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OpenMath Content Dictionary: limit1

Canonical URL:
http://www.openmath.org/cd/limit1.ocd
CD Base:
http://www.openmath.org/cd
CD File:
limit1.ocd
CD as XML Encoded OpenMath:
limit1.omcd
Defines:
above, below, both_sides, limit, null
Date:
2004-03-30
Version:
3
Review Date:
2006-03-30
Status:
official

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but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

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both this document and the derived work are Content Dictionaries
then the derived work must include a different CDName element,
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Dictionary Group whose name is, for example, `math' containing
Content Dictionaries named `math1', `math2' etc., then you should
not name a derived Content Dictionary `mathN' where N is an integer.
However you are free to name it `private_mathN' or some such.  This
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This CD holds the basic notion of the limits of unary functions whilst its variable tend (either from above, below or both sides) to a particular value.

limit

Role:
application
Description:

This symbol is used to denote the limit of a unary function. It takes 3 arguments: the limiting value of the argument, the method of approach (either null, above, below or both_sides) and the function.

Example:
The limit of sin(x) as x tends to 0
$\underset{x\to 0}{\mathrm{limit}}\phantom{\rule{0.1em}{0ex}}\mathrm{sin}\left(x\right)$
Signatures:
sts

 [Next: both_sides] [Last: null] [Top]

both_sides

Role:
constant
Description:

This symbol is used within a limit construct to show the limit is being approached from both sides. It takes no arguments.

Example:
The limit of sin(x) as x tends to 0 from either direction
$\underset{x\to 0}{\mathrm{limit}}\phantom{\rule{0.1em}{0ex}}\mathrm{sin}\left(x\right)$
Signatures:
sts

 [Next: above] [Previous: limit] [Top]

above

Role:
constant
Description:

This symbol is used within a limit construct to show the limit is being approached from above. It takes no arguments.

Example:
The limit of sin(x) as x tends to 0 from above
$\underset{x\to {0}^{+}}{\mathrm{limit}}\phantom{\rule{0.1em}{0ex}}\mathrm{sin}\left(x\right)$
Signatures:
sts

 [Next: below] [Previous: both_sides] [Top]

below

Role:
constant
Description:

This symbol is used within a limit construct to show the limit is being approached from below. It takes no arguments.

Example:
The limit of sin(x) as x tends to 0 from below
$\underset{x\to {0}_{-}}{\mathrm{limit}}\phantom{\rule{0.1em}{0ex}}\mathrm{sin}\left(x\right)$
Signatures:
sts

 [Next: null] [Previous: above] [Top]

null

Role:
constant
Description:

This symbol is used within a limit construct to avoid specifying the method of approach to the limit. It takes no arguments.

Example:
The limit of sin(x) as x tends to 0
$\underset{x\to 0}{\mathrm{limit}}\phantom{\rule{0.1em}{0ex}}\mathrm{sin}\left(x\right)$
Signatures:
sts

 [First: limit] [Previous: below] [Top]

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Circle - A question about filling it.

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Hello :) let's say i have a circle with a center x,y and radius r. i can manually "draw" it into an array, for later display etc. what i am having trouble with is filling the circle with a certain color. there is the "cheap" way, of simply drawing lines from the center, to the calculated circle borders in a certain color, but it's less than perfect. i clearly remember there was a recursive way to do it, but cannot remember it. it is also important i do not use any api to draw the circle and fill it, but do it on my own. if you can help in any way, it would be great :) thank you :)

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If you already have an algorithm for drawing the border of the circle, just find each pair of pixels on each row on the border, and you know that every pixel between them is inside the circle and thus should be colored in.

So, if (X1, Y) and (X2, Y) are on the circle, everything between (X1, Y) and (X2, Y) are in the circle.

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Thinking about how you could to do this recursively, the first thing that comes to my mind is this:

Since you already have the border, you can start at the center pixel and fill it with your color. Check the color of every of the four direct neighbors of that pixel and if the neighboring pixel isn't colored yet, fill it with your color and check every of the four neighbors of that pixel and so on...

This algorithm is easily implemented recursively and will fill your entire circle until it hits the border.

[Edited by - DraganO on January 21, 2009 2:51:47 AM]

Thanks a lot :)

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I would use a rasterization technique to fill the circle, since you already have an equation describing the border and it's a convex region. A recursive approach requires a ton of stack state, is slower, and more suited for arbitrary enclosed regions where you can't easily use other techniques and brute force is the best option.

Like andur mentioned, start at the top of your circle, which is y' = y + r (or y - r is your vertical axis is inverted), and calculate both values of x' using the circle equation. They should be the same at the top and bottom of the circle, but not in the middle. Continue on to the next row, y'-1, and calculate two more values for x'. Fill in the row between those values. Keep going until you reach y' = y - r, which is the bottom of the circle, and you'll be done!

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One easy way is just to frame the circle with an imaginary square and only colour in the region of the square that is within the bounds of the circle:

def drawFilledCircle((x, y), radius, colour):    radiusSquared = radius * radius    for u in range(-radius, radius + 1):        for v in range(-radius, radius + 1):            distanceSquared = u * u + v * v            if distanceSquared < radiusSquared:                plot(x + u, y + v, colour)

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What I'd do is start with a square around the circle, then recursively check whether the square intersected the circle (In which case I'd quarter the square and recurse over each quarter), or if the square was inside the circle (In which case I'd fill the square), or if the square was outside of the circle (In which case I'd stop recursing on it).

It's a simple way to quickly fill a circle, or any other shape you can check against a square. Of course you want to stop at a certain point; probably once the square has a width of 1.

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# Deriving Rodrigues' formula

Consider the function $$f_n(x)=(x^2-1)^n$$..........(20)

Differentiating this equation we get the second order differential equation, $$(1-x^2)f_n''+2(n-1)xf_n'+2nf_n=0$$..................(22)

We wish to differentiate this n times by use of Leibniz's formula, $$\frac{d^n}{dx^n}A(x)B(x)=\sum^n_{k=0}\frac{n!}{k!(n-k)!}\frac{d^kA}{dx^k}\frac{d^{n-k}B}{dx^{n-k}}$$......................(23)

Applying this to (22) we easily get $$(1-x^2)f_n^{(n+2)}-2xf_n^{(n+1)}+n(n+1)f_n^{(n)}=0$$......................................(24)

which is exactly Lergendre's differential equation (1-49). This equation is therefore satisfied by the polynomials $$y=\frac{d^n}{dx^n}(x^2-1)^n$$.....................(25)

The Legendre polynomials $P_n(x)$ are normalized by the requirement $P_n(1)=1$. Using $$y=2^nn!$$...............(26) for x=1,

We get $$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$...........................(27)

How does the author relate (24) with (25)? Where did he get equation (25)? And how do you do this normalization? Is this normalization the same as that in physics whereby we ensure that the probability does not exceed 1? This is the source.

• Have you plugged $(25)$ into $(24)$ to see what happens? – Olivier Oloa Oct 9 '16 at 6:10
• I will get the Legendre DE. But where did the author come up with (25)? – newbie125 Oct 9 '16 at 6:38
• In many problems in analysis, we proceed by trial and error. – Olivier Oloa Oct 9 '16 at 8:09
• The only term that survives, in the expansion of $\frac{d^n}{dx^n}(x^2-1)^n$ by the Cauchy formula you mentioned above, is when $k=n$, in this case we have $\binom n k=1$, $\frac{d^k}{dx^k}(x-1)^n=n!$ and $\frac{d^{n-k}}{dx^{n-k}}(x+1)^n=(1+1)^n=2^n.$ Do you see it? – Olivier Oloa Oct 9 '16 at 10:20
• It is not equal to $0$. You have $\frac{d^1}{dx^1}(x-1)^1=1$, $\frac{d^2}{dx^2}(x-1)^2=2 \times 1$ (you differentiate twice), $\frac{d^3}{dx^3}(x-1)^3=3\times 2 \times 1$, and so on. – Olivier Oloa Oct 9 '16 at 13:20

The author noticed that $f_n^{(n)}$ is a solution of Legendre's equation, according to (24), as when you take Legendre's equation $(1-x^2)y{''} -2xy{'} + n(n+1)y + 0$ and substitute $y=f_n^{(n)}$, you get (24). Normalizing is not at all like in quantum physics; it is just a simple choice to restrict the values of the functions between -1 and +1. Now apply rule (23) to $$f_n^{(n)} = ((x^2-1)^n)^{(n)} = ((x-1)^n \cdot (x+1)^n)^{(n)}$$, leading to $$\begin{split} f_n^{(n)} (x) &= [(x+1)^n\cdot(x-1)^n]^{(n)} \\ &= \sum_{k=0}^{n}\binom{n}{k}[(x+1)^n]^{(k)} \cdot [(x-1)^n]^{(n-k)} \\ &= \sum_{k=0}^{n}\binom{n}{k}\frac{n!}{(n-k)!}(x+1)^{n-k} \cdot \frac{n!}{k!}(x-1)^k \\ &= n!\sum_{k=0}^{n}\binom{n}{k}^2(x+1)^{n-k}(x-1)^k \end{split}$$
Plugging in $x=1$, we get $$f_n^{(n)} (1) = n!\binom{n}{0}^2 \cdot 2^n= 2^nn!$$, so in order to get 1 we need to divide $f_n$ by $2^nn!$


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# Theoretical background¶

## Definitions of correlation functions¶

### Positional correlation functions¶

In this section the correlation functions computed by dynasor are introduced. For a more a detailed description and derivations of all relations see [JPB80].

The density of particles is denoted $$\rho(\boldsymbol{r},t)$$ and defined as

$\rho(\boldsymbol{r},t) = \sum _i ^N \delta (\boldsymbol{r} - \boldsymbol{r}_i(t)),$

where $$N$$ is the number of particles and $$\boldsymbol{r}_i(t)$$ is the position of particle $$i$$ at time $$t$$. The intermediate scattering function $$F(\boldsymbol{q},t)$$ is defined in terms of the Fourier transform of the particle density

$F(\boldsymbol{q},t)=\frac{1}{N}\left<\rho(\boldsymbol{q},t)\rho(-\boldsymbol{q},0)\right> \quad= \quad \frac{1}{N}\sum _n ^N \sum _m ^N \left< \mathrm{exp} \left[i\boldsymbol{q}\cdot(\boldsymbol{r}_n(t)-\boldsymbol{r}_m(0))\right] \right>,$

where $$\boldsymbol{q}$$ is a wave vector. The brackets denote an ensemble average, which in the case of dynasor is replaced by a time average. (The system under study should be ergodic in order for the time average to be a suitable substitute for the ensemble average.)

The self part of the scattering function, i.e. $$n=m$$, is defined as

$F_s(\boldsymbol{q},t)=\frac{1}{N} \sum _n ^N \left < \mathrm{exp}\left[i\boldsymbol{q}\cdot(\boldsymbol{r}_n(t)-\boldsymbol{r}_n(0))\right] \right >$

and describes single particle motion, i.e. diffusion. $$F_s(\boldsymbol{q},t)$$ is commonly referred to as the incoherent scattering function. The structure factor is given by

$S(\boldsymbol{q}) = F(\boldsymbol{q},0)$

whereas the dynamical structure factor $$S(\boldsymbol{q},\omega)$$ is the Fourier transform of $$F(\boldsymbol{q},t)$$

$S(\boldsymbol{q},\omega) = \int _{-\infty} ^\infty F(\boldsymbol{q},t)e^{-iwt} \mathrm{dt}$

where $$\omega$$ is the angular frequency. $$S(\boldsymbol{q},\omega)$$ exhibits peaks in the $$(\boldsymbol{q},\omega)$$ plane corresponding to the dispersion of lattice vibrations (phonons). The broadening of these peaks are related to the phonon lifetimes and thus the anharmonicity of the system. Computing $$S(\boldsymbol{q},\omega)$$ from MD has the advantage of fully including anharmonic effects and allowing one to easily study the temperature dependence of phonon dispersions.

### Velocity correlation functions¶

It is often convenient to also consider current correlations based on the particle velocities. The current density $$\boldsymbol{j}(\boldsymbol{r},t)$$ is given by

$\boldsymbol{j}(\boldsymbol{r},t) = \sum _i ^N \boldsymbol{v}_i(t) \delta (\boldsymbol{r} - \boldsymbol{r}_i(t))$

where $$\boldsymbol{v}_i(t)$$ is the velocity of particle $$i$$ at time $$t$$. The current correlation function $$J_{\alpha \beta}(\boldsymbol{q},t)$$ is defined as

$J_{\alpha \beta}(\boldsymbol{q},t) = \frac{1}{N} \left < j_\alpha(-\boldsymbol{q},0) j_\beta(-\boldsymbol{q},t) \right >$

where $$\alpha$$ and $$\beta$$ refer to Cartesian coordinates. The current correlation can be split into a longitudinal and a transverse part, $$J_l(\boldsymbol{q},t)$$ and $$J_t(\boldsymbol{q},t)$$, respectivley. As in the case of $$F(\boldsymbol{q},t)$$ the current correlation function can be Fourier transformed to obtain $$J_{\alpha \beta}(\boldsymbol{q},\omega)$$. The longitudinal current correlation $$J_l(\boldsymbol{q},\omega)$$ is related to the dynamical structure factor by

$J_l(\boldsymbol{q},\omega) = \frac{\omega^2}{q^2} S(\boldsymbol{q},\omega),$

which means some features can be easier to resolve in one function rather than the other. The current correlation functions can be thought of as spatially-dependent generalization of the velocity correlation function.

In multi-component systems one can introduce partial correlations functions, which enable one to separate the contributions by groups of particles, e.g., by type or site symmetry. For example in the case of a binary system, one can define $$F_{AA}(\boldsymbol{q},t)$$, $$F_{BB}(\boldsymbol{q},t)$$, $$F_{AB}(\boldsymbol{q},t)$$ and so on for all correlation functions introduced above. In some cases, instead of analyzing the partial functions directly, it is interesting to consider linear combinations of them. This will be discussed in more depth in the liquid sodium chloride example.

In solids, it is furtheremo frequently desirable to determine the above mentioned quantites along specific paths between high symmetry $$vec{q}$$-points. In isotropic samples, such as for example liquids, it is, on the other hand, usually preferable to compute these functions with respect to $$q=|\boldsymbol{q}|$$, i.e. a spherical average over wave vectors.

## Analyzing correlation functions¶

The correlation functions can be convoluted with atomic form factors in order to obtain predictions for various types of neutron and X-ray scattering experiments. In addition, they contain crucial information about the structure and dynamics of the system. In this section, we provide a concise description of some of the information that can be obtained in this fashion.

### Positional correlation functions¶

If we assume that the particle density $$\rho(\boldsymbol{q},t)$$ oscillates as a damped harmonic oscillator we obtain

$\ddot{\rho}(\boldsymbol{q},t) + \Gamma \dot{\rho}(\boldsymbol{q},t) + \omega_0^2{\rho}(\boldsymbol{q},t)=f(\boldsymbol{q},t),$

which yields

$\frac{\mathrm{d}^2 }{\mathrm{d} t^2}F(\boldsymbol{q},t) + \Gamma \frac{\mathrm{d} }{\mathrm{d} t}F(\boldsymbol{q},t) + \omega_0^2 F(\boldsymbol{q},t) = 0$

This equation in turn describes a damped harmonic oscillator with frequency $$\omega_0$$ and damping factor $$\Gamma$$. Assuming weak damping and $$\frac{\mathrm{d} }{\mathrm{d} t}F(\boldsymbol{q},t=0) = 0$$ the solution can be found by Laplace transformation

$F(\boldsymbol{q},t) = S(\boldsymbol{q}) \mathrm{e}^{-\frac{\Gamma}{2} |t|} \Big [\cos\Big(\sqrt{\omega_0^2-\frac{\Gamma^2}{4}}t \Big) + \frac{\Gamma}{2\sqrt{\omega_0^2-\frac{\Gamma^2}{4}}}\sin \Big(\sqrt{\omega_0^2-\frac{\Gamma^2}{4}}|t| \Big) \Big],$

where the absolute value of the time is used to make the function even in time. While the structure factor $$S(\boldsymbol{q})$$ is known for simplicity when fitting we can use

$F(\boldsymbol{q},t) = A\mathrm{e}^{-B|t|}\Big(cos(\omega_et)+\frac{B}{\omega_e}\sin(\omega_e|t|)\Big),$

where $$A$$, $$B$$ and $$\omega_e$$ are fitting parameters. They are related to the original parameters by

$\omega_0 = \sqrt{\omega_e^2 + \frac{\Gamma^2}{4}}$
$\Gamma = 2B$
$\tau = 2\pi\frac{2}{\Gamma},$

where $$\tau$$ is the lifetime of the oscillation with frequency $$\omega_0$$. The factor of $$2\pi$$ comes from the fact that $$\Gamma$$ has units rad/s

One can also carry out the fitting in the frequency domain. The fitting function for the dynamical structure factor is given by the Fourier transform of the above function for the intermediate scattering function

$S(\boldsymbol{q},\omega) = S(\boldsymbol{q}) \frac{2\Gamma\omega_0^2}{(\omega^2-\omega_0^2)^2 + \Gamma^2 \omega^2}.$

Again for simplicity, $$S(\boldsymbol{q})$$ can be replaced with a constant for the purpose of fitting.

### Velocity correlation functions¶

The relation between the dynamical structure factor and the longitudinal current is known in the frequency domain and we obtain

$C_l(\boldsymbol{q},\omega) = \frac{\omega^2}{|\boldsymbol{q}|^2}S(\boldsymbol{q}) \frac{2\Gamma\omega_0^2}{(\omega^2-\omega_0^2)^2 + \Gamma^2 \omega^2}$

Fourier transforming this gives us $$C_l(\boldsymbol{q},t)$$ as

$C_l(\boldsymbol{q},t) = 0$

These functions can be fitted in a similarly to the intermediate scattering function and the dynamical structure factor. The transverse part of the current correlation is assumed to have the same functional form. However sometimes the transverse correlation function is made up by two oscillations not one.

Todo

This last part needs to be checked and finished.

Todo

Check all expressions for harmonic oscillators FTs in Göran


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## Chemistry and Chemical Reactivity (9th Edition)

$0.16218\ L$
The mass of $m=0.20389 g$ corresponds to nitrogen (with argon) that remains after oxygen, water and carbon dioxide are removed, and this gas has a density of $\rho= 1.25718\ g/L$, hence the volume of the globe is given by: $V= m/\rho=0.16218\ L$


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# Why is the word 'simultaneously' important in stating Heisenberg's uncertainty principle?

The Heisenberg's uncertainty principle states that a particle cannot have a precise value of its position and conjugate momentum simultaneously.

If these uncertainties are intrinsic properties of a state why is the word 'simultaneously' important? Is this important only for those states which have non-trivial time-dependence? For trivial time-dependence i.e., energy eigenstates $\Delta x$ and $\Delta p_x$ are fixed in time and it appears that one can measure them at separate times. But this is not true for other states. Am I correct?

Just look at the formal version of the Heisenberg uncertainty principle: $$\sigma_x(\psi) \sigma_p(\psi) \geq \hbar/2,$$ where $\sigma_A(\psi) = \sqrt{\langle\psi\vert A^2\vert \psi\rangle - \langle \psi \vert A\rvert \psi\rangle^2}$ is the standard deviation of an operator $A$ for the state $\vert \psi \rangle$.

When we say a state has a "well-defined" or "precise" value of the observable $A$, we mean it is an eigenstate. It is straightforward to check that $\sigma_A(\psi) = 0$ in an eigenstate. So no state can be both an eigenstate of $x$ and of $p$, since that would mean $0\geq \hbar / 2$, which is clearly false.

The "simultaneously" means precisely that: A (time-dependent) state $\psi(t)$ may be an eigenstate of $x$ in one instance, and an eigenstate of $p$ in another, but it is impossible that $\psi(t_0)$ for any fixed $t_0$ is both.

The word "simultaneously" is there because you can obtain a precise measurement of an particle's position and afterward obtain a precise measurement of the momentum. But you can't do both at the same time. Further, when you make the second measurement (the measurement of the momentum), you disturb the object's position, so it in fact no longer has a definite position, and the position measurement you had made at first no longer contains any meaningful information about the particles position.

For a concrete example, consider a particle in the ground state of the quantum harmonic oscillator. In this state, $\Delta x$ and $\Delta p$ are both non-zero and constant in time, so neither the position nor momentum is known. Now if you measure the position, the particle's wave function collapses to a delta function, and you observe a precise position for the particle. But now when you measure the momentum, the wavefunction collapses into a plane wave state, which has infinite extent. So while you obtain a precise value for the momentum, you have lost all the information about the particle's position.

• Everything you describe here is correct, but it seems to be more a description of how projective measurements work, and not about the uncertainty principle. You've hidden the application of the uncertainty principle behind the assertion that a $\delta$-function doesn't have a definite momentum, essentially. – ACuriousMind Aug 24 '17 at 16:34
• @ACuriousMind I didn't talk about the uncertainty principle because I assumed they already knew it. I was trying to address why simultaneity was important. To do this, I gave a counter-example, where at $\Delta x$ from one time was zero and a $\Delta p$ from another time was zero. I think they understood that the uncertainty principle applies when $\Delta x$ and $\Delta p$ are considered simultaneously, and therefore it applies to constant states even when $\Delta x$ and $\Delta p$ aren't considered simultaneously, but they had a hard time thinking about states that change with time. – Brian Moths Aug 24 '17 at 16:52
• @ACuriousMind Accordingly, I gave an example of a state which had a time dependence (owing to the fact that it is being measured), and therefore had varying $\Delta x$'s and $\Delta p$'s, such that a non-simultaneous product came out to be zero. This shows why the word "simultaneous" is necessary, and that he is correct that time-varying states play an important role. – Brian Moths Aug 24 '17 at 16:55

In principle it holds that when you decide to measure the momentum the more uncertain of the position and vice versa but surely this leads to there is a simultaneity in these states but this principle means we are unable to measure the two occuring variables without interference. it doesn't detract from the actual physical occurrence of the two, we just ain't able to measure the two simultaneously in principle. Well that's what I think it means

## protected by Qmechanic♦Aug 25 '17 at 5:30

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


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Weekly Papers on Quantum Foundations (39)

 下午12:34 | quant-ph updates on arXiv.org

Authors: Selman IpekAriel Caticha

In the Entropic Dynamics (ED) framework quantum theory is derived as an application of entropic methods of inference. The physics is introduced through appropriate choices of variables and of constraints that codify the relevant physical information. In previous work, a manifestly covariant ED of quantum scalar fields in a fixed background spacetime was developed. Manifest relativistic covariance was achieved by imposing constraints in the form of Poisson brackets and of intial conditions to be satisfied by a set of local Hamiltonian generators. Our approach succeeded in extending to the quantum domain the classical framework that originated with Dirac and was later developed by Teitelboim and Kuchar. In the present work the ED of quantum fields is extended further by allowing the geometry of spacetime to fully partake in the dynamics. The result is a first-principles ED model that in one limit reproduces quantum mechanics and in another limit reproduces classical general relativity. Our model shares some formal features with the so-called semi-classical approach to gravity.

 下午12:34 | quant-ph updates on arXiv.org

We introduce a fidelity-based measure of nonclassicality Q to quantify the differences between the dynamics of classical and quantum continous-time walks over a graph. We provide universal, graph-independent, analytic expressions of Q, showing that at short times nonclassicality of quantum walks is due to the appearance of coherence, whereas for long times it depends only on the size of the graph, with quantumness contributing only partially to the overall nonclassicality. At intermediate times, nonclassicality does instead depend on the graph topology through its algebraic connectivity.

 下午12:34 | quant-ph updates on arXiv.org

The phase diagram of a material is of central importance to describe the properties and behaviour of a condensed matter system. We prove that the general task of determining the quantum phase diagram of a many-body Hamiltonian is uncomputable, by explicitly constructing a one-parameter family of Hamiltonians for which this is the case. This work builds off recent results from Cubitt et al. and Bausch et al., proving undecidability of the spectral gap problem. However, in all previous constructions, the Hamiltonian was necessarily a discontinuous function of its parameters, making it difficult to derive rigorous implications for phase diagrams or related condensed matter questions. Our main technical contribution is to prove undecidability of the spectral gap for a continuous, single-parameter family of translationally invariant, nearest-neighbour spin-lattice Hamiltonians on a 2D square lattice: $H(\varphi)$ where $\varphi\in \mathbb R$. As well as implying uncomputablity of phase diagrams, our result also proves that undecidability can hold for a set of positive measure of a Hamiltonian’s parameter space, whereas previous results only implied undecidability on a zero measure set.

The Schr\”odinger-Newton equation is a proposed model to explain the localisation of macroscopic particles by suppressing quantum dispersion with the particle’s own gravitational attraction. On cosmic scales, however, dark energy also acts repulsively, as witnessed by the accelerating rate of universal expansion. Here, we introduce the effects of dark energy in the form of cosmological constant $\Lambda$, that drives the late-time acceleration of the universe, into the Schr\”odinger-Newton approach. We then ask in which regime dark energy dominates both canonical quantum diffusion and gravitational self-attraction. It turns out that this happens for sufficiently delocalised objects with arbitrary mass and that there exists a minimal delocalisation width of about $67$ meters. While extremely macroscopic from a quantum perspective, the value is in principle accessible to laboratories on Earth. Hence, we analyse, numerically, how the dynamics of an initially spherical Gaussian wave packet is modified in the presence of $\Lambda > 0$. A notable feature is the gravitational collapse of part of the wave packet, in the core region close to the centre of mass, accompanied by the accelerated expansion of the more distant shell surrounding it. The order of magnitude of the distance separating collapse from expansion matches analytical estimates of the classical turnaround radius for a spherically symmetric body in the presence of dark energy. However, the time required to observe these modifications is astronomical. They can potentially be measured only in physical systems simulating a high effective cosmological constant, or, possibly, via their effects on the inflationary universe.

Authors: Rubén ArjonaSavvas Nesseris

Machine learning algorithms have revolutionized the way we interpret data in astronomy, particle physics, biology and even economics, since they can remove biases due to a priori chosen models. Here we apply a specific machine learning method, the genetic algorithms (GA), to cosmological data that describes the background expansion of the Universe, the Pantheon Type Ia supernovae and the Hubble expansion history $H(z)$ datasets. We obtain model independent and non-parametric reconstructions of the luminosity distance $d_L(z)$ and Hubble parameter $H(z)$ without assuming any dark energy model or a flat Universe. We then estimate the deceleration parameter $q(z)$, a measure of the acceleration of the Universe, and we make a $\sim4.5\sigma$ model independent detection of the accelerated expansion, but we also place constraints on the transition redshift of the acceleration phase $(z_{\textrm{tr}}=0.662\pm0.027)$. We also confirm a recently reported mild tension between the SnIa/quasar data and the cosmological constant $\Lambda$CDM model at high redshifts $(z\gtrsim1.5)$ and finally, we show that the GA can be used in complementary null tests of the $\Lambda$CDM via reconstructions of the Hubble parameter and the luminosity distance.

 2019年10月4日 星期五 上午8:00 | Latest Results for Foundations of Physics

Abstract

The Hole Argument is primarily about the meaning of general covariance in general relativity. As such it raises many deep issues about identity in mathematics and physics, the ontology of space–time, and how scientific representation works. This paper is about the application of a new foundational programme in mathematics, namely homotopy type theory (HoTT), to the Hole Argument. It is argued that the framework of HoTT provides a natural resolution of the Hole Argument. The role of the Univalence Axiom in the treatment of the Hole Argument in HoTT is clarified.

 2019年10月4日 星期五 上午8:00 | Latest Results for Foundations of Physics

Abstract

The usual representation of quantum algorithms is limited to the process of solving the problem. We extend it to the process of setting the problem. Bob, the problem setter, selects a problem-setting by the initial measurement. Alice, the problem solver, unitarily computes the corresponding solution and reads it by the final measurement. This simple extension creates a new perspective from which to see the quantum algorithm. First, it highlights the relevance of time-symmetric quantum mechanics to quantum computation: the problem-setting and problem solution, in their quantum version, constitute pre- and post-selection, hence the process as a whole is bound to be affected by both boundary conditions. Second, it forces us to enter into relational quantum mechanics. There must be a representation of the quantum algorithm with respect to Bob, and another one with respect to Alice, from whom the outcome of the initial measurement, specifying the setting and thus the solution of the problem, must be concealed. Time-symmetrizing the quantum algorithm to take into account both boundary conditions leaves the representation to Bob unaltered. It shows that the representation to Alice is a sum over histories in each of which she remains shielded from the information coming to her from the initial measurement, not from that coming to her backwards in time from the final measurement. In retrospect, all is as if she knew in advance, before performing her problem-solving action, half of the information that specifies the solution of the problem she will read in the future and could use this information to reach the solution with fewer computation steps (oracle queries). This elucidates the quantum computational speedup in all the quantum algorithms examined.

 2019年10月4日 星期五 上午3:28 | Philsci-Archive: No conditions. Results ordered -Date Deposited.
Gomes, Henrique (2019) Holism as the significance of gauge symmetries. [Preprint]
 2019年10月4日 星期五 上午3:21 | Philsci-Archive: No conditions. Results ordered -Date Deposited.
Dougherty, John (2019) Large gauge transformations and the strong CP problem. Studies in History and Philosophy of Modern Physics. ISSN 1355-2198
 2019年10月3日 星期四 下午6:00 | Alejandro Pozas-Kerstjens, Rafael Rabelo, Łukasz Rudnicki, Rafael Chaves, Daniel Cavalcanti, Miguel Navascués, and Antonio Acín | PRL: General Physics: Statistical and Quantum Mechanics, Quantum Information, etc.

Author(s): Alejandro Pozas-Kerstjens, Rafael Rabelo, Łukasz Rudnicki, Rafael Chaves, Daniel Cavalcanti, Miguel Navascués, and Antonio Acín

We present a method that allows the study of classical and quantum correlations in networks with causally independent parties, such as the scenario underlying entanglement swapping. By imposing relaxations of factorization constraints in a form compatible with semidefinite programing, it enables the…

[Phys. Rev. Lett. 123, 140503] Published Thu Oct 03, 2019

 2019年10月1日 星期二 上午8:00 | Swanson N. | The British Journal for the Philosophy of Science Advance Access
Abstract

Candidates for fundamental physical laws rarely, if ever, employ higher than second time derivatives. Easwaran ([2014]) sketches an enticing story that purports to explain away this puzzling fact and thereby provides indirect evidence for a particular set of metaphysical theses used in the explanation. I object to both the scope and coherence of Easwaran’s account, before going on to defend an alternative, more metaphysically deflationary explanation: in interacting Lagrangian field theories, it is either impossible or very hard to incorporate higher than second time derivatives without rendering the vacuum state unstable. The so-called Ostrogradski instability represents a powerful constraint on the construction of new field theories and supplies a novel, largely overlooked example of non-causal explanation in physics.

The Information Geometry of Space-time

The method of maximum entropy is used to model curved physical space in terms of points defined with a finite resolution. Such a blurred space is automatically endowed with a metric given by information geometry. The corresponding space-time is such that the geometry of any embedded spacelike surface is given by its information geometry. The dynamics of blurred space, its geometrodynamics, is constructed by requiring that as space undergoes the deformations associated with evolution in local time, it sweeps a four-dimensional space-time. This reproduces Einstein’s equations for vacuum gravity. We conclude with brief comments on some of the peculiar properties of blurred space: There is a minimum length and blurred points have a finite volume. There is a relativistic “blur dilation”. The volume of space is a measure of its entropy.

The Entropic Dynamics approach to Quantum Mechanics

Entropic Dynamics (ED) is a framework in which Quantum Mechanics is derived as an application of entropic methods of inference. In ED the dynamics of the probability distribution is driven by entropy subject to constraints that are codified into a quantity later identified as the phase of the wave function. The central challenge is to specify how those constraints are themselves updated. In this paper we review and extend the ED framework in several directions. A new version of ED is introduced in which particles follow smooth differentiable Brownian trajectories (as opposed to non-differentiable Brownian paths). To construct the ED we make use of the fact that the space of probabilities and phases has a natural symplectic structure (i.e., it is a phase space with Hamiltonian flows and Poisson brackets). Then, using an argument based on information geometry, a metric structure is introduced. It is shown that the ED that preserves the symplectic and metric structures — which is a Hamilton-Killing flow in phase space — is the linear Schrödinger equation. These developments allow us to discuss why wave functions are complex and the connections between the superposition principle, the single-valuedness of wave functions, and the quantization of electric charges. Finally, it is observed that Hilbert spaces are not necessary ingredients in this construction. They are a clever but merely optional trick that turns out to be convenient for practical calculations.


text:
# Shan had the following gains and losses from Sec 1231 properties in 2014. She does not have any n...

Shan had the following gains and losses from Sec 1231 properties in 2014. She does not have any net sec 1231 loss from the last 5 years.
Casualty gains: $7,000 Casualty losses: ($8,000)
Sale gains: $5,000 Sale losses: ($3,000)

(a) What is the net ordinary gain or loss from these transactions?

(b) What is the net Sec 1231 gain or loss from these transactions?
(c) Matt had the following gains and losses from Sec 1231 properties. He does not have any net sec 1231 loss from the last 5 years. What is the net Sec 1231 gain or loss from these transactions?
Casualty gains: $25,000 Casualty losses: ($15,000)
Sale gains: $5,000 Sale losses: ($7,000)


text:
## May 12, 2008

### Superconnections for Dummies

For inexplicable reasons, I got involved in a discussion with Urs Schreiber about Quillen superconnections. Urs was enamoured of the idea that Quillen superconnections might be relevant to Garrett Lisi’s “Theory of Everything.” Dubious applications, aside, Urs wanted to construct Quillen superconnections from $\mathbb{Z}_2$ graded Lie algebras, instead of Lie superalgebras. This just plain doesn’t work. So while there is, perhaps, some kind of superconnection of the sort he’s after, it’s certainly not the Quillen superconnection.

So here’s an elementary, “For Dummies,” guide to superconnections.

Let $\mathfrak{g}=\mathfrak{g}^0 + \mathfrak{g}^1$ be a Lie superalgebra. The generators satisfy (anti)commutation relations

(1)\begin{aligned} [T_a, T_b] &= \tensor{f}{_a_b_^c}T_c\\ [T_a, \tau_\alpha] &= \tensor{c}{_a_\alpha_^\beta}\tau_\beta\\ \{\tau_\alpha, \tau_\beta\} &= \tensor{d}{_\alpha_\beta_^a}T_a \end{aligned}

which, in turn, satisfy certain associativity conditions called the super-Jacobi identity.

$\mathfrak{g}^0$ (spanned by the $T_a$) is an ordinary Lie algebra. Let $G_0$ be the corresponding Lie group, and $P\to X$ a $G_0$ principal bundle. $\mathfrak{g}$ furnishes a representation of $G_0$. Let $\mathbb{G}= \mathbb{G}^0 \oplus \mathbb{G}^1$ be the corresponding $\mathbb{Z}_2$ graded vector bundle associated to $P$. The main applications, to K-theory (or KO-theory) will involve $\mathfrak{g}= u(n|m)$ (or $so(n|m)$), but we can write the formulæ below for any Lie superalgebra.

Now let $V = V^0 \oplus V^1$ be any representation of $\mathfrak{g}$ (a $\mathbb{Z}_2$ graded vector space). Let $E=E^0\oplus E^1$ be the corresponding $\mathbb{Z}_2$ graded vector bundle associated to $P$. A Quillen superconnection on $E$ is given by the formula

(2)$\mathbb{D} = d + dx^\mu A^a_\mu(x) T_a + \phi^\alpha(x) \tau_\alpha$

where $A$ is an ordinary $G_0$ connection on $E$, and $\phi\in \Gamma(X, \mathbb{G}^1)$. Note that, though $\phi$ is an “odd” section of $\mathbb{G}$, both $A_\mu^a(x)$ and $\phi^\alpha(x)$ are bosonic fields.

The curvature of $\mathbb{D}$ is

(3)\begin{aligned} \mathbb{F} =\mathbb{D}^2= &\tfrac{1}{2} dx^\mu \wedge dx^\nu (\partial_\mu A_\nu-\partial_\nu A_\mu + \tensor{f}{_b_c_^a} A^b_\mu A^c_\nu)T_a\\ & +\, dx^\mu (\partial_\mu \phi^\alpha + \tensor{c}{_a_\beta_^\alpha}A^a_\mu \phi^\beta)\tau_\alpha\\ & +\, \tfrac{1}{2} \tensor{d}{_\alpha_\beta_^a} \phi^\alpha\phi^\beta T_a \end{aligned}

Note that the 0-form term (the last line of (3)) only makes sense because (1) is a Lie superalgebra, and hence $\tensor{d}{_\alpha_\beta_^a}$ is symmetric on $\alpha,\beta$.

In the applications to K-theory, $E^0$ and $E^1$ are the bundles living on the $D9$ and $\overline{D9}$ branes, and $\phi$ are the tachyons that condense between them.

Urs wanted to replace the Lie superalgebra (1) by a $\mathbb{Z}_2$-graded Lie algebra,

(4)\begin{aligned} [T_a, T_b] &= \tensor{f}{_a_b_^c}T_c\\ [T_a, t_\alpha] &= \tensor{c}{_a_\alpha_^\beta}t_\beta\\ [t_\alpha, t_\beta] &= \tensor{\tilde{d}}{_\alpha_\beta_^a}T_a \end{aligned}

The reason he wants to do this is that $e_8$ is a Lie algebra, not a Lie superalgebra, and it admits various $\mathbb{Z}_2$ gradings.

However, because $\tensor{\tilde{d}}{_\alpha_\beta_^a}$ is now antisymmetric on $\alpha,\beta$, we can’t use this to construct a Quillen superconnection (there’s a longer-winded explanation in my discussion with Urs).

But there is another kind of superconnection, let’s call it a “Schreiber superconnection”, which we can build, based on a $\mathbb{Z}_2$-graded Lie algebra. Let

(5)$\mathbb{D} = d + dx^\mu A^a_\mu(x) T_a + \psi^\alpha(x) t_\alpha$

where now $\psi$ is a fermionic field. Its curvature

(6)\begin{aligned} \mathbb{F} =\mathbb{D}^2=&\tfrac{1}{2} dx^\mu\wedge dx^\nu (\partial_\mu A_\nu-\partial_\nu A_\mu + \tensor{f}{_b_c_^a} A^b_\mu A^c_\nu)T_a\\ & +\, dx^\mu (\partial_\mu \psi^\alpha + \tensor{c}{_a_\beta_^\alpha}A^a_\mu \psi^\beta)t_\alpha\\ & +\, \tfrac{1}{2} \tensor{\tilde{d}}{_\alpha_\beta_^a} \psi^\alpha\psi^\beta T_a \end{aligned}

makes sense because, though $\tensor{\tilde{d}}{_\alpha_\beta_^a}$ is antisymmetric on $\alpha,\beta$, the $\psi$ are fermionic.

In Quillen’s case, “$dx^\mu$” and “$\tau_\alpha$” are anti-commuting (“fermionic”), so that $\mathbb{D}$ is overall fermionic, and $\mathbb{F}$ is bosonic. For the “Schreiber superconnection,” “$dx^\mu$” and “$\psi^\alpha(x)$” are anticommuting, with the same net parities for $\mathbb{D}$ and $\mathbb{F}$ as before.

Now, when all the dust settles, the Schreiber superconnection is equally useless for Lisi’s purposes as the Quillen superconnection (though for different reasons). But, at least we’re not taking the name of Quillen in vain.

Posted by distler at May 12, 2008 12:01 AM

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1680

### Re: Superconnections for Dummies

Hi Jacques,

A small typo: in the LHS of the second line of eq (1), $\tau_\beta$ should be $\tau_\alpha$.

Posted by: Lord Sidious on May 12, 2008 12:39 AM | Permalink | Reply to this

### Typos

Not the only such typo. Hopefully they’re all fixed now.

Thanks!

Posted by: Jacques Distler on May 12, 2008 4:44 AM | Permalink | PGP Sig | Reply to this

### Re: Superconnections for Dummies

Hi Jacques,

in my post and in the comments I suggested to interpret the expression

$d + dx^\mu A^a_\mu(x) \rho(T_a) + \phi^a(x) \rho(t^a)$

with (!)

$\rho : g \to end(V)$

a $\mathbb{Z}_2$-graded representation as a Quillen superconnection proper, simply observing that any sum of an ordinary connection on a $\mathbb{Z}_2$-graded vector bundle with a 0-form with values in odd endomorphisms can be read as a Quillen superconnection (Dumitrescu takes that as the definition). In this context your $g$ equals $end(V)$, but not my $g$.

The main point here was that it is not as weird to consider such mixed sums as some discussion around the blogs and elsewhere have implied.

What you call a Schreiber superconnection is more like what Garrett Lisi seemed to have in mind himself, who addresses his “$\psi$” as Grassmannian. But it’s hard to tell, because the difference in the Quillen and the Schreiber superconnection is just in the quadratic term, which he doesn’t write.

But in any case, sorry if indeed I added to the confusion where I was hoping to reduce it. Seems to be one of the things which are easier to clarify in personal discussion than in (inter-)blog discussion.

Finally, I should add that I am not after justifying any problematic speculations here, but that I do find it interesting to ponder (if only for my own humble education) in general what connections “with values in $e_8$” are like, if we allow the “fermionic contribution” to the connection (under your favorite splitting) to come from 0-forms (either way).

Posted by: Urs Schreiber on May 12, 2008 2:52 PM | Permalink | Reply to this

### Re: Superconnections for Dummies

In this context your $g$ equals $end(V)$, but not my $g$.

No, my $\mathfrak{g}$ is a subalgebra of $end(V)$. My point is that you really do need the algebra structure, not just $\mathfrak{g}$ as a $\mathbb{Z}_2$-graded vector space.

For instance, in constructing the curvature, $\mathbb{F}$, from the superconnection, $\mathbb{D}$, we used the algebra structure. And (as I hope my exposition makes clear) a Lie superalgebra is the structure that we need.

What you call a Schreiber superconnection is more like what Garrett Lisi seemed to have in mind…

I agree.

Which is why we should be pleased by the result, because whereas the Quillen superconnection is naturally associated with Lie superalgebras, the Schreiber superconnection (with its fermion field) is naturally associated to $\mathbb{Z}_2$-graded Lie algebras.

… the quadratic term, which he doesn’t write.

He claims (incorrectly, as usual) that it vanishes by Fermi statistics.

I do find it interesting to ponder … if we allow the “fermionic contribution” to the connection … to come from 0-forms.

There’s an important point here, so let me reiterate it. Lisi is not interested in just any old $\mathbb{Z}_2$-grading of $e_8$. He wants a very particular one. Namely, choose some embedding of $SL(2,\mathbb{C})\hookrightarrow E_8$. This defines an action of $SL(2,\mathbb{C})$ on $e_8$. He wants the $\mathbb{Z}_2$-grading that comes from the action of the ($\mathbb{Z}_2$) center of $SL(2,\mathbb{C})$ on $e_8$.

Then it’s automatic that the $\mathbb{Z}_2$-odd generators transform as spinors of $SL(2,\mathbb{C})$.

Posted by: Jacques Distler on May 12, 2008 3:32 PM | Permalink | PGP Sig | Reply to this
Read the post HIM Trimester on Geometry and Physics, Week 4
Weblog: The n-Category Café
Excerpt: Talk in Stanford on nonabelian differential cohomology.
Tracked: May 29, 2008 5:25 PM

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text:
# Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"

## Problem

Four siblings BRYAN, BARRY, SARAH and SHANA are having their names monogrammed on their towels. Different letters may cost different amounts to monogram. If it costs $\textdollar{21}$ to monogram BRYAN, $\textdollar{25}$ to monogram BARRY and $\textdollar{18}$ to monogram SARAH, how much does it cost to monogram SHANA?

## Solution

We set up the following equations:

$b+r+y+a+n= 21$

$b+a+2r+y= 25$

$s+2a+r+h= 18$

We are asked to find the price of "Shana", or $s+h+2a+n$. We notice that this expression has no $b$ or $y$, so we subtract the first equation from the second to eliminate those variables:

$r-n=4$

$r=n+4$

Which we substitute into the third equation

$s+2a+n+4+h=18$.

$s+2a+n+h=14$.

This matches the expression we need to solve for, so we are done. $\boxed{14}$.

## Slick Solution

BRYAN, SARAH is an anagram of BARRY, SHANA. Let SHANA be $x$. So we have the following equation: $$21+18=25+x\rightarrow \boxed{x=14}$$

~yofro


text:
# Latex cannot find my .bib file

I am using Texmaker in combination with Miktex and I am writing an article, but Latex can't find my .bib file. The .bib file is located in the same directory as my article latex file. I made .bib files in several ways: by using Mendeley, Jabref and by bibtex itself. In all three the cases I get warnings in Latex: "there were undefined references" and "citation 'citation1' on page 5 undefined".

I guess there is something wrong with my Latex?

I made a MWEB of the report and a MWEB of the bibliography file.

 \documentclass[11pt]{article}

\usepackage[utf8]{inputenc}

\usepackage{cite}
%\usepackage{natbib}

\begin{document}
\pagenumbering{gobble}
\title{Report}
\author{Charlie}
\maketitle
\newpage
\mbox{}
\pagebreak

\section{Theory}
Blabla \cite{Gazibegovic2017}

\pagebreak

\section{Bibliography}

\bibliography{biblio.bib}
\bibliographystyle{plain}

\end{document}

and the mweb from the bibliography

 @article{Gazibegovic2017,
author = {Gazibegovic, Sasa. \textit(et al)},
journal = {Nature},
number = {7668},
pages = {434-438},
title = {{Epitaxy of advanced nanowire quantum devices}},
volume = {548},
year = {2017}
}

I hope this information is sufficient for you to help me.

Kind regards

• Welcome to TeX.SX! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}.
– NBur
Oct 18 '18 at 10:08
• Have you run the entire path pdflatex - bibtex - pdflatex?
– NBur
Oct 18 '18 at 10:10
• At the moment the most likely explanation is that you did not run BibTeX, see tex.stackexchange.com/q/63852/35864. If you did run BibTeX and still get warnings and errors, please consider adding a full MWE (tex.meta.stackexchange.com/q/228/35864 and tex.meta.stackexchange.com/q/4407/35864) and show us the exact messages from the .log and .blg files (LaTeX and BibTeX log, respectively). Oct 18 '18 at 10:15

Compiling the following MWEB with

pdflatex document.tex
bibtex document.aux
pdflatex document.tex
pdflatex document.tex

gives the right result.

\documentclass{article}

\usepackage{filecontents}

\begin{filecontents}{\jobname.bib}
@book{key,
author = {Author, A.},
year = {2001},
title = {Title},
publisher = {Publisher},
}
\end{filecontents}

\begin{document}

\cite{key}

\bibliographystyle{plain}
\bibliography{\jobname}

\end{document}

• A comment to a working example on wikibooks would have had the same amount of information. Oct 18 '18 at 10:43
• @Johannes_B that's only to develop my previous comment… Do I have to delete it?
– NBur
Oct 18 '18 at 11:00
• Thanks for your response. I try to run my .bib-file to a bibTex, but i get an error message: "couldn't open file name 'biblio1.aux' " , while I saved my bibliography as a .bib-file. I don't know what an .aux-file is and why it says its an .aux-file while I saved it as a .bib-file. Oct 18 '18 at 12:00
• Let say you have source_tex.tex and biblio.bib. You have to run pdflatex source_tex.tex, then bibtex source_tex.aux and twice again pdflatex source_tex.tex.
– NBur
Oct 18 '18 at 13:20
• Hi, when I first run report.tex (source_tex.tex) I get the same errors as before: citation 'citation1' undefined. Then I run report.aux (source_tex.aux) I get two error messages and the file seems to have been dissapeared and then I run report.tex (source_tex.tex) twice I get the same errors: 'citation1' undefined. Do you know what is happening? Oct 18 '18 at 13:55


text:
explain the mechanism of esterification of carboxylic acid

The full text of this article hosted at iucr.org is unavailable due to technical difficulties. The 13C‐NMR spectra showed typical signals for highly substituted starch esters.

The loss of proton from second hydroxyl group gives ester. colleges and universities in the state. The procedure was repeated without imidazole and without the first heating cycle to check if propionic acid could react with lauric acid imidazolide, too. Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones containing a benzene ring). and Inverse Proportions, Areas Furthermore, the DStot exceeds the DSpropionate of the educt, precluding a simple transesterification. Uttarakhand CM announces free WI-FI service in all Govt. The authors would like to acknowledge the NMR platform at the Friedrich Schiller University Jena for support in the NMR spectroscopy. In industrial applications, starch is chemically modified to products of low DS exclusively. Esterification involves ANDN mechanism which is basically attack of one nucleophile and the departure of another from the same molecule.here OR- group attack the carboxylic acid at the carbonyl carbon causing OH- group to leave. Open access funding enabled and organized by Projekt DEAL. Step 3:Transfer of proton A proton is transferred to one of the hydroxyl groups to form a good leaving group. अल्कोहल के लिए उत्प्रेरित एसिड के तंत्र को संबंधित शराब बनाने के लिए समझाएं।, एस्टेरिफिकेशन (नीचे दिखाया गया है) एक प्रतिक्रिया है जो अपने एस्टर को कार्बोक्जिलिक एसिड परिवर्तित करता है। इसमें केवल कार्बोनिल कार्बन शामिल है। मेथनॉल उपज (+) -मेथाइल लैक्टेट के साथ (-) -लैक्टिक एसिड का स्थिरीकरण। यह मानते हुए कि कोई प्रतिक्रिया नहीं है, (ए) एक एल्डिहाइड या कीटोन के कार्बोनिल समूह पर न्यूक्लियोफिलिक हमले के तंत्र की व्याख्या करें।
(b) एक ऑर्गेनिक कमपाउंड (A) (आणविक सूत्र), CBSE Scholarship for Single Girl Child - More Power to Girls. The DSlaurate was calculated by integrating the signals of the modified AGU and the methyl group of the laurate according to Equation (2). Check BSEB sent up exam 2021 time table, sent up exam paper pattern, & other details. Carboxylic acid imidazolides can be transformed into each other due to a chemical equilibrium between carboxylic acids or their carboxylates and the imidazolides. This forms a carboxylic cation and an alcoholic nucleophile. However, they are also difficult to handle, may undergo undesirable hydrolysis, and lead to the formation of corrosive hydrogen chloride during the reaction with hydroxyl groups. Arham qualifying the Microsoft certification exam at Pearson VUE test centre. to Trigonometry, Complex Check detailed report on CBSE Exams 2021 dates! The only reason why it works in ester formation is that at the same time $\ce{C=O}$ is regenerated which provides the required energy. In general, higher DStot leads to lower melting temperature. On the other hand, no lauric anhydride species and, thus, no lauric acid imidazolide would be formed via this reaction path (Figure 2, reactions E and F). Bihar Board Sent Up Exam Starts from Today, 11th Nov 2020. What is the lowest level character that can unfailingly beat the Lost Mine of Phandelver starting encounter? The mixed anhydride could react with lauric acid yielding the symmetric anhydride (reaction C). These hydrolyses are nucleophilic substitution reactions that proceed by the addition-elimination mechanism. of Derivatives, Application The solutions were mixed in the NMR tube and instantly transferred to the spectrometer that was pre‐cooled to –40 °C and 1H‐NMR spectra were recorded every 2 min. Working off-campus? In the 1H‐NMR spectrum, only tiny signals corresponding to lauric acid imidazolide could be seen (Figure 4), i.e., this reaction can be neglected. and you may need to create a new Wiley Online Library account. Yield: 3.93 g (8.56 mmol; 69%); DSlaurate = 1.02; DSpropionate = 1.98. High DS values can be obtained by the reaction of the biopolymer with carboxylic acid chlorides in pyridine. Finally, there could be an equilibrium between carboxylic acid imidazolides and carboxylic acids or probably their carboxylates (reaction G). In summary, the components lauric acid and imidazole form the imidazolium laurate only. Why this happens is due to the reaction mechanism. Why is there 5GB of unallocated space on my disk on Windows 10 machine? of Integrals, Continuity All the steps in Fischer esterification are reversible and, in general, it is a slow reaction carried out in reflux using a strong acid such as sulfuric or phosphoric acids. Essentially, we are drawing the reverse order of Fischer esterificationso, in the first step the ester is protonated promoting the nucleophilic attack of water: Notice that just like the Fischer esterification, the process is an equilibrium which makes the reaction a little challenging as it may require higher temperaturesand removal of the alcohol as it is formed to push … The esterification system containing a short‐chain carboxylic acid anhydride and a long‐chain fatty acid seems to be a universal path for efficient esterification of polysaccharides that is studied at present. Use MathJax to format equations. CBSE Board Exams 2021 Could be Held Early, Know the Reason. Would the Millennium Falcon have been carried along on the hyperspace jump if it stayed attached to the Star Destroyer? If $\ce{HOR+}$ is expelled, the starting materials are regenerated. The mechanism of ester hydrolysis in acid consists of the addition of the nucleophile and the elimination of the leaving group: The first part of the mechanism is the addition of water. CBSE likely to conduct the class 12 & class 10 board exam 2021 earlier than expected. Explain the mechanism of acid catalysed of an alkene to form corresponding alcohol. We have discussed the steps below: Step 1:Cation formation Step 2:Delocalized carbocation Carboxyl oxygen gets protonated to give delocalized carbocation making the carbocation a better electrophile. to Euclids Geometry, Areas Meet 6-Year-Old Ahmedabad Boy World’s Youngest Computer Programmer. While starch decomposes under heating up to 300 °C without melting, starch esters may exhibit thermoplastic behavior, which is also desirable for use as plastic material. The completely converted position 6 of the AGU causes the signal at 61.46 ppm compared to a non‐esterified C‐6 that would lead to a signal at 60.00 ppm. Thus, the reaction of propionic anhydride with lauric acid proceeds slowly. The hydrolysis of esters in aqueous acid is a reversible equilibrium reaction that is driven to the right by using an excess of water. Moreover, melting could be expected in agreement with other studies about fatty acid esters of starch. [1] Furthermore, the amount of available fossil resources is limited. The removed groups react with each other to form water. Subsequently, lauric acid imidazolide was added and the reaction was allowed to continue for another 2 h. In the 1H‐NMR spectrum of the resulting mixture, signals for the solvent imidazole can be observed at 7.83 and 7.10 ppm (Figure 7a). The formation of carboxylic acid imidazolides from the carboxylic anhydride and imidazole proceeds rapidly and afterwards there is an equilibrium between carboxylic acid imidazolides and carboxylic acids. and propionyl chloride (2.5 mL, 28.37 mmol; 2.30 equiv.) Apart from starch, the reaction mixture consists of propionic anhydride, lauric acid, and imidazole that may undergo various reactions (Figure 2). However, as can be seen at the DS values summarized in Table 1, higher values can be reached. Assuming that there are no side reactions, what is true about this reactin? The synthesis of starch esters is an appropriate approach to get hydrophobic products with good thermal characteristics. Figure 6 shows the 1H‐NMR spectrum acquired after ≈2 min after mixing the components. In the FTIR spectrum, the signal at 1696 cm–1 attributed to the νCO vibration of lauric acid disappears completely (Figure 3). and Differentiability. The product was dried at 40 °C in vacuum to give a beige powder. The con… 13C‐NMR (62.90 MHz, 297 K, CDCl3): δ = 174.50–173.70 (C‐7), 173.50–172.50 (C‐7propionate), 95.68 (C‐1′), 72.12 (C‐3s), 70.43 (C‐2s), 69.27 (C‐5), 61.53 (C‐6s), 34.28 (C‐8), 32.02 (C‐16), 29.80–29.40 (C‐10–C‐15), 27.49 (C‐8propionate), 25.06–24.77 (C‐9), 22.79 (C‐17), 14.21 (C‐18), 9.21–8.82 (C‐9propionate). Uttarakhand CM Announces Free WI-FI Service in all Govt. If you do not receive an email within 10 minutes, your email address may not be registered, 13C‐NMR (62.90 MHz, 297 K, CDCl3): δ = 174.34 (C‐7), 97.71 (C‐1′), 78.19 (C‐4), 73.51 (C‐3s), 71.69 (C‐2s), 69.75 (C‐5), 63.92 (C‐6s), 27.77 (C‐8), 9.50–9.19 (C‐9). Fischer Esterification is an organic reaction which is employed to convert carboxylic acids in the presence of excess alcohol and a strong acid catalyst to give an ester as the final product. \begin{align}\ce{R-CO-OH + HO{*}-R &<=>> R-CO-O{*}-R + H2O}\\[0.5em] Besides, it shows signals caused by propionic acid, the mixed anhydride, and the symmetric lauric anhydride. of Parallelograms and Triangles, Introduction First, a protonated alcohol is more acidic than a carboxylic acid by six or seven orders of magnitude, which means that the initial proton transfer would be slow and unfavorable. Starch esters have become increasingly attractive as sustainable materials. bhi. Thus, various reactions need to be investigated to get a reasonable image of the processes that may occur. The values of the DSlaurate and DSpropionate are dependent on the molar ratio of the AGU to lauric acid and propionic anhydride applied (Table 1). The ATR‐FTIR spectra were recorded with a Shimadzu IRAffinity‐1. Unfortunately, it has poor mechanical properties and is hydrophilic. CBSE scholarship for single girl child to the deserving candidates, who are the only child of their parents.


text:
Chapter 22

The peritoneal cavity is lined by the parietal peritoneum, a mesothelial lining. This lining is called the visceral peritoneum where it is reflected onto the enclosed abdominal organs. Its relationship to intraperitoneal structures defines discrete compartments within which abscesses may form (see Intra-abdominal Abscesses). The peritoneal surface area is a semipermeable membrane with an area comparable to that of the cutaneous body surface. Nearly 1 m2 of the total 1.7 m2 area participates in fluid exchange with the extracellular fluid space at rates of 500 mL or more per hour. Normally, there is less than 50 mL of free peritoneal fluid, a transudate with the following characteristics: specific gravity below 1.016, protein concentration less than 3 g/dL, white blood cell count less than 3000/μL, complement-mediated antibacterial activity, and lack of fibrinogen-related clot formation. The circulation of peritoneal fluid is directed toward lymphatics in the undersurface of the diaphragm. There, particulate matter—including bacteria up to 20 μm in size—is cleared via stomas in the diaphragmatic mesothelium and lymphatics and discharged mainly into the right thoracic duct.

The peritoneal cavity is normally sterile. Small numbers of bacteria can be efficiently disposed of, but peritonitis ensues if the defense mechanisms are overwhelmed by massive or continued contamination. In response to tissue damage, mast cells in the delicate mesothelial lining discharge histamine and other vasoactive substances that enhance vascular permeability. The resulting fibrinogen-rich plasma exudate supplies complement and opsonic proteins that promote bacterial destruction. Tissue thromboplastin released by injured mesothelial cells converts fibrinogen into fibrin, which may in turn lead to collagen deposition and formation of fibrous adhesions. In health, this reaction is limited by a plasminogen activator in the cell lining, but the plasminogen activator is inactivated by injury or infection. Bacterial lipopolysaccharide (endotoxin) and cytokines can stimulate production of tumor necrosis factor (TNF). TNF, in turn, mediates the release of plasminogen activator inhibitor produced by inflamed peritoneal mesothelial cells, which can lead to persistence of fibrin. Fibrin clots segregate bacterial deposits, a source of endotoxins that contribute to sepsis, but segregation may also inadvertently shield bacteria from bacteria-clearing mechanisms.

The omentum is a well-vascularized, pliable, mobile double fold of peritoneum and fat that participates actively in the control of peritoneal inflammation and infection. Its composition is well suited to sealing off a leaking viscus (eg, perforated ulcer) or area of infection (eg, resulting from a ruptured appendix) and for carrying a collateral blood supply to ischemic viscera. Its bacteria scavenger functions include absorption of small particles and delivery of phagocytes that destroy unopsonized bacteria.

Acute Secondary Bacterial Peritonitis

Pathophysiology

Peritonitis is an inflammatory or suppurative response of the peritoneal lining to direct irritation. Peritonitis can occur after perforating, inflammatory, infectious, or ischemic injuries of the gastrointestinal or genitourinary system. Common examples are listed in Table 22–1. Secondary peritonitis results from bacterial contamination originating from within viscera or from external ...

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text:
# Find the most-connected nodes in a graph

I feel like this is a simple problem that I have overcomplicated. Any tips or different approach would be greatly appreciated.

## Problem and goal

Given a list of coordinates (each line represents an 'edge' containing 'nodes') my goal is to obtain (half of the amount of the total) nodes that occur the most. I hope my example would clarify this problem:

### Input file

Each line represents a 'coordinate' (in reality an 'Edge'), that consists of 2 nodes (integers).

20   1
7   15
7   1
7   20
7   4
7   19

### Desired output

Preferably a vector that contains: [7,20,1]

Why 7,20,1? Because 7, 20 and 1 occur the most if you look at the input file. This is also half (3) of the total amount (6) of coordinates. Note: the output vector doesn't have to be sorted.

My goal is to increase the performance of my current approach (any other approach however is very appreciated (but please check my constraints in the next section). The current performance seems to be slow for 1 million edges (lines). An example of such a large coordinates file can be obtained from here (8 MB file).

### Constraints

• The input file is being read and the results are added as a vector of struct with type Coordinate (as my code will show). Please note that I can not change this (due to constraints that are beyond the scope of my question). In other words, the input parameter to my function is a vector<Coordinate>.
• I would like to obtain an array/vector as an output of the highest occurrences.

## My current approach and code

Essentially, I am 'looping' through my coordinates in the select_coordinates_highest_occurrence function and add each coordinate (along with its first and second element) to a map (function add_node_to_node_degree_map) that keeps track of the amount of the occurrences. If I do a simple benchmark, this seems to also be the main bottleneck of the code.

I then proceed by converting the map to a vector, since I am unable to sort the vector by the value.

After I have done that, I sort the vector (ascending) and 'cut' it in half so that I obtain the highest amount of coordinates.

#include <iostream>
#include <sstream>
#include <vector>
#include <fstream>
#include <string>
#include <unordered_map>
#include <algorithm>

typedef struct Coordinate {
int x;
int y;
} Coordinate;

void add_node_to_node_degree_map(std::unordered_map<int, int>& node_degree, int node) {
if (!node_degree.count(node)) {
node_degree[node] = 1;
}
else {
node_degree[node] = node_degree[node] + 1;
}
}

void select_coordinates_highest_occurrence(std::vector<Coordinate>& coordinates) {
// Map all vertices onto a map along with their degree
std::unordered_map<int, int> node_degree;

for (auto &edge : coordinates) {
}

// Convert the map to a vector
std::vector<std::pair<int, int>> node_degree_vect;
for (auto it = node_degree.cbegin(); it != node_degree.cend(); ++it)
{
node_degree_vect.push_back(std::make_pair(it->first, it->second));
}

// Sort the vector (ascending, high degree nodes are on top)
std::sort(node_degree_vect.begin(), node_degree_vect.end(), [](const std::pair<int, int> &left, const std::pair<int, int> &right) {
return left.second > right.second;
});

// Only collect the high occurrences (delete the 'other' half of the vector with lower occurrences)
std::vector<int> high_occurrences;

for (int i = 0; i < node_degree_vect.size() / 2; i++) {
high_occurrences.push_back(node_degree_vect[i].first);
}

for (int i = 0; i < high_occurrences.size(); i++) {
printf("\n%d", high_occurrences[i]);
}
}

/*
This method can be ignored and is only added to load a graph in case you want to test it on your system.
*/
void read_coordinates_into_vector(std::vector<Coordinate>& coordinates, std::string file_path) {
std::ifstream infile(file_path);
std::cout << file_path;
int vertex_source;
int vertex_destination;
while (infile >> vertex_source >> vertex_destination) {
Coordinate edge;
edge.x = vertex_source;
edge.y = vertex_destination;
coordinates.push_back(edge);
}

infile.close();
}

int main() {

std::vector<Coordinate> coordinates;
std::string file_path = "C:\\Users\\USERNAMEHERE\\Desktop\\coordinates_list.txt";

select_coordinates_highest_occurrence(coordinates);

return 0;
}

I really feel like I overcomplicated this solution.

• When I run this against the small example provided in your question it prints 2 1 and not 2 0 1. Is this desired or did I break something? – yuri Jun 7 '17 at 15:06
• @yuri You probably broke something. It prints 2 0 1 for me too. – Rakete1111 Jun 7 '17 at 15:07
• @yuri hi yuri, it prints 2 0 1 for me, but the 1 might be a little bit harder to see since it is next to the 'Press any key to continue' message to exit the command line terminal (on Windows at least) – CarrotCake Jun 7 '17 at 15:09
• Oh yeah indeed, there was no trailing newline and I did not notice the 0. Apologies. – yuri Jun 7 '17 at 15:11
• Are the node-numbers guaranteed to be consecutive, or at least a reasonable percentage of all node-ids from 0 to last guaranteed used? – Deduplicator Jun 7 '17 at 15:11

Here are some things that may help you improve your program.

## Clarify your specification

In the text, you describe the desired quantity of nodes as "half the amount of coordinates" but what the code actually does is choose half the number of unique edges. There could well be a difference if, say, we had a thousand coordinate pairs that each referred to the same two edges.

## Understand standard structures

There is a much simpler way to create your unordered_map that doesn't require any helper function. Because each value is initialized to zero, we can rely on this and simplify the code considerably:

std::unordered_map<int, int> node_degree;
for (const auto &coord : coordinates) {
++node_degree[coord.x];
++node_degree[coord.y];
}

## Use appropriate data structures

Your use of an unordered_map is a sound idea, but I'm not so sure about the conversion to a std::vector. It seems to me that a std::priority_queue more directly matches what you're trying to accomplish. Here's the equivalent to your select_coordinates_highest_occurrence (which I find a bit wordy):

void popularEdges(const std::vector<Coordinate> & coordinates) {
std::unordered_map<int, int> m;
for (const auto &coord : coordinates) {
++m[coord.x];
++m[coord.y];
}
std::priority_queue<std::pair<int, int>> pq;
for (const auto &node : m) {
pq.emplace(node.second, node.first);
}
for (auto n=pq.size()/2; n; --n) {
std::cout << pq.top().second << '\n';
pq.pop();
}
}

In my testing on my computer, this version is 25% faster than the original.

## Separate I/O from processing

It may depend on your particular needs, but I'd suggest that passing back a vector would be much more generally useful than simply printing from within the function. Similarly, the read_coordinates_into_vector prints the file name back to the user. I'd expect instead that it would be silent. The caller can just as easily print that string if needed.

## Make return variables return variables

The read_coordinates_into_vector currently requires the user to pass in a vector, but what would likely make more sense would be to have the code return a newly created vector instead.

## Don't use spurious typedef

C++ is not C. The typedef here is not needed:

typedef struct Coordinate { /*...*/ };

because in C++, Coordinate is a proper type without the additional keyword.

## Prefer std::istream to file names

The current code for read_coordinates_into_vector is only capable of handing a named file and not, say, a std::stringstream. It could be made more generic and, I think, more durable, if the function prototype were this:

std::vector<Coordinate> read_coordinates_into_vector(std::istream &in);

## Use a friend function and custom inserter

Use a friend function and custom inserter to simplify reading coordinates from a file:

struct Coordinate {
int x;
int y;
friend std::istream &operator>>(std::istream &in, Coordinate &c) {
return in >> c.x >> c.y;
}
};

std::vector<Coordinate> read(std::istream &in) {
std::vector<Coordinate> v;
Coordinate pair;
while (in >> pair) {
v.push_back(edge);
}
return v;
}

• Node: If a key doesn't exist in an std::unordered_map, it is not zero initialized, but value initialized. Which is technically not the same thing. – Rakete1111 Jun 7 '17 at 15:32
• I've changed the wording to simply say that it is "initialized to zero" without specifying the particular mechanism. – Edward Jun 7 '17 at 15:44
• I do believe typedef struct Coordinate { /*...*/ }; is an error. You dropped a second Coordinate. – Deduplicator Jun 7 '17 at 21:47
• It's not exactly an error; it's just weird. The syntax in the original is legal but strange C++ and the version I show above is legal C++ and not strange. Dropping the second Coordinate is required or it would be a syntax error when Coordinate is used later. – Edward Jun 7 '17 at 21:50

add_node_to_node_degree_map always does double the work needed, specifically two lookups (the second might be inserting) where one would suffice.
Easily fixed:

void add_node_to_node_degree_map(std::unordered_map<int, int>& node_degree, int node) {
++node_degree[node];
}

You can directly initialize your node_degree_vect from node_degree:

std::vector<std::pair<int, int>> node_degree_vect(node_degree.begin(), node_degree.end());

Anyway, I wonder why you didn't at least use a for-range-loop there.

If you only collect the result in a std::vector to immediately output it, I wonder why you didn't just output it and skipped the intermediate step.

I wonder why you explicitly close a file manually just for RAII to kick in and do that in the destructor...

On my machine (i7 6700HQ running gcc 7.1.1 and compiled with -O3) your code runs in the following time:

./main  0.57s user 0.18s system 89% cpu 0.834 total

That's pretty good. No matter what I tried, I always got a value around 0.5s, so I can't provide anything that can speed up your code, sorry.

1. Compile with every warning turned on. In my case, g++ reports 2 things:

main.cpp: In function 'void select_coordinates_highest_occurrence(std::vector<Coordinate>&)':
main.cpp:48:23: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
for (int i = 0; i < node_degree_vect.size() / 2; i++) {
~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
main.cpp:52:23: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
for (int i = 0; i < high_occurrences.size(); i++) {
~~^~~~~~~~~~~~~~~~~~~~~~~~~

2. I don't know if this goes against your requirement, but typedefing structs is only a thing in C. In C++, there is no difference, so just leave it out.

3. Mark structs that shouldn't be used as base classes final, to prevent any potential bugs about a non-virtual destructor.

4. My opinion: Using integers as booleans is confusing. So, I would change

if (!node_degree.count(node))

to

if (node_degree.count(node) == 0)

5. If you read std::unordered_map::operator[], you'll see that it inserts a default constructed element if it doesn't exist. So, add_node_to_node_degree_map can be simplified to:

++node_degree[node];

Which also means that you don't need a function for it.

6. Every binary arithmetic operator (let's call one %%) can be rewritten if it has the form

num = num %% a; // 'a' is another number

Using %%=:

num %%= a;

Example, node_degree[node] = node_degree[node] + 1; can become node_degree[node] += 1;, or in this case, even ++node_degree[node]; because you're only adding one.

7. Mark functions that do not throw as noexcept to help the compiler optimize (if you compile without exceptions, ignore this.)

8. Your naming is a bit too specific, and as a result, they are long, especially function names. Try to find shorter names, but not too short :)

9. I can only see one printf, so I'm not sure on this one:

• If you always use printf, even if you are outputting an integer, then don't, because you are using the C function, which is not guaranteed to be included. In this case, #include <cstdio> and change it to std::printf.

• If you removed it because you know that ADL will find printf anyways in std in this case, good job :) Still, you need to #include <cstdio>.

10. Use more range-based for loops:

for (const auto &node : node_degree)
node_degree_vect.push_back(std::make_pair(node.first, node.second));

11. You are inconsistent with braces, but only in one case. The one replaced above is the only one with opening braces being on a separate line.

12. int is not guaranteed to cover the whole range of std::size_t, which is returned by most .size() calls. So, you'll need to use std::size_t instead of int in loops:

for (std::size_t i = 0; i < node_degree_vect.size() / 2; ++i) {
high_occurrences.push_back(node_degree_vect[i].first);
}

13. Prefer pre-increment to post-increment. Even though for ints this is negligible and the compiler will optimize the temporary away, this might not be the case for other heavier types (like iterators.)

14. Use auto.

[](const std::pair<int, int> &left, const std::pair<int, int> &right) {
return left.second > right.second;
}

can become

[](const auto &left, const auto &right) {
return left.second > right.second;
}

15. I think you didn't write this, but if you did: std::ifstream's destructor closes the file, so you don't need to at the end.

16. A function returning void and taking a reference to a type is not very nice. Consider returning the variable instead of taking it by reference. In this case, you can initialize coordinates in 1 line instead of 2.

17. Print a trailing new line please.

18. select_coordinates_highest_occurrence doesn't modify coordinates. So please mark it const to prevent any accidental modifications.

19. You don't need return 0; for main, you can safely remove it.


text:
## What I learned today

Personal blog about programming and all that stuff

Following the guide in my last post you can safely install NodeJS on a Raspberrry Pi. If you try this on a Raspberrry Pi Zero W, this will take a very long time, because it builds NodeJS from source and this device has a very limited CPU performance. Here is a better and faster solution.

The Raspberrry Pi Zero W features an ARM11 CPU, which is fairly slow, compared to the Raspberrry Pi 3B+ or Raspberrry Pi 4. This makes the device suitable for application, where the focus is on power consumption. But running CPU intensive tasks on the Pi Zero is not so much fun.

I recently described how to set up NodeJS on a RaspebrryPi using nvm. The disadvantage of that method is, that ist takes a huge amount of time when run on a Raspberrry Pi Zero.

## Here is a better solution

### Find out, what type of Raspberrry Pi you own

To determine the type of Raspberrry Pi you own, you need an ssh connection to the device. Open the SSH console to run the following command:

uname -a

With this command I can determine the CPU architecture. This way I find out for which architecture NodeJS must be built. Look for a string in the results that starts with armv. Example:

Linux raspberrypi 5.4.51+ #1333 Mon Aug 10 16:38:02 BST 2020 armv6l GNU/Linux

Here I executed this command on a Raspberrry Pi Zero W. The message tells me that this device’s CPU arcitecture is armv61. In this example I need to download NodeJS built for armv6l. (For Raspberrry Pi 3 or 4 you’ll get armv7l as the result).

The full archive of all releases of NodeJS can be found here. There you can look for the appropriate package for your machine. For the Raspberrry Pi Zero (remember: armv6l) the latest release of NodeJS is v11.15.0 (at the time of writing):

wget https://nodejs.org/dist/latest-v11.x/node-v11.15.0-linux-armv6l.tar.gz

Next extract the files:

tar -xzf node-v1.15.0-linux-armv6l.tar.gz

Copy to /usr/local:

sudo cp -R node-v11.15.0-linux-armv6l/* /usr/local/

To find out if everything went well, check the version of NodeJS:

node --version
v11.15.0

Success! Now clean up your home folder. You don’t need these files any longer.

rm -rf node-v*


text:
# iFinger Oxford Dictionary of English

## How to uninstall iFinger Oxford Dictionary of English from your PC

iFinger Oxford Dictionary of English is a software application. This page holds details on how to remove it from your PC. It was developed for Windows by iFinger LTD. Take a look here for more info on iFinger LTD. Detailed information about iFinger Oxford Dictionary of English can be seen at . The program is often placed in the C:\Program Files (x86)\iFinger directory. Keep in mind that this location can differ being determined by the user's choice. MsiExec.exe /I{C6E6A038-5208-42B6-9434-D5423DAB6575} is the full command line if you want to remove iFinger Oxford Dictionary of English. iFinger.exe is the iFinger Oxford Dictionary of English's primary executable file and it occupies close to 4.56 MB (4783104 bytes) on disk.

The executables below are part of iFinger Oxford Dictionary of English. They occupy about 4.56 MB (4783104 bytes) on disk.

• iFinger.exe (4.56 MB)

The current page applies to iFinger Oxford Dictionary of English version 4.0.1 alone. You can find below info on other releases of iFinger Oxford Dictionary of English:

After the uninstall process, the application leaves leftovers on the computer. Part_A few of these are listed below.

Directories found on disk:
• C:\Program Files\iFinger

Check for and delete the following files from your disk when you uninstall iFinger Oxford Dictionary of English:
• C:\Program Files\iFinger\Data\OUP-ODE2.ifd
• C:\Program Files\iFinger\Data\OUP-ODE2.swx
• C:\Program Files\iFinger\Help\1030.chm
• C:\Program Files\iFinger\Help\1031.chm
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• C:\Program Files\iFinger\Help\4148.chm
• C:\Program Files\iFinger\iFinger.exe
• C:\Program Files\iFinger\iFinger.NO
• C:\Program Files\iFinger\Impborl.dll
• C:\Program Files\iFinger\LangConf.ini
• C:\Program Files\iFinger\lx\Default.ini
• C:\Program Files\iFinger\lx\english\english.clg
• C:\Program Files\iFinger\lx\english\english.dct
• C:\Program Files\iFinger\lx\english\english.ds
• C:\Program Files\iFinger\lx\english\english.hmm
• C:\Program Files\iFinger\lx\english\english.ia
• C:\Program Files\iFinger\lx\english\English.ini
• C:\Program Files\iFinger\lx\english\english.npr
• C:\Program Files\iFinger\lx\english\english.tok
• C:\Program Files\iFinger\lxrt23.dll
• C:\Program Files\iFinger\pcre3.dll
• C:\Program Files\iFinger\plugins\GenWin.ifp
• C:\Program Files\iFinger\resource\CharisSILR.ttf
• C:\Program Files\iFinger\resource\ip457788.TTF
• C:\Program Files\iFinger\resource\Wikipedia.bmp
• C:\Program Files\iFinger\WinHook.dll
• C:\Windows\Installer\{C6E6A038-5208-42B6-9434-D5423DAB6575}\controlPanelIcon.exe

Generally the following registry keys will not be removed:
• HKEY_LOCAL_MACHINE\SOFTWARE\Classes\Installer\Products\830A6E6C80256B2449435D24D3BA5657
• HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Uninstall\{C6E6A038-5208-42B6-9434-D5423DAB6575}

Additional registry values that you should clean:
• HKEY_CLASSES_ROOT\Local Settings\Software\Microsoft\Windows\Shell\MuiCache\C:\Program Files\iFinger\iFinger.exe
• HKEY_LOCAL_MACHINE\SOFTWARE\Classes\Installer\Products\830A6E6C80256B2449435D24D3BA5657\ProductName
• HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Installer\Folders\C:\Program Files\iFinger\
• HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Installer\Folders\C:\windows\Installer\{C6E6A038-5208-42B6-9434-D5423DAB6575}\

## How to erase iFinger Oxford Dictionary of English with the help of Advanced Uninstaller PRO

iFinger Oxford Dictionary of English is a program marketed by the software company iFinger LTD. Some people try to uninstall this program. Sometimes this is hard because deleting this by hand requires some experience regarding removing Windows programs manually. The best EASY manner to uninstall iFinger Oxford Dictionary of English is to use Advanced Uninstaller PRO. Here are some detailed instructions about how to do this:

1. If you don't have Advanced Uninstaller PRO already installed on your PC, add it. This is good because Advanced Uninstaller PRO is a very potent uninstaller and all around utility to maximize the performance of your system.

• set up Advanced Uninstaller PRO
2. Run Advanced Uninstaller PRO. Take some time to admire Advanced Uninstaller PRO's design and number of tools available. Advanced Uninstaller PRO is a powerful package of utilities.

3. Press the General Tools button

4. Click on the Uninstall Programs button

5. A list of the programs installed on your PC will be made available to you

6. Scroll the list of programs until you find iFinger Oxford Dictionary of English or simply activate the Search field and type in "iFinger Oxford Dictionary of English". If it is installed on your PC the iFinger Oxford Dictionary of English program will be found automatically. When you click iFinger Oxford Dictionary of English in the list of apps, the following data regarding the program is shown to you:

• Safety rating (in the lower left corner). The star rating tells you the opinion other people have regarding iFinger Oxford Dictionary of English, from "Highly recommended" to "Very dangerous".
• Reviews by other people - Press the Read reviews button.
• Technical information regarding the application you want to uninstall, by pressing the Properties button.
For instance you can see that for iFinger Oxford Dictionary of English:

• The web site of the application is: http://www.ifinger.com
• The uninstall string is: MsiExec.exe /I{C6E6A038-5208-42B6-9434-D5423DAB6575}
7. Press the Uninstall button. A confirmation window will show up. accept the uninstall by clicking Uninstall. Advanced Uninstaller PRO will then uninstall iFinger Oxford Dictionary of English.

8. After removing iFinger Oxford Dictionary of English, Advanced Uninstaller PRO will ask you to run a cleanup. Press Next to go ahead with the cleanup. All the items that belong iFinger Oxford Dictionary of English that have been left behind will be detected and you will be asked if you want to delete them. By uninstalling iFinger Oxford Dictionary of English with Advanced Uninstaller PRO, you are assured that no registry entries, files or directories are left behind on your system.

Your PC will remain clean, speedy and ready to run without errors or problems.

## Disclaimer

The text above is not a piece of advice to uninstall iFinger Oxford Dictionary of English by iFinger LTD from your computer, we are not saying that iFinger Oxford Dictionary of English by iFinger LTD is not a good application for your computer. This page simply contains detailed instructions on how to uninstall iFinger Oxford Dictionary of English supposing you decide this is what you want to do. The information above contains registry and disk entries that other software left behind and Advanced Uninstaller PRO stumbled upon and classified as "leftovers" on other users' computers.

2016-06-22 / Written by Andreea Kartman for Advanced Uninstaller PRO


text:
sir,
regarding the b.arch exam conducted by iitjee, do we have to send a request letter or simply register and go to give the test.
also this time registering date(counselling date) is from 30th may to 12 th june. whereas test is on 10tj june. how is that possible????
my iitjee 2010 rank is 8510. can I get b.arch in kharagpur or roorkee????

4 years ago

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### Answers : (1)

                                        Dear kumarCandidates can register for this Aptitude test On-line after May 26, 2010. They must appear for the test on June 10,2010 at the institutes where counselling will be held.You should also bring your Admit card.First Go to the online Choice filling site .once you complete the registration then you will be provided a login ID and password.use this id and password to login .in next step of online filling ,you will also get the option for registration for B.Arch cource /B.Deshttp://jee.iitm.ac.in/counseling/counsel_help-desk/online_2010.html.  ( for online  filling)http://jee.iitm.ac.in/counseling/  ( here you will get last year seat allotement in different IITs ,vacancy in cources. councelling Broucher)Read all the information carefully given in above link .Please feel free to post as many doubts on our discussion forum as you can.we will get you the answer and detailed  solution very  quickly. All the best. Regards,Askiitians ExpertsBadiuddin

4 years ago

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text:
# How to Create Square Thumbnails in K2 for Joomla! 1.5

K2 is definitely one of my favorite Joomla! components. It does everything so well that I rarely need to make edits to anything other than template files. The one exception is creating square thumbnails. Most of my customers incapable of creating square thumbnails, so we need the system to be as user friendly as possible.

This is a quick tutorial on creating square cropped thumbnails with K2. We are going to make the small images square, but you could apply this code to any image size.

*This modification will be lost with each upgrade to k2, so be sure to keep a backup of this code handy.

Search for “//Small image”

We are going to replace the entire block of code responsible for the image resizing.

Replace this:

//Small image
$handle->image_resize = true;$handle->image_ratio_y = true;
$handle->image_convert = 'jpg';$handle->jpeg_quality = $params->get('imagesQuality');$handle->file_auto_rename = false;
$handle->file_overwrite = true;$handle->file_new_name_body = $filename.'_S'; if (JRequest::getInt('itemImageS')) {$imageWidth = JRequest::getInt('itemImageS');
} else {
$imageWidth =$params->get('itemImageS', '200');
}
$handle->image_x =$imageWidth;
$handle->Process($savepath);

With:

//Small image
$handle->image_resize = true;$handle->image_convert = 'jpg';
$handle->jpeg_quality =$params->get('imagesQuality');
$handle->file_auto_rename = false;$handle->file_overwrite = true;
$handle->file_new_name_body =$filename.'_S';
if (JRequest::getInt('itemImageS')) {
$imageWidth = JRequest::getInt('itemImageS'); } else {$imageWidth = $params->get('itemImageS', '200'); }$handle->image_x = $imageWidth;$handle->image_y = $imageWidth;$handle->image_ratio_crop = true;
$handle->Process($savepath);

That’s it!

You can now upload your images to your items. All small images (images set as small) will now be cropped square. The width and height will match the default size for the small image size, or the custom size set by you in the category parameters.

Be Sociable, Share!
Tarik Assagai
Tarik (@tarik916) is a web developer and owner of ACS Digital Media, a Sacramento based web development agency specializing in open source CMS systems. If you are looking for support for your current or upcoming projects, visit www.acs-digital.com.

1. I hate mixed thumbnail ratios.
This is one of those little things which make a huge difference!!! Should be standard and optional for all images. Hope to see something like that in K2 future releases!

2. This is really sweet. Agreeing that it would be a nice radio button option for core K2. Thanks!

3. Very nice tip for K2. This works very well for clients who manage the content of their own sites and they don’t have any clue about crop images from Photoshop or any other programs. Thank you.

4. Krakaos /

5. Jafar /

wow, thanks for this.

It’s so easy to crop an image in the code, why wouldn’t this feature be added to the admin?

6. How do you apply this to K2_v2.5.1, can’t seem to find this in the item.php file?

• tarik916 /

Hi Jason,
Are you searching under the administrator folder (administrator\components\com_k2\models\item.php)? In version 2.5.1, the image resize code starts around line 198.

• Raphael /

I did that in the latest k2 version and I’m afraid it doesn’t work. any clues??

• tarik916 /

I am pretty sure I made the same change in the latest version and it worked properly. Do you not see the resize code inside of the item.php file?

7. dalibor /

Is there any way how to crop images in K2 2.5.4?
thank you

8. Hello here

I am trying to add this code to my website.

Unfortunately, image doesnt wants to resize..

It can be seen here:

Why could be this problem ?

Thank You.

Tom

9. Debbie /

Nice improvement to k2 but i couldn’t help wondering…
is it possible to do this with the category image?

• tarik916 /

This is possible, but I have not had to do it yet since categories are added less frequently than content items. Most people just upload a square image. If you want to do it through the code, you can edit administrator/components/com_k2/models/category.php.

• PeterK /

Hi Tarek,
I found the code at the category.php but it is another one then the code at the item.php. How is the right syntax to do the same crop as item.php?

10. mark /

Cant find the code in the image.php of k2 version 2.5.7.

has it moved and if so where to i cant find it?

Thank you

11. Hmmm, I found item.php but it’s not resizing my thumbnails here:
http://penelopes.xyonet.com/index.php/unique-products/carriers

Can’t wait for Joomla/K2 to build this in.

12. hi tariq! assalamo alekum
thanks for the best code …

13. i’m trying it on k2 2.6.2 but it doesn’t work at all. Have you check it on this version?