentially larger as$k$grows, so we want to keep$k$relatively small. But on the other hand, the standard deviation$\sigma$should be much larger than$n$to keep LWE secure. This is effectively what we’re trying to suppress: error that grows like$O(n)$is small enough to deal with, but error that grows like$\omega(n)$is problematic. Increasing$k$gives us a meager (but nontrivial) means to reduce the constant coefficient on that part of the error in exchange for$\Theta(n)$growth with in the other term. I admit, as of the time of this writing I still don’t understand how to set production security parameters for LWE. Is it still linear in$n$? Super-linear? Not sure. I’m betting future Jeremy will clarify this to me in another article. Even if it were linear in$n$, the right term multiplies$\sigma$by$\sqrt{n \log n}$which makes the whole thing super-linear, whereas the left term adds a square root factor. So the tradeoff in$k$should still help. Until I understand LWE security, I won’t have the asymptotics I need to analyze this further. Moreover, the allowed values of$B, k$are so small that we can brute force evaluate all options. For example, if$B = 16$then$k$can be between 0 and 7. And realistically, if$n \approx 2^{10}$, then letting$k = 4$makes the first term roughly$2^{26}\$, which leaves only 6 bits left for the message (further reduced by any error introduced by the second term). Thanks to Cathie Yun and Asra Ali for providing feedback on an early draft of this article. Until next time!	2023-03-30 16:47:38	{"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": 0.9998722076416016, "perplexity": 2950.2491687820884}, "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/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00647.warc.gz"}
https://wikieducator.org/Thread:Thank_you_for_collaborating_on_the_MathGloss_project_(10)	# Thank you for collaborating on the MathGloss project Hi Deepti, Done. These are the changes I made: 1. Created the CompGloss Template. This will be the Navigation Bar for all the pages of the Computing Terms Glossary 2. Modified the CompGloss Main Page 3. Modified the CompGloss/A page and redirected the original "A" list to it. Important: • For each page you create in Level 2 or Level 3, please insert this code (for consistency and usability): {{CompGloss}}{{MyTitle\|}} • Please remember this: If you want to create a link for a sub-level new page, you should write this code (slashes "/" create sub-levels): '''[[/ NewPage /]]''' 4. You can change Titles and Image, by editing the CompGloss Template Cheers, Gladys Gahona --chela5808 17:02, 21 January 2009 (UTC) 05:02, 22 January 2009	2021-04-15 18:27:43	{"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": 0.35263049602508545, "perplexity": 7028.191060108428}, "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-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00142.warc.gz"}
https://math.stackexchange.com/questions/1524157/independent-coin-tosses-double-or-halve-current-sum	Independent coin tosses , double or halve current sum (Quant job Interviews - Questions and Answers - Joshi et al, Question 3.5) Suppose you have a fair coin. You start with 1 dollar, and if you toss a H your position doubles, if you toss a T your position halves. What is the expected value of the money you have if you toss the coin to infinity ? Now the answer is stated as follows: We work out what happens with one toss, then $$n$$ tosses and then let $$n$$ tend to infinity. Let $$X$$ denote a toss then: $$\mathbb E (X) = \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot 0.5= {5\over4}$$ Provided the tosses are independent, the product of expectations is the expectation of the product. Let $$X_j$$ be the effect of toss $$j$$. This means that $$\mathbb E \left(\prod_{j=1}^n X_j\right) = \prod_{j=1}^n \mathbb E (X_j) = \left({5\over4}\right)^n$$ this clearly tends to infinity as $$n$$ tends to infinity Now, I don't understand this answer :( First, the way the answer is written out, surely the $${5\over4}$$ is the expectation of the outcome of the first toss $$X_1$$ , not that of a toss $$X_j , j \ge 1$$ ? Secondly, whilst I do understand that the tosses are independent, it would seem that the $$X_{j+1}$$ is actually quite heavily dependent on the $$X_{j}$$ before it ? So then why is it so obvious that $$\mathbb E ( X_{j+1} ) = \mathbb E ( X_j)$$ ? • Let $X_i$ be the outcome of the $i$th toss translated into $\{-1,1\}$. Then $M_n=2^{X_1+\dots+X_n}$. The sum in the exponent is symmetric around $0$ and spreads out as $n\to \infty$, hence $E(M_n)\to \infty$. – A.S. Nov 11, 2015 at 14:38 You're right, the given answer is not very clear. I'll try to provide a proof, but I warn you I'm far from being a probabilist so it might be not perfectly correct. Let $S_{i}$ be the random variable that denotes your dollars after the toss $i$. It's clear that $E(S_{1}) = \frac{1}{2}\times 2 + \frac{1}{2} \times \frac{1}{2} = \frac{5}{4}$. Then, $S_{2}$ takes two "values" : $2S_{1}$ or $\frac{S_{1}}{2}$. Thus $E(S_{2}) = \frac{1}{2}\times 2E(S_{1}) + \frac{1}{2} \times \frac{E(S_{1})}{2} = \frac{5}{4} \times E(S_{1}) = (\frac{5}{4})^{2}$. By recurrence $$E(S_{n}) = (\frac{5}{4})^{n}$$ • Also solved it by recursion, official answer seemed fuzzy. I found it easier to formulate as $$\Pi_{t}=\Pi_{t-1} \cdot B$$ $$\text{E}(\Pi_{t})=\text{E}(\Pi_{t-1} \cdot B)$$ by independence $$\text{E}(\Pi_{t})=\text{E}(\Pi_{t-1}) \cdot \text{E}(B)$$ $$\text{E}(\Pi_{t})=\text{E}(\Pi_{t-1}) \cdot 1.25$$ Which then leaves me with a system of equations $$\begin{cases} \text{E}(\Pi_{t})=\text{E}(\Pi_{t-1}) \cdot 1.25 \\ \text{E}(\Pi_{0})=1 \end{cases}$$ to solve, yielding $$\text{E}(\Pi_{t})=1.25^{t}$$ Apr 28, 2020 at 15:23 $$X_j = \begin{cases} 2 & \text{with probability } 1/2, \\ 1/2 & \text{with probability } 1/2. \end{cases}$$ You said "Let $$X_j$$ be the effect of toss $$j.$$" If you think these are not independent, I wonder if you thought this meant "Let $$X_j$$ be your position after toss $$j.$$" or something like that. But your position after toss $$j$$ is $$Y_j = \prod_{k=1}^j X_k.$$ Certainly $$Y_{j+1}$$ is, in your words, "quite heavily dependent" on $$Y_j.$$	2022-10-03 07:40:28	{"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": 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": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9621609449386597, "perplexity": 256.4467531305056}, "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/1664030337404.30/warc/CC-MAIN-20221003070342-20221003100342-00184.warc.gz"}
https://www.incredigeek.com/home/tag/windows/	# VirtualBox – Failed to acquire the VirtualBox COM object. Under the Details it was complaining about VirtualBox.xml Looking in Windows Explorer in the .VirtualBox folder it shows that the VirtualBox.xml file being empty. Delete the file. Reinstall VirtualBox. Now go to your VM’s in Open up the VM folder and double click on the “Virtual Machine Definition” file to “reimport” them into VirtualBox. # Enable or Install Group Policy Editor on Windows 10/11 Home Normally you can’t run the Group Policy Editor on Windows Home editions. But there is a way to enable it. First, open up a Command Prompt (Not Terminal) as Administrator Now copy and paste each of the commands. FOR %F IN ("%SystemRoot%\servicing\Packages\Microsoft-Windows-GroupPolicy-ClientTools-Package~.mum") DO (DISM /Online /NoRestart /Add-Package:"%F") FOR %F IN ("%SystemRoot%\servicing\Packages\Microsoft-Windows-GroupPolicy-ClientExtensions-Package~.mum") DO (DISM /Online /NoRestart /Add-Package:"%F") Now we can launch Group Policy Editor by typing in the following gpedit.msc More details can be found at the following sites https://linustechtips.com/topic/1482968-windows-doesn%E2%80%99t-suck-microsoft-just-wants-you-to-think-so%E2%80%A6/#comment-15762053 https://www.itechtics.com/enable-gpedit-msc-windows-11/ # Disable Telemetry on Windows 10/11 If you are on Windows Home, you will need to enable the Group Policy Editor Enable Group Policy Editor on Windows 10 or 11 Open up gpedit.msc Computer Configuration -> Windows Components -> Data Collection and Preview Builds -> Allow Diagnostic Data Double click on Allow Diagnostic Data to bring up the window. Set it to Enabled, and then under Options, set Diagnostic data off Hit Apply and your good to go. # SSH “Permission denied” to local IP address – Windows This could potentially be a number of issues. But Windows can show this error if you have any VPNs running that have the “Kill Switch” enabled. You can either add the local range as an exception, or not enable the Kill Switch. # SCP can’t copy file “.//cap.pcap: Broken pipe” This error can show up on Windows if you have already copied a file with the same name. To fix the issue, just copy it to a file with a different name. Change cap.pcap to cap1.pcap and so forth as needed. scp user@192.168.1.20:/tmp/tcp.pcap .\cap1.pcap # Import cert.pem on Windows First thing you will need You will need the .pem certification. We’ll be using the certutil.exe utility to import the certificate. .\certutil.exe -addstore -f "Root" 'C:\Users\path\to\cert.pem' Example output for importing a self signed UniFi certificate. PS C:\Windows\system32> certutil.exe -addstore -f "Root" 'C:\Users\path\to\cert.pem' Root "Trusted Root Certification Authorities" Signature matches Public Key PS C:\Windows\system32> # How to setup a Chia Harvester on Windows The following instructions are for setting up a Windows computer as a Chia Harvester. # Prerequisites Before we get started you will need the following 1. Have a current Chia farmer 2. You will need the \ca folder from your main farmer. The ca folder should be located in %homepath%\.chia\mainnet\config\ssl\ You should be able to copy and paste the above in File Explorer. Copy the ca folder to a USB drive or share it via a network share. # Setting up harvester 1. Copy the ca folder to an easily accessible place on your harvester 3. Close chia 4. Open PowerShell and paste the cfollowing commands in. Change the sections in bold to reflect your settings/options cd \$env:APPDATA..\local\chia-blockchain\app-1.2.5\resources\app.asar.unpacked\daemon .\chia.exe init -c D:\ca\ .\chia.exe stop all .\chia.exe configure --set-farmer-peer 192.168.188.2:8447 .\chia.exe configure --enable-upnp false .\chia.exe start harvester -r Check the main Farmer to verify the Harvester connected. # Disable Telemetry for DotNet SDK First option is to open a Power Shell or Command Prompt and type, think it may need to be an admin prompt. set DOTNET_CLI_TELEMETRY_OPTOUT=1 As a secondary option you should also be able to do this from the GUI by doing the following. Search for Environment Variables Edit Environment Variables Create Variable named DOTNET_CLI_TELEMETRY_OPTOUT with a Variable value of 1 Save by Hitting OK and OK again.	2023-03-22 12:18:26	{"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": 0.18307757377624512, "perplexity": 11617.143496639623}, "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/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00584.warc.gz"}
http://ricerca.mat.uniroma3.it/ipparco/pagine/deposito/1967-1979/roma77.html	1978 G. Gallavotti, F. Guerra, S. Miracle-Sol\'e A comment to the talk by E. Seiler In {\it Mathematical problems in theoretical Physics}, ed. G. Dell'Antonio, S.Doplicher, G. Jona--Lasinio, Lecture notes in Physics, {\bf 80}, 436--438, 1978 Abstract A comment providing a proof that the $Z^2$ gauge theory had a phase transition area-perimeter for the Wilson's loop expectation	2019-04-25 15:45:44	{"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": 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": 0.9788331389427185, "perplexity": 7473.745813757919}, "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/1555578727587.83/warc/CC-MAIN-20190425154024-20190425180024-00471.warc.gz"}
https://wiki.math.uwaterloo.ca/statwiki/index.php?title=stat946w18/Implicit_Causal_Models_for_Genome-wide_Association_Studies&direction=next&oldid=34344	# stat946w18/Implicit Causal Models for Genome-wide Association Studies ## Introduction and Motivation There is progression in probabilistic models which could develop rich generative models. The models have been expanded with neural network, implicit densities, and with scalable algorithms to very large data for their Bayesian inference. However, most of the models are focus on capturing statistical relationships rather than causal relationships. Causal models give us a sense on how manipulate the generative process could change the final results. Genome-wide association studies (GWAS) are examples of causal relationship. Specifically, GWAS is about figuring out how genetic factors cause disease among humans. Here the genetic factors we are referring to is single nucleotide polymorphisms (SNPs), and getting a particular disease is treated as a trait, i.e., the outcome. In order to know about the reason of developing a disease and to cure it, the causation between SNPs and diseases is interested: first, predict which one or multiple SNPs cause the disease; second, target the selected SNPs to cure the disease. This paper dealt with two questions. The first one is how to build rich causal models with specific needs by GWAS. In general, probabilistic causal models involve a function $f$ and a noise $n$. For the working simplicity, we usually assume $f$ as a linear model with a Gaussian noise. However, proof has shown that in GWAS, it is necessary to accommodate non-linearity and interactions between multiple genes into the models. The second accomplishment of this paper is that it addressed the problem caused by latent confounders. Latent confounders are issues when we apply the causal models since we cannot observe them nor knowing the underlying structure. In this paper, they developed implicit causal models which can adjust for confounders. There has been growing works on causal models which focus on causal discovery and typically have strong assumptions such as Gaussian processes on noise variable or nonlinearities for the main function. ## Implicit Causal Models Implicit causal models are an extension of probabilistic causal models. Probabilistic causal models will be introduced first. ### Probabilistic Causal Models Probabilistic causal models have two parts: deterministic functions of noise and other variables. Consider a global variable $\beta$ and noise $\epsilon$, where Each $\beta$ and $x$ is a function of noise; $y$ is a function of noise and $x$ The target is the causal mechanism $f_y$ so that the causal effect $p(y\|do(X=x),\beta)$ can be calculated. $do(X=x)$ means that we specify a value of $X$ under the fixed structure $\beta$. By other paper’s work, it is assumed that $p(y\|do(x),\beta) = p(y\|x, \beta)$. An example of probabilistic causal models is additive noise model. $f(.)$ is usually a linear function or spline functions for nonlinearities. $\epsilon$ is assumed to be standard normal, as well as $y$. Thus the posterior $p(\theta \| x, y, \beta)$ can be represented as where $p(\theta)$ is the prior which is known. Then, variational inference or MCMC can be applied to calculate the posterior distribution. ### Implicit Causal Models The difference between implicit causal models and probabilistic causal models is the noise variable. Instead of an additive noise term, implicit causal models directly take noise $\epsilon$ into a neural network and output $x$. The causal diagram has changed to: They used fully connected neural network with a fair amount of hidden units to approximate each causal mechanism. Below is the formal description: ## Implicit Causal Models with Latent Confounders Previously, they assumed the global structure is observed. Next, the unobserved scenario is being considered. ### Causal Inference with a Latent Confounder Same as before, the interest is the causal effect $p(y\|do(x_m), x_{-m})$. Here, the SNPs other than $x_m$ is also under consideration. However, it is confounded by the unobserved confounder $z_n$. As a result, the standard inference method cannot be used in this case. The paper proposed a new method which include the latent confounders. For each subject $n=1,…,N$ and each SNP $m=1,…,M$, The mechanism for latent confounder $z_n$ is assumed to be known. SNPs depend on the confounders and the trait depends on all the SNPs and the confounders as well. The posterior of $\theta$ is needed to be calculate in order to estimate the mechanism $g_y$ as well as the causal effect $p(y\|do(x_m), x_{-m})$, so that it can be explained how changes to each SNP $X_m$ cause changes to the trait $Y$. Note that the latent structure $p(z\|x, y)$ is assumed known. ### Implicit Causal Model with a Latent Confounder This section is the algorithm and functions to implementing an implicit causal model for GWAS. #### Generative Process of Confounders $z_n$. The distribution of confounders is set as standard normal. $z_n \in R^K$ , where $K$ is the dimension of $z_n$ and $K$ should make the latent space as close as possible to the true population structural. #### Generative Process of SNPs $x_{nm}$. Given SNP is coded for, The authors defined a $Binomial(2,\pi_{nm})$ distribution on $x_{nm}$. And used logistic factor analysis to design the SNP matrix. A SNP matrix looks like this: Since logistic factor analysis makes strong assumptions, this paper suggests using a neural network to relax these assumptions, This renders the outputs to be a full $N*M$ matrix due the the variables $w_m$, which act as principal component in PCA. #### Generative Process of Traits $y_n$. Previously, each trait is modeled by a linear regression, This also has very strong assumptions on SNPs, interactions, and additive noise. It can also be replaced by a neural network which only outputs a scalar, ## Likelihood-free Variational Inference Calculating the posterior of $\theta$ is the key of applying the implicit causal model with latent confounders. could be reduces to However, with implicit models, integrating over a nonlinear function could be suffered. The authors applied likelihood-free variational inference (LFVI). LFVI proposes a family of distribution over the latent variables. Here the variables $w_m$ and $z_n$ are all assumed to be Normal, ## Empirical Study The authors performed simulation on 100,000 SNPs, 940 to 5,000 individuals, and across 100 replications of 11 settings. Four methods were compared: • implicit causal model (ICM); • PCA with linear regression (PCA); • a linear mixed model (LMM); • logistic factor analysis with inverse regression (GCAT). The feedforward neural networks for traits and SNPs are fully connected with two hidden layers using ReLU activation function, and batch normalization. ### Simulation Study Based on real genomic data, a true model is applied to generate the SNPs and traits for each configuration. There are four datasets used in this simulation study: 1. HapMap [Balding-Nichols model] 2. 1000 Genomes Project (TGP) [PCA] • Human Genome Diversity project (HGDP) [PCA] • HGDP [Pritchard-Stephens-Donelly model] 3. A latent spatial position of individuals for population structure [spatial] The table shows the prediction accuracy. The accuracy is calculated by the rate of the number of true positives divide the number of true positives plus false positives. True positives measure the proportion of positives that are correctly identified as such (e.g. the percentage of SNPs which are correctly identified as having the causal relation with the trait). In contrast, false positives state the SNPs has the causal relation with the trait when they don’t. The closer the rate to 1, the better the model is since false positives are considered as the wrong prediction. The result represented above shows that the implicit causal model has the best performance among these four models in every situation. Especially, other models tend to do poorly on PSD and Spatial when $a$ is small, but the ICM achieved a significantly high rate. The only comparable method to ICM is GCAT, when applying to simpler configurations. ### Real-data Analysis They also applied ICM to a real-world GWAS of Northern Finland Birth Cohorts which contain 324,160 SNPs and 5,027 individuals. Ten implicit causal models were fitted and the 2 neural networks both with two hidden layers were used for SNP and trait. The numbers in the above table are the number of significant loci for each of the 10 traits. The number for other methods, such as GCAT, LMM, PCA, and "uncorrected" are obtained from other papers. By comparison, the ICM reached the level of the best previous model for each trait. ## Conclusion This paper introduced implicit causal models in order to account for nonlinear complex causal relationships, and applied the method to GWAS. It can not only capture important interactions between genes within an individual and among population level, but also can adjust for latent confounders by taking account of the latent variables into the model. By the simulation study, the authors proved that the implicit causal model could beat other methods by 15-45.3% on a variety of datasets with variations on parameters. The authors also believed this GWAS application is only a start of the usage of implicit causal models. It might also be used in physics or economics. ## Critique I think this paper is an interesting and novel work. The main contribution of this paper is to connect the statistical genetics and the machine learning methodology. The method is technically sound and does indeed generalize techniques currently used in statistical genetics. The neural network used in this paper is a very simple feedforward 2 hidden layers neural network, but the idea of where to use the neural network is crucial and might be significant in GWAS. It has limitations as well. The empirical example in this paper is too easy, and far away from the realistic situation. Despite the simulation study showed some competing results, the Northern Finland Birth Cohort Data application did not demonstrate the advantage of using implicit causal model whether are better than the previous methods, such as GCAT or LMM. Another limitation is about linkage disequilibrium as the authors stated as well. SNPs are not completely independent of each other; usually they have correlations when the alleles at close locus. They did not consider this complex case, rather they only considered the simplest case where they assumed all the SNPs are independent. Furthermore, one SNP maybe does not have enough power to explain the causal relationship. Recent papers indicate that causation to a trait may involve multiple SNPs. This could be a future work as well. ## References Tran D, Blei D M. Implicit Causal Models for Genome-wide Association Studies[J]. arXiv preprint arXiv:1710.10742, 2017. Patrik O Hoyer, Dominik Janzing, Joris M Mooij, Jonas Peters, and Prof Bernhard Schölkopf. Non- linear causal discovery with additive noise models. In Neural Information Processing Systems, 2009. Alkes L Price, Nick J Patterson, Robert M Plenge, Michael E Weinblatt, Nancy A Shadick, and David Reich. Principal components analysis corrects for stratification in genome-wide association studies. Nature Genetics, 38(8):904–909, 2006. Minsun Song, Wei Hao, and John D Storey. Testing for genetic associations in arbitrarily structured populations. Nature, 47(5):550–554, 2015. Dustin Tran, Rajesh Ranganath, and David M Blei. Hierarchical implicit models and likelihood-free variational inference. 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https://www.meritnation.com/cbse-class-12-science/math/math-part-i-ncert-solutions/application-of-derivatives/ncert-solutions/46_1_1373_235_197_7086	NCERT Solutions for Class 12 Science Math Chapter 6 Application Of Derivatives are provided here with simple step-by-step explanations. These solutions for Application Of Derivatives are extremely popular among Class 12 Science students for Math Application Of Derivatives Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate. #### Question 1: Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm The area of a circle (A)with radius (r) is given by, Now, the rate of change of the area with respect to its radius is given by, 1. When r = 3 cm, Hence, the area of the circle is changing at the rate of 6π cm when its radius is 3 cm. 1. When r = 4 cm, Hence, the area of the circle is changing at the rate of 8π cm when its radius is 4 cm. #### Question 2: The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm? Let x be the length of a side, V be the volume, and s be the surface area of the cube. Then, V = x3 and S = 6x2 where x is a function of time t. It is given that. Then, by using the chain rule, we have: Thus, when x = 12 cm, Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of cm2/s. #### Question 3: The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm. The area of a circle (A) with radius (r) is given by, Now, the rate of change of area (A) with respect to time (t) is given by, It is given that, Thus, when r = 10 cm, Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm2/s. #### Question 4: An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long? Let x be the length of a side and V be the volume of the cube. Then, V = x3. (By chain rule) It is given that, Thus, when x = 10 cm, Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long. #### Question 5: A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing? The area of a circle (A) with radius (r) is given by. Therefore, the rate of change of area (A) with respect to time (t) is given by, [By chain rule] It is given that. Thus, when r = 8 cm, Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm2/s. #### Question 6: The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? The circumference of a circle (C) with radius (r) is given by C = 2πr. Therefore, the rate of change of circumference (C) with respect to time (t) is given by, (By chain rule) It is given that. Hence, the rate of increase of the circumference #### Question 7: The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle. Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have: and (a) The perimeter (P) of a rectangle is given by, P = 2(x + y) Hence, the perimeter is decreasing at the rate of 2 cm/min. (b) The area (A) of a rectangle is given by, A = x y When x = 8 cm and y = 6 cm, Hence, the area of the rectangle is increasing at the rate of 2 cm2/min. #### Question 8: A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. The volume of a sphere (V) with radius (r) is given by, Rate of change of volume (V) with respect to time (t) is given by, [By chain rule] It is given that. Therefore, when radius = 15 cm, Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is #### Question 9: A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm. The volume of a sphere (V) with radius (r) is given by. Rate of change of volume (V) with respect to its radius (r) is given by, Therefore, when radius = 10 cm, Hence, the volume of the balloon is increasing at the rate of 400π cm2. #### Question 10: A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall? Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall. Then, by Pythagoras theorem, we have: x2 + y2 = 25 [Length of the ladder = 5 m] Then, the rate of change of height (y) with respect to time (t) is given by, It is given that. Now, when x = 4 m, we have: Hence, the height of the ladder on the wall is decreasing at the rate of. #### Question 11: A particle moves along the curve. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate. The equation of the curve is given as: The rate of change of the position of the particle with respect to time (t) is given by, When the y-coordinate of the particle changes 8 times as fast as the x-coordinate i.e.,, we have: When. When. Hence, the points required on the curve are (4, 11) and #### Question 12: The radius of an air bubble is increasing at the rate of cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm? The air bubble is in the shape of a sphere. Now, the volume of an air bubble (V) with radius (r) is given by, The rate of change of volume (V) with respect to time (t) is given by, It is given that. Therefore, when r = 1 cm, Hence, the rate at which the volume of the bubble increases is 2π cm3/s. #### Question 13: A balloon, which always remains spherical, has a variable diameter Find the rate of change of its volume with respect to x. The volume of a sphere (V) with radius (r) is given by, It is given that: Diameter Hence, the rate of change of volume with respect to x is as #### Question 14: Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm? The volume of a cone (V) with radius (r) and height (h) is given by, It is given that, The rate of change of volume with respect to time (t) is given by, [By chain rule] It is also given that. Therefore, when h = 4 cm, we have: Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of. #### Question 15: The total cost C (x) in Rupees associated with the production of x units of an item is given by Find the marginal cost when 17 units are produced. Marginal cost is the rate of change of total cost with respect to output. Marginal cost (MC) When x = 17, MC = 0.021 (172) − 0.006 (17) + 15 = 0.021(289) − 0.006(17) + 15 = 6.069 − 0.102 + 15 = 20.967 Hence, when 17 units are produced, the marginal cost is Rs. 20.967. #### Question 16: The total revenue in Rupees received from the sale of x units of a product is given by Find the marginal revenue when x = 7. Marginal revenue is the rate of change of total revenue with respect to the number of units sold. Marginal Revenue (MR) = 13(2x) + 26 = 26x + 26 When x = 7, MR = 26(7) + 26 = 182 + 26 = 208 Hence, the required marginal revenue is Rs 208. #### Question 17: The rate of change of the area of a circle with respect to its radius r at r = 6 cm is (A) 10π (B) 12π (C) 8π (D) 11π The area of a circle (A) with radius (r) is given by, Therefore, the rate of change of the area with respect to its radius r is . When r = 6 cm, Hence, the required rate of change of the area of a circle is 12π cm2/s. The correct answer is B. #### Question 18: The total revenue in Rupees received from the sale of x units of a product is given by . The marginal revenue, when is (A) 116 (B) 96 (C) 90 (D) 126 Marginal revenue is the rate of change of total revenue with respect to the number of units sold. Marginal Revenue (MR)= 3(2x) + 36 = 6x + 36 When x = 15, MR = 6(15) + 36 = 90 + 36 = 126 Hence, the required marginal revenue is Rs 126. The correct answer is D. #### Question 1: Show that the function given by f(x) = 3x + 17 is strictly increasing on R. Letbe any two numbers in R. Then, we have: Hence, f is strictly increasing on R. #### Question 2: Show that the function given by f(x) = e2x is strictly increasing on R. Letbe any two numbers in R. Then, we have: Hence, f is strictly increasing on R. #### Question 3: Show that the function given by f(x) = sin x is (a) strictly increasing in (b) strictly decreasing in (c) neither increasing nor decreasing in (0, π) The given function is f(x) = sin x. (a) Since for eachwe have. Hence, f is strictly increasing in. (b) Since for each, we have. Hence, f is strictly decreasing in. (c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π). #### Question 4: Find the intervals in which the function f given by f(x) = 2x2 − 3x is (a) strictly increasing (b) strictly decreasing The given function is f(x) = 2x2 − 3x. Now, the pointdivides the real line into two disjoint intervals i.e., and In interval Hence, the given function (f) is strictly decreasing in interval. In interval Hence, the given function (f) is strictly increasing in interval. #### Question 5: Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is (a) strictly increasing (b) strictly decreasing The given function is f(x) = 2x3 − 3x2 − 36x + 7. x = − 2, 3 The points x = −2 and x = 3 divide the real line into three disjoint intervals i.e., In intervalsis positive while in interval (−2, 3), is negative. Hence, the given function (f) is strictly increasing in intervals , while function (f) is strictly decreasing in interval (−2, 3). #### Question 6: Find the intervals in which the following functions are strictly increasing or decreasing: (a) x2 + 2x − 5 (b) 10 − 6x − 2x2 (c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9xx2 (e) (x + 1)3 (x − 3)3 (a) We have, Now, x = −1 Point x = −1 divides the real line into two disjoint intervals i.e., In interval f is strictly decreasing in interval Thus, f is strictly decreasing for x < −1. In interval f is strictly increasing in interval Thus, f is strictly increasing for x > −1. (b) We have, f(x) = 10 − 6x − 2x2 The pointdivides the real line into two disjoint intervals i.e., In interval i.e., when,$f\text{'}\left(x\right)=-6-4x>0$. f is strictly increasing for . In interval i.e., when, f is strictly decreasing for . (c) We have, f(x) = −2x3 − 9x2 − 12x + 1 Points x = −1 and x = −2 divide the real line into three disjoint intervals i.e., In intervals i.e., when x < −2 and x > −1, . f is strictly decreasing for x < −2 and x > −1. Now, in interval (−2, −1) i.e., when −2 < x < −1, . f is strictly increasing for . (d) We have, The pointdivides the real line into two disjoint intervals i.e., . In interval i.e., for, . f is strictly increasing for. In interval i.e., for, f is strictly decreasing for. (e) We have, f(x) = (x + 1)3 (x − 3)3 The points x = −1, x = 1, and x = 3 divide the real line into four disjoint intervals i.e.,, (−1, 1), (1, 3), and. In intervalsand (−1, 1), . f is strictly decreasing in intervalsand (−1, 1). In intervals (1, 3) and, . f is strictly increasing in intervals (1, 3) and. #### Question 7: Show that, is an increasing function of x throughout its domain. We have, $\frac{dy}{dx}=\frac{1}{1+x}-\frac{\left(2+x\right)\left(2\right)-2x\left(1\right)}{\left(2+x{\right)}^{2}}=\frac{1}{1+x}-\frac{4}{\left(2+x{\right)}^{2}}=\frac{{x}^{2}}{\left(1+x\right)\left(2+x{\right)}^{2}}$ Now, $\frac{dy}{dx}=0$ Since x > −1, point x = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x < 0 and x > 0. When −1 < x < 0, we have: $x<0⇒{x}^{2}>0\phantom{\rule{0ex}{0ex}}x>-1⇒\left(2+x\right)>0⇒\left(2+{x}^{2}\right)>0$ ${y}^{\text{'}}=\frac{{x}^{2}}{\left(1+x\right)\left(2+x{\right)}^{2}}>0$ Also, when x > 0: ${y}^{\text{'}}=\frac{{x}^{2}}{\left(1+x\right)\left(2+x{\right)}^{2}}>0$ Hence, function f is increasing throughout this domain. #### Question 8: Find the values of x for whichis an increasing function. We have, The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e., In intervals, . y is strictly decreasing in intervals . However, in intervals (0, 1) and (2, ∞), y is strictly increasing in intervals (0, 1) and (2, ∞). y is strictly increasing for 0 < x < 1 and x > 2. #### Question 9: Prove that is an increasing function of θ in. We have, Since cos θ ≠ 4, cos θ = 0. Now, In interval, we have cos θ > 0. Also, 4 > cos θ ⇒ 4 − cos θ > 0. Therefore, y is strictly increasing in interval. Also, the given function is continuous at Hence, y is increasing in interval. #### Question 10: Prove that the logarithmic function is strictly increasing on (0, ∞). It is clear that for x > 0, Hence, f(x) = log x is strictly increasing in interval (0, ∞). #### Question 11: Prove that the function f given by f(x) = x2x + 1 is neither strictly increasing nor strictly decreasing on (−1, 1). The given function is f(x) = x2x + 1. The pointdivides the interval (−1, 1) into two disjoint intervals i.e., Now, in interval Therefore, f is strictly decreasing in interval. However, in interval Therefore, f is strictly increasing in interval. Hence, f is neither strictly increasing nor decreasing in interval (−1, 1). #### Question 12: Which of the following functions are strictly decreasing on? (A) cos x (B) cos 2x (C) cos 3x (D) tan x (A) Let In interval is strictly decreasing in interval. (B) Let is strictly decreasing in interval. (C) Let The point divides the intervalinto two disjoint intervals i.e., 0 f3 is strictly increasing in interval. Hence, f3 is neither increasing nor decreasing in interval. (D) Let In interval f4 is strictly increasing in interval Therefore, functions cos x and cos 2x are strictly decreasing in Hence, the correct answers are A and B. #### Question 13: On which of the following intervals is the function f given by strictly decreasing? (A) (B) (C) (D) None of these We have, In interval Thus, function f is strictly increasing in interval (0, 1). In interval Thus, function f is strictly increasing in interval. f is strictly increasing in interval. Hence, function f is strictly decreasing in none of the intervals. The correct answer is D. #### Question 14: Find the least value of a such that the function f given is strictly increasing on [1, 2]. We have, Now, function f is increasing on [1,2]. #### Question 15: Let I be any interval disjoint from (−1, 1). Prove that the function f given by is strictly increasing on I. We have, The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e., . In interval (−1, 1), it is observed that: f is strictly decreasing on . In intervals, it is observed that: f is strictly increasing on. Hence, function f is strictly increasing in interval I disjoint from (−1, 1). Hence, the given result is proved. #### Question 16: Prove that the function f given by f(x) = log sin x is strictly increasing on and strictly decreasing on We have, In interval f is strictly increasing in. In interval f is strictly decreasing in #### Question 17: Prove that the function f given by f(x) = log cos x is strictly decreasing on and strictly increasing on We have, In interval f is strictly decreasing on. In interval f is strictly increasing on. #### Question 18: Prove that the function given by is increasing in R. We have, For any xR, (x − 1)2 > 0. Thus, is always positive in R. Hence, the given function (f) is increasing in R. #### Question 19: The interval in which is increasing is (A) (B) (−2, 0) (C) (D) (0, 2) We have, The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e., In intervalsis always positive. f is decreasing on In interval (0, 2), f is strictly increasing on (0, 2). Hence, f is strictly increasing in interval (0, 2). The correct answer is D. #### Question 1: Find the slope of the tangent to the curve y = 3x4 − 4x at x = 4. The given curve is y = 3x4 − 4x. Then, the slope of the tangent to the given curve at x = 4 is given by, #### Question 2: Find the slope of the tangent to the curve, x ≠ 2 at x = 10. The given curve is. Thus, the slope of the tangent at x = 10 is given by, Hence, the slope of the tangent at x = 10 is #### Question 3: Find the slope of the tangent to curve y = x3x + 1 at the point whose x-coordinate is 2. The given curve is. The slope of the tangent to a curve at (x0, y0) is. It is given that x0 = 2. Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by, #### Question 4: Find the slope of the tangent to the curve y = x3 − 3x + 2 at the point whose x-coordinate is 3. The given curve is. The slope of the tangent to a curve at (x0, y0) is. Hence, the slope of the tangent at the point where the x-coordinate is 3 is given by, #### Question 5: Find the slope of the normal to the curve x = acos3θ, y = asin3θ at. It is given that x = acos3θ and y = asin3θ. Therefore, the slope of the tangent at is given by, Hence, the slope of the normal at #### Question 6: Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos2θ at . It is given that x = 1 − a sin θ and y = b cos2θ. Therefore, the slope of the tangent at is given by, Hence, the slope of the normal at #### Question 7: Find points at which the tangent to the curve y = x3 − 3x2 − 9x + 7 is parallel to the x-axis. The equation of the given curve is Now, the tangent is parallel to the x-axis if the slope of the tangent is zero. When x = 3, y = (3)3 − 3 (3)2 − 9 (3) + 7 = 27 − 27 − 27 + 7 = −20. When x = −1, y = (−1)3 − 3 (−1)2 − 9 (−1) + 7 = −1 − 3 + 9 + 7 = 12. Hence, the points at which the tangent is parallel to the x-axis are (3, −20) and (−1, 12). #### Question 8: Find a point on the curve y = (x − 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4). If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord. The slope of the chord is Now, the slope of the tangent to the given curve at a point (x, y) is given by, Since the slope of the tangent = slope of the chord, we have: Hence, the required point is (3, 1). #### Question 9: Find the point on the curve y = x3 − 11x + 5 at which the tangent is y = x − 11. The equation of the given curve is y = x3 − 11x + 5. The equation of the tangent to the given curve is given as y = x − 11 (which is of the form y = mx + c). ∴Slope of the tangent = 1 Now, the slope of the tangent to the given curve at the point (x, y) is given by, Then, we have: When x = 2, y = (2)3 − 11 (2) + 5 = 8 − 22 + 5 = −9. When x = −2, y = (−2)3 − 11 (−2) + 5 = −8 + 22 + 5 = 19. Hence, the required points are (2, −9) and (−2, 19). But, both these points should satisfy the equation of the tangent as there would be point of contact between tangent and the curve. ∴ (2, −9) is the required point as (−2, 19) is not satisfying the given equation of tangent. #### Question 10: Find the equation of all lines having slope −1 that are tangents to the curve . The equation of the given curve is. The slope of the tangents to the given curve at any point (x, y) is given by, If the slope of the tangent is −1, then we have: When x = 0, y = −1 and when x = 2, y = 1. Thus, there are two tangents to the given curve having slope −1. These are passing through the points (0, −1) and (2, 1). ∴The equation of the tangent through (0, −1) is given by, ∴The equation of the tangent through (2, 1) is given by, y − 1 = −1 (x − 2) y − 1 = − x + 2 y + x − 3 = 0 Hence, the equations of the required lines are y + x + 1 = 0 and y + x − 3 = 0. #### Question 11: Find the equation of all lines having slope 2 which are tangents to the curve. The equation of the given curve is. The slope of the tangent to the given curve at any point (x, y) is given by, If the slope of the tangent is 2, then we have: Hence, there is no tangent to the given curve having slope 2. #### Question 12: Find the equations of all lines having slope 0 which are tangent to the curve . The equation of the given curve is. The slope of the tangent to the given curve at any point (x, y) is given by, If the slope of the tangent is 0, then we have: When x = 1, ∴The equation of the tangent throughis given by, Hence, the equation of the required line is #### Question 13: Find points on the curve at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis The equation of the given curve is. On differentiating both sides with respect to x, we have: (i) The tangent is parallel to the x-axis if the slope of the tangent is i.e., 0 which is possible if x = 0. Then, for x = 0 Hence, the points at which the tangents are parallel to the x-axis are (0, 4) and (0, − 4). (ii) The tangent is parallel to the y-axis if the slope of the normal is 0, which givesy = 0. Then, for y = 0. Hence, the points at which the tangents are parallel to the y-axis are (3, 0) and (− 3, 0). #### Question 14: Find the equations of the tangent and normal to the given curves at the indicated points: (i) y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5) (ii) y = x4 − 6x3 + 13x2 − 10x + 5 at (1, 3) (iii) y = x3 at (1, 1) (iv) y = x2 at (0, 0) (v) x = cos t, y = sin t at (i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5. On differentiating with respect to x, we get: Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as: y − 5 = − 10(x − 0) y − 5 = − 10x ⇒ 10x + y = 5 The slope of the normal at (0, 5) is Therefore, the equation of the normal at (0, 5) is given as: (ii) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5. On differentiating with respect to x, we get: Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as: The slope of the normal at (1, 3) is Therefore, the equation of the normal at (1, 3) is given as: (iii) The equation of the curve is y = x3. On differentiating with respect to x, we get: Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as: The slope of the normal at (1, 1) is Therefore, the equation of the normal at (1, 1) is given as: (iv) The equation of the curve is y = x2. On differentiating with respect to x, we get: Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as: y − 0 = 0 (x − 0) y = 0 The slope of the normal at (0, 0) is , which is not defined. Therefore, the equation of the normal at (x0, y0) = (0, 0) is given by (v) The equation of the curve is x = cos t, y = sin t. ∴The slope of the tangent atis −1. When Thus, the equation of the tangent to the given curve at is The slope of the normal atis Therefore, the equation of the normal to the given curve at is #### Question 15: Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is (a) parallel to the line 2xy + 9 = 0 (b) perpendicular to the line 5y − 15x = 13. The equation of the given curve is. On differentiating with respect to x, we get: (a) The equation of the line is 2xy + 9 = 0. 2xy + 9 = 0 ⇒ y = 2x + 9 This is of the form y = mx + c. ∴Slope of the line = 2 If a tangent is parallel to the line 2xy + 9 = 0, then the slope of the tangent is equal to the slope of the line. Therefore, we have: 2 = 2x − 2 Now, x = 2 y = 4 − 4 + 7 = 7 Thus, the equation of the tangent passing through (2, 7) is given by, Hence, the equation of the tangent line to the given curve (which is parallel to line 2xy + 9 = 0) is. (b) The equation of the line is 5y − 15x = 13. 5y − 15x = 13 ⇒ This is of the form y = mx + c. ∴Slope of the line = 3 If a tangent is perpendicular to the line 5y − 15x = 13, then the slope of the tangent is Thus, the equation of the tangent passing throughis given by, Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y − 15x = 13) is. #### Question 16: Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −2 are parallel. The equation of the given curve is y = 7x3 + 11. The slope of the tangent to a curve at (x0, y0) is. Therefore, the slope of the tangent at the point where x = 2 is given by, It is observed that the slopes of the tangents at the points where x = 2 and x = −2 are equal. Hence, the two tangents are parallel. #### Question 17: Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point. The equation of the given curve is y = x3. The slope of the tangent at the point (x, y) is given by, When the slope of the tangent is equal to the y-coordinate of the point, then y = 3x2. Also, we have y = x3. ∴3x2 = x3 x2 (x − 3) = 0 x = 0, x = 3 When x = 0, then y = 0 and when x = 3, then y = 3(3)2 = 27. Hence, the required points are (0, 0) and (3, 27). #### Question 18: For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin. The equation of the given curve is y = 4x3 − 2x5. Therefore, the slope of the tangent at a point (x, y) is 12x2 − 10x4. The equation of the tangent at (x, y) is given by, When the tangent passes through the origin (0, 0), then X = Y = 0. Therefore, equation (1) reduces to: Also, we have When x = 0, y = When x = 1, y = 4 (1)3 − 2 (1)5 = 2. When x = −1, y = 4 (−1)3 − 2 (−1)5 = −2. Hence, the required points are (0, 0), (1, 2), and (−1, −2). #### Question 19: Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis. The equation of the given curve is x2 + y2 − 2x − 3 = 0. On differentiating with respect to x, we have: Now, the tangents are parallel to the x-axis if the slope of the tangent is 0. But, x2 + y2 − 2x − 3 = 0 for x = 1. y2 = 4 ⇒ Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, −2). #### Question 20: Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3. The equation of the given curve is ay2 = x3. On differentiating with respect to x, we have: The slope of a tangent to the curve at (x0, y0) is. The slope of the tangent to the given curve at (am2, am3) is ∴ Slope of normal at (am2, am3) = Hence, the equation of the normal at (am2, am3) is given by, yam3 = #### Question 21: Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0. The equation of the given curve is y = x3 + 2x + 6. The slope of the tangent to the given curve at any point (x, y) is given by, ∴ Slope of the normal to the given curve at any point (x, y) = The equation of the given line is x + 14y + 4 = 0. x + 14y + 4 = 0 ⇒ (which is of the form y = mx + c) ∴Slope of the given line = If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line. When x = 2, y = 8 + 4 + 6 = 18. When x = −2, y = − 8 − 4 + 6 = −6. Therefore, there are two normals to the given curve with slopeand passing through the points (2, 18) and (−2, −6). Thus, the equation of the normal through (2, 18) is given by, And, the equation of the normal through (−2, −6) is given by, Hence, the equations of the normals to the given curve (which are parallel to the given line) are #### Question 22: Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2, 2at). The equation of the given parabola is y2 = 4ax. On differentiating y2 = 4ax with respect to x, we have: ∴The slope of the tangent atis Then, the equation of the tangent atis given by, y − 2at = Now, the slope of the normal atis given by, Thus, the equation of the normal at (at2, 2at) is given as: #### Question 23: Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.] The equations of the given curves are given as Putting x = y2 in xy = k, we get: Thus, the point of intersection of the given curves is. Differentiating x = y2 with respect to x, we have: Therefore, the slope of the tangent to the curve x = y2 atis On differentiating xy = k with respect to x, we have: ∴ Slope of the tangent to the curve xy = k atis given by, We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at are perpendicular to each other. This implies that we should have the product of the tangents as − 1. Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at is −1. Hence, the given two curves cut at right angels if 8k2 = 1. #### Question 24: Find the equations of the tangent and normal to the hyperbola at the point. Differentiatingwith respect to x, we have: Therefore, the slope of the tangent atis . Then, the equation of the tangent atis given by, Now, the slope of the normal atis given by, Hence, the equation of the normal atis given by, #### Question 25: Find the equation of the tangent to the curve which is parallel to the line 4x − 2y + 5 = 0. The equation of the given curve is The slope of the tangent to the given curve at any point (x, y) is given by, The equation of the given line is 4x − 2y + 5 = 0. 4x − 2y + 5 = 0 ⇒ (which is of the form ∴Slope of the line = 2 Now, the tangent to the given curve is parallel to the line 4x − 2y − 5 = 0 if the slope of the tangent is equal to the slope of the line. ∴Equation of the tangent passing through the point is given by, Hence, the equation of the required tangent is. #### Question 26: The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is (A) 3 (B) (C) −3 (D) The equation of the given curve is. Slope of the tangent to the given curve at x = 0 is given by, Hence, the slope of the normal to the given curve at x = 0 is The correct answer is D. #### Question 27: The line y = x + 1 is a tangent to the curve y2 = 4x at the point (A) (1, 2) (B) (2, 1) (C) (1, −2) (D) (−1, 2) The equation of the given curve is. Differentiating with respect to x, we have: Therefore, the slope of the tangent to the given curve at any point (x, y) is given by, The given line is y = x + 1 (which is of the form y = mx + c) ∴ Slope of the line = 1 The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve. Thus, we must have: Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2). The correct answer is A. #### Question 1: 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) (i) Consider. Let x = 25 and Δx = 0.3. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 0.03 + 5 = 5.03. (ii) Consider. Let x = 49 and Δx = 0.5. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 7 + 0.035 = 7.035. (iii) Consider. Let x = 1 and Δx = − 0.4. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 1 + (−0.2) = 1 − 0.2 = 0.8. (iv) Consider. Let x = 0.008 and Δx = 0.001. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 0.2 + 0.008 = 0.208. (v) Consider. Let x = 1 and Δx = −0.001. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 1 + (−0.0001) = 0.9999. (vi) Consider. Let x = 16 and Δx = −1. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 2 + (−0.03125) = 1.96875. (vii) Consider. Let x = 27 and Δx = −1. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 3 + (−0.0370) = 2.9629. (viii) Consider. Let x = 256 and Δx = −1. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 4 + (−0.0039) = 3.9961. (ix) Consider. Let x = 81 and Δx = 1. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 3 + 0.009 = 3.009. (x) Consider. Let x = 400 and Δx = 1. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 20 + 0.025 = 20.025. (xi) Consider. Let x = 0.0036 and Δx = 0.0001. Then, Now, dy is approximately equal to Δy and is given by, Thus, the approximate value ofis 0.06 + 0.00083 = 0.06083. (xii) Consider. Let x = 27 and Δx = −0.43. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 3 + (−0.015) = 2.984. (xiii) Consider. Let x = 81 and Δx = 0.5. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 3 + 0.0046 = 3.0046. (xiv) Consider. Let x = 4 and Δx = − 0.032. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 8 + (−0.096) = 7.904. (xv) Consider. Let x = 32 and Δx = 0.15. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 2 + 0.00187 = 2.00187. #### Question 2: Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2 Let x = 2 and Δx = 0.01. Then, we have: f(2.01) = f(x + Δx) = 4(x + Δx)2 + 5(x + Δx) + 2 Now, Δy = f(x + Δx) − f(x) f(x + Δx) = f(x) + Δy Hence, the approximate value of f (2.01) is 28.21. #### Question 3: Find the approximate value of f (5.001), where f (x) = x3 − 7x2 + 15. Let x = 5 and Δx = 0.001. Then, we have: Hence, the approximate value of f (5.001) is −34.995. #### Question 4: Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%. The volume of a cube (V) of side x is given by V = x3. Hence, the approximate change in the volume of the cube is 0.03x3 m3. #### Question 5: Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1% The surface area of a cube (S) of side x is given by S = 6x2. Hence, the approximate change in the surface area of the cube is 0.12x2 m2. #### Question 6: If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume. Let r be the radius of the sphere and Δr be the error in measuring the radius. Then, r = 7 m and Δr = 0.02 m Now, the volume V of the sphere is given by, Hence, the approximate error in calculating the volume is 3.92 π m3. #### Question 7: If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating in surface area. Let r be the radius of the sphere and Δr be the error in measuring the radius. Then, r = 9 m and Δr = 0.03 m Now, the surface area of the sphere (S) is given by, S = 4πr2 Hence, the approximate error in calculating the surface area is 2.16π m2. #### Question 8: If f (x) = 3x2 + 15x + 5, then the approximate value of f (3.02) is A. 47.66 B. 57.66 C. 67.66 D. 77.66 Let x = 3 and Δx = 0.02. Then, we have: Hence, the approximate value of f(3.02) is 77.66. The correct answer is D. #### Question 9: The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is A. 0.06 x3 m3 B. 0.6 x3 m3 C. 0.09 x3 m3 D. 0.9 x3 m3 The volume of a cube (V) of side x is given by V = x3. Hence, the approximate change in the volume of the cube is 0.09x3 m3. The correct answer is C. #### Question 1: Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = (2x − 1)2 + 3 (ii) f(x) = 9x2 + 12x + 2 (iii) f(x) = −(x − 1)2 + 10 (iv) g(x) = x3 + 1 (i) The given function is f(x) = (2x − 1)2 + 3. It can be observed that (2x − 1)2 ≥ 0 for every xR. Therefore, f(x) = (2x − 1)2 + 3 ≥ 3 for every xR. The minimum value of f is attained when 2x − 1 = 0. 2x − 1 = 0 ⇒ ∴Minimum value of f == 3 Hence, function f does not have a maximum value. (ii) The given function is f(x) = 9x2 + 12x + 2 = (3x + 2)2 − 2. It can be observed that (3x + 2)2 ≥ 0 for every xR. Therefore, f(x) = (3x + 2)2 − 2 ≥ −2 for every xR. The minimum value of f is attained when 3x + 2 = 0. 3x + 2 = 0 ⇒ ∴Minimum value of f = Hence, function f does not have a maximum value. (iii) The given function is f(x) = − (x − 1)2 + 10. It can be observed that (x − 1)2 ≥ 0 for every xR. Therefore, f(x) = − (x − 1)2 + 10 ≤ 10 for every xR. The maximum value of f is attained when (x − 1) = 0. (x − 1) = 0 ⇒ x = 1 ∴Maximum value of f = f(1) = − (1 − 1)2 + 10 = 10 Hence, function f does not have a minimum value. (iv) The given function is g(x) = x3 + 1. Hence, function g neither has a maximum value nor a minimum value. #### Question 2: Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = \|x + 2\| − 1 (ii) g(x) = − \|x + 1\| + 3 (iii) h(x) = sin(2x) + 5 (iv) f(x) = \|sin 4x + 3\| (v) h(x) = x + 1, x (−1, 1) (i) f(x) = We know that for every xR. Therefore, f(x) = for every xR. The minimum value of f is attained when. ∴Minimum value of f = f(−2) = Hence, function f does not have a maximum value. (ii) g(x) = We know that for every xR. Therefore, g(x) = for every xR. The maximum value of g is attained when. ∴Maximum value of g = g(−1) = Hence, function g does not have a minimum value. (iii) h(x) = sin2x + 5 We know that − 1 ≤ sin 2x ≤ 1. ⇒ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 ⇒ 4 ≤ sin 2x + 5 ≤ 6 Hence, the maximum and minimum values of h are 6 and 4 respectively. (iv) f(x) = We know that −1 ≤ sin 4x ≤ 1. ⇒ 2 ≤ sin 4x + 3 ≤ 4 ⇒ 2 ≤≤ 4 Hence, the maximum and minimum values of f are 4 and 2 respectively. (v) h(x) = x + 1, x ∈ (−1, 1) Here, if a point x0 is closest to −1, then we find $\frac{{x}_{0}}{2}+1>{x}_{0}+1$ for all x0 (−1, 1). Also, if x1 is closest to 1, then for all x1 (−1, 1). Hence, function h(x) has neither maximum nor minimum value in (−1, 1). #### Question 3: Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (i). f(x) = x2 (ii). g(x) = x3 − 3x (iii). h(x) = sinx + cosx, 0 < (iv). f(x) = sinx − cos x, 0 < x < 2π (v). f(x) = x3 − 6x2 + 9x + 15 (vi). (vii). (viii). (i) f(x) = x2 Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f. We have, which is positive. Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = 0. (ii) g(x) = x3 − 3x g'(x) = 3x2 − 3 Now, g'(x) = 0 ⇒ 3x2 = 3 ⇒ x$±$1 g''(x) = 6x g''(1) = 6 > 0 g''(−1) = −6 < 0 By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1 is g(1) = 13 − 3 = 1 − 3 = −2. However, x = −1 is a point of local maxima and local maximum value of g at x = −1 is g(−1) = (−1)3 − 3 (− 1) = − 1 + 3 = 2. (iii) h(x) = sinx + cosx, 0 < x < Therefore, by second derivative test, is a point of local maxima and the local maximum value of h at is (iv) f(x) = sin x − cos x, 0 < x < 2π $f\text{'}\left(x\right)=0⇒\mathrm{cos}x=-\mathrm{sin}x⇒\mathrm{tan}x=-1⇒x=\frac{3\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\in \left(0,2\mathrm{\pi }\right)$ $f\text{'}\text{'}\left(x\right)=-\mathrm{sin}x+\mathrm{cos}x$ $f\text{'}\text{'}\left(\frac{3\mathrm{\pi }}{4}\right)=-\mathrm{sin}\frac{3\mathrm{\pi }}{4}+\mathrm{cos}\frac{3\mathrm{\pi }}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0$ $f\text{'}\text{'}\left(\frac{7\mathrm{\pi }}{4}\right)=-\mathrm{sin}\frac{7\mathrm{\pi }}{4}+\mathrm{cos}\frac{7\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0$ Therefore, by second derivative test, is a point of local maxima and the local maximum value of f at is However, is a point of local minima and the local minimum value of f at is . (v) f(x) = x3 − 6x2 + 9x + 15 Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is f(1) = 1 − 6 + 9 + 15 = 19. However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is f(3) = 27 − 54 + 27 + 15 = 15. (vi) Since x > 0, we take x = 2. Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is g(2) = (vii) Now, for values close to x = 0 and to the left of 0,Also, for values close to x = 0 and to the right of 0,. Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of. (viii) Therefore, by second derivative test,is a point of local maxima and the local maximum value of f at is #### Question 4: Prove that the following functions do not have maxima or minima: (i) f(x) = ex (ii) g(x) = logx (iii) h(x) = x3 + x2 + x + 1 1. We have, f(x) = ex Now, if. But, the exponential function can never assume 0 for any value of x. Therefore, there does not exist cR such that Hence, function f does not have maxima or minima. 1. We have, g(x) = log x Therefore, there does not exist cR such that. Hence, function g does not have maxima or minima. 1. We have, h(x) = x3 + x2 + x + 1 Now, h(x) = 0 ⇒ 3x2 + 2x + 1 = 0 ⇒ Therefore, there does not exist cR such that. Hence, function h does not have maxima or minima. #### Question 5: Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) (ii) (iii) (iv) (i) The given function is f(x) = x3. Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [−2, 2]. f(0) = 0 f(−2) = (−2)3 = −8 f(2) = (2)3 = 8 Hence, we can conclude that the absolute maximum value of f on [−2, 2] is 8 occurring at x = 2. Also, the absolute minimum value of f on [−2, 2] is −8 occurring at x = −2. (ii) The given function is f(x) = sin x + cos x. Then, we evaluate the value of f at critical point and at the end points of the interval [0, π]. Hence, we can conclude that the absolute maximum value of f on [0, π] is occurring atand the absolute minimum value of f on [0, π] is −1 occurring at x = π. (iii) The given function is Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval. Hence, we can conclude that the absolute maximum value of f onis 8 occurring at x = 4 and the absolute minimum value of f on is −10 occurring at x = −2. (iv) The given function is Now, 2(x − 1) = 0 ⇒ x = 1 Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval [−3, 1]. Hence, we can conclude that the absolute maximum value of f on [−3, 1] is 19 occurring at x = −3 and the minimum value of f on [−3, 1] is 3 occurring at x = 1. #### Question 6: Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 − 72x − 18x2 The profit function is given as p(x) = 41 − 72x − 18x2. $p\text{'}\left(x\right)=-72-36x$ $⇒x=-\frac{72}{36}=-2$ Also, $p\text{'}\text{'}\left(-2\right)=-36<0$ By second derivative test, $x=-2$ is the point of local maxima of p. $=41-72\left(-2\right)-18\left(-2{\right)}^{2}=41+144-72=113$ Hence, the maximum profit that the company can make is 113 units. The solution given in the book has some error. The solution is created according to the question given in the book. #### Question 7: Find both the maximum value and the minimum value of 3x4 − 8x3 + 12x2 − 48x + 25 on the interval [0, 3] Let f(x) = 3x4 − 8x3 + 12x2 − 48x + 25. Now,gives x = 2 or x2+ 2 = 0 for which there are no real roots. Therefore, we consider only x = 2 ∈[0, 3]. Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3]. Hence, we can conclude that the absolute maximum value of f on [0, 3] is 25 occurring at x = 0 and the absolute minimum value of f at [0, 3] is − 39 occurring at x = 2. #### Question 8: At what points in the interval [0, 2π], does the function sin 2x attain its maximum value? Let f(x) = sin 2x. Then, we evaluate the values of f at critical pointsand at the end points of the interval [0, 2π]. Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring atand. #### Question 9: What is the maximum value of the function sin x + cos x? Let f(x) = sin x + cos x. Now, will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive. Also, we know that sin x and cos x both are positive in the first quadrant. Then, will be negative when. Thus, we consider. ∴By second derivative test, f will be the maximum atand the maximum value of f is. #### Question 10: Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1]. Let f(x) = 2x3 − 24x + 107. We first consider the interval [1, 3]. Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3]. f(2) = 2(8) − 24(2) + 107 = 16 − 48 + 107 = 75 f(1) = 2(1) − 24(1) + 107 = 2 − 24 + 107 = 85 f(3) = 2(27) − 24(3) + 107 = 54 − 72 + 107 = 89 Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3. Next, we consider the interval [−3, −1]. Evaluate the value of f at the critical point x = −2 ∈ [−3, −1] and at the end points of the interval [1, 3]. f(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125 f(−1) = 2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129 f(−2) = 2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139 Hence, the absolute maximum value of f(x) in the interval [−3, −1] is 139 occurring at x = −2. #### Question 11: It is given that at x = 1, the function x4− 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a. Let f(x) = x4 − 62x2 + ax + 9. It is given that function f attains its maximum value on the interval [0, 2] at x = 1. Hence, the value of a is 120. #### Question 12: Find the maximum and minimum values of x + sin 2x on [0, 2π]. Let f(x) = x + sin 2x. Then, we evaluate the value of f at critical points and at the end points of the interval [0, 2π]. Hence, we can conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π occurring at x = 2π and the absolute minimum value of f(x) in the interval [0, 2π] is 0 occurring at x = 0. #### Question 13: Find two numbers whose sum is 24 and whose product is as large as possible. Let one number be x. Then, the other number is (24 − x). Let P(x) denote the product of the two numbers. Thus, we have: ∴By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12. #### Question 14: Find two positive numbers x and y such that x + y = 60 and xy3 is maximum. The two numbers are x and y such that x + y = 60. y = 60 − x Let f(x) = xy3. ∴By second derivative test, x = 15 is a point of local maxima of f. Thus, function xy3 is maximum when x = 15 and y = 60 − 15 = 45. Hence, the required numbers are 15 and 45. #### Question 15: Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum Let one number be x. Then, the other number is y = (35 − x). Let P(x) = x2y5. Then, we have: x = 0, x = 35, x = 10 When x = 35, and y = 35 − 35 = 0. This will make the product x2 y5 equal to 0. When x = 0, y = 35 − 0 = 35 and the product x2y2 will be 0. x = 0 and x = 35 cannot be the possible values of x. When x = 10, we have: ∴ By second derivative test, P(x) will be the maximum when x = 10 and y = 35 − 10 = 25. Hence, the required numbers are 10 and 25. #### Question 16: Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum. Let one number be x. Then, the other number is (16 − x). Let the sum of the cubes of these numbers be denoted by S(x). Then, Now, ∴ By second derivative test, x = 8 is the point of local minima of S. Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 − 8 = 8. #### Question 17: A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (18 − 2x) cm each and the height of the box is x cm. Therefore, the volume V(x) of the box is given by, V(x) = x(18 − 2x)2 x = 9 or x = 3 If x = 9, then the length and the breadth will become 0. x ≠ 9. x = 3. Now, By second derivative test, x = 3 is the point of maxima of V. Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible. #### Question 18: A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is 45 − 2x, and the breadth is 24 − 2x. Therefore, the volume V(x) of the box is given by, Now, x = 18 and x = 5 It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. Thus, x cannot be equal to 18. x = 5 Now, By second derivative test, x = 5 is the point of maxima. Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm. #### Question 19: Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. Let a rectangle of length l and breadth b be inscribed in the given circle of radius a. Then, the diagonal passes through the centre and is of length 2a cm. Now, by applying the Pythagoras theorem, we have: ∴Area of the rectangle, By the second derivative test, when, then the area of the rectangle is the maximum. Since, the rectangle is a square. Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area. #### Question 20: Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base. Let r and h be the radius and height of the cylinder respectively. Then, the surface area (S) of the cylinder is given by, Let V be the volume of the cylinder. Then, ∴ By second derivative test, the volume is the maximum when. Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter. #### Question 21: Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area? Let r and h be the radius and height of the cylinder respectively. Then, volume (V) of the cylinder is given by, Surface area (S) of the cylinder is given by, ∴By second derivative test, the surface area is the minimum when the radius of the cylinder is. Hence, the required dimensions of the can which has the minimum surface area is given by radius = and height = #### Question 22: A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? Let a piece of length l be cut from the given wire to make a square. Then, the other piece of wire to be made into a circle is of length (28 − l) m. Now, side of square =. Let r be the radius of the circle. Then, The combined areas of the square and the circle (A) is given by, Thus, when By second derivative test, the area (A) is the minimum when. Hence, the combined area is the minimum when the length of the wire in making the square iscm while the length of the wire in making the circle is. #### Question 23: Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is of the volume of the sphere. Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R. Let V be the volume of the cone. Then, Height of the cone is given by, h = R + AB ∴ By second derivative test, the volume of the cone is the maximum when #### Question 24: Show that the right circular cone of least curved surface and given volume has an altitude equal to time the radius of the base. Let r and h be the radius and the height (altitude) of the cone respectively. Then, the volume (V) of the cone is given as: $V=\frac{1}{3}\pi {r}^{2}h\mathit{⇒}h\mathit{=}\frac{\mathit{3}V}{\mathrm{\pi }{r}^{\mathit{2}}}$ The surface area (S) of the cone is given by, S = πrl (where l is the slant height) Thus, it can be easily verified that when ∴ By second derivative test, the surface area of the cone is the least when Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to times the radius of the base. #### Question 25: Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is. Let θ be the semi-vertical angle of the cone. It is clear that Let r, h, and l be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant. Now, r = l sin θ and h = l cos θ The volume (V) of the cone is given by, ∴By second derivative test, the volume (V) is the maximum when. Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is. #### Question 26: Show that semi-vertical angle of right circular cone of given surface area and maximum volume is . Let r be the radius, be the slant height and h be the height of the cone of given surface area, S. Also, let α be the semi-vertical angle of the cone. Then S = πrl + πr2 Let V be the volume of the cone. Differentiating (2) with respect to r, we get For maximum or minimum, put Differentiating again with respect to r, we get Thus, V is maximum when S= r2 As S = πrl + πr2 r2 = πrl + πr2 3πr2 = πrl l = 3r Now, in ΔCOB, #### Question 27: The point on the curve x2 = 2y which is nearest to the point (0, 5) is (A) (B) (C) (0, 0) (D) (2, 2) The given curve is x2 = 2y. For each value of x, the position of the point will be Let P and A(0, 5) are the given points. Now distance between the points P and A is given by, Let us denote PA2 by Z. Then, Differentiating both sides with respect to y, we get For maxima or minima, we have So, Z is minimum at . Or, PA2 is minimum at . Or, PA is minimum at . So, distance between the points is minimum at . So, the correct answer is A. #### Question 28: For all real values of x, the minimum value of is (A) 0 (B) 1 (C) 3 (D) Let ∴By second derivative test, f is the minimum at x = 1 and the minimum value is given by . The correct answer is D. #### Question 29: The maximum value of is (A) (B) (C) 1 (D) 0 Let Then, we evaluate the value of f at critical pointand at the end points of the interval [0, 1] {i.e., at x = 0 and x = 1}. Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1. The correct answer is C. #### Question 1: Using differentials, find the approximate value of each of the following. (a) (b) (a) Consider Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value of = 0.667 + 0.010 = 0.677. (b) Consider. Let x = 32 and Δx = 1. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value of = 0.5 − 0.003 = 0.497. #### Question 2: Show that the function given byhas maximum at x = e. Now, 1 − log x = 0 #### Question 3: The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base? Let ΔABC be isosceles where BC is the base of fixed length b. Let the length of the two equal sides of ΔABC be a. Now, in ΔADC, by applying the Pythagoras theorem, we have: ∴ Area of triangle The rate of change of the area with respect to time (t) is given by, It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. Then, when a = b, we have: Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of. #### Question 4: Find the equation of the normal to curve y2 = 4x at the point (1, 2). The equation of the given curve is. Differentiating with respect to x, we have: Now, the slope of the normal at point (1, 2) is ∴Equation of the normal at (1, 2) is y − 2 = −1(x − 1). y − 2 = − x + 1 x + y − 3 = 0 #### Question 5: Show that the normal at any point θ to the curve is at a constant distance from the origin. We have x = a cos θ + a θ sin θ. ∴ Slope of the normal at any point θ is. The equation of the normal at a given point (x, y) is given by, Now, the perpendicular distance of the normal from the origin is Hence, the perpendicular distance of the normal from the origin is constant. #### Question 6: Find the intervals in which the function f given by is (i) increasing (ii) decreasing Now, cos x = 0 or cos x = 4 But, cos x ≠ 4 ∴cos x = 0 divides (0, 2π) into three disjoint intervals i.e., In intervals, Thus, f(x) is increasing for In the interval Thus, f(x) is decreasing for. #### Question 7: Find the intervals in which the function f given byis (i) increasing (ii) decreasing Now, the points x = 1 and x = −1 divide the real line into three disjoint intervals i.e., In intervals i.e., when x < −1 and x > 1, Thus, when x < −1 and x > 1, f is increasing. In interval (−1, 1) i.e., when −1 < x < 1, Thus, when −1 < x < 1, f is decreasing. #### Question 8: Find the maximum area of an isosceles triangle inscribed in the ellipse with its vertex at one end of the major axis. The given ellipse is. Let the major axis be along the x −axis. Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0). Since the ellipse is symmetrical with respect to the x−axis and y −axis, we can assume the coordinates of A to be (−x1, y1) and the coordinates of B to be (−x1, −y1). Now, we have. Coordinates of A are and the coordinates of B are As the point (x1, y1) lies on the ellipse, the area of triangle ABC (A) is given by, But, x1 cannot be equal to a. Also, when, then Thus, the area is the maximum when Maximum area of the triangle is given by, #### Question 9: A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq meters for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank? Let l, b, and h represent the length, breadth, and height of the tank respectively. Then, we have height (h) = 2 m Volume of the tank = 8m3 Volume of the tank = l × b × h 8 = l × b × 2 Now, area of the base = lb = 4 Area of the 4 walls (A) = 2h (l + b) However, the length cannot be negative. Therefore, we have l = 4. Thus, by second derivative test, the area is the minimum when l = 2. We have l = b = h = 2. Cost of building the base = Rs 70 × (lb) = Rs 70 (4) = Rs 280 Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2) = Rs 8 (90) = Rs 720 Required total cost = Rs (280 + 720) = Rs 1000 Hence, the total cost of the tank will be Rs 1000. #### Question 10: The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle. Let r be the radius of the circle and a be the side of the square. Then, we have: The sum of the areas of the circle and the square (A) is given by, Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle. #### Question 11: A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. Let x and y be the length and breadth of the rectangular window. Radius of the semicircular opening It is given that the perimeter of the window is 10 m. Area of the window (A) is given by, Thus, when Therefore, by second derivative test, the area is the maximum when length. Hence, the required dimensions of the window to admit maximum light is given by #### Question 12: A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is Let ΔABC be right-angled at B. Let AB = x and BC = y. Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively. Let C = θ. We have, Now, PC = b cosec θ And, AP = a sec θ AC = AP + PC AC = b cosec θ + a sec θ … (1) Therefore, by second derivative test, the length of the hypotenuse is the maximum when Now, when, we have: Hence, the maximum length of the hypotenuses is. #### Question 13: Find the points at which the function f given byhas (i) local maxima (ii) local minima (ii) point of inflexion The given function is Now, for values of x close toand to the left of Also, for values of x close to and to the right of Thus, is the point of local maxima. Now, for values of x close to 2 and to the left of Also, for values of x close to 2 and to the right of 2, Thus, x = 2 is the point of local minima. Now, as the value of x varies through −1,does not changes its sign. Thus, x = −1 is the point of inflexion. #### Question 14: Find the absolute maximum and minimum values of the function f given by Now, evaluating the value of f at critical pointsand at the end points of the interval (i.e., at x = 0 and x = π), we have: Hence, the absolute maximum value of f is occurring at and the absolute minimum value of f is 1 occurring at #### Question 15: Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is. A sphere of fixed radius (r) is given. Let R and h be the radius and the height of the cone respectively. The volume (V) of the cone is given by, Now, from the right triangle BCD, we have: h ∴ The volume is the maximum when Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is. #### Question 16: Let f be a function defined on [a, b] such that f '(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b). Let such that ${x}_{1}<{x}_{2}$. Consider the sub-interval []. Since f (x) is differentiable on (a, b) and . Therefore, f(x) is continous on [] and differentiable on . By the Lagrange's mean value theorm, there exists such that Since f'(x) > 0 for all $x\in \left(a,b\right)$, so in particular, f'(c) > 0 [∵ ] $⇒f\left({x}_{2}\right)>f\left({x}_{1}\right)⇒f\left({x}_{1}\right) Since are arbitrary points in $\left(a,b\right)$. Therefore, Hence, f (x) is increasing on (a,b). #### Question 17: Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is. Also find the maximum volume. A sphere of fixed radius (R) is given. Let r and h be the radius and the height of the cylinder respectively. From the given figure, we have The volume (V) of the cylinder is given by, Now, it can be observed that at. ∴The volume is the maximum when When, the height of the cylinder is Hence, the volume of the cylinder is the maximum when the height of the cylinder is. #### Question 18: Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder istan2α. The given right circular cone of fixed height (h) and semi-vertical angle (α) can be drawn as: Here, a cylinder of radius R and height H is inscribed in the cone. Then, ∠GAO = α, OG = r, OA = h, OE = R, and CE = H. We have, r = h tan α Now, since ΔAOG is similar to ΔCEG, we have: Now, the volume (V) of the cylinder is given by, And, for, we have: ∴By second derivative test, the volume of the cylinder is the greatest when Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest. Now, the maximum volume of the cylinder can be obtained as: Hence, the given result is proved. #### Question 19: A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per hour. Then the depth of the wheat is increasing at the rate of (A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h Let r be the radius of the cylinder. Then, volume (V) of the cylinder is given by, Differentiating with respect to time t, we have: The tank is being filled with wheat at the rate of 314 cubic metres per hour. Thus, we have: Hence, the depth of wheat is increasing at the rate of 1 m/h. The correct answer is A. #### Question 20: The slope of the tangent to the curveat the point (2, −1) is (A) (B) (C) (D) The given curve is The given point is (2, −1). At x = 2, we have: The common value of t is 2. Hence, the slope of the tangent to the given curve at point (2, −1) is The correct answer is B. #### Question 21: The line y = mx + 1 is a tangent to the curve y2 = 4x if the value of m is (A) 1 (B) 2 (C) 3 (D) The equation of the tangent to the given curve is y = mx + 1. Now, substituting y = mx + 1 in y2 = 4x, we get: Since a tangent touches the curve at one point, the roots of equation (i) must be equal. Therefore, we have: Hence, the required value of m is 1. The correct answer is A. #### Question 22: The normal at the point (1, 1) on the curve 2y + x2 = 3 is (A) x + y = 0 (B) xy = 0 (C) x + y + 1 = 0 (D) x y = 1 The equation of the given curve is 2y + x2 = 3. Differentiating with respect to x, we have: The slope of the normal to the given curve at point (1, 1) is Hence, the equation of the normal to the given curve at (1, 1) is given as: The correct answer is B. #### Question 23: The normal to the curve x2 = 4y passing (1, 2) is (A) x + y = 3 (B) xy = 3 (C) x + y = 1 (D) xy = 1 The equation of the given curve is x2 = 4y. Differentiating with respect to x, we have: The slope of the normal to the given curve at point (h, k) is given by, ∴Equation of the normal at point (h, k) is given as: Now, it is given that the normal passes through the point (1, 2). Therefore, we have: Since (h, k) lies on the curve x2 = 4y, we have h2 = 4k. From equation (i), we have: Hence, the equation of the normal is given as: The correct answer is A. #### Question 24: The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes are (A) (B) (C) (D) The equation of the given curve is 9y2 = x3. Differentiating with respect to x, we have: The slope of the normal to the given curve at point is ∴ The equation of the normal to the curve at is It is given that the normal makes equal intercepts with the axes. Therefore, We have: Also, the pointlies on the curve, so we have From (i) and (ii), we have: From (ii), we have: Hence, the required points are The correct answer is A. View NCERT Solutions for all chapters of Class 12	2019-11-13 21:36:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "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": 0.8085046410560608, "perplexity": 535.5742731440861}, "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-2019-47/segments/1573496667333.2/warc/CC-MAIN-20191113191653-20191113215653-00481.warc.gz"}
https://itectec.com/ubuntu/ubuntu-extend-partition-on-freed-space-resize-or-moveresize-or-create-new-partition/	# Ubuntu – Extend partition on freed space: resize OR move&resize OR create new partition gpartedpartitioning I have Windows and Ubuntu partitions on my computer, and what I want is to use some space taken from a Windows partition for my file partition on Ubuntu. I booted a USB-live Ubuntu, opened Gparted, and here is a picture of my partitions: /dev/sda7 is the system partition of my Ubuntu. I would like to use the freed space (38.19GB) for the /dev/sda8 partition. I see three options: 1) extend /dev/sda7 to the left to occupy the non-allocated space, shrink it by the same amount (38.19GB) to the right, then extend /dev/sda8 to the left by the same amount. OR 2) move /dev/sda7 to the left to occupy the non-allocated space, then extend /dev/sda8 to the left to recover the 38.19GB. PROBLEM: When doing it in Gparted (without confirming), I get a warning message saying that I could probably be unable to boot on my Ubuntu (which is on /dev/sda7) if I move it. OR 3) Do not touch /dev/sda7 and /dev/sda8 but create a new partition at the non-allocated space, which I would use as additional partition when /dev/sda8 is full. Which of these three options should I use? Which one is advised or not, and why? By "advised" I mean safe for my Ubuntu system and data (I want to have my usual Ubuntu exactly as before after rebooting, but with more space available for my files). In option 1, would the extension to the left of my system partition /dev/sda7 be a problem? (in terms of booting, performance, or other…) Option 2 is probably not advised because I would end up with a non-bootable Ubuntu, but I mention it in case the other options have more drawbacks. EDITS: • I backed up my files, but not my Ubuntu system, and I am worried about Gparted not recommending to move partition /dev/sda7 (Ubuntu system) in option 2. Any explanation on the risks when moving this partition would be appreciated. • My plan is to remove /dev/sda5 in the future. • # Make sure that you have a good backup of your important Ubuntu files, as this procedure can corrupt or loose data. Keep these things in mind: • always start the entire procedure with issuing a swapoff on any mounted swap partitions, and end the entire procedure with issuing a swapon on that same swap partition • a move is done by pointing the mouse pointer at the center of a partition and dragging it left/right with the hand cursor • a resize is done by dragging the left/right side of a partition to the left/right with the directional arrow cursor • if any partition can't be moved/resized graphically, you may have to manually enter the specific required numeric data (don't do this unless I instruct you to) • you begin any move/resize by right-clicking on the partition in the lower part of the main window, and selecting the desired action from the popup menu, then finishing that action in the new move/resize window Do the following... Note: if the procedure doesn't work exactly as I outline, STOP immediately and DO NOT continue. • boot the a Ubuntu Live DVD/USB • start gparted • move sda7 partition all the way left • move sda8 partition all the way left • resize sda8 right side all the way to the right If it looks good, click the Apply button.	2021-06-18 15:01:07	{"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": 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": 0.3241359293460846, "perplexity": 2812.1577793446168}, "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/1623487637721.34/warc/CC-MAIN-20210618134943-20210618164943-00219.warc.gz"}
https://www.gamedev.net/blogs/entry/2250103-new-parameter-management-system-for-hieroglyph-3/	• entries 316 485 • views 321839 # New Parameter Management System for Hieroglyph 3 971 views As described in my last entry (which seems like it was ages ago), I have been overhauling my parameter system in an attempt to speed things up when it comes to configuring the rendering pipeline for a particular draw call. The overall process is fairly automated already, but I wanted to try to minimize the number of string comparisons needed to perform a rendering pass. As a side note, this was identified as one of the bigger performance wasters in the multithreaded rendering sample for our book - so this isn't a wild goose chase The tricky part is to design a system which isn't totally different than the current system, supports multiple threads using the parameters in parallel on the CPU, and allows for very fast lookups of parameters for both writing and reading of parameters for every object in every frame. Due to the frequency that this operation is performed, there is the potential for a good speed improvement if a useful design can be found. The existing system actually provided for multiple parameter manager instances to be used in parallel, with one instance for each rendering thread to use. This allowed for individual rendering passes to have their own 'ecosystem' of parameters that wouldn't be affected by anything being used by other threads --> for example, the view matrix could be used by multiple threads in isolation without any problem. This works, but the big drawback is that any client that wants to write a parameter uses a function call such as this: ParameterManager::SetVectorParameter( std::wstring name, Vector4f value ); The name is used to look up a parameter instance in a std::map, then later on the user of the parameters (i.e. the renderer) reads the parameter data in the same way. This amplifies the number of lookups needed, and they are performed for any named parameter that is used in a shader. For D3D11, there are five programmable stages (plus the compute shader) which can add up to be lots of lookups... My initial thoughts on how to solve this problem was to completely eliminate the use of a lookup at all. Instead, any clients that write a parameter value would initialize their own copy of the parameter and then just directly reference the parameter themselves. This does indeed eliminate the cost of the lookup and has an amortized cost of zero, which is pretty hard to beat . However, if every client has a direct reference to a named parameter, then it messes up the parallelizability of the whole system. I could have a render view reference a unique copy of each parameter per thread, but I don't guarantee that a render view will be run on the same thread from frame to frame... I needed something a little better. Instead, I decided to have multiple parameter manager instances referencing a single std::map of parameters, and have multiple copies of the data stored in the parameter objects - one copy for each thread. Then the parameter managers could be assigned an ID, the rendering clients could use the parameter manager for a given thread without knowing which copy of the data they were using (the parameter manager handles the selection of the right array index transparently). This allows for parallel access to the data from multiple threads, while still providing the "pre-lookup" benefits mentioned above, and actually reducing the overall complexity of the system. I am quite pleased with this solution, and it also has the nice side benefit that it allows existing code to be run in a single threaded mode unmodified - so I am backwards compatible with my own sample programs. The implementation part took some time, the better part of a month (that is mostly due to lack of availability, but still...). Now that it has been implemented, I can run some speed tests and see how much of a difference it makes. As mentioned above, I wanted to improve the performance of the multithreaded sample program from our book. This makes it a natural choice for performing a before and after benchmark. So the sample essentially maximizes the number of state changes by rendering reflective paraboloid maps with many surrounding objects. The demo is run in 512x512 output mode with three reflective objects and a varying number of objects around them (ranging from 200 - 1000 objects), with 4 threads on a quad core machine. The following chart shows the improvement with FPS along the y-axis and the number of objects on the x-axis. As you can see, the new method runs at roughly 150% of the framerate of the old method. To say the least, I am happy with the results . With this improvement, I think the parameter system can be considered good enough and just needs to be cleaned up before I push it to the Codeplex repository. So the point is this - don't be afraid to fundamentally change the way that you do a very frequent task, since it can be quite easy to get a big performance win with it! Even if it takes lots of work and time (and for hobbyists this is tough to swallow), it can really be worth it in the end. There are no comments to display. ## Create an account Register a new account	2017-10-18 11:19:46	{"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": 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": 0.37126806378364563, "perplexity": 595.8991702525024}, "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/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00582.warc.gz"}
http://mymathforum.com/number-theory/333351-fermat-last-i-hope.html	My Math Forum Fermat the Last (I hope) Number Theory Number Theory Math Forum June 27th, 2016, 12:39 AM #1 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 Fermat the Last (I hope) Here the last turn for FLT: (for clarify what Step Sum, step $1/K$ are and works, pls check my several post on) Giving for true under Fermat's conditions (for $n>2$, $A2$ can be summarized as: it's due to mixed terms of higher degree (from 2). And a short cut will be: first derivate = curve. I know one or more bugs can affect this "process" I won't call "proof", so I'll wait for an expert check... (sub judice) It was a very hard work for me (8 years long)... I hope it's close to the full stop (for many reasons). Thanks Stefano June 27th, 2016, 10:45 AM #2 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry, still 2 computation bugs, but result don't change in the behaviour.... I'll check better asap... Thanks Stefano June 27th, 2016, 09:34 PM #3 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 Here the last version with fixed develope bugs: On the right hand we will have (here first error fixed): $3B (C-B-1/K)^2 + 3B^2/K (C-B-1/K) = 3BC^2-3CB^2-6BC/K+3B^2/K+3B/K^2$ Reordering (here the second error fixed): $(A+B-C)^3 =$ $-3C(A+B-C)^2+3B(A+B-C)^2+3/K(A+B-C)^2-(-6ABC/K+3AB^2/K+3AC^2/K+9BC^2/K-9CB^2/K+ 3B^3/K-3C^3/K+6AB/K^2-6AC/K^2-12BC/K^2+6B^2/K^2+6C^2/K^2+3A/K^2+3B/K^2-3C/K^2)+ 3BC^2-3CB^2-6BC/K+3B^2/K+3B/K^2$ $= -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3+6ABC/K+12CB^2/K-3AB^2/K- 3AC^2/K-9BC^2/K+3C^3/K-6B^3/K+6AB/K-6AC/K-12BC/K+3A^2/K+3C^2/K+9B^2/K+6AC/K^2+ 12BC/K^2-6AB/K^2-6C^2/K^2-9B^2/K^2+3C/K^2-3A/K^2$ that if $K\to\infty$ becomes: $(A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3$ So it's clear that the right hand is NOT the right tri-nomial develop, so this is the absurdum we are looking for. Thanks Ciao Stefano June 29th, 2016, 11:17 PM #4 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 ...I hope someone kindly answer... The point is if it's possible to works with the limits for $K\to\infty$ in this way or not, so if the turn it's true, or false... I hope it's true since moving parts from left to right or viceversa we respect the inequality, just respecting the sing rule, and for the limits since we have just Sums nothing has to change... I already did another trip, just doble longer, where will be clear that assuming $A^=C^n-B^n$ for true in the integers there is no continuty if we assume, and computate in this way from right (+1/K) or from the left side (-1/K), since the two limits, once reduced in this way are different. But I still be prepared to the worst... Thanks Ciao Stefano July 7th, 2016, 12:36 AM #5 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 OK, I've a nice discussion with italian friends... (that just erase the topics ...) It's clear that: $(A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3$ is NOT the right tri-nomial develop, but it has, unfotunately, some integer solution... so what we have to do is to compare this result to the result coming from the limit from the right side so of: (3) $\displaystyle (A+1/K)^n= (C+1/K)^n-B^n$ At the moment it (once reduced with sum and pushed to the limit) give to me another result, so seems the two limits are different and this is the right absurdum since we know they must be the same on a continuos function... I will check the computation to avoid errors, and I've also to check witch of the possible cooling, or warming, is the shortest one to show the absurdum. ...it's a long way to tipperary...it's a long way to go... Thanks Ciao Stefano July 7th, 2016, 02:10 AM #6 Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 I check: (3) $\displaystyle (A-1/K)^n= (C)^n-(B-1/K)^n$ and sems to finish at: $\displaystyle (A+B-C)^3 = 3BC^2-6CB^2+3B^3$ So if all computation is right I'm alowd to say that this prove that the initial equation must be wrong ? Thanks Ciao Stefano Tags fermat, hope Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post jeneferhokins New Users 9 July 12th, 2012 08:35 PM JOANA Abstract Algebra 1 May 5th, 2010 12:20 PM MathChallenged2010 New Users 7 February 25th, 2010 10:15 PM Rooonaldinho Algebra 1 April 7th, 2008 01:52 PM Contact - Home - Forums - Cryptocurrency Forum - Top	2018-09-19 18:13:42	{"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": 0.7615224719047546, "perplexity": 2785.1273624143764}, "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/1537267156270.42/warc/CC-MAIN-20180919180955-20180919200955-00423.warc.gz"}
https://www.shaalaa.com/question-bank-solutions/a-bag-contains-5-white-and-3-black-balls-two-balls-are-drawn-at-random-then-the-probability-that-out-of-the-two-one-ball-is-red-and-other-is-black-will-be-probability-entrance-exam_105016	# A Bag Contains 5 White and 3 Black Balls. Two Balls Are Drawn at Random Then the Probability that Out of the Two One Ball is Red and Other is Black Will Be: - Mathematics MCQ A bag contains 5 white and 3 black balls. Two balls are drawn at random then the probability that out of the two one ball is red and other is black will be: #### Options • 15/16 • 11/28 • 15/28 • 2/7 #### Solution 15/28 Concept: Probability (Entrance Exam) Is there an error in this question or solution?	2022-07-07 16:45:09	{"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": 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": 0.7490358352661133, "perplexity": 235.98061878346502}, "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-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00750.warc.gz"}
https://math.stackexchange.com/questions/2676969/prove-that-if-a-equiv-b-pmod-p2-p-then-aa-equiv-bb-pmodp-whe	# Prove that if $a\equiv b \pmod {p^{2}-p}$, then $a^{a}\equiv b^{b} \pmod{p}$ where $p$ is any prime and $a$ and $b$ are nonzero integers. I have shown the case in which $p=2$, so now I'm considering $p\geq 3$. I see that $p^{2}-p = p(p-1)$, in which case we can apply the Chinese Remainder Theorem to obtain $a\equiv [b,b] \mod{[p,\phi(p)]}$, since $p$ and $(p-1)$ are coprime and $\phi(p)=(p-1)$ where $\phi$ is Euler's totient function. Also, I see that $a\equiv b \pmod{\phi(p)}$ implies that $x^a\equiv x^b \pmod{p}$ for some $x\in \Phi(p)$ by the Fermat-Euler theorem. However, I'm not sure if this is the best approach or how to continue from here. Any help is appreciated, thanks! Edit: Now I see that $x^a\equiv x^b \pmod{p}$ is true for all $x\in \mathbb{Z}$. Because $p$ is prime, $x$ will either be coprime, or a multiple of $p$. In each case the statement is still true. This is why we can let $x = b$. Then we can use $a^a\equiv b^a \pmod{p}$ and $b^a\equiv b^b \pmod{p}$, and we're done. • You are basically done. By $a\equiv_p b$ we have $a^a\equiv_p b^a$. Then, as you said yourself (set $x=b$), we have $b^a\equiv_p b^b$. – Arthur Mar 4 '18 at 22:48 By the Chinese remainder theorem, the hypothesis is equivalent to $a\equiv b\mod p$ AND $a\equiv b\mod p-1$. Now lil' Fermat states that $\;a^a\equiv \bigl((a\bmod p)^{a\bmod p-1}\bigr)\bmod p$, and similarly for $b$. Fix $p\ge 3$ and note that $p(p-1)\mid a-b$ iff $p\mid a-b$ and $p-1 \mid a-b$. So $a$ and $b$ have the same remainder modulo $p$ and modulo $p-1$. Moreover, the sequences $(a^n)_{n\ge 1}$ and $(b^n)_{n\ge 1}$ are periodic modulo $p-1$, which implies that $$a^x \equiv b^y \bmod{p}$$ whenever $x \equiv y \bmod{p-1}$. But $x=a$ and $y=b$ verify this condition by hypothesis.	2019-07-17 07:18:03	{"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": 0.9438063502311707, "perplexity": 37.21275985713651}, "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-20190717061451-20190717083451-00017.warc.gz"}
https://www.collegeguide4parents.com/2016/10/	## ACT Math – Pre-Algebra Quiz Parents – Did you know that the questions on the ACT Math test vary greatly in difficulty? 20 to 25% of the questions are on concepts learned in Junior High while a few very hard questions test concepts students learn in Algebra II and Trigonometry classes Your child may be thinking that there is no need to study before taking the ACT or worrying that there is too much to review. In reality, preparing for it will increase your child’s score by several points, and CollegeGuide4Parents can help speed up the process. CollegeGuide4Parents’ Math tools allow students to quickly identify her/his weaknesses and direct them to online educational resources offering a clear explanation of the concepts as well as examples. If your child takes the 30-question Pre-Algebra quiz on this page, he/she will get a customized report showing the topics she/he should work on. CollegeGuide4Parents’ Pre-Algebra review page list links for will take her/him directly to online providing links to reputable educational websites where she/he will find the information needed. The easiest Math questions on the ACT represent 20 to 25% of the Math portion of the test. Your child learned the skills necessary to do well on these questions in Junior High in Pre-Algebra class. She/He has probably forgotten a lot of what she/he has learned then, but a quick review is most likely all that is necessary for her/him to do well on these questions. The 30-question quiz below will allow your child to quickly identify what she/he needs to focus on. She/He can also go directly to the review age to go over the main Pre-Algebra skills. If you remember your child struggled in that class, the online worksheets linked from our review page will give her/him the extra practice she/he needs. Students: Take our Pre-Algebra Quiz to identify your weaknesses in pre-algebra.Pre-Algebra problems represent 20-25% of the ACT Math score so it is essential to be well prepared. You learned these concepts in Pre-Algebra class. How does it work? The Pre-Algebra Quiz consists of 33 questions. Answer as many as you feel comfortable doing so and click on Correct my quiz. You will get a page (that you can print) that will show, for each incorrect answer, your answer, the actual answer and the section to review. There is a lot of websites out there explaining math concepts, and the post “ACT Math – Pre-Algebra Review” will direct you to online resources for each of the section you should review. You can retake the quiz when you are ready to work on the questions you skipped. To take another CollegeGuide4Parents’ quiz, check the complete list at the bottom of this post. Note: Want to the review before finishing the quiz? Go to post ACT Math – Pre-Algebra Review #### Pre-Algebra Quiz $Question\ 1.\ Which\ are\ multiples\ of\ 6?$ 1, 3, 6 6, 12, 18 6, 9, 12 2, 3 4, 5, 6 » Read more	2021-04-13 10:10:39	{"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": 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": 0.24690955877304077, "perplexity": 1412.8307030781684}, "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-2021-17/segments/1618038072180.33/warc/CC-MAIN-20210413092418-20210413122418-00639.warc.gz"}
https://community.wolfram.com/groups/-/m/t/1506499	# Easy generation, testing, and export of audio loops Posted 9 months ago 565 Views \| 0 Replies \| 4 Total Likes \| I wanted to build some samples where the individual notes in a chord would repeat at various rates, and I needed to do it quickly (I had a performance coming up and of course I procrastinated). The traditional method—getting the sounds into a DAW, setting up the individual tracks for each note, setting the fade-ins and fade-outs, testing the loops, exporting, and all the rest—would have been too time consuming. The Wolfram Language made the process a breeze. Plus, I could make adjustments and test them much quicker.This is the chord:First, I made a list of keys for when I use Map[] to build "the chord" and export the files: keys = {"1", "2", "3", "4", "5", "6"}; Next, I set up the fade-in time and the durations of my loops. The tempo will be 80 beats per minute, so I needed to convert the number of beats into seconds for SoundNote[]: fadein = N[2/80] 60; (2 beats) time1 = N[24/80] 60; (6 measures) time2 = N[20/80] 60; (5 measures) time3 = N[18/80] 60; (4.5 measures) time4 = N[28/80] 60; (7 measures) time5 = N[16/80] 60; (4 measures) time6 = N[22/80] 60; (5.5 measures*) Then, I set up the individual notes to be looped. The fade-in time was defined above, and I wanted the fade-out time to be half the length of the SoundNote[]. There was also no need for the final files to be in stereo, so I used AudioChannelMix[]: note1 = AudioChannelMix[AudioFade[Sound[SoundNote["D4", time1, "BlownBottle"]], {fadein, N[time1/2]}, Method -> "Exp"], "Mono"]; note2 = AudioChannelMix[AudioFade[Sound[SoundNote["A4", time2, "BlownBottle"]], {fadein, N[time2/2]}, Method -> "Exp"], "Mono"]; note3 = AudioChannelMix[AudioFade[Sound[SoundNote["E5", time3, "BlownBottle"]], {fadein, N[time3/2]}, Method -> "Exp"], "Mono"]; note4 = AudioChannelMix[AudioFade[Sound[SoundNote["G5", time4, "BlownBottle"]], {fadein, N[time4/2]}, Method -> "Exp"], "Mono"]; note5 = AudioChannelMix[AudioFade[Sound[SoundNote["B5", time5, "BlownBottle"]], {fadein, N[time5/2]}, Method -> "Exp"], "Mono"]; note6 = AudioChannelMix[AudioFade[Sound[SoundNote["E6", time6, "BlownBottle"]], {fadein, N[time6/2]}, Method -> "Exp"], "Mono"]; Time to play! This one will start as a giant chord and then the individual notes will fade in and out: Map[AudioPlay[ToExpression[StringJoin["note", #]], AudioLooping -> True] &, keys] Or evaluate one at a time, in any order, and any pause between notes. AudioPlay[note1, AudioLooping -> True] AudioPlay[note2, AudioLooping -> True] AudioPlay[note3, AudioLooping -> True] AudioPlay[note4, AudioLooping -> True] AudioPlay[note5, AudioLooping -> True] AudioPlay[note6, AudioLooping -> True] After enough testing or listening enjoyment, stop all sounds: AudioStop[] After trying out both, I decided to go with the "big chord" approach. It was nice to be able to test both before exporting. It was also nice to be able to quickly adjust the times of the individual notes by going up to my initial definitions.And now, to export: Map[Export[StringJoin["loop", #, ".wav"], ToExpression[StringJoin["note", #]]] &, keys] This gave me a set of similarly named wav files (loop1.wav, loop2.wav, etc), so I could find them easily in my sample pad. Attachments:	2019-06-27 00:59:34	{"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": 0.21448548138141632, "perplexity": 13863.316588849013}, "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/1560628000609.90/warc/CC-MAIN-20190626234958-20190627020958-00469.warc.gz"}
https://www.esaral.com/q/in-the-experimental-set-up-of-metre-bridge-shown-in-the-figure-18533/	In the experimental set up of metre bridge shown in the figure, Question: In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of $40 \mathrm{~cm}$ from A. If a $10 \Omega$ resistor is connected in series with $\mathrm{R}_{1}$, the null point shifts by $10 \mathrm{~cm}$. The resistance that should be connected in parallel with $\left(\mathrm{R}_{1}+10\right) \Omega$ such that the null point shifts back to its initial position is : 1. (1) $20 \Omega$ 2. (2) $40 \Omega$ 3. (3) $60 \Omega$ 4. (4) $30 \Omega$ Correct Option: , 3 Solution: (3) Initially at null deflection $\frac{R_{1}}{R_{2}}=\frac{2}{3}$ …(1) Finally at null deflection, when null point is shifted $\frac{\mathrm{R}_{1}+10}{\mathrm{R}_{2}}=1 \Rightarrow \mathrm{R}_{1}+10=\mathrm{R}_{2} \ldots$ Solving equations (i) and (ii) we get $\frac{2 R_{2}}{3}+10=R_{2}$ $10=\frac{R_{2}}{3} \Rightarrow R_{2}=30 \Omega$ $\& \mathrm{R}_{1}=20 \Omega$ Now if required resistance is $\mathrm{R}$ then $\frac{\frac{30 \times R}{30+R}}{30}=\frac{2}{3}$ $R=60 \Omega$	2022-12-09 21:58:00	{"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": 0.8001872301101685, "perplexity": 1246.3320603914617}, "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-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00303.warc.gz"}
http://codeforces.com/blog/entry/65569	### RubenAshughyan's blog By RubenAshughyan, history, 23 months ago, Hi everyone. Today I encountered a data structures problem, which is a slight swist on standart segment tree problem. You are given an array A of size N. N ≤ 10^5, A[i] ≤ 10^6, You need to process 3 types of queries 1. Print the sum of all integers ai where L ≤ i ≤ R. 2. Add x to each integer ai where L ≤ i ≤ R. 3. For each integer A[i] where L ≤ i ≤ R, replace it with floor(sqrt(A[i])). Where sqrt(a) is the square root of a, and floor(b) is the integer value of b after removing everything on the right of the decimal point. There can be Q≤20000 queries. If we didn't have type 3 query, then we are left with standard lazy segment tree problem. How to handle 3rd type queries? • +46 » 23 months ago, # \| +24 I think the problem you are talking about was in ACPC 2016, this problem to be exact.I didn't solve it, however, I was told by one of the judges during the contest that it can be solved by holding the maximum and minimum in the range of the segment tree and using lazy propagation in the form of vectors and since any range will become equal after not that much operations when the maximum becomes equal to minimum it's easy to note that lazy propagation becomes easier to do in O(1). » 23 months ago, # \| +42 I don't know exact time complexity of this solution. Firstly we store maximum and minimum in our segment tree. We will have three options. If sqrt(maximum) == sqrt(minimum) then this equals to assigning sqrt(maximum) to this segment If maximum — minimum == 1 then after taking the square root of every element they will differ maximum for 1. This is equal to making minimum = sqrt(minimum) and maximum = sqrt(minimum) + 1. We can obtain this by adding (sqrt(minimum) — minimum) to this segment elements. If no condition above meets we will solve it recursively. • » » 23 months ago, # ^ \| 0 Can you please explain the 2nd option?How can we obtain it(the new version of the segment, after applying sqrt) by adding (sqrt(minimum) — minimum) to this segment elements? • » » » 23 months ago, # ^ \| 0 It can be done using segment tree with lazy propagation which supports adding and assigning. We can support it just only using one segment tree. • » » » » 23 months ago, # ^ \| 0 I mean the following:Suppose the segment is [11,5,7,10,6]After applying sqrt we should get [3,2,2,3,2] right?But if we do it your way i.e. by adding (sqrt(minimum) — minimum) to this segment — we getsqrt(minimum) — minimum = sqrt(5) — 5 = -3So we add -3 to the range and get [8,2,4,7,3]. Which is different than [3,2,2,3,2]. • » » » » » 23 months ago, # ^ \| 0 We will have an array [11, 5, 7, 10, 6]. No condition meets. (11 — 5 != 1)So, we will divide it into two [11, 5], [7, 10, 6]. Also no condition meets.[11], [5], [7], [10, 6][11], [5], [7], [10], [6] => [3,2,2,3,2] • » » » » » » 23 months ago, # ^ \| 0 Thanks Zharaskhan got your point. Can you provide intuition why this works fast? In this query example we saw that it is linear. • » » » 23 months ago, # ^ \| +13 I think the idea is if max - min ≥ 2, then it will decrease proportional to sqrt(N). So 2nd type queries won't ruin everything because the difference will still decrease quickly. But it's possible that max - min = 1 and still will be after taking square roots for each element. For example [3, 4] to [1, 2]. So if we're able to do such updates without solving recursively, we can solve the problem. If we know min = max - 1 then we can, for example, use counts for min or max and update the interval without the need of recursive. • » » 23 months ago, # ^ \| 0 That's nice solution.What is the complexity of this solution?Thank you! • » » 23 months ago, # ^ \| ← Rev. 4 → 0 Would the solution be considerably worse if we solve option 2 recursively?EDIT: Nevermind, I finally understood the corner case. » 23 months ago, # \| ← Rev. 2 → +8 For query type 3:when we get the range modification entry, we can modify each element in the range one by one, and hence perform (R — L + 1) single element update queries. This approach would be to slow, as it makes the complexity of range update query to O ((R — L + 1) lg N). But many of these single element update queries can be ignored, and the number of actual performed queries is much lower. Note that, if the value of an element A[x] is 1, then update query on A[x] does not have any effect on the segment tree.Let call such queries as degenerate queries, and We can discard them. Now think A[x]>1, then how many update operation is required on that position to make the A[x] is 1. It will require less than lg(A[x]) operation approximately.So,every node of segment tree gets updated at most lg(10^6) time . However, now the problem is how to find the non-degenerate single element update queries for the range [L, R]. For this purpose, we maintain another information in the segment tree node, which is the number of elements in the range which are larger than one.Similar problem You can see editorial for better explanation . • » » 23 months ago, # ^ \| 0 What if we have lots of 2nd type queries, such that A[i] is never becoming 1? This way the complexity may be upto O(n) for each query. » 23 months ago, # \| +3 My guess is that you can do it with a small modification to the lazy segment tree. For each node in the tree keep track of whether or not its corresponding interval is constant. Then when you get a query of type 3, you can lazily take care of it if the interval is constant, otherwise let the children nodes handle the query.The reason why I think that this works is that floor(sqrt( )) is really destructive. Suppose you start out with A[i] ≥ 1, then after you apply query 3 enough times to it, the new value is independent of its original value. You would have gotten the exact same result had A[i] = 1 in the beginning. This in turn makes intervals become constant, which allows queries of type 3 to be taken care of lazily. • » » 23 months ago, # ^ \| 0 It seems like a good idea. Btw do you have an intuitive reasoning why should the invervals become constant quickly? We have 2nd type queries which may ruin everything.Also do you have an estimate of time complexity? In the worst case it may be linear per query, this can happen if all internal(non-leaf) segments aren't constant. • » » » 23 months ago, # ^ \| +8 (Note there is a small error in my approach, so just use Zharaskhan approach, still here is the analysis for the running time)Assuming that x > y ≥ 1, from .one can show that. This describes how quickly the difference between two values x, y decreases when you apply a floor(sqrt( )) on the pair. It will pretty much decrease with a factor of , so it decreases very quickly.I did some calculations given the second inequality and from that I concluded that it takes at most 5 iterations for the difference to become ≤ 1.Finally, about the running time. For each node in the tree you need to apply type3 queries 5 times to reach max - min ≤ 1. Also whenever you do a type2 or type3 query you will update nodes, which after 5 type3 updates will go back to have max - min ≤ 1. So from this I conclude that the running time should be something like . • » » 23 months ago, # ^ \| ← Rev. 2 → +5 If you have 3s and 4s and alternate sqrts and +=2 I think you'll never get constant and you will have linear sqrt updates • » » » 23 months ago, # ^ \| 0 Thanks for pointing that out, that is a problem.I think it is possible to fix it by keeping track of whether or not the interval only contains at most two different values (and keep a count for both). But if you do that, then you might as well just do the min/max approach that Zharaskhan mentioned. » 23 months ago, # \| ← Rev. 2 → +3 Can I submit this problem somewhere? It seems the ICPC Live Archive doesn't have any input (my assertion assert(ntest > 0) failed).I can't think of an Segment Tree solution so here goes my shitty sqrt one: SpoilerLet's process one block of T queries at a time (we will choose T later). We divide the N numbers into intervals, with starting points being L[1], L[2], ..., L[T], R[1] + 1, R[2] + 1, ..., R[T] + 1. This way we have at most 2T intervals, and each query cover some intervals completely. For each interval, we store those values: min, max, sum, offset, count of elements equal to min/max.For the "add X" queries, we simply increase the offset of the affected intervals by X.For the "take sqrt" queries, we notice that after at most 6 "sqrt" queries, max[b] - min[b] ≤ 1 will hold, because the value of A[i] won't exceed 226 and the "add" queries won't change max[b] - min[b]. Thus, if max[b] - min[b] > 1 holds at the moment, we just recompute all elements of block b, knowing we won't do that more than 6 times per block, and update the min, max, and sum accordingly. Otherwise, we can set min[b] = sqrt(min[b] + offset[b]), max[b] = sqrt(max[b] + offset[b]); and ignore value sum because it can be deduced (see below).For the "compute sum" queries, we compute the sum of each block. For block b:If min[b] = max[b] then sum[b] = min[b] * cntmin[b] + offset[b] * length[b].If min[b] + 1 = max[b] then sum[b] = min[b] * cntmin[b] + max[b] * cntmax[b] + offset[b] * length[b].Otherwise, we just return sum[b].After processing T queries, we need to retrieve the new array A (I did that in O(N * log(N)), not sure if it can be done faster).The time complexity is , where 6 = log2(log2(maxA)). • » » 23 months ago, # ^ \| 0 Probably you can download here: http://acmacpc.org/archive/y2017/ • » » » 23 months ago, # ^ \| 0 I tried but it says "404 page not found"... • » » 23 months ago, # ^ \| 0 I think you can submit the problem from here : https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=804&page=show_problem&problem=6491 • » » » 23 months ago, # ^ \| 0 Already did but my assertion assert(ntest > 0); fails, which likely means the judge has no input.	2021-01-21 17:59:48	{"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": 0.45451948046684265, "perplexity": 852.3707757324607}, "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/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00261.warc.gz"}
https://en.wikipedia.org/wiki/Talk:Stirling_numbers_of_the_second_kind	# Talk:Stirling numbers of the second kind WikiProject Mathematics (Rated C-class, Low-importance) This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks. Mathematics rating: C Class Low Importance Field: Discrete mathematics ## Unclear I'm removing a large block of text from this article, because it is unclear. The problems are: • Use of blackpage symbols without any attempt to explain their meaning • Use of the term "EGF" without explaining what it means. Is this, perhaps, the exponential generating function? or something else? • Use of the symbol ${\displaystyle [x^{n}]}$ without explanation. Clearly, its not equal to ${\displaystyle x^{n}}$, but I can't figure out what its supposed to be. (edit: ${\displaystyle [x^{n}]}$ denotes the coefficient of ${\displaystyle x^{n}}$ in the power series expansion of the specified generating function) • Most of the text appears to be a long proof, and, as such, is not appropriate for article space. See Category:Article proofs for examples on how to deal with proofs. linas 02:16, 29 December 2006 (UTC) #### Generating functions These numbers count the number of partitions of ${\displaystyle [n]}$ into k nonempty subsets. First consider the total number of partitions, i.e. ${\displaystyle B_{n}}$ where ${\displaystyle B_{n}=\sum _{k=1}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}{\mbox{ and }}B_{0}=1,}$ i.e. the Bell numbers. The Symbolic combinatorics#The Flajolet–Sedgewick fundamental theorem applies (labelled case). The set ${\displaystyle {\mathcal {B}}\,}$ of partitions into non-empty subsets is given by ("set of non-empty sets of singletons") ${\displaystyle {\mathcal {B}}={\mathfrak {P}}({\mathfrak {P}}_{\geq 1}({\mathcal {Z}})).}$ This decomposition is entirely analogous to the construction of the set ${\displaystyle {\mathcal {P}}\,}$ of permutations from cycles, which is given by ${\displaystyle {\mathcal {P}}={\mathfrak {P}}({\mathfrak {C}}({\mathcal {Z}})).}$ and yields the Stirling numbers of the first kind. Hence the name "Stirling numbers of the second kind." The decomposition is equivalent to the EGF ${\displaystyle B(z)=\exp \left(\exp z-1\right).}$ Differentiate to obtain ${\displaystyle {\frac {d}{dz}}B(z)=\exp \left(\exp z-1\right)\exp z=B(z)\exp z,}$ which implies that ${\displaystyle B_{n+1}=\sum _{k=0}^{n}{n \choose k}B_{k},}$ by convolution of exponential generating functions and because differentiating an EGF drops the first coefficient and shifts ${\displaystyle B_{n+1}}$ to ${\displaystyle z^{n}/n!.}$ The EGF of the Stirling numbers of the second kind is obtained by marking every subset that goes into the partition with the term ${\displaystyle {\mathcal {U}}\,}$, giving ${\displaystyle {\mathcal {B}}={\mathfrak {P}}({\mathcal {U}}\;{\mathfrak {P}}_{\geq 1}({\mathcal {Z}})).}$ Translating to generating functions, we obtain ${\displaystyle B(z,u)=\exp \left(u\left(\exp z-1\right)\right).}$ This EGF yields the formula for the Stirling numbers of the second kind: ${\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}=n![u^{k}][z^{n}]B(z,u)=n![z^{n}]{\frac {(\exp z-1)^{k}}{k!}}}$ or ${\displaystyle n![z^{n}]{\frac {1}{k!}}\sum _{j=0}^{k}{k \choose j}\exp(jz)(-1)^{k-j}}$ which simplifies to ${\displaystyle {\frac {n!}{k!}}\sum _{j=0}^{k}{k \choose j}(-1)^{k-j}{\frac {j^{n}}{n!}}={\frac {1}{k!}}\sum _{j=0}^{k}{k \choose j}(-1)^{k-j}j^{n}.}$ We can use ${\displaystyle B(z,u)}$ to evaluate the sum ${\displaystyle \sum _{k=0}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}(x)_{k}.}$ This is equal to ${\displaystyle \sum _{k=0}^{n}n![u^{k}][z^{n}]B(z,u)(x)_{k}=n![z^{n}]\sum _{k=0}^{n}[u^{k}]B(z,u)(x)_{k}}$ or ${\displaystyle n![z^{n}]\sum _{k=0}^{n}{\frac {(\exp z-1)^{k}}{k!}}(x)_{k}=n![z^{n}]\sum _{k=0}^{\infty }{\frac {(\exp z-1)^{k}}{k!}}(x)_{k}}$ where the last equality occurs because ${\displaystyle [z^{n}](\exp z-1)^{k}=0\,}$ when ${\displaystyle n Now consider the Taylor series of ${\displaystyle y^{x}}$ at ${\displaystyle y=1:}$ ${\displaystyle y^{x}=\sum _{k=0}^{\infty }\left(\left({\frac {d}{dy}}\right)^{k}y^{x}{\Bigg \|}_{y=1}\right){\frac {(y-1)^{k}}{k!}}=\sum _{k=0}^{\infty }(x)_{k}{\frac {(y-1)^{k}}{k!}}.}$ Hence ${\displaystyle \sum _{k=0}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}(x)_{k}=n![z^{n}]\exp(xz)=n!{\frac {x^{n}}{n!}}=x^{n}.}$ ## Explicit formula I have noticed that the index j in the sum of the explicit formula starts at 0 in all textbooks I have had a look on this topic. I understand that it does not matter, as the jn part will make the whole thing zero anyway when j=0. Was just wondering if there is standard notation, since wikipedia is the first place where I have seen it starting at 1. —Preceding unsigned comment added by 130.95.29.113 (talkcontribs) 08:54, 25 May 2007 (UTC) Its starts at j=0 to cover the case when n=0, because then we get ${\displaystyle 0^{0}=1}$. --Ray andrew 21:48, 26 September 2007 (UTC) ## n < k ? I think the article could use a mention of whether or not n < k is allowed in ${\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}}$ Btyner (talk) 01:02, 21 January 2008 (UTC) ## Combinatorial argument for explicit formula for S(n,k) The article states that ${\displaystyle S(n,k)}$ is the number of k-partitions of an n-set and gives an explicit formula, but it is not clear how the explicit formula relates to partitions. Some exposition giving a combinatorial argument would be a welcome addition to the page. The ${\displaystyle 1/k!}$ in the formula is clear since permuting the subsets in the partition does not give a different partition. The other factors in each term are less obvious to me. Steve Checkoway (talk) 00:52, 6 August 2008 (UTC) There is a nice combinatorial argument for the recurrence (which I will add soon), but I've not seen one for the explicit formula. As is often the case in determining closed formulas, the solution lies in dry, analytic methods that rarely shed light on the underlying combinatorial structure. Austinmohr (talk) 07:44, 12 October 2008 (UTC) It's an uncomplicated application of inclusion/exclusion. That proof is quite comprehensible. Zaslav (talk) 03:05, 15 November 2013 (UTC) ## Cerial Box Problem? This is either not well defined or just wrong. Since both, boxes and prices are distinguishable, the events (1,1,2,3) and (2,2,1,3) should be counted individually. So the number of cases when you get all three prizes after buying 4 boxes is ${\displaystyle 3!S(4,3)}$, or in general ${\displaystyle k!S(n,k)}$. The problem is related to the Coupon collector's problem which asks how many boxes you need to collect all prizes (coupons), i.e. the stopping time for the renewal process ${\displaystyle k(n)}$ of having k different prizes after buying n boxes. —Preceding unsigned comment added by 133.65.54.177 (talk) 04:07, 3 March 2010 (UTC) Right, I'll remove that section. Marc van Leeuwen (talk) 13:25, 12 September 2011 (UTC) I dispute this as it doesn't matter which order you get the prizes – and they are often unordered, and because in fact cereal boxes in a given promotion are typically indistinguishable. Should the OP's complaint apply to the previous example, in which both lines of poetry and rhyming schemes are well-ordered? Regregex (talk) 14:20, 12 September 2011 (UTC) No there is an essential difference between trinket collection and rhyming schemes. In a rhyming scheme, only the matching of rhymes is counted, not the actual rhymes. Thus, if you take a rhyme using an ABAB rhyming scheme, and you (for instance) reverse the lines, then you don't get a different BABA rhyming scheme, but again an ABAB scheme. However, in trinket collection the trinkets themselves are distinguished and collected; otherwise there can be no notion of getting all the trinkets (compare how strange it would be to require a scheme to involve all possible rhymes). If there are trinkets A,B,C, then finding A,B,A,C is different from finding B,C,B,A (one even has a different multi-set of trinkets at the end) even though in both cases one (only) finds identical trinkets in the first and third cerial box. The section is unhelpful and should go. Marc van Leeuwen (talk) 15:56, 12 September 2011 (UTC) I intended to defend it's inclusion, but after looking at it more closely, I vote to remove the section. The Stirling numbers count the number of ways to arrange $n$ labeled balls into $k$ unlabeled buckets. This cereal box problem can either be interpreted as having labeled or unlabeled balls (i.e. cereal box openings) depending on whether you want to consider the order of the openings as important. In either scenario, however, the buckets (i.e. trinkets) are certainly labeled, and so the Stirling numbers are an inappropriate model. Austinmohr (talk) 19:31, 13 September 2011 (UTC) I'll respect the consensus, but I maintain that a Monty Hall-style error has occurred. Consider the case where you only know the number of prizes in advance, not their names or appearances (their location in a packshot or gallery might impose an order). You can tell when your collection is complete, but not that the prize in the first box is trinket B. Regregex (talk) 14:25, 16 September 2011 (UTC) ## Reduced variations The source listed as [1] here does not seem to be a peer-reviewed or expertly edited document. Has major typographic error even on first page. This should probably be removed awaiting further documentation. —Preceding unsigned comment added by 65.209.156.130 (talkcontribs) 2010-04-28T16:30:20Z There are two sources listed as [1], but to be clear the second one, the one in the section on "reduced" is pretty clearly self-published by the editor who inserted this section, in this edit. Similar concerns were raised at Talk:Distributive lattice#External links, so I removed the section as original research and conflict of interest. JackSchmidt (talk) 17:01, 28 April 2010 (UTC) I am one of the authors of the paper in question. The article had not yet been published when I originally uploaded it (though it had already been reviewed), but now appears in the Journal of Combinatorial Mathematics and Combinatorial Computing, Vol. 70 (2009), pg. 57 - 64. I am presently reverting the edit, but am open to discussing the issue further. Austinmohr (talk) 05:14, 6 May 2010 (UTC) I think that section in the article should be removed; it does not say much about the subject of this article. Marc van Leeuwen (talk) 13:27, 12 September 2011 (UTC) It is a natural generalization of the topic at hand, so I do not think it is off-topic, nor is it a particularly lengthy addition to the article. Austinmohr (talk) 20:16, 12 September 2011 (UTC) ## Generating function The identity given in this section is not really the generating function in the usual sense .... Summsumm (talk) 10:13, 17 September 2010 (UTC) ## Typo in definition The lead in the Definition section: I suppose, "nonempty" is correct, but "nondistinct" may have been originated as a typo. I suppose, it was originally meant as "nonempty and disjoint". Physis (talk) 10:02, 8 January 2011 (UTC) A partition of a set is a collection of disjoint subsets whose union is the original set. Thus, disjointness is very explicit in the definition of a partition. Slightly less explicit is the fact that the subsets are nondistinct. That is, there is no ordering placed on the subsets. For example, if the original set is {1, 2, 3}, the partition {{1},{2,3}} is considered to be the same as the partition {{2,3},{1}}. Perhaps it is less confusing to just remove the mention of distinctness altogether (since it is already a part of the definition of a partition). Austinmohr (talk) 04:21, 9 January 2011 (UTC) "nondistinct" is not the right adjective; if anything it would mean "identical" which is wrong. It should say "unlabeled", which is the conventional term in combinatorics in such situations. A surjection to {1,2, ..., k} would define a partition of the original set, but with the parts labeled (by their common image). See also Twelvefold way for more of this. Marc van Leeuwen (talk) 07:40, 14 September 2011 (UTC) ## Explicit formula for S(n,3) is wrong The explicit formula for S(n,3) is incorrect. I don't know what causes the problem, but the formula is translated by 2 from the correct one. — Preceding unsigned comment added by 77.254.110.81 (talk) 03:37, 13 May 2013 (UTC) Which formula? (There are several.) The ones I checked looked fine. --JBL (talk) 13:31, 13 May 2013 (UTC) ## Pronunciation Can someone comment on how you would read ${\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}}$ aloud? — Preceding unsigned comment added by Garfieldnate (talkcontribs) 14:22, 16 September 2014 (UTC) ## Citation for exact formula The citation for the exact formula in the Stirling_numbers_of_the_second_kind#Definitiondefinition section was to the following article: I deleted the citation because it does not support the claim at all. Sharp's paper doesn't contain the formula and, in fact, it doesn't even mention Stirling numbers. Since the paper is completely irrelevant to the article, I followed the advice on the Failed Verification template page and replaced the citation with "Citation needed" rather than "Failed Verification". Dricherby (talk) 12:01, 17 June 2016 (UTC) Well-caught. Here's one possible reference: Richard Stanley, EC1, second edition, equation (1.94a) (page 82 of the online edition http://math.mit.edu/~rstan/ec/ec1.pdf ). --JBL (talk) 16:23, 17 June 2016 (UTC)	2016-08-27 07:43:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 40, "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": 0.8640305995941162, "perplexity": 1035.9907281486464}, "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-2016-36/segments/1471982298551.3/warc/CC-MAIN-20160823195818-00037-ip-10-153-172-175.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/3203299/hamiltonian-cycle-with-n-vertex-graph	# Hamiltonian Cycle with n vertex graph Let a n-vertex graph such that every pair of not adjacent vertices a & b has degree(x) + degree(y) $$\geq$$ n. Show the graph contains a Hamiltonian cycle. By dirac's thm, a simple graph with n vertices (n ≥ 3) is Hamiltonian if every vertex has degree n / 2 or greater. How would i relate this thm to the question? • Try to mimic the proof of Dirac's theorem using the given weaker degree condition. The key point is finding a vertex in the middle of the longest path that both end points are adjacent to. – Jeremy Dover Apr 26 '19 at 13:50 • how do i sub the new conidition to the proof of dirac in the step of finding vertex j in the longest path? – james black Apr 26 '19 at 14:51 • That is the crux of the problem. See how the condition $deg(v) \ge \frac{n}{2}$ is used in the proof of Dirac's theorem to ensure two sets of vertices intersect, and try to extend it to the weaker condition in the problem. – Jeremy Dover Apr 26 '19 at 15:01 I guess non-adjacent vertices should be named $$x$$ and $$y$$. Then you ask to prove Ore’s theorem.	2020-12-03 17:36:10	{"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6406251788139343, "perplexity": 329.5135124122957}, "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-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00713.warc.gz"}
https://www.projecteuclid.org/euclid.pjm/1102720111	## Pacific Journal of Mathematics ### Compact connected Lie groups acting on simply connected $4$-manifolds. Hae Soo Oh #### Article information Source Pacific J. Math., Volume 109, Number 2 (1983), 425-435. Dates First available in Project Euclid: 8 December 2004 https://projecteuclid.org/euclid.pjm/1102720111 Mathematical Reviews number (MathSciNet) MR721931 Zentralblatt MATH identifier 0548.57020 Subjects Primary: 57S15: Compact Lie groups of differentiable transformations Secondary: 57S25: Groups acting on specific manifolds #### Citation Oh, Hae Soo. Compact connected Lie groups acting on simply connected $4$-manifolds. Pacific J. Math. 109 (1983), no. 2, 425--435. https://projecteuclid.org/euclid.pjm/1102720111 #### References • [A] J. F. Adams, Lectures on Lie Groups, W. A. Benjamin, (1969). • [B] G. E. Bredon, Introduction to Compact Transformation Groups, Academic Press, (1972). • [E] L. P. Eisenhart, Riemannian Geometry, Princeton University Press, (1949). • [F] R. Fintushel, Circle actions on simply connected 4-manifolds, Trans. Amer. Math. Soc, 230(1977), 147-171. • [F7] R. Fintushel, Classification of Circle Actions on A-Manifolds, Trans. Amer. Math. Soc, 242(1978), 377-390. • [Ma] L. N. Mann, Gaps in the Dimensions of Compact Transformation Groups, pro- ceedings of the conference on Transformation Groups, Springer-Verlag, (1967), 293-296. • [M-Z] D. Montgomery and L. Zippin, Topological Transformation Groups, Interscience Publishers, (1955). • [Mo] P. S. Mostert, On a compact Lie group acting on a manifold, Annals, of Math., 65 (1957), 447-455; Errata, Annals of Math., 66 (1957), 589, Math. Annalen, 167 (1966), 224. • [O-R] P. Orlik and F. Raymond, Actions of the torus on A-manifolds I, Trans. Amer. Math. Soc, 152 (1970), 531-559. • [P] J. Pak, Actions of the torus T" on (n + \)-manifolds M"+\Pacific J. Math., 44 (1973), 671-674. • [R] R. W. Richardson, Groups acting on the 4-sphere, Illinois J. Math., 5 (1961), 474-485. • [Wa] H. C. Wang, On Finsler spaces with completely integrable equations of killing, J. London Math. Soc, 22 (1947), 5-9. • [Wo] J. A. Wolf, Spaces of constant curvature, Publish or Perish, (1973).	2019-08-17 11:41:27	{"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": 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": 0.4772357642650604, "perplexity": 4236.34865623198}, "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/1566027312128.3/warc/CC-MAIN-20190817102624-20190817124624-00273.warc.gz"}
http://hackage.haskell.org/package/hgeometry-0.10.0.0/docs/Algorithms-Geometry-WellSeparatedPairDecomposition-WSPD.html	hgeometry-0.10.0.0: Geometric Algorithms, Data structures, and Data types. Copyright (C) Frank Staals see the LICENSE file Frank Staals None Haskell2010 Algorithms.Geometry.WellSeparatedPairDecomposition.WSPD Description Algorithm to construct a well separated pair decomposition (wspd). Synopsis # Documentation fairSplitTree :: (Fractional r, Ord r, Arity d, 1 <= d, Show r, Show p) => NonEmpty (Point d r :+ p) -> SplitTree d p r () Source # Construct a split tree running time: $$O(n \log n)$$ wellSeparatedPairs :: (Floating r, Ord r, Arity d, Arity (d + 1)) => r -> SplitTree d p r a -> [WSP d p r a] Source # Given a split tree, generate the Well separated pairs running time: $$O(s^d n)$$ # Building the split tree fairSplitTree' :: (Fractional r, Ord r, Arity d, 1 <= d, Show r, Show p) => Int -> Vector d (PointSeq d (Idx :+ p) r) -> BinLeafTree Int (Point d r :+ p) Source # Given the points, sorted in every dimension, recursively build a split tree The algorithm works in rounds. Each round takes O(n) time, and halves the number of points. Thus, the total running time is O(n log n). The algorithm essentially builds a path in the split tree; at every node on the path that we construct, we split the point set into two sets (L,R) according to the longest side of the bounding box. The smaller set is "assigned" to the current node and set asside. We continue to build the path with the larger set until the total number of items remaining is less than n/2. To start the next round, each node on the path needs to have the points assigned to that node, sorted in each dimension (i.e. the Vector (PointSeq))'s. Since we have the level assignment, we can compute these lists by traversing each original input list (i.e. one for every dimension) once, and partition the points based on their level assignment. distributePoints :: (Arity d, Show r, Show p) => Int -> Vector (Maybe Level) -> Vector d (PointSeq d (Idx :+ p) r) -> Vector (Vector d (PointSeq d (Idx :+ p) r)) Source # Assign the points to their the correct class. The Nothing class is considered the last class transpose :: Arity d => Vector d (Vector a) -> Vector (Vector d a) Source # Arguments :: Int number of classes -> Vector (Maybe Level) level assignment -> PointSeq d (Idx :+ p) r input points -> Vector (PointSeq d (Idx :+ p) r) Assign the points to their the correct class. The Nothing class is considered the last class reIndexPoints :: (Arity d, 1 <= d) => Vector d (PointSeq d (Idx :+ p) r) -> Vector d (PointSeq d (Idx :+ p) r) Source # Given a sequence of points, whose index is increasing in the first dimension, i.e. if idx p < idx q, then p[0] < q[0]. Reindex the points so that they again have an index in the range [0,..,n'], where n' is the new number of points. running time: O(n' * d) (more or less; we are actually using an intmap for the lookups) alternatively: I can unsafe freeze and thaw an existing vector to pass it along to use as mapping. Except then I would have to force the evaluation order, i.e. we cannot be in reIndexPoints for two of the nodes at the same time. so, basically, run reIndex points in ST as well. type RST s = ReaderT (MVector s (Maybe Level)) (ST s) Source # ST monad with access to the vector storign the level of the points. Arguments :: (Fractional r, Ord r, Arity d, Show r, Show p) => Int Number of items we need to collect -> Int Number of items we collected so far -> Vector d (PointSeq d (Idx :+ p) r) -> Level next level to use -> [Level] Levels used so far -> RST s (NonEmpty Level) Assigns the points to a level. Returns the list of levels used. The first level in the list is the level assigned to the rest of the nodes. Their level is actually still set to Nothing in the underlying array. compactEnds :: Arity d => Vector d (PointSeq d (Idx :+ p) r) -> RST s (Vector d (PointSeq d (Idx :+ p) r)) Source # Remove already assigned pts from the ends of all vectors. assignLevel :: (c :+ (Idx :+ p)) -> Level -> RST s () Source # Assign level l to point p levelOf :: (c :+ (Idx :+ p)) -> RST s (Maybe Level) Source # Get the level of a point hasLevel :: (c :+ (Idx :+ p)) -> RST s Bool Source # Test if the point already has a level assigned to it. compactEnds' :: PointSeq d (Idx :+ p) r -> RST s (PointSeq d (Idx :+ p) r) Source # Remove allready assigned points from the sequence pre: there are points remaining Arguments :: (Ord r, Arity d, Show r, Show p) => Int the dimension we are in, i.e. so that we know which coordinate of the point to compare -> PointSeq d (Idx :+ p) r -> r the mid point -> RST s (PointSeq d (Idx :+ p) r, PointSeq d (Idx :+ p) r) Given the points, ordered by their j^th coordinate, split the point set into a "left" and a "right" half, i.e. the points whose j^th coordinate is at most the given mid point m, and the points whose j^th coordinate is larger than m. We return a pair (Largest set, Smallest set) fi ndAndCompact works by simultaneously traversing the points from left to right, and from right to left. As soon as we find a point crossing the mid point we stop and return. Thus, in principle this takes only O(\|Smallest set\|) time. running time: O(\|Smallest set\|) + R, where R is the number of old points (i.e. points that should have been removed) in the list. widestDimension :: (Num r, Ord r, Arity d) => Vector d (PointSeq d p r) -> Int Source # Find the widest dimension of the point set pre: points are sorted according to their dimension widths :: (Num r, Arity d) => Vector d (PointSeq d p r) -> Vector d r Source # extends :: Arity d => Vector d (PointSeq d p r) -> Vector d (Range r) Source # get the extends of the set of points in every dimension, i.e. the left and right boundaries. pre: points are sorted according to their dimension # Finding Well Separated Pairs findPairs :: (Floating r, Ord r, Arity d, Arity (d + 1)) => r -> SplitTree d p r a -> SplitTree d p r a -> [WSP d p r a] Source # Arguments :: (Arity d, Arity (d + 1), Fractional r, Ord r) => r separation factor -> SplitTree d p r a -> SplitTree d p r a -> Bool Test if the two sets are well separated with param s boxBox :: (Fractional r, Ord r, Arity d, Arity (d + 1)) => r -> Box d p r -> Box d p r -> Bool Source # Test if the two boxes are sufficiently far appart # Alternative def if wellSeparated that uses fractional areWellSeparated' :: (Floating r, Ord r, Arity d) => r -> SplitTree d p r a -> SplitTree d p r a -> Bool Source # boxBox1 :: (Floating r, Ord r, Arity d) => r -> Box d p r -> Box d p r -> Bool Source # # Helper stuff maxWidth :: (Arity d, Num r) => SplitTree d p r a -> r Source # Computes the maximum width of a splitTree bbOf :: Ord r => SplitTree d p r a -> Box d () r Source # Computes the bounding box of a split tree ix' :: (Arity d, KnownNat d) => Int -> Lens' (Vector d a) a Source # Turn a traversal into lens dropIdx :: (core :+ (t :+ extra)) -> core :+ extra Source #	2020-03-31 00:04:13	{"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": 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": 0.5297203063964844, "perplexity": 7388.274672231315}, "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-2020-16/segments/1585370497309.31/warc/CC-MAIN-20200330212722-20200331002722-00130.warc.gz"}
http://ibmmainframes.com/post-343029.html	Portal \| Manuals \| References \| Downloads \| Info \| Programs \| JCLs \| Mainframe wiki \| Quick Ref Author Message anatol Active User Joined: 20 May 2010 Posts: 121 Posted: Thu Mar 15, 2018 3:43 pm Post subject: Replace last name by first name Hello, I have file with users Last name ( position 1-10) First name ( position 11-21) and second file, where in field positions 40 - 200 may be user Last name in any of this field position. Could you please give me idea to create output file where Last name from second file changed to the user first name. Output file should have only records from second file that was changed. sergeyken Active Member Joined: 29 Apr 2008 Posts: 609 Location: Maryland Posted: Thu Mar 15, 2018 7:43 pm Post subject: Try to do something YOURSELF, then ask here for any advice. This forum is not supposed to MAKE YOUR JOB instead of you. sergeyken Active Member Joined: 29 Apr 2008 Posts: 609 Location: Maryland Posted: Thu Mar 15, 2018 7:52 pm Post subject: Re: Replace last name by first name anatol wrote: Could you please give me idea to create output file where Last name from second file changed to the user first name. The idea is, using the parameters PARSE=, and BUILD= Another thing is - using SS to perform string comparison within varying positions. enrico-sorichetti Senior Member Joined: 14 Mar 2007 Posts: 10632 Location: italy Posted: Thu Mar 15, 2018 9:48 pm Post subject: Reply to: Replace last name by first name changing data in some dataset based on the content of another dataset can be done easily enough with a JOINKEY operation but JOINKEY needs the keys in a fixed position since one of the keys is not in a FIXED position, the approach is impossible anatol Active User Joined: 20 May 2010 Posts: 121 Posted: Thu Mar 15, 2018 9:59 pm Post subject: Reply to: Replace last name by first name Thank you all. Rohit Umarjikar Senior Member Joined: 21 Sep 2010 Posts: 2299 Location: NY,USA Posted: Thu Mar 15, 2018 11:38 pm Post subject: You haven't told us clearly at first place as to what exactly is expected with sample data. Do you need to match two DS and if last name from second ds= first name from first ds then write else skip? enrico-sorichetti Senior Member Joined: 14 Mar 2007 Posts: 10632 Location: italy Posted: Fri Mar 16, 2018 1:39 am Post subject: Reply to: Replace last name by first name no need for any detailed description of the data Quote: ... where in field positions 40 - 200 may be user Last name in any of this field position. ... that' s the show stopper for a joinkey solution sergeyken Active Member Joined: 29 Apr 2008 Posts: 609 Location: Maryland Posted: Fri Mar 16, 2018 2:13 am Post subject: Re: Reply to: Replace last name by first name enrico-sorichetti wrote: but JOINKEY needs the keys in a fixed position since one of the keys is not in a FIXED position, the approach is impossible Using PARSE=, and BUILD= before JOINKEYS may help with this approach, too. Also other approaches are available. enrico-sorichetti Senior Member Joined: 14 Mar 2007 Posts: 10632 Location: italy Posted: Fri Mar 16, 2018 2:40 am Post subject: Reply to: Replace last name by first name parse on what ? sergeyken Active Member Joined: 29 Apr 2008 Posts: 609 Location: Maryland Posted: Fri Mar 16, 2018 9:30 pm Post subject: Re: Reply to: Replace last name by first name enrico-sorichetti wrote: parse on what ? in field positions 40 - 200, using INREC for one of joined files before joining them anatol Active User Joined: 20 May 2010 Posts: 121	2019-11-16 23:45:35	{"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": 0.9241878390312195, "perplexity": 8606.50086454247}, "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/1573496668772.53/warc/CC-MAIN-20191116231644-20191117015644-00145.warc.gz"}
https://mathematica.stackexchange.com/questions/120334/collect-all-inputs-from-documentation-center	# Collect all inputs from Documentation Center Is it possible to collect all inputs programmatically from every function in the Documentation Center in a list and wrap them in a HoldComplete? My case: With some fundamental functions modified, I would like run a test of all functions to see what function will fail to execute. • Use NotebookImport on Documentation directory restricting it to InputCells. It is quite slow but you only have to do it once. – Kuba Jul 9 '16 at 9:35 • keep in mind that some examples are meant to fail to show some issues or are just Messages related. – Kuba Jul 9 '16 at 9:36 • @Kuba thanks for the hint. In this case I may try to avoid importing the inputs from "possible issues", etc – vapor Jul 9 '16 at 9:38 • You can but I bet it is not the only section where messages are generated. – Kuba Jul 9 '16 at 9:39 You may use the "DocumentationExampleInputs" property of the "WolframLanguageSymbol" entity or WolframLanguageData function. With "WolframLanguageSymbol" entity: EntityValue[Entity["WolframLanguageSymbol", "Round"], "DocumentationExampleInputs"] For more than one at a time you can use "EntityAssociation" to have the entities as keys. EntityValue[RandomEntity["WolframLanguageSymbol", 2], "DocumentationExampleInputs", "EntityAssociation"] Similarly for WolframLanguageData function: WolframLanguageData["Round", "DocumentationExampleInputs"] EntityValue[WolframLanguageData[], "DocumentationExampleInputs",	2020-06-05 12:29:29	{"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": 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": 0.3330574035644531, "perplexity": 2375.84030190669}, "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/1590348500712.83/warc/CC-MAIN-20200605111910-20200605141910-00003.warc.gz"}
https://www.transtutors.com/questions/the-records-at-the-end-of-january-of-the-current-year-for-young-company-showed-the-f-2568144.htm	# The records at the end of January of the current year for Young Company showed the following for ... The records at the end of January of the current year for Young Company showed the following for a particular kind of merchandise Beginning Inventory at FIFO: 15 Units @$19 =$285 Beginning inventory at LIFO: 15 Units @$15 =$225 Compute the inventory turnover ratio for the month of January under the FIFO and LIFO inventory costing methods. Which costing method is the more accurate indicator of the efficiency of inventory management? FIFO LIFO No accuracy difference	2018-10-22 23:24:36	{"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": 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": 0.17682918906211853, "perplexity": 3306.7645493188143}, "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-2018-43/segments/1539583515555.58/warc/CC-MAIN-20181022222133-20181023003633-00406.warc.gz"}
http://www.theinfolist.com/html/ALL/s/transitive_relation.html	TheInfoList In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry), and quantities and their changes (cal ... , a relation on a set is transitive if, for all elements , , in , whenever relates to and to , then also relates to . Each partial order upright=1.15, Fig.1 The set of all subsets of a three-element set \, ordered by set inclusion">inclusion Inclusion or Include may refer to: Sociology * Social inclusion, affirmative action to change the circumstances and habits that leads to s ... as well as each equivalence relation In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ... needs to be transitive. # Definition A homogeneous relation Homogeneity and heterogeneity are concepts often used in the sciences and statistics Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data. In applying statistics to a sc ... on the set is a ''transitive relation'' if, :for all , if and , then . Or in terms of first-order logic First-order logic—also known as predicate logic, quantificational logic, and first-order predicate calculus—is a collection of formal system A formal system is an used for inferring theorems from axioms according to a set of rules. These rul ... : :$\forall a,b,c \in X: \left(aRb \wedge bRc\right) \Rightarrow aRc,$ where is the infix notation Infix notation is the notation commonly used in arithmetical and logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, label=none, lit=possessed of reason, intellectual, dialectical, argumentative, translit=logikḗ) ... for . # Examples As a nonmathematical example, the relation "is an ancestor of" is transitive. For example, if Amy is an ancestor of Becky, and Becky is an ancestor of Carrie, then Amy, too, is an ancestor of Carrie. On the other hand, "is the birth parent of" is not a transitive relation, because if Alice is the birth parent of Brenda, and Brenda is the birth parent of Claire, then Alice is not the birth parent of Claire. What is more, it is antitransitive In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ... : Alice can ''never'' be the birth parent of Claire. "Is greater than", "is at least as great as", and "is equal to" ( equality) are transitive relations on various sets, for instance, the set of real numbers or the set of natural numbers: : whenever ''x'' > ''y'' and ''y'' > ''z'', then also ''x'' > ''z'' : whenever ''x'' ≥ ''y'' and ''y'' ≥ ''z'', then also ''x'' ≥ ''z'' : whenever ''x'' = ''y'' and ''y'' = ''z'', then also ''x'' = ''z''. More examples of transitive relations: * "is a subset In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ... of" (set inclusion, a relation on sets) * "divides" ( divisibility In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ... , a relation on natural numbers) * "implies" ( implication, symbolized by "⇒", a relation on proposition In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, lab ... s) Examples of non-transitive relations: * "is the successor Successor is someone who, or something which succeeds or comes after (see success (disambiguation), success and Succession (disambiguation), succession) Film and TV * The Successor (film), ''The Successor'' (film), a 1996 film including Laura Girli ... of" (a relation on natural numbers) * "is a member of the set" (symbolized as "∈") * "is perpendicular In elementary geometry Geometry (from the grc, γεωμετρία; ' "earth", ' "measurement") is, with , one of the oldest branches of . It is concerned with properties of space that are related with distance, shape, size, and relativ ... to" (a relation on lines in Euclidean geometry Euclidean geometry is a mathematical system attributed to Alexandria Alexandria ( or ; ar, الإسكندرية ; arz, اسكندرية ; Coptic Coptic may refer to: Afro-Asia * Copts, an ethnoreligious group mainly in the area of modern ... ) The empty relation Empty may refer to: ‍ Music Albums * ''Empty'' (God Lives Underwater album) or the title song, 1995 * ''Empty'' (Nils Frahm album), 2020 * ''Empty'' (Tait album) or the title song, 2001 Songs * "Empty" (The Click Five song), 2007 * ... on any set $X$ is transitive because there are no elements $a,b,c \in X$ such that $aRb$ and $bRc$, and hence the transitivity condition is vacuously trueIn mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ha ... . A relation containing only one ordered pair In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ... is also transitive: if the ordered pair is of the form $\left(x, x\right)$ for some $x \in X$ the only such elements $a,b,c \in X$ are $a=b=c=x$, and indeed in this case $aRc$, while if the ordered pair is not of the form $\left(x, x\right)$ then there are no such elements $a,b,c \in X$ and hence $R$ is vacuously transitive. # Properties ## Closure properties * The converse Converse may refer to: Mathematics and logic * Converse (logic), the result of reversing the two parts of a categorical or implicational statement Converse implication, the converse of a material implication Converse nonimplication, a logical ... (inverse) of a transitive relation is always transitive. For instance, knowing that "is a subset In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ... of" is transitive and "is a superset In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ... of" is its converse, one can conclude that the latter is transitive as well. * The intersection of two transitive relations is always transitive. For instance, knowing that "was born before" and "has the same first name as" are transitive, one can conclude that "was born before and also has the same first name as" is also transitive. * The union of two transitive relations need not be transitive. For instance, "was born before or has the same first name as" is not a transitive relation, since e.g. Herbert Hoover Herbert Clark Hoover (August 10, 1874 – October 20, 1964) was an American politician and engineer who served as the 31st president of the United States The president of the United States (POTUS) is the head of state and head of gove ... is related to Franklin D. Roosevelt Franklin Delano Roosevelt (, ; January 30, 1882April 12, 1945), often referred to by his initials FDR, was an American politician who served as the 32nd president of the United States from 1933 until his death in 1945. A member of the De ... , which is in turn related to Franklin Pierce Franklin Pierce (November 23, 1804October 8, 1869) was the 14th president of the United States The president of the United States (POTUS) is the and of the . The president directs the of the and is the of the . The power of the ... , while Hoover is not related to Franklin Pierce. * The complement of a transitive relation need not be transitive. For instance, while "equal to" is transitive, "not equal to" is only transitive on sets with at most one element. ## Other properties A transitive relation is asymmetric if and only if it is irreflexive In mathematics, a homogeneous binary relation ''R'' over a set (mathematics), set ''X'' is reflexive if it relates every element of ''X'' to itself. An example of a reflexive relation is the relation "equality (mathematics), is equal to" on the se ... . A transitive relation need not be reflexive. When it is, it is called a preorder In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry), and quantities a ... . For example, on set ''X'' = : * ''R'' = is reflexive, but not transitive, as the pair (1,2) is absent, * ''R'' = is reflexive as well as transitive, so it is a preorder, * ''R'' = is reflexive as well as transitive, another preorder. # Transitive extensions and transitive closure Let be a binary relation on set . The ''transitive extension'' of , denoted , is the smallest binary relation on such that contains , and if and then . For example, suppose is a set of towns, some of which are connected by roads. Let be the relation on towns where if there is a road directly linking town and town . This relation need not be transitive. The transitive extension of this relation can be defined by if you can travel between towns and by using at most two roads. If a relation is transitive then its transitive extension is itself, that is, if is a transitive relation then . The transitive extension of would be denoted by , and continuing in this way, in general, the transitive extension of would be . The ''transitive closure'' of , denoted by or is the set union of , , , ... . The transitive closure of a relation is a transitive relation. The relation "is the birth parent of" on a set of people is not a transitive relation. However, in biology the need often arises to consider birth parenthood over an arbitrary number of generations: the relation "is a birth ancestor of" ''is'' a transitive relation and it is the transitive closure of the relation "is the birth parent of". For the example of towns and roads above, provided you can travel between towns and using any number of roads. # Relation properties that require transitivity * Preorder In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry), and quantities a ... – a reflexive and transitive relation * Partial order upright=1.15, Fig.1 The set of all subsets of a three-element set \, ordered by set inclusion">inclusion Inclusion or Include may refer to: Sociology * Social inclusion, affirmative action to change the circumstances and habits that leads to s ... – an antisymmetric preorder * Total preorder The 13 possible strict weak orderings on a set of three elements . The only total orders are shown in black. Two orderings are connected by an edge if they differ by a single dichotomy. In mathematics Mathematics (from Ancient Greek, Gre ... – a connected (formerly called total) preorder * Equivalence relation In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ... – a symmetric Symmetry (from Greek συμμετρία ''symmetria'' "agreement in dimensions, due proportion, arrangement") in everyday language refers to a sense of harmonious and beautiful proportion and balance. In mathematics, "symmetry" has a more pre ... preorder * Strict weak ordering The 13 possible strict weak orderings on a set of three elements . The only total orders are shown in black. Two orderings are connected by an edge if they differ by a single dichotomy. In mathematics Mathematics (from Ancient Greek, Gre ... – a strict partial order in which incomparability is an equivalence relation * Total ordering In mathematics, a total order, simple order, linear order, connex order, or full order is a binary relation on some Set (mathematics), set X, which is Antisymmetric relation, antisymmetric, Transitive relation, transitive, and a connex relation. ... – a connected (total), antisymmetric, and transitive relation # Counting transitive relations No general formula that counts the number of transitive relations on a finite set is known. However, there is a formula for finding the number of relations that are simultaneously reflexive, symmetric, and transitive – in other words, equivalence relation In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ... s – , those that are symmetric and transitive, those that are symmetric, transitive, and antisymmetric, and those that are total, transitive, and antisymmetric. Pfeiffer has made some progress in this direction, expressing relations with combinations of these properties in terms of each other, but still calculating any one is difficult. See also Brinkmann and McKay (2005). Mala showed that no polynomial with integer coefficients can represent a formula for the number of transitive relations on a set, and found certain recursive relations that provide lower bounds for that number. He also showed that that number is a polynomial of degree two if contains exactly two . # Related properties A relation ''R'' is called '' intransitive In grammar In linguistics, the grammar (from Ancient Greek ''grammatikḗ'') of a natural language is its set of structure, structural constraints on speakers' or writers' composition of clause (linguistics), clauses, phrases, and words. Th ... '' if it is not transitive, that is, if ''xRy'' and ''yRz'', but not ''xRz'', for some ''x'', ''y'', ''z''. In contrast, a relation ''R'' is called '' antitransitive In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ... '' if ''xRy'' and ''yRz'' always implies that ''xRz'' does not hold. For example, the relation defined by ''xRy'' if ''xy'' is an even number In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no ge ... is intransitive, but not antitransitive. The relation defined by ''xRy'' if ''x'' is even and ''y'' is odd is both transitive and antitransitive. The relation defined by ''xRy'' if ''x'' is the successor Successor is someone who, or something which succeeds or comes after (see success (disambiguation), success and Succession (disambiguation), succession) Film and TV * The Successor (film), ''The Successor'' (film), a 1996 film including Laura Girli ... number of ''y'' is both intransitive and antitransitive. Unexpected examples of intransitivity arise in situations such as political questions or group preferences. Generalized to stochastic versions ('' stochastic transitivityStochastic transitivity models are stochastic Stochastic () refers to the property of being well described by a random In common parlance, randomness is the apparent or actual lack of pattern or predictability in events. A random sequence ... ''), the study of transitivity finds applications of in decision theory Decision theory (or the theory of choice not to be confused with choice theory) is the study of an agent's choices. Decision theory can be broken into two branches: normative Normative generally means relating to an evaluative standard. Normativi ... , psychometrics Psychometrics is a field of study within psychology Psychology is the scientific Science () is a systematic enterprise that builds and organizes knowledge Knowledge is a familiarity or awareness, of someone or something, s ... and utility models. A '' quasitransitive relation The mathematical notion of quasitransitivity is a weakened version of transitive relation, transitivity that is used in social choice theory and microeconomics. Informally, a relation is quasitransitive if it is symmetric relation, symmetric for som ... '' is another generalization; it is required to be transitive only on its non-symmetric part. Such relations are used in social choice theory Social choice theory or social choice is a theoretical A theory is a rational Rationality is the quality or state of being rational – that is, being based on or agreeable to reason Reason is the capacity of consciously making sense o ... or microeconomics Microeconomics is a branch of mainstream economics Mainstream economics is the body of knowledge, theories, and models of economics, as taught by universities worldwide, that are generally accepted by economists as a basis for discussion. Als ... . # See also * Transitive reductionIn mathematics, a transitive reduction of a directed graph ''D'' is another directed graph with the same vertices and as few edges as possible, such that for all pairs of vertices ''v'', ''w'' a (directed) path from ''v'' to ''w'' in ''D'' exists if ... * Intransitive diceA set of dice is intransitive (or nontransitive) if it contains three dice, ''A'', ''B'', and ''C'', with the property that ''A'' rolls higher than ''B'' more than half the time, and ''B'' rolls higher than ''C'' more than half the time, but it is ... * Rational choice theory Rational choice theory refers to a set of guidelines that help understand economic and social behaviour. The theory postulates that an individual will perform a cost-benefit analysis to determine whether an option is right for them. It also sugge ... * Hypothetical syllogism In classical logic Classical logic (or standard logic) is the intensively studied and most widely used class of deductive logic Deductive reasoning, also deductive logic, is the process of reasoning from one or more statements (premises) to reach ... — transitivity of the material conditional # References * * * Gunther Schmidt Gunther Schmidt (born 1939, Rüdersdorf) is a Germans, German mathematician who works also in informatics. Life Schmidt began studying Mathematics in 1957 at Göttingen University. His academic teachers were in particular Kurt Reidemeister, Wilhe ... , 2010. ''Relational Mathematics''. Cambridge University Press, . * # External links * {{springer, title=Transitivity, id=p/t093810 Transitivity in Action at cut-the-knot Alexander Bogomolny (January 4, 1948 July 7, 2018) was a Soviet Union, Soviet-born Israeli American mathematician. He was Professor Emeritus of Mathematics at the University of Iowa, and formerly research fellow at the Moscow Institute of Electron ... Binary relations Elementary algebra	2022-09-30 06:02:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "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": 0.7317402362823486, "perplexity": 2542.174455396958}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2022-40/segments/1664030335444.58/warc/CC-MAIN-20220930051717-20220930081717-00426.warc.gz"}
https://codeforces.cc/blog/entry/95287	### oleksg's blog By oleksg, history, 5 weeks ago, • +8 » 5 weeks ago, # \| ← Rev. 3 → +36 It's n times a partial sum of [harmonic series](https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Partial_sums), so it's nlogn. • » » 5 weeks ago, # ^ \| 0 Oh, I didn't notice that. Thanks for the answer! • » » » 5 weeks ago, # ^ \| 0 There is also proofs for why partial sum of harmonic series is like that. » 5 weeks ago, # \| ← Rev. 4 → +14 If you meant that you want to get the complexity of computing this (with floor division), you can do so in $O(\sqrt n)$ by noting that there can be at most $2 \sqrt n$ distinct terms in the sum. • » » 5 weeks ago, # ^ \| +8 No, I was specifically looking at this problem — https://codeforces.cc/contest/1558/problem/B. I solved it with that complexity with a seemingly simpler solution than what the editorial offered. At least I think it was that complexity. • » » 5 weeks ago, # ^ \| ← Rev. 2 → 0 Also you can also calculate the floor sum function in $O(\sqrt[3]{n})$One of the way is to approximate $O(\sqrt[3]{n})$ slopes in the function $y = f(x) = \left \lfloor \frac{n}{x} \right \rfloor$The other way is to split into $O(\sqrt[3]{n})$ parts of $[A, 2A)$ • » » 5 weeks ago, # ^ \| +8 yes, my favorite Dirichlet hyperbola method » 5 weeks ago, # \| 0 That's n(1/2 + 1/3 + ...) 1/2 + 1/3 + ... + 1/n is the harmonic series and equals with aproximative log(n). So the complexitate will be O(nlog(n)) » 5 weeks ago, # \| ← Rev. 2 → +46 Here's how you can approximate that it is $nlogn$: $\frac{n}{1} + \frac{n}{2} + ... + \frac{n}{n} =$ $n(\frac{1}{1} + \frac{1}{2} +\frac{1}{3}+ \frac{1}{4} + \frac{1}{5} + \frac{1}{6}+... +\frac{1}{n} \leq$ $\leq n(\frac{1}{1} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} ... + \frac{1}{2^k} =$ $= n(\frac{1}{1} + \frac{1}{2^1} + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^2} + \frac{1}{2^2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^3} ... + \frac{1}{2^k}$ $2^k \approx n \Rightarrow k \approx log_2n$ If we add all the components with the same power of 2, we get $1$. Because there are $\approx log_2n$ powers, we get that the sum is $\approx nlog_2n$ » 5 weeks ago, # \| ← Rev. 2 → 0 I didn't actually do this myself, but there's an interesting comment about bounding the harmonic numbers here. Jensen's inequality + some elementary calculus should do the trick (it's quite nice)You can always bound with trapezoidal rule and some calc too, if you prefer » 5 weeks ago, # \| 0 galen_colin explained it here here	2021-10-28 01:37:21	{"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": 0.535046398639679, "perplexity": 1001.9858691998371}, "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-2021-43/segments/1634323588246.79/warc/CC-MAIN-20211028003812-20211028033812-00145.warc.gz"}
https://www.physicsforums.com/threads/centripetal-force-acc-problem.158319/	# Centripetal force/acc problem ## Homework Statement What angular velocity would cause a centripetal acceleration of 3 g's if the radius were 2.5 m? i have a test 2moro on this stuff... yea i need to know =D thx ## Homework Equations i guess m(v^2)/r or mW^2r no idea. srry ## Homework Equations i guess m(v^2)/r or mW^2r What is the W? And what do you calculate with mW^2r? W is rotational velocity.. sryy for not making that clear. W^2R= Centripetal Acceleration I just plugged numbers in 29.4= W^2(2.5) and i got 3.4radian/sec for W. this looks 2 easy... if someoen could verify please =D Yes, very good, your answer is correct. But you should also write down the units, for example acceleration is in m/s^2 and radius in m. This allows you to check whether your result is correct. For example if you calculated W as something with 1/s^2, you know you made a mistake because the unit must be 1/s (or radian/sec). Last edited: thanks tons edgardo!	2022-05-21 18:20:57	{"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": 0.8297269344329834, "perplexity": 5169.431111486022}, "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/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00345.warc.gz"}
https://socratic.org/questions/find-the-volume-of-the-solid-by-using-cylindrical-shells#621229	# Find the volume of the solid by using cylindrical shells? May 28, 2018 $\textcolor{b l u e}{V = 2 \pi {\int}_{0}^{1} \left(y\right) \left(\sqrt{y} - {y}^{3}\right) \cdot \mathrm{dy} = \frac{2 \cdot \pi}{5} {\left(u n i t e\right)}^{3}}$ #### Explanation: the volume by cylindrical shell method when the curve rotating about the $\text{x-axis}$ is given by: $\textcolor{red}{V = 2 \pi {\int}_{a}^{b} \left(y\right) \left({x}_{2} - {x}_{1}\right) \cdot \mathrm{dy}}$ ${x}_{2} = \pm \sqrt{y}$ but the area enclosed by the curve lies in the first quadrant. so ${x}_{2} = \sqrt{y}$ ${x}_{1} = {y}^{3}$ lets find the interval of integral. $\sqrt{y} = {y}^{3}$ square both sides ${y}^{6} = y \Rightarrow {y}^{6} - y = 0$ $y \cdot \left({y}^{5} - 1\right) = 0$ $y = 0 \mathmr{and} y = 1$ the interval of the integral $y \in \left[0 , 1\right]$ now let set up the integral: $V = 2 \pi {\int}_{0}^{1} \left(y\right) \left(\sqrt{y} - {y}^{3}\right) \cdot \mathrm{dy}$ $V = 2 \pi {\int}_{0}^{1} \left({y}^{\frac{3}{2}} - {y}^{4}\right) \cdot \mathrm{dy} = {\left[- \frac{2 \cdot \pi \cdot \left({y}^{5} - 2 \cdot {y}^{\frac{5}{2}}\right)}{5}\right]}_{0}^{1}$ $= \frac{2 \cdot \pi}{5} {\left(u n i t e\right)}^{3}$ show below the region revolving (shaded region) : show the link below that will help you to understand how to find the volume by cylindrical shell method: cylindrical shell method	2021-10-24 12:54:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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": 0.91490638256073, "perplexity": 678.836379185008}, "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/1634323585997.77/warc/CC-MAIN-20211024111905-20211024141905-00132.warc.gz"}
http://clay6.com/qa/26380/for-a-perfect-correlation-between-the-variables-x-and-y-the-line-of-regress	# For a perfect correlation between the variables $x$ and $y$ the line of regression is $ax+by+c=0$ where $a,b,c > 0$ then $r$ is $\begin {array} {1 1} (A)\;0 & \quad (B)\;-1 \\ (C)\;1 & \quad (D)\;None \: of \: these \end {array}$ $ax+by+c=0$ has slope $-\large\frac{a}{b} <0$ So the two variables are in perfect negative correlation Hence $r = -1$ Ans : (B)	2017-09-24 05:08:14	{"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": 0.646418571472168, "perplexity": 405.48737543854054}, "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-2017-39/segments/1505818689874.50/warc/CC-MAIN-20170924044206-20170924064206-00046.warc.gz"}
https://www.aimsciences.org/article/doi/10.3934/dcds.2012.32.4111	# American Institute of Mathematical Sciences December 2012, 32(12): 4111-4131. doi: 10.3934/dcds.2012.32.4111 ## Noninvertible cocycles: Robustness of exponential dichotomies 1 Departamento de Matemática, Instituto Superior Técnico, Universidade de Lisboa, 1049-001 Lisboa 2 Departamento de Matemática, Instituto Superior Técnico, 1049-001 Lisboa Received June 2010 Revised May 2012 Published August 2012 For the dynamics defined by a sequence of bounded linear operators in a Banach space, we establish the robustness of the notion of exponential dichotomy. This means that an exponential dichotomy persists under sufficiently small linear perturbations. We consider the general cases of a nonuniform exponential dichotomy, which requires much less than a uniform exponential dichotomy, and of a noninvertible dynamics or, more precisely, of a dynamics that may not be invertible in the stable direction. Citation: Luis Barreira, Claudia Valls. Noninvertible cocycles: Robustness of exponential dichotomies. Discrete & Continuous Dynamical Systems, 2012, 32 (12) : 4111-4131. doi: 10.3934/dcds.2012.32.4111 ##### References: [1] L. Barreira and Ya. Pesin, "Nonuniform Hyperbolicity," Encyclopedia of Math. and Its Appl. 115, Cambridge Univ. Press, 2007. Google Scholar [2] L. Barreira and C. Valls, Stability theory and Lyapunov regularity, J. Differential Equations, 232 (2007), 675-701. doi: 10.1016/j.jde.2006.09.021. Google Scholar [3] L. Barreira and C. Valls, Robustness of nonuniform exponential dichotomies in Banach spaces, J. Differential Equations, 244 (2008), 2407-2447. doi: 10.1016/j.jde.2008.02.028. Google Scholar [4] L. Barreira and C. Valls, "Stability of Nonautonomous Differential Equations," Lect. Notes in Math. 1926, Springer, 2008. Google Scholar [5] L. Barreira and C. Valls, Robustness of discrete dynamics via Lyapunov sequences, Comm. Math. 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Kreĭn, "Stability of Solutions of Differential Equations in Banach Space," Translations of Mathematical Monographs 43, Amer. Math. Soc., 1974. Google Scholar [12] J. Hale, "Asymptotic Behavior of Dissipative Systems," Mathematical Surveys and Monographs 25, Amer. Math. Soc., 1988. Google Scholar [13] D. Henry, "Geometric Theory of Semilinear Parabolic Equations," Lect. Notes in Math. 840, Springer, 1981. Google Scholar [14] N. Huy, Exponential dichotomy of evolution equations and admissibility of function spaces on a half-line, J. Funct. Anal., 235 (2006), 330-354. doi: 10.1016/j.jfa.2005.11.002. Google Scholar [15] J. Massera and J. Schäffer, Linear differential equations and functional analysis. I, Ann. of Math., 67 (1958), 517-573. doi: 10.2307/1969871. Google Scholar [16] J. Massera and J. Schäffer, "Linear Differential Equations and Function Spaces," Pure and Applied Mathematics, 21, Academic Press, 1966. Google Scholar [17] R. Naulin and M. 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Sasu and A. Sasu, Input-output conditions for the asymptotic behavior of linear skew-product flows and applications, Commun. Pure Appl. Anal., 5 (2006), 551-569. Google Scholar [24] G. Sell and Y. You, "Dynamics of Evolutionary Equations," Applied Mathematical Sciences 143, Springer, 2002. Google Scholar show all references ##### References: [1] L. Barreira and Ya. Pesin, "Nonuniform Hyperbolicity," Encyclopedia of Math. and Its Appl. 115, Cambridge Univ. Press, 2007. Google Scholar [2] L. Barreira and C. Valls, Stability theory and Lyapunov regularity, J. Differential Equations, 232 (2007), 675-701. doi: 10.1016/j.jde.2006.09.021. Google Scholar [3] L. Barreira and C. Valls, Robustness of nonuniform exponential dichotomies in Banach spaces, J. Differential Equations, 244 (2008), 2407-2447. doi: 10.1016/j.jde.2008.02.028. Google Scholar [4] L. Barreira and C. Valls, "Stability of Nonautonomous Differential Equations," Lect. Notes in Math. 1926, Springer, 2008. Google Scholar [5] L. Barreira and C. Valls, Robustness of discrete dynamics via Lyapunov sequences, Comm. Math. Phys., 290 (2009), 219-238. doi: 10.1007/s00220-009-0762-z. Google Scholar [6] L. Barreira and C. Valls, Robust nonuniform dichotomies and parameter dependence, J. Math. Anal. Appl., 373 (2011), 690-708. doi: 10.1016/j.jmaa.2010.08.026. Google Scholar [7] C. Chicone and Yu. Latushkin, "Evolution Semigroups in Dynamical Systems and Differential Equations," Mathematical Surveys and Monographs 70, Amer. Math. Soc., 1999. Google Scholar [8] S.-N. Chow and H. Leiva, Existence and roughness of the exponential dichotomy for skew-product semiflow in Banach spaces, J. Differential Equations, 120 (1995), 429-477. doi: 10.1006/jdeq.1995.1117. Google Scholar [9] W. Coppel, Dichotomies and reducibility, J. Differential Equations, 3 (1967), 500-521. Google Scholar [10] W. Coppel, "Dichotomies in Stability Theory," Lect. Notes in Math. 629, Springer, 1978. Google Scholar [11] Ju. Dalec$'$kiĭ and M. Kreĭn, "Stability of Solutions of Differential Equations in Banach Space," Translations of Mathematical Monographs 43, Amer. Math. Soc., 1974. Google Scholar [12] J. Hale, "Asymptotic Behavior of Dissipative Systems," Mathematical Surveys and Monographs 25, Amer. Math. Soc., 1988. Google Scholar [13] D. Henry, "Geometric Theory of Semilinear Parabolic Equations," Lect. Notes in Math. 840, Springer, 1981. Google Scholar [14] N. Huy, Exponential dichotomy of evolution equations and admissibility of function spaces on a half-line, J. Funct. Anal., 235 (2006), 330-354. doi: 10.1016/j.jfa.2005.11.002. Google Scholar [15] J. Massera and J. Schäffer, Linear differential equations and functional analysis. I, Ann. of Math., 67 (1958), 517-573. doi: 10.2307/1969871. Google Scholar [16] J. Massera and J. Schäffer, "Linear Differential Equations and Function Spaces," Pure and Applied Mathematics, 21, Academic Press, 1966. Google Scholar [17] R. Naulin and M. Pinto, Stability of discrete dichotomies for linear difference systems, J. Differ. Equations Appl., 3 (1997), 101-123. Google Scholar [18] R. Naulin and M. Pinto, Admissible perturbations of exponential dichotomy roughness, Nonlinear Anal., 31 (1998), 559-571. doi: 10.1016/S0362-546X(97)00423-9. Google Scholar [19] O. Perron, Die Stabilit\"atsfrage bei Differentialgleichungen, Math. Z., 32 (1930), 703-728. doi: 10.1007/BF01194662. Google Scholar [20] V. Pliss and G. Sell, Robustness of exponential dichotomies ininfinite-dimensional dynamical systems, J. Dynam. Differential Equations, 11 (1999), 471-513. doi: 10.1023/A:1021913903923. Google Scholar [21] L. Popescu, Exponential dichotomy roughness on Banach spaces, J. Math. Anal. Appl., 314 (2006), 436-454. doi: 10.1016/j.jmaa.2005.04.011. Google Scholar [22] A. Sasu, Exponential dichotomy and dichotomy radius for difference equations, J. Math. Anal. Appl., 344 (2008), 906-920. doi: 10.1016/j.jmaa.2008.03.019. Google Scholar [23] B. 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Discrete & Continuous Dynamical Systems - B, 2006, 6 (5) : 1051-1076. doi: 10.3934/dcdsb.2006.6.1051 2019 Impact Factor: 1.338	2021-06-22 14:24:09	{"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": 0.7874794602394104, "perplexity": 5355.223939367019}, "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/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00430.warc.gz"}
https://informationtransfereconomics.blogspot.com/2015/08/the-dungeons-and-dragons-approach-to.html	## Tuesday, August 18, 2015 ### The Dungeons and Dragons approach to economics Dice roll with six-sided and twenty-sided dice. Editor's note: This post has been languishing in the drafts folder for several months, so I gave it a quick finish and posted it. In a footnote to the previous post I mentioned flipping a coin versus rolling a twenty sided die as a different way of thinking about the solution to paradox of value -- why diamonds cost more than water (we need water to survive, yet water is cheaper than diamonds). The 19th century solution is marginal utility. And it works well. If we look at the picture above, we need a total of about 33 bits to specify this particular result $$11 \cdot \log_{2} 6 + 1 \cdot \log_{2} 20 \simeq 32.8 \text{ bits}$$ Each of the six-sided dice reveals about 2.6 bits, while the 20-sided die reveals about 4.3 bits. In equilibrium, the supply effectively knows the demand's "roll". 'Quanta' of 4.3 bits are flowing back and forth in the diamond market and quanta of 2.6 bits are flowing back and forth in the water market. However if there is a change, a change in the 20-sided die in the demand's roll, it requires a flow of 4.3 bits to the supply side. A change in one of the six-sided dice requires a flow of 2.6 bits. Therefore the information transfer index $k_{d}$ for diamonds is going to be larger than the index for water $k_{w}$. If there is something that functions as money (see here and here, Ed. note: these are later posts that more clearly demonstrate my point), then the information transfer index in terms of money will also be larger. Therefore, the price of diamonds will grow much faster than the price of water since for some $m$: $$p_{d} \sim m^{k_{d} - 1} > p_{w} \sim m^{k_{w} - 1}$$ and therefore $$\frac{d}{dm} p_{d} > \frac{d}{dm} p_{w}$$ if $k_{d} > k_{w}$. The absolute price is not knowable in the information transfer model, but by simply being rarer (i.e. lower probability so that more information is revealed by specifying its allocation) the price will grow much faster. Thus eventually, regardless of the starting point, diamonds will be more valuable than water. The one caveat (assumption) is that both things must continue to have a market. ... Update: fixed typos of 4.6 bits where it should read 4.3 bits (H/T Tom Brown in comments below). 1. Interesting. Twice in the text you mention "4.6 bits" but did you mean to write "4.3 bits?" 1. Yes, thanks. Fixed it. 2. Yes very interesting, makes something of intuitive sense, but as you state "...that both things must continue to have a market." Diamonds are rare, and expensive, because there is a market for them. One could imaging that certain types of natural minerals are as rare as diamonds, but not attractive when placed on fingers, and thus would not be very expensive, as the information from its rareness does not reach equilibrium with demand. 1. There are other factors to consider -- diamonds are crystals prized partially for their size and e.g. metals can be melted down. But even industrial diamonds are about as expensive, if not more expensive, than gold even without looking nice. I should add that part of the reason non-industrial diamonds are more expensive than they 'should' be is that there is something of a cartel controlling supply. But as there are markets now for some rare-earth metals that have uses in electronics, we might expect their price to outstrip diamonds eventually ...	2018-02-25 05:42:37	{"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": 0.6812832951545715, "perplexity": 1195.6394973424538}, "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/1518891816138.91/warc/CC-MAIN-20180225051024-20180225071024-00729.warc.gz"}
http://www.sjbhschool.org/curriculum/	Curriculum St. John the Baptist follows the curriculum guidelines of the Wisconsin Department of Instruction and the Diocesan Department of Education. GRACE schools’ in-services evaluate curriculum on a regular basis. MAP ASSESSMENT Students in grades K-8 take our computerized adaptive testing introduced in the 2014 – 2015 school year. This assessment is compatible with our curriculum. The tests are administered throughout the year and are meant to help teachers make informed decisions to differentiate instruction and adapt learning as needed DAILY SCHOOL PROGRAM 7:30 A.M. . . . . . . . Teacher Preparation Time 7:50 A.M. . . . . . . . Students to enter building 8:30 A.M. . . . . . . . Liturgy K-4 Wednesdays and grades 5 – 8 on Thursdays 2:45 P.M. . . . . . . . .4K Dismissal 2:55 P.M. . . . . . . . .K-4 Dismissal 3:00 P.M. . . . . . . . .5-8 Dismissal 11:00 – 12:00 P.M. Rotating Lunch • 11:00 – 11:20 PK – 1 • 11:30 – 11:50 Grades 5-8 • 11:40 – 12:00 Grades 2-4 There is one 15-minute recess for grades PK-5.	2017-10-21 06:35:03	{"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": 0.9301182627677917, "perplexity": 5033.829107410096}, "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-2017-43/segments/1508187824618.72/warc/CC-MAIN-20171021062002-20171021082002-00656.warc.gz"}
http://nrich.maths.org/public/leg.php?code=12&cl=3&cldcmpid=455	Search by Topic Resources tagged with Factors and multiples similar to Magic Caterpillars: Filter by: Content type: Stage: Challenge level: There are 92 results Broad Topics > Numbers and the Number System > Factors and multiples LCM Sudoku Stage: 4 Challenge Level: Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it. LCM Sudoku II Stage: 3, 4 and 5 Challenge Level: You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku. Multiplication Equation Sudoku Stage: 4 and 5 Challenge Level: The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid. Product Sudoku Stage: 3, 4 and 5 Challenge Level: The clues for this Sudoku are the product of the numbers in adjacent squares. Transposition Cipher Stage: 3 and 4 Challenge Level: Can you work out what size grid you need to read our secret message? Product Sudoku 2 Stage: 3 and 4 Challenge Level: Given the products of diagonally opposite cells - can you complete this Sudoku? Substitution Transposed Stage: 3 and 4 Challenge Level: Substitution and Transposition all in one! How fiendish can these codes get? Substitution Cipher Stage: 3 and 4 Challenge Level: Find the frequency distribution for ordinary English, and use it to help you crack the code. Remainder Stage: 3 Challenge Level: What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2? 14 Divisors Stage: 3 Challenge Level: What is the smallest number with exactly 14 divisors? Cuboids Stage: 3 Challenge Level: Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all? The Remainders Game Stage: 2 and 3 Challenge Level: A game that tests your understanding of remainders. X Marks the Spot Stage: 3 Challenge Level: When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" . GOT IT Now Stage: 2 and 3 Challenge Level: For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target? Factorial Stage: 4 Challenge Level: How many zeros are there at the end of the number which is the product of first hundred positive integers? AB Search Stage: 3 Challenge Level: The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B? A First Product Sudoku Stage: 3 Challenge Level: Given the products of adjacent cells, can you complete this Sudoku? Data Chunks Stage: 4 Challenge Level: Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . . Stage: 3 Challenge Level: A mathematician goes into a supermarket and buys four items. Using a calculator she multiplies the cost instead of adding them. How can her answer be the same as the total at the till? Squaresearch Stage: 4 Challenge Level: Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares? Factors and Multiples Game Stage: 2, 3 and 4 Challenge Level: A game in which players take it in turns to choose a number. Can you block your opponent? Shifting Times Tables Stage: 3 Challenge Level: Can you find a way to identify times tables after they have been shifted up? Stage: 3 Challenge Level: Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . . Remainders Stage: 3 Challenge Level: I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number? Reverse to Order Stage: 3 Challenge Level: Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number? Different by One Stage: 4 Challenge Level: Make a line of green and a line of yellow rods so that the lines differ in length by one (a white rod) Three Times Seven Stage: 3 Challenge Level: A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why? What a Joke Stage: 4 Challenge Level: Each letter represents a different positive digit AHHAAH / JOKE = HA What are the values of each of the letters? Factoring Factorials Stage: 3 Challenge Level: Find the highest power of 11 that will divide into 1000! exactly. N000ughty Thoughts Stage: 4 Challenge Level: Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the. . . . Ben's Game Stage: 3 Challenge Level: Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters. Inclusion Exclusion Stage: 3 Challenge Level: How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5? Thirty Six Exactly Stage: 3 Challenge Level: The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors? Hot Pursuit Stage: 3 Challenge Level: The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed? American Billions Stage: 3 Challenge Level: Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3... Even So Stage: 3 Challenge Level: Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why? Eminit Stage: 3 Challenge Level: The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M? Oh! Hidden Inside? Stage: 3 Challenge Level: Find the number which has 8 divisors, such that the product of the divisors is 331776. Phew I'm Factored Stage: 4 Challenge Level: Explore the factors of the numbers which are written as 10101 in different number bases. Prove that the numbers 10201, 11011 and 10101 are composite in any base. How Old Are the Children? Stage: 3 Challenge Level: A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?" Two Much Stage: 3 Challenge Level: Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears. Factor Track Stage: 2 and 3 Challenge Level: Factor track is not a race but a game of skill. The idea is to go round the track in as few moves as possible, keeping to the rules. Stage: 3 Challenge Level: List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it? Helen's Conjecture Stage: 3 Challenge Level: Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true? Special Sums and Products Stage: 3 Challenge Level: Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48. Gaxinta Stage: 3 Challenge Level: A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N? Expenses Stage: 4 Challenge Level: What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time? Dozens Stage: 3 Challenge Level: Do you know a quick way to check if a number is a multiple of two? How about three, four or six? Diggits Stage: 3 Challenge Level: Can you find what the last two digits of the number $4^{1999}$ are?	2014-12-19 17:36:35	{"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": 0.42801281809806824, "perplexity": 1041.4689406473951}, "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-2014-52/segments/1418802768741.99/warc/CC-MAIN-20141217075248-00133-ip-10-231-17-201.ec2.internal.warc.gz"}
https://chemistry.stackexchange.com/questions/35564/name-the-following-coordination-complex	Name the following coordination complex What are the IUPAC rules when naming a coordination complex like: $$\ce{[Pt(NH3)4][Pt(NH3)Cl3]}$$ Now according to my textbook, it is named as tetraammineplatinum(II) amminetrichloridoplatinate(II). Both the $\ce{Pt}$ atoms have the oxidation state +2 .There are no charges assigned to the two complex ions in the formula, so, my assumption is that the entire complex is neutral and thus the oxidation states for $\ce{Pt}$ should be assigned accordingly. However, on assigning the oxidation states as mentioned in the answer the assumption is proven wrong. I'm confused about the mentioned oxidation numbers. • How do you calculate the oxidation numbers of the metal atoms in such a complex? • Shouldn't the charge on each complex ion be mentioned in the formula if the oxidation states given in the answer are correct? • This is a salt of two complex ions. Consider it as two separate complexes, happened to be in same salt: one is cationic $\ce{[Pt(NH3)_4]^{2+}}$ and another is anionic $\ce{[Pt(NH3)Cl3]^-}$. However, it seems that it should have two anions, with formula $\ce{[Pt(NH3)_4] [Pt(NH3)Cl3]2}$ – permeakra Aug 27 '15 at 13:09 • That's exactly what I was thinking, but I asked an expert who told me that the given formula was correct. He was kinda' unsure though. – Tabish Mir Aug 27 '15 at 13:12 • He is incorrect in this particular regard. Howevere, there is $\ce{[PtCl4]^{2-}}$ anion, wich could ... messed with his line of though. – permeakra Aug 27 '15 at 13:31 Your textbook contains an error. Either the formula is written incorrectly or the name is. Since according to comments the name is correct, it must be an incorrect formula. Let’s break it down: If the formula is correct We have two complex ions that together create a doubly-complex salt. According to the rules of (pseudo-) binary ionic compounds, one ion must be positively charged and the other negatively. By convention, the cation is written first so it is safe to assume that $\ce{[Pt(NH3)4]^{?}}$ be the cation — which fits in well with chemical intuition because $\ce{NH3}$ is a neutral ligand and platinum is a metal. Therefore, $\ce{[Pt(NH3)Cl3]^{?}}$ must be the anion — very nice, because whe have chlorido ligands that are commonly negative. Since every chlorido ligand adds $-1$ charge to the complex, the anion can be anything from $-1$ to $-3$ with decreasing probability. The complex needs to be a $1\,:\,1$ binary complex, so the negative charge needs to be equivalent to the positive charge. From here onwards it gets difficult. We have the following possibilities: • $\ce{[Pt^{I}(NH3)4]^{+} [Pt^{II}(NH3)Cl3]^-}$ • $\ce{[Pt^{II}(NH3)4]^2+ [Pt^{I}(NH3)Cl3]^2-}$ • $\ce{[Pt^{III}(NH3)4]^3+ [Pt^{0}(NH3)Cl3]^3-}$ They all differ in the assumed oxidative states of platinum. We cannot decide a priori which one of these three is correct, although the last one is by far less probable. Technically, we can even assume further diverging oxidation states but they get proportionally less probable. Only if the charge were clearly stated could we distinguish between the cases. For the sake of consistency, here are the names: • tetraammineplatinum(I) amminetrichloridoplatinate(II) • tetraammineplatinum(II) amminetrichloridoplatinate(I) • tetraammineplatinum(III) amminetrichloridoplatinate(0) If the name is correct Then we can logically build up the resulting salt from the name. Let’s start with the cation: • It is platinum(II), so we get $\ce{Pt^2+}$ • adding four ammine ligands gives us $\ce{[Pt(NH3)4]^2+}$ For the anion: • It is still platinum(II) (although rendered as platinate, so we need to end up with an anion): $\ce{Pt^2+}$ • We have one ammine ligand: $\ce{[Pt(NH3)]^2+}$ • and three chlorido ($\ce{Cl-}$) ligands: $\ce{[Pt(NH3)Cl3]-}$ Since we have a cation with $+2$ charge and an anion with $-1$ charge, we need two anions per cation, giving us the final formula: $$\ce{[Pt(NH3)4][Pt(NH3)Cl3]2}$$ So what is correct? Going by chemical intuition (platinum(II) being a well-known $\mathrm{d}^8$ system that forms square-planar complexes), I would say that the name is correct and the formula isn’t. This also has the implication that they messed up the question by missing a subscripted 2 (rather likely) rather than messing up the answer by getting a Roman numeral wrong (not quite as likely). Hence there is nothing miraculous or confusing about assigning oxidation states to metals in complexes — and the IUPAC even formally defines oxidation states for promiscuous ligands such as nitrosyl (which is always considered to be a radical neutral ligand, never $\ce{NO^+}$ or $\ce{NO-}$) allowing for a clean-cut naming of complex compounds without knowing the chemistry involved. The charge does not have to be mentioned in doubly-complex salts for any component. However, it always helps students to give them and in non-trivial cases (like the assumed correct formula) it might be vital to know the overall charges. If there are no charges written, though, the final product has to be assumed neutral.	2020-09-26 11:54:02	{"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": 0.8777813911437988, "perplexity": 1222.3719228657133}, "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/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00045.warc.gz"}
https://math.stackexchange.com/questions/646428/mean-and-variance-of-reciprocal-normal-distribution/646467	# mean and variance of reciprocal normal distribution If $X$ is a normal distributed with mean $\mu$ and variance $\sigma^2$. What would be the mean and variance of $Y = \dfrac{1}{X}$ • What makes you think that the mean and/or variance exist? Both are defined by improper integrals; are you sure these integrals actually converge? Have you tried it explicitly with mean $0$ and variance $1$? – John Hughes Jan 21 '14 at 17:26 • I didnot do that mathematically. However, I ran this code in the MATLAB: a = randn(1000000,1); b = 1./a; The mean and variance of b does have some value. So I was interested in if there is a closed mathematical term, anyone has ever worked out. – Sam Jan 21 '14 at 17:31 • Please accept answers to your other questions by pressing tick button. Many people do not like helping people who do not accept answers. – Lost1 Jan 21 '14 at 17:38 • Which question, specifically? – Sam Jan 21 '14 at 17:39 • Sorry didnt realise tey got no answers except self answers. – Lost1 Jan 21 '14 at 17:45 Mean and variance do not exist. For the mean to exist, the integral $\int^\infty_{-\infty} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\frac{1}{\|x\|} \text{d}x$ needs to be finite. This is clearly not the case. Note it is necessary that mean exists for variance to exist. See http://en.wikipedia.org/wiki/Inverse_distribution#Reciprocal_normal_distribution Note inverse gaussian is something completely different. It is connected to brownian motion hitting a level. I changed the title. The thing you are referring to is a reciprocal normal. • The integral is not finite since, denoting by $p(x)$ the density of a standard normal, $$\begin{array}{rl}\displaystyle\int_{-\infty}^{\infty} \frac{1}{\|x\|} p(x)dx &=& 2 \int_0^{\infty} \frac{1}{x} p(x)dx \\ & = & 2 \int_0^1 \frac{1}{x} p(x) dx + 2\int_1^{\infty} \frac{1}{x}p(x)dx \\ & \ge & 2 p(1)\int_0^1 \frac{1}{x}dx + 2 \int_1^{\infty} \frac{1}{x}p(x)dx = \infty\end{array}$$ since $\int_0^1 \frac{1}{x}dx$ is divergent and $p(x)$ is decreasing on $[0,1]$, so bounded below by $p(1)$. (I assume the answerer knows this, I am writing it for future reference of anyone else slow like me.) – Chill2Macht Oct 31 '17 at 0:35	2019-08-20 11:53:13	{"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": 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": 0.8098312616348267, "perplexity": 406.29291300275645}, "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/1566027315329.55/warc/CC-MAIN-20190820113425-20190820135425-00449.warc.gz"}
https://zbmath.org/?q=an:06403388	## Small $$k$$-pyramids and the complexity of determining $$k$$.(English)Zbl 1320.05130 Summary: Motivated by the computational complexity of determining whether a graph is Hamiltonian, we study under algorithmic aspects a class of polyhedra called $$k$$-pyramids, introduced in [C. T. Zamfirescu and T. I. Zamfirescu, Math. Nachr. 284, No. 13, 1739–1747 (2011; Zbl 1236.05121)], and discuss related applications. We prove that determining whether a given graph is the 1-skeleton of a $$k$$-pyramid, and if so whether it is belted or not, can be done in polynomial time for $$k \leq 3$$. The impact on Hamiltonicity follows from the traceability of all 2-pyramids and non-belted 3-pyramids, and from the hamiltonicity of all non-belted 2-pyramids. The algorithm can also be used to determine the outcome for larger values of $$k$$, but the complexity increases exponentially with $$k$$. Lastly, we present applications of the algorithm, and improve the known bounds for the minimal cardinality of systems of bases called foundations in graph families with interesting properties concerning traceability and Hamiltonicity. ### MSC: 05C85 Graph algorithms (graph-theoretic aspects) 05C45 Eulerian and Hamiltonian graphs ### Keywords: pyramid; prism; Halin graph; Hamiltonian Zbl 1236.05121 Full Text: ### References: [1] Aldred, R. E.L.; Bau, S.; Holton, D. A.; McKay, B. D., Nonhamiltonian 3-connected cubic planar graphs, SIAM J. Discrete Math., 13, 1, 25-32, (2000) · Zbl 0941.05041 [2] Berge, C., Theory of graphs and its applications, (1962), Methuen London · Zbl 0097.38903 [3] Bondy, J. A., Pancyclic graphs: recent results, Colloq. Math. Soc. János Bolyai, 10, 181-187, (1975) · Zbl 0324.05115 [4] Boyer, J. M.; Myrvold, W. J., On the cutting edge: simplified $$O(n)$$ planarity by edge addition, J. Graph Algorithms Appl., 8, 3, 241-273, (2004) · Zbl 1086.05067 [5] Brandstädt, A.; Chepoi, V. D.; Dragan, F. F., The algorithmic use of hypertree structure and maximum neighbourhood orderings, Discrete Appl. Math., 82, 1-3, 43-77, (1998) · Zbl 0893.05018 [6] Broersma, H. J.; Kloks, T.; Kratsch, D.; Müller, H., Independent sets in asteroidal triple-free graphs, (Proc. 24th Int. Colloq. Automata, Languages and Programming, Lect. Notes Comput. Sci., vol. 1256, (1997), Springer), 760-770 · Zbl 1401.05278 [7] Chang, M. S., Efficient algorithms for the domination problems on interval and circular-arc graphs, SIAM J. Comput., 27, 6, 1671-1694, (1998) · Zbl 0911.05051 [8] Cormen, T. H.; Leiserson, C. E.; Rivest, R. L.; Stein, C., Introduction to algorithms, (2009), MIT Press and McGraw-Hill · Zbl 1187.68679 [9] Corneil, D. G.; Perl, Y., Clustering and domination in perfect graphs, Discrete Appl. Math., 9, 27-40, (1984) · Zbl 0581.05053 [10] Farber, M., Independent domination in chordal graphs, Oper. Res. Lett., 1, 134-138, (1982) · Zbl 0495.05053 [11] Garey, M. R.; Johnson, D. S., Computers and intractability: A guide to the theory of NP-completeness, (1979), Freeman New York · Zbl 0411.68039 [12] Garey, M. R.; Johnson, D. S.; Tarjan, R. E., The planar Hamiltonian circuit problem is NP-complete, SIAM J. Comput., 5, 704-714, (1976) · Zbl 0346.05110 [13] Gaspers, S.; Liedloff, M., A branch-and-reduce algorithm for finding a minimum independent dominating set in graphs, (2007), Univ. Bergen, Technical Report no. 344 [14] Grinstead, D. L.; Slater, P. J., A recurrence template for several parameters in series-parallel graphs, Discrete Appl. Math., 54, 151-168, (1994) · Zbl 0812.68099 [15] Holton, D. A.; McKay, B. D., The smallest non-Hamiltonian 3-connected cubic planar graphs have 38 vertices, J. Comb. Theory, Ser. B, 45, 305-319, (1989) · Zbl 0607.05051 [16] Irving, R. W., On approximating the minimum independent dominating set, Inf. Process. Lett., 37, 197-200, (1991) · Zbl 0713.68033 [17] Kahng, A. B.; Park, C.-H.; Xu, X., Fast dual-graph-based hotspot filtering, IEEE Trans. Comput.-Aided Des. Integr. Circuits Syst., 27, 9, (2008) [18] Karp, R. M., Reducibility among combinatorial problems, 85-103, (1972), Plenum Press New York [19] Knuth, D., The art of computer programming, vol. 3: sorting and searching, (1997), Addison-Wesley, Section 5.2.4: Sorting by Merging. Section 5.3.2: Minimum-Comparison Merging · Zbl 0883.68015 [20] Kratsch, D.; Stewart, L., Domination on cocomparability graphs, SIAM J. Discrete Math., 6, 400-417, (1993) · Zbl 0780.05032 [21] Liu, C.; Song, Y., Exact algorithms for finding the minimum independent dominating set in graphs, (ISAAC 2006, Lect. Notes Comput. Sci., vol. 4288, (2006), Springer), 439-448 · Zbl 1135.05315 [22] Malik, S.; Qureshi, A. M.; Zamfirescu, T., Hamiltonian properties of generalized halin graphs, Can. Math. Bull., 52, 3, 416-423, (2009) · Zbl 1171.05031 [23] Manlove, D. F., On the algorithmic complexity of twelve covering and independence parameters of graphs, Discrete Appl. Math., 91, 1-3, 155-175, (1999) · Zbl 0922.05041 [24] Pfaff, J.; Laskar, R.; Hedetniemi, S. T., Linear algorithms for independent domination and total domination in series-parallel graphs, Congr. Numer., 45, 71-82, (1984) · Zbl 0571.05045 [25] Skowrońska, M., Hamiltonian properties of halin-like graphs, Ars Comb., 16(B), 97-109, (1983) · Zbl 0545.05043 [26] Skowrońska, M.; Sysło, M. M., Hamiltonian cycles in skirted trees, Zastos. Mat., 19, 3-4, 599-610, (1987) · Zbl 0719.05045 [27] Skupień, Z., Crowned trees and planar highly Hamiltonian graphs, 537-555, (1990), Wissenschaftsverlag Mannheim · Zbl 0717.05029 [28] Steinitz, E., Polyeder und raumeinteilungen, Geometrie, vol. 3, 1-139, (1922) [29] Wiener, G.; Araya, M., On cubic planar Hypohamiltonian and hypotraceable graphs, Electron. J. Comb., P85, 1-11, (2011) · Zbl 1217.05065 [30] Yannakakis, M.; Gavril, F., Edge dominating sets in graphs, SIAM J. Appl. Math., 38, 364-372, (1980) · Zbl 0455.05047 [31] Zamfirescu, C. T.; Zamfirescu, T. I., Hamiltonian properties of generalized pyramids, Math. Nachr., 284, 13, 1739-1747, (2011) · Zbl 1236.05121 [32] Zamfirescu, T., Three small cubic graphs with interesting Hamiltonian properties, J. Graph Theory, 4, 287-292, (1980) · Zbl 0442.05047 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.	2022-05-20 07:51:10	{"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": 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": 0.7780361771583557, "perplexity": 7428.460087699899}, "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-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00573.warc.gz"}
https://codereview.stackexchange.com/questions/166168/scraping-website-for-book-listings-checking-if-database-contains-relevant-listi	# Scraping website for book listings, checking if database contains relevant listings, and displaying result in a Django view This code works but is very slow. I am scraping a website for urls for various books, and then checking to see if we have complete records of those books in our database. After I process the URL to get the book info, I see if we have it in our database. If we don't, I move on, that's fine. If we do, I need to check that we have a matching BookTextVer type; that foreign keys to the book, and filters on an attribute from the URL. If we have it, great; if not, I add it to my list of dictionaries to eventually pass to the django template. def book_missing(request): data = [] books = book.objects.all() sitemap = "www.sitemap.com" page = requests.get(sitemap) tree = objectify.fromstring(page.content) regex = re.compile("books-(\d+)([A-Za-z]+)(\d+)([A-Za-z]+)") for node in tree.getchildren(): book_url = node.book.text # e.g. https://www.sitemap.com/fdsys/pkg/books-27hr200nf/content-detail.html m = regex.search(book_url) object_type_cd = m.groups()[1] # hr book_num = m.groups()[2] # 200 book_type = m.groups()[3] # nf try: book = books.get(book_num=book_num, object_type_cd_id=object_type_cd.upper()) except book.DoesNotExist: pass if (BookTextVer.objects.filter( book=book, book_text_ver_type__short_name=book_type.upper()).exists()): pass # matching book type exists, no processing needed else: data.append({"book": object_type_cd.upper() + book_num, "book_type": book_type.upper()}) context = { 'data': data, } return render(request, 'etl/reports/gpo-book-missing.html', context) After I process the URL to get the book info, I see if we have it in our database. If we don't, I move on, that's fine. I guess you're referring to this part of the code: try: book = books.get(book_num=book_num, object_type_cd_id=object_type_cd.upper()) except book.DoesNotExist: pass This doesn't do what you described, because after the pass, the rest of the loop body is executed instead of moving on the next iteration. Perhaps you meant to use continue there instead of pass. Instead of having an empty if block and an else block here, it would be better to flip the condition and have only an if block without an else: if (BookTextVer.objects.filter( book=book, book_text_ver_type__short_name=book_type.upper()).exists()): pass # matching book type exists, no processing needed else: data.append({"book": object_type_cd.upper() + book_num, "book_type": book_type.upper()}) The query on BookTextVer tries to match by short_name. Is that field indexed in the database. If not, the query will be inevitably slow, and indexing should be a significant improvement. The current implementation runs 1 or 2 queries for every node in tree.getchildren(). You might get better performance if you first collect all the book_num, book_type, object_type_cd values from the document, and formulate a single database query or constant number of database queries. That would be a major change, and possibly you might need to forego Django's OR-mapping benefits, but the reduced number of database queries could make a big difference. • You're right on the pass vs continue... I don't know why that didn't trigger an error since I use that (non-existent) book in the next part of the query – thumbtackthief Jun 22 '17 at 20:41	2020-07-09 02:45:55	{"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": 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": 0.23363566398620605, "perplexity": 2245.161560623815}, "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-2020-29/segments/1593655897844.44/warc/CC-MAIN-20200709002952-20200709032952-00031.warc.gz"}
https://bindsnet-docs.readthedocs.io/	# Welcome to BindsNET’s documentation!¶ BindsNET is built on top of the PyTorch deep learning platform. It is used for the simulation of spiking neural networks (SNNs) and is geared towards machine learning and reinforcement learning. BindsNET takes advantage of the torch.Tensor object to build spiking neurons and connections between them, and simulate them on CPUs or GPUs (for strong acceleration / parallelization) without any extra work. Recently, torchvision.datasets has been integrated into the library to allow the use of popular vision datasets in training SNNs for computer vision tasks. Neural network functionality contained in torch.nn.functional module is used to implement more complex connections between populations of spiking neurons. Spiking neural networks are sometimes referred to as the third generation of neural networks. Rather than the simple linear layers and nonlinear activation functions of deep learning neural networks, SNNs are composed of neural units which more accurately capture properties of their biological counterparts. An important difference between spiking neurons and the artificial neurons of deep learning are the former’s integration of input in time; they are naturally short-term memory devices by their maintenance of a (possibly decaying) membrane voltage. As a result, some have argued that SNNs are particularly well-suited to model time-varying data. Neurons are connected together with directed edges (synapses) which are (in general) plastic. Synapses may have their own dynamics as well, which may or may not depend on pre- and post-synaptic neural activity https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3395004/ or other biological signals https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4717313/. The modification of synaptic strengths is thought to be an important mechanism by which organisms learn. Accordingly, BindsNET provides a module (bindsnet.learning) which contains functions used for the updating of synapse weights. At its core, BindsNET provides software objects and methods which support the simulation of groups of different types of neurons (bindsnet.network.nodes), as well as different types of connections between them (bindsnet.network.topology). These may be arbitrarily combined together under a single bindsnet.network.Network object, which is responsible for the coordination of the simulation logic of all underlying components. On creation of a network, the user can specify a simulation timestep constant, $$dt$$, which determines the granularity of the simulation. Choosing this parameter induces a trade-off between simulation speed and numerical precision: large values result in fast simulation, but poor simulation accuracy, and vice versa. Monitors (bindsnet.network.monitors) are available for recording state variables from arbitrary network components (e.g., the voltage $$v$$ of a group of neurons). The development of BindsNET is supported by the Defense Advanced Research Project Agency Grant DARPA/MTO HR0011-16-l-0006. Package reference	2022-12-05 18:00:06	{"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": 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": 0.26177918910980225, "perplexity": 1176.9761219876493}, "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/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00804.warc.gz"}
http://publish.uwo.ca/~pzsambok/calc%201000a%20002%2016au/notes/notes/15%20-%20Derivatives%20of%20trigonometric%20functions.html	# Derivatives of trigonometric functions ## Correction about $$G(t)=\frac{1-2t}{3+t}$$. • I should have really calculated those infinite limits • Let's look at the situation as $$t\to(-3)^-$$. • The nominator $$1-2t$$ has a nonzero, positive value at $$t=-3$$. • As $$t$$ approaches $$(-3)$$ from the left, the denominator $$3+t$$ gets negative values indefinitely close to zero. • Therefore, the fractional $$G(t)=\frac{1-2t}{3+t}$$ gets indefinitely small as $$t\to(-3)^-$$, which means $$\lim_{t\to(-3)^-}G(t)=-\infty$$. • The shape of the graph confirms this. ## Derivatives of $$\sin(x)$$ and $$\cos(x)$$ • If you're interested, you can find a geometric argument for these derivatives in the textbook. • We'll apply Euler formula magic, which will be much quicker! • Of course, you just have to remember the formulas themselves, but you have to see how much shorter this proof is. • Recall that we have the imaginary number $$i$$, which has the wonderful property $$i^2=-1$$. • Euler's formula was the following. For a real number $$x$$, we have that $e^{ix}=\cos(x)+i\sin(x).$ • Let's derivate this exponential function. • In general, for a constant $$a$$, we have $$\frac{\mathrm d}{\mathrm dx}e^{ax}=ae^{ax}$$. • In this particular case, we get $(e^{ix})'=ie^{ix}.$ • Applying Euler's formula on the left side, we get $(e^{ix})'=(\cos(x)+i\sin(x))'=\cos'(x)+i\sin'(x).$ • On the right side, we get $ie^{ix}=i(\cos(x)+i\sin(x))=i\cos(x)+i^2\sin(x)=i\cos(x)-\sin(x).$ ## Derivatives of $$\sin(x)$$ and $$\cos(x)$$ • We equate what we've gotten on the two sides to get: $\cos'(x)+i\sin'(x)=-\sin(x)+i\cos(x).$ • Voilà! Equating the real parts gives $\cos'(x)=-\sin(x),$ and equating the imaginary parts gives $\sin'(x)=\cos(x).$ • Note that these derivatives give limits we couldn't verify before. • Recall that we've made the guess $$\lim_{x\to0}\frac{\sin(x)}{x}=1$$ based on numerical evidence. • Note that we have $\lim_{x\to 0}\frac{\sin(x)}{x}=\sin'(x=0).$ • And now we know that. $\sin'(x=0)=\cos(x=0)=1.$ • Exercise. Find $$\lim_{x\to0}\frac{\cos(x)-1}{x}$$. ## Derivatives of the other trigonometric functions. • Recall the quotient rule: $\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}.$ • Using this, we get the following. $\tan'(x)=\left(\frac{\sin(x)}{\cos(x)}\right)'=\frac{\sin'(x)\cos(x)-\sin(x)\cos'(x)}{\cos(x)^2} =\frac{\cos(x)^2+\sin(x)^2}{\cos(x)^2}=\frac{1}{\cos(x)^2}.$ • Exercises. 3.3. 2, 4, 12, 16, 34, 35, 40, 42, 44, 50	2018-10-18 21:05:35	{"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": 0.9934030175209045, "perplexity": 583.9720369452608}, "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-2018-43/segments/1539583512014.61/warc/CC-MAIN-20181018194005-20181018215505-00151.warc.gz"}
http://stackoverflow.com/questions/4616132/how-to-append-line-into-rtf-using-richtextbox-control	# How to append \line into RTF using RichTextBox control When using the Microsoft RichTextBox control it is possible to add new lines like this... richtextbox.AppendText(System.Environment.NewLine); // appends \r\n However, if you now view the generated rtf the \r\n characters are converted to \par not \line How do I insert a \line control code into the generated RTF? What does't work: Token Replacement Hacks like inserting a token at the end of the string and then replacing it after the fact, so something like this: string text = "my text"; text = text.Replace("\|\|" "\|"); // replace any '\|' chars with a double '\|\|' so they aren't confused in the output. text = text.Replace("\r\n", "_\|0\|_"); // replace \r\n with a placeholder of \|0\| richtextbox.AppendText(text); string rtf = richtextbox.Rtf; rtf.Replace("_\|0\|_", "\\line"); // replace placeholder with \line rtf.Replace("\|\|", "\|"); // set back any \|\| chars to \| This almost worked, it breaks down if you have to support right to left text as the right to left control sequence always ends up in the middle of the placeholder. Sending Key Messages public void AppendNewLine() { Keys[] keys = new Keys[] {Keys.Shift, Keys.Return}; SendKeys(keys); } private void SendKeys(Keys[] keys) { foreach(Keys key in keys) { SendKeyDown(key); } } private void SendKeyDown(Keys key) { user32.SendMessage(this.Handle, Messages.WM_KEYDOWN, (int)key, 0); } private void SendKeyUp(Keys key) { user32.SendMessage(this.Handle, Messages.WM_KEYUP, (int)key, 0); } This also ends up being converted to a \par Is there a way to post a messaged directly to the msftedit control to insert a control character? I am totally stumped, any ideas guys? Thanks for your help! - Adding a Unicode "Line Separator" (U+2028) does work as far as my testing showed: private void Form_Load(object sender, EventArgs e) { richText.AppendText("Hello, World!\u2028"); richText.AppendText("Hello, World!\u2028"); string rtf = richText.Rtf; richText.AppendText(rtf); } When I run the program, I get: Hello, World! Hello, World! {\rtf1\ansi\ansicpg1252\deff0\deflang1031{\fonttbl{\f0\fnil\fcharset0 Courier New;}} {\colortbl ;\red255\green255\blue255;} \viewkind4\uc1\pard\cf1\f0\fs17 Hello, World!\line Hello, World!\line\par } It did add \line instead of \par. - A note, though, mono's RichTextBox emulation is rather broken and, among other imcompatibilities, doesn't understand the line separator. It appears as a box character in the text. – Peter Remmers Jan 15 '12 at 11:42 This worked like a charm for me, it allowed me to remove the horrible hack we had in place. Our customers will thank you! – Steve Sheldon Jan 25 '12 at 20:25 Since you want to use a different RTF code, I think you may need to forget about the simplistic AppendText() method and manipulate the .Rtf property of your RichTextBox directly instead. Here is a sample (tested) to demonstrate: RichTextBox rtb = new RichTextBox(); //this just gets the textbox to populate its Rtf property... may not be necessary in typical usage rtb.AppendText("blah"); rtb.Clear(); string rtf = rtb.Rtf; //exclude the final } and anything after it so we can use Append instead of Insert StringBuilder richText = new StringBuilder(rtf, 0, rtf.LastIndexOf('}'), rtf.Length /* this capacity should be selected for the specific application */); for (int i = 0; i < 5; i++) { string lineText = "example text" + i; richText.Append(lineText); //add a \line and CRLF to separate this line of text from the next one richText.AppendLine(@"\line"); } //Add back the final } and newline richText.AppendLine("}"); System.Diagnostics.Debug.WriteLine("Original RTF data:"); System.Diagnostics.Debug.WriteLine(rtf); System.Diagnostics.Debug.WriteLine("New Data:"); System.Diagnostics.Debug.WriteLine(richText.ToString()); //Write the RTF data back into the RichTextBox. //WARNING - .NET will reformat the data to its liking at this point, removing //any unused colors from the color table and simplifying/standardizing the RTF. rtb.Rtf = richText.ToString(); //Print out the resulting Rtf data after .NET (potentially) reformats it System.Diagnostics.Debug.WriteLine("Resulting Data:"); System.Diagnostics.Debug.WriteLine(rtb.Rtf); Output: Original RTF data: {\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset0 Microsoft Sans Serif;}} \viewkind4\uc1\pard\f0\fs17\par } New RTF Data: {\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset0 Microsoft Sans Serif;}} \viewkind4\uc1\pard\f0\fs17\par example text0\line example text1\line example text2\line example text3\line example text4\line } Resulting RTF Data: {\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset0 Microsoft Sans Serif;}} \viewkind4\uc1\pard\f0\fs17\par example text0\line example text1\line example text2\line example text3\line example text4\par } - if you are using paragraphs to write to richtextbox you can use the LineBreak() same code shown below Paragraph myParagraph = new Paragraph(); FlowDocument myFlowDocument = new FlowDocument(); // Add some Bold text to the paragraph	2016-02-05 23:05:46	{"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": 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": 0.5883851647377014, "perplexity": 13922.488658928714}, "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-2016-07/segments/1454701145519.33/warc/CC-MAIN-20160205193905-00059-ip-10-236-182-209.ec2.internal.warc.gz"}
https://ai.stackexchange.com/questions/28289/what-are-the-noisy-factors-leading-to-overfitting	# What are the 'noisy factors' leading to overfitting? Consider the following excerpt from section 5.5 Regularization (p. 13) of this chapter Logistic Regression. There is a problem with learning weights that make the model perfectly match the training data. If a feature is perfectly predictive of the outcome because it happens to only occur in one class, it will be assigned a very high weight. The weights for features will attempt to perfectly fit details of the training set, in fact too perfectly, modeling noisy factors that just accidentally correlate with the class. This problem is called overfitting. What are the 'noisy factors' here? Does it refer to the features that are irrelevant to the class label? Or does it mean the noise/errors in the values taken by features that accidentally correlate with the class label? • I am only referring the decision boundary to be a line for simplicity, more often than not it is a hyperplane which is difficult to visualize and spans over n dimensions where n is the dimensionality of your feature space. • The explanation is toned in a more general way for emphasizing explainability. What are the 'noisy factors' here? Does it refers to the features that are irrelevant to the class label? • Not Necessarily. Or does it mean the noise/errors in the values taken by features that accidentally correlate with the class label? • I'm not quite sure I understand. # However Noisy factors as the literature puts it are outliers in a finite data class. Imagine a dataset where we are asked to calculate the average of a set of numbers. Lets say the numbers are the set S = {2,2,2,2,2,2,2,1000}. The mean value in the case mentioned is 2 if it wasn't for the 1000 at the end. • 1000 could be an outlier when you are trying to approximate the mean of the set with some algorithm. The algorithm is more likely to encounter a 2 in the unseen test set than the 1000 it encountered once in a training set. • When an algorithm like Linear Regressions "learns" something it is actually learning the weights and intercepts which modify the position of the line in the decision space. • The modifications are Translation(Additive operations) and Rotation(Multiplicative operations) to the said line. It is commonly referred to as the "weights" and "biases" respectively in case of a simple line Y = mx + c where m is the slope and c is the intercept. • The idea behind noise in the above text is that this: When an algorithm is trying to "learn" these weights you would want it to ignore the outliers - "Generalize" but you also want it to not ignore the feature data completely and introduce randomness - "Specialization". • How well the line is placed in your feature space is what determines your Classification Effectiveness. • In an ideal world you would want your decision boundary to ignore such outliers which are called noise in the above literature. # Pictures (because everyone likes them) • In the picture below the left image is what a good decision boundary is and the right image is what an overfitted Decision boundary is. • In the left-image case you are trading misclassifications at the cost of better generalizability(Consider that more likely you are going to see an x in the 2nd quadrant and the o was maybe an outlier). You do not fit your training set completely with a 100% accuracy. # TL;DR You want the model to learn that the x(s) are on the right and the o(s) are on the left but not so specifically as to which x was where. The x in the 4th quadrant and the o in the second quadrant are noisy factors.	2021-10-28 09:11:21	{"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": 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": 0.692541241645813, "perplexity": 599.3793035141517}, "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/1634323588282.80/warc/CC-MAIN-20211028065732-20211028095732-00011.warc.gz"}
http://gatkforums.broadinstitute.org/gatk/discussion/1886/significant-difference-in-vqsr-or-variantannotation-between-v1-6-and-v2-2-10	The current GATK version is 3.7-0 Examples: Monday, today, last week, Mar 26, 3/26/04 #### Howdy, Stranger! It looks like you're new here. If you want to get involved, click one of these buttons! You can opt in to receive email notifications, for example when your questions get answered or when there are new announcements, by following the instructions given here. #### ☞ Did you remember to? 1. Search using the upper-right search box, e.g. using the error message. 3. Include tool and Java versions. 4. Tell us whether you are following GATK Best Practices. 5. Include relevant details, e.g. platform, DNA- or RNA-Seq, WES (+capture kit) or WGS (PCR-free or PCR+), paired- or single-end, read length, expected average coverage, somatic data, etc. 6. For tool errors, include the error stacktrace as well as the exact command. 7. For format issues, include the result of running ValidateSamFile for BAMs or ValidateVariants for VCFs. 8. For weird results, include an illustrative example, e.g. attach IGV screenshots according to Article#5484. 9. For a seeming variant that is uncalled, include results of following Article#1235. #### ☞ Formatting tip! Wrap blocks of code, error messages and BAM/VCF snippets--especially content with hashes (#)--with lines with three backticks ( ) each to make a code block as demonstrated here. GATK 3.7 is here! Be sure to read the Version Highlights and optionally the full Release Notes. # Significant difference in VQSR or VariantAnnotation between v1.6 and v2.2-10 Posts: 14 edited November 2012 Hi, I observed a significant difference of the variant call sets from the same exomes between v1.6 and v2.2(-10). In fact, I observed a significant decrease in the overall novel TiTv in the latter call sets from around 2.6 to 2.1 at TruthSensitivity threshold at 99.0. When I looked at a sample to compare variant sites using VariantEval, it showed that Filter JexlExpression Novelty nTi nTv tiTvRatio called Intersection known 14624 4563 3.2 called Intersection novel 856 312 2.74 called filterIngatk22-gatk16 known 264 132 2 called filterIngatk22-gatk16 novel 28 18 1.56 called gatk16 known 3 1 3 called gatk16 novel 1 1 1 called gatk22-filterIngatk16 known 258 94 2.74 called gatk22-filterIngatk16 novel 144 425 0.34 called gatk22 known 2 2 1 called gatk22 novel 17 30 0.57 filtered FilteredInAll known 1344 649 2.07 filtered FilteredInAll novel 1076 1642 0.66 The novel TiTv of new calls in v2.2 not found in v1.6 or called in v2.2 but filtered in v1.6 demonstrated novel TiTv around 0.5. So I suspect that VQSLOD scoring (or ranking) of SNPs was changed substantially in somewhat an unfavorable way. The major updates in v2.2 affecting my result were BQSRv2, ReduceReads, UG and VariantAnnotation. (Too many things to pin-point the culprit...) The previous BAM processing and variant calls were made using v1.6. For the new call set, I used v2.1-9 (so after serious bug fix in ReduceReads, thank you for the fix) for BQSRv2 and ReduceReads and v2.2-10 for UG and VQSR. As a first clue, I found that distribution of FS values changed dramatically from the v1.6 (please see attached plots). Although I recognized that FS value calculations were recently updated, the distribution of previous FS values (please see attached) makes more sense for me because the current FS values do not seem to provide us information to classify true positives and false positives. Katsuhito Tagged: • Posts: 14 Let me try it. Thanks. • Posts: 14 Hi Geraldine, I've just realized that I also have a variant call set that were called and filtered by using GATK v2.1-9 even though this set included other samples as well. I'm attaching a figure that compares 3 versions of FS vs QD plots. I think it suggests that ReduceReads is fine but FS calculations are different between v2.1-9 and v2.2-10. Thanks OK, that makes sense -- we recently made some changes to how the FS is calculated; seems to be a likely culprit. We'll take a closer look and get back to you once we have a clear idea of the problem. Geraldine Van der Auwera, PhD Hi Katsuhito, could you please submit a detailed bug report as explained here? Thanks! Geraldine Van der Auwera, PhD • Posts: 14 Hi Geraldine, For this issue, it's ambiguous which region I should select because I didn't see any error messages and FS values differ across many (if not all) regions as can be seen in the figure attached previously. Do you think that some portion of VCF files for the sites shared between callsets v1.6 and v2.1-10 would be enough to see the difference or do you really need snippets of BAMs? Let me know. Thanks Ah, fair point. We're definitely going to need BAM snippets, preferably containing regions where you find some different sites and some shared sites, if you can identify a region of reasonable length that contains all that. Geraldine Van der Auwera, PhD • Posts: 14 Hi Geraldine, I've uploaded the requested BAM snippets, named bam_for_FS_check.tar.gz. I confirmed that v2.1-9 and v2.2-10 UGs produce very different FS values. Hope it helps to solve the issue. Thanks It looks like you've uploaded an invalid bam file. samtools view bam_for_FS_check.bam ` Can you please upload a valid bam file? Also, if you are not using a standard reference from our bundle, I'll need you to upload the reference files (fasta, fai, and dict) too. Eric Banks, PhD -- Director, Data Sciences and Data Engineering, Broad Institute of Harvard and MIT • Posts: 14 Let me check it again. I could run variant calling for the bam file with the same name. Thanks (PS. I'm using humang_g1k_v37_decoy.fasta in the bundle.) • Posts: 14 Sorry, no, I uploaded a wrong one. • Posts: 14 I've uploaded the file (with the same name, overwritten). Thanks • Posts: 14 Ok, in that case, another questions come to my mind: can the FisherStrand annotation from v2.1-9 be reliable? Does this issue affect only to v2.2 series? Thanks	2017-05-28 04:45:37	{"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": 0.41512516140937805, "perplexity": 8228.003564271665}, "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-2017-22/segments/1495463609598.5/warc/CC-MAIN-20170528043426-20170528063426-00040.warc.gz"}
https://ask.sagemath.org/answers/27008/revisions/	# Revision history [back] You can just modify the layout to list auto_update in the second column. I believe this is what you are looking for. Based on the post by kcrisman at http://ask.sagemath.org/question/9999/using-layout-option-with-auto_updatefalse/, you have: @interact(layout={'top': [["ansA",'auto_update', "ansB"]]}) def _(ansA=4,ansB=5,auto_update=False): print ansA+ansB You can just modify the layout to list auto_update in the second column. I believe this is what you are looking for. Based on the post by kcrisman at http://ask.sagemath.org/question/9999/using-layout-option-with-auto_updatefalse/, you have: @interact(layout={'top': [["ansA",'auto_update', "ansB"]]}) def _(ansA=4,ansB=5,auto_update=False): print ansA+ansB This works for both sagecell and sagenb (version 6.7) in the same way. In SMC, the result is a checkbox in the second column. Checking or unchecking the box results in evaluation.	2020-11-27 00:35:22	{"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": 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": 0.32401931285858154, "perplexity": 2635.6180047087123}, "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-2020-50/segments/1606141189030.27/warc/CC-MAIN-20201126230216-20201127020216-00413.warc.gz"}
https://www.edaboard.com/threads/what-is-the-integration-of-square-wave.338297/	What is the integration of square wave ? Status Not open for further replies. vetriece86@gmail.com Junior Member level 2 I know that integration of Square wave will get triangular wave. Can anyone explain me with an example with necessary derivations? It will be very much helpful to me. Thank you. Tahmid Integrating a constant gives you a linear relationship: ie integrating a constant k (dx) results in kx (+constant of integration). A square wave is just two constant levels (eg +1, -1). So, that results in the triangle wave. chuckey integration is finding how the area under a curve varies with time. So with a square wave in the first little time interval (t), you have V X t Voltseconds which has some value. now at the end of the next time interval you have another V X t. So for every t, you clock up an extra (V X t) volt seconds, so you have a relationship where the voltage at time T, is V X (T/t) which is similar to Y = mX where m is the slope (T/t). Frank upand_at_them Points: 2	2022-07-04 11:37:57	{"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": 0.8140356540679932, "perplexity": 852.0040186506984}, "config": {"markdown_headings": false, "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-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00386.warc.gz"}
https://zbmath.org/?q=an:0867.58048	# zbMATH — the first resource for mathematics Induced expansion for quadratic polynomials. (English) Zbl 0867.58048 The present paper deals with a class $$\mathcal F$$ of maps defined as follows. A map $$f:[-1,1]\to [-1,1]$$ is said to be unimodal if $$f(-1)=-1$$ and there is an orientation-reversing diffeomorphism $$h:[0,1]\to [-1,a]$$ (with $$0<a<1$$) such that $$f(x)=h(x^2)$$. Further, let $$\eta>0$$. We say that a unimodal map $$f(x)=h(x^2)$$ belongs to the class $${\mathcal F}_\eta$$ if $$h^{-1}$$ has a conformal extension which maps the upper half-plane into the lower half-plane, and $$h$$ has a real analytic extension to some open interval $$U\supset [0,1]$$ as a diffeomorphism onto $$(-1-\eta,1+\eta)$$. Finally we define $${\mathcal F}=\bigcup_{\eta>0}{\mathcal F}_\eta$$. All maps from $${\mathcal F}$$ have non-positive Schwarzian derivative. In particular, $$\mathcal F$$ includes the family of quadratic polynomials $$f_a(x)=a-(a+1)x^2$$. The main results of the paper are Theorems 1 and 2 below. Theorem 1. Let $$f\in {\mathcal F}_\eta$$ be non-renormalizable (that is, there is no proper subinterval $$I$$ of $$[-1,1]$$ containing 0 with the properties $$f^n(I)\subset I$$ and $$f^n(\partial I)\subset \partial I$$ for some $$n>1$$). Then $$f$$ is expansion inducing, that is, there are an open, dense and having full measure subset $$O$$ of $$(-q,q)$$ and a map $$F:O\to (1-q,q)$$ satisfying the following properties (here $$q$$ is the only fixed point of $$f$$ in $$[0,1]$$): (i) If $$O_i$$ is a connected component of $$O$$ then $$F\|_{O_i}=f^{n_i}\|_{O_i}$$ for some $$n_i$$ and $$F(O_i)=(1-q,q)$$. (ii) $$F$$ is expanding and has bounded distortion, which means that there are constants $$K>1$$ and $$D>0$$ such that $$\|F'(x)\|>K$$ and $$\|(\log F')'(x)\|<D$$ for any $$x\in O$$. Moreover, $$D$$ depends only on $$\eta$$. Theorem 2. Let $$f\in {\mathcal F}_\eta$$ be renormalizable and let $$I\ni 0$$ be a maximal subinterval of $$[-1,1]$$ such that $$f^n(I)\subset I$$ and $$f^n(\partial I)\subset \partial I$$ for some $$n>1$$. Define a point $$x\in [-1,1]$$ to be almost parabolic with period $$m$$ and depth $$k$$ provided that the derivative of $$f^m$$ at $$x$$ is one, $$f^m$$ is monotone between $$x$$ and $$0$$, and $$k$$ consecutive images $$f^m(0),\dots,f^{km}(0)$$ are between $$x$$ and $$0$$, and denote by $$k(n)$$ the maximum of depths of almost parabolic points with periods less than $$n$$. Specify a number $$D>0$$. Then, for every given $$k$$, there is a number $$N(\eta,D,k)$$ not depending on $$f$$ so that if $$n>N(\eta,D,k)$$ and $$k(n)\leq k$$, then $$f^n\|_I$$ is conjugate to a map from $${\mathcal F}_D$$ via a linear transformation. Theorem 1 implies that if $$f\in {\mathcal F}$$ is non-renormalizable then the $$\omega$$-limit set of almost all points from $$[-1,1]$$ is the whole interval $$[f^2(0),f(0)]$$; this result was also previously proved (in the more general setting of maps with non-positive Schwarzian derivative whose only critical point $$c$$ satisfies $$f''(c)\neq 0$$) by M. Lyubich [Ann. Math., II. Ser. 140, No. 2, 347-404 (1994; Zbl 0821.58014)]. Theorem 2 is used by the second author [‘Hyperbolicity is dense in the real quadratic family’, to appear in Ann. Math.], where it is shown that the set of parameters $$a$$ such that the polynomial $$f_a(x)=a-(a+1)x^2$$ has a hyperbolic periodic attractor is dense in $$[-1,1]$$. Both Theorems 1 and 2 are consequences of a technical Theorem C, where, roughly speaking, so-called “decaying of box geometry” is demonstrated for a special type of maps called box mappings. The proof uses a mixture of real and complex tools. While the significance of the results of the paper is obvious, in the reviewer’s opinion they are presented in a somewhat involved way (besides, there are some misprints: in the paper it is written “upper half-plane” instead of “lower half-plane” in the definition of $${\mathcal F}_\eta$$, and $$1/2$$ instead of $$0$$ in Theorem 2). Hence an interested but non-truly specialist reader may also wish to check the introduction to M. Jakobson and G. Świątek [Ergodic Theory Dyn. Syst. 14, No. 4, 721-755 (1994; Zbl 0830.58019)] (which is generalized by the paper under review) and Chapter V from the monograph by W. de Melo and S. van Strien [‘One-dimensional dynamics’. Berlin: Springer-Verlag (1993; Zbl 0791.58003)]. ##### MSC: 37D99 Dynamical systems with hyperbolic behavior 37F99 Dynamical systems over complex numbers 37A99 Ergodic theory 37C70 Attractors and repellers of smooth dynamical systems and their topological structure Full Text: ##### References: [1] A. BLOKH and M. LYUBICH , Non-existence of wandering intervals and structure of topological attractors for one dimensional dynamical systems (Erg. Th. and Dyn. Sys., Vol. 9, 1989 , pp. 751-758). MR 91e:58101 \| Zbl 0665.58024 · Zbl 0665.58024 · doi:10.1017/S0143385700005319 [2] B. BRANNER and J. H. HUBBARD , The iteration of cubic polynomials , Part II : patterns and parapatterns (Acta Math., Vol. 169, 1992 , pp. 229-325). MR 94d:30044 \| Zbl 0812.30008 · Zbl 0812.30008 · doi:10.1007/BF02392761 [3] A. DOUADY and J. H. HUBBARD , On the dynamics of polynomial-like mappings (Ann. Sci. Ec. Norm. Sup. (Paris), Vol. 18, 1985 , pp. 287-343). Numdam \| MR 87f:58083 \| Zbl 0587.30028 · Zbl 0587.30028 · numdam:ASENS_1985_4_18_2_287_0 · eudml:82160 [4] J. GRACZYK , Ph. D. Thesis (Mathematics Department of Warsaw University ( 1990 ) ; also : Dynamics of non-degenerate upper maps , preprint of Queen’s University at Kingston, Canada, 1991 ). [5] J. GRACZYK and G. ŚWIATEK , Critical circle maps near bifurcation (Stony Brook IMS preprint, 1991 , Proposition 2). [6] J. GUCKENHEIMER , Limit sets of S-unimodal maps with zero entropy (Commun. Math. Phys., Vol. 110, 1987 , pp. 655-659). Article \| MR 88i:58111 \| Zbl 0625.58027 · Zbl 0625.58027 · doi:10.1007/BF01205554 · minidml.mathdoc.fr [7] J. GUCKENHEIMER and S. JOHNSON , Distortion of S-unimodal maps (Annals of Math., Vol. 132, 1990 , pp. 71-130). MR 91g:58157 \| Zbl 0708.58007 · Zbl 0708.58007 · doi:10.2307/1971501 [8] F. HOFBAUER , F. and G. KELLER , Some remarks about recent results on S-unimodal maps (Annales de l’Institut Henri Poincaré, Physique Théorique, Vol. 53, 1990 , pp. 413-425). Numdam \| Zbl 0721.58018 · Zbl 0721.58018 · numdam:AIHPA_1990__53_4_413_0 · eudml:76513 [9] M. JAKOBSON , Absolutely continuous invariant measures for one-parameter families of one-dimensional maps (Commun. Math. Phys., Vol. 81, 1981 , pp. 39-88). Article \| MR 83j:58070 \| Zbl 0497.58017 · Zbl 0497.58017 · doi:10.1007/BF01941800 · minidml.mathdoc.fr [10] M. JAKOBSON and G. ŚWIATEK , Metric properties of non-renormalizable S-unimodal maps (preprint IHES, no. IHES/M/91/16, 1991 ). [11] M. JAKOBSON and G. ŚWIATEK , Quasisymmetric conjugacies between unimodal maps (Stony Brook preprint, Vol. 16, 1991 ). [12] G. KELLER and T. NOWICKI , Fibonacci maps revisited (manuscript, 1992 ). [13] O. LEHTO and K. VIRTANEN , Quasikonforme Abbildungen (Springer-Verlag, Berlin-Heidelberg-New York, 1965 ). MR 32 #5872 \| Zbl 0138.30301 · Zbl 0138.30301 [14] M. LYUBICH , Milnor’s attractors, persistent recurrence and renormalization , in (Topological methods in modern mathematics, Publish or Perish, Inc., Houston TX, 1993 ). MR 94e:58082 \| Zbl 0797.58050 · Zbl 0797.58050 [15] M. LYUBICH and J. MILNOR , The dynamics of the Fibonacci polynomial (Jour. of the AMS, Vol. 6, 1993 , pp. 425-457). MR 93h:58080 \| Zbl 0778.58040 · Zbl 0778.58040 · doi:10.2307/2152804 [16] M. MARTENS , Ph. D. thesis (Math. Department of Delf University of Technology, 1990 ; also : IMS preprint, Vol. 17, 1992 ). [17] J. MILNOR , The Yoccoz theorem on local connectivity of Julia sets . A proof with pictures (class notes, Stony Brook, 1991 - 1992 ). [18] W. DE MELO and S. VAN STRIEN , One-Dimensional Dynamics (Springer-Verlag, New York, 1993 ). MR 95a:58035 \| Zbl 0791.58003 · Zbl 0791.58003 [19] C. PRESTON , Iterates of maps on an interval (Lecture Notes in Mathematics, Vol. 999, Berlin, Heidelberg, New York : Springer, 1983 ). MR 85c:58058 \| Zbl 0582.58001 · Zbl 0582.58001 [20] D. SULLIVAN , Bounds, quadratic differentials and renormalization conjectures (to appear in American Mathematical Society Centennial Publications, Vol. 2, American Mathematical Society, Providence, R.I., 1991 ). Zbl 0936.37016 · Zbl 0936.37016 [21] G. ŚWIATEK , Hyperbolicity is dense in the real quadratic family (preprint Stony Brook, 1992 ). [22] O. TEICHMÜLLER , Untersuchungen über konforme und quasikonforme Abbildung (Deutsche Mathematik, Vol. 3, pp. 621-678). Zbl 0020.23801 \| JFM 64.0313.06 · Zbl 0020.23801 · www.emis.de [23] J.-C. YOCCOZ , unpublished results. [24] J. J. P. VEERMAN and F. M. TANGERMAN , Scalings in circle maps (1) (Commun. in Math. Phys., Vol. 134, 1990 , pp. 89-107). Article \| MR 92h:58106 \| Zbl 0723.58026 · Zbl 0723.58026 · doi:10.1007/BF02102091 · minidml.mathdoc.fr 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.	2021-04-21 18:34:19	{"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": 0.842013418674469, "perplexity": 1115.3215131021677}, "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-2021-17/segments/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00551.warc.gz"}
http://openstudy.com/updates/50fd9144e4b010aceb33ae53	## anonymous 3 years ago Rearange this equation to isolate a (a+b/c)(d-e) = f Help !! 1. Mertsj $\frac{a+b}{c}(d-e)=f$ 2. Mertsj Is that the problem? 3. anonymous umm the "a" is isolated only b/c 4. anonymous \|dw:1358802569854:dw\| 5. anonymous @Mertsj 6. Mertsj First divide both sides by d-e 7. anonymous i got this one, do you mind take a look at my next problem ? 8. Mertsj What is it?	2016-12-09 04:06:58	{"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": 0.5832317471504211, "perplexity": 11398.174876330473}, "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-2016-50/segments/1480698542680.76/warc/CC-MAIN-20161202170902-00481-ip-10-31-129-80.ec2.internal.warc.gz"}
http://html.rhhz.net/HEBGCDXXB/html/201805001.htm	«上一篇文章快速检索高级检索哈尔滨工程大学学报 2019, Vol. 40 Issue (4): 676-682 DOI: 10.11990/jheu.201805001 0 ### 引用本文 DING Hu. Free vibration analysis of a functionally graded rectangular plate with central circular cutout[J]. Journal of Harbin Engineering University, 2019, 40(4), 676-682. DOI: 10.11990/jheu.201805001. ### 文章历史 Free vibration analysis of a functionally graded rectangular plate with central circular cutout DING Hu No. 704 Research Institute, China Shipbuilding Industry Corporation, Shanghai 200031, China Abstract: Based on first-order shear deformation theory and isogeometric analysis, a model of functionally graded rectangular plate with a circular cutout was built to study its free vibration characteristics.The non-uniform rational B-spline (NURBS) basis functions used to precisely model the geometry also described the solution field, which achieved seamless integration of computer-aided design and finite element analysis.The accuracy of the present method was validated through comparison with other solutions.The influence of boundary conditions, gradient index, aspect ratio, plate thickness, and radius of circular cutout on the natural frequency of the plate was investigated in detail.Numerical examples show that the isogeometric analysis has relatively high accuracy, and the method can effectively solve the free vibration of a functionally graded rectangular plate with a central circular cutout. Keywords: isogeometric analysis functionally graded first-order shear deformation theory rectangular plate circular cutout NURBS basis functions free vibration 1 功能梯度Mindlin板的基本方程 1.1 带圆开孔的功能梯度矩形板 Download: 图 1 带圆开孔的功能梯度矩形板 Fig. 1 Geometry of FG rectangle plate with a circular cutout ${P_e} = {P_c}{V_c} + {P_m}\left( {1 - {V_c}} \right)$ (1) ${V_c} = {\left( {1/2 + z/h} \right)^n}$ (2) Download: 图 2 体积分数随厚度的变化 Fig. 2 Variation of volume fraction Vc with the plate thickness 1.2 Mindlin板理论 $\left\{ \begin{array}{l} U\left( {x,y,z} \right) = {u_0}\left( {x,y} \right) + z{\theta _x}\left( {x,y} \right)\\ V\left( {x,y,z} \right) = {v_0}\left( {x,y} \right) + z{\theta _y}\left( {x,y} \right)\\ W\left( {x,y,z} \right) = {w_0}\left( {x,y} \right) \end{array} \right.$ (3) $\left\{ \begin{array}{l} \mathit{\boldsymbol{\varepsilon }} = \left[ {\begin{array}{{20}{c}} {{\varepsilon _x}}\\ {{\varepsilon _y}}\\ {{\gamma _{xy}}} \end{array}} \right] = {\mathit{\boldsymbol{\varepsilon }}_m} + z{\mathit{\boldsymbol{\varepsilon }}_b} = \left[ {\begin{array}{{20}{c}} {{u_{0,x}}}\\ {{v_{0,y}}}\\ {{u_{0,y}} + {v_{0,x}}} \end{array}} \right] + z\left[ {\begin{array}{{20}{c}} {{\theta _{x,x}}}\\ {{\theta _{y,y}}}\\ {{\theta _{x,y}} + {\theta _{y,x}}} \end{array}} \right]\\ \mathit{\boldsymbol{\gamma }} = \left[ {\begin{array}{{20}{c}} {{\gamma _{xz}}}\\ {{\gamma _{yz}}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {{\theta _x} + {w_{0,x}}}\\ {{\theta _y} + {w_{0,y}}} \end{array}} \right] \end{array} \right.$ (5) $\mathit{\boldsymbol{\sigma }} = \left[ {\begin{array}{{20}{c}} {{\sigma _x}}\\ {{\sigma _y}}\\ {{\tau _{xy}}} \end{array}} \right] = \frac{{{E_e}}}{{1 - \nu _e^2}}\left[ {\begin{array}{{20}{c}} 1&{{\nu _e}}&0\\ {{\nu _e}}&1&0\\ 0&0&{\frac{{1 - {\nu _e}}}{2}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {{\varepsilon _x}}\\ {{\varepsilon _y}}\\ {{\gamma _{xy}}} \end{array}} \right] = \mathit{\boldsymbol{D\varepsilon }}$ (6) $\mathit{\boldsymbol{\tau }} = \left[ {\begin{array}{{20}{c}} {{\tau _{xz}}}\\ {{\tau _{yz}}} \end{array}} \right] = \frac{{k{E_e}}}{{2\left( {1 + {\nu _e}} \right)}}\left[ {\begin{array}{{20}{c}} 1&0\\ 0&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} {{\gamma _{xz}}}\\ {{\gamma _{yz}}} \end{array}} \right] = {\mathit{\boldsymbol{D}}_s}\mathit{\boldsymbol{\gamma }}$ (7) 2 Mindlin板自由振动的等几何分析 2.1 B样条基函数 p=0时， ${N_{i,0}}\left( \xi \right) = \left\{ \begin{array}{l} 1,\;\;\;\;{\xi _i} \le \xi \le {\xi _{i + 1}}\\ 0,\;\;\;\;其他 \end{array} \right.$ (8) p>0时， ${N_{i,p}}\left( \xi \right) = \frac{{\xi - {\xi _i}}}{{{\xi _{i + p}} - {\xi _i}}}{N_{i,p - 1}}\left( \xi \right) + \frac{{{\xi _{i + p + 1}} - \xi }}{{{\xi _{i + p + 1}} - {\xi _{i + 1}}}}{N_{i + 1,p - 1}}\left( \xi \right)$ (9) 2.2 NURBS基本概念 NURBS基函数是B样条基函数的有理形式，其定义为[19] $R_i^p\left( {\xi ,\eta } \right) = \frac{{{w_i}{N_{i,p}}\left( \xi \right)}}{{\sum\limits_{i = 1}^n {{w_i}{N_{i,p}}\left( \xi \right)} }}$ (10) $R_{i,j}^{p,q}\left( {\xi ,\eta } \right) = \frac{{{w_{i,j}}{N_{i,p}}\left( \xi \right){M_{j,q}}\left( \eta \right)}}{{\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^m {{w_{i,j}}{N_{i,p}}\left( \xi \right){M_{j,q}}\left( \eta \right)} } }}$ (11) NURBS曲面可由控制点Bi, j与NURBS基函数得[19] $S\left( {\xi ,\eta } \right) = \sum\limits_{i = 1}^n {\sum\limits_{j = 1}^m {R_{i,j}^{p,q}\left( {\xi ,\eta } \right){B_{i,j}}} }$ (12) 2.3 基于NURBS的等几何分析 ${\mathit{\boldsymbol{x}}^h}\left( {\xi ,\eta } \right) = \sum\limits_{A = 1}^N {{R_A}\left( {\xi ,\eta } \right){\mathit{\boldsymbol{x}}_A}}$ (13) ${\mathit{\boldsymbol{u}}^h}\left( {\xi ,\eta } \right) = \sum\limits_{A = 1}^N {{R_A}\left( {\xi ,\eta } \right){\mathit{\boldsymbol{u}}_A}}$ (14) ${\left[ {\begin{array}{{20}{c}} {\mathit{\boldsymbol{\varepsilon }}_m^{\rm{T}}}&{\mathit{\boldsymbol{\varepsilon }}_b^{\rm{T}}}&{{\mathit{\boldsymbol{\gamma }}^{\rm{T}}}} \end{array}} \right]^{\rm{T}}} = \sum\limits_{A = 1}^N {\left[ {\begin{array}{{20}{c}} {{{\left( {\mathit{\boldsymbol{B}}_A^m} \right)}^{\rm{T}}}}&{{{\left( {\mathit{\boldsymbol{B}}_A^b} \right)}^{\rm{T}}}}&{{{\left( {\mathit{\boldsymbol{B}}_A^s} \right)}^{\rm{T}}}} \end{array}} \right]{\mathit{\boldsymbol{u}}_A}}$ (15) $\mathit{\boldsymbol{B}}_A^m = \left[ {\begin{array}{{20}{c}} {{R_{A,x}}}&0&0&0&0\\ 0&{{R_{A,y}}}&0&0&0\\ {{R_{A,y}}}&{{R_{A,x}}}&0&0&0 \end{array}} \right]$ (16) $\mathit{\boldsymbol{B}}_A^b = \left[ {\begin{array}{{20}{c}} 0&0&0&{{R_{A,x}}}&0\\ 0&0&0&0&{{R_{A,y}}}\\ 0&0&0&{{R_{A,y}}}&{{R_{A,x}}} \end{array}} \right]$ (17) $\mathit{\boldsymbol{B}}_A^s = \left[ {\begin{array}{{20}{c}} 0&0&{{R_{A,x}}}&{{R_A}}&0\\ 0&0&{{R_{A,y}}}&0&{{R_A}} \end{array}} \right]$ (18) $\left( {\mathit{\boldsymbol{K}} - {\omega ^2}\mathit{\boldsymbol{M}}} \right)\mathit{\boldsymbol{X}} = 0$ (19) $\begin{array}{{20}{c}} {\mathit{\boldsymbol{K}} = \int\limits_\Omega {\left[ {\begin{array}{{20}{c}} {{\mathit{\boldsymbol{B}}^m}}&{{\mathit{\boldsymbol{B}}^b}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} \mathit{\boldsymbol{D}}&{z\mathit{\boldsymbol{D}}}\\ {z\mathit{\boldsymbol{D}}}&{{z^2}\mathit{\boldsymbol{D}}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {{\mathit{\boldsymbol{B}}^m}}\\ {{\mathit{\boldsymbol{B}}^b}} \end{array}} \right]{\rm{d}}\mathit{\Omega }} + }\\ {\int\limits_\mathit{\Omega } {{{\left( {{\mathit{\boldsymbol{B}}^s}} \right)}^{\rm{T}}}{\mathit{\boldsymbol{D}}_s}{\mathit{\boldsymbol{B}}^s}{\rm{d}}\mathit{\Omega }} } \end{array}$ (20) $\begin{array}{{20}{c}} {\mathit{\boldsymbol{M}} = \int\limits_\mathit{\Omega } {{\mathit{\boldsymbol{N}}^{\rm{T}}}\left[ {\begin{array}{{20}{c}} {{\rho _e}}&0&0&{z{\rho _e}}&0\\ 0&{{\rho _e}}&0&0&{z{\rho _e}}\\ 0&0&{{\rho _e}}&0&0\\ {z{\rho _e}}&0&0&{{z^2}{\rho _e}}&0\\ 0&{z{\rho _e}}&0&0&{{z^2}{\rho _e}} \end{array}} \right]\mathit{\boldsymbol{N}}{\rm{d}}\mathit{\Omega }} }\\ {\mathit{\boldsymbol{N}} = \left[ {\begin{array}{*{20}{c}} {{R_{\rm{A}}}}&0&0&0&0\\ 0&{{R_{\rm{A}}}}&0&0&0\\ 0&0&{{R_{\rm{A}}}}&0&0\\ 0&0&0&{{R_{\rm{A}}}}&0\\ 0&0&0&0&{{R_{\rm{A}}}} \end{array}} \right]} \end{array}$ (22) 3 数值验证 $\begin{array}{l} {\rm{SSSS}}:\\ \;\;\;\;\;\;\;\;x = 0\;与\;x = a:{v_0} = {w_0} = {\theta _y} = 0 \end{array}$ (23) $y = 0\;与\;y = b:{u_0} = {w_0} = {\theta _x} = 0$ (24) $\begin{array}{l} {\rm{CSCS}}:\\ \;\;\;\;\;\;\;\;x = 0\;与\;x = a:{v_0} = {w_0} = {\theta _y} = 0 \end{array}$ (25) $y = 0\;与\;y = b:{u_0} = {v_0} = {w_0} = {\theta _x} = {\theta _y} = 0$ (26) $\begin{array}{l} {\rm{CCCC:}}\\ \;\;\;\;\;\;\;\;四边:{u_0} = {v_0} = {w_0} = {\theta _x} = {\theta _y} = 0 \end{array}$ (27) 3.1 各向同性的带圆开孔矩形板 3.2 边界条件、梯度指数及长宽比对固有频率的影响 Download: 图 3 CCCC矩形板的前6阶模态振型 Fig. 3 The first six mode shapes of CCCC rectangle plate 3.3 板厚度及圆孔半径对固有频率的影响 Download: 图 4 固支矩形板前四阶无量纲频率与圆孔半径r的关系 Fig. 4 The variation of the first four non-dimensional frequencies for clamped rectangle plate with radius r 4 结论 1) 边界的约束越大，对应的频率越高；随着梯度指数的增加，前6阶固有频率逐渐降低；随着长宽比的增大，前6阶固有频率逐渐增大。 2) 前6阶固有频率随板厚度的升高而减小；开孔半径对前四阶频率的影响较为复杂。 3) 借助NURBS强大的几何建模能力，本方法适用于更加复杂的工程实际问题；此外，可以进一步研究力、热、电等多场耦合作用下的功能梯度板的动力学行为。 [1] MIYAMOTO Y, KAYSSER W A, RABIN B H, et al. 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(0)	2019-04-22 17:06:36	{"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": 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": 0.5197581648826599, "perplexity": 12569.837981991972}, "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/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00284.warc.gz"}
http://math.stackexchange.com/questions/95439/find-lim-limits-n-to-infty-frac1n3-sum-limits-k-0n-1k2e-frack	# Find $\lim\limits_{n \to \infty}\frac1{n^3}\sum\limits_{k=0}^{n-1}k^2e^{-\frac{k}{n}}$ Can you help me find $$\lim_{n \to \infty}\frac1{n^3}\sum_{k=0}^{n-1}k^2e^{-\frac{k}{n}} \ \ ?$$ Using Riemann sums, I found it is equal to $2-\frac3{e}$. Is that correct ? Thank you in advance. - uhm - since you know the answer -- what do you want to know, then? – user20266 Dec 31 '11 at 16:51 Just to make sure my answer is right :-) Wolfram Alpha doesn't seem to accept the command. – user20010 Dec 31 '11 at 16:55 The devil is often in the details, so it's possible for your work to be subtly incorrect even if you got the right answer. It would be more beneficial to you, and quicker for the answerers, if you include the entire work rather than just the final answer. – Srivatsan Dec 31 '11 at 17:00 ## 1 Answer This is just $$\lim_{n \to \infty} \frac 1{n^3} \sum_{k=0}^{n-1} k^2 e^{-k/n} = \lim_{n \to \infty} \sum_{k=0}^{n-1} \left( \frac {k+1}n - \frac kn \right) \left( \frac kn \right)^2 e^{- \left( \frac kn \right) } = \int_0^1 x^2 e^{-x} \, dx.$$ Do that and you got your answer (integrate by parts twice). Looks like you already did so I won't detail more. Hope that helps, - Actually, the correct answer is $2-\frac{5}{e}$ and Srivatasan was right ! – user20010 Dec 31 '11 at 17:11 Well I didn't even bother computing the integral... it looks quite elementary. If OP wanted the answer he could've just computed this integral in Wolfram Alpha if he didn't trust himself. =) – Patrick Da Silva Dec 31 '11 at 17:48	2015-01-30 01:17:59	{"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": 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": 0.8443180918693542, "perplexity": 506.82549989887735}, "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-2015-06/segments/1422122192267.50/warc/CC-MAIN-20150124175632-00087-ip-10-180-212-252.ec2.internal.warc.gz"}
http://www.futilitycloset.com/category/puzzles/page/5/	# Footwork A poser from Penn State mathematician Mark Levi’s Why Cats Land on Their Feet (2012): Using only a stopwatch and a sneaker, how can you find an approximate value for $\sqrt{2}$? # Hat Check A puzzle from MIT Technology Review, July/August 2008: Each of three logicians, A, B, and C, wears a hat that displays a positive integer. The number on one of the hats is the sum of the numbers on the other two. They make the following statements: A: “I don’t know my number.” B: “My number is 15.” What numbers appear on hats A and C? # Person to Person The president of a 100-member society receives word that the meeting place must be changed, and he needs to inform the rest of the members. He starts a telephone tree: He informs three members, each of whom informs another three members, and so on until all 100 members have received the news. Using this method, what is the greatest number of members who don’t have to make a call? # Black and White From a Florentine manuscript, 1600. White to mate in two moves. # Moving Day Is it possible to pack six 1 × 2 × 2 blocks and three 1 × 1 × 1 blocks into a 3 × 3 × 3 box? # Cross Purposes A perplexing problem from the Pi Mu Epsilon Journal, Spring 1983: In the little hamlet of Abacinia, the people use two base systems. One resident says, “26 people use my base, base 10, and only 22 people speak base 14.” Another says, “Of the 25 residents, 13 are bilingual and 1 is illiterate.” All the residents speak the truth, but each (naturally) expresses numbers in her own base. How many residents are there? # Coming and Going In 1978 the Chronicle of Higher Education mentioned an old exam question: Q. How far can a dog run into the woods? A. Halfway. The rest of the time he is running out. Harvard’s Richard E. Baym wrote in to take issue with the answer: The correct answer is ‘All the way’. Certainly we understand that the dog is running ‘in’ only until he reaches the middle of the forest, but this is in fact, all the way in. If the dog ran only half ‘in’, he would not yet be at the middle. Indeed if the dog ran halfway in and then ran halfway out, he would still be in the woods. The editors noted, “It occurs to us that the dog’s continued presence there would be useful, in case something happens to that tree that we’ve been hearing about since high school physics — the one that falls when no one is in the forest and since there is no eardum to register sound waves, makes no noise. You know what a fine sense of hearing a dog has. Let him run halfway in (or as Mr. Baym argues, all the way), settle there, and keep an ear cocked for that tree.” (from Robert L. Weber, ed., Science With a Smile, 1992.) # Technicalities In presenting the rules of chess, some writers carelessly say that a pawn that reaches the eighth rank can be promoted to any piece that the player chooses. That’s a bit too generous, as a couple of puzzle composers have noted. In 1941 Leonid Kubbel presented this problem — White is to mate in two moves: It’s not immediately clear how to release Black from his stalemate and still mate him on the next move. The solution is to promote the e7 pawn to a black king! Now it’s Black’s move — he has to play 1. … Kd8, and White can mate both kings with 2. Qd7#! The Polish master Johannes Zukertort offered this one: White is to mate on the move: Here White promotes the pawn to a black knight, ending the game. (Note that it must be a knight — crazy as it seems, this is the only black piece that produces mate.) # Divide and Conquer Facing dental surgery one day, mathematician Matt Parker asked Twitter for a math puzzle to distract him. A friend challenged him to put the digits 1-9 in order so that the first two digits formed a number that was a multiple of 2, the first three digits were a multiple of 3, and so on. Leaving the digits in the conventional order 1234356789 doesn’t work: 12 is divisible by 2 and 123 by 3, but 1234 isn’t evenly divisible by 4. “By the end of my dental procedure, I had some but not all of the digits worked out, but, apparently, you’re not allowed to stay in the dentist’s chair after they’re finished.” At home he finished working out the solution, which is unique. What is it? # Not So Fast José Paluzie offered this chess poser in 1910. White is to move and mate in 1: The key is to notice that the position is illegal: There’s no legal way for the black king to have arrived at a2. Black, desperate to avoid mate, must have put it there when White wasn’t looking. Where did it come from? It doesn’t matter: White can place the black king on any legal square and mate in 1. (From Burt Hochberg’s Chess Braintwisters, 1999.)	2017-02-27 13:36:04	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "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": 0.2503468096256256, "perplexity": 1659.078034550526}, "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-2017-09/segments/1487501172831.37/warc/CC-MAIN-20170219104612-00581-ip-10-171-10-108.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/133012/moving-arguments-inside-group	Moving arguments inside group I have a setup similar to the MWE below in which I have a macro that needs to be called in the document, and also regularly becomes a moving argument (in this case, inserted into the index, though it could really be anything). This macro (\arobustcmd in the example) simply performs some operation on another macro that's hidden away from the end user (making \someothercmd capital or lowercase in this example). It occasionally needs to be changed in a scoped fashion (eg. inside the \begin{@empty}). This scoping works as expected when printing the command (inside the environment it's the new text, outside the environment it's the old text), however, since \arobustcmd gets passed to the index and expanded later, the index entry is always the old text. I'd like to have two index entries, one for the old text, and one for the text set inside the \begin{@empty}. Is there a way to force the \arobustcmd to expand before being written to the index file? I can always rework it to declare two commands (eg. \arobustcmd and \anotherrobustcmd and redefine both of them with \protected@edef inside the \setthetext command instead of the way I've done it here, but I'm curious if there's a way to make my setup work. \documentclass{article} \usepackage[splitindex]{imakeidx} \makeindex[name=anidx] \def\someothercmd{Some text} \newcommand{\arobustcmdStar}{\expandafter\MakeUppercase\someothercmd} \newcommand{\arobustcmdNoStar}{\expandafter\MakeLowercase\someothercmd} \makeatletter \DeclareRobustCommand{\arobustcmd}{\@ifstar\arobustcmdStar\arobustcmdNoStar} \makeatother \newcommand{\setthetext}[1]{\def\someothercmd{#1}} \begin{document} \begin{@empty} \setthetext{new text} \makeatletter\imki@wrindexentry{anidx}{\arobustcmd}{3}\makeatother Some output: \arobustcmd \end{@empty} \makeatletter\imki@wrindexentry{anidx}{\arobustcmd}{3}\makeatother Some output: \arobustcmd \printindex[anidx] \end{document} • Usage of \DeclareRobustCommand is already a problem; the -variant is another; no expansion can take you past it. – egreg Sep 12 '13 at 17:28 • The star variant doesn't really matter, I can remove that. I kind of figured that would be the answer (since that's the whole point of \protected and \DeclareRobustCommand), but I was hoping their would be some other non-obvious solution that I was missing. Thanks anyways though. – Sam Whited Sep 12 '13 at 18:01 • A \protected macro is insensitive to \edef, but it isn't to \expandafter. I didn't really understand your motivation, though. Be careful in using internal commands such as \imki@wrindexentry in a document. – egreg Sep 12 '13 at 18:10 • @egreg It's not actually in a document; it's in a DTX that defines a package and a class. I just made it a document for the purpose of the MWE. – Sam Whited Sep 12 '13 at 18:27 • If someone wants to write a nice explanation of why my original idea doesn't work, I'll happily accept that as the answer. Otherwise I've posted my reworked example which does essentially the same thing (but without the star command). – Sam Whited Sep 12 '13 at 20:00 Since this is probably impossible; I've modified my example to look something like this which works for my use case (even if it isn't quite the same): \documentclass{article} \usepackage[splitindex]{imakeidx} \makeindex[name=anidx] \def\arobustcmd{Some text} \def\anotherrobustcmd{some text} \newcommand{\setthetext}[1]{% \protected@edef\arobustcmd{\expandafter\MakeUppercase#1}% \protected@edef\anotherrobustcmd{\expandafter\MakeLowercase#2}% } \begin{document} \begin{@empty} \setthetext{new text} \makeatletter\imki@wrindexentry{anidx}{\arobustcmd}{3}\makeatother Some output: \arobustcmd \end{@empty} \makeatletter\imki@wrindexentry{anidx}{\arobustcmd}{3}\makeatother Some output: \arobustcmd \printindex[anidx] \end{document} • Why the starting \protected@edef\arobustcmd, if the replacement text is just some text? – egreg Sep 12 '13 at 20:34 • Because of a copy-pasta error (fixed) – Sam Whited Sep 12 '13 at 20:39	2020-02-18 08:14:48	{"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": 0.8078393340110779, "perplexity": 1516.7200532130232}, "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-2020-10/segments/1581875143635.54/warc/CC-MAIN-20200218055414-20200218085414-00527.warc.gz"}
https://nathanielpark.com/notes-template-dad/stress-and-strain-dd4cb5	# stress and strain 10.3. Stress & Strain Materials for this chapter are taken from : 1. CHAPTER 1 CONCEPT OF STRESS & STRAIN. Stress, Strain and Young's Modulus are all factors linked to the performance of a material in a particular setting. 4.Strain is an objectâs response to stress while stress is the force that can cause strain â¦ Steady state, transient, accelerated. Stress, strain, thermal conductivity, magnetic susceptibility and electrical permittivity are all second rank tensors. The function stress_strain_diagram is used to visualize how the stress and strain vary with successive load cycles as described by the hysteresis curve equation. If stress levels are ok, deformations are acceptable and there is no unacceptable strain, then this is a good design. Compressive stress: It is defined as the decrease in length of the body due to applied force. QUESTION: 18. A third rank tensor would look like a three-dimensional matrix; a cube of numbers. We can say that a body is strained due to stress. Elastic and non elastic materials . simple stress is defined as the internal resistance force that opposes the external force per unit area.Tensile Stresses, Compressive Stresses, Shear Stresses, Bending Stresses, Torsion Stresses. The cross-sectional area at the time of necking (in mm 2) is _____ (correct to two decimal places) 1 (b) shows the same bar in compression.The applied forces F are in line and are normal (perpendicular) to the cross-sectional area of the bar.Therefore the bar is said to be subject to direct stress.Direct stress is given the symbol Ï (Greek letter sigma). Strain cannot occur without stress, but stress can occur without strain. The slope of the stress-strain curve in the elastic region is defined as the elastic modulus , E. Structures should be designed so that any applied load would not cause the stress in the structure to be greater than s y . Stress and strain are normalized so that they do not depend on the geometry of the part. For example, when a solid vertical bar is supporting an overhead weight, each particle in the bar pushes on the particles immediately below it. What you are experiencing then is bulk stress, or in other words, pressure.Bulk stress always tends to decrease the volume enclosed by â¦ The cross-sectional area at the start of a test (when the stress and strain values are equal to zero) is 100 mm 2. Chapter-1. Mathematically: E= Stress/Strain Youngâs Modulus E, is generally assumed to be same in tension or Compression and for most of engineering application has high Numerical value. Now we can use our Hook's Law, tau is equal to g times gamma, or rearranging, g is equal to tau divided by gamma, is the shear stress we've calculated is 474 times ten cubed divided by the strain is 0.249 is equal to 1.9 times 10 to the 6th pascals or 1.9 mega pascals and the closest answer is D. Stress, strain, and modulus of elasticity. Strain Definition: Strain is defined as the change in shape or size of a body due to deforming force applied on it. Stress is the resisting tension and strain is the deformity. Strain = extension / original length where, Îµ = strain, lo = the original length e = extension = (l-lo), and l = stretched length Strain has no units because it is a ratio of lengths. The other three types of stress, tension, compression and shear, are non-uniform, or directed, stresses.All rocks in the earth experience a uniform stress at all times. Did you know that the typical stress-strain curve obtained from a uniaxial tensile test is just an approximation? Strain is dimensionless quantity. This is all about the difference between stress and strain in strength of materials. Figure 1 (a) shows a cylindrical bar of cross-sectional area A in tension, whilst Fig. The initial moduli of these elastomers, which depend on the amine content, are much greater than those of many other TPEs in the same hardness range. strain: The amount by which a material deforms under stress or force, given as a ratio of the deformation to the initial dimension of the material and typically symbolized by Îµ is termed the engineering strain. Strain is the deformation resulting from the Stress is the cause, and strain is the result. Young's modulus of elasticity. The unit is newton per square meter (N/m^2), kilogram (force) per square centimeter (kg/cm^2) or pascal. It is written in the belief that a sound introduction to the mechanics of continu ous bodies is essential for students of structural geology and tectonics, just as a sound introduction to physical chemistry is necessary for STRESS AND STRAIN DIAGRAM These properties relate the stresses to the strains and can only be determined by experiment. Stress: The force of resistance per unit area, offered by a body against deformation is known as stress. The correct sequence of creep deformation in a creep curve in order of their elongation is : A. Simple Engineering Stress is similar to Pressure, in that in this instance it is calculated as force per unit area. Stress and Strain is the first topic in Strength of Materials which consist of various types of stresses, strains and different properties of materials which are important while working on them. Stress-Strain diagram¶. The greater the stress, the greater the strain; however, the relation between strain and stress does not need to be linear. Direct Stress and Strain. Typical stressâstrain behavior of several formulations of polyamide thermoplastic elastomers is depicted in Fig. Solved example: Stress and strain. B. Next lesson. The true stress (in MPa) versus true strain relationship for a metal is given by $\sigma=1020\varepsilon^{0.4}$ . Page- 1. Stress and Strain Theory at a Glance (for IES, GATE, PSU) 1.1 Stress When a material is subjected to an external force, a resisting force is set up within the component. Feasibility of strain and SR analysis during stress in this study of almost 200 patients was slightly lower than conventional WMA, but the sensitivity of the test increased with deformation imaging from 75% for WMA alone to 84â88% with the addition of the new techniques. Deformation and Failure; 2 Stress and Strain. In such case, you can simply ignore this partâ¦ unless you want to optimize the element. On the stress strain curve, point B is the breaking stress point. Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek Mechanics of Materials 5th Edition in SI units 2. Fig.1: Stress. It is denoted by a symbol âÏâ. Ductile Material: Ductile materials are materials that can be plastically twisted with no crack. Outline. When you dive into water, you feel a force pressing on every part of your body from all directions. The stress-strain curve is the simplest way to describe the mechanical properties of the material. Strain is any change in volume or shape.There are four general types of stress. 1. Stress â Strain Relationships Tensile Testing One basic ingredient in the study of the mechanics of deformable bodies is the resistive properties of materials. Tensile stress is measured in units of force per unit area. A simple Stress and strain are produced due to any of the following type of actions done on the machine parts. Introduction to Stress and Strain in a Tensile Test; Stress and Strain Curves or Diagram: This curve is a behavior of the material when it is subjected to load. Piezoelectricity is described by a third rank tensor. Therefore, strain is a dimensionless number. One type of stress is uniform, which means the force applies equally on all sides of a body of rock. Strain under a tensile stress is called tensile strain, strain under bulk stress is called bulk strain (or volume strain), and that caused by shear stress is called shear strain. Modulus of Resilience Modulus of resilience is the work done on a unit volume of material as the force is â¦ This is an elementary book on stress and strain theory for geologists. The stress is denoted by the symbol âÏâ and strain by âeâ. The unit of stress is N/m2 or N/mm2 whereas the strain does not have units. 3.Stress can be measured and has a unit of measure while strain does not have any unit and, therefore, cannot be measured. An elastic bar of lenght L, uniform cross sectional area A, coefficient of thermal expansion $\style{font-family:'Times New Roman'}\alpha$, and Young's modulus E is fixed at the two ends. Stress & strain . This stress is called the yield stress, s y. The constant is known as Modulus of elasticity or Youngâs Modulus or Elastic Modulus. Stress vs strain curve . Title: Stress and Strain 1 Stress and Strain. Stress and Strain. Solved example: strength of femur. The true strain is defined as the natural logarithm of the ratio of the final dimension to the initial dimension. In continuum mechanics, stress is a physical quantity that expresses the internal forces that neighbouring particles of a continuous material exert on each other, while strain is the measure of the deformation of the material. A fourth rank â¦ A true stress-strain curve is called flow curve as it gives the stress required to cause the material to flow plastically to certain strain. A 1 inch ( 25mm) diameter rod can take a much higher load than a 1/8â³ (3mm) rod before yeilding (stretching to the point of permanent deformation). FACULTY OF MECHANICAL ENGINEERING DIVISION OF ENGINEERING MECHANICS. Within its elastic limits, strain is directly proportional to the stress applied, but stress cannot be caused directly by strain. Tangential stress: It is defined as the deforming force applied per unit area. The stress-strain curve can provide information about a materialâs strength, toughness, stiffness, ductility, and more. The stress can exist without strain but the existence of strain without stress in not possible. Bulk Stress, Strain, and Modulus. V{tx D Concept of. Shear and bulk stress. The internal resistance force per unit area acting on a material or intensity of the forces distributed over a given section is called the stress at a point. Stress is a directed force distributed over an area. We can use the above definitions of stress and strain for forces causing tension or compression. 2.Stress can occur without strain, but strain cannot occur with the absence of stress. The stress-strain curve depends on two types of material.. 1. Typically E=210×109 N/m2 for steel 18. This is the currently selected item. Due to residual tensile and compressive stresses, the stress and strain â¦ Stress. Rank tensors simplest way to describe the mechanical properties of materials second tensors. To cause the material kg/cm^2 ) or pascal such case, you feel a pressing. 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Kilogram ( force ) per square meter ( N/m^2 ), kilogram ( force ) per square centimeter kg/cm^2! In such case, you can simply ignore this partâ¦ unless you want to optimize the element for geologists forces! Are four general types of stress and strain by âeâ the relation between strain and stress does need! Ductility, and more the hysteresis curve equation stress required to cause the material to plastically.	2023-04-01 23:49:22	{"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": 0.5886496901512146, "perplexity": 1025.6291422915024}, "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/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00788.warc.gz"}
https://sisu.ut.ee/simple-mathmodel/components-mathematical-model	# Components of Mathematical Model Components of Mathematical Model are variables or decision parameters; constants and calibration parameters; input parameters, data; phase parameters; output parameters; noise and random parameters. Process of mathematical modelling contains stages of problem identification and formulation; choice of experts; elaboration of model; development and clarification of model; investigation and implementation; numerical experiments; interpretation of results. The latter action is crucial and in these both, mathematicians and practitioners must participate. Mathematical modelling needs collaboration between industrials and academicals. This collaboration is called Industrial Mathematics; see https://ecmiindmath.org/. Main forms of Mathematical Models are; numbers, data; functions; equations and inequalities; systems of equations; differential equation, ordinary and partial.	2020-05-29 03:57:07	{"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": 0.8055816292762756, "perplexity": 9825.58718471294}, "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-2020-24/segments/1590347401260.16/warc/CC-MAIN-20200529023731-20200529053731-00272.warc.gz"}
https://math.stackexchange.com/questions/1605964/show-that-the-angles-satisfy-xy-z	# Show that the angles satisfy $x+y=z$ How can I show that $x+y=z$ in the figure without using trigonometry? I have tried to solve it with analytic geometry, but it doesn't work out for me. • Are all figures in the image intended to be squares? – Edward Jiang Jan 9 '16 at 20:00 • Yes! I forgot to mention that. – Hamid Mohammad Jan 9 '16 at 20:01 • See the solution detailed here. It also comes with a nice extension problem which can be approached similarly. – πr8 Jan 9 '16 at 20:10 (This space intentionally left blank.) Hint: denoting the points like on the picture below, triangles $EGD$ and $DGF$ are similar (why?). • Very neat indeed. – Lubin Jan 9 '16 at 20:40 • Please could you explain how these triangles are similar? – Fly by Night Jan 9 '16 at 22:46 • @FlybyNight: Compare the ratios of corresponding sides of $\angle G$ in both triangles. (This is effectively the product relation provided in Batominovski's answer.) – Blue Jan 9 '16 at 23:53 • @Blue Hi Blue. My comment was aimed at wojowu, in the hope that s/he might add the detail to his/her original answer. – Fly by Night Jan 10 '16 at 0:13 Let $X$, $Y$, and $Z$ be the apexes of the angles $x$, $y$, and $z$, respectively. Also, let $P$ be the common intersection of the red lines. Show that $ZP^2=ZX\cdot ZY$. Thus, the circumcircle of the triangle $PXY$ is tangent to $PZ$ at $P$. This will prove that $\angle YPZ=\angle YXP=x$. Equivalently, you have to prove that $$\underbrace{\mathrm{arctan}(1/3)}_{x}+\underbrace{\mathrm{arctan}(1/2)}_{y}=\underbrace{\mathrm{arctan}(1)}_{z}.$$ This is pretty clear by addition of tangents, indeed $$\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}=1 \implies x+y=\frac{\pi}{4}.$$ • The OP clearly states he wanted a non-trigonometric solution. – Edward Jiang Jan 9 '16 at 20:09 Imagine that all the the squares are 1 by 1 and so the rectangle has base 3 and a height of $1$. There are three right-angled triangles in the diagram. The one with angle $x$ has base 3, and height 1. The triangle with angle $y$ has base 2 and height 1. The triangle with angle $z$ has base 1 and height 1. Using the standard trig' ratio $\tan \theta = \frac{\mathrm{opp}}{\mathrm{adj}}$, we get $\tan x = \frac{1}{3}$, $\tan y = \frac{1}{2}$ and $\tan z= \frac{1}{1}=1$. There is a well-know formula for angle addition: $$\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$ Applying this formula to the case of $\alpha =x$ and $\beta = y$ gives: $$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}=1$$ It follows that $\tan(x+y)=\tan z$. Since $0^{\circ} < x<y<z < 90^{\circ}$ it follows that $$\tan(x+y) = \tan z \iff x+y = z$$ • "... without using trigonometry?" – Blue Jan 9 '16 at 22:36 • @Blue All of the answers given so far involve trigonometry to some degree, even yours. The fact that you don't mention sine, cosine or tangent is irrelevant. The answer to the OP's question should be: no such answer exists. Trigonometry is the science of triangles using lengths, angles and their ratios. I could re-phrase my answer in terms of lengths and proprtion, but that's really just trigonometry, but without the nice simple place-holders of sine, cosine and tangent. – Fly by Night Jan 9 '16 at 22:40 • @fleablood You're absolutely right. I hadn't read the "without trigonometry" until after I'd written my answer. However, being asked to solve this problem without trigonometry is, like I said, being asked to solve it without using angles, lengths, and their ratios. I'll leave my answer here because I think it might help people who Google this question in the future, and who are willing to use trigonometry to solve problems involving triangles. – Fly by Night Jan 9 '16 at 22:55 • @Blue I agree. Your solution is great! – Fly by Night Jan 9 '16 at 23:42 • @fleablood Be careful with the accessibility statement. We teach children trigonometry from the ages of 13/14 in the UK. The angle addition formulae come along at around 16/17. My answer is perfectly accessible for someone able to generate the graphic like the OP did. It's not like I'm using cohomology. – Fly by Night Jan 9 '16 at 23:45	2019-07-17 06:38:02	{"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": 0.7231584191322327, "perplexity": 528.3562535881307}, "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-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00552.warc.gz"}
https://gitter.im/dreamnettech/community?utm_source=redirect&utm_medium=link&utm_campaign=redirect/	by ## Where communities thrive • Join over 1.5M+ people • Join over 100K+ communities • Free without limits ##### Activity • Jul 09 18:10 kolessios on canary Fixed invisible Queuebar. Fixed… (compare) • Jul 09 16:49 kolessios on canary CI/CD fix. (compare) • Jul 09 15:56 kolessios on canary CI/CD fix. (compare) • Jul 09 15:32 kolessios on canary A lot of changes. v1.5 RC 1 (compare) • Jul 09 14:47 kolessios on master DreamTrack update. (compare) • Jul 09 14:47 kolessios on master [DreamTrack] Preferences sharin… (compare) • Jul 09 13:56 kolessios on v1.2.6 • Jul 09 13:55 kolessios on master Removed DreamLink Cluster from … (compare) • Jul 09 13:54 kolessios on v1.2.6 • Jul 09 12:19 kolessios on v1.2.6 • Jul 09 12:18 kolessios on master Merge tag 'v1.2.5' into canary … Directory structure and minor c… New --masks-path argument. Remo… and 9 more (compare) • Jul 09 12:16 kolessios on canary CI/CD fix. Merge remote-tracking branch 'o… (compare) • Jul 07 12:38 kolessios on canary CI/CD fix. (compare) • Jul 07 12:01 kolessios on canary CI/CD fix. (compare) • Jul 07 11:58 kolessios on canary CI/CD fix. (compare) • Jul 07 10:27 kolessios on canary CI/CD fix. (compare) • Jul 07 08:44 kolessios on canary Added required hidden import. P… (compare) • Jul 07 04:07 kolessios on canary CI/CD fix. (compare) • Jul 07 03:54 kolessios on canary CI/CD fix. (compare) • Jul 07 03:00 kolessios on canary Directory structure and minor c… New --masks-path argument. Remo… (compare) xrebel1 @xrebel1 Ok let me run DreamPower with this padding and I'm going to send you the step images there Iván Bravo Bravo @kolessios Thanks! hyunsoengbae @hyunsoengbae are there any manuals to learn how to use the editor in dreamtime? It would be really useful Patrick092 @Patrick092 Hi,, when this software going to be available on Android?? Iván Bravo Bravo @kolessios @hyunsoengbae I suppose you mean the photo editor, do you have any specific problem? I think it is very easy to use, almost like Paint. @Patrick092 With DreamTime v2 it will be possible to implement it on Android and iPhone, however not natively and free, more information here: https://time.dreamnet.tech/docs/support/faq#will-there-be-a-version-for-androidios Iván Bravo Bravo @kolessios @Hurzo6 Go to Settings -> Processing and change the Device option to CPU. It will take a few more minutes to create the nudes, but unfortunately your GPU does not have enough memory to be used. slitherysnake12365 @slitherysnake12365 could anyone assist me on using dreamtime on google colab? I read on the websites FAQ's that overlay is the most recommended resize mode. I dont know how to fill out this information. "If selecting Overlay, then specify the top left corner and width of the processing region. overlay_x: 0 overlay_y: 0 overlay_size: 512 deeppppp @deeppppp the required input width x height is 512px by 512px if your image does not have those dimensions, it should be resized. several resizing options are available, and the 'overlay' option has the least quality loss. but also assume you can grab a 512x512 pixel area from the input image as the new input that still features the model you want nudified. image it as a 512x512 area selection.. the parameters required are the top-left corner.. by an (x,y) coordinate and the width, that should be 512px that gives enough info to 'cut' out the 512x512px area and use that as input the other resize option can apply 'padding' by resizing the bigger of width/height to 512px and then apply padding to get to the square 512x512 size.. I would use that one, as it's easier to implement for colab Iván Bravo Bravo @kolessios @slitherysnake12365 As @deeppppp has said, Overlay requires that you have some external application that can give you the X and Y coordinates to overlay your photo. The website recommends Overlay if you are using DreamTime since it has the necessary tool for that, when using Google Colab it is better to use the other scaling methods. John Daze It is okay to use GTX1060 6 GB for this? deeppppp @deeppppp should be just fine, you can always switch to CPU mode.. the results are the same as with GPU, it just takes a bit longer to compute. stealthxphanoms @stealthxphanoms hi if i send a pic of a woman can someone make her nude for me? Iván Bravo Bravo @kolessios @stealthxphanoms Sorry but this chat is only to talk about the technical part of DreamTime/DreamPower, not to request photos. Egle11 @Egle11 Hello, I have a problem, I have 24 GB of RAM, but it says "You have run out of RAM on your system! Please install more RAM on your system before using DreamTime." how can this be fixed? Iván Bravo Bravo @kolessios @Egle11 Hmm weird. Could you open the terminal for any of the results and send me the text? Egle11 @Egle11 @kolessios Error: [DEBUG] C:\XONE\checkpoints [DEBUG] STREAM b'IHDR' 16 13 [DEBUG] STREAM b'IDAT' 41 8192 [DEBUG] Namespace(altered=None, asize=1, auto_rescale=False, auto_resize=False, auto_resize_crop=False, bsize=1, checkpoints={'correct_to_mask': 'C:\XONE\checkpoints\cm.lib', 'maskref_to_maskdet': 'C:\XONE\checkpoints\mm.lib', 'maskfin_to_nude': 'C:\XONE\checkpoints\mn.lib', 'checkpoints_path': 'C:\XONE\checkpoints'}, color_transfer=False, cpu=True, debug=True, disable_persistent_gan=False, export_step=None, export_step_path=None, func=<function main at 0x0000028506360EE8>, gpu=None, gpu_ids=None, hsize=0, ignore_size=False, input='C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', json_args=None, json_folder_name='settings.json', mode='run', n_cores=12, n_runs=1, nsize=1, output='C:/XONE/Uncategorized/women-underwear-inti-RUN1-1593081346873-dreamtime.jpg', overlay=(6, 6, 800, 800), prefs={'titsize': 1, 'aursize': 1, 'nipsize': 1, 'vagsize': 1, 'hairsize': 0}, steps=(5, 6), vsize=1) [INFO] Welcome to DreamPower [INFO] GAN Processing Will Use CPU [DEBUG] Process to execute : <processing.image.ImageProcessing object at 0x00000285063B8DC8> [INFO] Executing Image Processing [DEBUG] [INFO] Processing on ['C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', 'C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', 'C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', 'C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', 'C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', 'C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', 'C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg', 'C:/XONE/40e6903baa1d7c4d4a68122a234d1784-editor.jpg'] [DEBUG] wk [] [DEBUG] Model load done in 6.67 seconds [DEBUG] GAN Processing will use CPU Traceback (most recent call last): File "main.py", line 82, in <module> File "argv__init.py", line 35, in run File "main.py", line 25, in main File "processing\ init__.py", line 23, in run File "processing\image.py", line 96, in _execute File "processing\worker.py", line 49, in run_worker File "transform__init.py", line 26, in run File "processing\ init__.py", line 23, in run File "transform\gan__init__.py", line 74, in _execute File "transform\gan\model.py", line 154, in inference File "transform\gan\generator.py", line 99, in forward File "site-packages\torch\nn\modules\module.py", line 493, in call File "site-packages\torch\nn\modules\container.py", line 92, in forward File "site-packages\torch\nn\modules\module.py", line 493, in call File "site-packages\torch\nn\modules\conv.py", line 338, in forward RuntimeError: [enforce fail at ..\c10\core\CPUAllocator.cpp:62] data. DefaultCPUAllocator: not enough memory: you tried to allocate %dGB. Buy new RAM!7 [22940] Failed to execute script main at kr.getPowerError (file:///C:/Users/%D0%A1%D0%BE%D1%81%D0%B8%D1%81%D0%BA%D0%B0/AppData/Local/Programs/DreamTime/resources/app.asar/dist/assets/1efca3d6a830f66f106a.js:1:8910) at file:///C:/Users/%D0%A1%D0%BE%D1%81%D0%B8%D1%81%D0%BA%D0%B0/AppData/Local/Programs/DreamTime/resources/app.asar/dist/assets/1efca3d6a830f66f106a.js:1:8184 at CallbacksR Iván Bravo Bravo @kolessios @Egle11 Sorry but the algorithm is reporting that it was unable to allocate memory and there is no more information about it. Could you try again with the task manager opened and see if DreamTime RAM consumption is high? (Consumes all your RAM) Also keep in mind that it is recommended to have at least 12 GB of free RAM to use DreamTime. deeppppp @deeppppp @Egle11 you could try the cloud based version at https://www.xxxpaint.com if you can't get it to work as an alternative. xxxPaint chat is at: https://keybase.io/xxxpaint Egle11 @Egle11 @kolessios @deeppppp Thanks you edenzaraf @edenzaraf [ERROR] File is not a zip file dmcnsins @dmcnsins what happened to old chat ? dmcnsins @dmcnsins and is there any way that ı can show aı genitals ? Martin @xkpx64 Hello gentlemens I found last night this project and start working with it i see error in my CMD line like this [ERROR] File is not a zip file Iván Bravo Bravo @kolessios Martin @xkpx64 Thank You Sir ! Iván Bravo Bravo @kolessios No problem 🙂 @xkpx64 Oh, the instructions link is more for DreamTime (the user interface) than just for DreamPower. I think it was easier to tell you to extract the checkpoints folder from the zip into the folder where DreamPower is located. 😅 Martin @xkpx64 I will do that Thank you again Iván Bravo Bravo @kolessios After so much time.... • DreamTime v1.5 • DreamPower v1.2.6 • Waifu2X v0.1.0 They are being compiled for release very soon... 🕺 orangesponge32 @orangesponge32 Oooh exciting! How on track to release is it? When do you think it will be online? Iván Bravo Bravo @kolessios @orangesponge32 The first Release Candidate will be available to the Patrons in probably less than an hour. If everything goes well, the final version will begin to be distributed as usual for tomorrow or the day after tomorrow. Meanwhile I will update the documentation on the website, add a tutorial for Custom Masks with photos and more details :) orangesponge32 @orangesponge32 guess i have a reason to patron ^_^ Iván Bravo Bravo @kolessios @orangesponge32 Thank you very much for your interest 💖 reymond555 @reymond555 Hola después de tanto tiempo seria bueno que pasaran el link de la versión 1.5 por este canal Seria* Iván Bravo Bravo @kolessios @/all 🎉 DreamTime v1.5 RC 1 has been released for Patrons! Be the first to experience the new version and report problems or suggestions. https://www.patreon.com/posts/dreamtime-v1-5-39143853 Iván Bravo Bravo @kolessios I was waiting as some issues occurred during compilation and caused problems like invisible Queuebar. The new files should work. I will be away for a few hours and when I come back I will work on RC 2 fixing problems and suggestions reported by the Patrons. 😊 reymond555 @reymond555 Yo crei que la versión 1.5 hiba a ser free pero bueno.... no importa	2020-07-10 01:25:36	{"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": 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": 0.2505428194999695, "perplexity": 12184.165498624527}, "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/1593655902377.71/warc/CC-MAIN-20200709224746-20200710014746-00478.warc.gz"}
https://www.mooseframework.org/newsletter/2020_05.html	## INL Applications and Remote Access link Automatic differentiation (AD) capabilities in MOOSE have been refactored. Application developers no longer have to use templating to take advantage of AD; AD object APIs should appear nearly identical to their non-AD counterparts. This change is meant to lower the barrier to entry to AD for application developers as well as ease maintenance in the framework. This doesn't come without some tradeoff; users who employ solve_type = PJFNK and extensively use AD will experience a slow-down in their simulations of about 1.5-2. However, for solve_type = NEWTON, which is the recommmended solve_type for AD applications, no appreciable slow-down should be observed. One other important change is that a material property that is declared AD must be retrieved as an AD property; similary for a regular property. E.g. these are the allowed pairings of declare/get: declareADProperty/getADMaterialProperty and declareProperty/getMaterialProperty. An overview of AD can be found here. ## Stochastic Tools 1. The surrogate system within the stochastic tools modules was enhanced to support the creation of trainer and evaluation objects. This includes the ability to store and load arbitrary training data for both in memory evaluation or loading from a file. 2. The documentation for the stochastic tools module was improved to include more examples, including using the new train/evaluate object structure, see Stochastic Tools. ## Python parameters package A new python package for input parameters was created in the python directory. This new package is thoroughly tested and will replace the various other parameter systems within the MOOSE python tools in the near future. Please consider using this new package if you are creating python tools in an application and desire a parameter system. ## Finite Volumes Experimental support for solving PDEs using the finite volume method has been added to MOOSE. ## Conda Environment Installation The traditional moose-environment packages have been replaced with conda-packages, see Conda MOOSE Environment.	2020-09-19 22:37:56	{"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": 0.2360469251871109, "perplexity": 2969.6268160384734}, "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/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00596.warc.gz"}
http://www.intpforum.com/threads/the-random-thoughts-thread.2235/page-96	#### EndogenousRebel ##### We're all trying our best. Aren't we? Women are put on a pedestal honestly. I was discussing with a friend once about this, and he said that women were better than men. I didn't entirely agree with him so I just said "Well yeah, but they're still human, not by much." and he was adamant about them being better. I think it mostly on the basis of sex drive and abusiveness, but any fucking human abuses things almost whenever they can, so it's pretty moot. I feel as women mature (the ones that need to) they'll realize how fucking uninteresting they are, or just blame it all on men because they can't keep a good one around. #### peoplesuck ##### caretaker of machines My mom just found out about my date, lmao, she was so worried that I stayed the night with her. Like shes going to kill me or something, moms. Yay I reached the point where I have to start using intpf in incognito, and like, hiding it. exciting. #### moody ##### Well-Known Member speaking of girls, I'm kinda tired of them. If I do everything correctly, take initiative, do all the legwork etc etc then I get what I want. But in the end what do I really get? Pussy is only mildly interesting at this point, and their converstation is as inspiring as filing taxes. And what's interesting they love to proclaim that they want a guy like x and not like y, and if you're doing online dating you're only allowed to contact them if you "have something more interesting to say than 'Hi'", etc (and ironically, btw, it's always the barely-mediocre ones who are the most vocal about their supposed high standards). Meanwhile they themselves have exactly zero game, and speaking of 'Hi' that's the only message girls themselves ever send online. Girls want – and in fact need – the guy to chase them, but at this point in history what's the return one gets from chasing them? Maybe you should go to Amsterdam? I've heard in the Netherlands, girls take more initiative and do the chasing. Not that I would personally know, but it's relatively close to Norway. @peoplesuck The very first posts I read from Serac were his qualms about the women species in general. He's just jaded and honest about it. #### moody ##### Well-Known Member @Rebis It's time. Did you go on a hike??!?!?!? #### Rebis ##### Blessed are the hearts that can bend @moody It's still in the plans, so I've got a programming assignment for friday that everybody seems to be struggling with so that's my occupation until friday, holiday seasons initiate on friday too. Saturday and sunday is work, possibly monday the week following. #### Rebis ##### Blessed are the hearts that can bend @Serac Maybe spice up your profile a bit? I've had a decent amount of girls message me first time, my profile consists of my MBTI, the course I'm studying and a quote from a poem. In addition to my photos which show my clothing style, DMs here, funky shirts there and a few photos from gigs. My spotify link shows a range of music that I listen to frequently. All of these disparate interests add an eere of mystery in the person viewing the profile and people are generally attracted to novelties. This isn't to be deceptively disengenuous all of the content represents an aspect of myself but I'd guess you're not one to "bloat" your profile. People are attracted to options, right? Give them conversational tangents, I'm sure you have a few disparate interests yourself. yeah man, I get it. Enjoy it and don't listen to me, I'm a cynic. Maintain this self-awareness so you don't fall into the pithole of resentment! I used to know this person on a forum, initially he was very jolly, then cynical and eventually just hated people. Whether the resentment was genuine or just a farce to be percieved as that I'm not sure. He's like 44 now, you're late 20s so you still have time Spooky ghostwarning sounds #### Ex-User (14663) ##### Prolific Member @Rebis good pointers. But you gotta understand I’ve been doing this stuff for many years and with decent results. That’s not the issue. #### Rebis ##### Blessed are the hearts that can bend I'm sure you've already went through that but it seems to be a paradigm you can't change, you're constrained to playing the chasing game. The only power you have is over your domain (pictures bio, music etc) so if you can optimize or recalibrate your bio further it's less wishful/cynical thinking. They've got work cut out for them and there's no reason for them to take the initiative with the volume of messages they receive. I guess the proclivity of guys to be impatient and make the first move has lead to a culture of females waiting for attention. You could probably link that to civilization being patriarchal for the most part, they were operative while the female was not. I don't see a paradigm shift happening in the immediate future. Have you considered being the lead guitarist in a rock band? It should surely confer the burden of conversation onto the ladies chasing you. However I do think females put more conversational effort when they're actually in a relationship. If relationships were rocket fuel you're the combustion chamber and they're the after thrusters. Sent from my VOG-L09 using Tapatalk #### Ex-User (14663) ##### Prolific Member I have considered becoming rich and famous, yes. That might be a solution tbh. In the end I love women, but I don’t like chasing them. #### Rebis ##### Blessed are the hearts that can bend I have considered becoming rich and famous, yes. That might be a solution tbh. In the end I love women, but I don’t like chasing them. Have you been following AMDs rise and Nvidia's downfall? I suggest investing in amd stock, they're going to be killing it over the next few years, they're down to 7nm while intel is at 14, their GPUs are a few hundred pounds less than nvidias and the ps5 and xbox are coming out next year. They've been expanding into the PC and laptop space. Sent from my VOG-L09 using Tapatalk #### Ex-User (14663) ##### Prolific Member I have considered becoming rich and famous, yes. That might be a solution tbh. In the end I love women, but I don’t like chasing them. Have you been following AMDs rise and Nvidia's downfall? I suggest investing in amd stock, they're going to be killing it over the next few years, they're down to 7nm while intel is at 14, their GPUs are a few hundred pounds less than nvidias and the ps5 and xbox are coming out next year. They've been expanding into the PC and laptop space. Sent from my VOG-L09 using Tapatalk Yeah but what if the stock has been going up because of that information, then maybe I should sell Amd and buy intel. #### Rebis ##### Blessed are the hearts that can bend It's true communication influences stock prices, some renowned expert will be like "Invest in this" and because everyone does short selling occurs. Having said that, it's an early resugence for AMD: They've got high performance low power, they've been working on integrating the GPU into the CPU. I think their stock is going up naturally based on the units they're selling and how they're catching up to nvidia and intel in terms of performance but also on top of their niche of low power consumption. Intel has never designed GPUs for the general PC market, Nvidia hasn't created a successful line of CPUs either, AMD is playing on both fronts but coming out on top as we approach the end of moore's law. A lot of the tech world is a spec game and AMD is dominating in that respect, you still here of people talking in terms of i9s like it's the golden standard while ryzen 9 is approaching similar performance for cheaper. Buzz is a factor but their technological improvements are concrete. I think they'll have a flourishing 5 years of market growth. Their growth is similar to armv8 vs x86: x86 is CISC, armv8 is RISC. armv8 consumed significantly less power but since the main increase of performance was moore's law that didn't matter, now that we're approaching the end of this performance increase laptops are moving to arm instruction sets, faster clock speeds and power efficiency. Investing in stocks is always a risk but one that is based on technological improvements, which ultimately most people care about is something you can measure unlike the coke/pepsi wars. #### Rebis ##### Blessed are the hearts that can bend They're still just a small portion of the market so they have much more of the market to capitalize on, their stock price has increased from $2 to$39 in 4 years, \$10 in the last 3 months. the 2018 rise and fill was probably a product of bitcoin draining NVIDIA of GPUs. #### Ex-User (14663) ##### Prolific Member It’s an interesting analysis but the problem is that if one only has this kind of broad-picture narrative it is not useful for making trading decisions. One would have to do some math and look at eg what’s the expected earnings in various future scenarios, then combine that with a valuation of the company’s assets, cash flow, debt, expected dividends etc, end then eventually come up with a number that you think the company will be worth at point t in the future and compare that to the current market price of its stock. Otherwise you might end up being perfectly correct in your narrative yet make a losing bet because the market already discounted all the information #### Rebis ##### Blessed are the hearts that can bend It’s an interesting analysis but the problem is that if one only has this kind of broad-picture narrative it is not useful for making trading decisions. One would have to do some math and look at eg what’s the expected earnings in various future scenarios, then combine that with a valuation of the company’s assets, cash flow, debt, expected dividends etc, end then eventually come up with a number that you think the company will be worth at point t in the future and compare that to the current market price of its stock. I mean that would be the stringent economical way, but a small investment without the need of forecasting supply, demand and profit margins is ok based on the upwards trend and forecasting the future landscape. Expected earnings: before they were the cheap low-end option, intel and Nvidia didn't have any real competition at high-end equipment so they could get significant returns from each, now that amd is quite close in performance while being significantly cheaper and a lower power consumption. They're putting competitors out to these high end models and they're able to siphon cash from these two companies which have historicslly got the fat returns. They've went from having a small portion of the console and pc market to dominating the console market and increasing their market share in the pc market. Cash flow: Going into the next console generation production will be high for their low performance integrated graphics. Microsoft and sony will have set aside a certain number of units for launch sales while continuing to meet demand. If sony dominates sony will produce more consoles, if microsoft dominates they will produce more consoles. Either way amd wins in both fronts, there is no possibility of losing as people are still very much interested in consoles at their attractive price point. If household income is increasing and people want to spend a bit more money to penetrate into the PC market they'll find AMD all the way from low-high range in CPUs and GPUs, it isn't likely someone will go from a 500 dollar console to a high end 1500 dollar pc after all. AMD have no competition at low-end components and they're effectively competing in the high end. Debt, equity and their ratio: Sent from my VOG-L09 using Tapatalk ##### Well-Known Member It’s an interesting analysis but the problem is that if one only has this kind of broad-picture narrative it is not useful for making trading decisions. One would have to do some math and look at eg what’s the expected earnings in various future scenarios, then combine that with a valuation of the company’s assets, cash flow, debt, expected dividends etc, end then eventually come up with a number that you think the company will be worth at point t in the future and compare that to the current market price of its stock. Otherwise you might end up being perfectly correct in your narrative yet make a losing bet because the market already discounted all the information My dad is an effective investor and I’m pretty sure he doesn’t do any kind of complicated mathematical equations. He has an instinct for it - but he told me once that when he was a kid, he learned to keep a diary of stocks and watch for patterns. I don’t think it’s as complicated as people make it out to be, but you have to have a sense of business, maybe, and I think you have to be the kind of person who is very steady, uneasily shocked or phased, and carefully considerate of your assets and risks. According to dad, dividends are the most solid investments. Although, he makes more from rentals than he does from stocks, I believe. So maybe consider that approach. #### Ex-User (14663) ##### Prolific Member It’s an interesting analysis but the problem is that if one only has this kind of broad-picture narrative it is not useful for making trading decisions. One would have to do some math and look at eg what’s the expected earnings in various future scenarios, then combine that with a valuation of the company’s assets, cash flow, debt, expected dividends etc, end then eventually come up with a number that you think the company will be worth at point t in the future and compare that to the current market price of its stock. Otherwise you might end up being perfectly correct in your narrative yet make a losing bet because the market already discounted all the information My dad is an effective investor and I’m pretty sure he doesn’t do any kind of complicated mathematical equations. He has an instinct for it - but he told me once that when he was a kid, he learned to keep a diary of stocks and watch for patterns. I don’t think it’s as complicated as people make it out to be, but you have to have a sense of business, maybe, and I think you have to be the kind of person who is very steady, uneasily shocked or phased, and carefully considerate of your assets and risks. According to dad, dividends are the most solid investments. Although, he makes more from rentals than he does from stocks, I believe. So maybe consider that approach. high-dividend stocks is not a bad strategy. In fact when it comes to passive investing my research showed that low-volatility, high-dividend stocks is arguably the best strategy (historically at least) having said that I don't believe what anyone says about their results unless I get to see their exact historical portfolio returns lol (after subtracting commission and everything). Like I've mentioned before, I've worked in equity markets (hedge funds, asset management etc) and I can tell you that making money beyond just tracking the performance of a broad index like S&P500 is extremely, extremely difficult. and when it comes to predicting the direction of individual stocks, I would say that is impossible for an amateur investor (and 99% of all professional investors) #### Ex-User (14663) ##### Prolific Member yo @Rebis let's make a bet then. Let's pretend you bought AMD today and I bought Intel and let's see how gets the highest return in 6 months. #### Rebis ##### Blessed are the hearts that can bend having said that I don't believe what anyone says about their results unless I get to see their exact historical portfolio returns lol (after subtracting commission and everything). Like I've mentioned before, I've worked in equity markets (hedge funds, asset management etc) and I can tell you that making money beyond just tracking the performance of a broad index like S&P500 is extremely, extremely difficult. and when it comes to predicting the direction of individual stocks, I would say that is impossible for an amateur investor (and 99% of all professional investors) I think it's hard to predict the direction of individual stocks in the short term, if you were to measure sales quarterly but I think in the medium-long term, especially in long-established technology markets such as semi-conductor components, transistor wafers and camera lens there's a high entry market for these specialized components so it's less subject to volatility. While technology is ever-changing, revolutions are rare: there are many technologies that don't break the mould. Some companies that have high entry markets, like FABs, Processors and semi-conductor components. The likes of sony creating vaccum tube TVs which massively reduced the price would be considered a technological revolution. It's all a bet, it just depends on how much perspectives you can view it from. FABs for ASIICs, Chiplets and CPUs require a lot of upfront profit and aren't that profitable. I don't see any other company emerging into the CPU markets due to this upfront cost. yo @Rebis let's make a bet then. Let's pretend you bought AMD today and I bought Intel and let's see how gets the highest return in 6 months. I think the market share will have a higher baseline in 2-3 years, but we can't be assured we'll reach a conclusion in that time frame, 6 months it is! I've got it down in my calendar, see ya june 11th. AMD may dig into their pockets short term for better expansion into the market. Intel may cut back in R & D and drain the brand by selling their currently releasted CPUs at a discount price #### Rebis ##### Blessed are the hearts that can bend @Serac The calm and collected pokerface you've accrued playing high-stake tables won't work against me, as all I can see is a picture of King Schultz. You can't get the best of me, cowboy. Is that a snake or a quake in your boots? Only time will tell. #### Rebis ##### Blessed are the hearts that can bend I intentionally didn't pick up to one of my best friends phone call, this is the first time. Feels a bit rude and disengenious, ooff. If you're reading this (there is a chance he could be) soz bbe, ain't drinking no more and don't wanna kill the vibe. Explaining this over the phone would just be a boring exchange, plus the programming assignment is due. ##### Well-Known Member I intentionally didn't pick up to one of my best friends phone call, this is the first time. Feels a bit rude and disengenious, ooff. If you're reading this (there is a chance he could be) soz bbe, ain't drinking no more and don't wanna kill the vibe. Explaining this over the phone would just be a boring exchange, plus the programming assignment is due. I do that all the time. My friends know to just text me >_> Yeah, your strategy sounds similar to my dad’s. He’s not a fan of short-term investment. Anything he chooses to invest in is for the long haul. @Serac I didn’t actually read everything you said about your experience - I just came in on the tail end of the conversation. Sorry about that! Sounds like you and my dad would get along though. You refuse to take advice without evidence - and dad refuses to give advice! XD He will discuss almost anything related to business, from the standpoint of giving helpful tips and such , except for the stock market. All I know is he is of the opinion that one of the best ways to make a reliable profit is through stocks that pay good dividends, that he is cautious about investing in stocks and when he does, it’s something of a long-term investment, and...hmm let me think...oh! He doesn’t believe in day trading and he thinks you should only invest in industries you are very familiar with. I don’t think he really invests too much in technology...although I could be wrong. I’ve heard him talk more about infrastructure-related things, such as gas, and I know he has a pretty good investment he made in Tiffany (the jewelry store) from way, way back in the day when Tiffany was small. You would probably know more, though, because like I said - I believe he said that there’s more money in rentals for him. He doesn’t rely on investing as a means of regular income, I don’t think. I was curious about it for a while - but it’s a steep learning curve! I don’t think I have the head for business that would be required to make wise investments, and my ex lost a lot of money from letting Blackrock his investment portfolio...so I’m pretty sure that if I got into this hobby, I would drive myself broke. If I did though, I would do technology. #### Rebis ##### Blessed are the hearts that can bend @Serac from a previous quote of yours relating to no fap where you humorously said you could smell someone I'm getting to that point. Sent from my VOG-L09 using Tapatalk #### Ex-User (14663) ##### Prolific Member @Inexorable Username your father sounds like a wise man. I too have the philosophy that I never give recommendations on specific investments, I only say what I currently have in my own portfolio. It’s the principle of “skin in the game”; if someone takes my advice and loses money I should also lose money. #### Rebis ##### Blessed are the hearts that can bend Can someone tell me what the icon is for this forum, the cross in the display bar? It reminds me of nazis, I can't pinpoint the link. #### peoplesuck ##### caretaker of machines Job application in 2020: Hello sir, please take the next four hours to chat with our psychologist AI, jill. Jill is going to ask you very personal questions that you dont have to answer, but you should. Your personality traits, emotional stability, and sociability, will all be tested to make sure you can perform this job : cashier, effectively. We will begin with a calculus test, to insure you can count to 100. wai job applications are so long, fo real? It takes over an hour each time and, if you have as little experience as I do, you always get a no. CAN YOU AT LEAST START YOUR "NO" EMAIL DIFFERENTLY THEN EVERY OTHER COMPANY, OR SEND A CUTE CAT SAYING NO? dying sounding so naic rit aobut now. but people would be sad. ugh, people. Not valuing life really shows how little you do, that you like. Or maybe I take everything for granted, I personally just think im lonely. Trying your ass off, and making the slightest bit of progress, is better by so many magnitudes than getting nowhere. Just waiting for that slight bit of progress. THERE IS A VOCAB TEST never in my life have I seen the word rue. you got me, im not capable. #### peoplesuck ##### caretaker of machines ITS ASKING IF I HAVE FRIENDS, HOW DARE YOU Yo if I hear one more person confuse sex and gender, im gonna die inside, then look at them, crying, try my best to telepathically communicate, the difference, in these two words. for the fuck of sakes people cmmon #### Perfectly Normal Beast ##### Well-Known Member @Inexorable Username your father sounds like a wise man. I too have the philosophy that I never give recommendations on specific investments, I only say what I currently have in my own portfolio. It’s the principle of “skin in the game”; if someone takes my advice and loses money I should also lose money. He's pretty good with money. I had him take the test and he's an ENTJ - makes sense. Unfortunately, I inherited none of that gift. I'm good with business, and with marketing, but when it comes to using money to grow money - that's not something I excel at. #### Rebis ##### Blessed are the hearts that can bend I ate 70% of a cheesecake today and felt sick. Wow, this could be my body finally rejecting sugar. Diabetes is losing its chance to creep over the horizon. Sent from my VOG-L09 using Tapatalk #### moody ##### Well-Known Member He's pretty good with money. I had him take the test and he's an ENTJ - makes sense. Unfortunately, I inherited none of that gift. I'm good with business, and with marketing, but when it comes to using money to grow money - that's not something I excel at. I have to admit, I am somewhat jealous of those who are innately practical with money and common sense. None of my immediately family was good with money, my siblings and I aren’t either. We are good in academia, sports and art, just not anything fiscal. You sound like you have a good head on your shoulders, so even if you’re not good necessarily good at maximizing profits, I can’t imagine you being dumb with personal finances. It helps if you have family members who are good at it; hopefully it will eventually start rubbing off. #### peoplesuck ##### caretaker of machines Why dont I naturally smile back at people? why is my face always so flat? is this something I learned? A google search said that It means you arent attached to your emotions. I could see that being true. anyone have an opinion? #### Rebis ##### Blessed are the hearts that can bend Current stack is 200mg modafinil, 150mg of caffeine I'm finding modafinil and caffeine to stack really well, people report headaches at that dose but I feel great. I haven't got l theanine to mellow out unfortunately, still, good combo #### Rebis ##### Blessed are the hearts that can bend Just uploaded a daunting assignment, pressure from all angles. Finished for christmas break, now that I have time to reflect, the question remains: Did the pressure turn the shit into diamonds? We'll find out, quite literally; after the break. #### peoplesuck ##### caretaker of machines I had a calculus test that I knew I wasnt ready for, so I decided to stay up all night and study, I had about 600mg of caffeine and went straight to bed. Slept really well. Its amazing im alive, considering how much I abuse my body. We are some sturdy animals. ##### Well-Known Member I had a calculus test that I knew I wasnt ready for, so I decided to stay up all night and study, I had about 600mg of caffeine and went straight to bed. Slept really well. Its amazing im alive, considering how much I abuse my body. We are some sturdy animals. I used to not smile back at people when I was a kid. I saw smiles as being fake. In my case, I didn’t really think people had my best interest at heart and I didn’t think that I was anything worth smiling at, so when they smiled, I saw it as primarily manipulative, I think. Or else I might have seen them as weak - I don’t specifically remember. I was a dark child. Lol. When I got older and found it in myself to think about the feelings of other people more, and feel their pain, I started feeling like people were more relatable. I started caring more about how my behavior made other people feel, because I genuinely didn’t want to cause anyone to feel insecure because of me. In retrospect, looking back, I think I was too depressed as a kid to have any room for the feelings of other people. I spent a lot of time drawing or writing and isolating. As my outlook on life became happier and healthier, I found that I could afford to have compassion in ways I couldn’t have done before. Anyways...I can’t help you with your smiling thing. All I can do is talk about my personal experiences. But maybe if you hear enough stories from other people about why they’ve had a similar symptom, some bits and pieces of personal significance will emerge, and you’ll be able to put the puzzle pieces together and figure yourself out. ##### Well-Known Member He's pretty good with money. I had him take the test and he's an ENTJ - makes sense. Unfortunately, I inherited none of that gift. I'm good with business, and with marketing, but when it comes to using money to grow money - that's not something I excel at. I have to admit, I am somewhat jealous of those who are innately practical with money and common sense. None of my immediately family was good with money, my siblings and I aren’t either. We are good in academia, sports and art, just not anything fiscal. You sound like you have a good head on your shoulders, so even if you’re not good necessarily good at maximizing profits, I can’t imagine you being dumb with personal finances. It helps if you have family members who are good at it; hopefully it will eventually start rubbing off. I’m really bad at finances XD... My solution to finances is to just never spend any money. That’s like, one of the biggest red flags of someone who sucks at managing money, apparently, because people who can actually manage it aren’t averse to spending it in moderation. Still, my parents did help me some - even though dad doesn’t really give much advice in the money department. I learned not to take loans. That was something. I learned that eating out and getting things like coffee every day is insanely expensive, and that the goal of being financially savvy is to make your money make money. My issue is that, for the life of me, I can never seem to get myself to actually care about money. That instinct is just absent in me. It’s like I’m a psychopath, but instead of having no empathy, I have no financial drive. Only in about the last few years have I really started to get an understanding for how much things cost/should cost to where I can get whether X price for X thing is expensive or not. This is one of my most crippling defects as a person. I’ve been trying so hard to fix this issue with myself...but it’s tough. Visualizing having fancy things does not excite me. Everything I want/need is available to me in a single room, more or less. I just really struggle to develop financially-driven aspirations. I think a downside of growing up with parents that are good with money is that I developed an idea that money is something a hassle, and a chore, and that spending it frivolously is irresponsible...so I don’t think I really got the dopamine circuitry for experiencing the excitement of obtaining money. When I make it, I just think “Well...now I can add this number to the other number I guess, and I’ve succeeded slightly more as a human. Woot.” And when I spend it, I just feel this sense of like...dull guilt, and reservations. Do I really need this thing? Is it worth it? Does it matter if socks have holes if nobody ever sees them anyways? I dunno. Not having a proper love of money makes me feel like I’m not a good representative of the species sometimes. Plus, now I have this non-profit concept, which legitimately needs a significant amount of money to happen. So I have to do a bit of a for-profit project to fund it, and I need to get investors...which means I have to appreciate the way investors think and not take their generosity for granted. Which means I need to learn to have more respect for money. I’m working on it. I might be getting a couple of great courses in law and business for Christmas #### Rebis ##### Blessed are the hearts that can bend A friend of mine is looking me to add comments to his code for £30, upon questioning him he put it up to £40. What a strange offer. #### Ex-User (14663) ##### Prolific Member I'll write this as a little note I can go back to in about a month and compare the outcome to the current sense of suspense nowadays a lot of things are at stake. Large sums of money, reputations, and trust in machines as a means to predict the future. In particular machines that I have designed (not really "machines" but rather mathematical models and machine-learning algorithms). Huge bets have been placed based on the predictions from these models. Problem for me is that the people at the company don't really know what these things really are, so people are divided between two groups: 1) old-schoolers who are generally quite skeptical to them , 2) people who trust the models blindly and don't really understand the concept of randomness, probability etc. Another problem is that I don't think the models are particularly good myself – they were made under deadlines and weren't really properly tested before we started using them to take actual risk. What's more is that currently, the model makes predictions which are the diametrical opposite of the beliefs of everyone else in the company. by the end of December we will know who was right – half-baked ML models or humans. outcome: flawless victory. man vs machine: 0-1. but of course the ball keeps rolling. New bets have been placed for the next month. It's a weird-ass feeling though, to have a machine tell you something about the future and nobody knows why it says what it says, yet one has to commit to it. #### Rebis ##### Blessed are the hearts that can bend I'll write this as a little note I can go back to in about a month and compare the outcome to the current sense of suspense nowadays a lot of things are at stake. Large sums of money, reputations, and trust in machines as a means to predict the future. In particular machines that I have designed (not really "machines" but rather mathematical models and machine-learning algorithms). Huge bets have been placed based on the predictions from these models. Problem for me is that the people at the company don't really know what these things really are, so people are divided between two groups: 1) old-schoolers who are generally quite skeptical to them , 2) people who trust the models blindly and don't really understand the concept of randomness, probability etc. Another problem is that I don't think the models are particularly good myself – they were made under deadlines and weren't really properly tested before we started using them to take actual risk. What's more is that currently, the model makes predictions which are the diametrical opposite of the beliefs of everyone else in the company. by the end of December we will know who was right – half-baked ML models or humans. outcome: flawless victory. man vs machine: 0-1. but of course the ball keeps rolling. New bets have been placed for the next month. It's a weird-ass feeling though, to have a machine tell you something about the future and nobody knows why it says what it says, yet one has to commit to it. It is weird, though most are attuned to that type of blind faith. To those who aren't interested in the complexity that produces a result it has always been reliant on a feeling of intuition gathered from biases rather than analyzing the methodlogy. It must feel weird to be in the opposite end, one that always had to understand the concept for themself and now you have to have blind faith. It's your turn for blindness Sent from my VOG-L09 using Tapatalk #### peoplesuck ##### caretaker of machines Im buying film for my old polaroid. A long time ago I decided I wanted to take pictures of the people and things that made me who I am. I dont want to forget the things that got me to where ever I end up, the thought of forgetting scares me deeply. Maybe its silly not wanting to forget, but its my story, and I want to know every piece. I suppose its my insane memory that makes me scared of forgetting, everyone else has come to terms with it. Well I have not. Not sure whats going on with my head lately, cannot stop thinking. I watched one 2 minute youtube video, other than that Ive just been listening to music and thinking in my free time. I cant sleep either, I suppose this is hypomania, FEELIN GOOD THO. #### Minuend ##### pat pat @peoplesuck The very first posts I read from Serac were his qualms about the women species in general. He's just jaded and honest about it. [/QUOTE] How far are you willing to go to being okay with people beliefs because you consider them "just jaded and honest"? #### Rebis ##### Blessed are the hearts that can bend I Fking hate men they are GY FGGTS Even you , macho man #### Minuend ##### pat pat I Fking hate men they are GY FGGTS Even you , macho man Of course, an individual man has no flaws or beliefs to question. If you do, you're an emotional feminist. If a man believe rape is right, then questioning that is only being an irrational feminist. And how better to adress that than showing you consider your opponent is emotional and stupid. #### Rebis ##### Blessed are the hearts that can bend I F*king hate men they are GY FGGTS Even you , macho man Of course, an individual man has no flaws or beliefs to question. If you do, you're an emotional feminist. If a man believe rape is right, then questioning that is only being an irrational feminist. And how better to adress that than showing you consider your opponent is emotional and stupid. If you're a FCKIN MN you need to SHT UP, DUMBSS #### Minuend ##### pat pat If you're a FCKIN MN you need to SHT UP, DUMBSS I would never fuck MN #### Rebis ##### Blessed are the hearts that can bend EDIT:Oh my god watching that video scares me, these type of people are the scariest people I'll ever encounter. #### Minuend ##### pat pat I recommend seeing every person you dislike as her. That way, the world becomes a bit simpler and nicer #### Rebis ##### Blessed are the hearts that can bend @moody I think I woke up to the mouse again in my room, it was sleeping on my arm and I slightly threw it off. It felt furry, I didn't actually see it. It could be a sock! But I don't see how it reached my hand: 6 ft 2, I take my socks off at the bottom of the bed, they usually fall off the bed. Having said that, I don't see how a mouse can migrate to the front of my arm to sleep. I wish I had night vision to determine this haha Sent from my VOG-L09 using Tapatalk #### moody ##### Well-Known Member @moody I think I woke up to the mouse again in my room, it was sleeping on my arm and I slightly threw it off. It felt furry, I didn't actually see it. It could be a sock! But I don't see how it reached my hand: 6 ft 2, I take my socks off at the bottom of the bed, they usually fall off the bed. Having said that, I don't see how a mouse can migrate to the front of my arm to sleep. I wish I had night vision to determine this haha Sent from my VOG-L09 using Tapatalk Those damn corporeal mice. A few times I was 100% comvinced they were falling from my ceiling while I was sleeping.	2020-06-03 18:45:24	{"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": 0.25099265575408936, "perplexity": 2078.192681714322}, "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/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00475.warc.gz"}
https://www.coursehero.com/sg/general-chemistry/thermodynamics/vocabulary/	# Thermodynamics ## Vocabulary ### absolute zero minimum possible temperature theoretically achievable, equal to 0 K (–273.15°C), at which there is no particle motion ### entropy (S) measure of the disorder of a system ### equilibrium state in which the forward and reverse reactions are equal ### first law of thermodynamics law that states that energy cannot be created or destroyed, only transformed from one type of energy to another type of energy; given as $\Delta U=q+w$ ### free energy capacity of a system to do work ### Gibbs free energy (G) amount of work done by a system, expressed as $G=H-TS$, where H is enthalpy, T is temperature, and S is entropy ### macrostate measurable macroscopic properties of a system ### microstate possible energy and positional configuration of the particles of a system ### nonspontaneous change process in which energy must be added to the system for the change to occur ### second law of thermodynamics law that states that the total entropy of an isolated system only increases over time ### spontaneous change process in which energy is released as a system changes ### standard entropy (S°) entropy of one mole of a substance under standard state conditions, expressed in units J/(K mol) ### standard state set of specific conditions under which reactions are measured, typically 0°C and 1 atm pressure ### thermodynamics branch of physical science that investigates the energy and work of systems ### third law of thermodynamics law that states that the total entropy of a system approaches zero as the temperature of the system approaches absolute zero ### work (w) energy that is transferred when a force acts on an object over a distance ### zeroth law of thermodynamics law that states that when two thermodynamic systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other	2022-05-19 12:24:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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": 0.6987576484680176, "perplexity": 1096.1648076381296}, "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/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00444.warc.gz"}
https://db0nus869y26v.cloudfront.net/en/Landau_distribution	Parameters Probability density function ${\displaystyle \mu =0,\;c=\pi /2}$ ${\displaystyle c\in (0,\infty )}$ — scale parameter ${\displaystyle \mu \in (-\infty ,\infty )}$ — location parameter ${\displaystyle \mathbb {R} }$ ${\displaystyle {\frac {1}{\pi c))\int _{0}^{\infty }e^{-t}\cos \left(t\left({\frac {x-\mu }{c))\right)+{\frac {2t}{\pi ))\log \left({\frac {t}{c))\right)\right)\,dt}$ Undefined Undefined Undefined ${\displaystyle \exp \left(it\mu -{\frac {2ict}{\pi ))\log \|t\|-c\|t\|\right)}$ In probability theory, the Landau distribution[1] is a probability distribution named after Lev Landau. Because of the distribution's "fat" tail, the moments of the distribution, like mean or variance, are undefined. The distribution is a particular case of stable distribution. ## Definition The probability density function, as written originally by Landau, is defined by the complex integral: ${\displaystyle p(x)={\frac {1}{2\pi i))\int _{a-i\infty }^{a+i\infty }e^{s\log(s)+xs}\,ds,}$ where a is an arbitrary positive real number, meaning that the integration path can be any parallel to the imaginary axis, intersecting the real positive semi-axis, and ${\displaystyle \log }$ refers to the natural logarithm. In other words it is the Laplace transform of the function ${\displaystyle s^{s))$. The following real integral is equivalent to the above: ${\displaystyle p(x)={\frac {1}{\pi ))\int _{0}^{\infty }e^{-t\log(t)-xt}\sin(\pi t)\,dt.}$ The full family of Landau distributions is obtained by extending the original distribution to a location-scale family of stable distributions with parameters ${\displaystyle \alpha =1}$ and ${\displaystyle \beta =1}$,[2] with characteristic function:[3] ${\displaystyle \varphi (t;\mu ,c)=\exp \left(it\mu -{\tfrac {2ict}{\pi ))\log \|t\|-c\|t\|\right)}$ where ${\displaystyle c\in (0,\infty )}$ and ${\displaystyle \mu \in (-\infty ,\infty )}$, which yields a density function: ${\displaystyle p(x;\mu ,c)={\frac {1}{\pi c))\int _{0}^{\infty }e^{-t}\cos \left(t\left({\frac {x-\mu }{c))\right)+{\frac {2t}{\pi ))\log \left({\frac {t}{c))\right)\right)\,dt,}$ Taking ${\displaystyle \mu =0}$ and ${\displaystyle c={\frac {\pi }{2))}$ we get the original form of ${\displaystyle p(x)}$ above. ## Properties The approximation function for ${\displaystyle \mu =0,\,c=1}$ • Translation: If ${\displaystyle X\sim {\textrm {Landau))(\mu ,c)\,}$ then ${\displaystyle X+m\sim {\textrm {Landau))(\mu +m,c)\,}$. • Scaling: If ${\displaystyle X\sim {\textrm {Landau))(\mu ,c)\,}$ then ${\displaystyle aX\sim {\textrm {Landau))(a\mu -{\tfrac {2ac\log(a)}{\pi )),ac)\,}$. • Sum: If ${\displaystyle X\sim {\textrm {Landau))(\mu _{1},c_{1})}$ and ${\displaystyle Y\sim {\textrm {Landau))(\mu _{2},c_{2})\,}$ then ${\displaystyle X+Y\sim {\textrm {Landau))(\mu _{1}+\mu _{2},c_{1}+c_{2})}$. These properties can all be derived from the characteristic function. Together they imply that the Landau distribution is closed under affine transformations. ### Approximations In the "standard" case ${\displaystyle \mu =0}$ and ${\displaystyle c=\pi /2}$, the pdf can be approximated[4] using Lindhard theory which says: ${\displaystyle p(x+\log(x)-1+\gamma )\approx {\frac {\exp(-1/x)}{x(1+x))),}$ where ${\displaystyle \gamma }$ is Euler's constant. A similar approximation [5] of ${\displaystyle p(x;\mu ,c)}$ for ${\displaystyle \mu =0}$ and ${\displaystyle c=1}$ is: ${\displaystyle p(x)\approx {\frac {1}{\sqrt {2\pi ))}\exp \left(-{\frac {x+e^{-x)){2))\right).}$ ## Related distributions • The Landau distribution is a stable distribution with stability parameter ${\displaystyle \alpha }$ and skewness parameter ${\displaystyle \beta }$ both equal to 1. ## References 1. ^ Landau, L. (1944). "On the energy loss of fast particles by ionization". J. Phys. (USSR). 8: 201. 2. ^ Gentle, James E. (2003). Random Number Generation and Monte Carlo Methods. Statistics and Computing (2nd ed.). New York, NY: Springer. p. 196. doi:10.1007/b97336. ISBN 978-0-387-00178-4. 3. ^ Zolotarev, V.M. (1986). One-dimensional stable distributions. Providence, R.I.: American Mathematical Society. ISBN 0-8218-4519-5. 4. ^ 5. ^ Behrens, S. E.; Melissinos, A.C. Univ. of Rochester Preprint UR-776 (1981).	2023-03-30 06:26:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 37, "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": 0.9495326280593872, "perplexity": 1009.5659171165647}, "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/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00070.warc.gz"}
http://events.berkeley.edu/index.php/calendar/sn/?event_ID=112643&date=2017-11-01&tab=academic	## Topology Seminar (Main Talk): Boundary amenability of Out$(F_n)$ Seminar \| November 1 \| 4-5 p.m. \| 3 Evans Hall I will discuss boundary amenability and how to prove it for basic groups for most of the hour. The main interest in boundary amenability is that it implies the Novikov conjecture in manifold theory. I will then outline the main ideas in the proof of boundary amenability of Out$(F_n)$. This is joint work with Vincent Guirardel and Camille Horbez.	2019-07-19 20:56:27	{"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": 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": 0.5304599404335022, "perplexity": 1482.9832193480788}, "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/1563195526359.16/warc/CC-MAIN-20190719202605-20190719224605-00068.warc.gz"}
https://byjus.com/question-answer/find-all-the-roots-of-the-root-of-the-equation-displaystyle-1-cos-2x-sin/	Question Find all the roots of the root of the equation $$\displaystyle (1 \, - \, cos \, 2x) \, sin \, 2x \, = \, \sqrt{3} \, sin^2 \, x$$ which lie in the interval $$\displaystyle \left [ 0, \, \frac{\pi}{3} \, \right ] .$$ Solution $$\left(1-\cos{2x}\right)\sin{2x}=\sqrt{3}{\sin}^{2}{x}$$$$\Rightarrow 2{\sin}^{2}{x}\sin{2x}=\sqrt{3}{\sin}^{2}{x}$$$$\Rightarrow 2\sin{2x}=\sqrt{3}$$$$\Rightarrow \sin{2x}=\dfrac{\sqrt{3}}{2}$$$$\Rightarrow 2x=n\pi+\dfrac{\pi}{3}$$ or $$x=\dfrac{n\pi}{2}+\dfrac{\pi}{6}$$ is the general solutionThe principal solution between $$\left[0,\dfrac{\pi}{3}\right]$$ is $$n=0\Rightarrow x=\dfrac{\pi}{6}$$$$\therefore x=\dfrac{\pi}{6}$$ lies between $$\left[0,\dfrac{\pi}{3}\right]$$Maths Suggest Corrections 0 Similar questions View More People also searched for View More	2022-01-26 14:54:46	{"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": 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": 0.4719621539115906, "perplexity": 3071.2616458360217}, "config": {"markdown_headings": false, "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-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00105.warc.gz"}
https://scicomp.stackexchange.com/questions/19082/strong-coupling-of-a-non-linear-multiphysic-problem-failure-with-newton-raphson/19205	# Strong coupling of a non-linear multiphysic problem: failure with Newton Raphson method I am trying to solve a multiphysic problem using finite elements and a Newton Raphson solution scheme. I have two non-linear subsystems that are coupled bi-directionally. The first subsystem includes an equation that I solve for $p_f$: $\nabla u - \nabla a^3 \nabla p_f = \nabla a^3 B$ This equation becomes very non-linear, due to the parameter $a$ with an exponent of 3! $B$ are body forces and $u$ is the solution of the second subsystem that I am trying to solve. This second subsystem includes the Stokes equation that depends on the gradient of $p_f$. Therefore, both system are coupled bi-directionally. I am using a Newton-Raphson solution scheme for each subsystem. First, I solve for $p_f$. After convergence occurred, I pass this solution to the second subsystem, which I solve using a Newton Raphson scheme, too. Then, the solution $u$ is passed to the first subsystem ... until the solutions of both subsystems do not change anymore. See the flow chart below. This procedure works very well so far. However, if the values of the parameter $a$ become very low (thus the gradient of $a$ becomes very, very low), there is no convergence anymore! The reason seems to be that the coupling between both subsystem becomes very strong. A small change in one subsystem creates a significant changes in the other one (especially where the gradient of $a$ is very low). I tried a variety of good initial conditions, however, since the first subsystem reacts very sensitiv, convergence has been impossible so far. In the figure below, you can see how the solution of this approach ("Segregated Approach"-blue line) is different from the "true" solution ("Fully Coupled"-red line). The "true solution" is the result of a benchmark, which I calculated by solving a single system of equations that includes the equations of both subsystems. The greatest differences between both solutions are located where the gradient of $a$ becomes very small. See the figure below (in this figure the values of $p_f$ and $a$ are normalized to range between 0 and 1): So: Is a Newton-Raphson solution scheme the wrong method for solving a strongly coupled system, if there are low gradients in a subsystem? Could my method be fixed to solve the problem that I described? If yes, how? Are there alternative solution schemes for this problem? • It's probably the segregation scheme that's killing you not the inner non-linear solver. Why don't you want to solve the fully coupled problem? – Bill Barth Mar 5 '15 at 14:18 • 1. The model that I'm modifying makes it very hard to add a new subsystem to the existing system of equations 2. Later, the model should run on a larger scale, and it seemed less time and memory consuming to use a segregation scheme – Johann Mar 5 '15 at 14:23 • Find a new program to extend? I think the odds are good that the segregated version is converging to the wrong answer. If the inner Newton steps appear converged, and the outer iterations between the segregated models appear converged, then segregation may not be working. How many different initial guesses have you tried? It's possible that the fully coupled version has an easier time hitting what you think of as the right solution. How did you generate the fully coupled solution if the model is hard to extend, btw? Have you tried the method of manufactured solutions on both models? – Bill Barth Mar 5 '15 at 16:27 • The fully coupled solution is from a simple model I wrote in MATLAB (much too simple for later usage, but good enough to run this benchmark). I used the solution of the fully-coupled code as initial condition and did not get any convergence! I have not tried the method of manufactured solutions yet. – Johann Mar 5 '15 at 22:06 • The nonlinear term is $\nabla (a^3)$ or $(\nabla a)^3$? – nicoguaro Mar 19 '15 at 22:26 The issues you're running into now are not a failing of Newton-Raphson, but a question of coupling. You're doing iterated sequential coupling -- solving each equation sequentially and then iterating until (hopeful) convergence. No solver choice in place of NR is going to fix this lack of convergence, as long as you are doing iterated sequential coupling. Instead, you want to consider different coupling schemes that use more information about the coupling terms. What @paul suggests is effectively equivalent to doing a quasi-Newton method using an approximation of the Jacobian of the coupled system that is block diagonal. Note that this is NOT Newton-Raphson any more, because you do not have the true Jacobian of the coupled system -- instead you form an approximation of the Jacobian using fewer terms than the true Jacobian and use this instead. This is a good starting point because it requires no additional knowledge of the physics -- you can build this entirely from pieces you already have. Norms for the coupled system become a combination of the norms for the two systems you have. Residuals for the coupled system become the vector of both systems' residuals. And the approximate inverse is block diagonal. It is possible, but unlikely, that this will help much. The issue is that the approximate (block diagonal) inverse will be too far from the true inverse. One potential option is to try Jacobian-Free Newton Krylov to get a better inverse. This is likely to not work all that much better until you start to add off-diagonal blocks to the approximate inverse. From a practical standpoint -- your best bet is to bite the bullet and find a way to start to include at least some information about how the equations couple in your approximate inverse, i.e. including off-diagonal blocks in your approximate Jacobian for use in a fully implicit coupling scheme. Instead of running Newton-Raphson to convergence of each subsystem, try 1 iteration on the first subsystem followed by 1 iteration on the second subsystem. This may keep the subsystems more coupled, not going to "distant" unrelated sub solutions. Repeat this two step iteration and see if it converges. This is a short answer and mostly a workaround. This could hit you back again later on, but can provide you with a temporary solution. You can try to relax the solution to your Newton-Raphson solver. What I mean by this is do not take the full solution of your non-linear system, but just take a factor of the previous solution and of the new solution (say 0.1 new solution, 0.9 old solution), and use this intermediate solution in your other solver. This means that you would be relaxing your global system of equation (and is similar conceptually to the above answer of using a single Newton-Raphson iteration). This usually leads to more robust convergence, but it can turn out to be extremely slow.	2021-01-27 09:44:29	{"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": 0.6158246397972107, "perplexity": 460.92579223651506}, "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/1610704821381.83/warc/CC-MAIN-20210127090152-20210127120152-00649.warc.gz"}
http://www.ams.org/mathscinet-getitem?mr=1623032	MathSciNet bibliographic data MR1623032 (2000c:03026) 03C20 (03E05) Lipparini, Paolo Every $(\lambda\sp +,\kappa\sp +)$$(\lambda\sp +,\kappa\sp +)$-regular ultrafilter is $(\lambda,\kappa)$$(\lambda,\kappa)$-regular. Proc. Amer. Math. Soc. 128 (2000), no. 2, 605–609. Article For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.	2015-10-10 02:16:40	{"extraction_info": {"found_math": true, "script_math_tex": 2, "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": 0.9985269904136658, "perplexity": 9742.773600031996}, "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-2015-40/segments/1443737937342.66/warc/CC-MAIN-20151001221857-00104-ip-10-137-6-227.ec2.internal.warc.gz"}
https://www.vcalc.com/wiki/KurtHeckman/lever+%28beam+length%29	# lever (beam length) Not Reviewed y = Tags: Rating Copied from ID KurtHeckman.lever (beam length) UUID 84550150-d427-11e5-9770-bc764e2038f2 The Length of a Lever calculator (y = x • W/F ) Lever with beam (x+y) and fulcrum.computes the length (y) on the force side of the fulcrum to lift an object by leverage. INSTRUCTIONS: Choose your preferred units and enter the following: • (F) The downward force (weight) opposite the fulcrum from the object. • (W) The weight of the object being lifted by the lever. • (x) The length of the beam on the object's side of the fulcrum (see diagram) The calculator computes the length of the beam (y) in meters. However this can be automatically converted to other length units via the pull-down menu. ### Use If you set up a lever (beam and fulcrum), this formula will tell you the length needed to lift the mass. Assume you put your full weight on the end of the beam, 200 lbs, and that you want to lift 1,400 lbs. This formula will tell you that if there is 6" of the beam on the mass side of the fulcrum, you weight needs to be at the 3.5 foot (42 inches) mark on the other side to lift the object. ##### Simple Machine Calculators This equation, lever (beam length), is listed in 2 Collections.	2019-07-15 18:41:52	{"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": 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": 0.6917091608047485, "perplexity": 2434.4683840982225}, "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/1563195523840.34/warc/CC-MAIN-20190715175205-20190715201205-00263.warc.gz"}
https://iranjo.com/page-id243.php	# C calculator ### Description Solve any math problem with a basic & scientific calc. Use it at work or at school on your iPhone & iPad. Handy and convenient. Count on it whenever, wherever. Featured by Digital Trends, Educational App Store, and many more in their top calculator apps. Free in The Calculator • Basic & Scientific Calculator • History Tape: Save, Copy & Share Calculations • 70+ Amazing Themes • Edit Equation • Dictate Equations • Memory Banks • Calculate with Degrees or Radians • Set the Number of Decimals The Calculator supports: • Voice Control • Dark Mode • Apple Watch • Drag and Drop • Split View Have any questions? Please check out if the answers are available in the Frequently Asked Questions in the app Settings. Have suggestions or unable to find the answers to your questions? Our support team is ready and happy to help at [email protected]! • 3-Day Free Trial • Ad-Free • Unlimited Fraction Calculations • Unlimited Currency Conversions (Bitcoin included) • Premium Customer Support If you choose to purchase a subscription, payment will be charged to your Apple ID account, and your account will be charged within hours prior to the end of the free trial or current period. Subscription with a free trial period will automatically renew to a paid subscription. Any unused portion of a free trial period (if offered) will be forfeited when you purchase a premium subscription during the free trial period. You can cancel the automatic renewal of your subscription at any time by going to your settings in the iTunes & App Store after purchase. The cancellation will take effect the day after the last day of the current subscription period, and you will be downgraded to the free service. LIST OF MATHEMATICAL OPERATIONS Addition, Subtraction, Division, Multiplication, Percentages, Square, Cubic Exponent, Nth Exponent, Square Root, Cubic Root, Nth Root, Engineering Exponent, Factorial, Power of Two, Power of Ten, Power of E, Natural Logarithm, Common Logarithm, Binary Logarithm, Sine, Cosine, Tangent, Arc Sine, Arc Cosine, Arc Tangent, Hyperbolic Sine, Hyperbolic Cosine, Hyperbolic Tangent, Hyperbolic ArcSine, Hyperbolic Arc Cosine, Hyperbolic ArcTangent, Multiplicative Inverse, Random Number Input SHARE YOUR CALCULATION, WHEREVER YOU ARE Having a great result with your calculation or currency conversion? Save time and send it to your contacts fast & simple… by sharing with the calculator! EDIT EQUATION You can edit both digits and operators of the current equation. You can place the edit cursor by tapping or using the arrows. DICTATE & VOICE CONTROL Make your daily work tasks faster and easier. Simply use your voice to dictate equations by tapping on the mic icon. Dictate works best when you speak naturally as if you were talking to another person. Currently, Dictate supports only basic calculations. This feature requires ‘allow microphone access’ & Siri being enabled in Settings. • WE LOVE FEEDBACK • >> START CALCULATING The Calculator (C) Impala Studios. Version Your handy calculator just got handier. We've done performance improvements to make sure you have a smooth experience. Enjoying the update? Rate us! out of 5 K Ratings ### Currency calculator Currency calculator stopped working, said I was out of uses What the heck ! Thank you for your feedback on our app and subscription model. We do our best to offer good value for our premium subscribers while still offering the regular app for free. We will take your feedback into account with this, and will consider it for future improvements. If you have more feedback regarding the subscription model or the app in general, please feel free to contact us at [email protected] ### Seriously? Ok my iPhone have a free apple calc but I have to pay for one on my iPad? Or be harassed once this monstrosity loads with non stop analytical ads with geolocates you guys put a whole lotta thought into making things way more difficult than they have to be. I'm inconvenienced to say the least I hope iPhone x craps out, and apple shares plummet lol. I like my iPad, not impressed with apple right now tho. ### Ads are so prevalent, it's impossible to use This is an example of a very aggressive model for ads that ruins the user experience. Yes, I could purchase the ad-free version, but within 5 minutes I was so annoyed that I removed the application. I can appreciate that the developers need to make money and I would gladly pay if they weren't so outrageously aggressive in forcibly placing ads that block calculations on screen. Hi, Our apologies for the advertisements in the ads. We are constantly working hard on maintaining and improving our app, so we can offer the best product to our users for free. Unfortunately we have to use advertisement in order to fund this. We do want to thank you for your feedback, we will use this to balance the amount of ads better in the future. I would also like to point out that our ads should not block out any calculations on screen, so if this happened then it's most likely a bug. We would like to fix this as soon as possible, and could use your help in this. If could send us a screenshot of this happening to [email protected] that would be greatly appreciated. ### Subscriptions Free Trial The developer, International Travel Weather Calculator, indicated that the app’s privacy practices may include handling of data as described below. For more information, see the developer's privacy policy. ### Data Used to Track You The following data may be used to track you across apps and websites owned by other companies: • Identifiers • Usage Data • Diagnostics Privacy practices may vary, for example, based on the features you use or your age. Learn More ### Information Seller International Travel Weather Calculator Size MB Category Utilities Compatibility iPhone Requires iOS or later. iPod touch Requires iOS or later. Mac Requires macOS or later and a Mac with Apple M1 chip. Languages English, Danish, Dutch, Finnish, French, German, Greek, Indonesian, Italian, Japanese, Korean, Malay, Norwegian Bokmål, Norwegian Nynorsk, Portuguese, Russian, Simplified Chinese, Spanish, Swedish, Thai, Traditional Chinese, Vietnamese Age Rating 4+ Price Free In-App Purchases 1. 1 Month Premium Access$2. Yearly Premium Member$ 3. Monthly Premium Member$4. Smart Currency & Fraction Calc$ 5. 6 Months Premium Access$6. Premium Member Special Offer$ 7. Premium Member Special Offer$8. Premium Member Special Offer$ 9. Premium Membership Yearly$10. Half-Yearly Premium Member$ ### You Might Also Like Sours: https://apps.apple.com/ca/app/the-calculator/id ## Erlang Calculator - for Call Centre Staffing (Online Version ) ### Statistics from All Erlang Calculations Number of Erlang Calculations performed Past 24 HoursPast Month 3,73, Average Values Entered AHT (Seconds) AHT (Minutes) Average Service Level% Average Shrinkage27% Average Max Occupancy% Above figures include calls and other work tasks ### How To Use This Erlang Calculator 1. If you have calls per hour, then enter the number of incoming contacts as and the period is 60 minutes. 2. The Average Handling Time is the amount of time that a person (an agent) takes to handle a phone contact. This includes the talk time as well any paperwork time (wrap-up time) before they are able to answer the next call. This should be in seconds. 3. Put in your Service Level target and time. So if you wanted to handle 90% of calls in 15 seconds, put in 90 and If you are uncertain of this the industry "average" is 80% of calls answered in 20 seconds. 4. This contact centre staffing Erlang calculator is a hybrid model based on both the Erlang C formula the Erlang A formula. The Erlang C formula was invented by the Danish Mathematician A.K. Erlang and is used to calculate the number of advisors and the service level. Call Abandons are calculated using the Erlang A formula which was devised by Swedish statistician Conny Palm in This assumes an Average Patience - also know as Average Time to Abandon (ATA). 5. We also have a more flexible Microsoft Excel based version of this calculator. You can download the free Excel Erlang Calculator 6. This calculator works on probabilities, so may appear to overstate the number of agents needed at low levels. So for example if you enter 0 calls per hour it will say that you need 1 agent. This is quite correct, as there may be a possibility that one call may come in. In practicality, you may decide to not schedule any staff. 7. The maximum number of agents that the calculator can calculate before shrinkage is applied is 10, Agents. 9. The maximum occupancy is designed to improve accuracy. If you take Occupancy over 85% - 90% for long periods you will find that it gets hidden in a longer AHT figure, and agent burn out happens. 10. Call Abandons are calculated using the Erlang A formula, which assumes an Average Patience -also know as Average Time to Abandon (ATA). 11. The Calculator can deal with up to 10, agents, thanks to some help with the maths from Philip Wright CEng – (Former Technical Director & CTO Europe at Aspect Telecommunications/Communications ). ### Need To Include WebChat And Emails? Need a Multi-Channel calculator? Then use our free Multi Channel Calculator ### Terms and Conditions Use of the Erlang C Calculator is subject to our standard terms and conditions. Sours: https://www.callcentrehelper.com/tools/erlang-calculator/ Percentage calculator The percentage calculator allows you to find out what the amount, the percentage, the percent amount, the percent increase or the percent decrease is, if you know any of the two. Simple Tiling Calculator This calculator allows you to calculate how many tiles you need to cover a simple rectangular area. Works both in metric and imperial units. Works for both rectangular and square tiles. ### Financial Bond Valuation The purpose of this calculator is to provide calculations and details for bond valuation problems. It is assumed that all bonds pay interest semi-annually. Future versions of this calculator will allow for different interest frequency. Car Lease This calculator lets you calculate your estimated lease payments. General Loan This calculator allows you to calculate monthly payment, interest rate, total payments, and total interest of your loan. Loan Length Calculator Enter how much you want to spend each month, an interest rate, and a loan amount and the calculator tells you how long it will take you to pay it off! Stocks vs. Bonds How much do I keep in stocks and bonds? ### Financial Savings Future Value/Annuity Calculation Find out how much to put away tax deferred to get a certain amount of money in the future, and how much you could expect to draw out of that money. Just put in some numbers to see! How long until you are a millionaire? This calculator allows you to calculate how long it will take until you reach your desired account balance. IRA/(k)/(b) Retirement Calculation Find out how much to put away tax deferred to get a certain amount of money in the future, and how much you could expect to draw out of that money. Just put in some numbers to see! ### Financial Home Home Equity Calculator Use this calculator to see how much you may be eligible to borrow. How much house can I afford? This calculator allows you to calculate the amount you can afford to pay for a mortgage. Mortgage Payment The Mortgage Payment Calculator allows you to calculate monthly payments, average monthly interest, total interest, and total payment. This calculator allows you to compare renting versus buying by entering how much you want to spend a month and how much down you would put into your house. Simple Mortgage Payment Calculator This calculator allows you to calculate monthly payment, average monthly interest, total interest, and total payment of your mortgage. ### Fun Astrological Signs Calculator Find out your Western Sun Sign, and your Eastern or Chinese Sign. Days until Christmas Use this calculator to find out how many days are left until Christmas. The Love Calculator Fill in your name and his/her name and find out what are the chances for the two of you! Love Compatibility Calculator To find out what the chances for you and your dream partner are, just fill in your and his or her Info and click Calculate. ### Sports & Hobbies High Performance Bass Fishing Boat Speed Calculator Online High Performance Bass Fishing Boat Speed Calculator Miles per Hour Calculator This calculator allows you to calculate miles per hour. ### Health & Fitness Body Mass Index Calculator This calculator allows you to calculate your Body Mass Index (BMI). Body Mass Index (BMI) is a person's weight in kilograms divided by the square of height in meters. Find out your BMI and what it means for your health. Pregnancy Calculator This calculator allows you to calculate your due date, conception date, and fetal age. ### Converters Currency calculator Currency calculator allows you to convert between currencies. Exchange between dollars, pounds, euro, yen, yuan, and many more. Data Size Calculator This calculator allows you to calculate data size in bits, nibbles, bytes, kilobytes, megabytes, gigabytes, and terabytes. Temperature calculator The temperature calculator allows you to convert temperature degrees between Celsius, Fahrenheit, Kelvin, Rankine, Delisle, Newton, Réaumur and Rømer. Enter a value in any field to see the other values. ### Math Perches to square meters and square feet Calculator Online Perches to meters and feet Calculator. Stem and Leaf Plot This calculator allows you to create a special table where each data value is split into a stem (the first digit or digits) and a leaf (usually the last digit). Sours: https://calculator.com/ Building a Basic Calculator - C - Tutorial 13 ## Combinations Calculator (nCr) ### Calculator Use The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Basically, it shows how many different possible subsets can be made from the larger set. For this calculator, the order of the items chosen in the subset does not matter. Factorial There are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r. Combination The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed. Permutation The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed. When n = r this reduces to n!, a simple factorial of n. Combination Replacement The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are allowed. Permutation Replacement The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are allowed. n the set or population r subset of n or sample set ### Combinations Formula: $$C(n,r) = \dfrac{n!}{( r! (n - r)! )}$$ For n ≥ r ≥ 0. The formula show us the number of ways a sample of “r” elements can be obtained from a larger set of “n” distinguishable objects where order does not matter and repetitions are not allowed. [1] "The number of ways of picking r unordered outcomes from n possibilities." [2] Also referred to as r-combination or "n choose r" or the binomial coefficient. In some resources the notation uses k instead of r so you may see these referred to as k-combination or "n choose k." ### Combination Problem 1 Choose 2 Prizes from a Set of 6 Prizes You have won first place in a contest and are allowed to choose 2 prizes from a table that has 6 prizes numbered 1 through 6. How many different combinations of 2 prizes could you possibly choose? In this example, we are taking a subset of 2 prizes (r) from a larger set of 6 prizes (n). Looking at the formula, we must calculate “6 choose 2.” C (6,2)= 6!/(2! ()!) = 6!/(2! * 4!) = 15 Possible Prize Combinations The 15 potential combinations are {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, {5,6} ### Combination Problem 2 Choose 3 Students from a Class of 25 A teacher is going to choose 3 students from her class to compete in the spelling bee. She wants to figure out how many unique teams of 3 can be created from her class of In this example, we are taking a subset of 3 students (r) from a larger set of 25 students (n). Looking at the formula, we must calculate “25 choose 3.” C (25,3)= 25!/(3! * ()!)= 2, Possible Teams ### Combination Problem 3 A restaurant asks some of its frequent customers to choose their favorite 4 items on the menu. If the menu has 18 items to choose from, how many different answers could the customers give? Here we take a 4 item subset (r) from the larger 18 item menu (n). Therefore, we must simply find “18 choose 4.” C (18,4)= 18!/(4! * ()!)= 3, Possible Answers ### Handshake Problem In a group of n people, how many different handshakes are possible? First, let's find the total handshakes that are possible. That is to say, if each person shook hands once with every other person in the group, what is the total number of handshakes that occur? A way of considering this is that each person in the group will make a total of n-1 handshakes. Since there are n people, there would be n times (n-1) total handshakes. In other words, the total number of people multiplied by the number of handshakes that each can make will be the total handshakes. A group of 3 would make a total of 3() = 3 * 2 = 6. Each person registers 2 handshakes with the other 2 people in the group; 3 * 2. Total Handshakes = n(n-1) However, this includes each handshake twice (1 with 2, 2 with 1, 1 with 3, 3 with 1, 2 with 3 and 3 with 2) and since the orginal question wants to know how many different handshakes are possible we must divide by 2 to get the correct answer. Total Different Handshakes = n(n-1)/2 ### Handshake Problem as a Combinations Problem We can also solve this Handshake Problem as a combinations problem as C(n,2). n (objects) = number of people in the group r (sample) = 2, the number of people involved in each different handshake The order of the items chosen in the subset does not matter so for a group of 3 it will count 1 with 2, 1 with 3, and 2 with 3 but ignore 2 with 1, 3 with 1, and 3 with 2 because these last 3 are duplicates of the first 3 respectively. $$C(n,r) = \dfrac{n!}{( r! (n - r)! )}$$ $$C(n,2) = \dfrac{n!}{( 2! (n - 2)! )}$$ expanding the factorials, $$= \dfrac{1\times2\times\times(n-2)\times(n-1)\times(n)}{( 2\times1\times(1\times2\times\times(n-2)) )}$$ cancelling and simplifying, $$= \dfrac{(n-1)\times(n)}{2} = \dfrac{n(n-1)}{2}$$ which is the same as the equation above. ### References [1] Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. , Sours: https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php ## Permutation and Combination Calculator ### Result Permutations, nPr = = 30 Combinations, nCr = = 15 Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. A typical combination lock for example, should technically be called a permutation lock by mathematical standards, since the order of the numbers entered is important; is not the same as , whereas for a combination, any order of those three numbers would suffice. There are different types of permutations and combinations, but the calculator above only considers the case without replacement, also referred to as without repetition. This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example, ### Permutations The calculator provided computes one of the most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the "combination" lock, where a value can occur multiple times, such as For example, in trying to determine the number of ways that a team captain and goalkeeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goalkeeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team: A B C D E F G H I J K 11 members; A is chosen as captain B C D E F G H I J K 10 members; B is chosen as keeper As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goalkeeper, A was removed from the set before the second choice of the goalkeeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 &#; 10 &#; 9 &#; 8 &#; 7 &#; &#; 2 &#; 1, or 11 factorial, written as 11!. However, since only the team captain and goalkeeper being chosen was important in this case, only the first two choices, 11 &#; 10 = are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 &#; 8 &#; 7 &#; &#; 2 &#; 1, or 9!. Thus, the generalized equation for a permutation can be written as: Or in this case specifically: 11P2 = = = 11 &#; 10 = Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below: nPr = nr ### Combinations Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways, including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply . As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of Unlike the case given in the permutation example, where the captain was chosen first, then the goalkeeper, the order in which the strikers are chosen does not matter, since they will both be strikers. Referring again to the soccer team as the letters Athrough K, it does not matter whether Aand then Bor Band then Aare chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all npeople, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations ( from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, Athen Bor Bthen A, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient: nCr = Or in this case specifically: 11C2 = = = 55 It makes sense that there are fewer choices for a combination than a permutation, since the redundancies are being removed. Again for the curious, the equation for combinations with replacement is provided below: nCr (r + n -1)! r! &#; (n - 1)! Sours: https://www.calculator.net/permutation-and-combination-calculator.html Create Calculator in C programming Language in just 10 minutes with line by line code explanation Sleep breathing began to increase. I knew that I would not be able to pull it out for a long time. I stood with my back to the rest, and could not see what they were doing. Everything was so mediocre in my stretched cunt. ### You will also like: The freedom-loving courtesan, ready to become a slave of her friend, could not bear the beatings, from who-who-feeds-and-dresses. Equality, brotherhood and friendship were not like that in her understanding. And the author needs a pumpkin for such long lyrical digressions. In short, having crawled to the 3-storey building in which their. 241 242 243 244 245	2022-05-18 06:37:52	{"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": 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": 0.4088977873325348, "perplexity": 1394.694975580292}, "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/1652662521152.22/warc/CC-MAIN-20220518052503-20220518082503-00723.warc.gz"}
https://www.cheenta.com/visualizing-complex-line-integral/	Visualizing Complex Line Integral [et_pb_section fb_built="1" _builder_version="3.0.47"][et_pb_row _builder_version="3.0.47" background_size="initial" background_position="top_left" background_repeat="repeat"][et_pb_column type="4_4" _builder_version="3.0.106" parallax="off" parallax_method="on"][et_pb_text _builder_version="3.0.106" custom_padding="\|20px\|\|20px" box_shadow_style="preset1"] There is nothing complex about complex line integral. It is just vector addition (and taking a limit of that sum). Let's take a concrete example: $$\oint_{\lambda} \frac {1}{\zeta} d \zeta = 2 \pi i$$ Here, let $\lambda$ be the unit circle centered at the origin. Then we pick $\zeta$ from the circumference of the circle. Suppose we work with polar coordinates. Then the coordinate of a typical $\zeta$ is $(1, \theta)$. Watch the Lecture [/et_pb_text][/et_pb_column][/et_pb_row][/et_pb_section] Cheenta. Passion for Mathematics Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject. CAREERTEAM support@cheenta.com	2021-06-22 21:17:58	{"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": 0.9879630208015442, "perplexity": 2201.5599807203607}, "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-2021-25/segments/1623488519735.70/warc/CC-MAIN-20210622190124-20210622220124-00624.warc.gz"}
http://www.avrfreaks.net/comment/2293981	## Terminal for printing outputs? 65 posts / 0 new Author Message Hey people, I would like to see what a program is printing on a terminal. Does AS7 have any inbuilt terminals? If yes where? Otherwise, what are the options I have? Azim azimmali wrote: what are the options I have TeraTerm PuTTY Bray RealTerm etc, etc, ... EDIT This is a very frequently recurring question; eg, http://www.avrfreaks.net/comment... Last Edited: Thu. Oct 12, 2017 - 02:44 PM I have TeraTerm installed. Can I print data without the hardware? I need to see what certain line of code are printing. If yes, how? You mean you're running your code in a simulator, and want to see its UART output ? EDIT Is that specifically the Atmel Studio simulator? Last Edited: Thu. Oct 12, 2017 - 02:46 PM So, I have this code that uses a circular buffer. It receives incoming bytes in these UART buffers. I want to know what these incoming bytes are. In my previous post I had inquired on how I can assign dummy values to those buffers but in vain. The project does a dummy capture. I am frustrated with this now. I just dont know what to do and whom to approach for such (maybe) trivial questions. I do not know much about the ATMEL simulator. I was debugging the project to find out the flow of the program. But as usual its stuck in a while loop. So are you running your code in the simulator or not? Why not use the DEBUGGER to do this - stop at any moment and you can actually examine what is in the buffers in actual AVR memory - no need to "print" the contents No..:/ AVR memory? whats that? I am using the debugger. So this AVR memory is where I can see what's in those buffers? azimmali wrote: AVR memory? whats that? Are you serious?! The AVR has 3 types of memory: 1. RAM - where your variables, stack, etc, go; 2. Flash - where your code goes; 3. EEPROM. EDIT EDIT 2 typo #AVRMemories #AVRArchitecture Last Edited: Mon. Oct 16, 2017 - 08:26 AM Oh no. I am sorry. Lol. I know that. I thought AVR memory is a tool on AS7 . So how do I examine whats in the AVR Memory? ref: clawson azimmali wrote: how do I examine whats in the AVR Memory? That's one of the fundamental tools of the AS7 debugger - in fact, of any debugger. I think you need to spend some time reviewing the documentation and available tutorials for Atmel Studio ... There's a whole load of resources - including videos - linked from the Atmel Studio pages on the Atmel & Microchip websites. EDIT Last Edited: Thu. Oct 12, 2017 - 03:53 PM Ok. Done. AS7 has a terminal. That's what the documentation says. Under View. But I dont see any on my AS7 here. Ugh!!! There used to be a 'terminal' program in AS, but it was never any good! You can only use a terminal program(see list above) with real hardware, not with a simulator. It will require some interface hardware as well, most likely a USB to serial TTL cable like this: http://www.ebay.com/itm/USB-To-R... Note this has TTL outputs, not RS-232 outputs, you also need one in which the TTL signal levels match your VCC level, either 3.3v or 5.0v, the cables come in both versions, use the correct one for your project. These will create a COM port when plugged into a PC, set your terminal program to use that COM port, connect the TX,RX and GND pins only to your AVR, like this: PC - AVR TX - RX RX - TX gnd-gnd Now you should have serial comms between your PC and your AVR. NOTE: the AVR needs a xtal in order to have an accurate and stable clock for async comms to work well! Jim ki0bk wrote: You can only use a terminal program(see list above) with real hardware, not with a simulator. No, that's not necessarily true. eg, Keil has a simulator[1] which can send the simulated UART output to a COM port. So you could loop that back to a terminal on another real COM port, or use something like com0com to do it "virtually" http://com0com.sourceforge.net/ However, given chips with on-board debug, I found it just as easy to use a real devboard - and that was over ten years ago. Nowadays, given devboards with the debugger and virtual COM port built-in, it would be even more so. [1] Of course, not an AVR simulator - but the principle applies. EDIT But, as pointed out in #8, a serial port and terminal is not the way to do what the OP is asking anyhow. Last Edited: Thu. Oct 12, 2017 - 09:45 PM In case it's not clear what I am suggesting is that in either hardware debugging or simulation you use: http://www.atmel.com/webdoc/GUID... If you direct this to the location of "Rx_Buffer" or whatever your UART buffer is called you can immediately see the characters/bytes in the buffer. You do not need to employ a UART to send them "out" of the AVR or a terminal program to take them "in" and display them. The memory view in AS7 will let you "see" the buffer contents directly. You also have facilities to view your variables by name: http://www.atmel.com/webdoc/GUID... EDIT eg, Last Edited: Fri. Oct 13, 2017 - 09:07 AM I found this yesterday. Thanks Neil! :) I will keep you posted. I guess it's horses for courses but for "buffers" I think you are going to find a memory window rather than a variable watch is more useful for this kind of thing. (well I do). I will keep track of both. clawson wrote: for "buffers" I think you are going to find a memory window rather than a variable watch is more useful for this kind of thing. Probably a combination of both - a memory view to see the content, and watches for the read & write indexes. A watch can also be useful when you're interested in a particular buffer entry - eg, my_buffer[87] - rather than having to manually count through a memory view. I want to use a printf() function for serial printing. I added the stdio.h header. I was trying to define a stream based on a reference online. But its' showing me this error. azimmali wrote: I want to use a printf() function for serial printing Why?? After it's been amply demonstrated that you really do not need to do that!! Does this AVR have two UARTs then? . As to the error, do you know what a "function declaration" is in C and why we need/use them? Last Edited: Sat. Oct 14, 2017 - 01:36 PM or, in fact, what any declaration is? Here's the UART definitions from the header //UART define //------------------------------------------------------------------- //#define UART_DBG //debug UART0 #ifdef UART_DBG #define UART_RECV_VECTOR SIG_UART0_RECV #define UART_TX_VECTOR SIG_UART0_DATA #define UART_UDR UDR0 #define UART_UDRE (1<<UDRE0) #define UART_UDRIE (1<<UDRIE0) #define UART_UCSRA UCSR0A #define UART_UCSRB UCSR0B #define UART_UCSRC UCSR0C #define UART_UBRRL UBRR0L #define UART_TXC TXC0 #define UART_RXC RXC0 #define UART_TXEN TXEN0 #define UART_RXEN RXEN0 #define UART_ERRORS ((1<<FE0)\|(1<<DOR0)\|(1<<UPE0)) #define UART_UCSRB_INIT ((1<<RXCIE0) \| (0<<UDRIE0) \| (1<<RXEN0) \| (1<<TXEN0)) #define UART_UCSRC_INIT ((1<<UCSZ01) \| (1<<UCSZ00)) #else #define UART_RECV_VECTOR USART1_RX_vect #define UART_TX_VECTOR USART1_UDRE_vect #define UART_UDR UDR1 #define UART_UDRE (1<<UDRE1) #define UART_UCSRA UCSR1A #define UART_UCSRB UCSR1B #define UART_UCSRC UCSR1C #define UART_UBRRL UBRR1L #define UART_SPEED 51 #define UART_ERRORS ((1<<FE1)\|(1<<DOR1)\|(1<<UPE1)) #define UART_UCSRA_INIT ((1<<TXC1) \| (1<<U2X1)) #define UART_UCSRB_INIT ((1<<RXCIE1) \| (0<<UDRIE1) \| (1<<RXEN1) \| (1<<TXEN1)) #define BL_UCSRB_INIT ((1<<RXEN1) \| (1<<TXEN1)) #define UART_UCSRC_INIT ((1<<UCSZ11) \| (1<<UCSZ10)) #define UART_TXC TXC1 #define UART_RXC RXC1 #endif I could only locate UART definitions defined under the else condition in the project. Last Edited: Mon. Oct 16, 2017 - 03:47 PM It tells the compiler what the function parameters are, what return type, etc., I wanted to play with code. I am able to see whats going in the memory. So the firmware in the embedded board starts functioning based on commands from an application software (AS). So, the AS issues an opcode 02 03 11 12 03. I want to make the board independent of the application software. Forget the board, atleast the firmware. For now the firmware is doing a dummy capture. I would like the firmware to do a real time operation and be able to identify multiple modulation schemes and process that data accordingly. I tried inputting the opcode on a terminal like termite and its displaying some random values; all numbers. In your screenshot in your post #25 you try to use usart_putchar_printf() (in the macro call to FDEV_SETUP_STREAM) before you actually define it. Swap those two. First the definition, then do the macro call. Any function must be "seen", either by a definition (i.e. complete implementation) or by its declaration (its definition or just its prototype) before it is being used (referenced). This is absolutely basic C knowledge. IIRC we've advised you in the past to get a C textbook and read. That advice still stands. "He used to carry his guitar in a gunny sack, or sit beneath the tree by the railroad track. Oh the engineers would see him sitting in the shade, Strumming with the rhythm that the drivers made. People passing by, they would stop and say, "Oh, my, what that little country boy could play!" [Chuck Berry] "Some questions have no answers."[C Baird] "There comes a point where the spoon-feeding has to stop and the independent thinking has to start." [C Lawson] "There are always ways to disagree, without being disagreeable."[E Weddington] "Words represent concepts. Use the wrong words, communicate the wrong concept." [J Morin] "Persistence only goes so far if you set yourself up for failure." [Kartman] Ok. I changed that. static int uart_putchar(char c, FILE stream); static FILE mystdout = FDEV_SETUP_STREAM(uart_putchar, NULL, _FDEV_SETUP_WRITE); static int uart_putchar(char c, FILE stream) { if (c == '\n') uart_putchar('\r', stream); loop_until_bit_is_set(UCSRA, UDRE); UDR = c; return 0; } I have these two errors. I would shamelessly accept that my exposure to C in real world is less (I would say programming in general). This project has got me hooked on it now. Yes, I also have a text book on C. but Embedded C is completely new to me. Update: I changed the UCSRA and UDRE to how they are defined to in headers. RE: post #34 7 of the warnings there say "This header file is obsolete" talking about your use of <avr/signal.h>. The message even says use <avr/interrupt.h> instead. Why on earth are you ignoring things like this?!? As for UCSRA being undeclared. It should be (a) if you use the <avr/io.h> as you should and (b) you are actually building for a model of AVR that has a register called UCSRA This is a complete guess but I think you probably do include the io.h header so presumably the AVR you are using is one that does not have UCSRA. Is it a "modern" AVR like mega48/88/168/328 or perhaps mega164/324/644/1284? Those models have UCSR0A not UCSRA for the first (sometimes only) UART. The same is true if UDR. The register is probably UDR0 in fact. In #29 you posted some code (without any explanation) that mentions things like UCSR0A, UCSR1A, UDR0 and UDR1. That seems to confirm that you may be building for a "modern" (like last 10 years) model of AVR. The code you have found using UCSRA/UDR is probably from a 10+ year old AVR Just start by telling us which model of AVR it is you are trying to use. Last Edited: Mon. Oct 16, 2017 - 05:03 PM clawson wrote: The message even says use <avr/interrupt.h> instead. Why on earth are you ignoring things like this?!? +1 . Never ignore warnings. Just because they are warnings (rather than errors) does not mean they are always benign. "He used to carry his guitar in a gunny sack, or sit beneath the tree by the railroad track. Oh the engineers would see him sitting in the shade, Strumming with the rhythm that the drivers made. People passing by, they would stop and say, "Oh, my, what that little country boy could play!" [Chuck Berry] "Some questions have no answers."[C Baird] "There comes a point where the spoon-feeding has to stop and the independent thinking has to start." [C Lawson] "There are always ways to disagree, without being disagreeable."[E Weddington] "Words represent concepts. Use the wrong words, communicate the wrong concept." [J Morin] "Persistence only goes so far if you set yourself up for failure." [Kartman] Last Edited: Mon. Oct 16, 2017 - 05:24 PM Are you referring to the MCU? If yes, atmega64, 8 bit. The avr/interrupt.h warning. I have replaced the avr/signal.h with the avr/interrupt.h in all the program files. It still shows this error SOMETIMES, which is kind of weird. I have been noting this for sometime : whenever I add a new piece of code to the program, I get that error. I changed UDR to UART_UDR and now I have one warning - mystdout defined but not used <-Wunused-variable>. Never ignore warnings That's a warning that most programmers always ignore. When in the dark remember-the future looks brighter than ever. Try actually reading the data sheet. The datasheet for atmega 64 tells you that it has two UART so it has UCSR0A and UCSR1A. Equally it has UDR0 and UDR1. I was going through the avr library manual and came across #defines in uppercase and lowercase like this one below. Can anyone tell me what's the difference? #define fdev_setup_stream(stream, put, get, rwflag) #define FDEV_SETUP_STREAM(put, get, rwflag) Thanks two things: 1) you do know there is a user manual? http://www.nongnu.org/avr-libc/u... 2) so what you are looking at has to be read in context: #if defined(__DOXYGEN__) /** \brief Setup a user-supplied buffer as an stdio stream This macro takes a user-supplied buffer \c stream, and sets it up as a stream that is valid for stdio operations, similar to one that has been obtained dynamically from fdevopen(). The buffer to setup must be of type FILE. The arguments \c put and \c get are identical to those that need to be passed to fdevopen(). The \c rwflag argument can take one of the values _FDEV_SETUP_READ, intent, respectively. \note No assignments to the standard streams will be performed by fdev_setup_stream(). If standard streams are to be used, these \ref stdio_without_malloc "Running stdio without malloc()". / #define fdev_setup_stream(stream, put, get, rwflag) #else / !DOXYGEN / #define fdev_setup_stream(stream, p, g, f) \ do { \ (stream)->put = p; \ (stream)->get = g; \ (stream)->flags = f; \ (stream)->udata = 0; \ } while(0) #endif / DOXYGEN / DOXYGEN is the program that creates the documentation in (1) from this header file. What the above section of code says is that "when parsing this header, if generating the documentation then use the first part, but for building your "normal" AVR programs use the second part". So the #define you referred to is simply a dummy one for documentation purposes. The true definition that is used is: #define fdev_setup_stream(stream, p, g, f) \ do { \ (stream)->put = p; \ (stream)->get = g; \ (stream)->flags = f; \ (stream)->udata = 0; \ } while(0) So you create a FILE stream object then use this function to populate the function pointers and flags within it. As to the difference between the upper case and lower case versions. While there are subtle differences the actual intention is that FDEV_SETUP_STREAM is used in C and fdev_setup_stream() in C++. The latter has to exist because the former uses a feature of C (Named initialisers) that is not in C++ Thanks clawson. :) One more question: Should I be making these changes in the main.c or UART source file? It's not clear to me what "changes" you are talking about. Changes for a user supplied stdio stream. I would be tempted to keep that with the UART stuff. Maybe just make it part of uart_init() that stdin/stdout are connected to the local uart_getchar() and uart_putchar() routines so you main can look like: #include "uart.h" int main(void) { uart_init(); // maybe pass a baud rate here? printf("hello"); } That keeps main() very "clean" so there's no knowledge of the connection to stdout here. OTOH you might want to retain control of whether stdout connects to the UART output channel. Some libraries choose to do it more like: #include "uart.h" int main(void) { uart_init(); // maybe pass a baud rate here? uart_connect_to_stdio(); printf("hello"); } so that the connection is more obvious and also so it can optionally be removed. So yesterday, I put a main in the main.c file and the obvious happened; the program didn't compile. I am guessing defining the function for serial printing could be done in the uart source file. The init_uart() you are talking about is defined as UARTIni() under level4.c source file with its declarations and macros defined in level4.h. I am guessing I have create that function for serial printing in UART source file. You do know about things like extern and function declarations etc? Ultimately the code in a single C file can be split in any way you choose. But often it takes a little effort to get "cross links" between files working. For data you will need "extern" declarations so the non-implementing file can "see" the data in the other. For functions you do it by a split of function declaration and definition with the declarations usually being placed in a shared header file. Were you doing this? I am getting a feel of it now. There are a lot of such declarations especially extern in the project. I am doing this on AS7 (is that what you meant?) and all the files are in C(with reference to the io stream declaration - does it matter if I go for the uppercase or lowercase declarations - I read that the upper case is used as an initializer for a variable type 'FILE' Edited: I am getting a feel of it now. There are a lot of such declarations especially extern in the project. I haven't made much changes to the code yet - I will try doing it and keep it posted. and all the files are in C(with reference to the io stream declaration - does it matter if I go for the uppercase or lowercase declarations - I read that the upper case is used as an initializer for a variable type 'FILE' How do you delete a comment on here? Last Edited: Tue. Oct 17, 2017 - 04:48 PM Any views on this one? http://www.avrfreaks.net/forum/o... azimmali wrote: Any views on this one? http://www.avrfreaks.net/forum/o... Do you have specific questions about it? As Cliff says in that thread, the first post in it is more or less a repetition of what is written in the official avrlibc documentation. "He used to carry his guitar in a gunny sack, or sit beneath the tree by the railroad track. Oh the engineers would see him sitting in the shade, Strumming with the rhythm that the drivers made. People passing by, they would stop and say, "Oh, my, what that little country boy could play!" [Chuck Berry] "Some questions have no answers."[C Baird] "There comes a point where the spoon-feeding has to stop and the independent thinking has to start." [C Lawson] "There are always ways to disagree, without being disagreeable."[E Weddington] "Words represent concepts. Use the wrong words, communicate the wrong concept." [J Morin] "Persistence only goes so far if you set yourself up for failure." [Kartman] In my previous attempt, I had done this: static int uart_putchar(char c, FILE stream); static FILE mystdout = FDEV_SETUP_STREAM(uart_putchar, NULL, _FDEV_SETUP_WRITE); static int uart_putchar(char c, FILE stream) { if (c == '\n') uart_putchar('\r', stream); loop_until_bit_is_set(UART_UCSRA, UART_UDRE); UART_UDR = c; return 0; } int main(void) { //some other stuff UARTIni(); stdout = &mystdout; printf("Hello World!\n"); return 0; //more other stuff }/ Errors: int uart_putchar(char c, FILE stream) { if (c == '\a') //'\a' { fputs("ring\n", stderr); return 0; } if (c == '\n') // uart_putchar('\r', stream); loop_until_bit_is_set(UART_UCSRA, UDRE); UART_UDR = c; return 0; } int uart_getchar(FILE stream) { uint8_t c; char cp, cp2; static char b[UART_IN_BUFFER_SIZE]; static char rxp; if (rxp == 0) for (cp = b;;) { loop_until_bit_is_set(UART_UCSRA, RXC); if (UART_UCSRA & _BV(FE)) return _FDEV_EOF; if (UART_UCSRA & _BV(DOR)) return _FDEV_ERR; c = UART_UDR; //behaviour similar to Unix stty ICRNL if (c == '\r') c = '\n'; if (c == '\n') { cp = c; uart_putchar(c, stream); rxp = b; break; } else if (c == '\t') c = ' '; if ((c >= (uint8_t)' ' && c <= (uint8_t)'\x7e') \|\| c >= (uint8_t)'\xa0') { if (cp == b + UART_IN_BUFFER_SIZE - 1) uart_putchar('\a', stream); else { cp++ = c; uart_putchar(c, stream); } continue; } switch (c) { case 'c' & 0x1f: return -1; case '\b': case '\x7f': if (cp > b) { uart_putchar('\b', stream); uart_putchar(' ', stream); uart_putchar('\b', stream); cp--; } break; case 'r' & 0x1f: uart_putchar('\r', stream); for (cp2 = b; cp2 < cp; cp2++) uart_putchar(cp2, stream); break; case 'u' & 0x1f: while (cp > b) { uart_putchar('\b', stream); uart_putchar(' ', stream); uart_putchar('\b', stream); cp--; } break; case 'w' & 0x1f: while (cp > b && cp[-1] != ' ') { uart_putchar('\b', stream); uart_putchar(' ', stream); uart_putchar('\b', stream); cp--; } break; } } c = rxp++; if (c == '\n') rxp = 0; return c; } FILE uart_str = FDEV_SETUP_STREAM(uart_putchar, uart_getchar, _FDEV_SETUP_RW); stdout = stdin = &uart_str; I still get errors: Question: What I am doing wrong here? When does one use vfprintf and vfscanf? azimmali wrote: When does one use vfprintf and vfscanf? Possibly never (explicitly). the vfprintf(...) function is primarily the "worker" behind both printf(...) and sprintf(...). When documentation mentions vsprintf(...) it most likely does so to document that behavior that printf(...) and sprintf(...) shares, rather than documenting it for both of those two functions. azimmali wrote: What I am doing wrong here? We can't tell. You've shown a snippet of code, of about 100 lines. Yet all error messages in your screen dump are at line number 410, except for the venerable "recipe failed" - which is a stupid error message, better explained by the complete build output. Speaking of that, haven't we told you before to switch to the output tab and post complete build output rather than those meaningless screen shots? If we've missed that, here it comes: Don't post screen shots. For build errors, post complete build output (switch to output tab, mark everything, copy and paste in a post here. When posting such errors, also post the complete source file(s) (perhaps as attachments). "He used to carry his guitar in a gunny sack, or sit beneath the tree by the railroad track. Oh the engineers would see him sitting in the shade, Strumming with the rhythm that the drivers made. People passing by, they would stop and say, "Oh, my, what that little country boy could play!" [Chuck Berry] "Some questions have no answers."[C Baird] "There comes a point where the spoon-feeding has to stop and the independent thinking has to start." [C Lawson] "There are always ways to disagree, without being disagreeable."[E Weddington] "Words represent concepts. Use the wrong words, communicate the wrong concept." [J Morin] "Persistence only goes so far if you set yourself up for failure." [Kartman] Last Edited: Tue. Oct 17, 2017 - 08:08 PM I wanted a green signal to do that. Thanks Johan. ------ Build started: Project: EMTestRun, Configuration: Debug AVR ------ Build started. Project "EMTestRun.cproj" (default targets): Target "PreBuildEvent" skipped, due to false condition; ('$(PreBuildEvent)'!='') was evaluated as (''!=''). Target "CoreBuild" in file "C:\Program Files (x86)\Atmel\Studio\7.0\Vs\Compiler.targets" from project "E:\EMTestRunDev\EMTestRun\EMTestRun\EMTestRun.cproj" (target "Build" depends on it): Task "RunCompilerTask" Shell Utils Path C:\Program Files (x86)\Atmel\Studio\7.0\shellUtils C:\Program Files (x86)\Atmel\Studio\7.0\shellUtils\make.exe all --jobs 4 --output-sync Building file: .././level1_4026.c Invoking: AVR/GNU C Compiler : 5.4.0 In file included from .././level1_4026.c:61:0: c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h(36,2): warning: #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] #warning "This header file is obsolete. Use <avr/interrupt.h>." ^ "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level1_4026.d" -MT"level1_4026.d" -MT"level1_4026.o" -o "level1_4026.o" ".././level1_4026.c" Finished building: .././level1_4026.c Building file: .././level1_41xx.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level1_41xx.d" -MT"level1_41xx.d" -MT"level1_41xx.o" -o "level1_41xx.o" ".././level1_41xx.c" Finished building: .././level1_41xx.c In file included from .././level1_41xx.c:63:0: c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h(36,2): warning: #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] #warning "This header file is obsolete. Use <avr/interrupt.h>." ^ Building file: .././boot_ld.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "boot_ld.d" -MT"boot_ld.d" -MT"boot_ld.o" -o "boot_ld.o" ".././boot_ld.c" Finished building: .././boot_ld.c Building file: .././level2.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level2.d" -MT"level2.d" -MT"level2.o" -o "level2.o" ".././level2.c" Finished building: .././level2.c .././level2.c: In function 'ExtractData': E:\EMTestRunDev\EMTestRun\EMTestRun\level2.c(581,20): warning: variable 'column_parity' set but not used [-Wunused-but-set-variable] register uint8_t column_parity; ^ E:\EMTestRunDev\EMTestRun\EMTestRun\level2.c(580,20): warning: variable 'line_parity' set but not used [-Wunused-but-set-variable] register uint8_t line_parity; ^ Building file: .././level2_4026.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level2_4026.d" -MT"level2_4026.d" -MT"level2_4026.o" -o "level2_4026.o" ".././level2_4026.c" Finished building: .././level2_4026.c In file included from .././level2_4026.c:61:0: c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h(36,2): warning: #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] #warning "This header file is obsolete. Use <avr/interrupt.h>." ^ Building file: .././level2_41xx.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level2_41xx.d" -MT"level2_41xx.d" -MT"level2_41xx.o" -o "level2_41xx.o" ".././level2_41xx.c" Finished building: .././level2_41xx.c In file included from .././level2_41xx.c:61:0: c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h(36,2): warning: #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] #warning "This header file is obsolete. Use <avr/interrupt.h>." ^ Building file: .././level3_4026.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level3_4026.d" -MT"level3_4026.d" -MT"level3_4026.o" -o "level3_4026.o" ".././level3_4026.c" Finished building: .././level3_4026.c In file included from .././level3_4026.c:64:0: c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h(36,2): warning: #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] #warning "This header file is obsolete. Use <avr/interrupt.h>." ^ Building file: .././level4.c Invoking: AVR/GNU C Compiler : 5.4.0 In file included from .././level4.c:73:0: E:\EMTestRunDev\EMTestRun\EMTestRun\level4.c(413,1): error: conflicting types for '__iob' stdout = stdin = &uart_str; ^ c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\stdio.h(407,23): info: previous declaration of '__iob' was here extern struct __file __iob[]; ^ E:\EMTestRunDev\EMTestRun\EMTestRun\level4.c(413,16): warning: assignment makes integer from pointer without a cast [-Wint-conversion] stdout = stdin = &uart_str; ^ In file included from .././level4.c:73:0: E:\EMTestRunDev\EMTestRun\EMTestRun\level4.c(413,10): error: invalid initializer stdout = stdin = &uart_str; ^ make: * [level4.o] Error 1 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level4.d" -MT"level4.d" -MT"level4.o" -o "level4.o" ".././level4.c" E:\EMTestRunDev\EMTestRun\EMTestRun\Debug\Makefile(149,1): error: recipe for target 'level4.o' failed make: * Waiting for unfinished jobs.... Building file: .././level3.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level3.d" -MT"level3.d" -MT"level3.o" -o "level3.o" ".././level3.c" Finished building: .././level3.c Building file: .././level3_41xx.c Invoking: AVR/GNU C Compiler : 5.4.0 In file included from .././level3_41xx.c:64:0: c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h(36,2): warning: #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] #warning "This header file is obsolete. Use <avr/interrupt.h>." ^ "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level3_41xx.d" -MT"level3_41xx.d" -MT"level3_41xx.o" -o "level3_41xx.o" ".././level3_41xx.c" Finished building: .././level3_41xx.c Building file: .././main.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "main.d" -MT"main.d" -MT"main.o" -o "main.o" ".././main.c" Finished building: .././main.c Done executing task "RunCompilerTask" -- FAILED. Done building target "CoreBuild" in project "EMTestRun.cproj" -- FAILED. Done building project "EMTestRun.cproj" -- FAILED. Build FAILED. ========== Build: 0 succeeded or up-to-date, 1 failed, 0 skipped ========== I am still getting the signal.h warning. I am using the interrupt.h Warning #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] EMTestRun c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h 36 ## Attachment(s): Last Edited: Tue. Oct 17, 2017 - 08:31 PM Total votes: 0 azimmali wrote: What I am doing wrong here? stdout = stdin = &uart_str; This is line 410, right? You can't do that outside of a function. Edit: in your latest post it is line 413 Stefan Ernst Last Edited: Tue. Oct 17, 2017 - 08:40 PM Total votes: 0 Ok. So in what function do I put that in? Total votes: 1 azimmali wrote: I am still getting the signal.h warning. Look at the lines just before it in the build output. Here's one example from your build output: Building file: .././level3_4026.c Invoking: AVR/GNU C Compiler : 5.4.0 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level3_4026.d" -MT"level3_4026.d" -MT"level3_4026.o" -o "level3_4026.o" ".././level3_4026.c" Finished building: .././level3_4026.c In file included from .././level3_4026.c:64:0: c:\program files (x86)\atmel\studio\7.0\toolchain\avr8\avr8-gnu-toolchain\avr\include\avr\signal.h(36,2): warning: #warning "This header file is obsolete. Use <avr/interrupt.h>." [-Wcpp] #warning "This header file is obsolete. Use <avr/interrupt.h>." ^ All those line together tell the true story: It starts with the build system saying it compiles .././level3_4026.c , and ends with the warning about avr/signal.h . So, there is no question about avr/signal.h being included in .././level3_4026.c . You have several such errors. "He used to carry his guitar in a gunny sack, or sit beneath the tree by the railroad track. Oh the engineers would see him sitting in the shade, Strumming with the rhythm that the drivers made. People passing by, they would stop and say, "Oh, my, what that little country boy could play!" [Chuck Berry] "Some questions have no answers."[C Baird] "There comes a point where the spoon-feeding has to stop and the independent thinking has to start." [C Lawson] "There are always ways to disagree, without being disagreeable."[E Weddington] "Words represent concepts. Use the wrong words, communicate the wrong concept." [J Morin] "Persistence only goes so far if you set yourself up for failure." [Kartman] Total votes: 1 azimmali wrote: Ok. So in what function do I put that in? UARTIni Stefan Ernst Total votes: 0 Solved that error. Thanks Johan, I hardly take a look at the build output. That was an eye opener. Total votes: 0 I did that. ------ Build started: Project: EMTestRun, Configuration: Debug AVR ------ Build started. Project "EMTestRun.cproj" (default targets): Target "PreBuildEvent" skipped, due to false condition; ('$(PreBuildEvent)'!='') was evaluated as (''!=''). Target "CoreBuild" in file "C:\Program Files (x86)\Atmel\Studio\7.0\Vs\Compiler.targets" from project "E:\EMTestRunDev\EMTestRun\EMTestRun\EMTestRun.cproj" (target "Build" depends on it): Shell Utils Path C:\Program Files (x86)\Atmel\Studio\7.0\shellUtils C:\Program Files (x86)\Atmel\Studio\7.0\shellUtils\make.exe all --jobs 4 --output-sync Building file: .././level4.c Invoking: AVR/GNU C Compiler : 5.4.0 .././level4.c: In function 'UARTIni': E:\EMTestRunDev\EMTestRun\EMTestRun\level4.c(294,21): error: 'uart_str' undeclared (first use in this function) stdout = stdin = &uart_str; ^ E:\EMTestRunDev\EMTestRun\EMTestRun\level4.c(294,21): info: each undeclared identifier is reported only once for each function it appears in .././level4.c: At top level: E:\EMTestRunDev\EMTestRun\EMTestRun\level4.c(409,13): warning: 'uart_str' defined but not used [-Wunused-variable] static FILE uart_str = FDEV_SETUP_STREAM(uart_putchar, uart_getchar, _FDEV_SETUP_RW); ^ make: *** [level4.o] Error 1 "C:\Program Files (x86)\Atmel\Studio\7.0\toolchain\avr8\avr8-gnu-toolchain\bin\avr-gcc.exe" -x c -funsigned-char -funsigned-bitfields -DDEBUG -I"C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\include" -O1 -ffunction-sections -fdata-sections -fpack-struct -fshort-enums -g2 -Wall -mmcu=atmega64 -B "C:\Program Files (x86)\Atmel\Studio\7.0\Packs\atmel\ATmega_DFP\1.2.132\gcc\dev\atmega64" -c -std=gnu99 -MD -MP -MF "level4.d" -MT"level4.d" -MT"level4.o" -o "level4.o" ".././level4.c" E:\EMTestRunDev\EMTestRun\EMTestRun\Debug\Makefile(149,1): error: recipe for target 'level4.o' failed Done building target "CoreBuild" in project "EMTestRun.cproj" -- FAILED. Done building project "EMTestRun.cproj" -- FAILED. Build FAILED. ========== Build: 0 succeeded or up-to-date, 1 failed, 0 skipped ========== uart_str - undeclared error You also need the static FILE uart_str = FDEV_SETUP_STREAM(uart_putchar, uart_getchar, _FDEV_SETUP_RW); prior to the stdout = stdin = &uart_str; David (aka frog_jr)	2017-10-18 00:14:15	{"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": 0.37590643763542175, "perplexity": 7479.164500172089}, "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-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00425.warc.gz"}
http://bib-pubdb1.desy.de/collection/Preprint?ln=en	# Preprints 2019-08-1214:12 [PUBDB-2019-02987] Preprint/Report et al Search for pair production of vector-like quarks in the fully hadronic final state [arXiv:1906.11903; CMS-B2G-18-005; CERN-EP-2019-129] The results of two searches for pair production of vector-like T or B quarks in fully hadronic final states are presented, using data from the CMS experiment at a center-of-mass energy of 13 TeV. The data were collected at the LHC during 2016 and correspond to an integrated luminosity of 35.9 fb$^{-1}$. [...] OpenAccess: PDF PDF (PDFA); 2019-08-1214:09 [PUBDB-2019-02986] Preprint/Report et al Combined search for supersymmetry with photons in proton-proton collisions at $\sqrt{s}=$ 13 TeV [arXiv:1907.00857; CMS-SUS-18-005; CERN-EP-2019-114] A combination of four searches for new physics involving signatures with at least one photon and large missing transverse momentum, motivated by generalized models of gauge-mediated supersymmetry (SUSY) breaking, is presented. All searches make use of proton-proton collision data at $\sqrt{s}=$ 13 TeV, which were recorded with the CMS detector at the LHC in 2016, and correspond to an integrated luminosity of 35.9 fb$^{-1}$. [...] OpenAccess: PDF PDF (PDFA); 2019-08-1214:03 [PUBDB-2019-02985] Preprint/Report et al Measurement of the top quark Yukawa coupling from $\mathrm{t\bar{t}}$ kinematic distributions in the lepton+jets final state in proton-proton collisions at $\sqrt{s} =$ 13 TeV [arXiv:1907.01590; CMS-TOP-17-004; CERN-EP-2019-119] Results are presented for an extraction of the top quark Yukawa coupling from top quark-antiquark ($\mathrm{t\bar{t}}$) kinematic distributions in the lepton plus jets final state in proton-proton collisions, based on data collected by the CMS experiment at the LHC at $\sqrt{s} =$ 13 TeV, corresponding to an integrated luminosity of 35.8 fb$^{-1}$. Corrections from weak boson exchange, including Higgs bosons, between the top quarks can produce large distortions of differential distributions near the energy threshold of $\mathrm{t\bar{t}}$ production. [...] OpenAccess: PDF PDF (PDFA); 2019-08-1213:59 [PUBDB-2019-02984] Preprint/Report et al Search for MSSM Higgs bosons decaying to $\mu^+\mu^-$ in proton-proton collisions at $\sqrt{s}=$ 13 TeV [arXiv:1907.03152; CMS-HIG-18-010; CERN-EP-2019-109] A search is performed for neutral non-standard-model Higgs bosons decaying to two muons in the context of the minimal supersymmetric standard model (MSSM). Proton-proton collision data recorded by the CMS experiment at the CERN Large Hadron Collider at a center-of-mass energy of 13 TeV were used, corresponding to an integrated luminosity of 35.9 fb$^{-1}$. [...] OpenAccess: PDF PDF (PDFA); 2019-08-1213:56 [PUBDB-2019-02983] Preprint/Report et al Measurement of the top quark polarization and $\mathrm{t\bar{t}}$ spin correlations using dilepton final states in proton-proton collisions at $\sqrt{s}=$ 13 TeV [arXiv:1907.03729; CMS-TOP-18-006; CERN-EP-2019-073] Measurement of the top quark polarization and $\mathrm{t\bar{t}}$ spin correlations are presented using events containing two oppositely charged leptons $(e^+ e^−, e^± μ^∓$, or $μ^+ μ^−)$ produced in proton-proton collisions at a center-of-mass energy of 13 TeV. The data were recorded by the CMS experiment at the LHC in 2016 and correspond to an integrated luminosity of 35.9 fb$^{−1}$ [...] OpenAccess: PDF PDF (PDFA); 2019-08-1213:51 [PUBDB-2019-02982] Preprint/Report et al Study of the B$^+ \to$J$/\psi\overline{\Lambda}$p decay in proton-proton collisions at $\sqrt{s}=$ 8 TeV [arXiv:1907.05461; CMS-BPH-18-005; CERN-EP-2019-128] A study of the B$^+\to$J$/\psi\overline{\Lambda}$p decay using proton-proton collision data collected at $\sqrt{s}=$ 8 TeV by the CMS experiment at the LHC, corresponding to an integrated luminosity of 19.6 fb$^{-1}$, is presented. The ratio of branching fractions $\mathcal{B}$(B$^+ \to$J$/\psi\overline{\Lambda}$p )$/\mathcal{B}$(B$^\to$J$/\psi$K$^$(892)$^+$) is measured to be (1.054 $\pm$ 0.057 (stat) $\pm$ 0.035 (syst) $\pm$ 0.011 ($\mathcal{B}$))%, where the last uncertainty reflects the uncertainties in the world-average branching fractions of $\overline{\Lambda}$ and K$^$(892)$^+$ decays to reconstructed final states. [...] OpenAccess: PDF PDF (PDFA); 2019-08-1213:46 [PUBDB-2019-02981] Preprint/Report et al Search for physics beyond the standard model in events with overlapping photons and jets [arXiv:1907.06275; CMS-B2G-18-007; CERN-EP-2019-135] Results are reported from a search for new particles that decay into a photon and two gluons, in events with jets. Novel jet substructure techniques are developed that allow photons to be identified in an environment densely populated with hadrons. [...] OpenAccess: PDF PDF (PDFA); 2019-08-1213:41 [PUBDB-2019-02980] Preprint/Report et al Measurements of triple-differential cross sections for inclusive isolated-photon+jet events in pp collisions at $\sqrt{s} =$ 8 TeV [arXiv:1907.08155; CMS-SMP-16-016; CERN-EP-2019-127] Measurements of triple-differential cross sections for inclusive isolated-photon+jet events in pp collisions at $\sqrt{s} =$ 8 TeV as a function of photon transverse momentum $(p_{T}^{\gamma})$, photon pseudorapidity $(η^{γ})$, and jet pseudorapidity $(η^{jet})$. The data correspond to an integrated luminosity of 19.7 fb$^{−1}$ that probe a broad range of the available phase space, for $\|η^{γ}\| <$ 1.44 and 1.57 $< \|η^{γ}\| <$ 2.50, $\|η^{jet}\| <$ 2.5, 40 25 GeV [...] OpenAccess: PDF PDF (PDFA); 2019-08-1213:34 [PUBDB-2019-02979] Preprint/Report et al Measurement of differential cross sections and charge ratios for $t$-channel single top quark production in proton-proton collisions at $\sqrt{s}=$ 13 TeV [arXiv:1907.08330; CMS-TOP-17-023; CERN-EP-2019-138] A measurement is presented of differential cross sections for $t$-channel single top quark and antiquark production in proton-proton collisions at a centre-of-mass energy of 13 TeV by the CMS experiment at the LHC. From a data set corresponding to an integrated luminosity of 35.9 fb$^{-1}$, events containing one muon or electron and two or three jets are analysed. [...] OpenAccess: PDF PDF (PDFA); 2019-08-1213:32 [PUBDB-2019-02978] Preprint/Report et al Search for anomalous triple gauge couplings in WW and WZ production in lepton + jet events in proton-proton collisions at $\sqrt{s} =$ 13 TeV [arXiv:1907.08354; CMS-SMP-18-008; CERN-EP-2019-137] A search is presented for three additional operators that would lead to anomalousWW$γ$ or WWZ couplings with respect to those in the standard model. They are constrained by studying events with two vector bosons; a W boson decaying to e$ν$ or $μν$, and a W or Z boson decaying hadronically, reconstructed as a single, massive, large-radius jet [...] OpenAccess: PDF PDF (PDFA);	2019-08-25 12:20:37	{"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": 0.9710344076156616, "perplexity": 2786.313523070253}, "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/1566027323328.16/warc/CC-MAIN-20190825105643-20190825131643-00308.warc.gz"}
http://mathhelpforum.com/calculus/279826-particle-moving.html	1. ## particle moving A particle of mass 6 kg starts from rest and is subject to a force which increases uniformly from 0N to 12N in a time of 20s. prove the acceleration of the particle is t/10 m/s^2 seconds after it begins to move. prove that after the particle has moved a distance s meters when its speed is v m/s then 9s^2 =20v^3 I am not sure how to set this up. I can not use f=ma or P=FV 2. ## Re: particle moving There are very few problems in physics where you can't use $F=m a$ This is not one of them. You're given mass. You're given force as a function of time. Therefore you are given acceleration as a function of time. The rest is a bit of straightforward calculus. 3. ## Re: particle moving So this is what I did so the force varies over time and we not one of the conditions. so I said F=Kt where k is a constant which we do not know and t is time. so the question says the force is 12N when t is20 so 12=20k 3/5 =k so then using F=ma 3/5 t =6a a=t/10 I then let dv/dt and I let the upper limits be t and v and I rearranged to get t=√(20v) so i then let t/10 = v dv/ds and I subbed in t=√(20v) and I integrated this and I got a useless expression. I thought I did every thing right and I can not see where I went wrong 4. ## Re: particle moving Originally Posted by edwardkiely So this is what I did so the force varies over time and we not one of the conditions. so I said F=Kt where k is a constant which we do not know and t is time. so the question says the force is 12N when t is20 so 12=20k 3/5 =k so then using F=ma 3/5 t =6a a=t/10 I then let dv/dt and I let the upper limits be t and v and I rearranged to get t=√(20v) so i then let t/10 = v dv/ds and I subbed in t=√(20v) and I integrated this and I got a useless expression. I thought I did every thing right and I can not see where I went wrong ok so you got $a(t) = \dfrac {t}{10}$ simple calculus gets you $v(t) = \dfrac{t^2}{20}$ $s(t) = \dfrac{t^3}{60}$ $v(T) = V$ $T^2 = 20V$ $T = \sqrt{20V}$ $s(T) = S$ $T^3 = 60S$ $T = (60S)^{1/3}$ $(20V)^{1/2} = (60S)^{1/3}$ $8000V^3 = 3600S^2$ $20V^3 = 9S^2$ thank you	2018-02-18 23:09:57	{"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": 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": 0.9025213122367859, "perplexity": 953.7896124111726}, "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/1518891812259.30/warc/CC-MAIN-20180218212626-20180218232626-00665.warc.gz"}
http://serverfault.com/questions/89342/dhcp-on-the-fly-block-assignment	# DHCP on-the-fly block assignment We have a large number of clients who connect to our DHCP machine. We want to assign some of them to a different IP block, which is routed with lower priority. Every DHCP lease renewal, we'd like to check a database and decide which IP block we'd like to assign a customer to. Is there a way to have a DHCP server execute a script, look in a database, or execute some dynamic code when deciding which address to assign in which pool? Each client is uniquely identified by option82, aka "DHCP Relay Agent Information Option". - Which DHCP server on what OS? – Zypher Jul 14 '10 at 18:35 This sounds like a carrier-type setup... is that the case? I would be surprised if something like this already existed in the open tools. I would probably look at writing it myself. Very interesting question though! – MikeyB Jul 14 '10 at 20:18 @Zypher: Debian OS, regular dhcpd @MikeyB: Yes, carrier setup – Andomar Jul 14 '10 at 21:15 Ideally, you'd modify dhcpd to support address assignment based on Option82, equivalent to the "hardware" lines in host objects. I've done it with the OpenBSD dhcpd when I worked at an ISP, which has a simpler internal structure to isc-dhcpd. If you're not in a position to do that, then look at omapi(3) and omshell(1); you'd use OMAPI to dynamically create "class" and "pool" objects, to implement Zypher's suggestion. I just checked dhcpd.h and the class struct has an OMAPI_OBJECT_PREAMBLE, so this should be possible. Beware that the documentation on OMAPI can be a little ... skimpy. - The solution Zypher suggested is what we currently have. And I don't think the sysadmins would appreciate an edited version of dhcpd. But OMAPI seems like a good alternative, thanks! – Andomar Jul 20 '10 at 22:31 So i havn't done this with option 82, but your best bet would be to use classing in isc dhcpd. What you would do is setup a class like: class "userclass1" { match if substring(option agent.circuit-id, 2, 2) = "<your_id1>"; } class "userclass2" { match if substring(option agent.circuit-id, 2, 2) = "<your_id2>"; } pool { allow members of "userclass1"; range 10.0.0.11 10.0.0.50; } pool { allow members of "userclass2"; range 10.0.0.51 10.0.0.100; } Reference: dhcpd.conf This should at least get you on the right track, i don't have my play server up to test it, but i've done something similar with other options. - Wouldn't we have to reset dhcpd every time there is a change? Is it possible to have pool membership determined dynamically by a script call? – Andomar Jul 15 '10 at 5:48 @Andomar: Yep you would. Unfortunately i don't think so - but i could be very wrong there – Zypher Jul 15 '10 at 13:30 Maybe you can start from here: http://blog.nominet.org.uk/tech/2005/12/21/using-omapi-object-management-application-programming-interface/ Never used, but with a bit of scripting I think it could work. EDIT man omshell(1) would give some other examples - One way to do this is to assign those clients to a separate VLAN, then the DHCP address those clients get will automatically be in a different pool. - Actually, switching VLAN is what we'd like to avoid. VLAN switches confuse the client's modems and take a long time to execute – Andomar Jul 14 '10 at 21:19	2015-01-27 12:42:02	{"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": 0.3043251633644104, "perplexity": 3219.07167478788}, "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-2015-06/segments/1422120842874.46/warc/CC-MAIN-20150124173402-00094-ip-10-180-212-252.ec2.internal.warc.gz"}
http://mathhelpforum.com/math-topics/15408-pairs-x-y.html	1. ## Pairs of (x,y) How many pairs (x,y) of positive integers satisfiy 2x+7y=1000 2. Originally Posted by ginafara The answer is all of the points x,y in Quadrant I, that lie on the line... y = (-2/7)x +1000 when y = 0, x = 500 so all of the integers from 0 -> 500, since it is defined over its domain, it would be 500 pair (?) Hello, I don't want to pick at you but from $2x + 7y = 1000 \ \Longleftrightarrow \ y=-\frac{2}{7}x + \frac{1000}{7}$ Since you were looking for integers for x and y you have to calculate how often 1000 is divided by 14. Thus I got 71 pairs. 3. Originally Posted by Dragon How many pairs (x,y) of positive integers satisfiy 2x+7y=1000 $ y=\frac{1000-2x}{7} $ so for $y$ to be an integer $1000-2x$ must be a multiple of $7$, but as this is even $500-x$ must be a multiple of $7$. But as $x$ ranges over the integers $1, .. , 500$ (which is the range that $x$ is constrained to) $500-x$ ranges over $500, .. , 1$. Of the numbers $500, .. , 1$ exactly $\lfloor 500/7 \rfloor=71$ are divisible by 7, so there are $71$ positive integer pairs $(x,y)$ that satisfy the given equation. RonL 4. Hello, Dragon! My approach is a variation of CaptainBlack's . . . How many pairs (x,y) of positive integers satisfy $2x+7y\:=\:1000$ We have: . $y \:=\:\frac{1000-2x}{7}$ Since $y$ is a positive integer, we have:. $y \;=\;\frac{\text{even number }\leq 1000}{\text{divisible by 7}}$ There are: $\left[\frac{1000}{7}\right] = 142$ multiples-of-seven which are less than 1000 . . and half of them are even. Therefore, there are 71 solutions. 5. Originally Posted by Dragon How many pairs (x,y) of positive integers satisfiy 2x+7y=1000 $2x+7y=1000$ We can solve this using Bezout's Identity. Trivially we see that, $2(-3)+7(1)=1$ Thus, $2(-3000)+7(1000)=1000$. Hence, all integer solutions are given by, $x=-3000+7t$ $y=1000-2t$. We want positive solutions, meaning, $x>0 \to -3000+7t >0 \to 7t > 3000 \to t> 428.5...$ $y>0 \to 1000-2t >0 \to 2t <1000 \to t<500$ Since, $t$ is an integer, we need to find all integers so that, $428.5... We see that, $429,430,431,....,499$ Are all these integers. In total we have 71 positive solutions.	2017-02-25 09:40:42	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "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": 0.8237219452857971, "perplexity": 684.4874972848105}, "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-2017-09/segments/1487501171706.94/warc/CC-MAIN-20170219104611-00297-ip-10-171-10-108.ec2.internal.warc.gz"}
https://www.homebuiltairplanes.com/forums/threads/a-new-all-electric-aircraft-with-a-range-up-to-600-miles-unveiled-at-paris-air-show.27918/page-3	# A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Show ### Help Support HomeBuiltAirplanes.com: #### pictsidhe ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh At 30:1, it's still around 200 mile range. Batteries are expensive, something with a decent L/D but less gross weight would be better. Range is proportional to battery capacity : drag, battery cost is proportional to capacity. Make it light! As a friend used to tell me, the lighter you make it, the lighter you can make it. #### BBerson ##### Well-Known Member HBA Supporter Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh There should be an optimal cruise altitude. Also, I think with a foldaway prop, periodic engine off porpoise gliding segments gives the best range. Last edited: #### cheapracer ##### Well-Known Member Log Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh Batteries are expensive, something with a decent L/D but less gross weight would be better. Range is proportional to battery capacity : drag, battery cost is proportional to capacity. Make it light! As a friend used to tell me, the lighter you make it, the lighter you can make it. My whole car sells for $15K retail, 2 to 3 hours range. There is no "make it light", 2 hours of batteries weigh what they weigh, that is set, the object is to build a craft around them, not vice versa, and that thinking is what is limiting electric planes now. I can only see something like a 2 place Facetmobile being a practical solution, something that will still have reasonable wing loading with simple packaging realities. Reading this article may convince people as to my proposition, specific to comparisons on page 23 and 28 .... http://www.wainfan.com/pavreport.pdf VB, no practical person wants to live with a 60ft wingspan plane, they want a "park and drive it". #### pictsidhe ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh My whole car sells for$15K retail, 2 to 3 hours range. There is no "make it light", 2 hours of batteries weigh what they weigh, that is set, the object is to build a craft around them, not vice versa, and that thinking is what is limiting electric planes now. I can only see something like a 2 place Facetmobile being a practical solution, something that will still have reasonable wing loading with simple packaging realities. Reading this article may convince people as to my proposition, specific to comparisons on page 23 and 28 .... http://www.wainfan.com/pavreport.pdf VB, no practical person wants to live with a 60ft wingspan plane, they want a "park and drive it". The facetmobile is exactly what I'm talking about, less structure weight for a given payload. I believe it also has benign flight characteristics. It's oddball appearance may actually work in it's favour for something as revolutionary as an electric plane. If you can reduce GW by 10% and keep the L/D, that's 10% less batteries to haul it. The trick is to not let the drag climb so much as result that you end up needing more thrust than the heavy version. Keeping total drag down with a small span means keeping the speed up. ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh The power and weight numbers look about right. However, the 40:1 best glide is with everything tucked in. With a motor-on-mast rig like a typical self-launch sailplane, I'd put the best glide at about 24:1, and with a motor-in-nose rig like FES I'd put the under power best glide at like 32:1. At issue with the latter is that the prop wake will be turbulent; I think that increases drag for the fuselage and for the wing/body junction. --Bob K. Put an outrunner around the tail boom. Zero extra drag in both modes #### pictsidhe ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh Put an outrunner around the tail boom. Zero extra drag in both modes That could even reduce cockpit drag. Goldschmeid and Glauert stuff comes to mind. #### 12notes ##### Well-Known Member Log Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh If you'd like to see what the prototype "unveiled" looks like, this article has a nice picture. You may notice some slight changes from the CAD drawings, in that it's the size of a medium RC plane, has a single pusher prop, no v-tail, a single canopy with little room for passengers or cargo if scaled up, completely different combination wing/tail design, is way shorter and has different relative sizes in every dimension. But it is painted white with silver accents, exactly like the drawings. They've certainly proved the color concept works. #### cheapracer ##### Well-Known Member Log Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh If you'd like to see what the prototype "unveiled" looks like, this article has a nice picture. Synergy telephoned, they want their model back :gig: #### pictsidhe ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh Aluminium refining is an environmentally dirty process using huge amounts of fossil fuels and electelectricity. Al-air batteries don't make sense for bulk energy storage. #### 12notes ##### Well-Known Member Log Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh If one of the wing-tip engines fail on takeoff or landing approach, I certainly wouldn't like to be the pilot. I think you misunderstand their target customers - they're not pilots, they're investors. The plane is not the product, the company is. #### pictsidhe ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh I think you misunderstand their target customers - they're not pilots, they're investors. The plane is not the product, the company is. D'oh! I'm just not cynical enough yet either! #### pictsidhe ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh "the plane weights 300 times less than others of the saame size" Hmmmm, sounds rather good. Looking at the art, the passenger compartment is mostly behind the cg. I think I may stop bothering with "seeking investors" Projects. #### Aviator168 ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh I'm not getting political at all, or taking either side, it just makes me laugh at all the climate change articles I see when 90% of the time they show a picture of steam pouring out! Steam is a worse green gas than CO2. I just read today where the Tesla battery pack manufacturing process puts out the same CO2 emissions as 8 years of driving a modern gasoline powered car, not to mention the generating CO2 footprint. Hate to burst your bubble. You have to compare the total CO2 footprints of making a comparable gasoline car and the Tesla. Until than, everything is just speculation or worse, propaganda. #### cheapracer ##### Well-Known Member Log Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh Steam is a worse green gas than CO2. I'm not that savvy on the knowledge, care to explain please? #### Aesquire ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh It has to do with the spectra of light in the infrared that is reflected back to the earth from the atmosphere. Water vapor is a Much more powerful greenhouse gas. So is Methane, produced by swamps, garbage dumps, and animal metabolism. ( you surely know jokes about lighting such gasses. YouTube has many videos of idjits lighting their pants on fire. ) There's a reason that farmers and waste disposal companies collect the Methane from garbage dumps and manure piles. It's not only free energy, you get tax credits for reducing greenhouse emissions. This forum does not allow religious arguments for good reason. So I will not comment on the religious propaganda & extremism surrounding this subject. #### Aesquire ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh I'm all for the electric car, bike, airplane, space drive. I might get an electric bike for recreation. ( to augment my 3 conventional bikes. Although I should sell my racing bike that I haven't ridden in years. ) I'm unlikely to get an electric car. I routinely drive longer than available ranges. There's a financial limit to my toys. Ditto for electric planes. Although I am very tempted to talk to BoKu about his self launch sailplane in development. It's all about the batteries. They are improving as we speak. Not, alas, as fast as we might like or are being sold by some people. Some of these folk are honestly optimistic. Others are con men. Sometimes the difference is easy to tell. Often it is not. I was fooled by one electric motor glider company that had lovely pictures of imaginary aircraft flying in exotic world locations. I assumed they were a con. I was wrong. They have actual flying hardware. I don't think after several years they have a commercial flying electric model but it's in progress. Pipistrel has a flying product you can buy. A few others do too. Last I looked the market was tiny and the aircraft fall into the self launch glider or glider sustainer category. But in your lifetime there most probably will be Cessna trainer versions and in your children's day, RV-X equivalent planes. The pretty picture folk from the OP want you to think their product is ready next week. Bet ya not. #### pictsidhe ##### Well-Known Member Re: A new all-electric aircraft with a range up to 600 miles unveiled at Paris Air Sh There seems to be a new electric wonderplane project every week. Most of them don't seem viable. This one doesn't even have a mockup and the weight and balance is visibly wrong. That doesn't scream cutting edge competence to me. It's also environmentally dirty. The Lilium looks a much better bet, though I suspect it depends on tomorrow's battery tech. But, they're electric man, wooooo! 2	2020-07-14 00:52:16	{"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": 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": 0.2562015652656555, "perplexity": 4840.493684850548}, "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/1593657147031.78/warc/CC-MAIN-20200713225620-20200714015620-00049.warc.gz"}
https://www.programsinformationpeople.org/runestone/static/publicPIP/Sequences/Exercises.html	# Exercises¶ 1. What is the result of each of the following: 1. ‘Python’[1] 2. “Strings are sequences of characters.”[5] 3. len(“wonderful”) 4. ‘Mystery’[:4] 5. ‘p’ in ‘Pineapple’ 6. ‘apple’ in ‘Pineapple’ 7. ‘pear’ not in ‘Pineapple’ 8. ‘apple’ > ‘pineapple’ 9. ‘pineapple’ < ‘Peach’ 1. ‘Python’[1] evaluates to ‘y’ 2. ‘Strings are sequences of characters.’[5] evaluates to ‘g’ 3. len(‘wonderful’) evaluates to 9 4. ‘Mystery’[:4] evaluates to ‘Myst’ 5. ‘p’ in ‘Pineapple’ evaluates to True 6. ‘apple’ in ‘Pineapple’ evaluates to True 7. ‘pear’ not in ‘Pineapple’ evaluates to True 8. ‘apple’ > ‘pineapple’ evaluates to False 9. ‘pineapple’ < ‘Peach’ evaluates to False 2. Write code that asks the user to type something and deletes all occurrences of the word “like”. 3. Write code that asks the user to type something and removes all the vowels from it, then prints it out. 4. Write code that transforms the list [3, 6, 9] into the list [3, 0, 9] and then prints it out Next Section - Extra Exercises	2018-10-22 08:34:45	{"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": 0.23049335181713104, "perplexity": 13531.605162802396}, "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-2018-43/segments/1539583514879.30/warc/CC-MAIN-20181022071304-20181022092804-00517.warc.gz"}
https://www.physicsforums.com/threads/quick-question.703215/	# Quick question 1. Jul 27, 2013 ### oldspice1212 Hi guys, I've been having a pretty hard time understanding exactly what acceleration is, the definitions I usually see are always different, so from what I've read, I know acceleration is a vector quantity and the definition is basically change in velocity? So if lets say a car is going 45 km/h east, and it changes its route to and speed (magnitude) to 35 km/h west, that would be change in acceleration right? Or do you have to account in the time as well, acceleration = velocity / time? I may be over thinking this concept, but would be nice if someone can really clarify this for me, thanks again, huge fan of the physics forums. 2. Jul 27, 2013 ### Mandelbroth No. It's change in velocity over a change in time. That would be a change in velocity. Not a change in acceleration. What do you know about calculus? It's easy to explain if you know derivatives. 3. Jul 27, 2013 ### oldspice1212 Ah right change in velocity over time lol. I've taken calculus one, so yes I know about derivatives and such :p. 4. Jul 27, 2013 ### voko This is essentially correct. To make it fully correct, say the rate of change in velocity. Rate implies that the "change" is divided by the time over which the "change" happens. Just like velocity itself is the rate of change in displacement. Mathematically, these "rate of change" things are known as "derivatives". Velocity is the derivative of displacement, and acceleration is the derivative of velocity (and the second derivative of displacement). Change in velocity. It is -80 km/h. Note the minus sign, I assume that the eastern direction is positive. Not exactly. You do not need just time, you need the time it took for the change to occur. In your example, you did not give that time. If it took the car 20 seconds to undergo that change (say, by going through a roundabout), then the rate of change in velocity is (80 km/h) / (20 s) = 1.1 m/s2. 5. Jul 27, 2013 ### Mandelbroth Alright. That's good. Then, I'll introduce you to a new concept. Since we can write a vector, let's say $\vec{x}=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}$, as a sum of the form $\vec{x}=x_1\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}+x_2\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}+x_3\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$, we can say that vectors are differentiated by components. That is, $\frac{d\vec{x}}{dt}=\frac{d}{dt}\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}=\begin{bmatrix}\frac{dx_1}{dt} \\ \frac{dx_2}{dt} \\ \frac{dx_3}{dt}\end{bmatrix}$. This is the definition of the derivative of a vector (in Euclidean space) with respect to a scalar. Thus, if we call $\vec{x}$ our displacement vector, then $\vec{v}=\frac{d\vec{x}}{dt}$ is the velocity vector and $\vec{a}=\frac{d^2\vec{x}}{dt^2}$ is the acceleration vector. 6. Jul 27, 2013 ### oldspice1212 @Voko Thanks Voko, I remember how the concept of derivatives relates to acceleration and other physics quantities now, your definitions were very helpful. @Mandelbroth I see somewhat of what you're saying, but at other places I feel lost lol, I'll have to fresh up on my linear algebra I think to fully understand the acceleration vector as you presented it. It is very much appreciated though, thank you! Thanks guys! 7. Jul 28, 2013 ### siddharth23 I'll put it in layman's terms. You're travelling on a cycle. You're speed after 1 second in 5 seconds is 0 second- 0 m/s 1 second- 2 m/s 2 seconds - 4 m/s 3 seconds - 6 m/s 4 seconds -8 m/s 5 seconds - 10 m/s So what is happening is, you are increasing your speed by 2 m/s after every second. This makes your acceleration 2m/s/s or 2m/s^s. This is just linear acceleration. For direction change, vector diagrams aer necessary.	2017-11-23 02:55:01	{"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": 0.8074655532836914, "perplexity": 764.532057184312}, "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-2017-47/segments/1510934806715.73/warc/CC-MAIN-20171123012207-20171123032207-00315.warc.gz"}
http://math.stackexchange.com/questions/360345/series-with-complex-terms-convergence	# Series with complex terms, convergence Could you tell me how to determine convergence of series with terms being products of real and complex numbers, like this: $\sum_{n=1} ^{\infty}\frac{n (2+i)^n}{2^n}$ , $\ \ \ \ \sum_{n=1} ^{\infty}\frac{1}{\sqrt{n} +i}$? I know that $\sum (a_n +ib_n)$ is convergent iff $\sum a_n$ converges and $\sum b_n$ converges. (How) can I use it here? - Well, if you can prove absolute convergence, for instance... – Guess who it is. Apr 13 '13 at 13:49 Just like for series of reals (the underlying theorems are in the complex numbers, after all). Your first series doesn't converge. By the ratio test, as $n \rightarrow \infty$: $$\lvert \frac{(n + 1) (2 + i)^{n + 1}/2^{n + 1}}{n (2 + i)^n / 2^n} \rvert = \lvert \frac{(n + 1) (2 + i)}{2 n} \rvert = \frac{(n + 1) \sqrt{5}}{2 n} \rightarrow \frac{\sqrt{5}}{2}$$ - We have $$\left\|\frac{n(2+i)^n}{2^n}\right\|=n\left(\frac{\sqrt{5}}{2}\right)^n\not\to0$$ so the series $\displaystyle\sum_{n=1}^\infty \frac{n(2+i)^n}{2^n}$ is divergent. For the second series we have $$\frac{1}{\sqrt{n}+i}\sim_\infty\frac{1}{\sqrt{n}}$$ then the series $\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n}+i}$ is also divergent. - Thank you. Could you tell me if I can use Cauchy's root test for $(\frac{n(2-i)+1}{n(3-2i)-3i})^n$? – Andrew Apr 13 '13 at 14:36 @Andrew Yes, if we denote by $u_n$ your given expression then we have $$\sqrt[n]{\|u_n\|}=\left\|\frac{n(2-i)+1}{n(3-2i)-3i}\right\|=\sqrt{\frac{5n^2+4n+1‌}{13n^2+12n+9}}\to\sqrt{\frac{5}{13}}<1$$so the series is convergent. – Sami Ben Romdhane Apr 13 '13 at 15:24	2015-06-30 13:03:33	{"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": 0.9588741064071655, "perplexity": 355.2428714602121}, "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-2015-27/segments/1435375093899.18/warc/CC-MAIN-20150627031813-00203-ip-10-179-60-89.ec2.internal.warc.gz"}
http://biomechanical.asmedigitalcollection.asme.org/article.aspx?articleid=1426266&resultClick=1	0 Technical Briefs # Photoacoustic Detection of Melanoma Micrometastasis in Sentinel Lymph Nodes [+] Author and Article Information Devin McCormack, Benjamin S. Goldschmidt, Kiran Bhattacharyya Department of Biological Engineering, University of Missouri, Columbia, MO 65211 Mays Al-Shaer, Paul S. Dale, Chris Papageorgio School of Medicine, University of Missouri, Columbia, MO 65212 Carolyn Henry College of Veterinary Medicine, University of Missouri, Columbia, MO 65211 John A. Viator Department of Biological Engineering, University of Missouri, Columbia, MO 65211; School of Medicine, University of Missouri, Columbia, MO 65212; Christopher S. Bond Life Sciences Center, University of Missouri, Columbia, MO 65211viatorj@missouri.edu J Biomech Eng 131(7), 074519 (Jul 16, 2009) (5 pages) doi:10.1115/1.3169247 History: Received October 15, 2008; Revised June 11, 2009; Published July 16, 2009 ## Abstract Melanoma is the deadliest form of skin cancer and has the fastest growth rate of all cancer types. Proper staging of melanoma is required for clinical management. One method of staging melanoma is performed by taking a sentinel node biopsy, in which the first node in the lymphatic drainage path of the primary lesion is removed and tested for the presence of melanoma cells. Current standard of care typically involves taking fewer than ten histologic sections of the node out of the hundreds of possible sections available in the tissue. We have developed a photoacoustic method that probes the entire intact node. We acquired a lymph node from a healthy canine subject. We cultured a malignant human melanoma cell line HS 936. Approximately $1×106$ cells were separated and injected into the lymph node. We also had a healthy lymph node in which no melanoma cells were implanted. We used a tunable laser system set at 532 nm to irradiate the lymph nodes. Three piezoelectric acoustic detectors were positioned near the lymph node to detect photoacoustic pulses generated within the lymph nodes. We also acquired lymph nodes from pigs and repeated the experiments with increased amplification and improved sensors. We detected photoacoustic responses from a lymph node with as few as 500 melanoma cells injected into the tissue, while normal lymph nodes showed no response. Photoacoustic generation can be used to detect melanoma micrometastasis in sentinel lymph nodes. This detection can be used to guide further histologic study of the node, increasing the accuracy of the sentinel lymph node biopsy. <> ## Figures Figure 1 A histological section of a human lymph node stained with hematoxylin and eosin. (left) The intersecting double arrows span the area, which is infiltrated with melanoma cells. This area is located at the periphery of this lymph node. (right) The indicated area is infiltrated heavily by melanoma cells, which are large cells with abundant cytoplasm. The arrow indicates a single melanoma cell. Figure 2 The photoacoustic setup for detecting micrometastasis in intact lymph nodes. The position of only one of three acoustic transducers is shown here. Figure 3 (left) Photoacoustic response from lymph nodes without melanoma for all three detectors show no signal with a noise floor of approximately 100 μV. (right) For the lymph nodes with melanoma, each sensor shows a distinct photoacoustic wave, with detector 1 showing one at 9.5 μs, detector 2 showing one at 5 μs, and detector 3 showing one at 4 μs. Figure 4 The average signals from irradiating pig lymph nodes are shown here. The error bars denote standard deviation of the measurements. The amplitudes are shown in millivolts and correspond approximately by 100 mV per bar of pressure. ## Errata Some tools below are only available to our subscribers or users with an online account. ### Related Content Customize your page view by dragging and repositioning the boxes below. Related Journal Articles Related Proceedings Articles Related eBook Content Topic Collections	2019-07-18 20:33:19	{"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": 0.2474350482225418, "perplexity": 7573.498434638582}, "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/1563195525793.19/warc/CC-MAIN-20190718190635-20190718212635-00512.warc.gz"}
https://rdrr.io/cran/orthopolynom/man/jacobi.p.weight.html	# jacobi.p.weight: Weight function for the Jacobi polynomial Description Usage Arguments Details Value Author(s) References Examples ### Description This function returns the value of the weight function for the order k Jacobi polynomial, P_k^{≤ft( {α ,β } \right)} ≤ft( x \right). ### Usage 1 jacobi.p.weight(x,alpha,beta) ### Arguments x the function argument which can be a vector alpha the first polynomial parameter beta the second polynomial parameter ### Details The function takes on non-zero values in the interval ≤ft( -1,1 \right) . The formula used to compute the weight function is as follows. w≤ft( x \right) = ≤ft( {1 - x} \right)^α \;≤ft( {1 + x} \right)^β ### Value The value of the weight function ### Author(s) Frederick Novomestky fnovomes@poly.edu ### References Abramowitz, M. and I. A. Stegun, 1968. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover Publications, Inc., New York. Courant, R., and D. Hilbert, 1989. Methods of Mathematical Physics, John Wiley, New York, NY. Press, W. H., S. A. Teukolsky, W. T. Vetterling, and B. P. Flannery, 1992. Numerical Recipes in C, Cambridge University Press, Cambridge, U.K. Szego, G., 1939. Orthogonal Polynomials, 23, American Mathematical Society Colloquium Publications, Providence, RI. ### Examples 1 2 3 4 5 6 ### ### compute the Jacobi P weight function for argument values ### between -1 and 1 ### x <- seq( -1, 1, .01 ) y <- jacobi.p.weight( x, 2, 2 ) Search within the orthopolynom package Search all R packages, documentation and source code Questions? Problems? Suggestions? or email at ian@mutexlabs.com. Please suggest features or report bugs with the GitHub issue tracker. All documentation is copyright its authors; we didn't write any of that.	2017-03-31 00:41:37	{"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": 0.746880292892456, "perplexity": 4008.828988997318}, "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-2017-13/segments/1490218205046.28/warc/CC-MAIN-20170322213005-00346-ip-10-233-31-227.ec2.internal.warc.gz"}
https://hsm.stackexchange.com/questions/6098/riemann-surfaces-and-covering	# Riemann surfaces and covering Assuming we have a Riemann surface $S$ of degree $n$ and we look at it as a covering of the projective line $\mathbb{P}^1$. If $B$ is the set of branch points of $S$ (when $B$ is a subset in $\mathbb{P}^1$), then there is an epimorphism from the free group $\pi_1(\mathbb{P}^1 - B)$ to $Sym_n$, determining this covering (birationally, I assume). I have therefore two questions: 1. Was Riemann already thinking in these terms (of epimorphism to the symmetric group), or was it a later development? I assume that Riemann thought about his surfaces otherwise (in terms of algebraic functions and Abelian integrals?), and not as explicitly covering surfaces with epimorphisms to this group. 2. Who actually proved first the "reverse" theorem, that given a set of points $B$, if there is an epimorphism $\pi_1(\mathbb{P}^1 - B)$ to $Sym_n$, then there is a Riemann surface of degree $n$ branched along $B$? === EDIT: as was remarked below: "the ideas of coverings, the fundamental group, and the symmetric group, [...] [were] developed together"; I think nevertheless the the idea of a symmetry group existed before Riemann, but the developpment of the ideas indeed happened same time. Dieudonne writes in his book "A History of Algebraic and Differential Topology 1900-1960", p. 293, that: "In modern mathematics three notions are interconnected in such a way that each one essentially determines the other two: fundamental group, covering space, and properly discontinuous group. Historically, they appeared in the reverse order." The chapter "Fundamental Group and Covering Spaces" (pp. 293-310) in this book gives the historical overview on these topics. It seems however, that it was Adolf Hurwitz who indeed connected these ideas together: "In 1891, Adolf Hurwitz published a paper on (closed) Riemann surfaces, understood as branched coverings of the complex number sphere with finitely many sheets and a finite number $n$ of branch points." (from: "History of Topology", ed. I. M. James, p. 327, from the paper of Moritz Epple, "Geometric aspects in the development of knot theory"). Hurwitz indeed refers there directly to the symmetric group. The 1891 paper of Hurwitz is called "Ueber Riemann'sche Flächen mit gegebenen Verzweigungspunkten". I assume that answers at least my first question. • Riemann certainly did not think in THESE TERMS. The notion of covering was formally introduced by Schwarz after Riemann's death, and the term "epimorphism" by Bourbaki in 1930s. – Alexandre Eremenko Jun 1 '17 at 13:02 • Indeed, but disregarding for a moment the term "epimorphism", I assume that at some point the connection between coverings, the fundamental group of the complement and the symmetric group was noticed. Was it Schwarz, who noticed these? Could you please indicate where Schwarz used the notion "covering"? thanks – David Jun 1 '17 at 14:02 • The question seems to be based on the assumption that there were some pre-existing ideas of coverings, the fundamental group, and the symmetric group, which somebody then "noticed". This assumption is likely erroneous, these ideas developed together, so at first there was nothing to notice, and by the time there was it came already noticed. I would rephrase the question into just asking how this field developed. – Conifold Jun 2 '17 at 1:54	2020-08-13 12:41:39	{"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": 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": 0.7913548350334167, "perplexity": 711.3765791626244}, "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-2020-34/segments/1596439738982.70/warc/CC-MAIN-20200813103121-20200813133121-00022.warc.gz"}
http://pballew.blogspot.com/2011/08/on-this-day-in-math-aug-21.html	## Sunday, 21 August 2011 ### On This Day in Math - Aug 21 As for methods I have sought to give them all the rigour that one requires in geometry, so as never to have recourse to the reasons drawn from the generality of algebra. ~Augustin-Louis Cauchy The 233rd day of the year; 233 is the only three digit prime that is also a Fibonacci number. EVENTS 1560 The occurrence at the predicted time of a solar eclipse in Copenhagen turned Tycho Brahe toward a life of observational astronomy. VFR 1609 Galileo demonstrates his telescope to the aristocrats of Venice. Renaissance Mathematicus, 1706 Jakob Hermann writes to Leibniz about proof that Machin's series converges to pi. In 1706 William Jones published a work Synopsis palmariorum matheseos or, A New Introduction to the Mathematics, Containing the Principles of Arithmetic and Geometry Demonstrated in a Short and Easie Method ... Designed for ... Beginners. (This is the book in which Jones first uses Pi in the mathematical sense it is now used) This contains on page 243 the following passage:- There are various other ways of finding the lengths or areas of particular curve lines, or planes, which may very much facilitate the practice; as for instance, in the circle, the diameter is to the circumference as 1 to (16/5- 4/239) - 1/3(16/53- 4/2393) &c. = 3.14159 &c. = π. This series (among others for the same purpose, and drawn from the same principle) I received from the excellent analyst, and my much esteemed friend Mr John Machin; and by means thereof, van Ceulen's number, or that in Art. 64.38 may be examined with all desirable ease and dispatch. Jones also reports that this formula allows π be calculated, "... to above 100 places; as computed by the accurate and ready pen of the truly ingenious Mr John Machin." No indication is given in Jones's work, however, as to how Machin discovered his series expansion for π so when de Moivre wrote to Johann Bernoulli on 8 July 1706 telling him about Machin's series for π he suggested that Johann Bernoulli might tell Jakob Hermann about Machin's unproved result. He did so and Hermann quickly discovered a proof that Machin's series converges to π. He produced techniques that show other similar series also converge rapidly to π and he wrote on 21 August 1706 to Leibniz giving details. Two years later, on 6 July 1708, de Moivre wrote again to Johann Bernoulli about Machin's series, on this occasion giving two different proofs that it converged to π. VFR 1776 First recorded use of dollar symbol $. (see cajori, vol II pg 24) but for much longer, it would not be common.. many textbooks used D, d, c for Dollar, Dime and Cents. VFR Ezra l'Hommedieu, a member of the New York Provencial Assembly had over a dozen different symbols in his diary beginning with a single vertical bar and proceeding to two vertical bars. F Carjori 1888 William Seward Burroughs of St. Louis obtained a patent for his adding machine, the ﬁrst successfully marketed. In January, 1886, he incorporated as the American Arithmometer Corporation. VFR He received patents on four adding machine applications (No. 388,116-388,119), the first U.S. patents for a "Calculating-Machine" that the inventor would continue to improve and successfully market. One year after making his first patent application on 10 Jan 1885, he incorporated his business as the American Arithmometer Corporation of St. Louis, in Jan 1886, with an authorized capitalization of $100,000.$ After Burrough's early death in 1898, after moving from St. Louis to Detroit, Michigan, that company reorganized as the Burroughs Adding Machine Co., incorporated in Jan 1905, with a capital of$5 million. The new name was in tribute to the inventor.TIS 1893 The zeroeth International Mathematical Congress with representatives of seven countries was held in conjunction with the Chicago World’s Fair on August 21–25. William E. Story of Clark University was president of the Congress. Felix Klein of Germany came at Kaiser Wilhelm’s personal request. Klein brought nearly all of the mathematical papers published by his countrymen and a superb collection of mathematic models. [AMS Semicentennial Publishers, vol 1, p. 74]. VFR I have read that Klein's models led to more frequent use of them in American Education. 1949 John Mauchly and J. Presper Eckert, Jr. demonstrate BINAC, a computer capable of calculating 12,000 times faster than a human being.VFR (I wonder how they decided how fast a human being could calculate?) 1972 Peru issued a Air Post Stamp picturing a Quipu. [Scott #C341]. VFR BIRTHS 1757 Josiah Beigs, meteorologist and mathematician, born. This freethinking Democrat left his professorship at Yale for political reasons and became president of the University of Georgia. He applied Galileo’s formula for fallen bodies to the nine day’s fall of Lucifer and his angels, to determine that Hell was 1,832,308,363 miles deep. [Struik, Origins of American Science, p. 370] VFR 1789 Augustin-Louis Cauchy (21 Aug 1789;23 May 1857) French mathematician who pioneered in analysis and the theory of substitution groups (groups whose elements are ordered sequences of a set of things). He was one of the greatest of modern mathematicians. TIS 1901 Edward Copson (21 Aug 1901; 16 Feb 1980) English mathematician known for his studies in classical analysis, differential and integral equations, and their use in mathematical physics. After graduating from Oxford University with a B.A. degree in 1922, he moved to Scotland where he spent the nearly all of his career. His first book, The Theory of Functions of a Complex Variable (1935) was immediately successful. He was a co-author for his next book, The Mathematical Theory of Huygens' Principle (1939). By 1975, he had published four more books, on asymptotic expansions, metric spaces and partial differential equations. Many of the papers he wrote bridged mathematics and physics, of which his last showed his interest in astrophysics, Electrostatics in a Gravitational Field (1978) which was relevant to Black Holes.TIS 1932 Louis de Branges de Bourcia (born August 21, 1932) is a French-American mathematician. He is the Edward C. Elliott Distinguished Professor of Mathematics at Purdue University in West Lafayette, Indiana. He is best known for proving the long-standing Bieberbach conjecture in 1984, now called de Branges' theorem. He claims to have proved several important conjectures in mathematics, including the generalized Riemann hypothesis.SAU 1940 Endre Szemerédi (August 21, 1940, ) is a Hungarian mathematician, working in the field of combinatorics and theoretical computer science. He is the State of New Jersey Professor of computer science at Rutgers University since 1986. He received his PhD from Moscow State University. His adviser was the late mathematician Israel Gelfand. He has published over 200 scientific articles in the fields of Discrete Mathematics, Theoretical Computer Science, Arithmetic Combinatorics and Discrete Geometry. He is best known for his proof from 1975 of an old conjecture of Paul Erdős and Paul Turán: if a sequence of natural numbers has positive upper density then it contains arbitrarily long arithmetic progressions. This is now known as Szemerédi's theorem. One of the key tools introduced in his proof is now known as the Szemerédi regularity lemma, which has become a very important tool in combinatorics, being used for instance in property testing for graphs and in the theory of graph limits. He is also known for the Szemerédi-Trotter theorem in incidence geometry and the Hajnal-Szemerédi theorem in graph theory. Ajtai and Szemerédi proved the corners theorem, an important step toward higher dimensional generalizations of the Szemerédi theorem. With Ajtai and Komlós he proved the ct2 /log t upper bound for the Ramsey number R(3,t), and constructed a sorting network of optimal depth. With Ajtai, Chvátal, and M. M. Newborn, Szemerédi proved the famous Crossing Lemma, that a graph with n vertices and m edges, where m greater than 4n has at least m3 / 64n2 crossings. With Paul Erdős, he proved the Erdős-Szemerédi theorem on the number of sums and products in a finite set. With Wolfgang Paul, Nick Pippenger, and William Trotter, he established a separation between nondeterministic linear time and deterministic linear time, in the spirit of the infamous P versus NP problem. With William Trotter, he established the Szemerédi–Trotter theorem obtaining an optimal bound on the number of incidences between finite collections of points and lines in the plane.Wik DEATHS 1757 Samuel König was a German mathematician who is best remembered for his part in a dispute with Maupertuis over the Principle of Least Action.SAU In the 17th century Pierre de Fermat postulated that "light travels between two given points along the path of shortest time," which is known as the principle of least time or Fermat's principle. Credit for the formulation of the principle of least action is commonly given to Pierre Louis Maupertuis, who wrote about it in 1744 and 1746. Maupertuis felt that "Nature is thrifty in all its actions", and applied the principle broadly. Wik 1814 Count Benjamin Thompson Rumford (26 Mar 1753, 21 Aug 1814) American-born British physicist, government administrator, and a founder of the Royal Institution of Great Britain, London. Because he was a Redcoat officer and an English spy during the American revolution, he moved into exile in England. Through his investigations of heat he became one of the first scientists to declare that heat is a form of motion rather than a material substance, as was popularly believed until the mid-19th century. Among his numerous scientific contributions are the development of a calorimeter and a photometer. He invented a double boiler, a kitchen stove and a drip coffee pot. TIS 1836 Claude-Louis Navier was a French mathematician best known for the Navier-Stokes equations describing the behaviour of a incompressible fluid. SAU 1927 William Burnside wrote the first treatise on groups in English and was the first to develop the theory of groups from a modern abstract point of view. SAU 1957 Harald Ulrik Sverdrup ( 15 Nov 1888; 21 Aug 1957)was a Norwegian meteorologist and oceanographer known for his studies of the physics, chemistry, and biology of the oceans. He explained the equatorial countercurrents and helped develop the method of predicting surf and breakers. As scientific director of Roald Amundsen's polar expedition on Maud (1918-1925), Sverdrup worked extensively on meteorology, magnetics, atmospheric electricity, physical oceanography, and tidal dynamics on the Siberian shelf, and even on the anthropology of Chukchi natives. In 1953, Sverdrup quantified the concept of "critical depth", explaining the onset of the spring phytoplankton bloom in newly stratified water columns.TIS 1995 Subrahmanyan Chandrasekhar (19 Oct 1910, 21 Aug 1995) Indian-born U.S. astrophysicist who shared with William A. Fowler the 1983 Nobel Prize for Physics for formulating the currently accepted theory on the later evolutionary stages of massive stars, work that subsequently led to the discovery of neutron stars and black holes. TIS Credits: VFR = V Frederick Rickey, USMA TIS= Today in Science History Wik = Wikipedia SAU=St Andrews Univ. Math History *CHM=Computer History Museum	2017-06-25 00:01:02	{"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": 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": 0.5429505109786987, "perplexity": 1992.3463494229618}, "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-2017-26/segments/1498128320368.57/warc/CC-MAIN-20170624235551-20170625015551-00081.warc.gz"}
http://math.stackexchange.com/questions/105172/what-is-the-conjugate-of-frac12-frac32i/105174	# What is the conjugate of $\frac{1}{2}+ \frac{3}{2}i$? What is the conjugate of $\dfrac{1}{2}+ \dfrac{3}{2}i$? Firstly, what is conjugation? And secondly, can you should the steps to doing this? "$i$" is the imaginary unit. - $\frac{1}{2}-\frac{3}{2}i$ – Potato Feb 3 '12 at 2:22 The conjugate of $a+bi$ is $a-bi$.	2015-07-05 06:38:28	{"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": 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": 0.9046341776847839, "perplexity": 1224.8377657696442}, "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-2015-27/segments/1435375097246.96/warc/CC-MAIN-20150627031817-00271-ip-10-179-60-89.ec2.internal.warc.gz"}
https://theotherdunwoody.blogspot.com/2010_09_01_archive.html	## Thursday, September 30, 2010 ### The Other Dunwoody Humbled There is a new satirist in town, and she is good! In fact she is so good, her current work not only got past the Thought Censors at the Dunwoody Fan Magazine, she scored the lead-off letter. And the vehicle for this satirical masterpiece? Chik-fil-a! This is the only tell that this is a satirical piece (all satire must have a tell). As we all know Chik-fil-a is run by Truett Cathy, who, if Dunwoody were in the business of handing out keys to the city, would surely warrant the first. Telling Mr. Cathy he cannot open a Chik-fil-a is like telling Bill Grant he cannot build million dollar clutter homes on Chamblee Dunwoody. But the selection of a chicken joint was simply brilliant. It reminds us of the recent Chicken Fiasco from the Dunwoody Home Owners Association City Council and it strikes terror into the hearts of all loyal Dunwoodians by evoking images of the kind of people who eat fried chicken. Not even tax hikes are more frightening than the prospect of those people visiting, or worse yet, staying in Dunwoody. But this author is a master of subtlety, foregoing any mention of watermelon smoothies or Cadillac-only drive through lanes. And with Chik-fil-a there would be no discussion of chicken bones littering the street--this author is a fencer, striking with the point, not the edge. So it should be of no surprise that she followed with a fluid transition to the slippery slope argument, long a favorite of those whose sole focus is impeding progress. The mere suggestion that a Chik-fil-a would be gateway drug to trashing the zoning of the adjacent, and long time empty, retail property surely had the same bobble-heads who opposed Goodwill (now in Sandy Springs and John's Creek) nodding in agreement. But the icing on the cake, tapping into a consistent whine emanating from NE Dunwoody, was traffic. Just imagine what was going through those NIMBY minds. Images of ungracefully aging Cadillacs sporting chrome spinners cruising our cobble-stone like streets flocking to and fleeing from yet another Atlanta Chik-fil-a. Could this be the only thing worse than day-hops at the community college? Indeed it could. So once again dear reader, there's a new satirist in town and she's damn good. Check it out for yourself. ## Friday, September 24, 2010 ### Rural Assault Continues In recent weeks the "Wendell Douglas" crowd has ramped up their terrorist attacks against suburban sensibilities. Rebounding from a stinging loss during the "Chicken Coup" they have regrouped and are now engaging in a Land-Sea-Air Assault: backyard aqua-culture, suburban "over-farming" and apiaries. Yes indeed, right here in DeKalb County, of which, to date, Dunwoody is still a part, we have folks raising catfish and talipia in backyards. Lord knows you're one leak away from a smelly mess, and those bird-eating waterfowl leave little bombs on neighbors' cars as they flee with their bounty. Yet another eco-terrorist has radicalized the suburban garden by eliminating his lawn in favor of vegetables. Say it isn't so! Can there be any greater insult to a "Yard of the Month" obsessed community? But DeKalb County, in a rare display of support for anything Dunwoody, has put a stop to this vegan anarchist, levying fines for this gent's clearly illegal operation and his arrogant disregard for the rule of law. Protection from the illegal fish farm cannot be far behind. Farther afield, we have a self-indulgent beekeeper who has virtually destroyed his neighbor's quality of life. They are no longer even able to enjoy an adult beverage by their pool! Again, that county's Code Enforcers came to the community's rescue, forcing out the bees by cleverly branding them "livestock".City Council take note. While many would like to avoid the necessary controversy surrounding a potential replay of the "Chicken Wars", Dunwoodians cannot be complacent. These looming quality of life endangerments will have greater impact on our daily lives than any road resurfacing, sidewalk or traffic calming problems. We must act and act now! As a first step towards ensuring the integrity of our neighborhoods the City must show citizens that they mean business--the bees at the Dunwoody Nature Center must go! Then and only then can we take swift legislative action to ensure our community does not suffer from an invasion of fish, vegetables and bees. ## Sunday, September 19, 2010 ### Valuing Schools If you, like many in Dunwoody, believe that good schools drive property values up then you should be quite concerned about living in DeKalb, and particularly in Dunwoody. Recent SAT scores show that Dunwoody High not only did not lead the county, but scored fifth place behind Chamblee, Lakeside, DeKalb School of Arts and Druid Hills. But it gets worse. Four of the top five Cobb County high schools scored above the best in DeKalb, and Cobb's fifth ranking high school would be listed in third place were it in DeKalb. And Cobb is not alone in besting Dunwoody High and DeKalb. Were someone looking to buy a suburban home in this depressed market and that parental bromide of "wanting the best schools for their kids" is really true, then a better option is had by looking only a little further to the west or east. Given this reality, perhaps it is time for Dunwoody to transform itself from a suburb "dedicated to the worship of children" to a modern urban oasis for young and old alike. ## Thursday, September 2, 2010 ### "Raising" a Ruckus To quote the source of all things factual, wikipedia, regarding some key Dunwoody stats: "The median income for a household in the CDP was $82,838, and the median income for a family was$100,796. Males had a median income of $70,460 versus$42,813 for females. The per capita income for the CDP was $62,523" Where CDP refers to Dunwoody as a "Census Designated Place". And what a place it is. The Other Dunwoody poses a challenge to neophyte watchdogs and gadflies alike: examine the payroll at city hall and compare to readily available statistics to determine whether or not city employees' pay is in line with the citizenry they are Liege Lord over. You'll be looking to see if High Lord of the Manor, recent recipient of a raise, is paid above the median for a resident of similar gender, an entire household, etc. For extra credit determine how many standard deviations off these medians that salary falls. But don't stop at one salary point. This game can go on and on. There is the Enforcer, and his minions (don't count the donuts or free breakfast with select council members) and whilst it is easy pickings, there is Consigliare, whose bankroll was enormously inflated in the previous budget with nary a peep from the public sheep herd. If time permits, work on aggregate numbers. As a whole, how does city payroll stack up against the taxpayers that subsidize it? Isn't it nice that the little people can pay so the folk at City Hall can live so large? And lest you think this is a limited time offer, or a one time effort, a new census is forthcoming and rest assured the council will issue raises next year and extend departmental budgets, so before you know it you will be playing the 2011 version of "Where's My Money?" This could go into syndication. ## Wednesday, September 1, 2010 ### Warren "5K" Scores His$5K It appears our city manager has managed to make last year's bonus permanent with the recent announcement of City raises. All the while taxpayers in the private sector are being told "you're lucky to have a job". The irony is the raise is justified by the manager's "success at saving money" which directly translates into "not providing City services." Perhaps there may come a day when we wish someone would pay him not to do his job. So, two years and millions of dollars later, we're not any safer, our roads are not any better, we have another taxing entity, and we have more intrusion into our private lives. To make up for this we get to hand out raises to folks who made this happen. Oh, lest we forget, we've got those lovely beige "farmhouse" signs letting anyone who cares know they are about to trespass on Dunwoody.	2017-06-25 17:27:46	{"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": 0.17161665856838226, "perplexity": 7905.428825967629}, "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-2017-26/segments/1498128320545.67/warc/CC-MAIN-20170625170634-20170625190634-00201.warc.gz"}
https://mail.haskell.org/pipermail/haskell/2006-April/017866.html	# [Haskell] lhs2TeX: lazy pattern "~" formatting kahl at cas.mcmaster.ca kahl at cas.mcmaster.ca Wed Apr 19 22:45:21 EDT 2006 > > Does anyone have a nice way to format the "~" of a lazy pattern in > lhs2TeX? I use > > %format ~ = "{} ^\sim " > > but the result isn't as pretty as I'd like. For one thing, no space > is inserted before the "~" after a lambda, "do", or whatever. Nice'' is in the eye of the beholder --- after limited fiddling, I use: \def\irref{{\lower0.3ex\hbox{\texttt{\bf\~{}}}}\kern-0.1em\strut} Wolfram	2015-05-23 05:51:06	{"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": 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": 0.43577855825424194, "perplexity": 11174.373453417647}, "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-2015-22/segments/1432207927245.60/warc/CC-MAIN-20150521113207-00202-ip-10-180-206-219.ec2.internal.warc.gz"}
https://fykos.org/year23/problems/series1	# 1. Series 23. Year ### 1. text scrolling in underground Information system in Prague underground has following feature: during scrolling text on the display to the left, the letters are tilted. Is there a simple way how to do this function on „hardware“ level? What effect will have this at text scrolling in vertical direction? Assume, that information display consists from LEDs spaced in orthogonal grid. Ze tmy tunelu přitáhl Byrot. ### 2. a rope on two wedges A rope is placed on two wedges (see figure) in such way, that in center is not touching the base. The situation is symmetric in left-right direction. Calculate maximal part of rope can be free standing in static equilibrium. Bonus: for which angle is the ratio biggest? S nití si rád hraje Aleš. The movement of point mass is restricted on straight line by 2 end points. At the beginning the point mass $m$ is moving at speed $v$. The end point starts to move away at speed $v_{1}<<v$. How will the energy of point mass change? Na Zajímavé teoretické fyzice nespala Janap. ### 4. emptying centrifuge Lets have a tall cylindrical vessel filled with water (radius $r$, height of water $h)$ and spin it around its axis at angular speed $ω$. At the centre of bottom is a small orifice of surface area $S$, while the vessel is still rotating. How much water will escape the vessel? Archivní víno. ### P. thermometer The capillary of medical thermometer is at its bottom thinner, to stop mercury to return back to the reservoir at the base, which allows us to read the maximum reached temperature. From June 2009 it is forbidden to sell such thermometers. At this historical occasion, take the opportunity and explain us, why the capillary narrowing works only in one direction – only when heating. Because while cooling down the mercury cannot go through the narrow point back. Při horečce chtěl podvádět Honza Prachař. ### E. antifreeze We are going to the north pole. Having snowmobile helping to bring them north, we need into the cooler some antifreeze. What is the ideal concentration of alcohol in water, more precisely what is dependence of melting point on alcohol concentration? If you have good freezer, then measure, what concentration will freeze at some specific temperature. Do not forgot, that this task is experimental. Ze svých cest po Sibiři přivezl Jarda. ### S. Petrin mirror maze • What will you see when standing between two vertical mirrors connected at right angle? Lets have plane mirror inclined at angle 45°, moving to the left at the speed $v$. From the right there is a light ray of speed $c$ (e.g. angle of incidence is 45°) and is reflected upwards. Using Huygens principle calculate angle between incoming and reflected ray, e.g. correct the law of reflection for mowing mirrors. Z dílny Dalimilovy.	2019-02-18 08:36:30	{"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": 0.6070911884307861, "perplexity": 2167.4375932018356}, "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-2019-09/segments/1550247484772.43/warc/CC-MAIN-20190218074121-20190218100121-00355.warc.gz"}
https://hal-ensta-paris.archives-ouvertes.fr/UPMC/hal-01102024v1	“Chinese & Match”, an alternative to Atkin's “Match and Sort” method used in the SEA algorithm - Archive ouverte HAL Access content directly Journal Articles Mathematics of Computation Year : 2001 ## “Chinese & Match”, an alternative to Atkin's “Match and Sort” method used in the SEA algorithm (1) , (2) 1 2 Antoine Joux Reynald Lercier #### Abstract A classical way to compute the number of points of elliptic curves defined over finite fields from partial data obtained in SEA (Schoof Elkies Atkin) algorithm is a so-called Match and Sort'' method due to Atkin. This method is a baby step/giant step'' way to find the number of points among $C$ candidates with $O(\sqrt{C})$ elliptic curve additions. Observing that the partial information modulo Atkin's primes is redundant, we propose to take advantage of this redundancy to eliminate the usual elliptic curve algebra in this phase of the SEA computation. This yields an algorithm of similar complexity, but the space needed is smaller than what Atkin's method requires. In practice, our technique amounts to an acceleration of Atkin's method, allowing us to count the number of points of an elliptic curve defined over $\mathbb{F}_{2^{1663}}$. As far as we know, this is the largest point-counting computation to date. Furthermore, the algorithm is easily parallelized. ### Dates and versions hal-01102024 , version 1 (11-01-2015) ### Identifiers • HAL Id : hal-01102024 , version 1 • DOI : ### Cite Antoine Joux, Reynald Lercier. “Chinese & Match”, an alternative to Atkin's “Match and Sort” method used in the SEA algorithm. Mathematics of Computation, 2001, 70 (234), pp.827-836. ⟨10.1090/S0025-5718-00-01200-X⟩. ⟨hal-01102024⟩ ### Export BibTeX TEI Dublin Core DC Terms EndNote Datacite 175 View	2023-01-27 00:52:54	{"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": 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": 0.4026249349117279, "perplexity": 3698.7151589513655}, "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-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00592.warc.gz"}
https://math.stackexchange.com/questions/linked/14970	614 views ### Intuition for a proof that the rationals are incomplete. [duplicate] Let A be a set of positive rationals $p$ such that $p^2<2$. Now this set contains no upper bound. To prove this, for every rational $p$, a number $p- \frac{p^2-2}{p+2}$ is associated. This number (... 288 views ### Choice of $\xi$ [duplicate] Possible Duplicate: Rational Numbers Suppose $\{x \in \mathbb{Q}\|x>0,x^2<2\}$ has a supremum. Call this supremum $c$. In order to show that this cannot be the case, we learned that we need ... 255 views ### Constructing a larger rational whose square is also less than two [duplicate] Possible Duplicate: Rational Numbers Baby Rudin has a very nice construction showing that, given a positive rational number whose square is less than (greater than) two, one can always find a ... I understand the following proof but how did the author come up with the following expression? $$q = p - \frac{p^2 -2}{p + 2}$$ $q$ being defined that way is crucial for the proof but what's so ...	2019-08-24 16:17:04	{"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": 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": 0.9382238984107971, "perplexity": 258.58305706076555}, "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/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00402.warc.gz"}
https://www.xarg.org/puzzle/codesignal/fibonacciword/	# Codesignal Solution: fibonacciWord Original Problem The Fibonacci word sequence of bit strings is defined as: F(0) = "0" F(1) = "1" F(n) = F(n − 1) + F(n − 2) if n >= 2 Where + is the string concatenation operation. The first few elements are: np 0 0 1 1 2 10 3 101 4 10110 5 10110101 6 1011010110110 Given a number n and a bit pattern p, return the number of occurrences of the bit pattern p in F(n). Example For n = 6 and p = "10", the output should be fibonacciWord(n, p) = 5. The Fibonacci word for n = 6 is "1011010110110", and the string "10" occurs 5 times in the string "1011010110110". Input/Output • [time limit] 4000ms (js) • [input] integer n Guaranteed constraints: 0 ≤ n ≤ 40. • [input] string p The bit pattern you are looking for. Guaranteed constraints: 0 < p.length ≤ 105. • [output] integer The number of occurrences of the bit pattern p in F(n). It is guaranteed that the answer fits in a 32-bit integer. ## Solution The given recursive definition of the fibonacci word sequence has a complexity of $$O(2^n)$$. However, a linearized dynamic programming algorithm can be derived. Taking the first two elements, every successive element concatenates the last two elements: n0 = "0" n1 = "1" n2 = n1 + n0 n3 = n2 + n1 n4 = n3 + n2 n5 = ... This algorithm can be implemented without recursion already: function fib(n) { a = "0" b = "1" t = "" + n for (; n > 1; n--) { t = b + a a = b b = t } return t } With this function the given table can be reproduced. Now to the interesting part, the counting of the overlapping. Using the derived function above, we could simply loop over the string and count the occurrences of $$p$$ at the current position. But with $$n=40$$ 165580141 iterations are already needed. So we have a new limiting factor. A better way is to use the point only, where b and a get connected. Looking at the $$P$$ last digits of the left string and the $$P$$ first digits of the right string - with $$P$$ being the length of $$p$$ - also prevents double-countings. Using the function above as a template, we can implement this idea as: function fibonacciWord(n, p) { a = "0" b = "1" c = p == a r = 0 s = 0 P = p.length - 1 for (; n > 1; n--) { r = s + c if (P > 0) r+= (b.slice(-P) + a.slice(0, P)).indexOf(p) != -1 c = s s = r t = b b = b + a a = t } return r } Which can be simplified a bit, but unfortunately JavaScript is quite expressive on string slicing and matching, so that the final solution isn't that much shorter: fibonacciWord = (n, p) => { a = "0" b = "1" c = p == a r = 0 P = p.length - 1 while (--n) { t = r r+= c + !!P * (b.slice(-P) + a.slice(0, P)).includes(p) c = t t = b b+= a a = t } return r } « Back to problem overview	2022-05-29 04:42:48	{"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": 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": 0.4133114814758301, "perplexity": 2212.8696663148908}, "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-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00604.warc.gz"}
http://www.birs.ca/events/2012/5-day-workshops/12w5067/videos/watch/201205071537-Meltzer.html	## Video From 12w5067: Linking representation theory, singularity theory and non-commutative algebraic geometry Monday, May 7, 2012 15:37 - 16:17 Rank-two vector bundles and Happel-Seidel symmetry	2016-10-25 12:13:31	{"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": 0.8716345429420471, "perplexity": 9851.41925709081}, "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-2016-44/segments/1476988720062.52/warc/CC-MAIN-20161020183840-00167-ip-10-171-6-4.ec2.internal.warc.gz"}