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# Representation Theories and Algebraic Geometry (Nato Science

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Presentation of the principal models: classical particle systems, lattice gases, and spin systems. Exercise 2. 3 .) Solution. 2. 4 ] =. which can be simpliﬁed to Part 4 as ( ).1 −. 1− −1 All of these values will produce a value equal to ( ). Suppose that maps the hyperbola > 0. based on boundedness. that is. Show that an arbitrary intersection of algebraic sets is an algebraic set.. There are different stability notions coming from different ideas of constructing the moduli space.

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Presentation of the principal models: classical particle systems, lattice gases, and spin systems. Exercise 2. 3 .) Solution. 2. 4 ] =. which can be simpliﬁed to Part 4 as ( ).1 −. 1− −1 All of these values will produce a value equal to ( ). Suppose that maps the hyperbola > 0. based on boundedness. that is. Show that an arbitrary intersection of algebraic sets is an algebraic set.. There are different stability notions coming from different ideas of constructing the moduli space.

# An Alpine Anthology of Homotopy Theory

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Numerical algebraic geometry is a growing area of algebraic geometry that involves describing solution sets of systems of polynomial equations. We assume = + 1 in the 0 = 1 patch and show that = + 1 in the = 1 patch. (. DRAFT COPY: Complied on February 4. 2010. The zeros of on the curve are the elements of these are (0: 0: 1). then div( ) = (0: 1: 0) + 2(0: 0: 1) − 3(0: 1: 0) = Exercise 3. The semigroups in this case have tractable structures, while the distribution of primes is more mysterious and related to some deep combinatorial conjectures on matroids. be a connected smooth Shimura variety.

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Numerical algebraic geometry is a growing area of algebraic geometry that involves describing solution sets of systems of polynomial equations. We assume = + 1 in the 0 = 1 patch and show that = + 1 in the = 1 patch. (. DRAFT COPY: Complied on February 4. 2010. The zeros of on the curve are the elements of these are (0: 0: 1). then div( ) = (0: 1: 0) + 2(0: 0: 1) − 3(0: 1: 0) = Exercise 3. The semigroups in this case have tractable structures, while the distribution of primes is more mysterious and related to some deep combinatorial conjectures on matroids. be a connected smooth Shimura variety.

# C#: Crash Course - The Ultimate Beginner's Course to

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Instead drawings and pictures that are meant to provide "geometric" feel are supplanted. For if = ( ) were a holomorphic function that expresses as its graph near. explain why ∂ coordinate at .266 Algebraic Geometry: A Problem Solving Approach Exercise 3. ) aﬃne plane. = (1: 0: 1). ) coordinates and give an alternate expression for as the graph of a function near. is a local coordinate by ∂ the Implicit Function Theorem.

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Instead drawings and pictures that are meant to provide "geometric" feel are supplanted. For if = ( ) were a holomorphic function that expresses as its graph near. explain why ∂ coordinate at .266 Algebraic Geometry: A Problem Solving Approach Exercise 3. ) aﬃne plane. = (1: 0: 1). ) coordinates and give an alternate expression for as the graph of a function near. is a local coordinate by ∂ the Implicit Function Theorem.

# Geometry Part 1

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There has been a great deal of work in various kinds of persistent homology (a methodology for inferring topological invariants of a geometric object from finite samples with error from the object), the homological properties of sensor networks and their implications for coverage and other questions, and the extension of algebraic topological tools for qualitative analysis of dynamical systems (Conley indices, for example) to tools in the finite approximation and stochastic settings. That is a bit more diﬃcult in this case.. ) and (. .. . which shows that (. ).

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There has been a great deal of work in various kinds of persistent homology (a methodology for inferring topological invariants of a geometric object from finite samples with error from the object), the homological properties of sensor networks and their implications for coverage and other questions, and the extension of algebraic topological tools for qualitative analysis of dynamical systems (Conley indices, for example) to tools in the finite approximation and stochastic settings. That is a bit more diﬃcult in this case.. ) and (. .. . which shows that (. ).

# Surfaces and Planar Discontinuous Groups (Lecture Notes in

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We know that by rearranging these four points. 3. [ 1. 2 ( − 1)2 which we called in section 2. −1 .5. 1. cubic curves? We call 0 the leading coeﬃcient of the polynomial 0.. . A complex aﬃne change of coordinates in the complex plane ℂ2 is given by = = where .3. depending on one’s mood. − ∕= 0.7.2. realaffineinverse Solution. ellihyper Exercise 1. = + + + ) ). This allowed us to deﬁne the ring of functions on.

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We know that by rearranging these four points. 3. [ 1. 2 ( − 1)2 which we called in section 2. −1 .5. 1. cubic curves? We call 0 the leading coeﬃcient of the polynomial 0.. . A complex aﬃne change of coordinates in the complex plane ℂ2 is given by = = where .3. depending on one’s mood. − ∕= 0.7.2. realaffineinverse Solution. ellihyper Exercise 1. = + + + ) ). This allowed us to deﬁne the ring of functions on.

# Mathematical Methods for Curves and Surfaces II:

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A divisor is said to be eﬀective (or positive) if ni ≥ 0 for all i. The last two-thirds of the semester concerns functional analysis: normed linear spaces, convexity, the Hahn-Banach Theorem, duality for Banach spaces, weak convergence, bounded linear operators, Baire category theorem, uniform boundedness principle, open mapping theorem, closed graph theorem, compact operators, Fredholm theory, interpolation theorems, L^p theory for the Fourier transform.

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A divisor is said to be eﬀective (or positive) if ni ≥ 0 for all i. The last two-thirds of the semester concerns functional analysis: normed linear spaces, convexity, the Hahn-Banach Theorem, duality for Banach spaces, weak convergence, bounded linear operators, Baire category theorem, uniform boundedness principle, open mapping theorem, closed graph theorem, compact operators, Fredholm theory, interpolation theorems, L^p theory for the Fourier transform.

# Curves and Abelian Varieties: International Conference March

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The material in the book (homology and cohomology theory, universal coefficient theorems, Kunneth theorem, duality in manifolds, applications to classical theorems of point-set topology) is for the most part solid. If ( ) is 2 2 a parabola in ℝ .20. and if ( ) is a hyperbola in ℝ2. then 2 −4 > 0.1classifycorollary + 2 + + + ℎ) is an Corollary 1. then a parabola or ellipse.e.1.19. )= 2 + 2 + 2 + + ℎ. and > 0. but if the coeﬃcient = ( 2 + = 0. any hyperbola into any other hyperbola. ) = ′ 2 + ′ + ′ each of whose coeﬃcients is a polynomial in.

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The material in the book (homology and cohomology theory, universal coefficient theorems, Kunneth theorem, duality in manifolds, applications to classical theorems of point-set topology) is for the most part solid. If ( ) is 2 2 a parabola in ℝ .20. and if ( ) is a hyperbola in ℝ2. then 2 −4 > 0.1classifycorollary + 2 + + + ℎ) is an Corollary 1. then a parabola or ellipse.e.1.19. )= 2 + 2 + 2 + + ℎ. and > 0. but if the coeﬃcient = ( 2 + = 0. any hyperbola into any other hyperbola. ) = ′ 2 + ′ + ′ each of whose coeﬃcients is a polynomial in.

# Rational Points: Seminar Bonn/Wuppertal 1983/84 (Aspects of

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Thus, the fundamental \htmladdnormallink{anabelian}{ http://planetphysics.us/encyclopedia/IsomorphismClass.html} question posed by Grothendieck was, and is: {\em how much information about the isomorphism class of the variety $X$ is contained in the knowledge of the etale fundamental group?} (on p. 2 in $$http://www.math.jussieu.fr/~leila/SchnepsLM.pdf$$ ). This OpenCourseWare includes lecture notes, problem sets and a final exam.

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Thus, the fundamental \htmladdnormallink{anabelian}{ http://planetphysics.us/encyclopedia/IsomorphismClass.html} question posed by Grothendieck was, and is: {\em how much information about the isomorphism class of the variety $X$ is contained in the knowledge of the etale fundamental group?} (on p. 2 in $$http://www.math.jussieu.fr/~leila/SchnepsLM.pdf$$ ). This OpenCourseWare includes lecture notes, problem sets and a final exam.

# Algebraic Threefolds: With Special Regard to Problems of

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Vector bundles on Riemann surfaces: classification. We have −1 1( 2 ( 0: 0 = ℳ( 0: 0: 0). consider the set Exercise 1. ( 0: 0: 0 )) ∈ Σ}. Thus, for spaces and maps, the classification up to homotopy equivalence precisely captures their qualitative features. One possibility would be to read some classic books on the arithmetic theory of elliptic curves, which has recently had exciting applications, such as in the proof of Fermat's Last Theorem, but other topics are also possible

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Vector bundles on Riemann surfaces: classification. We have −1 1( 2 ( 0: 0 = ℳ( 0: 0: 0). consider the set Exercise 1. ( 0: 0: 0 )) ∈ Σ}. Thus, for spaces and maps, the classification up to homotopy equivalence precisely captures their qualitative features. One possibility would be to read some classic books on the arithmetic theory of elliptic curves, which has recently had exciting applications, such as in the proof of Fermat's Last Theorem, but other topics are also possible

# Fibre Bundles

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Differentiable manifolds (of a given dimension) are all locally diffeomorphic (by definition), so there are no local invariants to a differentiable structure (beyond dimension). A polynomial is homogeneous if every monomial term has the same total degree.. has degree three. (1) Since the ﬁrst terms have degree two and the last term has degree one. We now want to show.17. + + 2 + = = = = ( + )2 + ( + )( + + ) + ( − )2 + ( − )( + + ) 2 +2 2 2 + 2 2 +2 ( + + )+ 2 2 2 −2 + 2 2 4 +2 +2 +2 +2 +2 +2 +2 Exercise 1. by Exercise 1.9.

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Browse Questions

# Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{−7}\; C?$ The radii of A and B are negligible compared to the distance of separation.

$1.5 \times 10^{-2} N$

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1. ## simple question about combinations

Hello,
I have an issue with my little experiment it's working but I'm trying to optimize it... to reduce number of operations... and similar things...

It's hard to explain, but let's say that it is similar to the betting systems... when you play system 10 of 20 (meaning you have to win at least 10 of 20 games to get paid... if more win, more you get paid ) System 10 of 20 have 184756 combinations (without repetition, and order is not important) and that's a really big number... Let's say that we have 20 variables and each one of them have some value $(x_1 = 10, x_2 = 20, .... x_{20} = 5)$ if i play system 10 of 20 (or any other) and i win 14 of those 20 games... that means from 184756 combinations, i have winning 1001 combination... and then comes problem...
I have to calculate each of those 1001 combinations and then add them all... I'm not interested in the values of every one of those 1001 combination, just total value at the end...

I have created file with all of the combinations 2-20, 3-20, .... 19-20 and it's really really big file (with couple of "for loops") ... but it's necessary so my program know how and what to multiply .... $(x_1 * x_2 * x_3...* x_{10}, .....)$ depending on let's say system....

I tried to optimize number of summations and multiplications, but any bigger system from 5 it's getting really messy....

(for smaller it's ok.... like 3-5...
$x_1 [x_2 (x_3 + x_4 + x_5) +x_3 (x_4 + x_5) + x_4 x_5] + x_2 (x_3 (x_4 + x_5)) + x_3 x_4 x_5$
with this i reduced calculation to from 9 to 8 sumations and from 20 to 6 multiplications... but for any bigger then 5-10 i got headache hehehehehehe )

So I wonder, is there any kind of formula to get that final result or i have to do it step by step... and go through all of those combinations?

Thanks,
any kind of help is really really welcome

P.S. Sorry if I put this at the wrong place... i haven't been here for a while

2. ## Re: simple question about combinations

You seem to be beating around the bush regarding details of your calculations and that makes it difficult to see if there are any patterns that collapse into a nice expression. Without more detail (i.e. how you arrived at the 3-5 sum), it is difficult to help you. Have you considered using matrices?


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# gpytorch.likelihoods¶

## Likelihood¶

class gpytorch.likelihoods.Likelihood(max_plate_nesting=1)[source]

A Likelihood in GPyTorch specifies the mapping from latent function values $$f(\mathbf X)$$ to observed labels $$y$$.

For example, in the case of regression this might be a Gaussian distribution, as $$y(\mathbf x)$$ is equal to $$f(\mathbf x)$$ plus Gaussian noise:

$y(\mathbf x) = f(\mathbf x) + \epsilon, \:\:\:\: \epsilon ~ N(0,\sigma^{2}_{n} \mathbf I)$

In the case of classification, this might be a Bernoulli distribution, where the probability that $$y=1$$ is given by the latent function passed through some sigmoid or probit function:

$\begin{split}y(\mathbf x) = \begin{cases} 1 & \text{w/ probability} \:\: \sigma(f(\mathbf x)) \\ 0 & \text{w/ probability} \:\: 1-\sigma(f(\mathbf x)) \end{cases}\end{split}$

In either case, to implement a likelihood function, GPyTorch only requires a forward method that computes the conditional distribution $$p(y \mid f(\mathbf x))$$.

Calling this object does one of two things:

• If likelihood is called with a torch.Tensor object, then it is assumed that the input is samples from $$f(\mathbf x)$$. This returns the conditional distribution p(y|f(mathbf x)).
• If likelihood is called with a MultivariateNormal object, then it is assumed that the input is the distribution $$f(\mathbf x)$$. This returns the marginal distribution p(y|mathbf x).
Args:
max_plate_nesting (int, default=1)
(For Pyro integration only). How many batch dimensions are in the function. This should be modified if the likelihood uses plated random variables.
expected_log_prob(observations, function_dist, *args, **kwargs)[source]

(Used by VariationalELBO for variational inference.)

Computes the expected log likelihood, where the expectation is over the GP variational distribution.

$\sum_{\mathbf x, y} \mathbb{E}_{q\left( f(\mathbf x) \right)} \left[ \log p \left( y \mid f(\mathbf x) \right) \right]$
Args:
observations (torch.Tensor)
Values of $$y$$.
function_dist (MultivariateNormal)
Distribution for $$f(x)$$.
args, kwargs
Passed to the forward function
Returns
torch.Tensor (log probability)
forward(function_samples, *args, data={}, **kwargs)[source]

Computes the conditional distribution $$p(\mathbf y \mid \mathbf f, \ldots)$$ that defines the likelihood.

Parameters: function_samples (torch.Tensor) – Samples from the function ($$\mathbf f$$) data (dict) – (Optional, Pyro integration only) Additional variables ($$\ldots$$) that the likelihood needs to condition on. The keys of the dictionary will correspond to Pyro sample sites in the likelihood’s model/guide. args – Additional args kwargs – Additional kwargs Distribution object (with same shape as function_samples) Distribution
get_fantasy_likelihood(**kwargs)[source]
log_marginal(observations, function_dist, *args, **kwargs)[source]

(Used by PredictiveLogLikelihood for approximate inference.)

Computes the log marginal likelihood of the approximate predictive distribution

$\sum_{\mathbf x, y} \log \mathbb{E}_{q\left( f(\mathbf x) \right)} \left[ p \left( y \mid f(\mathbf x) \right) \right]$

Note that this differs from expected_log_prob() because the $$log$$ is on the outside of the expectation.

Args:
observations (torch.Tensor)
Values of $$y$$.
function_dist (MultivariateNormal)
Distribution for $$f(x)$$.
args, kwargs
Passed to the forward function
Returns
torch.Tensor (log probability)
marginal(function_dist, *args, **kwargs)[source]

Computes a predictive distribution $$p(y^* | \mathbf x^*)$$ given either a posterior distribution $$p(\mathbf f | \mathcal D, \mathbf x)$$ or a prior distribution $$p(\mathbf f|\mathbf x)$$ as input.

With both exact inference and variational inference, the form of $$p(\mathbf f|\mathcal D, \mathbf x)$$ or $$p(\mathbf f| \mathbf x)$$ should usually be Gaussian. As a result, function_dist should usually be a MultivariateNormal specified by the mean and (co)variance of $$p(\mathbf f|...)$$.

Args:
function_dist (MultivariateNormal)
Distribution for $$f(x)$$.
args, kwargs
Passed to the forward function
Returns:
Distribution object (the marginal distribution, or samples from it)
pyro_guide(function_dist, target, *args, **kwargs)[source]

(For Pyro integration only).

Part of the guide function for the likelihood. This should be re-defined if the likelihood contains any latent variables that need to be infered.

Parameters: function_dist (MultivariateNormal) – Distribution of latent function $$q(\mathbf f)$$. target (torch.Tensor) – Observed $$\mathbf y$$. args – Additional args (for forward()). kwargs – Additional kwargs (for forward()).
pyro_model(function_dist, target, *args, **kwargs)[source]

(For Pyro integration only).

Part of the model function for the likelihood. It should return the This should be re-defined if the likelihood contains any latent variables that need to be infered.

Parameters: function_dist (MultivariateNormal) – Distribution of latent function $$p(\mathbf f)$$. target (torch.Tensor) – Observed $$\mathbf y$$. args – Additional args (for forward()). kwargs – Additional kwargs (for forward()).

## One-Dimensional Likelihoods¶

Likelihoods for GPs that are distributions of scalar functions. (I.e. for a specific $$\mathbf x$$ we expect that $$f(\mathbf x) \in \mathbb{R}$$.)

One-dimensional likelihoods should extend gpytoch.likelihoods._OneDimensionalLikelihood to reduce the variance when computing approximate GP objective functions. (Variance reduction is accomplished by using 1D Gauss-Hermite quadrature rather than MC-integration).

### GaussianLikelihood¶

class gpytorch.likelihoods.GaussianLikelihood(noise_prior=None, noise_constraint=None, batch_shape=<MagicMock name='mock()' id='140401726369464'>, **kwargs)[source]

### BernoulliLikelihood¶

class gpytorch.likelihoods.BernoulliLikelihood(*args, **kwargs)[source]

Implements the Bernoulli likelihood used for GP classification, using Probit regression (i.e., the latent function is warped to be in [0,1] using the standard Normal CDF Phi(x)). Given the identity Phi(-x) = 1-Phi(x), we can write the likelihood compactly as:

$\begin{equation*} p(Y=y|f)=\Phi(yf) \end{equation*}$

## Multi-Dimensional Likelihoods¶

Likelihoods for GPs that are distributions of vector-valued functions. (I.e. for a specific $$\mathbf x$$ we expect that $$f(\mathbf x) \in \mathbb{R}^t$$, where $$t$$ is the number of output dimensions.)

class gpytorch.likelihoods.MultitaskGaussianLikelihood(num_tasks, rank=0, task_correlation_prior=None, batch_shape=<MagicMock name='mock()' id='140401727320696'>, noise_prior=None, noise_constraint=None)[source]
A convenient extension of the gpytorch.likelihoods.GaussianLikelihood to the multitask setting that allows for a full cross-task covariance structure for the noise. The fitted covariance matrix has rank rank. If a strictly diagonal task noise covariance matrix is desired, then rank=0 should be set. (This option still allows for a different log_noise parameter for each task.). This likelihood assumes homoskedastic noise.
class gpytorch.likelihoods.SoftmaxLikelihood(num_features=None, num_classes=None, mixing_weights=True, mixing_weights_prior=None, **kwargs)[source]


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Chinese Journal of Chemical Physics  2018, Vol. 31 Issue (1): 27-32

#### The article information

Zhi-guo Zhang, Min Xin, Shu-tao Zhao, Yang Chen

Imaging Isocyanic Acid Photodissociation at 193 nm: the NH(a1Δ)+ CO(X1Σ+) Channel

Chinese Journal of Chemical Physics, 2018, 31(1): 27-32

http://dx.doi.org/10.1063/1674-0068/31/cjcp1706120

### Article history

Received on: June 15, 2017
Accepted on: July 26, 2017
Imaging Isocyanic Acid Photodissociation at 193 nm: the NH(a1Δ)+ CO(X1Σ+) Channel
Zhi-guo Zhanga, Min Xina, Shu-tao Zhaoa, Yang Chenb
Dated: Received on June 15, 2017; Accepted on July 26, 2017
a. School of Physics and Electronic Engineering, Fuyang Normal University, Fuyang 236041 China;
b. Hefei National Laboratory for Physical Sciences at the Microscale and Department of Chemical Physics, University of Science and Technology of China, Hefei 230026, China
*Author to whom correspondence should be addressed. Zhi-guo Zhang, E-mail:zhgzhang@mail.ustc.edu.cn; Min Xin, E-mail:xinmin_0916@163.com; Yang Chen, E-mail:yangchen@ustc.edu.cn
Abstract: The photodissociation dynamics of isocyanic acid (HNCO) has been studied by the timesliced velocity map ion imaging technique at 193 nm. The NH(aΔ) products were measured via (2+1) resonance enhanced multiphoton ionization. Images have been accumulated for the NH(aΔ) rotational states in the ground and vibrational excited state (v=0 and 1). The center-of-mass translational energy distribution derived from the NH(aΔ) images implies that the CO vibrational distributions are inverted for most of the measured NH(v|j) internal states. The anisotropic product angular distribution observed indicates a rapid dissociation process for the N-C bond cleavage. A bimodal rotational state distribution of CO(v) has been observed, this result implies that isocyanic acid dissociates in the S1 state in two different pathways.
Key words: Isocyanic acid    Photodissociation dynamics    Sliced velocity map ion imaging
Ⅰ. INTRODUCTION

The photochemistry and thermochemistry of isocyanic acid (HNCO) is fundamentally as well as practically interesting. It plays a significant role in the rapid reduction of nitrogen oxides in the atmosphere [1, 2]. It can serve as a benchmark system for understanding multiple decomposition pathways in a four-atom molecule, such as internal conversion (IC), intersystem crossing (ISC), and direct dissociation.

Its first UV absorption band (180$-$280 nm) is broad and continuous with weak superimposed vibrational bands above 220 nm. It has been analyzed by Dixon and Kirby [3] and by Rabalais et al. [4, 5] as an S$_1$($^1$A$''$)$\leftarrow$S$_0$($^1$A$'$) transition. Photodissociation dynamics of HNCO have been revealed by a great deal of experimental and theoretical studies and are found to be exceptional complicated on account of at least three potential surfaces (S$_0$, S$_1$, and T$_1$) participating in the dissociation [6$-$29]. There are three different dissociation channels:

 $\begin{eqnarray} \textrm{HNCO}(\textrm{S}_1) \rightarrow &&\hspace{-0.25cm}\textrm{NH} (X^3\Sigma^-) + \textrm{CO}(X^1\Sigma^+) \\ && D_0=(30060\pm30) \hspace{0.15cm}\textrm{cm}^{-1} \end{eqnarray}$ (1)
 $\begin{eqnarray} \textrm{HNCO}(\textrm{S}_1) \rightarrow&&\hspace{-0.25cm}\textrm{H}(^2\textrm{S})+\textrm{NCO} (X^2\Pi) \\ &&D_0=(38370\pm30) \hspace{0.15cm}\textrm{cm}^{-1} \end{eqnarray}$ (2)
 $\begin{eqnarray} \textrm{HNCO}(\textrm{S}_1) \rightarrow&&\hspace{-0.25cm}\textrm{NH} (a^1\Delta) + \textrm{CO}(X^1\Sigma^+) \\ && D_0=(42750\pm30) \hspace{0.15cm}\textrm{cm}^{-1} \end{eqnarray}$ (3)

In the following paragraphs, NH(X$^3$$\Sigma^-) and NH(a^1$$\Delta$) are denoted by $^3$NH and $^1$NH, respectively.

Channel (1) is a spin-forbidden dissociation pathway. It is observed from 280 nm to at least up to 217 nm. Zyrianov et al. [23] detected CO around 230 nm via ion imaging technique. Their results showed that the angular distribution of CO originating from channel (1) is essentially isotropic. The isotropic angular distributions indicate that the lifetime of the intermediate state exceeds 5 ps. They suggested that the dissociation of the channel follows the internal conversion (IC) from S$_1$ to S$_0$ and then ISC from S$_0$ to T$_1$. The exact quantum yield is indefinite, but channel (1) should still be the primary channel in the region just above the opening of channel (2).

Channel (2) is a spin-allowed dissociation pathway and has been investigated in recent years. Spiglanin et al. [10] detected NCO via laser induced fluorescence (LIF) technique. Zhang et al. [11] studied this channel at 193 nm using the high-Rydberg H atom time-of-flight method. They found that the NCO fragment is substantial bending excited and the anisotropy parameter $\beta$=$-$0.85. It indicated that the channel is a direct dissociation process on a repulsive surface. Zyrianov et al. [23] studied this channel in the photolysis wavelength range of 215$-$243 nm. Their results indicated that NCO was rotationally cold and the angular distribution was isotropic. Recently, Yu et al. [28] reinvestigated this channel in 200$-$240 nm photolysis by high resolution HRTOF technique. Their results indicated that at low photon energy excitation HNCO dissociates in the ground state S$_0$, following IC from S$_1$ to S$_0$, which is consistent with an indirect dissociation mechanism. While as the photon energy increases, a new pathway of direct dissociation through S$_1$ appears and is more and more important.

Channel (3) is also a spin-allowed dissociation pathway and has been studied at several different wavelengths. Fujimoto et al. [6] probed CO($v$) at 193.3 nm and levels up to $v$=4 were observed. Spiglanin and co-workers [7, 8] studied the internal state distributions of CO and rotational state distributions of $^1$NH at several wavelengths between 230 and 193 nm. They found that the $^1$NH rotational distribution is cold, but the CO distribution is hot. Wang et al. [27] investigated this channel at 210 nm via ion velocity slice imaging technique for CO product detection. Zyrianov et al. [15, 18, 23, 24] performed a series of experiments to study this channel in the wavelength region of 217$-$230 nm. This dissociation pathway is thought to evolve initially on S$_0$, but after exceeding a small barrier on S$_1$, estimated at 400$-$600 cm$^{-1}$, direct dissociation on this surface quickly dominates. Channel (3) becomes the major channel. Recently, we investigated channel (3) at 201 nm using sliced velocity map ion imaging technique. From the image of $^1$NH product, a bimodal rotational distribution has been observed in the vibrational ground state of CO.

In this work, we further study channel (3) of HNCO photodissociation dynamics at 193 nm by the sliced velocity map ion imaging the $^1$NH fragments. The center-of-mass total translational energy distributions and angular distributions were derived from the images. Experimental results show that both the vibrational ground state and the vibrational excited state of CO($v$) show a bimodal rotational distribution. This result further deepens our understanding of the photodissociation mechanism of HNCO.

Ⅱ. EXPERIMENTS

The photodissociation experiments were carried out on a time-sliced velocity map ion imaging apparatus described elsewhere [29, 30]. In brief, a skimmed molecular beam containing 2% HNCO seeded in He was produced by a pulsed valve (General valve series 9) with a 0.5 mm nozzle. At a distance of 22 mm downstream from the nozzle, the supersonic expanded beam was skimmed to form a well collimated molecular beam by a 1 mm diameter aperture skimmer. After passing through a 2 mm hole in the first electrode plate, the HNCO/He beam aligned along the time-of-flight axis was intersected by two counter-propagating (pump and probe) laser beams in the detection zone.

The pump laser was generated by an ArF excimer laser(EX50/250), the typical power of $\sim$0.3 mJ per pulse was focused by a spherical quartz lens with f=300 mm. The radiation could be polarized by an eight-plate stack of quartz slides placed at the Brewster angle, resulting in approximately 90% polarization. While the probe laser was produced by doubling the output of a tunable dye laser (Cobra-Stretch, Sirah), which was pumped by the third harmonic of a continuum Nd: YAG laser (Continuum PL8000), the typical power of $\sim$0.5 mJ per pulse was focused by a spherical quartz lens with $f$=200 mm.

The $^1$NH fragments were ionized by the $g^1\Delta$(3p$\pi)$ $\leftarrow\leftarrow$$a^1\Delta(2+1) REMPI scheme \sim265 nm for ^1NH(v=0|$$j$) and $\sim$267 nm for $^1$NH($v$=1$|$$j) [31]. The polarization direction of the photolysis laser was set to be parallel to the detector face, while that of the probe laser was set to be perpendicular to the detector face. The ^1NH ions were accelerated by the focusing electric fields and projected onto a 40 mm diameter Chevron multi-channel plates (MCP) coupled to a P-47 phosphor screen (APD 3040FM, Burle Electro-Optics). A fast high-voltage switch (PVM-4210, DEI; typical duration\approx50 ns) was used to gate the gain of the MCP's for mass selection as well as the time slicing of the ion packets. The transient images shown on the phosphor screen were captured by a charge-coupled device (CCD) camera (ImagerPro2 M 640\times480 pixels, LaVision) and transferred to a computer on an every shot basis for event counting [32] and data analysis. Meanwhile, the total fluorescence from the phosphor screen was monitored by a photomultiplier tube (PMT) to optimize the experimental conditions. Timing of the pulsed valve, pump and probe lasers, the gate pulse applied to the MCP detector was controlled by two multichannel digital delay pulse generators (DG645, SRS). Ⅲ. RESULTS AND DISCUSSION A series of ^1NH images were recorded by setting the probe wavelength at the resolved ^1NH(v=0 and 1|$$j$) rotational states. FIG. 1 shows $^1$NH($v$=0$|$$j) sliced images of HNCO photodissociation at 193 nm, obtained by accumulating the ^1NH^+ signals over 2\times10^4laser shots. Well-resolved anisotropic rings were obtained and these structures are assigned to the rovibrational states of the partner CO product in the ^1NH+CO binary dissociation process, channel (3). These results provide the correlated information of HNCO photodissociation at 193 nm.  FIG. 1 Raw sliced images of ^1NH(v=0|j) products after photodissociation of HNCO at 193 nm. The double arrow indicates the polarization direction of the photodissociation laser. (A) j=2, (B) j=5, (C) j=8, (D) j=10. As seen in the images of FIG. 1, the relative intensities of the structures in each image change with the ^1NH rotational states. With the rotational energy increasing, the number of rings gradually decreases, indicating clear pair-correlation between the ^1NH(v=0|$$j$) and CO($v$$|j) products. The corresponding total kinetic energy distributions for the different ^1NH(v=0|$$j$) products were derived from the images, results are depicted in FIG. 2. The structure in these distributions represents the internal energy distribution of CO formed in conjunction with the particular probed level of $^1$NH($v$=0$|$$j). As seen in FIG. 2(a), the CO vibrational excitation distributions extend to v=4, this result is consistent with Fujimoto et al.'s [6]. Clearly, the CO vibrational distributions are inverted and peak at v=2 for most of the measured ^1NH(v=0|$$j$) products. And the vibrational level of CO($v$) decreases as the $^1$NH($v$=0$|j$) rotational energy increases. Most of the vibrational ground state and the excited state of CO($v$) shown in FIG. 2 can be consistently described by the sum of two broad rotational distributions which peak at low-$j$ and high-$j$ values.

 FIG. 2 The product total kinetic energy distributions (black empty circles) from FIG. 1 for $^1$NH($v$=0$|j$). The red lines are the fitting results and the dash lines are the individual CO rotational components. (a) $j$=2, (b) $j$=5, (c) $j$=8, (d) $j$=10.

A qualitative fitting of the energy distributions were carried out for extracting the correlated vibrational distribution of the CO product. The structure was fitted in total kinetic energy space using a Gaussian line shape. The CO vibrational states correlated with $^1$NH were fitted by a bimodal distribution which could account for the obvious two peaks in the distribution. The simulated results are also depicted in FIG. 2.

FIG. 3 shows $^1$NH($v$=1$|j$) sliced images after photodissociation of HNCO at 193 nm. Compared with $^1$NH($v$=0$|j$), the ring sizes of vibrational excited $^1$NH($v$=1$|j$) are smaller. The correlated CO vibrational state distribution shifts to lower vibrational state, as shown in FIG. 4. As can be seen, the vibrational level of CO($v$) decreases as the $^1$NH rotational energy increases. This indicates that the vibrational excitation of the correlated CO product is anti-correlated to the $^1$NH rotational excitation.

 FIG. 3 Raw sliced images of $^1$NH($v$=1$|j$) products after photodissociation of HNCO at 193 nm. The double arrow indicates the polarization direction of the photodissociation laser. (A) $j$=2, (B) $j$=5, (C) $j$=7, (D) $j$=9.
 FIG. 4 The product total kinetic energy distributions (black empty circles) from FIG. 3 for $^1$NH($v$=1$|j$). The red lines are the fitting results and the dash lines are the individual CO rotational components. (a) $j$=2, (b) $j$=5, (c) $j$=7, (d) $j$=9.

By integrating the images over the relevant radius region, angular distribution of the $^1$NH($v$$|$$j$)+CO($v$) channel was obtained for the $^1$NH($v|j$) rotational states. FIG. 5(a) shows the corresponding product anisotropy parameters determined by fitting the angular distributions. It is easy to see that anisotropy parameters of the $^1$NH($v$=0$|j$)+CO($v$) are negative. The angular distribution becomes slightly less anisotropic as the $^1$NH($v$=0$|j$) rotational energy increases. The $^1$NH($v$=0$|j$)+CO($v$=0 and 4) anisotropy parameters become as low as $-$0.3, however, this low value should be due to the weak signal compared with the roughly isotropic background.

 FIG. 5 (a) Anisotropy parameter for individual CO vibrational state correlated to the $^1$NH($v$=0$|j$). (b) Anisotropy parameter for individual CO vibrational state correlated to the $^1$NH($v$=1$|j$).

As seen in FIG. 4 only $v$=0, 1 and 2 of CO can correspond with vibrational excited $^1$NH($v$=1$|j$). The $^1$NH($v$=1$|j$) products also display anisotropic angular distribution (FIG. 5(b)). With $j$ increasing, the angular distribution also becomes slightly less anisotropic.

The anisotropic product angular distribution (FIG. 5) suggests a perpendicular transition and a fast dissociation process which most likely occurs on a repulsive surface, which is consistent with the results of Reisler et al.'s [23]. HNCO has nearly a linear geometry in the S$_0$ state. When HNCO is excited from the ground state to the S$_1$ state, the N$-$C$-$O angle is strongly bent. At 193 nm excitation energy the barriers to all the channels are exceeded. In this condition, since excited HNCO on the S$_1$ surface has a strong repulsive gradient along the CN bond, preference should be given to the rapid dissociation on S$_1$. However, the anisotropy parameter $\beta$ becomes considerably smaller as $^1$NH rotational level increases. This phenomenon has been also seen previously in HNCO photodissociation at 210 and 201 nm [27, 29], which could be explained by a non-axial recoil impact model [33].

As shown in FIG. 2 and 4, both the vibrational ground state and the excited state of CO($v$) show a bimodal rotational distribution. And the CO low-$j$ component and the CO high-$j$ component both change as $^1$NH $j$ increases in both the $^1$NH($v$=0$|j$) and $^1$NH($v$=1$|j$) channels. By integrating the images over the relevant radius region, angular distributions of the corresponding CO low-$j$ and high-$j$ components were obtained for the $^1$NH($v|j$) rotational state. The anisotropy parameters are depicted in Table Ⅰ. Clearly, all the CO products display anisotropic angular distributions, and the values for the low-$j$ and high-$j$ components are only slightly different. This result is similar to our previous results [29]. It implies that both low-$j$ and high-$j$ components are from HNCO rapid dissociation via a repulsive surface on S$_1$. Recently, by full-dimensional theoretical calculation, Bonnet et al. [34] reproduced the bimodal rotational state distribution of CO. Our experimental results are further verified by theoretical calculation. Moreover, Bonnet et al. provided new insight into the dissociation mechanism. The observed bimodal rotational distribution of CO should be the consequence of an impulsive-deflective mechanism due to two repulsive walls of the 6D PES, the first one producing rotationally hot CO products and the second one cooling part of them [34].

Table Ⅰ Anisotropy parameters of the CO high-$j$ and low-$j$ components corresponding $^1$NH($v|j$).
Ⅳ. CONCLUSION

In this report, using the sliced velocity map ion imaging technique, the photodissociation dynamics of HNCO at 193 nm for NH($a^1$$\Delta)+CO(X^1$$\Sigma^+$) channel has been studied. The NH($a^1$$\Delta) images have been measured. The state-resolved imaging results show anisotropic product angular distribution and correlation between the NH(a^1$$\Delta$) and CO rovibrational state distributions. The vibrational excitation of the correlated CO product is anti-correlated to the $^1$NH rotational excitation. The anisotropic product angular distribution suggests that the NH($a^1$$\Delta)+CO(X^1$$\Sigma^+$) dissociation channel is a fast dissociation process. A bimodal rotational distribution of CO has been observed from the $^1$NH images. It should be the consequence of an impulsive-deflective mechanism. This experimental result further deepens our understanding of the HNCO photodissociation dynamics.

Ⅴ. ACKNOWLEDGMENTS

This work was supported by the National Natural Science Foundation of China (No.21573227, NO.11604052), the National Science Foundation of Anhui Province of China (No.1608085QA19), the Natural Science Research Project of Education Department of Anhui Province of China (No.2014KJ020), the Open Foundation of State Key Laboratory (No.SKLMRD-K201503, No.SKLMRD-K201611, and No.SKLMRD-K201711), and the Doctoral Foundation of Fuyang Normal University (No.FSB201501005).

Reference
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a. 阜阳师范学院物理与电子工程学院, 阜阳 236041;
b. 中国科学技术大学化学物理系, 合肥微尺度物质科学国家研究中心, 合肥 230026


text:
# Methane Bubbles

## Summary

Bubbles filled with methane rise when released. The rising bubbles are set on fire with a candle on a stick to demonstrate the combustion reaction.

## Hazards

Methane gas is flammable.

## Chemicals and Solutions

• Bubble Solution
• Methane source

## Materials

• Glass funnel with long stem
• Large crystallizing dish
• Long stick with candle attached
• Matches
• Methane gas outlet
• 2-3 feet of amber tubing

## Procedure

1. Place the soap solution in the crystallizing dish.
2. Attach one end of the tubing to the gas outlet and the other end to the stem of a funnel.
3. Place the mouth of the funnel into the soap solution.
4. Light the candle and hold the stick in one hand.
5. With the other hand, turn the gas on low such that bubbles form slowly.
6. Once the inside of the funnel is coated with soap, remove the funnel from the soap solution and hold it such that the mouth points upward. Once you have a large bubble, shake it loose from the funnel. As the bubble rises, try to ignite it with the candle.

Hint: Be careful to release the bubbles before lighting, or the funnel will catch fire. If the funnel does catch fire, turn off the methane to extinguish.

## Discussion

$$\ce{ CH4_{(g)} + 2O2_{(g)} -> 2H2O_{(g)} + CO2_{(g)} ~~~~ \Delta G^{{o}}_{Rxn} = -801.1 KJ/Mol }$$

## Disposal

The soap solution can be reused.


text:
# Cutting a cake into $2$ pieces of equal area

You have a large rectangular cake, and someone cuts out a smaller rectangular piece from the middle of the cake at a random size, angle and position in the cake (see the picture below). Without knowing the dimensions of either rectangle, using one straight (vertical) cut, how can you cut the cake into two pieces of equal area?

Here is my attempt at the solution. I assume that all the dimensions are known.

Area of large rectangular cake $=ab\text{ }(a,b\in \mathbb{R}^+,a>b)$

Area of smaller rectangular piece $=cd\text{ }(c\in [0,a],d\in [0,b],c>d)$

Total area $=ab-cd$

Area right of vertical cut $=mb$ (we need to determine $m$)

Area left of vertical cut $=(a-m)b-cd$

Equate both expressions: $mb=(a-m)b-cd$

$$cd=(a-2m)b$$

Therefore

$$m=\frac{1}{2}\left(a-\frac{cd}{b}\right)$$

Is there a way to answer the question without assuming we know the dimensions.

• The word "vertical" in the question is confusing. You might change it to, "one straight cut". – daniel Oct 8 '16 at 20:34
• @daniel, if we imagine a 3D picture of a rectangular cake, the problem is trying to say that we should not cut the cake horizontally (half way of its height) because cutting horizontally will also solve the problem and makes two equal parts. Therefore vertical cuts in here means to cut the cake from top surface to its bottom. – Seyed Oct 8 '16 at 22:49
• @Seyed: yes i know what the problem is trying to say. a 'vertical line' with reference to the picture is not what is meant. – daniel Oct 9 '16 at 5:13
• It's probably not necessary to even use the phrase "vertical cut"... Assuming we are dealing with separating the new shape into two equal parts in terms of area, we can just deal with a single "cut" along the provided plane. – Eliseo d'Annunzio Feb 13 at 1:10


text:
The improper filter

Idea

A filter is improper if it is the whole thing … whatever thing the filters are filters in.

Definitions

Definition (improper filter in a lattice or other proset)

If $L$ is a lattice or even a proset, then $L$ is an filter in itself, the improper filter.

Definition (improper filter on a set)

If $S$ is a set, then the improper filter on $S$ is the power set of $S$.

Of course, the improper filter on $S$ is simply the improper filter in the power set of $S$.

Properties

The improper filter contrasts with proper filters (all of the other filters).

The improper filter does not count as a prime filter or an ultrafilter, because it is too simple to be simple, although it may satisfy careless or naïve definitions of those concepts (which must be required to be proper).

Conversely, sometimes one defines ‘filter’ to exclude the improper filter, especially in analysis and topology; however, this makes the set of all filters less nice to work with.

Created on August 17, 2012 21:52:06 by Toby Bartels (98.19.44.121)


text:
# Appendix Title with scrreprt

I'm trying to get the word Appendix in front of my appendix chapter titles. I would like it to look like:

Appendix A My first appendix

I tried the appendix package, using \usepackage[title,titletoc]{appendix}. Somehow this works in the toc but not in the title on the page itself. How can add it for the title as well?

Here what I got:

\documentclass[pdftex,parskip=half,twoside, BCOR15mm]{scrreprt}%
\usepackage[title,titletoc]{appendix}
\usepackage{pdfpages}
\usepackage{amsmath, amssymb, fancyhdr}
\usepackage[english]{babel}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{here}

\begin{document}

\tableofcontents

\chapter{Chapter 1}
\chapter{Chapter 2}

\begin{appendices}

\chapter{Bli bla blubb}

asdasd
\end{appendices}

\end{document}

If you have nothing after the appendices, then this should do (I omitted the irrelevant packages and options). You don't need the appendix package.

\documentclass[
appendixprefix=true, % the title will have Appendix A
]{scrreprt}

\makeatletter
%% change \numberline so that it will print "Appendix A"
\newcommand\appendix@numberline[1]{\appendixname\ #1\autodot\ }
\let\protect\numberline\protect\appendix@numberline}%
}
\makeatother

\begin{document}

\tableofcontents

\chapter{Chapter 1}
\chapter{Chapter 2}

\appendix
\chapter{Bli bla blubb}

asdasd

\end{document}

1. Don't add the pdftex option. It's not needed and may cause problems.
2. Don't load the here package. It's obsolete and actually loads float. The [H] option to figure or table only makes typesetting worse.


text:
# How to count the additional lines?

A plane has 6 lines of which no two lines are parallel and no three are concurrent. Their points of intersection are joined, how many of additional lines are so formed?

I know that number of points of intersection for $n$ lines would be $\sum \limits_{i=1}^{n-1} i=\frac{n(n-1)}{2}$, but then how do I do the rest?

-
Are you sure that it doesn’t ask for the maximum possible number of additional lines? –  Brian M. Scott Nov 4 '11 at 22:35
What if three intersection points lie on a line? Do we call the resulting number of lines 1, 2, or 3? –  mixedmath Nov 4 '11 at 22:38

## 1 Answer

Every choice of two intersection points determines a line. You already have some of these -- how many pairs of intersection points will generate each of the original 6 lines?

There's a risk that three of the intersection points will lie on a common line that was not one of the originals, but you're probably supposed to ignore that possibility.

-
The case where three of the intersection points lie on a new line is much more interesting than the original problem! But I'm sure you are right that one was meant to ignore that possibility. –  Gerry Myerson Nov 5 '11 at 3:36


text:
# Long abstract for technical report orphaning Abstract title due to vfil

When using a custom template to prepare a report I noticed that long abstracts split across pages, fine, but it leaves the abstract title hanging by itself on it's own page.

How can I structure the abstract environment to have vertically centered abstracts if they fit on one page and top aligned in long abstracts? The \vfil is useful for centering short abstracts, but then means it orphans the title in long abstracts

## The MWE of abstract environment and usage

Normally the environment def is split into the template class.

\documentclass[11pt]{book}
\usepackage{lipsum}

\newenvironment{abstract}
{
\thispagestyle{empty}
\null\vfil %Toggling this line will align to the top in multipage abstract
\begin{center}
\setlength{\parskip}{0pt}
{\textsc\normalsize University \par}
\bigskip
{\underline{Abstract} \par}
\bigskip
{\textsc\normalsize Faculty \par}
{\textsc\normalsize Department \par}
\bigskip
{\normalsize \underline{A progress report submitted for continuation towards a PhD}\par}
\bigskip
{\normalsize\bf \title \par}
\medskip
{\normalsize by [Author] \par}
\bigskip
\end{center}
}
{
\vfil\vfil\vfil\null
\cleardoublepage
}

\title{\bf An MWE}
\author{for SO}
\date{\today}

\begin{document}
\frontmatter
\maketitle
\begin{abstract}
\lipsum[1-4]
\end{abstract}
\end{document}

I have found a simple approach without making many changes to your source code. By replacing the center environment \begin{center} ... \end{center} with the switch command \centering inside \begingroup ... \endgroup or between braces with \par gives the desired output.

\begingroup
\centering
...
\endgroup

or analogously

{\centering
...
\par
}

## But without breaking short abstracts

MWE (with \begingroup ... \endgroup):

\documentclass[11pt]{book}
\usepackage{lipsum}

\newenvironment{abstract}
{
\thispagestyle{empty}
\null\vfil\vfil
\begingroup
\centering
\setlength{\parskip}{0pt}
{\textsc\normalsize University \par}
\bigskip
{\underline{Abstract} \par}
\bigskip
{\textsc\normalsize Faculty \par}
{\textsc\normalsize Department \par}
\bigskip
{\normalsize \underline{A progress report submitted for continuation towards a PhD}\par}
\bigskip
{\normalsize\bf \title \par}
\medskip
{\normalsize by [Author] \par}
\bigskip
\endgroup
}
{
\vfil\vfil\null
\cleardoublepage
}

\title{\bf An MWE}
\author{for SO}
\date{\today}

\begin{document}
\frontmatter
\maketitle
\begin{abstract}
\lipsum[1-2]
%\lipsum[1-6]
\end{abstract}
\end{document}

• This is great, thank you. Was the original problem because of how LaTeX treats the ends of environments?
– EdL
Sep 21, 2020 at 8:17
• @EdL Honestly, I am not sure the reason why. Commenting the 1st \null (the same as \hbox{}) the issue with the long abstracts is solved, but the short ones are not vertically "centered". Just replacing the center environment with \centering it works, but the whole text is than centered. In the latter case, loading the ragged2e package lets you with \justify to typeset fully-justified back. The drawback is the same output as with the center environment. center is defined in terms of trivlist and leaves vertical space before and after it. This is not the case for \centering.
– Ñako
Sep 21, 2020 at 22:19
• I see, thanks for letting me know about how that center environment is defined. Perhaps it is to do with the spacing below it that makes the LaTeX engine split the page there. I think this answer clearly shows how the centre environment was the wrong implementation for my use case and centering is much better.
– EdL
Sep 23, 2020 at 9:00


text:
ビジネステクノロジー事業

データサイエンス事業

KSK's Efforts

PAGE TOP

## Hadley Wickham’s “Tidy Data” in RapidMiner (Part 6)

Hello everyone, and welcome back!  In my last post, I wrote about the third most common problem discussed by Hadley Wickham, where variables are stored in both rows and columns.  In this post, I will discuss the fourth problem discussed by Hadley Wickham, where multiple types of observational units are stored in the same table.  The data used for this, like the data used as an example by Hadley Wickham in his paper, is data from Billboard’s Hot 100 list.  If you would like to follow along in RapidMiner as you read, the “messy” data can be found here (clicking will begin download).

Problem 4: Multiple Types of Observational Units are Stored in the Same Table

First, let’s get a look at the “messy” version of the data I started with and the “tidy” version of the data that will result from the cleaning process.

“Messy” Data (Click to Open)

“Tidy” Song Information Data (Click to Open)

“Tidy” Rank Information Data (Click to Open)

The “messy” data this time isn’t messy in the same way as the previous posts.  In Hadley Wickham’s original paper, he says that data is tidy once it is in this form:

• Each variable forms a column (attribute)
• Each observation forms a row (example)
• Each type of observational unit forms a table (ExampleSet)

In our case, this data meets the first two criteria, but it fails in the third.  This is because it contains observations on two types of observational units: the song information (circled in red) and its rank information (circled in blue).  That is why the “tidy” data shown consists of two tables.  Why is this important?  Well, in his paper, Hadley Wickham states this:

Datasets often involve values collected at multiple levels, on different types of observational units.  During tidying, each type of observational unit should be stored in its own table.  This is closely related to the idea of database normalization, where each fact is expressed in only one place.  If this is not done, it is possible for inconsistencies to occur.

However, while this is useful for tidying data and eliminating inconsistencies, most data analysis tools can’t work with relational data.  As a result, data that is split up into observational units for tidying will most likely have to be merged back into one table for analysis.

The Process

(Click to Open)

First, let me quickly summarize the overall process (see above), and then I will explain each operator individually.  The “messy” data is multiplied into two separate branches, and then I create an “id” column by using the data in one of the branches.  Those branches are then merged back together so that the “id” column is added to the original dataset.  I then multiply the data again into two separate branches, with the top branch containing the “song information” and the bottom branch containing the “rank information”.  Now, let’s look at each operator individually.

Part 1: Generating IDs

To start, the “messy” data is fed into the “Multiply” operator, and this operator is very simple.  All it does is make copies of the data, and multiple copies can be created and used by connecting an output port to another operator.  In this case, I only make two copies of the data, so only two output ports are used.  The data that comes out of the top branch is used to generate an id.  To do this, I fist use the “Aggregate” operator.

Aggregate Parameters (Click to Open)

First, click on the operator to open the “Parameters” window shown above on the right, and then click on the “Edit List” button (circled in red) to open the “Select Attributes: group by attributes” window shown above on the left.  Next, find and click on the “title” attribute listed on the left in the “Attributes” box, and then move it to the “Selected Attributes” box on the right by using the right arrow button (circled in blue).  Finally, click the “Apply” button.

Generate ID Parameters (Click to Open)

Next, click on the “Generate ID” operator to open the “Parameters” window shown above.  In this case, there aren’t any settings you need to change.  The “create nominal ids” box is unchecked by default, so integer ids are created (1, 2, 3, etc…), but if you check the “create nominal ids” box the id’s will become nominal ids (id_1, id_2, id_3, etc…).  The data that comes out of this operator will look like the example below.

Generate ID Output Data (Click to Open)

Part 2: Joining Datasets

Join Parameters (Click to Open)

Now, let’s move on to the “Join” operator.  First, click on the operator to open the “Parameters” window shown above on the right.  Next, set “join type” to “left” (circled in red), and then make sure to uncheck the “use id attribute as key” parameter (circled in blue).  Finally, click on the “Edit List” button (circled in green) to bring up the “Edit Parameter List: key attributes” window shown above on the left, set the “left key attributes” and “right key attributes” to “title”, and click apply.  An example of the data that comes out of this operator can be seen below.

Join Output Data (Click to Open)

Par 3: Song Information Branch

Next, the data is multiplied again using the “Multiply” operator, and this results in two different branches.  The top branch will result in the “tidy” data that will contain only song information, and this is achieved by using a “Select Attributes” and “Remove Duplicates” operator.

Select Attributes Parameters (Click to Open)

Fist, click on the “Select Attributes operator to open the “Parameters” window shown above on the right, and then set the “attribute filter type” to “subset” (circled in red).  Next, click in the “Select Attributes…” button (circled in blue) to open the “Select Attributes: attributes” window shown above on the left.  Choose “artist”, “id”, and “title” from the “Attributes” box on the left, and then move them to the “Selected Attributes” box on the right using the right arrow key (circled in green).  Finally, click apply.  An example of the data that will come out of this operator can be seen below.

Select Attributes Output Data (Click to Open)

Remove Duplicates Parameters (Click to Open)

Next is the “Remove Duplicates” operator, but there is nothing that needs to be done with its parameters (see above).  There are many different options to choose from with the “attribute filter type” parameter, and extra parameters will show up depending on what you choose, but we want to make sure there aren’t any duplicates for all our attributes.  Also, special attributes can be dropped by checking the “include special attributes” box, but we want to keep our special “id” attribute, so leave it unchecked.  The final “tidy” data for this branch can be seen below.

“Tidy” Song Information Data (Click to Open)

Part 4: Rank Information Branch

The bottom branch that was created by the “Multiply” operator only contains a “Select Attributes” operator.  This branch will result in the “tidy” data that contains only rank information for each song.

Select Attributes Parameters (Click to Open)

The “Select Attributes” operator is used very similarly to before, but this time we are selecting different attributes.  First, click on the operator to open the “Parameters” window shown above on the right, and then set the “attribute filter type” to “subset”.  Next, make sure the “invert selection” box is checked (circled in red).  Doing that will save us work in the next step.  Finally, click on the “Select Attributes” button, move the “artist” and “title” attributes to the right side, and then click “Apply”.  Since we checked the “invert selection” box previously, the selected “artist” and “title” attributes will be dropped while all the others will be kept.  The final “tidy” data for this branch can be seen below.

“Tidy” Rank Information Data (Click to Open)

Conclusion

In summary, the original “messy” data was duplicated using the “Multiply” operator, and the top branch was used to generate an id using the “Aggregate” and “Generate ID” operators.  The two branches were then joined together using the “Join” operator.  The data was then duplicated again using the “Multiply” operator, and the top branch resulted in the “tidy” data containing only song information, while the bottom branch resulted in the “tidy” data containing only rank information.  The top branch created the “song information” data using the “Select Attributes” and “Remove Duplicates” operators, and the bottom branch created the “rank information” data by using a “Select Attributes” operator.

I hope this post has been helpful, and I hope that you will be able to use some of what you learned in your own analysis.  In the next post, I will discuss the fifth and final problem with “messy” data discussed by Hadley Wickham, where a single observational unit is stored in multiple tables.

See you then!

\begin{aligned} \newcommand\argmin{\mathop{\rm arg~min}\limits} \boldsymbol{\beta}_{\text{ridge}} & = \argmin_{\boldsymbol{\beta} \in \mathcal{R^p}} \biggl[ ||\boldsymbol{y}-\boldsymbol{X\beta}||^2 + \lambda ||\boldsymbol{\beta}||^2 \biggr] \\ & = (\boldsymbol{X}^T\boldsymbol{X} + \lambda\boldsymbol{I_{p+1}})^{-1}\boldsymbol{X}^T\boldsymbol{y} \end{aligned}
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text:
# Chapter 2 Good practices

## 2.1 Coding style

Good coding style is like correct punctuation: you can manage without it, butitsuremakesthingseasiertoread.

Please make your code readable by following e.g. this coding style (examples below come from this guide).

You can use package styler (that provides RStudio addins) to correct your style:

### 2.1.1 Naming

Be smart with your naming. I can’t tell the number of times I’ve seen df <- as.matrix(mtcars) on Stack Overflow.

### 2.1.2 Spacing

Put a space before and after = when naming arguments in function calls. Most infix operators (==, +, -, <-, etc.) are also surrounded by spaces, except those with relatively high precedence: ^, :, ::, and :::. Always put a space after a comma, and never before (just like in regular English).

# Good
average <- mean((feet / 12) + inches, na.rm = TRUE)
x <- 1:10
base::sum

average<-mean(feet/12+inches,na.rm=TRUE)
x <- 1 : 10
base :: sum

### 2.1.3 Indenting

Curly braces, {}, define the most important hierarchy of R code. To make this hierarchy easy to see, always indent the code inside {} by two spaces. This should be automatic in RStudio.

# Good
if (y < 0 && debug) {
message("y is negative")
}

if (y == 0) {
if (x > 0) {
log(x)
} else {
message("x is negative or zero")
}
} else {
y ^ x
}

if (y < 0 && debug)
message("Y is negative")

if (y == 0)
{
if (x > 0) {
log(x)
} else {
message("x is negative or zero")
}
} else { y ^ x }

### 2.1.4 Long lines

Strive to limit your code to 80 characters per line. This fits comfortably on a printed page with a reasonably sized font. If you find yourself running out of room, this is a good indication that you should encapsulate some of the work in a separate function.

If a function call is too long to fit on a single line, use one line for the function name, for each argument, and for the closing ). This makes the code easier to read and to modify later.

### 2.1.5 Other

• Use <-, not =, for assignment. Keep = for parameters.
# Good
x <- 5
system.time(
x <- rnorm(1e6)
)

x = 5
system.time(
x = rnorm(1e6)
)
• Don’t put ; at the end of a line, and avoid multiple commands on the same line.

• Only use return() for early returns. Otherwise rely on R to return the result of the last evaluated expression.

# Good
x + y
}

return(x + y)
}
• Use ", not ', for quoting text. The only exception is when the text already contains double quotes and no single quotes.
# Good
"Text"
'Text with "quotes"'

'Text'
'Text with "double" and \'single\' quotes'

## 2.2 RStudio

RStudio features:

• everything you can expect from a good IDE
• keyboard shortcuts I use
1. Ctrl + Space (auto-completion, better than Tab)
2. Ctrl + Up (command history & search)
3. Ctrl + Click (function source code)
4. Ctrl + Enter (execute line of code)
5. Ctrl + Shift + A (reformat code)
6. Ctrl + Shift + C (comment/uncomment selected lines)
7. Ctrl + Shift + K (knit)
8. Ctrl + Shift + B (build package, website or book)
9. Ctrl + Shift + M (pipe)
10. Alt + Shift + K to see all shortcuts…
• Panels (everything is integrated, including Git and a terminal)
• Interactive data importation from files and connections (see this webinar)

• R Projects:
• Meaningful structure in one folder
• The working directory automatically switches to the project’s folder
• The File tab displays the associated files and folders in the project
• History of R commands and open files
• Any settings associated with the project, such as Git settings, are loaded. Note that you can have a .Rprofile file in the project’s root directory to enable project-specific settings to be loaded each time people open the project.

Read more at https://www.tidyverse.org/articles/2017/12/workflow-vs-script/ and also see chapter Efficient set-up of book Efficient R programming.

## 2.3 Version control (Git)

### 2.3.1 Why use Git? You don’t use Git?

Have you ever:

• Made a change to code, realized it was a mistake and wanted to revert back?
• Lost code or had a backup that was too old?
• Wanted to submit a change to someone else’s code?
• Wanted to share your code, or let other people work on your code?

In these cases, and probably many others, a version control system should make your life easier (see https://stackoverflow.com/a/1408464/6103040).

• Version control for the researcher: don’t do that, use Git

• Version control for the data analyst: reproducible workflow
• Use version control to work from anywhere

• Working with GitHub can be a line on your CV (read more):

A lot of students have said to me later, even first-year undergraduates, that using GitHub has helped them a lot when they went for an internship or a research position interview.

They are able to say, “Oh, I already have worked with GitHub. I’m familiar with it. I know how it works.” So I think they are at least able to put that on their CV and go into a situation where there’s a research or data analysis team and say, “Yeah, sure. I am actually familiar with the same tools that you use.”

– Mine Cetinkaya-Rundel, Duke University, RStudio

• Main Git platforms (share your code, collaborate):
• 3 main commands:
• commit: snapshot of your code at a specified point in time
• push: merge your local modifications with the main project
• Simple (solo) use of git to prevent merge conflicts:
• after opening a project, always pull
• before closing a project, always commit/push
• How to link between an RStudio project and a GitHub repository?

## 2.4 Getting help

A basic solution is to print everything, but it’s usually not working well on complex problems. A convenient solution to see all the variables’ states in your code is to place some browser() from where you want to check the variables’ states. To debug functions, debugonce() is also very useful.

my_log <- function(x) log(x - 1)

my_fun <- function(a, b) {
# browser()
la <- my_log(a)
lb <- my_log(b)
la + lb
}
my_fun(1, 0)
#> Warning in log(x - 1): NaNs produced
#> [1] NaN

Try to uncomment browser() or use debugonce(my_fun):

debugonce(my_fun)
my_fun(1, 0)

Learn more with this book chapter, this other book chapter, this webinar and this RStudio article.

### 2.4.2 External help

Can’t remember useful functions? Use cheat sheets.

You can search for specific R stuff on https://rseek.org/. You should also read documentations carefully. If you’re using a package, search for vignettes and a GitHub repository.

You can also use Stack Overflow. The most common use of Stack Overflow is when you have an error or a question, you google it, and most of the times the first links are Q/A on Stack Overflow.

You can ask questions on Stack Overflow (using the tag r). You need to make a great R reproducible example if you want your question to be answered. Sometimes, while making this reproducible example, you find the answer to your problem.

If you’re confident enough with your R skills, you can take the next step and answer questions on Stack Overflow. It’s a good way to increase your skills, or just to procrastinate while writing a scientific manuscript.

You can also join communities, e.g. join the French-speaking R community or join the R-Ladies community. These are generally much friendlier and welcoming spaces compared to Stack Overflow.


text:
# Gr 3 C5 Lesson Introduction to Sorting Quadrilaterals

In this lesson, students explore various quadrilaterals by sorting them based on attributes such as opposite sides having equal lengths and square corners (right angles).


text:
# Simple probability question

• Dec 11th 2012, 07:58 AM
VII
Simple probability question
Hi There's a 15% chance to get a blue.

I've got 3 chances to hit blues. What's the chance I hit exactly 2/3 blues?
• Dec 11th 2012, 08:19 AM
Plato
Re: Simple probability question
Quote:

Originally Posted by VII
Hi There's a 15% chance to get a blue.
I've got 3 chances to hit blues. What's the chance I hit exactly 2/3 blues?

Please rewrite your question in much more complete details.
As written, it is too vague to answer.
• Dec 11th 2012, 08:28 AM
VII
Re: Simple probability question
There's some sweets in a bag. There's a 15% chance to get a blue sweet.
I will take a sweet out, and put it back in.
If take a sweet out and put it back 3 times, what's the chance 2/3 of those sweets will be blue?
• Dec 11th 2012, 08:36 AM
Plato
Re: Simple probability question
Quote:

Originally Posted by VII
There's some sweets in a bag. There's a 15% chance to get a blue sweet. I will take a sweet out, and put it back in.
If take a sweet out and put it back 3 times, what's the chance 2/3 of those sweets will be blue?

Simple binomial probability: $\binom{3}{2}(.15)^2(.85).$
• Dec 11th 2012, 08:52 AM
VII
Re: Simple probability question
I don't understand that >.>
What's the % chance of getting 2/3 blues?
• Dec 11th 2012, 10:30 AM
Plato
Re: Simple probability question
Quote:

Originally Posted by VII
I don't understand that >.>
What's the % chance of getting 2/3 blues?

If you do not understand how to evaluate a binomial coefficient, $\binom{N}{k}$ then you have no business trying this problem.


text:
# Day/night cycle in Greenland

I have heard that in Greenland there is day for 6 months and night for 6 months. Is this true? If so, how does it happen?

• This question appears to be off-topic because it is about day-night cycles and not physics. – Danu Jul 14 '14 at 14:50
• Isn't this on topic as it has to do with orbits and rotation axis inclinations of planets, i.e. astronomy? Pretty sure the astronomers of the past spent a good deal of time figuring out the answers to this and similar questions. Granted from our modern perspective this question isn't that interesting since the answer is a bit of fairly simple geometry, but it's hardly off topic, right? – Kyle Oman Jul 14 '14 at 16:08
• Buy a globe and go to a dark room with a flashlight. Now spin the globe as you shine on one side and see which parts of the earth are day or night and why. – John Alexiou Jul 14 '14 at 16:23

That's not quite correct.

You may have noticed that during summer the days are longer (and the nights shorter) than during winter. That is because the earth's axis is tilted about $23^o$ from the plane of it's orbit around the sun. With this tilt, as the earth travels around the sun the northern hemisphere gets longer days is the north pole is tilted towards the sun. Shorter days happen when it's the reverse. The change happens gradually, with the shortest day at midwinter (approximately 21 December) and the longest day around 21 June. At equinox time (21 March & 21 September) day and night are of equal length.

Once you go north of the Arctic Circle (or south of the Antarctic Circle), which is at $67^o$ north ($67=90-23$), you will find that there are periods of the year when the sun does not rise above the horizon during the day, because the bulge of the earth is in the way. At other times of the year, the sun does not set at all.

However, as in more temperate latitudes, the change is gradual. On 21 December there is indeed no sunlight. Then, as you go into the new year, there comes a day when the sky starts to brighten a bit around noon time. Still later, the sun may come up for a short period. The exact day this happens depends on your latitude (how far north you are). That period gets longer by the day, until on a certain day the sun no longer sets at all, and you have 24 hour daylight. The Arctic Circle is the latitude where there will be a day without sun, and a day without night. Further north more and more days are like that.

Greenland stretches from about $60^o$ to $85^o$ North. Hence the southern end of Greenland always gets at least some sun, even in the middle of winter. ANd, during summer, ther will still be some night.

To add a final bit of confusion: I've been talking all along about the northern hemisphere. If you live in Australia like me, it's the exact reverse. We have winter in June, summer in December.

• It also depends on which definition of night and day you use, which like mentioned in this video. And you could also take into account that atmospheric refraction cause longer days (and thus shorter nights), which would increase the average length of a day and shorten the average length of a night. – fibonatic Jul 14 '14 at 18:15
• I know it's more complicated than my explanation. For another variation, the solstice date varies from year to year. But hopefully, this gives the asker some idea. – hdhondt Jul 15 '14 at 4:07

The 6 months day/night cycle is exactly happening only at the poles (as pointed out in comments). Between the poles and the arctic circle you have a gradual change from 6 month cycle to the 24 hour cycle. Greenland is partially in this area (south Greenland is actually outside the arctic circle). It is happening because the Earth rotational exis is tilted, so the poles are tilted away from (or towards) the Sun every 6 months under such angle, that the Sun light in certain parts of the year does not get into the area beyond the arctic circle.

See here.

• This is not true. Only exactly at the poles there is 6 months of day/night. – pfnuesel Jul 14 '14 at 11:07
• @pfnuesel You are correct, thank you. I'm going to improve the answer. – mpv Jul 14 '14 at 11:59


text:
# Formula Cylindrical Capacitor Capacitance    Length    Inner radius    Outer radius    Relative permittivity    Vacuum Permittivity

## Capacitance

Unit
Capacitance is a measure of how much charge can be separated on the electrodes of the cylindrical capacitor, in other words - how good the cylinder can store electric charge.

## Length

Unit
Length of the cylinder. The longer the cylinder, the greater its electric capacitance.

Unit
Radius of the inner electrode of the charged cylindrical capacitor.

Relative permittivity depends on the dielectric medium between the outer and inner cylinder and indicates the permeability of the electric field. In vacuum it has the value: $$\varepsilon_r ~=~ 1$$. By using a different medium (e.g. air or water) between the electrodes of the cylinder capacitor, the capacitance of the cylinder can be increased.
Permittivity of free space is a physical constant and has the value: $$\varepsilon_0 = 8.854187817 ~\cdot~ 10^{-12} \, \frac{\text{As}}{\text{Vm}}$$.


text:
Browse Questions

# If $\overrightarrow a = \hat i + \hat j + \hat k, \overrightarrow b = 2\hat i - \hat j + 3\hat k$ and $\overrightarrow c = \hat i - 2\hat j + \hat k$, find a unit vector parallel to the vector $2\overrightarrow a - \overrightarrow b + 3\overrightarrow c$

$\begin{array}{1 1} (A) \large\frac{1}{\sqrt{22}}(3\hat i +3\hat j+2\hat k) \\ (B) \large\frac{1}{\sqrt{22}}(3\hat i -3\hat j-2\hat k) \\ (C) \large\frac{1}{\sqrt{22}}(3\hat i +3\hat j-2\hat k) \\ (D) \large\frac{1}{\sqrt{22}}(3\hat i -3\hat j+2\hat k) \end{array}$

Toolbox:
• Unit vector of $\overrightarrow a=\large\frac{\overrightarrow a}{|\overrightarrow a|}$
• $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^2}$ where $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k$
Step 1:
$\overrightarrow a=\hat i+\hat j+\hat k,\overrightarrow b=2\hat i-\hat j+3\hat k$ and $\overrightarrow c=\hat i-2\hat j+\hat k$
$2\overrightarrow a=2(\hat i+\hat j+\hat k)$
$\quad\;\;=2\hat i+2\hat j+2\hat k$
$3\overrightarrow c=3(\hat i-2\hat j+\hat k)$
$\quad\;\;=3\hat i-6\hat j+3\hat k$
Step 2:
We have to find $2\overrightarrow a-\overrightarrow b+3\overrightarrow c$
$2\overrightarrow a-\overrightarrow b+3\overrightarrow c=(2\hat i+2\hat j+2\hat k)-(2\hat i-\hat j+3\hat k)+(3\hat i-6\hat j+3\hat k)$
$\qquad\qquad\qquad=3\hat i-3\hat j+2\hat k$
Step 3:
Next we have to find the magnitude of this vector $|2\overrightarrow a-\overrightarrow b+3\overrightarrow c$|
$|2\overrightarrow a-\overrightarrow b+3\overrightarrow c|=\sqrt{(3)^2+(-3)^2+(2)^2}$
$\qquad\qquad\quad\;\;\;\;=\sqrt{9+9+4}$
$\qquad\qquad\quad\;\;\;\;=\sqrt{22}$
Step 4:
Hence the unit vector parallel to $2\overrightarrow a-\overrightarrow b+3\overrightarrow c=\large\frac{2\overrightarrow a-\overrightarrow b+3\overrightarrow c}{|2\overrightarrow a-\overrightarrow b+3\overrightarrow c|}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\large\frac{1}{\sqrt{22}}$$(3\hat i-3\hat j+2\hat k)$


text:
# How do you simplify sqrt(14q)*2sqrt(4q)?

##### 1 Answer
Jun 3, 2015

You can write it using a single square root:
$2 \sqrt{14 q \cdot 4 q} = 2 \sqrt{14 {q}^{2} 4} =$
You can take out from the square: $4$ and ${q}^{2}$ as:
$= 2 \cdot 2 \cdot q \sqrt{14} = 4 q \sqrt{14}$


text: 3. Alternatively, if $F\subseteq\mathbb{Q}$ is a subfield then $1\in F$, so that $1+1,1+1+1,\ldots\in F$ so $\mathbb{N}\subseteq F$. But then $-1,-1-1,\ldots \in F$ so that $\mathbb{Z}\subseteq F$. But for each $0\neq z\in\mathbb{Z}$ we have that $z^{-1} \in F$, so that $pq^{-1} \in F$ for any $p,q\in\mathbb{Z},q\neq 0$. Hence $\mathbb{Q}\subseteq F.$


text:
## Rajeshkumar 3 years ago A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?

1. dumbcow

The current is 1 so the man can row 4 kmph upstream and 6 kmph downstream distance = rate*time d = 4t1 d = 6t2 t1 + t2 = 1 use substitution t1 = 1 - t2 4(1-t2) = 6t2 solve for t2 4 - 4t2 = 6t2 4 = 10t2 0.4 = t2 then d = 6*0.4 = 2.4 The place is 2.4 km away

2. robtobey

t=d/r Assume the place is down stream. Let x = the time it takes to row downstream. Solve the following simultaneous equations for x and d:$\left\{x\text{=}\frac{d}{6},(1-x)\text{=}\frac{d}{4}\right\}$$\left\{x\to \frac{2}{5},d\to \frac{12}{5}\right\}$If the place is up stream then let x = the time it takes to row upstream.$\left\{x\text{=}\frac{d}{4},(1-x)\text{=}\frac{d}{6}\right\}$$\left\{x\to \frac{3}{5},d\to \frac{12}{5}\right\}$In either case the place is 12/5 or 2.4 km from the departure point.


text:
## Saturday, 23 February 2008

### INTEGRAL

At the last Cosmo Coffee, Celine Boehm was discussing the current status of the 511 keV gamma-ray line from the Milky Way center, in light of the recent results from the INTEGRAL satellite. Photons carrying 511 keV energy arise from annihilation of positrons and electrons that are more less at rest. It is not clear which mechanism is responsible for injecting enough positrons into the interstellar medium of the galactic bulge. One hypothesis is that the positrons are scattered remnants of the ILC (Interplanetary Linear Collider) - an unfinished project of a technologically advanced civilization from that region. More contrived explanations involve black holes, radioactive nuclei from supernovae, pulsars and other fluffy toys.

Yet another, quite exciting possibility is that the positrons come from annihilation of dark matter. At first sight this seems quite natural. We are pretty confident that a lot of dark matter is present in the galactic center and its distribution should be roughly spherical, something which agreed quite well with the earlier INTEGRAL observations. Furthermore, it is likely that dark matter can annihilate into ordinary matter so that its present abundance is a thermal relic. But a dark matter particle that could explain the INTEGRAL signal must be quite peculiar. In particular, it cannot be a WIMP: its mass should be in the MeV range. If it were heavier, annihilation would yield too energetic positrons and the 511 keV line would not stand so prominently over the continuum gamma-ray radiation.

On the microscopic level, MeV dark matter can be realized as a scalar particle (so as to avoid the Lee-Weinberg bound) annihilating via an exotic heavy fermion exchange, or via an exotic heavy gauge boson. See the artist's view above. In fact, both diagrams are needed to explain the INTEGRAL signal and, at the same time, derive the correct dark matter abundance from the thermal relic density computation. The first diagram leads to the cross section that goes to a constant at small velocities and this one is relevant for the annihilation of dark matter today. The second leads to the cross section that goes like $\sigma \sim b v^2$, and it's supposed to dominate in the early hot universe.

Recently, the INTEGRAL satellite announced new results which show an asymmetry of emission (by a factor of two) with respect to the central axis of the galaxy. Morevover, the asymmetry seems to be correlated with the distribution of low mass X-ray binaries (LMXB) - systems including a neutron star or a black hole that accretes matter from its companion. LMXBs have long been one of the suspects in the positron case. The general feeling is that the new results make the dark matter explanation unlikely; Julianne on CV readily flushed MeV dark matter down her toilet. Celine, on the other hand, is more reluctant to push the button (she is, in fact, responsible for much of the stuff being thrown into the toilet). She argued that: 1) We still don't know a conventional astrophysical explanation that could account for enough positron emission, 2) While part of the gamma-ray emission maybe due to boring astrophysics, there's still a large, roughly spherical component that could be due to MeV dark matter. In fact, if only a part of the emission is assigned to dark matter, the microscopic models can more readily satisfy constraints from other experiments, for example, from measurements of the electron anomalous magnetic moment.

I guess it's fair to say that, at the moment, a conventional astrophysical explanation seems far more likely. My faith in that signal is further diminished by the fact that astrophysics provides us with too many excesses (EGRET, for example), each one having its dark matter particle that's supposed to explain it. Finally, MeV dark matter is hard to accommodate in our favourite Beyond the Standard Model schemes. But this last argument should always be taken with a grain of salt, since all existing BSM models suck. It's better to keep our eyes open...

#### 2 comments:

Matti Pitkanen said...

I have tried many times to communicate an explanation of this and some other anomalies in terms of what I call leptohadron hypothesis emerging naturally in TGD framework.

Already at seventies a resonant production of electron positron pairs in heavy ion collisions just above Coulomb wall was observed and the proposal was that they result in a decay of a pseudoscalar state which has mass slightly above 2m_e. There is also a well-known anomaly in positronium decay which could be explained in terms of coupling of leptopion to photon pair by a universal coupling dictated by axial anomaly considerations.

About year ago also similar production of muon antimuon pairs was observed. In the blog of Lubos there were comments about a new particle. The finding has been published (Phys. Rev. D74) and (Phys. Rev. Lett. 98). The mass of the new particle, which is either scalar or pseudoscalar, is 214.4 MeV whereas muon mass is 105.6 MeV. The mass is about 1.5 per cent higher than two times muon mass. The proposed interpretation is as light Higgs. I do not resonate with this interpretation although p-adically scaled up variants of also Higgs bosons live happily in the fractal Universe of TGD.

My explanation for all these findings is in terms of leptohadron hypothesis. One of the basic deviations of TGD from standard model is the prediction of colored excitations of quarks and leptons. The reason is that color is not spin like quantum number but partial wave in CP_2 degrees of freedom and thus angular momentum like. Accordingly, new scaled variants of QCD are predicted. As a matter fact, dark matter hierarchy and p-adic length scale hierarchy populate many-sheeted Universe with fractal variants of standard model physics. The colored excitations of leptons can form hadron like states: electropions, muopions, and taupions.

I have proposed that the decays of electropions residing at magnetic flux tubes of the strong galactic magnetic fields, and indeed dark matter - but in these sense of TGD- are responsible for the observed narrow lines from Milky Way Center.

For more details about the leptohadron hypothesis see the chapter The Recent Status of Lepto-Hadron Hypothesis.

Below a list of references about early evidence for pionlike states which people for some reason probably related to shortcomings of ancient theories;-) want to call axions.

1. A.T. Goshaw et al(1979), Phys. Rev. Lett. 43, 1065.

2.J.Schweppe et al(1983), Phys. Rev. Lett. 51, 2261.

3. M. Clemente et al (1984), Phys. Rev. Lett. 137B, 41.

4. P.V. Chliapnikov et al(1984), Phys. Lett. B 141, 276.

5. L. Kraus and M. Zeller (1986), Phys. Rev. D 34, 3385.

6. A. Chodos (1987), Comments Nucl. Part. Phys., Vol 17, No 4, pp. 211, 223.

7. W. Koenig et al(1987), Zeitschrift fur Physik A, 3288, 1297.

8. C. I. Westbrook ,D. W Kidley, R. S. Gidley, R. S. Conti and A. Rich (1987), Phys. Rev. Lett. 58 , 1328.

Anonymous said...

Whoa, dude.

I want some of what he's smoking...


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# How did Elementary Teachers Handle Critical Experiments in Science Classrooms?

• Published : 2009.05.31

#### Abstract

Critical Experiments (CE) in science classrooms mean, tentatively, critical situations as comparable to anomalous cases in scientific revolutions where the results of science experiments in schools are unclear, differ from the theory, or students misunderstand the purpose of the experiments. The purpose of this research is to identify what CE occurred during science classes and to investigate how elementary teachers handled them. To analyze how teachers recognized and handled CE, we selected nine typical CE from the $7^{th}$ Korean science curriculum. 125 teachers were selected from 8 districts' elementary schools in a local city. A questionnaire with photos of the nine CE above-mentioned was distributed to these teachers. The focus in this research was the way that each teacher handled the CE. We discovered that there were three basic ways in which teachers handled CE. When CE occurred, 51% of elementary teachers explained the correct result of the experiment (what should have happened) to the students while 40.7% of the teachers repeated to get the correct results. The focuses of handling CE varied. 57 % of the teachers focused on the 'materials' while 30% of the teachers focused on the 'theory'. The other focus was 'thinking'. Only 7.6% of the teachers answered that they gave students a chance to think about the reasons why the CE happened. By analyzing our survey results, we could determine what each teacher did as a follow up to the CE and their focus and reasoning for handling the CE this way. When the CE happened in the science class, few handled the CE with the point of view about purpose of doing experiment. As a result, students could not gain educational experience from the CE. If we use CE as a new method to teach science, it will be a good subject incorporating the nature of science in science education.


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## At A Glance

### Limit of a sequence: L'Hopital's rule applied to $(\ln n)/n$

Free ebook http://tinyurl.com/EngMathYTI discuss and solve an example of a limit of a sequence where L'Hopital's rule is applied. The particular sequence is $(\ln n)/n$. This type of example is seen in 1st-year university mathematics. Note that I have used a 'big nib marker' to make everything clearer in full screen mode!
Length: 06:13

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# NVMREG Write to DFM

DFM is programmed using the following steps:
1. 1.Set the NVMREGS bit.
A single byte (0xFF) is written into the DFM location. The DFM circuitry automatically erases the memory location before performing the write operation. If the DFM address is write-protected, the WR bit will be cleared and the write operation will not take place.