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# American Institute of Mathematical Sciences

February  2009, 25(1): 123-132. doi: 10.3934/dcds.2009.25.123

## On the long-time limit of positive solutions to the degenerate logistic equation

 1 Department of Mathematics, School of Science and Technology, University of New England, Armidale, NSW2351, Australia 2 Department of Mathematics, Waseda University, 3-4-1 Ohkubo, Shinjuku-ku 169-8555, Tokyo

Received  November 2006 Revised  December 2007 Published  June 2009

We study the long-time behavior of positive solutions to the problem

$u_t-\Delta u=a u-b(x)u^p \mbox{ in } (0,\infty)\times \Omega, Bu=0 \mbox{ on } (0,\infty)\times \partial \Omega,$

where $a$ is a real parameter, $b\geq 0$ is in $C^\mu(\bar{\Omega})$ and $p>1$ is a constant, $\Omega$ is a $C^{2+\mu}$ bounded domain in $R^N$ ($N\geq 2$), the boundary operator $B$ is of the standard Dirichlet, Neumann or Robyn type. Under the assumption that $\overline\Omega_0$:=$\{x\in\Omega: b(x)=0\}$ has non-empty interior, is connected, has smooth boundary and is contained in $\Omega$, it is shown in [8] that when $a\geq \lambda_1^D(\Omega_0)$, for any fixed $x\in \overline{\Omega}_0$, $\overline{\lim}_{t\to\infty}u(t,x)$=$\infty$, and for any fixed $x\in \overline{\Omega}\setminus \overline{\Omega}_0$,

$\overline{\lim}_{t\to\infty}u(t,x)\leq \overline{U}_a(x), \underline{\lim}_{t\to\infty}u(t,x)\geq \underline{U}_a(x), where$\underline{U}_a$and$\overline{U}_a$denote respectively the minimal and maximal positive solutions of the boundary blow-up problem$-\Delta u=au-b(x)u^p \mbox{ in} \ \Omega\setminus\overline{\Omega}_0,\ Bu=0 \mbox{ on}\ \partial \Omega,\ \ u=\infty \mbox{ on}\ \partial \Omega_0.$The main purpose of this paper is to show that, under the above assumptions,$\lim_{t\to\infty} u(t,x)=\underline U_a(x),\forall x\in \overline\Omega\setminus \overline\Omega_0.\$

This proves a conjecture stated in [8]. Some extensions of this result are also discussed.

Citation: Yihong Du, Yoshio Yamada. On the long-time limit of positive solutions to the degenerate logistic equation. Discrete & Continuous Dynamical Systems - A, 2009, 25 (1) : 123-132. doi: 10.3934/dcds.2009.25.123
 [1] Jong-Shenq Guo. Blow-up behavior for a quasilinear parabolic equation with nonlinear boundary condition. Discrete & Continuous Dynamical Systems - A, 2007, 18 (1) : 71-84. doi: 10.3934/dcds.2007.18.71 [2] Yihong Du, Zongming Guo. The degenerate logistic model and a singularly mixed boundary blow-up problem. Discrete & Continuous Dynamical Systems - A, 2006, 14 (1) : 1-29. doi: 10.3934/dcds.2006.14.1 [3] István Győri, Yukihiko Nakata, Gergely Röst. Unbounded and blow-up solutions for a delay logistic equation with positive feedback. Communications on Pure & Applied Analysis, 2018, 17 (6) : 2845-2854. doi: 10.3934/cpaa.2018134 [4] Qiong Chen, Chunlai Mu, Zhaoyin Xiang. Blow-up and asymptotic behavior of solutions to a semilinear integrodifferential system. Communications on Pure & Applied Analysis, 2006, 5 (3) : 435-446. doi: 10.3934/cpaa.2006.5.435 [5] Keng Deng, Zhihua Dong. Blow-up for the heat equation with a general memory boundary condition. Communications on Pure & Applied Analysis, 2012, 11 (5) : 2147-2156. doi: 10.3934/cpaa.2012.11.2147 [6] Hongwei Zhang, Qingying Hu. Asymptotic behavior and nonexistence of wave equation with nonlinear boundary condition. Communications on Pure & Applied Analysis, 2005, 4 (4) : 861-869. doi: 10.3934/cpaa.2005.4.861 [7] Guanggan Chen, Jian Zhang. Asymptotic behavior for a stochastic wave equation with dynamical boundary conditions. Discrete & Continuous Dynamical Systems - B, 2012, 17 (5) : 1441-1453. doi: 10.3934/dcdsb.2012.17.1441 [8] Bhargav Kumar Kakumani, Suman Kumar Tumuluri. Asymptotic behavior of the solution of a diffusion equation with nonlocal boundary conditions. Discrete & Continuous Dynamical Systems - B, 2017, 22 (2) : 407-419. doi: 10.3934/dcdsb.2017019 [9] Haoyue Cui, Dongyi Liu, Genqi Xu. Asymptotic behavior of a Schrödinger equation under a constrained boundary feedback. Mathematical Control & Related Fields, 2018, 8 (2) : 383-395. doi: 10.3934/mcrf.2018015 [10] Alexander Gladkov. Blow-up problem for semilinear heat equation with nonlinear nonlocal Neumann boundary condition. Communications on Pure & Applied Analysis, 2017, 16 (6) : 2053-2068. doi: 10.3934/cpaa.2017101 [11] Zhiqing Liu, Zhong Bo Fang. Blow-up phenomena for a nonlocal quasilinear parabolic equation with time-dependent coefficients under nonlinear boundary flux. Discrete & Continuous Dynamical Systems - B, 2016, 21 (10) : 3619-3635. doi: 10.3934/dcdsb.2016113 [12] Shouming Zhou, Chunlai Mu, Yongsheng Mi, Fuchen Zhang. Blow-up for a non-local diffusion equation with exponential reaction term and Neumann boundary condition. Communications on Pure & Applied Analysis, 2013, 12 (6) : 2935-2946. doi: 10.3934/cpaa.2013.12.2935 [13] Haitao Yang, Yibin Chang. On the blow-up boundary solutions of the Monge -Ampére equation with singular weights. Communications on Pure & Applied Analysis, 2012, 11 (2) : 697-708. doi: 10.3934/cpaa.2012.11.697 [14] Joachim von Below, Gaëlle Pincet Mailly, Jean-François Rault. Growth order and blow up points for the parabolic Burgers' equation under dynamical boundary conditions. Discrete & Continuous Dynamical Systems - S, 2013, 6 (3) : 825-836. doi: 10.3934/dcdss.2013.6.825 [15] Ning-An Lai, Yi Zhou. Blow up for initial boundary value problem of critical semilinear wave equation in two space dimensions. Communications on Pure & Applied Analysis, 2018, 17 (4) : 1499-1510. doi: 10.3934/cpaa.2018072 [16] Petri Juutinen. Convexity of solutions to boundary blow-up problems. Communications on Pure & Applied Analysis, 2013, 12 (5) : 2267-2275. doi: 10.3934/cpaa.2013.12.2267 [17] Sanling Yuan, Xuehui Ji, Huaiping Zhu. Asymptotic behavior of a delayed stochastic logistic model with impulsive perturbations. Mathematical Biosciences & Engineering, 2017, 14 (5&6) : 1477-1498. doi: 10.3934/mbe.2017077 [18] Alberto Bressan, Massimo Fonte. On the blow-up for a discrete Boltzmann equation in the plane. Discrete & Continuous Dynamical Systems - A, 2005, 13 (1) : 1-12. doi: 10.3934/dcds.2005.13.1 [19] Yong Zhou, Zhengguang Guo. Blow up and propagation speed of solutions to the DGH equation. Discrete & Continuous Dynamical Systems - B, 2009, 12 (3) : 657-670. doi: 10.3934/dcdsb.2009.12.657 [20] Antonio Suárez. A logistic equation with degenerate diffusion and Robin boundary conditions. Communications on Pure & Applied Analysis, 2008, 7 (5) : 1255-1267. doi: 10.3934/cpaa.2008.7.1255

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# Solution for If X = Sin 3 T √ Cos 2 T , Y = Cos 3 T √ Cos T 2 T , Find D Y D X ? - CBSE (Science) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

If  $x = \frac{\sin^3 t}{\sqrt{\cos 2 t}}, y = \frac{\cos^3 t}{\sqrt{\cos t 2 t}}$ , find$\frac{dy}{dx}$ ?

#### Solution

$\text { We have, x } = \frac{\sin^3 t}{\sqrt{\cos2t}} \text { and y } = \frac{\cos^3 t}{\sqrt{\cos2t}}$
$\Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left[ \frac{\sin^3 t}{\sqrt{\cos2t}} \right]$
$\Rightarrow \frac{dx}{dt} = \frac{\sqrt{\cos2t}\frac{d}{dt}\left( \sin^3 t \right) - \sin^3 t\frac{d}{dt}\sqrt{\cos2t}}{\cos2t} \left[ \text { Using quotient rule } \right]$
$\Rightarrow \frac{dx}{dt} = \frac{\sqrt{\cos2t}\left( 3 \sin^2 t \right)\frac{d}{dt}\left( \sin t \right) - \sin^3 t \times \frac{1}{2\sqrt{\cos2t}}\frac{d}{dt}\left( \cos 2t \right)}{\cos2t}$
$\Rightarrow \frac{dx}{dt} = \frac{3\sqrt{\cos2t}\left( \sin^2 t \cos t \right) - \frac{\sin^3 t}{2\sqrt{\cos2t}}\left( - 2 \sin2t \right)}{\cos 2t}$
$\Rightarrow \frac{dx}{dt} = \frac{3\cos2t \sin^2 t \cos t + \sin^3 t \sin2t}{\cos2t\sqrt{\cos2t}}$
$Now, \frac{dy}{dt} = \frac{d}{dt}\left[ \frac{\cos^3 t}{\sqrt{\cos2t}} \right]$
$\Rightarrow \frac{dy}{dt} = \frac{\sqrt{\cos2t}\frac{d}{dt}\left( \cos^3 t \right) - \cos^3 t\frac{d}{dt}\sqrt{\cos2t}}{\cos2t} \left[ \text { Using quotient rule } \right]$
$\Rightarrow \frac{dy}{dt} = \frac{\sqrt{\cos2t}\left( 3 \cos^2 t \right)\frac{d}{dt}\left( \cos t \right) - \cos^3 t \times \frac{1}{2\sqrt{\cos2t}}\frac{d}{dt}\left( \cos 2t \right)}{\cos2t}$
$\Rightarrow \frac{dy}{dt} = \frac{3\sqrt{\cos2t} \cos^2 t \left( - \sin t \right) - \frac{\cos^3 t}{2\sqrt{\cos2t}}\left( - 2 \sin2t \right)}{\cos 2t}$
$\Rightarrow \frac{dy}{dt} = \frac{- 3\cos2t \cos^2 t \sin t + \cos^3 t \sin2t}{\cos2t\sqrt{\cos2t}}$
$\therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{- 3\cos2t \cos^2 t \sin t + \cos^3 t \sin2t}{\cos2t\sqrt{\cos2t}} \times \frac{\cos2t\sqrt{\cos2t}}{3\cos2t \sin^2 t \cos t + \sin^3 t \sin2t}$
$\Rightarrow \frac{dy}{dx} = \frac{- 3\cos2t \cos^2 t \sin t + \cos^3 t \sin2t}{3\cos2t \sin^2 t \cos t + \sin^3 t \sin2t}$
$\Rightarrow \frac{dy}{dx} = \frac{\sin t \cos t\left[ - 3\cos2t \cos t + 2 \cos^3 t \right]}{\sin t \cos t\left[ 3\cos2t \sin t + 2 \sin^3 t \right]}$
$\Rightarrow \frac{dy}{dx} = \frac{\left[ - 3\left( 2 \cos^2 t - 1 \right)\cos t + 2 \cos^3 t \right]}{\left[ 3\left( 1 - 2 \sin^2 t \right)\sin t + 2 \sin^3 t \right]} \binom{\cos2t = 2 \cos^2 t - 1}{\cos2t = 1 - 2 \sin^2 t}$
$\Rightarrow \frac{dy}{dx} = \frac{- 4 \cos^3 t + 3\cos t}{3\ sint - 4 \sin^3 t}$

$\Rightarrow \frac{dy}{dx} = \frac{- \cos3t}{\sin3t} \binom{\cos3t = 4 \cos^3 t - 3\cos t}{\sin3t = 3\sin t - 4 \sin^3 t}$
$\therefore \frac{dy}{dx} = - \cot3t$

Is there an error in this question or solution?

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Solution If X = Sin 3 T √ Cos 2 T , Y = Cos 3 T √ Cos T 2 T , Find D Y D X ? Concept: Simple Problems on Applications of Derivatives.
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text: Image text transcribed for accessibility: Figure shows a system consisting of a continuous-time LT1 system followed by a sampler, conversion to a sequence, and an LT1 discrete-time system. The continuous-time LT1 system is causal and satisfies the linear, constant-coefficient differential equation dyc(t)/dt + y.c(t) = xc(t). The input xc(t) is a unit impulse delta(t). Determine yc(t). Determine the frequency response H(ejw) and the impulse response h[ pi ] such that w[n] = delta [n].


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# Math Help - prove that τ is the topology on ℝ

1. ## prove that τ is the topology on ℝ

Define τ⊂ P(ℝ) as follows:
τ={φ}∪{u⊆ℝ:{-π,π}⊆u} .
prove that τ is a topology on ℝ .

2. Originally Posted by blbl
Define τ⊂ P(ℝ) as follows:
τ={φ}∪{u⊆ℝ:{-π,π}⊆u} .
prove that τ is the topology on ℝ .
Is this $T=\{\varnothing\}\cup\left\{U\subseteq\mathbb{R}-\pi,\pi)\subseteq U\right\}" alt="T=\{\varnothing\}\cup\left\{U\subseteq\mathbb{R}-\pi,\pi)\subseteq U\right\}" />? What's wrong? clearly $\varnothing,\mathbb{R}\in T$ if $(-\pi,\pi)\subseteq U_\alpha$ then $(-\pi,\pi)\subseteq\bigcup_{\alpha\in\mathcal{A}}U_\ alpha$ and similarly for the intersection. Or, is this $T=\{\varnothing\}\cup\left\{U\subseteq\mathbb{R}:\ {-\pi,\pi\}\subseteq U\right\}$? It's the same.


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# Polynomial division hardware implementation

I am beginning the implementation of the polynomial binary division algorithm now as I understood i will be checking the MSB bit if 1 to XOR and shift the sum if 0 i will only shift. What I am not getting well with is the numerator and denominator degree calculation and when to stop the the division in a case of a polynomial division with M=8 and an irreducible polynomial : X^8+X^4+X^3+X+1 will be divided by X^6+X^5+X^2+X at first I will have to pad with zeros to obtain the same size

I have found that shifting the numerator by L and denominator by K until hitting the first '1' will give me the degree being M-L and M-K-1 of each polynomial and the division will be of Deg(N)-Deg(D) cycles in this example: it will be: Numerator: 100011011 Denominator: 01100110

Numerator MSB is 1 then degree(N) is M-0=8 Denominator MSB is 0 : then it will be shifted to the left by 1 and the Degree(D)=8-1-1=6 the needed counter is 8-6=2 steps and 8-6+1=3 XOR operations

what I would like to ask is this a correct way to start and won't the following degree calculation step includes a large hardware delay since it is done sequentially and M might get to 233

What is the best way of calculating the : Degree of numerator and denominator and when can i know where to stop the division

Thank you

The method you describe is called the "pencil and paper algorithm" and is well described in Knuth's Book, Semi-Numerical Algorithm II.

In fact the number of steps can be easily determined by the size of operands. If Dividend D, is m-bit and Divisor d, is n-bit, then the Quotient q, will be (m-n+1)-bit, and the remainder in case of binary division will be at most (n-1)-bit.

Under these conditions, the algorithm il like that follow:

Shift the divisor (m-n)-bit left (this is called Normalization step by Knuth), We use some registers to be initiated by the operand
Reg_D <-- D: init with D, used as an Accumulator,
Reg_d <-- d: init with divisor $2^{m-n}$d; remains unchanged during all the process iteration,
Reg_q <-- 0: init with Operand 0,

for k from (m-n) DownTo 0, Do:
- Reg_q <-- 2.Reg_q;  //Shift Reg_q 1-bit Left (or equivalently q = 2.q)
- Reg_D = Reg_D XOR Reg_d;   //... XORing(Reg_D, Reg_d):
- q_k=MSB(Reg_D); //Get the bit which represent the partial bit of quotient.
- Reg_q = Reg_q + q_k; //... Inject bit q_k in Reg_q, or
- if( !k)
Reg_D = 2.Reg_D;  //... Shift  1-bit Left D excepted for last step!
End For;

NB: Observe:

• The Remainder is contained in Reg_D,

• To obtain the quotient, Complement the content of Reg_q.

To optimise the process, in case of multiple division with the same divisor you can try two methods described by Knuth:

• Normalization of the Divisor,
• Precomputation tables.


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#### Assignment - main entry

Whether you are starting from scratch or working with an assignment already created, using the assignment activity in a topic ideally consists of 5 stages, in a looped process.

##### 1. Plan  |  2. Build  |  3. Test  |  4. Administer  |  5. Review  || Support

This entry relates to the Assignment activity.

All student assignments generated in electronic form should be submitted by the student and returned to the student electronically (see sections 5.5 & 5.7 of the assessment policy).

The Assignment activity in FLO allows teachers to collect work from students, review it and provide feedback, including grades. The work students submit is visible only to the teacher, not to other students, unless a a group assignment is set up (see link below). For students, the assignment activity is generally referred to as the assignment dropbox, as this is where they upload their file/s (either though searching for it or dragging and dropping).

If students submit their assignment through FLO, there is no need for them to attach a cover sheet. FLO records their name, FAN, email address, exact date and time of submission, and by default provides them with an academic integrity statement that they must agree to before submitting their assignment (depending on the purpose of the assignment, you may want to untick this box).

#### 1. Plan

For considerations and questions you might ask when planning/designing for assignments submitted in FLO using the Assignment activity (dropbox):

#### 2. Build

Assignment – purpose and settings (group, iterative, take-home exam, offline, online text, video etc) will determine the settings you use for the assignment activity.

##### Feedback

How do you set up your assignment to reflect the feedback method/s you want to use? When reviewing assignments, teachers can leave feedback comments and upload files such as marked-up student submissions, documents with comments or spoken audio feedback (if marking in PDF format offline).

#### 3. Test

Once you have set up the assignment, ask your local eLearning support team to check it for you (especially if this is the first time you have set up this activity, or if you are trying a new approach to assignments). It is best to amend mistakes before students submit their assignments.

##### Mark assignments

A topic coordinator/teacher may provide a marking guide (this could also be called a rubric or feedback form) in the assignment activity by:

• adding it to the assignment activity as a file (Word/editable PDF) (for offline marking) OR
• creating an online marking guide, rubric or checklist (for online marking)
Check with the topic coordinator about the way the marking guide has been added if you are not sure, as this will determine your marking method.

Text-matching software (Turnitin)
Mark offline (no internet connection required)
Mark online (internet connection required)

#### 5. Review

How did your assignment activity go? Would you set up the activity differently next time round? Do you want to mark differently (eg using an online marking guide)? Talk to colleagues and/or your local eLearning support team to get ideas for improvement.

Training and support

Troubleshooting

Training

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eLearning support teams

You may have one of the following issues:

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What is the magnetic flux through a trefoil knot?

Imagine a closed loop in the shape of a trefoil knot (https://en.wikipedia.org/wiki/Trefoil_knot). How should one calculate the flux through this loop? Normally we define an arbitrary smooth surface, say, $$\mathcal{S}$$ whose boundary $$\partial{\mathcal{S}}$$ is the given loop and calculate the flux using its integral definition as $$\Phi_B = \int_{{\mathcal{S}}} \mathbf{B}\cdot d\mathbf{S}\tag{1}\label{1}$$ It is clear how to use $$\eqref{1}$$ when the loop is a simple loop and the surface is also a simple one, but how can one spread a surface on a trefoil and it be still true that for such surfaces the flux is always the same because $$\nabla \cdot \mathbf{B}=0$$, in other words how does Gauss' theorem hold for surfaces whose boundary is a trefoil?

Alternatively, one could introduce the vector potential $$\mathbf{B}=\nabla \times \mathbf{A}$$ and using Stokes' theorem derive from the definition of flux $$\eqref{1}$$ that $$\Phi_A = \int_{{\mathcal{S}}} \nabla \times \mathbf{A}\cdot d\mathbf{S}\\ =\oint_{\partial\mathbf{S}} \mathbf{A}\cdot d \ell \tag{2}\label{2}$$ So, whenever we can use Stokes' theorem we also have $$\Phi_A=\Phi_B$$. How does Stokes' theorem hold if the loop is a trefoil?

If in fact the application of either Gauss' or Stokes' theorem has a problem then does the fact that the line integral via $$\eqref{2}$$ can always be used to define the flux $$\Phi_A$$ mean that at least in this sense $$\mathbf{A}$$ is more fundamental than $$\mathbf{B}$$?

• $\mathbf{A}$ is more fundamental than $\mathbf{B}$? This is one interpretation of the Aharanov-Bohm effect. Wikipedia says “The Aharonov–Bohm effect shows that the local $\mathbf{E}$ and $\mathbf{B}$ fields do not contain full information about the electromagnetic field, and the electromagnetic four-potential, (Φ, $\mathbf{A}$), must be used instead.” – G. Smith Jul 31 at 21:31
• @G.Smith the intent of my question was if in case either Gauss or Stokes were to fail for a trefoil and its surface then would we justly consider A be more fundamental than B already in classical EM. As I have just learned from J.Murray and ChiralAnomaly that since there is always an orientable surface for any contour, Gauss/Stokes always hold hence that question is indeed irrelevant but not because Aharonov-Bohm. – hyportnex Jul 31 at 21:37

Every knot is the boundary of an orientable surface. Such a surface is called a Seifert surface.$$^\dagger$$ For any given knot (with a given embedding in 3-d space), the flux is the same through two such surfaces. As usual, the flux can be calculated either by integrating $$\mathbf{B}$$ over the surface, or by integrating $$\mathbf{A}$$ around the knot.

Figure 6 in "Visualization of Seifert Surfaces" by van Wijk and Cohen (link to pdf) shows this nice picture of an orientable surface whose boundary is a trefoil knot:

The boundary (the trefoil knot) is highlighted in yellow. To see that this really is a trefoil knot, imagine smoothing out the kinks and then looking down on the figure from above. The fact that the surface is orientable is clear by inspection (an insect on one side cannot walk to the other side without crossing the boundary), as is the fact that it does not intersect itself.

Intuitively, we can see that Stokes' theorem will still work in this case by subdividing the surface into small cells, each with the unknot as its boundary, and applying Stokes' theorem to each individual cell. The contributions from the cell-surfaces add up to the flux over the full surface, and the contributions from the cell-boundaries cancel each other wherever two boundaries are adjacent, leaving only the integral over the trefoil.

We can also see intuitively that the flux must be the same through any two such surfaces, because those two surfaces can be joined into a single closed surface over which the total flux must be zero because of $$\nabla\cdot\mathbf{B}=0$$. The fact that the closed surface might intersect itself is not a problem, just like it's not a problem for two intersecting surfaces sharing the same unknot as the boundary.

$$^\dagger$$ The idea behind the proof that a Seifert surface exists is sketched in "Seifert surfaces and genera of knots" by Landry (link to pdf).

For a generic, oriented knot, you can construct an oriented surface which has the knot as it's boundary through the Seifert algorithm. Stokes' theorem says that the flux through any two such surfaces which share the same boundary must be the same.

In principle, one could construct a Seifert surface for the trefoil knot, parameterize it, and then evaluate the flux integral. This might be tedious, but it is possible. However, it would be far simpler as you say to simply evaluate the line integral of $$\mathbf A$$ around the knot.

That being said, this is not an indicator that $$\mathbf A$$ is more fundamental than $$\mathbf B$$, because there is no problem defining those flux integrals. It would just be particularly difficult to evaluate them directly.

• is there always a Seifert surface for any (rectifiable) contour? In other words are all contours "knots"? – hyportnex Jul 31 at 20:56
• @hyportnex A knot is defined to be a closed, non self-intersecting curve embedded in $\mathbb R^3$, and it is a theorem in knot theory that every knot has at least one Seifert surface. – J. Murray Jul 31 at 21:05
• @hyportnex Chiral Anomaly has written a wonderful answer with more intuition and a nice visualization to go with it. I would not at all be offended if you accepted that one instead (and that's actually a good argument for waiting a while before accepting an answer). – J. Murray Jul 31 at 21:21


text:
SoC 2006: DruTeX

This group should probably have more organizers. See documentation on this recommendation.

This group is dedicated to the development of DruTeX, a powerfull LaTeX module for Drupal.

Features include:

• rendering of maths or arbitrary TeX via latex + ImageMagic / dvipng
• support of classical LaTeX-math environments like $x^2$, $x^2$
• some plain latex2html and html2latex facilities (no external program needed)
• pdf generation of nodes (with LaTeX)
• highly-configurable input formats
• security-enhanced modes (to allow even anonymous users to write latex)
• Drupal / LaTeX UTF-8 (utf8) support

Features in development (partly done):

• latex math 2 MathML via some external application, e.g. blahtex
• support to render gnuplot, eukleides, chemestry and so on
• using mimetex instead of latex to render images

Notice: Use of undefined constant filter - assumed 'filter' in preg_replace() (.../modules/drutex/drutex.module(363) : regexp code).

Dear All,
I try to use DruTex for the first time here : http://reza.rumahilmu.me - It is able to render my TeX code, but I still got error on my frontpage that is looked like this :

Notice: Use of undefined constant filter - assumed 'filter' in preg_replace() (line 1 of /home/rezaerv/public_html/modules/drutex/drutex.module(363) : regexp code).
Notice: Trying to get property of non-object in drutex_linebreak_node2html_postprocess() (line 49 of /home/rezaerv/public_html/modules/drutex/drutex_linebreak.inc).

What should I do ?

Thank You,

Regards,
Reza Ervani

Image conversion

I have just successfully installed Drutex on a shared server. As it turned out, for whatever reason my host does not support dvipng, but they do support ImageMagick. However, the convert binary does not accept a dvi input, and so a third program is required to run after latex and before convert. Luckily the binary dvips is included with the texlive package my host installed, making this not a very big hurdle. After creating the custom commands for Drutex the latex finally rendered correctly.

How to use CCK custom fields on DruTeX

Hi.

I just installed TeXlive and DruTeX on a VPS and they're working fine.

I even tweaked the DruTeX template to get the layout I need, but I can't figure out how to render to LaTeX custom fields created by CCK.

I need to render some text input by the user along with some headers and titles hardcoded in the template, son I need at least 15 fields to be filled by the user and render by DruTeX.

I don't know enough Drupal or PHP to fully understand the code, but I found some clues in the -- drutex.pdf.inc -- file, but I don't know how to tell drutex which CCK field to use.

Next Version

Ideas for the next version of DruTeX at: http://drupal.org/node/248364

Math server

Hello,
I'm relatively new to Drupal which I installed with ease. Now I'm trying to render mathematics within my pages, so I installed DruteX. My server can't handle latex requests since I can't install software onto it (except all kind of php of course).
I can access a math server, which transforms gracefully math formulae into images, e.g.
http://math.spip.org/txt.php?\int_0^1x^2dx
gives the expected result (a PNG image, of course).

How can I use that remote server ?

tth

I've just integrated tth into a filter, simply because somebody wanted to throw a load of latex onto my site.
It worked - it took 1/2 an hour to do.

I dont think it's exactly tryng to solve the same problem as drutex, I'm not sure, but it seems remarkably close, and would probably work for most people?

tth can be found here: http://hutchinson.belmont.ma.us/tth/

I'm not especially fixated on it, I'd appreciate hearing other peoples opinions.

If anybody wants the module, it's fairly trivial, I could check it in if it helps....

Issues on the PDF and TeX generation

I found two disturbing issues when creating a PDF or a Tex file from a node.

First, the PDF is not updated, once it is created the first time. The reason is this test located in the drutex_node2pdf_wrapper function:

  if (!is_file("$pdf_dir/{$node->nid}.pdf")) {    drutex_node2pdf($nid); } I just commented the test in my site. PDF should be generated each time it is requested, as node might have been updated in the meanwhile. Read more PDF Generation error Hi folks: I've been playing around with the module, and I found an error when converting html tags to latex commands (more precisely, when converting the tag <em> to the \emph{} command). For some reason I got a "\" before the opening curly bracket, so the command - \emph{ whatever } - was not properly recongized. Read more latexgen Hi, just posted a tiny module which "works for me" - it's not pretty and its not cleaver, but it might be helpful to others who use latex. It simply allows you to put some latex code, with embedded php code within it, onto a content type - and then gives you the ability to download the resulting PDF. The module itself is the trivial part - getting your php code and latex to work just the way you want it - thats the real problem. I'd guess most people would hack the module about a bit too... Anyway, it maybe helpful to some people. Read more Drutex 5.0 Hey everyone - I just got my own local copy of drutex working for drupal 5.0 (I still need to get an install of latex for my server; its a hosted server - I'm looking into server farms, but haven't been able to find anything reasonable yet) I just installed drupal for the first time, so this was sort of a get oriented with the system project. I expected it to take a long time.... it took 5 minutes! (granted, I still have to install latex on my server - but it appears drutex is working fine) Read more <equations> vs. \begin{eqnarray} A friend and I are starting a Drupal for mathematicians (who know LaTeX, but little else). Anyway, I am wondering if there is a good reason why Drutex uses <equations>...</equations> instead of just supporting the \begin{eqnarray}...\end{eqnarray} environment from LaTeX. At least for my target audience, the latter would be preferable. Thanks in advance for any comments/advice. Matt Read more Documentation D r u T e X  1. Installation 2. Configuration 3. Usage 4. Additional modules 5. Developer notes 6. Author Read more Installation Instructions 1) Download the DruTeX module from the Drupal website and unpack it to the modules directory. You now have a directory modules/drutex. 2) Go to the module admin section and enable "drutex". 3) Go to the input-filters admin section and create a new input format. You can choose an arbitrary name, but "DruTeX" would make sense. When you create a new format, you can select an input filter from a list: Just choose DruTeX. 4) Go again to the input-filters admin section, and select "configure" for the newly added input format. Click on the "configure" tab, and you are able to adjust many settings of DruTeX. Read more Drupal key to our non-profit progress Hi Everyone Keeping this short is tough (and off to a bad start ;) - at home (South Africa) there is huge social disparity which is perpetuated by only the well-off having access to resources that could potentially break the cycle. We are focused on education. Many students in poorer schools (>50%) do not even have sufficient textbooks. Enter our project - a note: without publishers', editors' and authors' royalties a hard cover 250 page books costs less than ZAR20 (<$2.5) to print:

<

blockquote>
FHSST (Free High School Science Texts) is a project that aims to provide free science and mathematics textbooks for Grades 10 to 12 science learners in South Africa.

Hello, I want to contribute to drutex

Hi! I'd like to start contributing to drutex. I've installed it on my server and guess what? It didn't work. So I'm asking myself: Why doesn't it work? I've added a couple of useful debugging features to my own drutex_render.inc, one which is related to an issue posted a while ago, stating that the message TeX Embedding failed! was not enough (http://drupal.org/node/71218). The other one was trying to access the latex error log. Both were quite simple.

In items 4.2 and 4.3 of our technical specification, we talked about links between \ref{equation-id} and <equation id="equation-id"></equation>. What plans do we have to include this functionality in DruTeX?

Mimetex.cgi hack

I've created a hack for converting Drutex from using LateX to render the pictures to using Mimetex.cgi to render them. The image quality isn't quite as good, but it works. The hack is currently in the really ugly stage where a lot of the Drutex stuff probably doesn't work (like multi-line equations), but I will test this stuff out, and see if I can modify the code to support this. I'm a real beginner in PhP so I'm not able to answer any fancy questions about it. Also, please note that you will want to change the URL to the Mimetex.cgi, and that Mimetex.cgi needs to be installed into the CGI-BIN directory of your webhost.

Any news here?

It would be nice to get an update on progress.

Technical specification

Our technical specification is available through Writeboard.com. (Ask for the password.) We finished the technical spec on May 30 and excised one line on June 6, 2006.


text:
# Properties

 Label 363.3.b.m Level $363$ Weight $3$ Character orbit 363.b Analytic conductor $9.891$ Analytic rank $0$ Dimension $8$ CM no Inner twists $2$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$363 = 3 \cdot 11^{2}$$ Weight: $$k$$ $$=$$ $$3$$ Character orbit: $$[\chi]$$ $$=$$ 363.b (of order $$2$$, degree $$1$$, minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$9.89103359628$$ Analytic rank: $$0$$ Dimension: $$8$$ Coefficient field: $$\mathbb{Q}[x]/(x^{8} + \cdots)$$ Defining polynomial: $$x^{8} + 29 x^{6} + 282 x^{4} + 1061 x^{2} + 1331$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 33) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{7}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + \beta_{1} q^{2} + ( -\beta_{2} + \beta_{4} + \beta_{5} ) q^{3} + ( -4 - \beta_{2} + \beta_{3} + \beta_{4} + \beta_{6} ) q^{4} + ( \beta_{1} - \beta_{3} + \beta_{5} - \beta_{7} ) q^{5} + ( 2 - \beta_{1} + 3 \beta_{2} - 2 \beta_{3} + \beta_{4} ) q^{6} + ( 5 + 2 \beta_{2} - \beta_{3} + \beta_{5} + \beta_{7} ) q^{7} + ( -2 \beta_{1} - 2 \beta_{3} - 3 \beta_{5} - 2 \beta_{7} ) q^{8} + ( 3 + 2 \beta_{2} + \beta_{4} + 4 \beta_{5} + 3 \beta_{7} ) q^{9} +O(q^{10})$$ $$q + \beta_{1} q^{2} + ( -\beta_{2} + \beta_{4} + \beta_{5} ) q^{3} + ( -4 - \beta_{2} + \beta_{3} + \beta_{4} + \beta_{6} ) q^{4} + ( \beta_{1} - \beta_{3} + \beta_{5} - \beta_{7} ) q^{5} + ( 2 - \beta_{1} + 3 \beta_{2} - 2 \beta_{3} + \beta_{4} ) q^{6} + ( 5 + 2 \beta_{2} - \beta_{3} + \beta_{5} + \beta_{7} ) q^{7} + ( -2 \beta_{1} - 2 \beta_{3} - 3 \beta_{5} - 2 \beta_{7} ) q^{8} + ( 3 + 2 \beta_{2} + \beta_{4} + 4 \beta_{5} + 3 \beta_{7} ) q^{9} + ( 2 + 3 \beta_{2} - 2 \beta_{3} + \beta_{4} + 3 \beta_{5} + \beta_{6} + 3 \beta_{7} ) q^{10} + ( 11 - \beta_{1} + 7 \beta_{2} + \beta_{3} - 3 \beta_{4} - \beta_{5} + 3 \beta_{7} ) q^{12} + ( 5 + 2 \beta_{2} + 3 \beta_{3} - 3 \beta_{5} - 3 \beta_{7} ) q^{13} + ( 3 \beta_{1} + 5 \beta_{3} - 2 \beta_{4} - 3 \beta_{5} + 2 \beta_{6} + 3 \beta_{7} ) q^{14} + ( -4 + 2 \beta_{1} + 5 \beta_{2} + \beta_{3} - \beta_{4} + \beta_{5} + 3 \beta_{6} ) q^{15} + ( 5 + 11 \beta_{2} - 4 \beta_{3} - 3 \beta_{4} + \beta_{5} - 3 \beta_{6} + \beta_{7} ) q^{16} + ( 5 \beta_{1} + 2 \beta_{3} - 5 \beta_{4} - 4 \beta_{5} + 5 \beta_{6} - 3 \beta_{7} ) q^{17} + ( -4 - \beta_{1} - 6 \beta_{2} + 10 \beta_{3} + \beta_{4} - 6 \beta_{5} + 6 \beta_{6} + 3 \beta_{7} ) q^{18} + ( 5 - 9 \beta_{2} - 3 \beta_{3} - 2 \beta_{4} + \beta_{5} - 2 \beta_{6} + \beta_{7} ) q^{19} + ( -2 \beta_{1} + 7 \beta_{3} - 4 \beta_{4} - 6 \beta_{5} + 4 \beta_{6} + 3 \beta_{7} ) q^{20} + ( -6 - 3 \beta_{1} - 9 \beta_{2} + 3 \beta_{3} + 3 \beta_{4} + 3 \beta_{6} ) q^{21} + ( 2 \beta_{1} - \beta_{3} + 6 \beta_{4} + 10 \beta_{5} - 6 \beta_{6} + 5 \beta_{7} ) q^{23} + ( -1 + 8 \beta_{1} - 7 \beta_{2} + 10 \beta_{3} - \beta_{4} - 8 \beta_{5} + 6 \beta_{6} ) q^{24} + ( 13 + 15 \beta_{2} + \beta_{4} + \beta_{5} + \beta_{6} + \beta_{7} ) q^{25} + ( 11 \beta_{1} - 7 \beta_{3} - 2 \beta_{4} + \beta_{5} + 2 \beta_{6} - 9 \beta_{7} ) q^{26} + ( -6 - 9 \beta_{1} - 7 \beta_{2} + \beta_{4} + 4 \beta_{5} + 3 \beta_{7} ) q^{27} + ( -30 - 23 \beta_{2} + 10 \beta_{3} + 5 \beta_{4} - 5 \beta_{5} + 5 \beta_{6} - 5 \beta_{7} ) q^{28} + ( 2 \beta_{1} + 2 \beta_{3} - 2 \beta_{4} - 2 \beta_{5} + 2 \beta_{6} ) q^{29} + ( -5 - 8 \beta_{1} - 13 \beta_{2} + 8 \beta_{3} - 3 \beta_{4} - 8 \beta_{5} + 9 \beta_{6} + 3 \beta_{7} ) q^{30} + ( -6 - 15 \beta_{2} + 5 \beta_{4} + 5 \beta_{5} + 5 \beta_{6} + 5 \beta_{7} ) q^{31} + ( \beta_{1} + 9 \beta_{3} - 8 \beta_{4} - 12 \beta_{5} + 8 \beta_{6} + \beta_{7} ) q^{32} + ( -21 - 26 \beta_{2} + \beta_{3} + 3 \beta_{4} + 2 \beta_{5} + 3 \beta_{6} + 2 \beta_{7} ) q^{34} + ( -3 \beta_{1} + \beta_{3} - 4 \beta_{4} - 3 \beta_{5} + 4 \beta_{6} - 3 \beta_{7} ) q^{35} + ( -16 - 10 \beta_{1} - 23 \beta_{2} + \beta_{3} + 6 \beta_{4} - 4 \beta_{5} - 3 \beta_{6} - 9 \beta_{7} ) q^{36} + ( 21 + 16 \beta_{2} + 9 \beta_{3} - 9 \beta_{5} - 9 \beta_{7} ) q^{37} + ( 7 \beta_{1} - 4 \beta_{3} + 11 \beta_{4} + 16 \beta_{5} - 11 \beta_{6} + 7 \beta_{7} ) q^{38} + ( 10 + \beta_{1} + 15 \beta_{2} - \beta_{3} + 11 \beta_{4} + 12 \beta_{5} - 9 \beta_{6} ) q^{39} + ( -7 - 27 \beta_{2} + 3 \beta_{3} + 3 \beta_{4} + 3 \beta_{6} ) q^{40} + ( -10 \beta_{1} + 6 \beta_{3} + 13 \beta_{4} + 14 \beta_{5} - 13 \beta_{6} + 19 \beta_{7} ) q^{41} + ( 24 - 12 \beta_{1} + 3 \beta_{2} - 15 \beta_{3} + 3 \beta_{4} - 3 \beta_{5} - 9 \beta_{6} - 6 \beta_{7} ) q^{42} + ( -15 - 6 \beta_{2} + 2 \beta_{3} - 5 \beta_{4} - 7 \beta_{5} - 5 \beta_{6} - 7 \beta_{7} ) q^{43} + ( 2 - 7 \beta_{1} + 14 \beta_{2} + 7 \beta_{3} - 10 \beta_{4} - 11 \beta_{5} + 6 \beta_{6} - 3 \beta_{7} ) q^{45} + ( -33 + 11 \beta_{2} + 11 \beta_{3} + 11 \beta_{4} + 11 \beta_{6} ) q^{46} + ( 23 \beta_{1} + 4 \beta_{3} - 18 \beta_{4} - 15 \beta_{5} + 18 \beta_{6} - 14 \beta_{7} ) q^{47} + ( -45 - 6 \beta_{1} - 12 \beta_{2} + 6 \beta_{3} - 15 \beta_{5} + 3 \beta_{6} - 9 \beta_{7} ) q^{48} + ( -8 + 12 \beta_{2} - 9 \beta_{3} - 4 \beta_{4} + 5 \beta_{5} - 4 \beta_{6} + 5 \beta_{7} ) q^{49} + ( 9 \beta_{1} + 17 \beta_{3} - 16 \beta_{4} - 20 \beta_{5} + 16 \beta_{6} + \beta_{7} ) q^{50} + ( 13 + 4 \beta_{1} + 14 \beta_{2} - \beta_{3} - 9 \beta_{4} - 11 \beta_{5} + 9 \beta_{6} - 15 \beta_{7} ) q^{51} + ( 2 + 25 \beta_{2} - 2 \beta_{3} + 5 \beta_{4} + 7 \beta_{5} + 5 \beta_{6} + 7 \beta_{7} ) q^{52} + ( 7 \beta_{1} + 6 \beta_{3} - 6 \beta_{4} - 15 \beta_{5} + 6 \beta_{6} ) q^{53} + ( 68 - 10 \beta_{1} + 3 \beta_{2} - 8 \beta_{3} + \beta_{4} + 3 \beta_{5} - 12 \beta_{6} + 3 \beta_{7} ) q^{54} + ( -18 \beta_{1} - 23 \beta_{3} + 10 \beta_{4} - 4 \beta_{5} - 10 \beta_{6} - 13 \beta_{7} ) q^{56} + ( -15 + 12 \beta_{1} - 28 \beta_{2} - 12 \beta_{3} + \beta_{4} + 7 \beta_{5} + 3 \beta_{6} - 6 \beta_{7} ) q^{57} + ( -18 - 16 \beta_{2} + 4 \beta_{3} + 2 \beta_{4} - 2 \beta_{5} + 2 \beta_{6} - 2 \beta_{7} ) q^{58} + ( 11 \beta_{1} + 7 \beta_{3} + 11 \beta_{4} + 12 \beta_{5} - 11 \beta_{6} + 18 \beta_{7} ) q^{59} + ( 23 - 7 \beta_{1} - 15 \beta_{2} - 5 \beta_{3} - 5 \beta_{4} - 9 \beta_{5} - 9 \beta_{6} - 12 \beta_{7} ) q^{60} + ( 15 + 17 \beta_{2} - 13 \beta_{3} - 11 \beta_{4} + 2 \beta_{5} - 11 \beta_{6} + 2 \beta_{7} ) q^{61} + ( -26 \beta_{1} - 5 \beta_{3} + 10 \beta_{4} - 10 \beta_{5} - 10 \beta_{6} + 5 \beta_{7} ) q^{62} + ( 33 + 6 \beta_{1} + 3 \beta_{3} - 6 \beta_{4} + 9 \beta_{5} + 9 \beta_{7} ) q^{63} + ( -15 - 14 \beta_{2} - 4 \beta_{3} - 14 \beta_{4} - 10 \beta_{5} - 14 \beta_{6} - 10 \beta_{7} ) q^{64} + ( 13 \beta_{1} - 15 \beta_{3} + 12 \beta_{4} + 13 \beta_{5} - 12 \beta_{6} - 3 \beta_{7} ) q^{65} + ( 17 - 21 \beta_{2} - 15 \beta_{3} - 12 \beta_{4} + 3 \beta_{5} - 12 \beta_{6} + 3 \beta_{7} ) q^{67} + ( -11 \beta_{1} - 15 \beta_{3} + 3 \beta_{4} - 4 \beta_{5} - 3 \beta_{6} - 12 \beta_{7} ) q^{68} + ( -11 - 17 \beta_{1} + 11 \beta_{2} - 19 \beta_{3} + 15 \beta_{4} + 25 \beta_{5} - 15 \beta_{6} + 18 \beta_{7} ) q^{69} + ( 44 - 11 \beta_{2} - 8 \beta_{3} - 5 \beta_{4} + 3 \beta_{5} - 5 \beta_{6} + 3 \beta_{7} ) q^{70} + ( 3 \beta_{1} - 3 \beta_{3} - 2 \beta_{4} - 11 \beta_{5} + 2 \beta_{6} - 5 \beta_{7} ) q^{71} + ( 80 - 10 \beta_{1} + 32 \beta_{2} - 26 \beta_{3} + 11 \beta_{4} + 7 \beta_{5} - 12 \beta_{6} - 3 \beta_{7} ) q^{72} + ( -56 - 18 \beta_{2} + \beta_{3} + 3 \beta_{4} + 2 \beta_{5} + 3 \beta_{6} + 2 \beta_{7} ) q^{73} + ( 39 \beta_{1} - 11 \beta_{3} - 16 \beta_{4} - 7 \beta_{5} + 16 \beta_{6} - 27 \beta_{7} ) q^{74} + ( -9 - 18 \beta_{1} + 18 \beta_{3} + 12 \beta_{4} - 3 \beta_{5} + 3 \beta_{6} + 3 \beta_{7} ) q^{75} + ( -59 - 6 \beta_{2} + 5 \beta_{3} + 11 \beta_{4} + 6 \beta_{5} + 11 \beta_{6} + 6 \beta_{7} ) q^{76} + ( -8 + 16 \beta_{1} + 43 \beta_{2} + 5 \beta_{3} - 5 \beta_{4} - 7 \beta_{5} + 27 \beta_{6} - 6 \beta_{7} ) q^{78} + ( -1 + 48 \beta_{2} - 17 \beta_{3} - 8 \beta_{4} + 9 \beta_{5} - 8 \beta_{6} + 9 \beta_{7} ) q^{79} + ( -21 \beta_{1} - 2 \beta_{3} + 8 \beta_{4} - 9 \beta_{5} - 8 \beta_{6} + 6 \beta_{7} ) q^{80} + ( -15 + 9 \beta_{1} - 34 \beta_{2} + 9 \beta_{3} - 17 \beta_{4} + 4 \beta_{5} + 3 \beta_{7} ) q^{81} + ( -37 + 5 \beta_{2} + 34 \beta_{3} + 10 \beta_{4} - 24 \beta_{5} + 10 \beta_{6} - 24 \beta_{7} ) q^{82} + ( -12 \beta_{1} + 6 \beta_{3} + 7 \beta_{4} + 10 \beta_{5} - 7 \beta_{6} + 13 \beta_{7} ) q^{83} + ( 93 + 18 \beta_{1} + 57 \beta_{2} - 18 \beta_{3} - 15 \beta_{4} + 18 \beta_{5} - 15 \beta_{6} + 15 \beta_{7} ) q^{84} + ( -7 + 28 \beta_{2} - \beta_{3} - 8 \beta_{4} - 7 \beta_{5} - 8 \beta_{6} - 7 \beta_{7} ) q^{85} + ( 9 \beta_{1} - 22 \beta_{3} + 11 \beta_{4} + 33 \beta_{5} - 11 \beta_{6} - 11 \beta_{7} ) q^{86} + ( 10 - 2 \beta_{1} + 4 \beta_{2} - 4 \beta_{3} - 2 \beta_{4} - 4 \beta_{5} - 6 \beta_{7} ) q^{87} + ( 16 \beta_{1} + 6 \beta_{3} - 5 \beta_{4} - 8 \beta_{5} + 5 \beta_{6} + \beta_{7} ) q^{89} + ( 52 + 10 \beta_{1} - 43 \beta_{2} + 8 \beta_{3} - 21 \beta_{4} - 11 \beta_{5} + 3 \beta_{6} - 9 \beta_{7} ) q^{90} + ( -7 + 28 \beta_{2} + 7 \beta_{3} + 4 \beta_{4} - 3 \beta_{5} + 4 \beta_{6} - 3 \beta_{7} ) q^{91} + ( -47 \beta_{1} - 4 \beta_{3} + 2 \beta_{4} - 15 \beta_{5} - 2 \beta_{6} - 2 \beta_{7} ) q^{92} + ( 45 - 19 \beta_{2} - 11 \beta_{4} + 4 \beta_{5} + 15 \beta_{6} + 15 \beta_{7} ) q^{93} + ( -95 - 82 \beta_{2} - \beta_{3} + 12 \beta_{4} + 13 \beta_{5} + 12 \beta_{6} + 13 \beta_{7} ) q^{94} + ( 19 \beta_{1} - 18 \beta_{3} + 4 \beta_{4} + 27 \beta_{5} - 4 \beta_{6} - 14 \beta_{7} ) q^{95} + ( 41 - 4 \beta_{1} - 10 \beta_{2} - 5 \beta_{3} - 10 \beta_{4} - 23 \beta_{5} - 3 \beta_{6} - 24 \beta_{7} ) q^{96} + ( -41 + 5 \beta_{2} + 10 \beta_{3} + 8 \beta_{4} - 2 \beta_{5} + 8 \beta_{6} - 2 \beta_{7} ) q^{97} + ( -10 \beta_{1} + 31 \beta_{3} - 8 \beta_{4} - \beta_{5} + 8 \beta_{6} + 23 \beta_{7} ) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$8 q + 5 q^{3} - 26 q^{4} + q^{6} + 28 q^{7} + 11 q^{9} + O(q^{10})$$ $$8 q + 5 q^{3} - 26 q^{4} + q^{6} + 28 q^{7} + 11 q^{9} - 6 q^{10} + 53 q^{12} + 44 q^{13} - 54 q^{15} - 14 q^{16} + q^{18} + 68 q^{19} - 6 q^{21} + 33 q^{24} + 42 q^{25} - 25 q^{27} - 118 q^{28} + 10 q^{30} + 2 q^{31} - 66 q^{34} - 7 q^{36} + 140 q^{37} + 38 q^{39} + 58 q^{40} + 174 q^{42} - 78 q^{43} - 36 q^{45} - 286 q^{46} - 285 q^{48} - 140 q^{49} + 58 q^{51} - 102 q^{52} + 523 q^{54} - 22 q^{57} - 68 q^{58} + 262 q^{60} + 22 q^{61} + 246 q^{63} - 52 q^{64} + 184 q^{67} - 176 q^{69} + 374 q^{70} + 489 q^{72} - 378 q^{73} - 33 q^{75} - 450 q^{76} - 246 q^{78} - 252 q^{79} + 11 q^{81} - 200 q^{82} + 450 q^{84} - 156 q^{85} + 66 q^{87} + 598 q^{90} - 148 q^{91} + 380 q^{93} - 460 q^{94} + 399 q^{96} - 324 q^{97} + O(q^{100})$$

Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{8} + 29 x^{6} + 282 x^{4} + 1061 x^{2} + 1331$$:

 $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$\nu^{6} + 30 \nu^{4} + 260 \nu^{2} + 541$$$$)/52$$ $$\beta_{3}$$ $$=$$ $$($$$$-2 \nu^{7} + 22 \nu^{6} - 47 \nu^{5} + 517 \nu^{4} - 377 \nu^{3} + 3575 \nu^{2} - 1121 \nu + 6897$$$$)/286$$ $$\beta_{4}$$ $$=$$ $$($$$$-27 \nu^{7} + 11 \nu^{6} - 706 \nu^{5} + 330 \nu^{4} - 5304 \nu^{3} + 3432 \nu^{2} - 9199 \nu + 10527$$$$)/1144$$ $$\beta_{5}$$ $$=$$ $$($$$$4 \nu^{7} + 94 \nu^{5} + 611 \nu^{3} + 812 \nu$$$$)/143$$ $$\beta_{6}$$ $$=$$ $$($$$$35 \nu^{7} - 77 \nu^{6} + 894 \nu^{5} - 1738 \nu^{4} + 6812 \nu^{3} - 10868 \nu^{2} + 13683 \nu - 17061$$$$)/1144$$ $$\beta_{7}$$ $$=$$ $$($$$$-10 \nu^{7} - 22 \nu^{6} - 235 \nu^{5} - 517 \nu^{4} - 1599 \nu^{3} - 3575 \nu^{2} - 2745 \nu - 6897$$$$)/286$$
 $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$\beta_{6} + \beta_{4} + \beta_{3} - \beta_{2} - 8$$ $$\nu^{3}$$ $$=$$ $$-2 \beta_{7} - 3 \beta_{5} - 2 \beta_{3} - 10 \beta_{1}$$ $$\nu^{4}$$ $$=$$ $$\beta_{7} - 15 \beta_{6} + \beta_{5} - 15 \beta_{4} - 16 \beta_{3} + 23 \beta_{2} + 85$$ $$\nu^{5}$$ $$=$$ $$33 \beta_{7} + 8 \beta_{6} + 36 \beta_{5} - 8 \beta_{4} + 41 \beta_{3} + 113 \beta_{1}$$ $$\nu^{6}$$ $$=$$ $$-30 \beta_{7} + 190 \beta_{6} - 30 \beta_{5} + 190 \beta_{4} + 220 \beta_{3} - 378 \beta_{2} - 1011$$ $$\nu^{7}$$ $$=$$ $$-470 \beta_{7} - 188 \beta_{6} - 352 \beta_{5} + 188 \beta_{4} - 658 \beta_{3} - 1331 \beta_{1}$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/363\mathbb{Z}\right)^\times$$.

 $$n$$ $$122$$ $$244$$ $$\chi(n)$$ $$-1$$ $$1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
122.1
 − 3.58727i − 3.06025i − 2.00420i − 1.65816i 1.65816i 2.00420i 3.06025i 3.58727i
3.58727i −2.74329 + 1.21424i −8.86854 2.08639i 4.35581 + 9.84093i 8.86303 17.4648i 6.05125 6.66201i 7.48447
122.2 3.06025i 0.126437 2.99733i −5.36516 6.63932i −9.17261 0.386930i −3.06298 4.17774i −8.96803 0.757950i −20.3180
122.3 2.00420i 2.80061 + 1.07544i −0.0168066 5.48438i 2.15539 5.61298i 5.59084 7.98311i 6.68687 + 6.02376i 10.9918
122.4 1.65816i 2.31624 + 1.90658i 1.25051 0.698503i 3.16141 3.84069i 2.60911 8.70618i 1.72990 + 8.83218i −1.15823
122.5 1.65816i 2.31624 1.90658i 1.25051 0.698503i 3.16141 + 3.84069i 2.60911 8.70618i 1.72990 8.83218i −1.15823
122.6 2.00420i 2.80061 1.07544i −0.0168066 5.48438i 2.15539 + 5.61298i 5.59084 7.98311i 6.68687 6.02376i 10.9918
122.7 3.06025i 0.126437 + 2.99733i −5.36516 6.63932i −9.17261 + 0.386930i −3.06298 4.17774i −8.96803 + 0.757950i −20.3180
122.8 3.58727i −2.74329 1.21424i −8.86854 2.08639i 4.35581 9.84093i 8.86303 17.4648i 6.05125 + 6.66201i 7.48447
 $$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 122.8 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 363.3.b.m 8
3.b odd 2 1 inner 363.3.b.m 8
11.b odd 2 1 363.3.b.l 8
11.c even 5 2 33.3.h.b 16
11.c even 5 2 363.3.h.o 16
11.d odd 10 2 363.3.h.j 16
11.d odd 10 2 363.3.h.n 16
33.d even 2 1 363.3.b.l 8
33.f even 10 2 363.3.h.j 16
33.f even 10 2 363.3.h.n 16
33.h odd 10 2 33.3.h.b 16
33.h odd 10 2 363.3.h.o 16

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
33.3.h.b 16 11.c even 5 2
33.3.h.b 16 33.h odd 10 2
363.3.b.l 8 11.b odd 2 1
363.3.b.l 8 33.d even 2 1
363.3.b.m 8 1.a even 1 1 trivial
363.3.b.m 8 3.b odd 2 1 inner
363.3.h.j 16 11.d odd 10 2
363.3.h.j 16 33.f even 10 2
363.3.h.n 16 11.d odd 10 2
363.3.h.n 16 33.f even 10 2
363.3.h.o 16 11.c even 5 2
363.3.h.o 16 33.h odd 10 2

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{3}^{\mathrm{new}}(363, [\chi])$$:

 $$T_{2}^{8} + 29 T_{2}^{6} + 282 T_{2}^{4} + 1061 T_{2}^{2} + 1331$$ $$T_{5}^{8} + 79 T_{5}^{6} + 1687 T_{5}^{4} + 6576 T_{5}^{2} + 2816$$ $$T_{7}^{4} - 14 T_{7}^{3} + 35 T_{7}^{2} + 138 T_{7} - 396$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$1331 + 1061 T^{2} + 282 T^{4} + 29 T^{6} + T^{8}$$
$3$ $$6561 - 3645 T + 567 T^{2} + 135 T^{3} - 72 T^{4} + 15 T^{5} + 7 T^{6} - 5 T^{7} + T^{8}$$
$5$ $$2816 + 6576 T^{2} + 1687 T^{4} + 79 T^{6} + T^{8}$$
$7$ $$( -396 + 138 T + 35 T^{2} - 14 T^{3} + T^{4} )^{2}$$
$11$ $$T^{8}$$
$13$ $$( 4444 + 2626 T - 85 T^{2} - 22 T^{3} + T^{4} )^{2}$$
$17$ $$8306771 + 12909824 T^{2} + 172761 T^{4} + 734 T^{6} + T^{8}$$
$19$ $$( -64999 + 8694 T + 7 T^{2} - 34 T^{3} + T^{4} )^{2}$$
$23$ $$25255373616 + 307285308 T^{2} + 1188583 T^{4} + 1848 T^{6} + T^{8}$$
$29$ $$340736 + 93696 T^{2} + 7312 T^{4} + 184 T^{6} + T^{8}$$
$31$ $$( -97204 - 30786 T - 1709 T^{2} - T^{3} + T^{4} )^{2}$$
$37$ $$( -698036 + 107680 T - 1111 T^{2} - 70 T^{3} + T^{4} )^{2}$$
$41$ $$3927187055891 + 26651833143 T^{2} + 27887368 T^{4} + 9433 T^{6} + T^{8}$$
$43$ $$( -684409 - 73219 T - 1528 T^{2} + 39 T^{3} + T^{4} )^{2}$$
$47$ $$14778028911536 + 51690193292 T^{2} + 40526643 T^{4} + 11237 T^{6} + T^{8}$$
$53$ $$1733384590256 + 8491109372 T^{2} + 11081859 T^{4} + 5633 T^{6} + T^{8}$$
$59$ $$162457590843251 + 315187108454 T^{2} + 162295647 T^{4} + 23636 T^{6} + T^{8}$$
$61$ $$( 1180476 - 62286 T - 4549 T^{2} - 11 T^{3} + T^{4} )^{2}$$
$67$ $$( -11977619 + 552568 T - 3927 T^{2} - 92 T^{3} + T^{4} )^{2}$$
$71$ $$1657701936 + 269041212 T^{2} + 2752883 T^{4} + 3227 T^{6} + T^{8}$$
$73$ $$( 2336356 + 316506 T + 12357 T^{2} + 189 T^{3} + T^{4} )^{2}$$
$79$ $$( -27267284 - 928896 T - 3903 T^{2} + 126 T^{3} + T^{4} )^{2}$$
$83$ $$46820268891 + 892608003 T^{2} + 4637264 T^{4} + 4757 T^{6} + T^{8}$$
$89$ $$3248250664131 + 15048846351 T^{2} + 19103392 T^{4} + 8529 T^{6} + T^{8}$$
$97$ $$( -3010169 + 18612 T + 7293 T^{2} + 162 T^{3} + T^{4} )^{2}$$


text:
## On the casualty ratio in GazaJanuary 19, 2009

Posted by dorigo in news, politics.
Tags: , , , , ,

In this short post I wish to discuss whether during the 22-day-long war in Gaza the israeli defence force (IDF) has been, as some commenter claimed here, “the most moral army in the world”. It is indeed reported by media that the IDF makes phone calls to civilian houses before those are bombed, that leaflets are spread on areas which will be targeted, etcetera. I think these are commendable actions, but on the other hand the media these days are full of news that report despicable episodes by the israeli army, which tend to demonstrate the opposite. Leaving aside the endless debate that such conflicting evidences would set up, let me stay on the easier ground of mathematics, and aseptically discuss the civilian casualty ratio in Gaza.

First of all, there are a couple of caveats to mention. The ratio is not too easy to ascertain, because -to some extent- many of the Gaza “civilians” might be considered supporters of the Hamas force. So we have to stick to numbers provided by Israel itself, rather than rely on the claims by Hamas that just 48 of its combatants have been killed in the attack. Israel sets that number at “at least 500”, and I will take 500 as a fair estimate, ignoring the Hamas claim and the rather propaganda-sounding “at least” by the IDF.  Let us give the number a 20% uncertainty: 500+-100. As for the total casualties, their number is less uncertain: most sources set it at 1300 or above, if we stick to palestinians who have lost their life until today. More can die of the wounds in the near future, but the ratio we are to compute should not be affected by that.

The other caveat is that one could argue that it is unfair to judge the actions of the IDF on the basis of the casualty ratio, because the action was aimed at eradicating the weapons potential of Hamas rather than Hamas itself, and the weapons had allegedly been concealed in homes, hospitals, schools, places of worship. This objection to me is not too valuable, because it amounts to declaring that the action had been decided in spite of that practice. Or, to put it another way: if Hamas uses the horrible practice of human shields, Israel uses the even more horrible practice of ignoring it!

Now, let us look at the numbers. $(1300-500)/500 = 1.6 \pm 0.5$, roughly. That means that for each Hamas fighter, more than one civilian has been killed.

Let us now look at some statistics reported one year ago by Haaretz  here, about the civilian casualty ratio of air strikes. It is reported at 1:30.

So I am led to conclude that the IDF considered not enough the air strikes, and needed a different strategy to counter the Kassam missiles; and that they did not really care about the civilian losses they would inflict on palestinians, this time.

And a warning: Of course, I am willing to hear other opinions on the matter; this blog is usually censorship-free. Since, however, I am bored by allegations of being “anti-semitic”, “communist”, of “caring zero for the other wars in the world”, “where were you when Serbia was bombed by your country”, etcetera, I will apply some measure of censorship here. Either leave your name and address, or refrain from posting meaningless, insulting comments, please.

## In support of UN sanctions to IsraelJanuary 15, 2009

Posted by dorigo in news, politics.
Tags: , , , , , , , ,

As the war in the Gaza strip is approaching its fourth week, the number of casualties has surpassed a thousand, 300 of them children -needless to say, all from the defending side. 1500 more children have been wounded, and the situation configures itself as a humanitarian catastrophe, of which the guilt is entirely of the offenders, the leaders of the state of Israel.

The Israeli army continues the attacks, denies humanitarian aid, bombs hospitals, uses white phosporus to bomb densely populated areas. “Intelligent bombs“, somebody insists calling them. In order to minimize their army’s losses, the Israeli prefer to kill civilians blindly. Today they bombed a UN building, home of the UNRWA. Later Olmert commented that Hamas militians were shooting from its roof.

The IDF needs to be stopped, because it is clear that they cannot get back to reason by themselves. I believe the only way that the UN can put real pressure on Israel is to sanction it economically. Israel is a small state and its economy is more frail than its army. There needs to be a response to the barbarian acts they are committing in Gaza.

Yes, barbarian acts. How else can a civilized person call the indiscriminate bombing of civilians ? Israeli supporters will argue that the death of those 300 children is Hamas fault, because they allegedly use children as human shields. Israeli supporters will show videos of small palestinian children educated to fight, dressed in mimetic suits, carrying guns. They believe those admittedly disturbing images are enough to justify the deliberate killing of children. This, to me, is barbarous, period.

The hundreds of rockets randomly fired by Hamas into Israeli territory this year are a drop in the sea in comparison to the destruction, the deaths, and the steps farther from peace that this senseless war has brought. Terrorist is Hamas, and terrorist is Israel. And since the war is ultimately the result of the need of the present Israeli government to strengthen itself in view of the forthcoming elections, economic sanctions by the UN are the best way to respond.

I hope President Obama will be able to use his power to put an end to this horrible situation. A change from the unconditioned support that the US have given to Israel would be the strongest message. The US needs to support Israel in its attempts at creating the conditions for a durable peace with the Arab countries, and sanction it in case this bloody foreign politics is furthered.

I know these are likely to remain delusions: despite the strong condemnation by the French government, the shock declared by Merkel and Brown, the indignant reactions by UN officials, and the other reactions that have been proclaimed today, there is little hope that anything more concrete than spelling words will be done. But Israel, unfortunately, will pay for this senseless attack, because it has worsened the hatred that arabs feel for its citizens. This is the sorry aftermath of any military action against civilians: a renewed, reinforced hatred.

The israeli fanatics who read this blog will argue that I am anti-semitic: this is false, of course, and it equates to an accusation of racism. I hate racism as much as I hate wars, whomever moves them, and whomever argues that the attacking side has the right on their side.

## Stop the bombs!January 2, 2009

Posted by dorigo in history, news, politics.
Tags: , , , ,

It is quite sad for me to see a country built on the ashes of millions of jews, victims of the nazis during WWII, show for the umpteenth time their leaders and soldiers are not altogether different from their grandparents’ executioners. We are all human, and that means we are all beasts. Israel feels threatened by the constant threat of palestinian attacks -not a big threat, to be honest: the wiping out of Israel off the middle east map is an oversold story which has lost all its meaning since the seventies- and decides unilaterally to unleash an attack on a strip of land so thin that planes would have trouble landing along its shorter side.

And hundreds die. More than 400 so far, plus 2000 wounded, most of them civilians of course. The acclaimed killing by israeli bombs of a Hamas leader, Nizar Rayyan, cost the life of his wives and 10 of his 12 children. Rayyan was no doubt an extremist and a fomenter of suicide attacks, but I ask your forgiveness if I can see no difference between the bombs that killed him and his family and those he tried to explode in israeli soil. Terrorist is Hamas, and terrorist is Tzipi Livni and her government.

What is striking to me is how plain it is to see, from a distance, that these bombs are only a political tool for power, and that they will but cause more suffering to the same israeli people that dropped them. Does Livni think that killing a few Hamas leaders and a few thousand civilians will cool down the minds of the arabs ? No, she does not. But she feels more powerful today.

## Your daily dose of violenceJuly 21, 2008

Posted by dorigo in news, politics.
Tags: , ,


text:
# Tag Info

Accepted

### Universal bases (dictionary) for image compression

This is a great and interesting question. There are 2 ways to look at it, empirically and analytically. But before we start, a major detail is that when dealing with images we mainly talk about the ...
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### Number of dimensions? Color image vs gray scale image? Colour video vs gray scale video? Especially in the context of MATLAB

Color images are usually modeled as a vector valued function of 2D: $$I : \mathbb{R}^{2} \to \mathbb{R}^{3}$$ Namely for 2D coordinates input it outputs 3 values (RGB). Hence images are 2D functions....
• 40.6k
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### Building a Pipeline for Image Classification / Clustering Tasks with Features Extractor and Dimensionality Reduction (Example on MNIST Data)

Feature Extraction There are many modern known features for images. Among them: BRISK Feature. FAST Feature. Harris Feature. KAZE Feature. MSER Feature. ORB Feature. SIFT Feature. SURF Feature. LBP ...
• 40.6k
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### When to Use Composite Filters and When to Use Separable Filters?

Image Processing Context In classic Image Processing the filters used are known. Hence being separable is a property of a given filter which is suitable to the task. In this context, separability only ...
• 40.6k
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### Understanding the Bilateral Filter (Image Filtering)

There are some good resources on our site: Understanding the Bilateral Filter - Neighbors and Sigma. How to Validate Bilateral Filter Implementation? What Is the Bilateral Filter Category: LPF, HPF, ...
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### Image Segmentation Using Deep Learning

Well, I think the best way to tackle this question is a little background and a code as an example. I chose MATLAB for this example though PyTorch / Keras would probably be as easy. This task requires ...
• 40.6k
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### Rectangle Segments of Image (Rectangle Super Pixels) per Pixel

One answer for your question would be limiting Super Pixels to rectangle forms. It requires changing the code of a Super Pixel algorithm to constraint the shape of the Super Pixel. Another approach ...
• 40.6k

### Universal bases (dictionary) for image compression

As a complement to the neat answer by @Royi, I would add that "sparsity" is originally a heuristic principle in science, that applies well to many interesting really world data and problems. ...
• 29.9k

### Slicing an Image into Tiles According to Content

Under the following assumptions (Written for the vertical lines, same for the horizontal): The background is uniform. There at least a gap of 2 background pixels between the closest objects on ...
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### Image Standardization for Image Classification (Machine / Deep Learning)

I will display image standardization using MATLAB: ...
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### Detecting bullet holes using Python with camera or sensors

I can't really comment on the machine vision part, other than any question that asks "How do I do <some signal processing task> in <some language>" is fairly naive. The way you ...
• 8,275

### Number of dimensions? Color image vs gray scale image? Colour video vs gray scale video? Especially in the context of MATLAB

My answer became quite long. So normally a digital signal is 1D, a digital image is 2D, a video is 3D. But it can get complicated. Long introduction on mathematics for the start. In mathematics, the ...
• 29.9k
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### Comparing distribution of vectors with different length?

From what you've said, you have two sample sets: $$x_n, n = 0 \ldots N-1$$ and $$y_m, m = 0 \ldots M-1$$ where $M \ne N$ and you want to compare the distributions of the underlying processes. Do you ...
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### Alternative to support vector machine?

I think that the best option, probably, would be Random Forests or any other Boosting / Bagging method based on decision trees. I would probably start with SK Learn Random Forests. As more advanced ...
• 740

### Edge Detection Convolution Intuition

If it helps, try a simpler version. Just look at two neighbouring pixels in a row. First example: There is 10 and 10. Difference between them is 0, no difference, no edge. Second example: In the ...
• 1,461

### Number of dimensions? Color image vs gray scale image? Colour video vs gray scale video? Especially in the context of MATLAB

Depends what "dimension" means, but I'll say 2D, and interpret in context of convolutions. A convolution operates on 1D: (channels, time) 2D: ...
• 5,066

### How to perform Spatial derivative calculation?

What you should do instead to calculate $I_x$:  \begin{align} I_{x}(x,y)&=\frac{\partial I(x,y)}{\partial x}\\ &=I(x+1,y)-I(x-1,y)\\ &= \begin{bmatrix}5& 7 & 11\\ 9 & 12 &...
• 22.4k

### Fixing distortion in photographic reproductions

I answered a related question in SO some time ago. Briefly, you can work your setup through a standard camera calibration procedure, and you have the option of "freezing" the focal length to ...
1 vote
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### Fixing distortion in photographic reproductions

In a general setting, multiple observations of a calibration target are needed because otherwise the focal length parameter(s) and principle point parameters are coupled so you would not be able to ...
• 146
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### Slicing an Image into Tiles According to Content

Thanks to user Ash for giving me a key insight: Measuring the HSL (hue, saturation, and lightness) of the pixels can help detect where the gaps are. If I take one horizontal row of pixels that goes ...
• 161
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### Universal bases (dictionary) for image compression

Families of basis functions that are roughly sinusoidal are efficient for images I think because they can efficiently encode an edge at any spatial location. In images it is typical to have edges ...
• 12.4k
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### If a camera optical axis is perpendicular to a flat surface, does the area covered by a pixel is the same for all pixels?

For an ideal pinhole camera model, your geometric reasoning is correct: the projected pixel area on an orthogonal flat world-surface will be independent of sensor pixel position on the flat image-...
• 26.7k
1 vote

### What transformations to go from top-down fisheye to occupancy (ground texture)

Allright well, @TimeWescott was essentially correct about what I had to do, though the theory took some investigation and thought before I realized what the parts to the puzzle are, and the inherent ...
• 146
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### Optical-flow visualizations explained

Yes, as your linked page shows: the idea is that the pixels move from $(x,y)$ in the first from to $(x+\Delta x, y + \Delta y)$ in the second. I believe the image you quote is just connecting a ...
• 22.4k
1 vote

### When to Use Composite Filters and When to Use Separable Filters?

For any given problem definition, there's a filter that -- if you ignore execution time and hardware expense -- is "best"*. In general, that "best" filter isn't separable. ...
• 8,275
1 vote

### Understanding the Bilateral Filter (Image Filtering)

A simple filter (convolution) replace value of a pixel (for simplicity assume its the pixel positioned at the center of kernel) with a weighted sum of its neighbors and the weights are constant ( the ...
• 1,306
1 vote

### Projecting a 3D point onto Multiple Cameras in the Same Projective Frame

It appears that my problem was normalization. The mentioned Algorithm for calculating the cameras makes use of the epipoles $e'$ and $e''$ (i.e., the images of the first camera's focal point in the ...
• 121
1 vote

### Help or suggestions with Line detection in microscopy images

Instead of using a homegrown amalgamation of algorithms, I suggest you look in the scientific literature for existing solutions. People have solved the same problem over and over again, there exist ...
• 1,002
1 vote

### How are line pairs and pixels related in image resolution?

This answer is to be read along with Fat32's answer. The whole electrooptical imaging thing is complicated, and there's always more details to stumble over. In this case, I think the following three ...
• 8,275
1 vote

### Counting vechicles in an image

This is a relatively easy method using modern Computer Vision, namely, Deep Learning. The easiest way would be using transfer learning one a model which was trained on face detection. The reason to ...
• 349

Only top scored, non community-wiki answers of a minimum length are eligible


text:
# Reynolds stress equation model

Reynolds stress equation model (RSM), also referred to as second moment closures are the most complete classical turbulence model. In these models, the eddy-viscosity hypothesis is avoided and the individual components of the Reynolds stress tensor are directly computed. These models use the exact Reynolds stress transport equation for their formulation. They account for the directional effects of the Reynolds stresses and the complex interactions in turbulent flows. Reynolds stress models offer significantly better accuracy than eddy-viscosity based turbulence models, while being computationally cheaper than Direct Numerical Simulations and Large Eddy Simulations.

## Shortcomings of Eddy-viscosity based models

Eddy-viscosity based models like the ${\displaystyle k-\epsilon }$ and the ${\displaystyle k-\omega }$ models have significant shortcomings in complex, real-life turbulent flows. For instance, in flows with streamline curvature, flow separation, flows with zones of re-circulating flow or flows influenced by mean rotational effects, the performance of such models is unsatisfactory.

Such one- and two-equation based closures cannot account for the return to isotropy of turbulence,[1] observed in decaying turbulent flows. Eddy-viscosity based models cannot replicate the behaviour of turbulent flows in the Rapid Distortion limit,[2] where the turbulent flow essentially behaves as an elastic medium (instead of viscous).

## Reynolds Stress Transport Equation

Reynolds Stress equation models rely on the Reynolds Stress Transport equation. The equation for the transport of kinematic Reynolds stress ${\displaystyle R_{ij}=u_{i}^{\prime }u_{j}^{\prime }=-\tau _{ij}/\rho }$ is [3]

  ${\displaystyle {\frac {DR_{ij}}{Dt}}=D_{ij}+P_{ij}+\Pi _{ij}+\Omega _{ij}-\varepsilon _{ij}}$

Rate of change of ${\displaystyle R_{ij}}$ + Transport of ${\displaystyle R_{ij}}$ by convection = Transport of ${\displaystyle R_{ij}}$ by diffusion + Rate of production of ${\displaystyle R_{ij}}$ + Transport of ${\displaystyle R_{ij}}$ due to turbulent pressure-strain interactions + Transport of ${\displaystyle R_{ij}}$ due to rotation + Rate of dissipation of ${\displaystyle R_{ij}}$.

The six partial differential equations above represent six independent Reynolds stresses. While the Production term (${\displaystyle P_{ij}}$) is closed and does not require modelling, the other terms, like pressure strain correlation (${\displaystyle \Pi _{ij}}$) and dissipation (${\displaystyle \varepsilon _{ij}}$) , are unclosed and require closure models.

## Production term

The Production term that is used in CFD computations with Reynolds stress transport equations is

   ${\displaystyle P_{ij}}$ =${\displaystyle -\left(R_{im}{\frac {\partial U_{j}}{\partial x_{m}}}+R_{jm}{\frac {\partial U_{i}}{\partial x_{m}}}\right)}$

Physically, the Production term represents the action of the mean velocity gradients working against the Reynolds stresses. This accounts for the transfer of kinetic energy from the mean flow to the fluctuating velocity field. It is responsible for sustaining the turbulence in the flow through this transfer of energy from the large scale mean motions to the small scale fluctuating motions.

## Slow Pressure-strain correlation term

The slow pressure-strain correlation term redistributes energy amongst the normal Reynolds stresses. This is responsible for the return to isotropy of decaying turbulence. Physically, this arises due to the self-interactions amongst the fluctuating field. The correction term is given as [4]

${\displaystyle \Pi _{ij}=-C_{1}{\frac {\epsilon }{k}}\left(R_{ij}-{\frac {2}{3}}k\delta _{ij}\right)-C_{2}\left(P_{ij}-{\frac {2}{3}}P\delta _{ij}\right)}$

## Dissipation term

The modelling of dissipation rate ${\displaystyle \epsilon _{\rm {ij}}}$ assumes that the small dissipative eddies are isotropic. This term affects only the normal Reynolds stresses. [5]

          ${\displaystyle \epsilon _{\rm {ij}}}$ = ${\displaystyle 2/3\epsilon \delta _{ij}}$

where ${\displaystyle \epsilon }$ is dissipation rate of turbulent kinetic energy, and ${\displaystyle \delta _{ij}}$ = 1 when i = j and 0 when i ≠ j

## Diffusion term

The modelling of diffusion term ${\displaystyle D_{ij}}$ is based on the assumption that the rate of transport of Reynolds stresses by diffusion is proportional to the gradients of Reynolds stresses. The simplest form of ${\displaystyle D_{ij}}$ that is followed by commercial CFD codes is

${\displaystyle D_{ij}}$ = ${\displaystyle {\frac {\partial }{\partial x_{m}}}\left({\frac {v_{t}}{\sigma _{k}}}{\frac {\partial R_{ij}}{\partial x_{m}}}\right)}$ = ${\displaystyle div\left({\frac {v_{t}}{\sigma _{k}}}\nabla (R_{ij})\right)}$

where ${\displaystyle \upsilon _{t}}$ = ${\displaystyle C_{\mu }{\frac {k^{2}}{\epsilon }}}$ , ${\displaystyle \sigma _{k}}$ = 1.0 and ${\displaystyle C_{\mu }}$ = 0.9

## Rotational term

The rotational term is given as [6]

 ${\displaystyle \Omega _{ij}=-2\omega _{k}\left(R_{jm}e_{ikm}+R_{im}e_{jkm}\right)}$

here${\displaystyle \omega _{k}}$ is the rotation vector, ${\displaystyle e_{ijk}}$=1 if i,j,k are in cyclic order and are different,${\displaystyle e_{ijk}}$=-1 if i,j,k are in anti-cyclic order and are different and ${\displaystyle e_{ijk}}$=0 in case any two indices are same.

1) Unlike the k-ε model which uses an isotropic eddy viscosity, RSM solves all components of the turbulent transport.
2) It is the most general of all turbulence models and works reasonably well for a large number of engineering flows.
3) It requires only the initial and/or boundary conditions to be supplied.
4) Since the production terms need not be modeled, it can selectively damp the stresses due to buoyancy, curvature effects etc.

## References

1. ^ Lumley, John; Newman, Gary (1977). "The return to isotropy of homogeneous turbulence". Journal of Fluid Mechanics. 82: 161–178.
2. ^ Mishra, Aashwin; Girimaji, Sharath (2013). "Intercomponent energy transfer in incompressible homogeneous turbulence: multi-point physics and amenability to one-point closures". Journal of Fluid Mechanics. 731: 639–681.
3. ^ Bengt Andersson , Ronnie Andersson s (2012). Computational Fluid Dynamics for Engineers (First ed.). Cambridge University Press, New York. p. 97. ISBN 9781107018952.
4. ^ Magnus Hallback (1996). Turbulence and Transition Modelling (First ed.). Kluwer Academic Publishers. p. 117. ISBN 0792340604.
5. ^ Peter S. Bernard & James M. Wallace (2002). Turbulent Flow: Analysis, Measurement & Prediction. John Wiley & Sons. p. 324. ISBN 0471332194.
6. ^ H.Versteeg & W.Malalasekera (2013). An Introduction to Computational Fluid Dynamics (Second ed.). Pearson Education Limited. p. 96. ISBN 9788131720486.

## Bibliography

• "Turbulent Flows", S. B. Pope, Cambridge University Press (2000).
• "Modelling Turbulence in Engineering and the Environment: Second-Moment Routes to Closure", Kemal Hanjalić and Brian Launder, Cambridge University Press (2011).


text:
# statsmodels.robust.norms.RobustNorm¶

class statsmodels.robust.norms.RobustNorm[source]

The parent class for the norms used for robust regression.

Lays out the methods expected of the robust norms to be used by statsmodels.RLM.

Notes

Currently only M-estimators are available.

References

PJ Huber. ‘Robust Statistics’ John Wiley and Sons, Inc., New York, 1981.

DC Montgomery, EA Peck. ‘Introduction to Linear Regression Analysis’,

John Wiley and Sons, Inc., New York, 2001.

R Venables, B Ripley. ‘Modern Applied Statistics in S’

Springer, New York, 2002.

Methods

 Derivative of rho. Derivative of psi. The robust criterion estimator function. Returns the value of psi(z) / z


text:
# Writing Christmas Trees, or Building A Cute Turing Tarpit

## December 27, 2012

My wife and I recently decided to enter this month’s contest over at PLT Games. We produced a variation of the lambda calculus called Tannenbaum in which every program looks like (or hopefully looks like) a Christmas tree. For example, here is omega, which is normally written ((lambda (x) (x x )) (lambda (x) (x x))). It’s just a cute little bush!

To be honest, the language interpretation itself isn’t difficult, once in the proper form—once everything is lambdas and application tags and variable and line references. But properly stripping it apart is a fancy trick. Luckily, most of the string processing tools I needed got written while I was writing that blogging software last week. The harder part, though, was properly picking apart application, and that’s the algorithm I’m going to spend this talking about.

The idea is simple: given application tags and the guarantee that every application is of one argument, properly group up the arguments, separated only by %. For example, (@ @  % $%$ ) is, in Scheme, (((var 1) (var 0)) (var 2)). But given a chain of application tags and delimiters, how does one group up and properly partition these?

I wouldn’t call my solution elegant, but it gets the job done. I have never done this before, so it was certainly a first for me. (I imagine that automatic currying follows a similar pattern, so maybe this is a more useful exercise than I initially suspected.) But I’ve pontificated enough; onto this algorithm:

The algorithm I have uses three helpers: app-splitter, app-builder, and app-parser (which really just calls app-builder on app-splitter’s output). The app-splitter function walks over the input, as shaped above, using two stacks. Any time it sees a symbol such as a +, \$, or 0, it simply pushes that symbol onto stack A. On a %, the current contents of stack A are moved to stack B as their own element (as a list). This means that the delimiters chunk up the input.

When an @—application—tag is found, if stack A is empty an app tag is pushed onto stack B. If stack A is empty, it is moved onto stack B before the app tag is pushed on. At the end of all of this, the string is split into tokens based on its application and argument delimiters, but still shaped incorrectly (and backwards).

This moves us to app-builder, which takes this reversed list and adds proper nesting by using its own stack as follows: it ‘pops’ something off the original stack handed to it (stack B from before), and if it’s not an app tag, the element gets pushed onto app-builder’s own stack C. An app tag pops the previous two elements off of stack C and pushes the whole (app [element 1] [element 2]) back onto the stack (in case it’s nested in another application tag). This repeats until stack B is empty, at which point (if the input was correct), stack C contains the properly-shaped result.

This completes the parsing, handing out the result. I thought this approach was neat (though possibly a little expensive); it requries two passes over the code for O(2n) complexity based on input size. It is certainly possible to do all of this in one pass, but it would require performing building up tokens while reshaping applications, and must be done back-to-front.

As an interesting aside, I find this entire exercise has led me to one interesting historical insight: I suspect the reason that APL is right-to-left precedence is for precisely this reason of application. Since this requires far less backtracking when done in reverse, if a computer with little power is reading from left to right, providing precedence from right to left allows you to perform this analysis while reading input.

(It’s worth noting that doing this in one fell swoop only requires one stack, and would probably be far more efficient after the initial reversal. That said, I feel like this implementation is more clear at the cost of efficiency.)

I hope this article helps someone somewhere; writing a parser like this is a nice academic endeavor, if nothing else, and I encourage you to write one for your favorite language some time.

### Using LaTeX for Programming Language Semantics

LaTex is a fantastic tool for typesetting, but there seem to be a serious gapin documentation for using it to lay out programming semanti...… Continue reading

#### A Small ALU in Haskell, Part I

Published on April 24, 2015

#### The Refined Gradual Guarantee and Compilation

Published on April 15, 2015


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# Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.$f(t) = 2.5t^2 + 6t$

## $5 t+6$

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Okay here we have a function. F F T. I'm going to use capital T. So as I'm doing all this work, my lower case T. S don't confuse you and look like plus times. I'm going to use capital T. So F F T. Is defined to be 2.5 times T squared plus six times T. The domain of this function. The domain of a function ah is in this case is the variables T. The domain would be all the values of T where this function is defined. Well, this function is defined for any number that you plug into T. So the domain is, you know, all the real numbers, the entire X axis. The next thing we want to do is we want to calculate uh F prime of T. The derivative of F with respect to T. Using uh the definition of derivative. So the definition of the derivative for F prime of T is going to be to limit of F of T plus delta T minus F of T. All divided by delta T. And we take the limit of this expression as delta T approaches zero. So F of T plus delta T. Wherever you see T. In the original function, we're gonna uh plug in T plus delta T. So F F T plus Delta T. is going to be 2.5. And instead of times T squared OK, we are doing T plus delta T to the second Plus six times T. But remember right now we're writing the expression for F of T plus delta T. Uh So instead of tea, we're going to be writing in T. Plus delta T. So plus six times this variable is really plus six times to T. Plus delta T. So that takes care of the F. Of T. Plus delta T. Now we have to subtract F. Of T. Which is the 2.5 T square plus 60. So we're going to subtract 2.5 T sq. And since we're subtracting FFT and this entire expressionist FFT, we're subtracting each of these terms. So I already wrote minus 2.5 times T squared. Now, since once again, since I'm subtracting FFT, I also had to subtract 2 60. So that's why we write -16. Now all of this gets divided by delta T. And I forgot to write the little limit. Um but so we'll just squeeze it in here. We want to take the limit of this whole expression as delta T approaches zero. So what I'm going to do next is I'm going to do T plus delta T to the second. Using algebra. If you think of foil basically this is a binomial. Um multiplying a binomial, you would really have T plus delta T. To the second is T. Plus delta T. Times itself. That's what it means to raise it to the second. So T plus delta T. Times T. Plus delta T. You can use foil to multiply two by no mules. And I'm gonna start changing the color just to help make it look a little less cluttered here. So we have 2.5 and that has two times two T. Plus delta T squared. Well, T plus delta T square T. Plus delta T. Times itself is going to be T squared plus two times T times delta T. Plus delta T squared. So all of that in princes has to be multiplied by 2.5. Uh And then we have to do, I'm just going to bring this down for now and then I'm going to scroll down on the screen and will continue simplifying this expression. At this point. It looks really confusing, but a lot of things will drop out nicely. So plus six times T plus delta T minus 2.5 T. Times T squared minus 16. And all of that has to be divided by delta T. So let's scroll down so that what we have in blue is near the top of this page so that we have some room to continue simplifying. And let's change the green for now. So we want to take the limit of this entire expression. This delta T approaches zero. Remember what we're trying to do? We're calculating the derivative of the given function. Using the definition of the rib there. It's a 2.5 has to be distributed as it multiplies each to three terms and Princes. So we're going to have 2.5 T squared plus 2.5 times to T delta T. 2.5 times two is five. So plus five times T delta take Then the 2.5 is going to talk to delta T squared plus 2.5 delta T squared. And then plus Now six has to multiply these two terms. So plus 60 plus six times delta T. Ah And then we're minus seeing 22.5 T squared. And we're minus ng or subtracting 2 16. And all of this once again is over the delta take. All right, let's move down a little bit more and bring what we have in green near the top. All right. This is where things will start simplifying a little bit because we'll be able to combine some terms. So, the limit of this entire expression is delta T approaches zero. Well, let's combine like terms. First of all. 2.5 T squared minus 2.5 T squared cancels. So that works out nice. Um we have plus 60. subtract 60 that cancels out lace. We have a five T Delta T. We have 2.5 delta T squared. And we have a six delta T. So, these are all unlike terms. So we're gonna write all three of these five, she delta T Plus 2.5 delta T square. Uh and then plus six Delta T. And all of that needs to be divided by delta. Take Now each of these three terms in the numerator are being divided by delta T. So we can separate ah these additions in the numerator and take each one of these three different terms and individually divided by delta T. So five tee times delta T. Over delta T. The delta teeth would cancel out. You would have five T. Here 2.5. Uh let's write down the plus time here. The 2.5 times delta T squared divided by delta T. Or delta T squared divided by delta T. Leads you just a delta T. Up here's a 2.5 delta T. And then here we have a plus and then six delta T divided by delta T. Two. Delta teas will cancel and you'll have just six. So now as we take the limit of this expression five T plus 2.5 times delta T. Plus six as delta T. Goes to zero. The delta T. Here is going towards zero out of these three terms. Uh This middle term to 2.5 delta T. Is the only term that has adult it T. So is the delta T approaches zero. Here's a delta T. It's approaching zero. So 2.5 uh times delta T. Is going to approach zero because if delta T. Is approaching 2.5 times delta T approaches zero. This five T. Is going to stay five T. Okay because it's delta T. That approaches zero, not the T. So five T stays five T. Likewise two plus six days plus six. So when we take this limit, when we take the limit of this expression, five T plus 2.5 times delta T plus six, as delta T approaches zero, we end up with five T Plus six. So that is the derivative of the function FF. T. Okay, so this just moved down a little bit. Yeah, this is our answer for F prime of T. And we evaluated it or we found it by using the definition, the limit definition of the derivative. So at prime of T equals five T plus six. Uh If we went to the top of the problem and we found a derivative, the old fashioned way, it should be five T plus six. All right. Uh If we found a derivative of this 2.5 times T squared would be two times 2.5 uh times T. Which would be five T plus derivative 60 is six. So derivative of this is five T plus six. And that's exactly what we found basically doing it. More complicated way using the definition of the derivative, the limit definition of the derivative last but not least. What is the domain of the F prime of T. The derivative function. Well, the derivative is F I'm sorry, the derivative is five times T plus six. This is defined for any real value of T. So basically, T could be any number. So the domain of F prime A. T once again is the entire X axis, any number on the X axis or basically any real number.

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# How do you find the limit of y=lnx/cscx  as x approaches 0 using l'hospital's rule?

Apr 28, 2016

${\lim}_{x \to {0}^{+}} \ln \frac{x}{\csc} \left(x\right) = 0$

#### Explanation:

First, assuming we are only consider real numbers, we can only take the limit of $\ln \left(x\right)$ as $x$ approaches $0$ from the positive direction, and so we will proceed as such.

Noting that ${\lim}_{x \to {0}^{+}} \ln \left(x\right) = - \infty$ and ${\lim}_{x \to {0}^{+}} \csc \left(x\right) = \infty$, we have ${\lim}_{x \to {0}^{+}} \ln \frac{x}{\csc} \left(x\right)$ as a $\frac{\infty}{\infty}$ indeterminate form. As such, we can apply L'hopital's rule.

${\lim}_{x \to {0}^{+}} \ln \frac{x}{\csc} \left(x\right) = {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \ln \left(x\right)}{\frac{d}{\mathrm{dx}} \csc \left(x\right)}$

$= {\lim}_{x \to {0}^{+}} \frac{\frac{1}{x}}{- \csc \left(x\right) \cot \left(x\right)}$

$= {\lim}_{x \to {0}^{+}} - \sin \frac{x}{x} \cdot \tan \left(x\right)$

If two functions converge at a point, then their product also converges at that point, and the limit of the product is the product of the limits of the original functions. Using that fact, together with the known limits ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$, and ${\lim}_{x \to 0} \tan \left(x\right) = 0$, we can factor our expression into the product of two limits:

${\lim}_{x \to {0}^{+}} \ln \frac{x}{\csc} \left(x\right) = {\lim}_{x \to {0}^{+}} - \sin \frac{x}{x} \cdot \tan \left(x\right)$

$= - \left({\lim}_{x \to {0}^{+}} \sin \frac{x}{x}\right) \left({\lim}_{x \to {0}^{+}} \tan \left(x\right)\right)$

$= - 1 \cdot 0$

$= 0$


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# Proof of the conservation of the energy functional for the Gross-Pitaevskii equation?

From the Gross-Pitaevskii equation $$i\hbar\frac{\partial\psi}{\partial t}=\left(-\frac{\hbar^2}{2m}\nabla^2+V+g|\psi|^2\right)\psi$$ using the variational relation $$i\hbar\frac{\partial\psi}{\partial t}=\frac{\partial\varepsilon}{\partial \psi^*}$$ we find the energy density $$\varepsilon=\frac{\hbar^2}{2m}|\nabla\psi|^2+V|\psi|^2+\frac{g}{2}|\psi|^4$$ The energy would be $E=\int d^3r \varepsilon$ and this is a prime integral of the motion, meaning it is a conserved quantity.

My questions are:

1) How do we get the variational relation?

2)How can we prove that $E$ is a conserved quantity?

• 1) You may read the chapter 7 of Pethick and Smith's book on BEC. 2) $\epsilon$ is energy density which is not conserve, only the total energy – unsym Jul 9 '14 at 23:24
• 1) Even if I read that book already I didn't remember that. Thank you. 2) Of course you are right, my mistake; I will amend the question – Semola Jul 9 '14 at 23:37
• Do you know a nice way to show the conservation of $E$ different from brute force computation? – Semola Jul 9 '14 at 23:41
• After an integration by parts, $\epsilon$ is almost the operator that gives the time-evolution, and one could perhaps try to copy the proof of the Ehrenfest theorem. – Robin Ekman Jul 10 '14 at 0:35
• I think it won't work because the exponential would depend on $|\psi|^2$ as well. – Semola Jul 11 '14 at 13:40

by computing $\partial_t E$ and using what given in the Gross-Pitaevskii for $\dot{\psi}$ and $\dot{\psi^*}$ one can check that all the terms cancel out so that $\partial_tE=0$.
• Do you think Dirac-Frenkel variational principle (which can be derived as a saddle point approximation for the coherent state Feynman path integral) with the wave functional ansatz $$|\{\psi(\bar{r},t)\}>=e^{\int_{V}d^3_{}\bar{r}\psi(\bar{r},t)\Psi_{}^{\dagger}(\bar{r})}|\ \text{vac}>$$ will lead to the Gross-Pitaevskii equation? – Sunyam Oct 6 '17 at 15:16


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# Wikipedia:Size of Wikipedia

 related topics {math, number, function} {rate, high, increase} {style, bgcolor, rowspan} {work, book, publish} {system, computer, user} {day, year, event}

This Wikipedia:Statistics page measures the size of the English-language edition of Wikipedia; mostly page and article count. There are currently 3,518,580 articles in the English Wikipedia.

Most of the earlier entries were extracted from Wikipedia:Announcements. Later entries are taken from observations of the new software's built-in article count features.

## Contents

### Wikipedia growth

Wikipedia's growth to date approximately follows a Gompertz growth model. This model is determined by the Gompertz function, $y(t)=ae^{be^{ct}}$, with parameters:

Some characteristics of this model are:

• a pivot point at which the growth is at its peak. For en.wikipedia.org this might have been in August 2006 with 60,000 new articles a month.
• a maximum to the number of articles of about 4.4 million (as determined by parameter a of the model). In reality there will always be new events and people to describe in the future, and these will lead to the creation of perhaps 12,000 new articles a month.

This model is related to the quantity (number of articles). The quality might still increase independently.

### Graphs of size and growth rate

The two graph images show: in the first graph, the projected total articles; and in the second graph, the monthly growth rate, as it has slowed (line sloping downward) since late 2006.

Detailed analysis of the data shows that from 2006 to 2009 the article growth rate followed a six-monthly cycle with faster growth in February and August than in May and November. This cycle does not appear in the growth-rate graph here, because the values shown in the graph have been averaged over periods of six months.

### Annual growth rate

For the English Wikipedia.


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# Convolution of exp(-a*norm(x)^2) and exp(-b*norm(x)^2) ?

1. Aug 14, 2010

### jefferson_lc

How do I compute convolution of exp(-a*norm(x)^2) and exp(-b*norm(x)^2) where a,b > 0 and x belongs to Rn?

I wonder if there is an easy way to compute this convolution using Fourier transform.

2. Aug 14, 2010

### Petr Mugver

The convolution becomes simple multiplication after Fourier transforming. The two functions you have are gaussian functions, their FT is still a gaussian function, so is the product, and finally so is the antitransform. All you need is the formula for gaussian transforms.

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# All Science Fair Projects

## Science Fair Project Encyclopedia for Schools!

 Search    Browse    Forum  Coach    Links    Editor    Help    Tell-a-Friend    Encyclopedia    Dictionary

# Science Fair Project Encyclopedia

For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.

# Predicate transformer semantics

Predicate transformer semantics is an extension of Floyd-Hoare Logic invented by Dijkstra and extended and refined by other researchers. It was first introduced in Dijkstra's paper "Guarded commands, nondeterminacy and formal derivation of programs". It is a method for defining the semantics of an imperative programming language by assigning to each command in the language a corresponding predicate transformer. A predicate transformer is a total function mapping between two predicates on the state space of a program.

The canonical predicate transformer of sequential imperative programming is the so-called "weakest pre-condition" wp(S,R). Here S denotes a list of commands and R denotes a predicate on the space, called the "post-condition". The result of applying this function gives the "weakest pre-condition" for S terminating with R true. An example is the following definition of the asignment statment:

$wp(x := E, R) = R_E^x$

This gives a predicate that is a copy of R with the value of x replaced by E.

An example of a valid calculation of wp for assignments with integer valued variables x and y is:

wp(x: = y - 5,x > 10) = (y - 5 > 10) = (y > 15)

This means that in order for the "post-condition" x > 10 to be true after the assignment, the "pre-condition" y > 15 must be true before the assignment. This is also the "weakest pre-condition", in that it is the "weakest" restriction on the value of y which makes x > 10 true after the assignment.

Dijkstra also defined alternative (if) and repetitive (do) constructs as well as a composition operator (;) using wp. The alternative and repetitive constructs used guarded commands to influence execution. Because of the rules he imposed on the definition of wp, both constructs allow for non-deterministic execution if the guards in the commands are non disjoint.

Unlike many other semantic formalisms, predicate transformer semantics was not designed as an investigation into foundations of computation. Rather, it is intended to provide programmers with a methodology to develop their programs as "correct by construction" in a "calculational style". This style was advocated by Dijkstra and others, and also developed further in a higher order logic setting by R.J Back in the Refinement Calculus.

Although the most common and most widely discussed because of their relevance to sequential programming, "weakest pre-conditions" are not the only predicate transformers. For example, Leslie Lamport has suggested win and sin as predicate transformers for concurrent programming.

## References

• Edsger W. Dijkstra. "Guarded commands, nondeterminacy and formal derivation of programs". Communications of the ACM, 18(8):453-457, August 1975. [1]
• Leslie Lamport. "win and sin: Predicate Transformers for Concurrency". ACM Transactions on Programming Languages and Systems, 12(3), July 1990. [2]
• Ralph-Johan J. Back et al. "Refinement Calculus: A Systematic Introduction, 1st edition". 1998. ISBN 0-387-98417-8
• Edsger W. Dijkstra: "A Discipline Of Programming" (A systematic introduction with examples.) ISBN 0-613-92411-8
03-10-2013 05:06:04


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0
Relationship Between Tool Deterioration and Cutting Force During Milling of a Nickel-Based Superalloy Using Cemented Carbide Tool

## Excerpt

Nickel-based superalloys have long been recognized as difficult-to-machine alloys. Chipping and breakage are understood as the dominant features of tool deterioration particularly in milling of these alloys. But how tool deterioration affects cutting force (Fc) has not been sufficiently established so that force model base deterioration detection methods can be practically developed. Thus, this study intends to better demonstrate the relationship between modes of tool deterioration and Fc. Milling experiments were conducted using 718Plus superalloy and uncoated cemented carbide tool and a commonly used set of milling parameters, with or without the use of coolant. During these experiments, Fc was recorded and in between milling passes the traditionally used flank wear value (VB) was precisely measured. We have observed that as milling progressed, along with the slow edge wear, edge chipping and fracturing intensified and Fc increased. However, before the final catastrophic failure, the rate of increase was relatively small in comparison to the scatters in Fc vs VBMax plot. Thus, a model based on Fc vs VBMax for detection of tool deterioration may not be suitable. We will explain the reasons for the scatters and relate the fracturing path to the effect of it to cutting edge contact area, thus to Fc.

Introduction
Experimental Procedures
Results and Discussion
Conclusions
Acknowledgments
References
\$25.00

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