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Regnskapsfører

Bookkeeper er ditt lokale regnskapsbyrå som tilbyr et bredt spekter av tjenester som regnskap, lønn, HR, rådgivning blant andre. Vår regnskapsførere vil ta seg av regnskap og finans jobb å sørge for at du lever et bekymringsløst liv. Vi sørger for at regnskap er enkelt, tilgjengelig og morsomt for hver kunde som bruker våre tjenester.

Secure Research Funding With Visuals

and 1 collaborator

Among the many challenges scientists face today, a major headache is **securing funding**. Generally, scientists receive funding based on how much attention their research is estimated to generate. The more popular the topic, the more likely it is to receive funding. For instance, research on cancer gene BRCA2 is more likely to gain traction than frog copulation processes... for now. Fishing in a smaller pool of money means that scientists need a competitive edge to get a bite.

Fear not! There are ways to increase attention and discussion of the research for popular and nonpopular topics alike. Infographics and interactive data allow researchers to communicate more effectively and engage readers in a refreshing way. Content with visuals get 94% more total views and is 40x more likely to get shared on social media (Lee). Thus, visualized data can be the path to funding.

Numerki

and 1 collaborator

*- Równania liniowe*

*Równanie Laplace*

Równanie to wyraża następującą własność pola potencjalnego: dywergencja (rozbieżność) wektorowego pola potencjalnego (czyli gradientu potencjału), pod nieobecność źródła jest równa zeru. Opisuje ono zatem wiele procesów zachodzących w przyrodzie, np. potencjał grawitacyjny poza punktami źródeł pola (czyli bez punktów materialnych), potencjał prędkości cieczy przy braku źródeł. Równanie Laplace’a jest szczególnym przypadkiem równania Poissona, wyrażającego analogiczny związek w przypadku istnienia źródeł pola.

$ \bigtriangleup u = \sum_{i=1}^{n} u_{x_ix_i} =0$

$ u_{x_i} = \frac{du}{dx_i}$

$ u_{x_i x_i} = \frac{d^2u}{dx_i^2}$

*Równanie Poisona*

Równanie Poissona opisuje wiele procesów zachodzących w przyrodzie, np. rozkład pola prędkości cieczy wypływającej ze źródła, potencjał pola grawitacyjnego w obecności źródeł, potencjał pola elekrostatycznego w obecności ładunków, temperaturę wewnątrz ciała przy stałym dopływie ciepła.

△*u* = *f*

*Liniowe równanie transportu*

$u_t + \sum_{i=1}^{n}b^iu_{x_i} = 0$

*przewodnictwa / dyfuzji*

*u*_{t} − △*u* = 0

*Schrodingera*

*u*_{t} + △*u* = 0

*falowe*

*u*_{tt} − △*u* = 0

*- Równania nieliniowe*

*nieliniowe równanie Poissona*

− △ *u* = *f*(*u*)

*p-harmoniczne*

*d**i**r*(|*D**u*|^{P − 2}*D**u*)=0

dir - dywergencja

$dir v = \sum_{i=1}^n \frac{dv_i}{dx_i}$

*D**u* = ▽*u*

Food resources - info about healthy eating, habits and more

and 1 collaborator

Brian Wansink - Mindless Eating - Why we eat more than we think

How To Submit Your Essay

- Using Authorea, write your submission.
- When complete, share your document with
__contest@authorea.com__

Essay Contest: How has social media enhanced your research?

Social media is generally discouraged in science today. A recent article, "I'm a serious academic, not a professional Instagrammer" castigated scholars with active online social lives. Most advisors won't ask you to "tweet out our paper" or "write a blog post about our findings" and it's likely that you'd close Twitter or Reddit if your colleague or advisor walked by. Some conferences have taken it so far as to enforce a "no tweeting" policy. But social networking is here to stay and will likely become even more integrated into our lives *and* research.

Welcome to Authorea!

@inproceedings{el2010two,

title={Two-step scheduling algorithm for IEEE802. 16 wireless networks},

author={El-Shinnawy, Ahmed H and Badawi, Ashraf H and Nassar, Amin M},

booktitle={Communication Technology (ICCT), 2010 12th IEEE International Conference on},

pages={620--623},

year={2010},

organization={IEEE}

}

Reinventing Peer Review

and 2 collaborators

Peer review is arguably necessary for effective communication amongst researchers. Authors, editors, and the public rely on peer review to ensure a first measure of trust in scientific communication. While peer review is considered to be integral in scholarly communication by most, its shortcomings are becoming evident. Former editor of *JAMA *and *NEJM* Drummond Rennie once said, "if peer review was a drug it would never be allowed onto the market." Is this true? Does peer review, as it is done today, cause more harm than good?

PHYS6562 F2 Daily Quaetion

First observe that the probability density at the origin at any time is zero.

\begin{equation} \rho(0,t) = 0 \end{equation}

And the probability density on *x* = *x*′ at *t* = 0 is

\begin{equation} \rho(x,0) = \delta(x-x') \end{equation}

We can assume that there is a image probability density at *t* = 0 on *x* = −*x*′. Therefore the total probability at *t* = 0 should be

\begin{equation} \rho(x,0) = \delta(x-x')-\delta(x+x') \end{equation}

The Green’s function for each delta function is a time-dependent Gaussian distribution,

\begin{equation} G(x,t) = \frac{1}{\sqrt{4\pi Dt}} e^{-x^2/4Dt} \end{equation}

with the particle originally at the origin. For this case, both delta functions diffuse, yielding the same Gaussian distribution.

\begin{equation} G(x,t) = \frac{1}{\sqrt{4\pi Dt}} \left( e^{-(x-x')^2/4Dt} - e^{-(x+x')^2/4Dt}\right) \end{equation}

The time evolution probability distribution is shown in fig.1.

PHYS6562 M3 Daily Question

Most of the volume of high dimensional spheres is near the surface. As we can approximate the volume as *V* ≈ *r*^{3N} in 3N dimensions, near the center, *r* ≈ 0 and *V* ≈ 0. Near the surface, where *r* becomes larger and contributes to most of the volume.

For statistical mechanics, instead of taking the whole E sphere into account, which might involve high-dimensional integral, it’s easier to focus on the shell of E sphere. Then the momenta can be considered to be $p = \sqrt{2mE}$ instead of doing

\begin{equation} \int_0^{\sqrt{2mE}} (....)dp \end{equation}

PHYS6562 W2 Daily Question

a. The density of tin atoms *ρ*_{Sn}(**x**) is locally conserved since there is no tin atom created or annihilated.

b. It depends. Under the constraints that one cannot change his/her political belief, yes, the density of democrats is locally conserved. However, if changing of political belief is allowed, the density of democrats is not locally conserved. There might be democrats changing to republicans or vise versa, which implies that the democrats or republicans can teleport, created or annihilated.

Suppose there is no external force. We can write the current as

\begin{equation} J = -D(\rho)\nabla\rho \end{equation}

And apply

\begin{equation} \frac{\partial\rho}{\partial t} = -\nabla\cdot J \end{equation}

\begin{equation} \frac{\partial\rho}{\partial t} = \frac{\partial D(\rho)}{\partial\rho} \nabla\rho\cdot\nabla\rho + D(\rho)\nabla^2\rho \end{equation}

\begin{equation} \frac{\partial\rho}{\partial t} = \frac{\partial D(\rho)}{\partial\rho} (\nabla\rho)^2 + D(\rho)\nabla^2\rho \end{equation}

This is the diffusion equation with density-dependent diffusion constant.

PHYS6562 W3 Daily Question

First of all, *Ω*_{1}(*E*_{1})*δ**E*_{1} and *Ω*_{2}(*E*_{2})*δ**E*_{2} are the phase space volume for these two subsystems. Since most of the volume in phase space is near the surface, we can instead write *Ω*_{1}(*E*_{1}) and *Ω*_{2}(*E*_{2}) to denote the two volumes in phase space. Second, it’s clear that the probability density of subsystem 1 having energy *E*_{1} is proportional to the ’allowed’ volume product of the two subsystems.

\begin{equation}
\rho_1(E_1) \propto \Omega_1(E_1)\Omega_2(E_2)
\end{equation} where *E*_{2} = *E* − *E*_{1}. And finally, we impose the normalization condition:

\begin{equation}
\int \rho_1(E_1)dE_1 = A \int \Omega_1(E_1)\Omega_2(E-E_1)dE_1 = A \Omega(E) = 1
\end{equation} A is the proportional constant and equal to 1/*Ω*(*E*). Therefore,

\begin{equation} \rho_1(E_1) = \frac{\Omega_1(E_1)\Omega_2(E_2)}{\Omega(E)} \end{equation}

If we maximize the probability at *E*_{1}:

\begin{equation} \frac{d\rho_1}{dE_1} = 0 \end{equation}

\begin{equation} \frac{1}{\Omega_1}\frac{d\Omega_1}{dE_1} - \frac{1}{\Omega_2}\frac{d\Omega_2}{dE_2} = 0 \end{equation}

By plugging the definition of entropy *S* = *k*_{B}*l**o**g*(*Ω*),

\begin{equation} \frac{1}{k_B}\frac{dS_1}{dE_1} - \frac{1}{k_B}\frac{dS_2}{dE_2} = 0 \end{equation}

\begin{equation} \frac{dS_1}{dE_1} + \frac{dS_2}{dE_1} = \frac{d}{dE_1}(S(E_1)+S(E_2)) = 0 \end{equation} ,which implies the maximization of the sum of entropies of both subsystems.

For (6), we see that if the probability density *ρ*(*E*_{1}) is maximized,

\begin{equation} \frac{1}{T_1} = \frac{1}{T_2} \end{equation}

PHYS6562 F3 Daily Quaetion

For N distinguishable particles, if at certain instant they occupy N states in the phase space, there are *N*! different choices of such configuration. The total volume for this system in phase space is the product of the volume for each particle in its own sub phase space.

\begin{equation} \Omega_D = N! \space \omega_1 \omega_2 \omega_3 ... \omega_N \end{equation}

If the particles are now undistinguished, there is only one choice of such configuration without permutation.

\begin{equation} \Omega_U = \omega_1 \omega_2 \omega_3 ... \omega_N \end{equation}

Therefore, $\Omega_D = N!\space \Omega_U $.

For the mixing problem of black and white particles of the same pressure on both sides, if you cannot tell the difference between black and white ones, you cannot extract work from the mixing. Actully, you won’t know if they’ve mixed already or not. However, suppose the black and white particles are of different pressure, and you still cannot tell them apart. If there were a membrane which allows particles of lower pressure to penetrate through, this membrane then can tell the difference between the two sides without telling the particles apart. Then work can be extracted.

Now if a door can distinguish the black from the white particles and let through white ones, after very long time, the white particles will distribute themselves equally in the two sides, i.e. there will be $\frac{1}{4} N\space white + 0 \space black$ in the left and $\frac{1}{4} N \space white + \frac{1}{2} N \space black$ in the right. And we can extract work from it due to the pressure difference between the two sides.

PHYS6562 W4 Daily Question

It’s not true that neighboring initial conditions will converge to the fixed state after long time. As we saw from the Jupiter problem, there are some regions where the initial conditions for chaotic and periodic trajectories lies infinitesimally close but lead to different results. The chaotic initial conditions will not make the system converge to certain state. Therefore, there is no attractor in this 3-body problem.

Stable Hamiltonian systems can withstand a minor perturbation *in the process* and the system will still end up where it’s supposed to.

Unstable Hamiltonian system will diverge if perturbed *in the process* and the system will not come to the original equilibrium regardless of whether the system becomes chaotic or not.

PHYS6562 F4 Daily Quaetion

The system of classical particles bouncing back and forth is not ergodic since each particles can only occupy certain close volume in phase space, with fixed *x*, *y* position, *P*_{x}, *P*_{y} momenta and ±*p*_{z}. Therefore, the given volume in phase space of the ensemble will not evolve with time.

(a) True.

(b) True, since the work in done on the gas.

(c) False, it’s a refrigerator.

(d) True.

\begin{equation}
W = (P_0)(3V_0) + \int_{4V_0}^{V_0} \frac{4P_0V_0}{V} dV = 3P_0V_0 - 4P_0V_0\log4
\end{equation} (e) True, a part does no work so *Δ**U* = *Q*. \begin{equation}
\Delta U = \frac{3}{2} nRT_c - \frac{3}{2} nRT_h = -\frac{3}{2} (3P_0V_0) = -\frac{9}{2}P_0V_0
\end{equation} where n is the number of particles in mole and *R* = 8.317 is the ideal gas constant. (f) True. The whole cycle *Δ**U* = 0, the heat input and the net work done *onto* the gas must equal the heat output. \begin{equation}
Q_h + W = Q_c
\end{equation}

PHYS6562 M5 Daily Question

(a) \begin{equation} \Omega = \frac{V^N}{N!} \end{equation}

\begin{equation}
S = k_B \space \log(\Omega) = k_B(N\log(V) - B\log(N) + N) = Nk_B\left(1+\log\left(\frac{V}{N}\right) \right)
\end{equation} The average volume of a region for single particle is *v* = *V*/*N* with the entropy of *s* = *k*_{B}log(*v*). Therefore, the configuration entropy is related to N times the entropy of a particle in its region with average volume.

(b) \begin{equation}
S = Nk_B\left(\frac{5}{2}-\log(\rho\lambda^3)\right) = Nk_B\left(\frac{5}{2}+\log\left(\frac{V}{N}\lambda^3\right)\right)
\end{equation} The configuration entropy for an ideal gas particle is also related to the small average volume *v*. \begin{equation}
S_c = Nk_B\log(v)
\end{equation}

But the ideal gas particles also occupy the momentum space and each particle has the average length of *λ* in each dimension. So

\begin{equation} S_p = 3Nk_B\log(\lambda) = Nk_B\log(\lambda^3) \end{equation}

Combine the configuration and momentum entropy for one particle, \begin{equation} S = S_c+S_p = Nk_B\log(v*\lambda^3) = Nk_B\log\left(\frac{V}{N}*\lambda^3\right) \end{equation} The total entropy is related to the N times the space occupied both in configuration and momentum space.

PHYS6562 W5 Daily Question

For a child’s room containing 100 things in the room, we only need to consider the configuration space since those things are stationary in the room. The room before tightening up, the number of configurations is

\begin{equation} \Omega = V^N = 5^{100} \end{equation}

Let’s consider the center-of-mass location accuracy of 1 cm. First of all, the center-of-mass can be located itself at any point of the room. Therefore,

\begin{equation} \Omega_{CM} = 5 \end{equation}

Then we can assume that the center of mass in *x* direction is at 0. For each randomly located thing *A*1 at *x*, there would be the another thing *A*1′ located at the −*x*. We now have only 50 things we can throw at will. For the *A*50′, since the deviation of 1cm is acceptable, *A*50′ can have roughly 1m of freedom in the *x* direction. Therefore,

\begin{equation} \Omega_{item} = (L^{50}*1)^3 = 5^{50} \end{equation}

where *L* = 5^{1/3}.

\begin{equation} \Omega' = \Omega_{CM}\times\Omega_{item} = 5^{51} \end{equation}

Entropy decrease can be found to be

\begin{equation} \Delta S = k_B\log{\Omega'} - k_B\log{\Omega} = -49k_B\log{5} \end{equation} \begin{equation} \Delta S_{total} = 10^9\times\Delta S = 1.09\times10^{-12} \end{equation}

For 1 liter baloon contracting by 1% in volume, \begin{equation}
\Delta S = Nk_B\log{0.99V} - Nk_B\log{V} = \frac{PV}{T}\log(0.99)
\end{equation} where *P* = 1*a**t**m*, *V* = 1*L* and *T* = 300*K*. \begin{equation}
\Delta S = -0.0034 (J/K)
\end{equation}

PHYS6562 F5 Daily Quaetion

(a)

For discrete system, the entropy \begin{equation}
S_{discrete} = -k_B \sum p_i\log p_i
\end{equation} \begin{equation}
S_{discrete} = \frac{3}{2}k_B \log 2
\end{equation}

(b)

Entropy for being hit by a comet with $\rho_c(\theta,\phi) = \frac{1}{4\pi}$: \begin{equation}
S_c = -k_B \iint \rho_s \log (\rho_s) d\Omega = k_B \log (4\pi)
\end{equation} Entropy for being hit by an asteroid with $\rho_a(\theta,\phi) = \frac{cos(\theta)}{\pi^2}$: \begin{equation}
S_a = -k_B \iint \rho_a \log (\rho_a) d\Omega = -k_B \int \left( \frac{2\cos^2\theta}{\pi}\log(\cos\theta) - \frac{4\cos^2\theta}{\pi}\log\pi \right)d\theta
\end{equation} \begin{equation}
S_a = -k_B \left(\frac{1}{2}(1-\log 4) - 2\log(\pi) \right) = k_B \log \left(\frac{2\pi^2}{\sqrt{e}} \right)
\end{equation}

The entropy difference \begin{equation} \Delta S = S_c - S_a = k_B\log\left(\frac{2\sqrt{e}}{\pi} \right) \end{equation}

How do I write chemical formulae?

The best way to type chemical formulae in your document is by using LaTeX syntax. You can use our powerful equation editor if you write in HTML mode, or if you already use LaTeX or Markdown, here’s how to do it. First, make a macro in your header file. Access header.tex via “Quick Edit” or via the Folder view and add the following line in it:

`\newcommand*\chem[1]{\ensuremath{\mathrm{#1}}}`

Then, go back to the article view and type some chemical formulae, for example:

```
\chem{AlBr_3}
$\chem{H_2} + (1/2)\,\chem{O_2} = \chem{H_2O}$
$2\,\chem{H_2} + \chem{O_2} \to 2\,\chem{H_2O}$
```

They will render as:

$\chem{H_2} + (1/2)\,\chem{O_2} = \chem{H_2O}$

$2\,\chem{H_2} + \chem{O_2} \to 2\,\chem{H_2O}$

Proposal

Online learning is viewed as an effective way for distributing knowledge ubiquitously, especially with the robust development of Massive Open Online Courses (MOOCs) by academic institutions such as Stanford University's edX and Coursera. From enrollment to managing class discussions and assignments, these MOOCs succeed in migrating the classroom entirely to the cloud. The same concept has been applied to distant or blended learning, where learners are able to interact with materials uploaded online. Through such interactions, learners undergo a process of self motivated learning.

Self motivated learning (SML) describes the way in which an eager learner incorporates various self-regulation processes with task strategies and self-motivational beliefs. These learners go through their learning processes in three cyclical phases, namely: forethought, performance control and self-reflection Zimmerman 2000. In the second phase, learners gather information to evaluate the effectiveness of the strategic plan imposed and improve future attempts of learning. This phase is crucial as research shows students who are trained in self regulation processes display high levels of achievement Schunk 1996. In light of this view, some educators pointed to knowledge graphs for facilitating sharing of learning content.

In experiments led by researchers, learners are encouraged to use cognitive structures (i.e. knowledge graphs) to help themselves interpret, organize and share their knowledge. The experimental results show increased qualitative performance as well as increased engagement among both learners and teachers Chu 2010 Hwang 2011.

While current learning management systems lack promotion on SML, learners tend to spend unnecessary amounts of time on skimming and searching for designated information. The most obvious reason is that those learning management systems serve merely as repositories of information, rather than a catalyst for fostering SML activity. Therefore, a comprehensive learning management system that has an interface to create knowledge graphs for course material is needed for motivating self motivated learners, and ultimately for improving online learning throughput.

This paper aims at two contributions. First, it provides an implementation of the knowledge graph cognitive structure that integrates as a learning management system. Second, it proposes several interactive plug-ins for the creation and management of learning assets and evaluation of learners' performance.

Genetics Template

This Authorea document template can be used to prepare documents according to a desired citation style and authoring guidelines. Abstracts are not always required, but most academic papers have one
and writers should know how to produce a useful abstract. An abstract
should be a very short, clear and concise summation of the entire paper.
An abstract should provide enough of a preview that a typical reader will
know whether or not they wish to read the paper. It should reveal both
the purpose and conclusions of the paper.