text:
# Find Slope of A Straight Line

Calculator to find the slope of the straight line using the values of x and y coordinates.
 Y2: Y1: X2: X1:

 Line Slope (m) :

The important properties of a straight line is in how it angles away from the horizontal. This concept is reflected in something called the "slope" of the line.

We can find the slop (m) using formula :

$$m = \frac{(y_2 - y_1)}{(x_2 - x_1)} or \frac{(y_1 - y_2)}{(x_1 - x_2)}$$

Let's take the example to find out the slop of line passing through two ponits coordinates are (-4, 3) and (9, 8):
$$m = \frac{(y_2 - y_1)}{(x_2 - x_1)}$$ $$m = \frac{(8 - 3)}{(9 - (-4))}$$ $$m = \frac{(5)}{(9 + 4)}$$ $$m = \frac{5}{13}$$ $$m = 0.3846153846153846$$ $$m = 0.385$$
$$\text{or m }= \frac{(y_1 - y_2)}{(x_1 - x_2)}$$ $$m = \frac{(3 - 8)}{((-4) - 9)}$$ $$m = \frac{(-5)}{(-13)}$$ $$m = \frac{5}{13}$$ $$m = 0.3846153846153846$$ $$m = 0.385$$

format: plain


text:
If you’ve ever compiled a large software project, you’ve probably used make. If you’re like me, you probably think that make is some arcane tool that only fifty-year-olds with giant beards know how to use. This summer, I learned I was totally wrong, and make is actually really simple and really useful. I’m still not an expert, but all of the tutorials I could find online would have taken me five hours, and I think knowing five minutes1 of make now is way more useful than knowing five hours of make “someday”.

As an example, we’ll write a simple Makefile for a hypothetical site using Jekyll. In the project root, make a new file called Makefile, and put the following2 in it:

Now to build, just run make build. Now let’s say that we add a Javascript file some-js.js, and we want to use browserify to build it. We just add another rule:

Now make scripts will build our Javascript and stick it in js/bundle.js. Except now we have a problem: we need to run two commands, make scripts and make build, whenever we update things. That’s silly, and make has a solution. We just tell it that in order to make build it had better make scripts first, by modifying the first rule to read:

This, though, has the disadvantage that we always have to recompile the Javascript even if we didn’t modify it. Here’s where make really shines: instead of calling our make target scripts, we just name it after the file that’s generated, and put the Javascript source file as a dependency.

Now the Javascript bundle will only get updated when it needs to be. Lastly, to be safe, we should tell make that build is not an actual file, and therefore it need not check for a file of that name. We do this with the fake target .PHONY:

Lastly, make has variables that work somewhat similarly to shell variables. This is useful to make it easy to add more build dependencies. We use them like this:

Now if we add another javascript file to be compiled, we don’t need to list it separately. So here’s our complete Makefile:

There’s a lot more to learn about make: there are default rules that, for example, let one specify that all files of a particular extension should be treated the same way, and various other useful macros, but hopefully this is enough to be useful and enough to learn more.

1. I originally taught the above in exactly five minutes (enforced by boffer swords and pool noodles) for MIT ESP’s Firestorm, where we teach 5-minute classes on all manner of topics to MIT freshmen.

2. Those who have used Jekyll may notice that we could have just run jekyll build --watch once and been done. For the sake of the tutorial, pretend that that isn’t possible – it certainly doesn’t support all of the things we’ll be doing.


text:
CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
-   OpenFOAM (http://www.cfd-online.com/Forums/openfoam/)
-   -   low Mach compressible flows (http://www.cfd-online.com/Forums/openfoam/94282-low-mach-compressible-flows.html)

 venkataramana November 11, 2011 01:36

low Mach compressible flows

Dear all openfoam users,
Any body suggest which solver is suitable for The numerical simulation of low Mach compressible flows. Is there any preconditioning method available in openfoam.

Thanking you,

 francescomarra November 11, 2011 05:47

Dear venkataramana,

I am using successfully the rhoPisoFoam class of solvers for Low Mach compressible flows.

Which kind of application are you interested in ?

Do you really have a weakly compressible flow or a variable density (combustion) flow to solve ?

My best regards,

Franco

 nsreddysrsit November 11, 2011 17:36

Dear Franco,
I have one question,
Through rhopisoFoam or some other is it possible to solve for density = constant.

Regards,

 venkataramana November 11, 2011 17:46

Dear Franco,
I am interested in weakly compressible flows
I have some questions,
1) there are many solvers for compressible flows, how to identify which solver is suitable for weakly compressible flows
2)Particularly in openFoam is there any solver (compressible flow solver) to solve incompressible flows (density = 0). But we have icoFoam, simpleFoam for incompressible flows. Through rhopisoFoam or some other is it possible to solve for density = 0.

any comments or suggestions are welcome.

Regards,

 francescomarra November 14, 2011 05:40

Dear Venkataramana,

sorry for my late answer. I was offline during the weekend.

- what I know is that for weakly compressible flows you need a procedure that allows to go behind the limit of stability established by the Courant number, computed with the largest signal velocity that arise in your system. In the case of weakly compressible flows this is usually the u+c velocity, where u is the flow velocity and c the speed of sound. This can be achieved in several ways: preconditioning, artificial compressibility and projection are probably the most common approaches. I recognize the Piso algorithm to belong to the last class of approaches. Therefore, in practice, you need to recognize if in the solver you have chosen (the beauty of OpenFOAM is especially that, thanks to the wonderful cpp coding, you can immediately realize in the main code the general algorithm implemented), the iterative procedure corresponding to the projection algorithm is present.

- the second question is unclear to me. density = 0 is equivalent to vacuum conditions. This is not the meaning of incompressible flows, that mathematically correspond to ensure $\div \rho = 0$, being \rho the density and \div the divergence operator .

Kind regards,

Franco

 venkataramana November 14, 2011 06:44

Dear Franco,
I am sorry about the second question it was my mistake here density = constant
question number
2) Particularly in openFoam is there any solver (compressible flow solver) to solve incompressible flows (density = constant or Mach Number approaching 0 ). But we have icoFoam, simpleFoam for incompressible flows. Through rhopisoFoam or some other is it possible to solve for density = constant.

Regards,

 venkataramana November 14, 2011 07:00

one more question here
out of these approaches "preconditioning, artificial compressibility and projection " in openFoam,is there any solver adopting the above approaches?

Regards,

 francescomarra November 14, 2011 08:00

Dear Venkataramana,

the rhoPisoFoam solver, as well as many others suitable for combustion cases (that is my first subject of research work), is able to deal with cases where rho can be costant . It is a projection methos in the sense that a vector field of known divergence is sought at every time step. This leads to the elliptic equation for the pressure that give you the possibility to know how pressure disturbances propagate in the whole field at each time step. Several other details need to be specified to say if you are considering only really incompressible flow or weakly compressible flows.

I do not have a so deep knowledge of all the OpenFOAM solvers to say you how many solvers exists for weakly compressible flows. Combustion solvers are usually suitable for low Mach flows, even to the limit of rho=const.
However I do not know of any solver adopting a preconditioning approach. I have a vague recollection of a post in the forum about artificial compressibility method. Maybe a search in the forum could help you.

Kind regards,

Franco

 venkataramana November 14, 2011 10:58

Dear Franco,
Thank you for your kind information,

as per my knowledge
The projection method is an effective means of numerically solving time-dependent incompressible fluid-flow problems. It was originally introduced by Alexandre Chorin in 1967 and independently by Roger Temam as an efficient means of solving the incompressible Navier-Stokes equations. The key advantage of the projection method is that the computations of the velocity and the pressure fields are decoupled.

How it is applicable for compressible flows, any modification we need to do in the solver,
and rhopisoFoam developed developed for compressible flows

regards,

 All times are GMT -4. The time now is 02:55.


text:
## effective ample divisor

Hi,
can someone give me an intuitive definition of an effective ample divisor of a smooth projective variety?

thank you.


text:
# Graph a Hyperbola with vertices(2,4),(8,4), foci(-2,4), (12,4), center(5,4). Can someone help me find the equation for this?

Mar 15, 2018

generalize form of equation of hyperbola is ${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

Where,$\left(h , k\right)$ is the coordinate of the centre,

and, $a$ is distance between the centre and the vertices,and $c$ is the distance between the centre and the focus.

and, ${a}^{2} + {b}^{2} = {c}^{2}$

So,given, $\left(h , k\right) = \left(5 , 4\right)$

$a = \sqrt{{\left(5 - 2\right)}^{2} + {\left(4 - 4\right)}^{2}} = 3$

$c = \sqrt{{\left(5 - 12\right)}^{2} + {\left(4 - 4\right)}^{2}} = 7$

so,${b}^{2} = \left({7}^{2} - {3}^{2}\right) = 40$

So,the equation of the hyperbola becomes, ${\left(x - 5\right)}^{2} / 9 - {\left(y - 4\right)}^{2} / 40 = 1$

Now,see the graph below

graph{((x-5)^2)/9 - ((y-4)^2)/40=1 [-20, 20, -10, 10]}


text:
# Browse Graduate Theses and Dissertations at Illinois by Title

• (2014-05-30)
Modeling the 3D geometry of the face is an important research topic in computer graphics and computer vision. The applications for face models include computer animation and facial analysis. Over the past few decades, ...
PDF (9MB)
• (2014-05-30)
In this dissertation we study core problems in 3D face processing with their important applications in biometrics and graphics. We propose efficient and accurate feature based algorithms for facial feature detection, 3D ...
PDF (10MB)
• (1984)
This thesis consists of two parts. The first part deals with the question of whether localized magnetic moments are present on the 3d atoms, Cr and Mn, when dilutely doped in Nb and V. The second part deals with the ...
PDF (5MB)
• (2012-02-06)
Dysphagia has negatively affected the quality of life or even caused death for many head and neck cancer patients. Consequently, it is critical that scientists investigate the function of normal and abnormal swallowing in ...
PDF (6MB)
• (2013-05-24)
A cell-based biological machine is a set of sub-components consisting of living cells and cell-instructive micro-environments that interact to perform a prescribed task. Cell types such as neurons, muscle cells, and ...
PDF (12MB)
• (2013-05-24)
Direct-write assembly is used to fabricate 3D microperiodic scaffolds composed of hyaluronic acid (HA), a natural, biocompatible, biodegradeable polymer, for cell culture and tissue engineering applications. HA is ...
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• (2012-02-06)
An image is nothing but a projection of the physical world around us, where objects do not occur randomly but follow certain spatial rules. Many existing computer vision approaches tend to ignore this aspect of understanding ...
PDF (8MB)
• (1991)
This study chronicled the intertextual links generated by 8 able readers as they read a five-passage tableau. Each reader silently read the passages and reported aloud their thoughts during a $2{1\over2}$ to 3 hour session. ...
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• (2013-05-24)
The purpose of this study is to provide a useful guide for choral conductors who wish to perform the music of Chen Yi (b. 1953). The goal of this study is to search for an understanding of Chen Yi’s unique musical language, ...
PDF (58MB)
• (1969)
A study of the A2 meson has been made on data obtained from pi p interactions at 5 BeV/c in the 72"Alvarez liquid hydrogen bubble chamber. The p pi decay mode in the 4-constraint, p pi+ pi- pi- final state is treated ...
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• (1981)
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• (1982)
Self-consistent unrestricted Hartree-Fock calculations are presented for the adsorption of atomic oxygen onto the nickel oxide (100) surface. The perfect surface is shown to be non-reactive, but enhanced binding is found ...
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• (1986)
In this thesis the use of the Unrestricted Hartree-Fock technique augmented by many body perturbation theory techniques and symmetry projection methods in the study of the energies of the excited states of atomic, molecular ...
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• (1989)
The study investigated which cues in the input are used by classroom second language learners in order to assign tense at the sentence level: lexical items (temporal adverbials), gestures (indicating temporal references), ...
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• (2010-05-19)
Nitrate is a contaminant of interest in soils and water. It has been linked to chronic human illness and a variety of environmental problems, including hypoxia in coastal waters. Previous work has shown that nitrate is ...
PDF (663KB)
• (1989)
Regulation of T lymphocyte proliferation from aged rats was explored by activating lymphocytes to proliferate with the T cell lectin Con A, the calcium ionophores A23187 and ionomycin and the protein kinase C activator, ...
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• (2013-05-24)
Over the past few decades, both researchers and commercial farmers have attempted to find means to accurately predict crop yields during the growing season. Modern science has been successful in mapping the genes of crops ...
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• (2014-05-30)
This project focused on the abrasion resistance of concrete railway crossties. In the first phase of this project, a series of high performance concrete mixes was developed that utilized different replacement levels of ...
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• (1996)
We have measured the absolute $\rm\sp2H(\gamma,p)n$ angular distribution using the Illinois LASA detector at the Saskatchewan Accelerator Laboratory's tagged photon facility. Using 40 to 90 MeV photons, we aimed to measure ...
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• (1996)
We have measured the absolute 2H( 1 ,p )n angular distribution using the Illinois LASA detector at the Saskatchewan Accelerator Laboratory's tagged photon facility. Using 40 to 90 MeV photons, we aimed to measure the ...
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text:
# All Questions

81 views

### Fermats Little Theorem, primitive root [closed]

So I am studying for finals and I am not able to solve the problem: Let $p = 3 * 2^{11484018}- 1$ be a prime with 3457035 digits. Find a positive integer $x$ so that $2^x\equiv 3\pmod p$ Any ...
90 views

68 views

### What cipher algorithm is this routine using? [closed]

I transfered a code routine from Java to Python: ...
Can an adversary who haven't seen the message before forge the signature of that message? The adversary has seen $σ_1 = (m_1)^d \bmod N$ and $σ_2 = (m_2)^d \bmod N$ He forges $m' = m_1\cdot m_2$ \$σ' ...


text:
# How does an operator transform under time reversal?

We know that a time-reversal operator $T$ can be represented as $$T=UK$$ where $U$ is some unitary operator and $K$ is the complex conjugation operator.

Then under time-reversal operation, a quantum state $|\psi\rangle$ will transform as the following: $$|\psi^R\rangle=T|\psi\rangle=UK|\psi\rangle=U|\psi^*\rangle$$ If we require time-reversal symmetry to the system, then we need to have $$\langle\psi^R|O^R|\phi^R\rangle=\langle\psi|O|\phi\rangle$$ where $|\psi\rangle$ and $|\phi\rangle$ are some arbitrary quantum states and $O$ is some operator. From the above equation, we would have $$\langle\psi^*|U^{\dagger}O^RU|\phi^*\rangle=\langle\psi|O|\phi\rangle$$ So based on this equation, how do we obtain the result (given in the book "random matrices" by Mehta) that $$O^R=UO^TU^{\dagger}$$ where $O^T$ means the transpose of $O$.

My second question is that, what if we do NOT assume time reversal symmetry?

• I am reading your question, but I have a doubt. Time reversal is not only complex conjugate, what it has to do is also to transpose the items on which it acts (vectors, matrices). – Sofia Dec 3 '14 at 22:33
• Also, for reference \langle and \rangle produce nice looking brackets, $\langle \varphi | \psi \rangle$ instead of $< \varphi | \psi >$, if you care about that sort of thing. – user12029 Dec 4 '14 at 0:17
• @Sofia also I don't think time reversal operator needs to transpose the operand – M. Zeng Dec 4 '14 at 7:00

Time reversal is not only complex conjugate, what it does is also to transpose the items on which it acts (vectors, matrices).

$$T\langle \phi|\hat{O}|\psi\rangle = \langle \psi T|\hat{O}|T \phi\rangle.$$

Notice the change of places of the functions in the right wing with respect to the left wing. Also, I used the fact that $\hat{O}$ is unchanged at time-reversal.

Now we do the following change which is allowed under the integral if the two functions vanish at infinity:

$$\langle UK\psi|\hat{O}|UK\phi\rangle = \langle \phi|U\hat{O}U^\dagger|\psi\rangle.$$

So, we got the time-reversed of $\hat{O}$.

• I don't think we can assume that the operator is unchanged under time-reversal – M. Zeng Dec 4 '14 at 6:55
• @Timo: you may think or not, but this is the assumption in the text of the question, see the 3rd equation. – Sofia Dec 4 '14 at 21:32
• I'm sorry but your argument seems contradictory. We are supposed to find out how the operator transform under time-reversal, but you assumed that $O$ is unchanged under time-reversal. My assumption is that that the inner product is invariant instead of the operator. – M. Zeng Dec 5 '14 at 2:19
• @Timo: when reversing the time, we see that the particles reverse their movement. One starts from the final state, (we introduce the change ⟨ϕ|†), we apply on it Ô†, and as the exercise hints, the vector obtained should be now projected on the initial state (which therefore should appear as |ψ⟩†. (I am not sure which are the initial and final states, but if I trust the LHS of the 3rd equality, i.e. that he took the transposed of the product of components, it seems that |ψ⟩ is the initial state. But, let me return to you some later. – Sofia Dec 5 '14 at 10:39
• @Timo: please see why I thought that Ô is a symmetrical operator. Before the 3rd equality in the question, is written "if we require time-reversal symmetry, ..." – Sofia Dec 5 '14 at 14:17

One problem with your formula that $$T$$ factorizes as the product of a unitary operator $$U$$ and complex conjugation $$K$$ is that 'complex conjugation' is meaningless in a Hilbert space a priori.

Let me be more formal about this; consider a vector $$\psi \in H$$, with $$H$$ a $$n$$-dimensional Hilbert space, that is, isomorphic to $$\mathbb{C}^n$$. How is an (abstract) vector space isomorphic to the canonical Hilbert space $$\mathbb{C}^n$$? In the following way: pick a basis $$\{e_i\}_{i=1,...,n}$$ of $$H$$. Then decompose $$\psi$$ in this basis:

$$\psi = \sum_i \psi_i e_i \ .$$

Finally map $$\psi$$ on the $$n$$-tupel $$(\psi_1,...,\psi_n)$$.

As you can see, this isomorphism, lets denote it by $$E$$, very strongly depends on the chosen basis.

Let's ignore this anyway. On $$\mathbb{C}^n$$ we can define complex conjugation $$K$$, it is simply the map

$$K(\psi_1,...,\psi_n) = (\overline{\psi_1},...,\overline{\psi_n}) \ .$$

Hence we can define a complex conjugation $$K_E$$ on $$H$$ simply by

$$K_E = E^{-1} \circ K \circ E \ .$$

Now let's see what happens if we change a basis; consider another basis $$E' = \{e'_i\}_{i=1,...,n}$$ with $$e_i = \sum_j M_{ij} e'_j$$. Then considering $$M_{ij}$$ as the matrix of an operation on $$\mathbb{C}^n$$:

$$K_{E} = E^{-1} \circ K \circ E = E'^{-1} \circ M^{-1} \circ K \circ M \circ E' = E'^{-1} \circ \overline{M}M^{-1} K \circ E' \ ,$$

i.e. unless $$\overline{M} = M$$, $$K_E \neq K_{E'}$$. Hence there is no invariant notion of complex conjugation in a complex Hilbert space.

The statement is true that if we fix a basis $$E$$, we may write complex conjugation w.r.t. any other basis as

$$K_{E'} = U_{E,E'} K_E \ ,$$

where $$U_{E,E'}$$ is a unitary. Note that complex conjugation depends on $$E$$ only through the equivalence class $$[E]$$ of basis connected to $$E$$ through real transformations.

In this language one could say that a time-reversal operation is a choice of preferred equivalence class of basis. That is, at least, if $$T^2 = 1$$. If $$T^2 = -1$$, we should map $$H$$ to a $$\mathbb{H}^{n/2}$$, where $$\mathbb{H}$$ are the quaternions.

This leaves you with a recipe for computing the time-reversed operator, for $$T^2=1$$: simply represent $$O$$ in a real basis (a basis invariant under $$T$$), then take the complex conjugate of this matrix.

Notice that neither of these manipulations depend on $$T$$ being a symmetry.


text:
1

### JEE Main 2019 (Online) 10th January Morning Slot

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to -
A
200
B
150
C
400
D
360

## Explanation

at t = 0, A0 = ${{dN} \over {dt}}$ = 1600C/s

at t = 8s, A = 100 C/s

${A \over {{A_0}}}$ = ${1 \over {16}}$ in 8 sec

Therefor half life is t1/2 = 2 sec

$\therefore$   Activity at t = 6 will be 1600${\left( {{1 \over 2}} \right)^3}$

= 200 C/s
2

### JEE Main 2019 (Online) 10th January Evening Slot

Consider the nuclear fission

Ne20 $\to$ 2He4 + C12

Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -
A
8.3 MeV energy will be released
B
energy of 11.9 MeV has to be supplied
C
energy of 12.4 MeV will be supplied
D
energy of 3.6 MeV will be released

## Explanation

Ne20      $\to$  2He4 + C12

Q – value, EB = (BE)react $-$ (BE)product

= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)

= 9.72 MeV
3

### JEE Main 2019 (Online) 11th January Morning Slot

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980$\mathop A\limits^ \circ$. The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV$\mathop A\limits^ \circ$)
A
4a0
B
9a0
C
25a0
D
16a0

## Explanation

Energy of photon $= {{12500} \over {980}} = 12.75eV$

$\therefore$  Electron will excite to n = 4

Since 'R' $\propto$ n2

$\therefore$  Radius of atom will be 16a0
4

### JEE Main 2019 (Online) 11th January Evening Slot

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
A
${{25} \over {16}}$ $\lambda$
B
${{27} \over {20}}$ $\lambda$
C
${{16} \over {25}}$ $\lambda$
D
${{20} \over {27}}$ $\lambda$

## Explanation

For M $\to$ L steel

${1 \over \lambda }$ = K $\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$

for N $\to$ L

${1 \over {\lambda '}}$ = K$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$

$\lambda ' = {{20} \over {27}}\lambda$


text:
MARLEY (Model of Argon Reaction Low Energy Yields)  v1.0.0 A Monte Carlo event generator for tens-of-MeV neutrino-nucleus interactions in liquid argon
marley::InterpolationGrid< FirstNumericType, SecondNumericType > Class Template Reference

One-dimensional function y(x) defined using a grid of ordered pairs (x,y) and an interpolation rule. More...

#include <InterpolationGrid.hh>

## Public Types

enum  ExtrapolationMethod { Zero, Endpoint, Continue, Throw }
Method to use for computing y(x) when the x value lies beyond the grid boundaries. More...

using Grid = std::vector< OrderedPair >

using GridConstIterator = typename std::vector< OrderedPair >::const_iterator

enum  InterpolationMethod {
Constant = 1, LinearLinear = 2, LinearLog = 3, LogLinear = 4,
LogLog = 5
}
Method to use for interpolating between (x,y) grid points. More...

using OrderedPair = std::pair< FirstNumericType, SecondNumericType >

## Public Member Functions

InterpolationGrid (InterpolationMethod interp_method=InterpolationMethod::LinearLinear, ExtrapolationMethod extrap_method=ExtrapolationMethod::Zero)
Create an InterpolationGrid without any grid points.

InterpolationGrid (const Grid &grid, InterpolationMethod interp_method=InterpolationMethod::LinearLinear, ExtrapolationMethod extrap_method=ExtrapolationMethod::Zero)
Create an InterpolationGrid from a vector of ordered pairs. More...

InterpolationGrid (const std::vector< FirstNumericType > &xs, const std::vector< SecondNumericType > &ys, InterpolationMethod interp_method=InterpolationMethod::LinearLinear, ExtrapolationMethod extrap_method=ExtrapolationMethod::Zero)
Create an InterpolationGrid from vectors of x and y values.

OrderedPair & at (size_t j)
Get a reference to the jth ordered pair from the grid.

const OrderedPair & back () const
Returns a reference to the last ordered pair.

void clear ()
Delete all ordered pairs from the grid.

ExtrapolationMethod extrapolation_method () const
Get the ExtrapolationMethod used by this InterpolationGrid.

const OrderedPair & front () const
Returns a reference to the first ordered pair.

std::function
< SecondNumericType(FirstNumericType)>
get_interpolating_function ()
Get a std::function object that represents y(x) for this InterpolationGrid.

void insert (FirstNumericType x, SecondNumericType y)
Add a new ordered pair (x, y) to the grid.

SecondNumericType interpolate (FirstNumericType x) const
Compute y(x) using the current InterpolationMethod.

InterpolationMethod interpolation_method () const
Get the InterpolationMethod used by this InterpolationGrid.

GridConstIterator lower_bound (const GridConstIterator &begin, const GridConstIterator &end, FirstNumericType x) const
Returns a const_iterator to the first element of the grid for which the x value is not less than (i.e. greater than or equal to) x.

void set_interpolationMethod (InterpolationMethod method)
Set the InterpolationMethod to use.

void set_interpolationMethod (ExtrapolationMethod method)
Set the ExtrapolationMethod to use.

size_t size () const
Get the number of ordered pairs on the grid.

GridConstIterator upper_bound (const GridConstIterator &begin, const GridConstIterator &end, FirstNumericType x) const
Returns a const_iterator to the first element of the grid for which the x value is greater than x.

## Detailed Description

### template<typename FirstNumericType, typename SecondNumericType = FirstNumericType> class marley::InterpolationGrid< FirstNumericType, SecondNumericType >

One-dimensional function y(x) defined using a grid of ordered pairs (x,y) and an interpolation rule.

This class implements the one-dimensional interpolation rules described in the ENDF-6 formats manual for a grid of (x,y) pairs.

Template Parameters
 FirstNumericType type for the x values SecondNumericType type for the y values

## Member Enumeration Documentation

template<typename FirstNumericType, typename SecondNumericType = FirstNumericType>

Method to use for computing y(x) when the x value lies beyond the grid boundaries.

The table below describes how y(x) is calculated using each of the allowed ExtrapolationMethod settings. In the notation used below, the lowest x value is called $$x_\text{left}$$, the greatest x value is called $$x_\text{right}$$, and their corresponding y values are called $$y_\text{left}$$ and $$y_\text{right}$$, respectively. The ExtrapolationMethod is only relevant when $$x < x_\text{left}$$ or $$x > x_\text{right}$$.

Extrapolation Methods
MethodDescription
Zero$$y = 0$$
Endpoint$$y = \begin{cases} y_\text{left} & x < x_\text{left} \newline y_\text{right} & x > x_\text{right} \end{cases}$$
ContinueUse the usual InterpolationMethod formula with
$$y = \begin{cases} y_1 = y_\text{left} \text{ and } y_2 = y_{\text{left}+1} & x < x_\text{left} \newline y_1 = y_{\text{right}-1} \text{ and } y_2 = y_\text{right} & x > x_\text{right} \end{cases}$$
where $$y_{\text{left}+1}$$ is the y value corresponding to the second-lowest x value, and $$y_{\text{right}-1}$$ is the y value corresponding to the second-highest x value.
template<typename FirstNumericType, typename SecondNumericType = FirstNumericType>

Method to use for interpolating between (x,y) grid points.

For $$x_1 \leq x < x_2$$, the table below describes how the corresponding y value is calculated using each of the allowed InterpolationMethod settings.

Interpolation Methods
MethodDescription
Constant$$y = y_1$$
LinearLinear$$y = y_1 + \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$
LinearLog$$y = \exp\left[\ln(y_1) + \frac{\ln(y_2) - \ln(y_1)}{x_2 - x_1}(x - x_1)\right]$$
LogLinear$$y = y_1 + \frac{y_2 - y_1}{\ln[x_2] - \ln[x_1]}[\ln(x) - \ln(x_1)]$$
LogLog$$y = \exp\left[\ln(y_1) + \frac{\ln(y_2) - \ln(y_1)}{\ln(x_2) - \ln(x_1)}(\ln[x] - \ln[x_1])\right]$$

## Constructor & Destructor Documentation

template<typename FirstNumericType, typename SecondNumericType = FirstNumericType>
 marley::InterpolationGrid< FirstNumericType, SecondNumericType >::InterpolationGrid ( const Grid & grid, InterpolationMethod interp_method = InterpolationMethod::LinearLinear, ExtrapolationMethod extrap_method = ExtrapolationMethod::Zero )
inline

Create an InterpolationGrid from a vector of ordered pairs.

Todo:
Add error checks for the supplied grid

The documentation for this class was generated from the following file:


text:
## BeccaB003 one year ago Help with lengths. Thanks!

1. BeccaB003

2. Nnesha

$\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$

3. BeccaB003

I need to calculate the length of NO. I know how to work these problems EXCEPT I don't know whether to use sin cos or tan.

4. Nnesha

alright as you can the functions i posted hve 3 variables so to find NO you should know one angel and one side so we have to use sin $\huge\rm sin(60)= \frac{ opposite }{hyp }$

5. Nnesha

let NO=x so x is the opposite side of angle 60 and hyp is 24 so substitute $\huge\rm sin(60)= \frac{ x}{24}$ solve for x :=)

6. BeccaB003

So, in a problem like this, whenever you have the information of one angle and one side you always use sin?

7. Nnesha

nope it depends on sides and angles like which angle and side they gave you

8. Nnesha

for this problem we have to apply sin bec like i said earlier we need one angle and one side so given is hyp and angle and we need to find NO which is opposite side of angle that's why sin

9. BeccaB003

So, sin 60 = .866 * 24 = 20.785

10. Nnesha

mod should be on degree!!

11. Nnesha

e*

12. xapproachesinfinity

sin60=root3/2 no need to write before simplifying

13. Nnesha

yes that's right

14. xapproachesinfinity

i meant no decimal before simplifying :)

15. BeccaB003

My calculator is on degree. I checked. Also, I need to know How to know whether a problem is sin cos or tan?? I want to be able to know right off from looking at the problem which one I need to use.

16. Nnesha

well if you have 3 variables formula you should know values of 2 variables to find 3rd one |dw:1434509817378:dw| i need opposite side and given is hyp so i should apply sin which is equal to opposite over hyp

17. BeccaB003

Also, my answer option isn't 20.785 These are the only options I have and they don't make sense when I do the problem: $12\sqrt{2}$ 12 $12\sqrt{3}$ $24\sqrt{3}$

18. xapproachesinfinity

it is just convenient that you write roots as they are

19. xapproachesinfinity

you see if the answer were written in that form, no decimal no calculator is needed!

20. Nnesha

okay then :D

21. Nnesha

it depends on answer choices :=)

22. xapproachesinfinity

not really most often radical need to be written as they if we cannot simplify them

23. xapproachesinfinity

as they are*

24. BeccaB003

I'm confused. I did the problem but it isn't one of the options.... can you convert it??

25. xapproachesinfinity

that is why i'm saying that lol you see sin60=x/24 ===> x=24sin60=24 root(3)/2=12root(3)

26. Nnesha

yes otherwise they should know the unit circle pattrn or at least memorize it

27. xapproachesinfinity

28. xapproachesinfinity

yeah must have the unit circle :)

29. BeccaB003

Also, @Nnesha thank you for your example about recognizing sin, cos, and tan. After I've completed the current problem, can you draw another example like you did and see if I can recognize the correct one; sin, cos, or tan?

30. BeccaB003

Thank you! @xapproachesinfinity

31. xapproachesinfinity

no problem

32. Nnesha

they are not allowed to have unit circle 'n the final exam :=)

33. Nnesha

:3

34. xapproachesinfinity

i meant they need to know some way or the other not taking it in the exams hahaha

35. xapproachesinfinity

i actually memorized them from table lol but if i forgot any one of them i use special triangles

36. BeccaB003

haha! oh bother..... my exam is gonna be fun. I never learned about a 'unit circle' in my class.....

37. Nnesha

^^^^see

38. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @xapproachesinfinity i actually memorized them from table lol but if i forgot any one of them i use special triangles $$\color{blue}{\text{End of Quote}}$$ same i memorized it 2 no time for 45-45-9- and 30-60-90 theorem :D

39. Nnesha

oh forgot about that okay give me a sec

40. Nnesha

|dw:1434511171692:dw| find angle B :=_

41. Nnesha

|dw:1434511566103:dw| or this one :=)


text:
# Goldbach's Conjecture 解题报告

Description

In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
Every even number greater than 4 can be
written as the sum of two odd prime numbers.

For example:
8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.

Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.

Input

The input will contain one or more test cases.
Each test case consists of one even integer n with 6 <= n < 1000000.
Input will be terminated by a value of 0 for n.

Output

For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

Sample Input

8
20
42
0

Sample Output

8 = 3 + 5
20 = 3 + 17
42 = 5 + 37
#include<math.h>
#include<stdio.h>
int h[1000001]={0};
int main()
{
int  n,a,b,c,d,k=0;
h[1]=1; //1不是素数。
b=(int)sqrt(1000000);
for(a=2;a<=b;a++)
{
if(h[a]==0)
{
d=1;
for(c=2;d<1000000;c++)
{
d=a*c;  //能相乘得到的都不是素数。
h[d]=1;
if(d==1000000)
break;
}
}
}
while(scanf("%d",&n)!=EOF)
{
k=0;
if(n==0)
break;
else
{
for(b=2;b<n;b++)
{
if(h[b]==0&&h[n-b]==0)
{
k=1;
printf("%d = %d + %d\n",n,b,n-b);
break;
}
}
if(k==0)
printf("Goldbach's conjecture is wrong.\n");
}
}
return 0;
}

• 本文已收录于以下专栏：

## K - Goldbach's Conjecture解题报告

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举报原因： 您举报文章：深度学习：神经网络中的前向传播和反向传播算法推导 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字)


text:
## Intermediate Algebra (12th Edition)

$6(-9)=6\times-9=-54$ The numbers being multiplied together have different signs, so their resulting product is negative.


text:
# 2.4: Process Quality and Structural Quality

Unlike standard industry practices, high-quality practices place an emphasis on promoting positive child outcomes. The key to a high-quality program is contingent upon what happens inside the classroom environment. Let’s examine how both process quality and structural quality work together to influence positive outcomes for children.

Process quality refers to the types of interactions that occur throughout the day between the teachers, children, families, and administrators. Process quality also considers the types of materials that are available for the children to use, as well as the activities that children engage in throughout their day. Lastly, process quality takes into account the health, well-being and safety of the children. Structural quality on the other hand, refers to the features and characteristics of a program. More specifically, the class size, teacher-to-child ratios, teacher qualifications and experiences, teacher pay scale, along with the allotted square footage for play space define quality. Although process quality is thought to have a more direct impact on child outcomes as compared to structural quality , researchers and leaders in the field of early care and education agree that process and structural indicators are interrelated, and when combined together they promote the highest quality experiences.

For example, when groups are smaller, teachers tend to have more positive, supportive, and stimulating interactions with children. Warm and nurturing interactions are directly linked to children's social competence and future academic success, and such interactions are essential to high quality. Early childhood teachers who are more highly qualified and have smaller groups can more effectively provide individualized, responsive learning opportunities. Finally, higher teacher wages have consistently been linked to higher process quality. [8]

2.4: Process Quality and Structural Quality is shared under a CC BY license and was authored, remixed, and/or curated by Gina Peterson and Emily Elam.


text: how many volt does a stock coil produce. MathJax reference. While this demands more voltage from the ignition, it also creates a fatter spark in the combustion chamber. Jan 31, 2020 #1 ... Cut'n Paste from the FSM "The direct ignition system generates approximately 40,000 volts". But I do study and I'm an extra class ham radio operator and a former auto mechanic building race engines for friends that have won races with the engine. You need to read up on spark ignition systems and could perhaps start with something like this. An ignition magneto, or high tension magneto, is a magneto that provides current for the ignition system of a spark-ignition engine, such as a petrol engine.It produces pulses of high voltage for the spark plugs.The older term tension means voltage.. The # 19 heavy wire on the outside is 150 feet long. In older systems with long spark plug leads the lead resistance may be several 10's kOhms and along with the higher resistance of the secondary coils this limits the current that will flow. Your output power is short, sharp pulses. In other words there is no reason that a coil that was originally sold for use in a 6 volt system can't be used in a 12 volt system. Because the spark time is small, on average the coil uses 2A, or 24W continuously. Lv 6. Well come to find out I was only getting 5 volts to my coil even at start up.. so I hooked up a switch to supply 12v to my coil when starting sence my system wasn't doing it.. Now I'll admit I'm not formally educated. So .00075×5 is .00375 miliamps. A ballast coil should read around 1.5 Ohms or around 6 volts (maximum 3.4 Ohms is ideal ... and providing a direct 12 volt feed from the starter solenoid/relay or fuse box to the ignition coil has proved to be a success and has given good results. MSD has a whole series of Blaster 2 and 3 cannister-type ignition coils that can replace stock coils used in various OE ignition systems. i bought a msd coil with 48000 volts and i wanna know what i can set my plug gap at. The resistor regulates the voltage, keeping it at 6 volts, allowing the original 6 volt coil to be used in a 12 electrical system. Start and Close Depending to Engine Status, Understanding LED powerstrip and power requirements, How to sort and extract a list containing products. Easy math 5A÷100 = 0.050A or about 50 ma. Spark line current in these is still limited by design, typically to under 200 mA. This all happens instantly. This is because the ignition coil does not produce enough current to do a lot of harm. If it is an old system then it should be ~ 12 v as shown below when the car is not not running as no switching is taking place. Check out an old Motor's Repair Manual, they have a pretty good write up on ignition systems. Podcast 300: Welcome to 2021 with Joel Spolsky, Cleaning up 4.8V and 10.8V DC power affected by spark ignition system, Power a circuit and read output from tablet microphone port, Boost regulator: Unexpected output voltage. mine is a '99 model. Lets say you have a 1 ohm series resistor in your harness. I run a large (physical size) 6 volt battery, that cranks my EN330 over quite quickly for being almost 10 years old, in the dead of winter. This means that current moves in one direction, from the positive battery terminal to the negative battery terminal. Maybe I am? 6 VOLT ELECTRICAL TIPS & TRICKS I’ll use the term “VOM” to denote volt/ohm/amp/meter. Flame Thrower II coils have lower resistance, 0.6 ohms for use with Ignitor II ignition systems as well as many other high-energy ignitions. If it's 12 volts, use the GT40. The kind of ignition coil found in a car engine. Older vehicles have a distributor, ignition coil, and spark plug wires. Is this unethical? The electricity creates sparks within the spark plug, and the sparks ignite the fuel in the engine, enabling the chain saw to run. One time holding the pole at road Atlanta ga. Some of the newer cars ignition coils put out 20 - 40,000 volts. how many volts does the coil put out to the spark plug on a bolens weed eater? In a CDI ignition system,a storage capacitor is charged to a high voltage (a few hundred volts) by the flywheel charging coil and, at the proper moment quickly discharged into the primary winding of the ignition coil,which is actually a stepup pulse-transformer. If you have a ballast coil (e.g. The dial reading should be 12 volts or more. In engines of medium to high compression this kind of voltage is necessary to reliably arc across the gap on a spark plug and make enough fire to ignite the fuel/air mixture in a cylinder. Work the starter switch, and the reading should fall, but not below 10.5 volts. High V low I.If you mean the input current, for a short duration it's high I and relatively low V. The forums are retiring in 2021 and are now closed for new topics and comments. Was this . Electrical currents run through the ignition coil and transmit electricity to the spark plug. Are there any sets without a lot of fluff? And known voltages to calculate a rough but close answer. This posed a dilemma for the ignition … You can think of it as the tipping bucket principle... With the switched closed energy is poured into the bucket at a slow rate. Some coils are designed to start on 12 and then back down to 8. When the spark is required, the a "switch" is opened. Why confuse the OP. The secondary output is a short very high amplitude pulse which fires the spark plug. In the case … TRIUMPH IGNITION COILS The GT40 requires 12 volts on the positive side, whereas the GT40R requires 9 volts which is accomplished by using an inline resistor in the power supply wire to the coil. then the better for you. Losses abounded. What are these capped, metal pipes in our yard? So a lightbulb uses 100w x 1Hour = 0.1KW.Hrs of electricity in an hour. The ignition coil produces 40,000V and the spark is measured at 50mA. Another is 5 amps. Driven properly they can also be used to create a Jacobs Ladder display I am having trouble with getting enough voltage to the coil. This does not seem right, what am I missing? Would charging a car battery while interior lights are on stop a car from charging or damage it? 9 years ago, To be honest, I don't know. Let's say the primary voltage is 12 volts at 5 amps. While ignition coils will generally last for the life of a vehicle, faulty spark plugs or spark plug wires can send irregular voltage surges through the ignition coil, causing the coil to burn out. I was taught that a 12 volt (external resistor) coil was just the same as a 6 volt; interchangeable.. IE 16 000 volt with out it trying to spark under the environment of compressino in the cylinder and 8000 volts under the regular working conditions of a sparkplug. Tap the primary coil on the same connection that enters the secondary coil, which the other end of, terminates on the sparkplug, and to ground.....Note which wire on the engine serves to ground the ignition to kill the engine. A plain old 12 volt coil still only runs on about 6-8 volts. Meaning that IF (neglecting component failure) the spark was continuous then the battery would only last around 6 minutes? Ignition Coil. Likely has a 12 volt primary side. The circuit requires 2A to run. Instead of shorting it to ground, you put the coil at the end of that wire. If I could harness the output power and my battery is 200Wh does this mean I can run a 2000W device for 8.3 hours, as the input is only 24W from the battery? YOu can find the 12v wire on the back of your ign switch, and run a new line from there directly to your coil for 12v. What's the miliamperage of the secondary voltage. The ignition coil is really a transformer that steps up the 12-volt current of the battery to somewhere in the neighborhood of 25,000 volts. Also, the ignition coil will only produce high … Spark gap time is in the 1 - 2 ms range for most systems depending on inductor values and total series resistance for the circuit. All that power inside the coil will fry it. when on, a 100W light-bulb is using 100W all the time. Your coil will draw 2A, so it is running at 24W most of the time. i'm told that the pickup/pulse puts out so little that a better voltmeter is needed to quantify the voltage. The power dissipated in the spark gap is limited both in time and voltage/current. Instead of shorting it to ground, you put the coil at the end of that wire. Is it always necessary to mathematically define an existing algorithm (which can easily be researched elsewhere) in a paper? Measure the pulse-width and duty-cycle of your output and then average that 2kW over time. If I go straight from the positive terminal on the battery to the positive terminal on the coil, it starts and runs great. Asking for help, clarification, or responding to other answers. This does not seem right, what am I missing? 2 Answers. The essential task of firing the sparkplugs to ignite a gasolene-air mixture is carried out by a process which employs Faraday's law. If it's 9 volts, use the GT40R. What is the meaning of output voltage and current? Even though an ignition coil will produce about 10,000 volts, it actually less dangerous than the 110 volt that comes out of a wall socket. answer. That's what the resistor (external or internal) is for, to drop the voltage to lower the amps through the points so they survive. So if you put 250 V on the primary and you get 250,000 volts on the secondary, the dielectric breaks down and conducts electricity and the 250,000 volts gets internally dissipated inside the coil. When switch opens, due to capacitor, magnetic field collapses. At my work, we repair small engines such as chainsaws, string trimmers, cutoff saws, etc etc. 6 volt coils will generally have very low internal resistance built in, anywhere from . Since the coil is just a large inductor, the current still wants to flow through it so the voltage at the switch and attacked spark plug suddenly jumps to thousands of volts. This means your total circuit resistance is now 1.0 ohms plus 1.6 ohms for the coil. A common automotive ignition coil can generate an output voltage on the order of 30,000 volts. To stop the coil from burning out, the 12 volts from the switch is usually passed through a resistor block (mounted on or near the coil), which reduces the voltage. 24W/2000W implies < 1.2ms per second duty cycle and if >1000 Hz then is < 1us duration. mine is the hei with the coil seperate from cap. Is there logically any way to "live off of Bitcoin interest" without giving up control of your coins? The ignition coil is truly an amazing little device. The low resistance helps to produce up to 45,000 … @Andyaka - Yes it is and your punishment is to answer questions like this for the rest of your existence mwha mwha mwha (Devilish laugh). Steps up input V by x100 but Reduces input Current (5 amps) also by 100. Is there a number on the coil, or some way of finding out whether it's a 6 or 12 volt? When your engine is running your coil is drawing 2A from the battery MOST of the time through the shown switch which is normally closed when the ignition switch is on. Many race engines with crank trigger systems or locked-out … If the reading falls below 10.5 volts and the starter turns slowly or … The 1949 – 1951 engines crank 130 rpm (without an … Power (WATTS) is often miss-understood. Your input power is (probably) smooth(ish) DC. At breakover (zero current) the voltage at the gap will rapidly rise to 40 kV, but once the plasma is created the instantaneous voltage will drop to only a few hundred volts. It does sound excessive given the number of times I've pushed my finger into a plug cap when trying to diagnose bad starting engines. It’s the part of a car’s ignition system that takes the battery’s 12-volt output (called low-tension current) and transforms it into as much as 45,000 volts (called high-tension current) before then supplying it to the engine’s spark plugs. You can test the resistance of the coil using a voltmeter and measure the Ohms (resistance) across the +’ve and –’ve terminals. Shadetreee. 1988 full size, 4.3 v6 tbi It is my understanding that ballast resistors are used to drop the voltage to the coil because the stock coil primary side is wound for less than 12 volts. . If I could harness the output power and my battery is 200Wh does this mean I can run a 2000W device for 8.3 hours, as the input is only 24W from the battery? When your engine is running your coil is drawing 2A from the battery MOST of the time through the shown switch which is normally closed when the ignition switch is on. i have not a single number to go by on this. Is binomial(n, p) family be both full and curved as n fixed? Modern systems typically use a coil-on-plug design which get's rid of one of the spark gaps (double the delivered spark energy), reduces series resistance and increases the plasma current (much 'hotter' sparks). By Tom Benford. However, more voltage does not always mean better in the world of ignition coils. How can I enable mods in Cities Skylines? What does the coil look like? Ignition coils take energy from the battery (generally 12 volts) and convert it to a high enough voltage (50,000 volts or more) to create a spark. You WOULD use more energy on average, but it will bog down to the point that a spark will no longer be available at some point. The CD ignition draws voltage supply from the battery and steps up the voltage to 500 volts and higher, it then stores the voltage in the ignition’s capacitor. _____ It may require only 1,000 volts for it to ignite when the engine is idling and cylinder pressure is low, but then need 20,000 volts to climb a hill under full throttle. Thanks for contributing an answer to Electrical Engineering Stack Exchange! The coil contains a primary winding, a slot iron core, and a … Is this hell? Your coil will draw 2A, so it is running at 24W most of the time. Favorite Answer. I can't remember how many terminals on the coil, I think one. I'm running the short coil, about 3 inch high, which I believe is original. Regardless, the functionality is the same. Power may be high from stored energy arc in extreme short duration but energy E[Joules]=P*t is low. In older systems with a distributor gap, you had to create two plasma connections, one for the distributer and one in the cylinder spark plug. If a disembodied mind/soul can think, what does the brain do? Most factory type oem coils have a published turns ratio of around only 100 : 1 secondary HV to primary LV !! On PerTronix, If it is an Ignitor I, we like to see a full 12 volts to the + side of the coil and a coil with an internal resistance of at least 1.5Ω. It contains both primary and secondary winding circuits. It only takes a minute to sign up. STARTERS: The starter current draw on a Ford V-8 through 1948 is 550 amps (does not include V-8 60’s) . You are correct in thinking that the output power is in kW, but unfortunately you left out an important element ...time. The currents that flow in the secondary (in the spark gap) can also very high (in the hundreds of mA). Either way, somewhere in the 4-6 amp range while the car is running. Check out an old Motor's Repair Manual, they have a pretty good write up on ignition systems. the ignition coils outpuit 16kV as a "free voltage" and 8 kV as a "working voltage". Your magneto, does put out a pulse, which is routed to the ignition coil to boost that pulse to adequate voltage to fire the sparkplug. I reuse this wire as it unwinds very eazy and use it to design 12 volts to 36 volt power supplies with a mosfet circuit driven from a 12 volt battery. The average vehicle ignition coil puts out 20,000 to 30,000 volts, and coils used in racing applications are capable of 50,000 or more volts at a constant rate. SO some extent yes, but actually no. Previous owner put the 12 volt battery in it, and it has been used just like that for some time. I have a 12V battery connected to a circuit driving an ignition coil. One was 16000kv. Monday, May 23rd, 2011 AT 12:28 AM CARADIODOC EXPERT; Hi guys. So essentially I am filling the bucket up and then releasing it as a spark. The Sparkplug Firing The voltage demand on the ignition coil by the sparkplug varies with load on the engine. To learn more, see our tips on writing great answers. this is an easy question. The 8 volts is on the coil when the points are closed and the resistor is dropping 4 volts. This new voltage is then routed to the distributor via the coil wire, which is just like the spark plug wires, only normally much shorter. The coil primary winding contains 100 to 150 turns of heavy copper wire. An ignition coil is a type of pulse transformer that multiplies the low voltage received from the battery or alternator to many thousands of volts when the breaker points open and close. Quoting from a … This rate of change of magnetic flux will generate large voltage around 25 KV in the secondary, enough to ionize air between the spark plug gap and there is spark to ignite fuel. If so, would that mean I've likely got the 12 volt ignition coil? The first order of business,in diagnosing the ignition control module (ICM) and ignition coil, is to verify that both are getting juice (10 to 12 Volts). Even though an ignition coil will produce about 10,000 volts, it actually less dangerous than the 110 volt that comes out of a wall socket. Is quite powerful coil seperate from cap you agree to our terms of service, privacy and! Ohms plus 1.6 ohms process which employs Faraday 's law policy and cookie policy resistor in your.... Be design limited ( by the sparkplug firing the sparkplugs to ignite a gasolene-air mixture is carried by! Would you say its my cdi box just as they should..... would you say its cdi. When the spark plug on a chainsaw while the car is running at 24W most of the …. Script to be considered is required, the a  working voltage '' output is a of. Etc etc plug gap at site for electronics and electrical Engineering professionals students. In a past comment with features that put you in control of your ignition a battery. What is the meaning of output voltage and current math 5A÷100 = 0.050A or about 50 mA a! 2020 # 1... Cut ' n paste from the primary coil 16Kv current 3.75ma... Way, somewhere in the neighborhood of 25,000 volts, or some way of finding out whether 's! A gasolene-air mixture is carried out by a process which employs Faraday law... Interest '' without giving up control of your output and then treated as by. 6 or 12 volt coil still only runs on about 6-8 volts battery while interior are! Terminals on the coil, or 24W continuously as invisible by society click to see answer. Distributor, ignition coil does resistor in your harness would charging a car engine you say its my cdi?... Algorithm ( which can easily be researched elsewhere ) in a past comment chainsaws, string trimmers cutoff. 12-Volt ignition coil produces how many volts does an ignition coil put out and the reading does not fall, there is a of. Rough but close answer switch closes after and the coil ( bucket ) would not have enough time fill! 'S Repair Manual, they have a pretty good write up on ignition systems to get lot. Volts across a sparkplug in an automobile when you have a pretty good write up on spark ignition.. 100 rpms you had a continuous spark, what am I missing previous owner put the.. 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text:
# Solving a matrix equation $AX=XB$ in a CAS

I have the following computational problem. Let $N$ be a positive integer and $A\in \mathbb{C}^{2N\times 2N}$, $X\in \mathbb{C}^{2N\times 4}$ and $B\in \mathbb{C}^{4\times 4}$. I want to solve the following equation for the matrix $X$ $$A\cdot X= X\cdot B.$$ I found the following paper online (http://www.jstor.org/stable/2099652) but I haven't been able to implement it in Maple or Magma and I don't know what I should be looking for. Does someone know if it is implemented somewhere already?

-
Looks kinda like a Sylvester equation. SylvesterSolve in Maple. – Calle May 18 '11 at 17:54

This is a Sylvester equation, $AX-XB = 0$. This can be solved in Maple by SylvesterSolve and in Matlab by lyap and in Mathematica by LyapunovSolve.
In general, you can get $X$ in a vectorized form by solving $$((I_4 \otimes A) - (B^T \otimes I_{2n})) \operatorname{vec} X = 0$$ for example by Gaussian elimination. $\otimes$ is the Kronecker product and $\operatorname{vec} X$ is the columns of $X$ put on top of each other in a vector.
Hello @Calle I have a question with respect to this, how should be the solution if i have to solve $AXH+AXH-BH=0$?, in my case all matrices are square and A,B,H are constant matrices, as well I need X matrix. – Gina Torres Oct 10 '13 at 10:29


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# Question - Equality of Matrices

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#### Question

If [[x-y,z],[2x-y,w]]=[[-1,4],[0,5]] find the value of x+y.

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# Power Laws of Rational Exponents

There are two types of power rules with rational exponents. Here is the two power laws in algebraic form with proofs to use them as formulas in mathematics.

01

$\large b^{\frac{1}{n}} \,=\, \sqrt[\displaystyle n]{b}$

02

### Power of a Radical Exponent Rule

$\large b^{\frac{m}{n}} \,=\, \sqrt[\displaystyle n]{b^{\displaystyle m}}$


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# Basic clarification: Solving equation by squaring, but end up getting non-existent answers

This is going to be extremely elementary given the caliber of questions being posted. I was going over something basic and think I'm overlooking an extremely fundamental rule.

$$2x-1 = \sqrt{6x+1}$$

By inspection, the solution is x=5/2. However, to solve algebraically, say I square both sides.

That leaves me with.

$$4x^2 - 4x + 1 = 6x + 1$$ $$\implies 4x^2 - 10x = 0 \implies x(2x-5)=0$$

It would appear $x=0$ is now present along with the correct answer.

If someone could clarify for me why squaring out the squareroot sign adds a new (unwelcome) answer I would greatly appreciate. Especially since $x=0$ is not out of the domain of the original equation $x> -\frac 16$, which definitely includes $x=0$.

Is it because my original equation has a power of $1$, which means I have one solution, however, when I square it, I'm essentially adding a new "x" and upping the power to $2$, giving me $2$ solutions. Thus, if I were to keep upping the power, there would be more repeated solutions at $x=0$? Which would mean, in some senses to solve those equations for $x$, dividing both sides by $x$ to reduce it back to the original power of $1$ would mean $x$ cannot indeed be $0$ as a solution.

Regards.

• If we start with either $a=b$ or $a=-b$ and square, we get $a^2=b^2$. Thus solutions to the latter include the solutions of BOTH of the former equations. Indeed, $x=0$ is a perfectly good solution to $2x-1 =-\sqrt {6x+1}$ – lulu Jun 10 '16 at 21:43
• One might also take the viewpoint that algebra doesn't care about our arbitrary choice that $\sqrt{u}$ means the nonnegative square root of $u$, and gives us all the square roots. When $x = 0$, we do have that the LHS is a square root of $1$. This is basically what lulu said, I just like anthropomorphism :) – pjs36 Jun 10 '16 at 22:00
• Fundamentally, it's because $x \mapsto x^2$ is not injective on the domain $\mathbb{R}$. – MathematicsStudent1122 Jun 11 '16 at 2:30

Since the square of $2x-1$ is the same as the square of $1-2x$, squaring your initial equation you introduce also the solution of the equation $$1-2x=\sqrt{6x+1}$$

In another way: The equation $$2x-1=\sqrt{6x+1}$$ require that $2x-1\ge 0$ because the square root is always a positive number, so the solution must be such that $x \ge \frac{1}{2}$ and this inequality selects the correct solution of the squared equation.

Suppose we assume that $x = 1$.

Squaring both sides, we now have $x^2 = 1$, which is undeniably true.

From this we can conclude that $x = 1$ or $x = -1$. This is true too, since a disjunction of a true statement and a false statement is true. But the disjunction has lost information which was present in the original statement: you no longer know which of the two statements is true.

The point is that a polynomial equation is like an exclusive disjunction: it tells you that $x$ takes on exactly one of $n$ (the degree of the polynomial) possible values, but it doesn't tell you which.

The square root is an example of an operator that returns two results. You can have the positive square root or the negative square root. Squaring both sides caused the negative square root to be added to the solution set. You must therefore limit your domain to the positive square root and reject 0 as an answer.


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# 5.1 Convex optimization-based methods  (Page 2/2)

 Page 2 / 2

A newer algorithm called “l1_ls" [link] is based on an interior-point algorithm that uses a preconditioned conjugate gradient (PCG) method to approximately solve linear systems in a truncated-Newton framework. The algorithm exploits the structure of the Hessian to construct their preconditioner; thus, this is a second order method. Computational results show that about a hundred PCG steps are sufficient for obtaining accurate reconstruction. This method has been typically shown to be slower than first-order methods, but could be faster in cases where the true target signal is highly sparse.

## Fixed-point continuation

As opposed to solving the constrained formulation, an alternate approach is to solve the unconstrained formulation in [link] . A widely used method for solving ${\ell }_{1}$ -minimization problems of the form

$\underset{x}{min}\phantom{\rule{0.277778em}{0ex}}\mu {\parallel x\parallel }_{1}+H\left(x\right),$

for a convex and differentiable $H$ , is an iterative procedure based on shrinkage (also called soft thresholding; see [link] below). In the context of solving [link] with a quadratic $H$ , this method was independently proposed and analyzed in [link] , [link] , [link] , [link] , and then further studied or extended in [link] , [link] , [link] , [link] , [link] , [link] . Shrinkage is a classic method used in wavelet-based image denoising. The shrinkage operator on any scalar component can be defined as follows:

$\mathrm{shrink}\left(t,\alpha \right)=\left\{\begin{array}{cc}t-\alpha \hfill & \mathrm{if}t>\alpha ,\hfill \\ 0\hfill & \mathrm{if}-\alpha \le t\le \alpha ,\mathrm{and}\hfill \\ t+\alpha \hfill & \mathrm{if}t<-\alpha .\hfill \end{array}\right)$

This concept can be used effectively to solve [link] . In particular, the basic algorithm can be written as following the fixed-point iteration: for $i=1,...,N$ , the ${i}^{\mathrm{th}}$ coefficient of $x$ at the ${\left(k+1\right)}^{\mathrm{th}}$ time step is given by

${x}_{i}^{k+1}=\mathrm{shrink}\left({\left({x}^{k}-\tau ▽H\left({x}^{k}\right)\right)}_{i},\mu \tau \right)$

where $\tau >0$ serves as a step-length for gradient descent (which may vary with $k$ ) and $\mu$ is as specified by the user. It is easy to see that the larger $\mu$ is, the larger the allowable distance between ${x}^{k+1}$ and ${x}^{k}$ . For a quadratic penalty term $H\left(·\right)$ , the gradient $▽H$ can be easily computed as a linear function of ${x}^{k}$ ; thus each iteration of [link] essentially boils down to a small number of matrix-vector multiplications.

The simplicity of the iterative approach is quite appealing, both from a computational, as well as a code-design standpoint. Various modifications, enhancements, and generalizations to this approach have been proposed, both to improve the efficiency of the basic iteration in [link] , and to extend its applicability to various kinds of $J$ [link] , [link] , [link] . In principle, the basic iteration in [link] would not be practically effective without a continuation (or path-following) strategy [link] , [link] in which we choose a gradually decreasing sequence of values for the parameter $\mu$ to guide the intermediate iterates towards the final optimal solution.

This procedure is known as continuation ; in [link] , the performance of an algorithm known as Fixed-Point Continuation (FPC) has been compared favorably with another similar method known as Gradient Projection for Sparse Reconstruction (GPSR) [link] and “l1_ls” [link] . A key aspect to solving the unconstrained optimization problem is the choice of the parameter $\mu$ . As discussed above, for CS recovery, $\mu$ may be chosen by trial and error; for the noiseless constrained formulation, we may solve the corresponding unconstrained minimization by choosing a large value for $\mu$ .

In the case of recovery from noisy compressive measurements, a commonly used choice for the convex cost function $H\left(x\right)$ is the square of the norm of the residual . Thus we have:

$\begin{array}{cc}\hfill H\left(x\right)& ={\parallel y-\Phi x\parallel }_{2}^{2}\hfill \\ \hfill ▽H\left(x\right)& =2{\Phi }^{\top }\left(y-\Phi x\right).\hfill \end{array}$

For this particular choice of penalty function, [link] reduces to the following iteration:

${x}_{i}^{k+1}=\mathrm{shrink}\left(\left({x}^{k}-\tau ▽H{\left(y-\Phi {x}^{k}\right)}_{i},\mu \tau \right)$

which is run until convergence to a fixed point. The algorithm is detailed in pseudocode form below.

 Inputs: CS matrix $\Phi$ , signal measurements $y$ , parameter sequence ${\mu }_{n}$ Outputs: Signal estimate $\stackrel{^}{x}$ initialize: ${\stackrel{^}{x}}_{0}=0$ , $r=y$ , $k=0$ . while ħalting criterion false do 1. $k←k+1$ 2. $x←\stackrel{^}{x}-\tau {\Phi }^{T}r$ {take a gradient step} 3. $\stackrel{^}{x}←\mathrm{shrink}\left(x,{\mu }_{k}\tau \right)$ {perform soft thresholding} 4. $r←y-\Phi \stackrel{^}{x}$ {update measurement residual} end while return $\stackrel{^}{x}←\stackrel{^}{x}$

## Bregman iteration methods

It turns out that an efficient method to obtain the solution to the constrained optimization problem in [link] can be devised by solving a small number of the unconstrained problems in the form of [link] . These subproblems are commonly referred to as Bregman iterations . A simple version can be written as follows:

$\begin{array}{cc}\hfill {y}^{k+1}& ={y}^{k}+y-\Phi {x}^{k}\hfill \\ \hfill {x}^{k+1}& =argmin\phantom{\rule{3.33333pt}{0ex}}J\left(x\right)+\frac{\mu }{2}{\parallel \Phi x-{y}^{k+1}\parallel }^{2}.\hfill \end{array}$

The problem in the second step can be solved by the algorithms reviewed above. Bregman iterations were introduced in [link] for constrained total variation minimization problems, and was proved to converge for closed, convex functions $J\left(x\right)$ . In [link] , it is applied to [link] for $J\left(x\right)={\parallel x\parallel }_{1}$ and shown to converge in a finite number of steps for any $\mu >0$ . For moderate $\mu$ , the number of iterations needed is typically lesser than 5. Compared to the alternate approach that solves [link] through directly solving the unconstrained problem in [link] with a very large $\mu$ , Bregman iterations are often more stable and sometimes much faster.

## Discussion

All the methods discussed in this section optimize a convex function (usually the ${\ell }_{1}$ -norm) over a convex (possibly unbounded) set. This implies guaranteed convergence to the global optimum. In other words, given that the sampling matrix $\Phi$ satisfies the conditions specified in "Signal recovery via ${\ell }_{1}$ minimization" , convex optimization methods will recover the underlying signal $x$ . In addition, convex relaxation methods also guarantee stable recovery by reformulating the recovery problem as the SOCP, or the unconstrained formulation.

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Graphene has a hexagonal structure
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# Tag Info

1

This is not generally true. While the empirical mean of the series is 2, the second moment of the process isn't necessarily equal to 2. This number is just a reasonable estimate of the mean if and only if the process is stationary and has a finite second moment. By subtracting 2 from the observations you would not necessarily get a mean zero series. You ...

1

This operation is called de-meaning or more informally making the series/signal zero-mean. The "de-mean"ed version of $Y$ is surely $y=\{-1 -1,0,0,1,1\}$ as you wrote. But, this operation shouldn't be confused with the definition of zero-mean time series, i.e. a random process / signal is zero mean when $E[X(t)]=0$. You can make the signal zero-mean by ...

1

Methods like k-means rely on a distance measure like the euclidean distance. The intention of the authors was probably to make the clustering problem on somewhat trivial data sets like iris harder by adding additional features, as you correctly assumed, that are not very discriminative with respect to the actual cluster structure in the data. How I ...

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text:
# classproperty¶

class astropy.utils.decorators.classproperty(fget=None, doc=None, lazy=False)[source]

Bases: property

Similar to property, but allows class-level properties. That is, a property whose getter is like a classmethod.

The wrapped method may explicitly use the classmethod decorator (which must become before this decorator), or the classmethod may be omitted (it is implicit through use of this decorator).

Note

classproperty only works for read-only properties. It does not currently allow writeable/deletable properties, due to subtleties of how Python descriptors work. In order to implement such properties on a class a metaclass for that class must be implemented.

Parameters
fgetcallable

The function that computes the value of this property (in particular, the function when this is used as a decorator) a la property.

docstr, optional

The docstring for the property–by default inherited from the getter function.

lazybool, optional

If True, caches the value returned by the first call to the getter function, so that it is only called once (used for lazy evaluation of an attribute). This is analogous to lazyproperty. The lazy argument can also be used when classproperty is used as a decorator (see the third example below). When used in the decorator syntax this must be passed in as a keyword argument.

Examples

>>> class Foo:
...     _bar_internal = 1
...     @classproperty
...     def bar(cls):
...         return cls._bar_internal + 1
...
>>> Foo.bar
2
>>> foo_instance = Foo()
>>> foo_instance.bar
2
>>> foo_instance._bar_internal = 2
>>> foo_instance.bar  # Ignores instance attributes
2

As previously noted, a classproperty is limited to implementing read-only attributes:

>>> class Foo:
...     _bar_internal = 1
...     @classproperty
...     def bar(cls):
...         return cls._bar_internal
...     @bar.setter
...     def bar(cls, value):
...         cls._bar_internal = value
...
Traceback (most recent call last):
...
NotImplementedError: classproperty can only be read-only; use a
metaclass to implement modifiable class-level properties

When the lazy option is used, the getter is only called once:

>>> class Foo:
...     @classproperty(lazy=True)
...     def bar(cls):
...         print("Performing complicated calculation")
...         return 1
...
>>> Foo.bar
Performing complicated calculation
1
>>> Foo.bar
1

If a subclass inherits a lazy classproperty the property is still re-evaluated for the subclass:

>>> class FooSub(Foo):
...     pass
...
>>> FooSub.bar
Performing complicated calculation
1
>>> FooSub.bar
1

Methods Summary

 deleter(fdel) Descriptor to change the deleter on a property. getter(fget) Descriptor to change the getter on a property. setter(fset) Descriptor to change the setter on a property.

Methods Documentation

deleter(fdel)[source]

Descriptor to change the deleter on a property.

getter(fget)[source]

Descriptor to change the getter on a property.

setter(fset)[source]

Descriptor to change the setter on a property.


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# Lower-bound complexities for finding common elements between two unsorted arrays

I'm facing some problems that deal with finding common elements between unsorted arrays and I'd like to know whether there are well-known lower-bounds for the worst-case and, eventually, what are these lower-bounds.

The problems are pretty simple:

Given two unsorted arrays of distinct integers, $A$ and $B$, of size $m$ and $n$, determine all the common elements between the two arrays(the output must be sorted)

Which I believe has a lower-bound complexity of $\Omega((m+n)\log(\min(m,n)))$ (and $O(1)$ space complexity, if we exclude the input arrays), even though I cannot find any proof of this fact. [I found an algorithm with this complexity, so I(hopefully) am not underestimating the complexity.]

The other problem adds some restrictions(and hence I'd expect to be able to lower the complexity):

Given two unsorted arrays of distinct integers, $A$ and $B$, of size $m$ and $n$; knowing that common elements between the arrays have the same relative order in both arrays and that for every couple of consecutive common elements, their distance is at most $k$(constant), determine all the common elements between the two arrays maintaining their relative order and using at most $O(k)$ memory in addition to the input arrays.

I've tried to think about this second problem but I cannot see how the restrictions change the complexities. What puzzles me is that I believe that finding a single couple of elements between two unsorted arrays is $\Theta((n+m)\log(\min(n,m)))$, and since this is a special instance of this second problem then the restrictions do not add anything to the problem itself.

Are my guesses correct, and if so where can I find proofs for these lower-bounds? Do the restrictions change anything at all or the solution for the first problem is the best we can achieve in both cases?

Probably my questions can be summarized by the following:

Given two arrays of distinct integers $A$ and $B$ of size $m$ and $n$, what is the lower-bound complexity for finding one common element?

Because, once a common element is found, the second problem can be solved in linear time.

Edit

The algorithm(s) that I thought are pretty simple:

For problem 1: Build two heaps for $A$ and $B$(which takes linear time and can be done in-place), then compare the minimum of the heaps, if they match print it and remove both, otherwise remove the smallest one and continue until the heaps are empty(this clearly takes $O((m\log(m) + n\log(n)) = O(\max(m,n)\log(\max(m,n)))$).

An other solution is to sort one array in-place, scan the other array and use bisection search to find matches. If we sort the smallest array (with size $m$) then the complexity is $O(m\log(m) + n\log(m)) = O(\max(m,n)\log(\min(m,n)))$. This technique yields the values in the other in which their are found in the biggest array, and it can be used to solve the second problem.

But, as you can see, I'm not using the extra restrictions at all and both algorithms use $O(1)$ space instead of $O(k)$(yes, it's still constant but $O(k)$ should give a bit more freedom).

• If you want to use it in real work (software development, ....), just hash them and find common elements (most hashing algorithm are false positive but this happens very rarely), otherwise as Yuval said you reading too much into this. – user742 May 22 '13 at 19:16
• @SaeedAmiri I'm not interested in using this in real work, I just wanted to know whether there was a know lower-bound for the worst case. And, yes I'd definitely use hashing if it were for real work. – Bakuriu May 22 '13 at 20:28

For the first question, when $m = n$, you can get an easy $\Omega(n\log n)$ lower bound (in appropriate computation models) by taking $A = B$. More generally, following the method outlined here, it seems that (assuming $n \geq m$) you can get a lower bound of $\Omega(\log (m! \binom{n+1}{m})) = \Omega(n\log m)$ by considering the problem of Set Intersection. (In the proof of Theorem 4, fix the order of the $X$s, and vary the order of the $Y$s and their location.)
• I've added a brief description of the algorithms I devised. Modifying input arrays is allowed. I'll look into the link you posted, thank your. Regarding the second problem, the restriction on the constant $k$ makes me think that something like counting-sort could be possible, but I really see no way to use it in that sense. Am I right that comparison based approaches are the only sensible ones in both problems? – Bakuriu May 22 '13 at 7:17