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FOOTWEAR Sandals on YOOXCOM Alto Gradimento YYRysi

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Chuck Grassley

United States Senator from Iowa

Parenting and motherhood in the midst of military life

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Ever dream of a morning where you are drinking a cup of coffee, checking e-mail, paying a few bills, and you are without interruption and it’s quiet?! This is no dream. Teaching children to play independently, while you get some business done around the house, is totally possible, and you can start from birth! Today I’m continuing on with my Babywise Basics series and talking about Independent Playtime (If you missed my previous Babywise Basics posts on Casadei Woman Leather Platform Sandals Neutral Size 375 Casadei j6tKz
and How to Sleep-Train , be sure to check them out).

The idea of children playing alone comes most commonly from the book Babywise II . Using the Babywise method or not, anyone can teach their children this really awesome, mutually beneficial skill with a little bit of parental diligence.

1. Independent playtime defined.

Independent play time (IPT) is a daily scheduled time when your child plays alone, without parents or other siblings around. You choose the time of the day and with which toys your child will play. Occurring at approximately the same time every day, IPT typically takes place in a pack ‘n play or play yard for younger babies/ toddler. Older toddlers and children will transition from the play yard to their rooms and have IPT in their room or other room of the house.

Mental focusing skills : Playpen time helps a child develop the ability to concentrate on an object and apply knowledge to the activity at hand without distraction.

Sustained attention span: The interval during which a child can concentrate on a single object or activity will gradually improve and lengthen over time.

Creativity: Absolute freedom eliminates the need for creative thinking, while boundaries facilitate creativity. The child will learn to find enjoyment out of what’s available to them. The child will create meaningful new methods and interpretations during play.

Power analysis is challenging because it depends completely on assumptions and requirements we specify in advance. While we can comfortably rely on industry standard requirements for the power and the significance level, there are no such options for the detectable difference. However, if we follow some basic guiding principles, we should be able to find a detectable difference that satisfied all constituents. Specifically, the chosen detectable difference should incorporate the business context of the tactic that is being tested – e.g., ROI, and should be discussed in a cross-functional setting.

We should note that the use case covered in this post is a simple A/B test with equal sample sizes for test and control groups. Obviously, experimental designs and sampling schemes can get more much complex, but the underlying philosophy and approach are the same. And in all cases you will benefit from considering the business context when choosing the detectable difference.

Last but not least, Bayesian approaches to sample size determination exist as well. However, depending on your level of comfort with random distributions and Monte Carlo techniques, it may also be more challenging to execute. For more on Bayesian versus frequentist statistics, see this on the Stitch Fix technology blog.

In the example above, we did a test of proportions since the underlying outcome variable is binary. When comparing averages of continuous variables across two samples (such as revenue), the simplest test statistic is the t-test . This test statistic is similar to the test we use for proportions – i.e., subtract the means and divide by the pooled standard error. However, the detectable differences are typically expressed in terms of the effect size given by:

where S $S$ is the pooled standard deviation, X 1 ${X}_{1}$ is the average of the control group, and X 2 ${X}_{2}$ is the average of the test group. When choosing the D $D$ , we can follow the suggested framework above for the detectable difference. Another option is to use Cohen’s guidelines for small, medium, and large effects ( D = 0.2 , 0.5 , 0.8 $D=0.2,0.5,0.8$ ). However, these values are tied to the historical variance of the data (the S $S$ in the denominator) as opposed to the business context.

Academics

One of the top-ranked film schools in the country, SFTV provides a comprehensive academic education through small class sizes, engaged learning and world-class opportunities.

Admissions

SFTV's selective programs welcome qualified candidates from diverse backgrounds and artistic viewpoints to join our community of dedicated storytellers, artists and scholars.

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Outcomes

SFTV graduates are in high demand. Our alumni are successfully pursing careers in every area of the film and television industry, from art direction to 3D animation.

About SFTV

SFTV delivers the whole package, combining up-to-the-minute media education with a strong liberal arts foundation rooted in 500 years of Jesuit tradition.

> Academics > Graduate Programs > Writing and Producing for TV

The Writing and Producing for Television program has one goal: to prepare you for a professional career in the television industry, be it network, basic or premium cable, international or web-based. LMU provides the perfect platform for developing your craft; groups of eight unite in a non-competitive workshop setting where experimentation is applauded and the imagination encouraged to roam. You'll engage in constant revision to create a body of work that precisely conveys your artistry. Extensive hands-on production experience rounds out your immersion in this process.

Coursework in the first two years' centers on the development of spec scripts, pilots and long-form teleplays. But you'll also get a thorough grounding in writing for plays, games and the Internet and how the writers' room operates. By your third year, the creative refinement and technical skills you've acquired are funneled into a pipeline to a professional career. A scrupulous portfolio review, intensive pitching practice, and interaction with industry heavyweights mean you'll be comfortable and confident entering the workplace.

You will leave with a portfolio of speculative scripts, pilots and a webisode or trailer for a series you will write and produce. Whatever small-screen stories you choose to tell, you'll graduate with an insider's understanding of the business and the know-how you need to leap into your television career.

Lizzie Germann '17
Balancing Life and Creativity

1 LMU Drive,Los Angeles, California 90045

310.338.2700

LAW SCHOOL CAMPUS

919 Albany Street,Los Angeles, California 90015

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# Modern reference on general Deep Learning theory

In the present, Deep Learning is experiencing lightspeed growth, with plethora of new architectures and radiant ideas emerging each month. Since the last few years several influential ideas, applied in different papers have been developed - among which most important are Transformer, EfficientNet, MobileNets. Is there some modern reference, in the spirit of the famous https://www.deeplearningbook.org/, in some way, generalizing the most prominent advances in Deep Learning?

• Can I know are you looking for Deep Learning latest realized resource? Feb 22 at 16:48
• @AIFahim as far, as I undestand, the latest edition doesn't contain these topics, and at least I am unaware of plans to include them. I ask, whether there exist a book of similar strucutre, but accounting for the recent advances Feb 22 at 17:05

## 1 Answer

For the latest practices of deep learning architectures, I follow the Kaggle latest Competition's notebooks.
Say, for example, In the recent finished Cassava Leaf Deasese Computer Vision Competition, People are sharing the experimental notebooks on different State of the art Architectures Like Vision Transformer(Various versions pretrained - vit_large_patch16_384, vit_base_patch16_384 etc. models), Facebook's Data-efficient Image Transformers , EfficientNet (Noisy Student version, Imagenet Version) and etc.
https://www.deeplearningbook.org/ this book undoubtedly one of the great book. But for the latest practices if you follow the Kaggle notebooks you get a good explanation with codes.

I suggest you, Explore this Notebooks Section of the Casava Leaf Disease Classification Contest to get more Models and Various recent new released Loss Functions and so on.

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# Photon hitting a solar sail

1. Sep 21, 2006

### JoAuSc

The problem is that I'm doing what seems to be perfectly reasonable algebra but I'm getting a result that makes no sense.

Let's say a photon travels from the left and hits a stationary solar sail which is perpendicular to the light. There is perfect reflection and a change in momenta for both objects. For momentum we have:

p_iphoton + p_iSS = p_fphoton + p_fSS =>
h/c*nu1 + m*(0)^2 = h/c*nu2 + mv
h/c*(nu1-nu2) = mv

energy:

KE_iphoton + KE_iSS = KE_fphoton + KE_fSS
h*nu1 + 0 = h*nu2 + mv^2
h*(nu1-nu2) = mv^2 = c*mv => v = c

(KE_iphoton is the photon's initial kinetic energy, p_iphoton is its initial momentum, KE_fphoton is the final KE, etc.; h = planck's constant, c = speed of light, m = mass of solar sail, v = final velocity of solar sail, nu1 = initial frequency of photon, nu2 = final frequency of photon.)

Obviously, the final speed of the solar sail isn't the speed of light. What did I do wrong?

2. Sep 22, 2006

### andrevdh

You did not take direction into account with your momentum conservation calculation.

3. Sep 22, 2006

### Andrew Mason

As Andrevh pointed out, the change in momentum is the sum of, not difference of, the magnitudes of the momentum before and after due to the 180 degree change in direction.

Also, you are using $h/c\nu$ for momentum of the photon which is not correct. Momentum of the photon is $E/c = h\nu/c = h/\lambda$.

I don't follow your math here. You have to divide the change in momentum of the photon by the sail's mass to find the speed change in the sail. The m in your equation is the sail's mass not the mass of the photon.

$$\Delta P_{ph}/m_s = v_s = P_s/m_s$$

$$KE_s = \frac{1}{2}m_sv_s^2 = P_s^2/2m_s$$

AM

4. Sep 23, 2006

### JoAuSc

Yes, I see it now. Thank you and Andrevh. I'm guessing you'd have a difference of frequencies in the energy equation, but a sum of frequencies in the momentum equation, so you can't substitute a change in photon energy with a change in photon momentum, at least not like I did.

That's what I had, except that rather than clarify things with parentheses like I should have, I relied on order of operations. I used $h/c*\nu = (h/c)*\nu = (h*\nu)/c.$

Let me clarify. In the last line of equations, I had
$$h*(\nu_1-\nu_2) = mv^2 = c*mv => v = c$$
I believe the first two terms are, in fact, equal to each other, right? Now, based on my erroneous assumption that
$$\Delta P_{ph} = (h/c) * (\nu_1 - \nu_2)$$
I got
[tex]\Delta P_{ph} = c * \Delta E_{ph}[/itex]
which implies that mv^2 = cmv.


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Dividing CDF rather than PDF equally in Latin Hypercube Sampling

Returning back to the basics of Latin Hypercube Sampling, I wanted to ask why LHS is a division of CDF into N intervals rather than PDF. I'm having some trouble understanding what the difference is between the division of CDF vs. PDF. Also, since CDF always ranges from 0 to 1, is that the reason LHS has the name hyper"cube" even when the different parameters have different ranges?

Latin hypercube samples are normally constructed uniformly on the (0,1) interval, and then transformed to the marginal desired distributions. So, you are correct, that is why they are referred to as "hyper" (multiple dimensions) "cubes" (sides of equal length). If $$f(x)$$ is the PDF, $$F(x)$$ is the CDF, and $$X$$ is a ($$n$$ x $$p$$) matrix that is a LHS, then $$F^{-1}(X[,i])$$ are the quantiles of the desired distribution, transforming the (0,1) interval to the range of the distribution of interest. The bin endpoints around $$F^{-1}(X[,i])$$ are also the points that divide the marginal PDF in to $$n$$ equal intervals.


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# How do you find the area of an irregular polygon within a square?

I'm trying to make a function that, given floating point coordinates/arguments, returns the area of a shape within a given pixel. The purpose is drawing anti-aliased shapes for a game's UI.

I know how to do rectangles, as it is very simple, but other shapes (ellipses, triangles, non-regular polygons, etc.) are a bit harder.

So, given a set of points for a polygon, if I loop through the shape on a grid with a cell size of 1x1, I want the area of the shape within said cell.

Picture example: Pixel in Circle

What is the area of the circle within the bounds of the red pixel, given the pixel coordinates, circle coordinates, and circle radius?

Better example: Zoomed Pixel in Circle

What is the area of the white within the non-shaded area, given the four coordinates of the shaded area, coordinate of circle, and radius of circle?

What about with any polygon, regular or irregular, given a list of coordinates, and a coordinate to find the area within?

What I mean by the checking coordinate is, area within given-coordinate-to-check -> given-coordinate-to-check+1

Sorry if my question is confusing!

To restate the question more clearly, what is the area of a square-shaped shaded region on any possible geometric shape?

• "The purpose is drawing anti-aliased shapes for a game's UI." Don't do this yourself! Surely your graphics library/API already provides a way to draw smooth anti-aliased shapes. – Rahul Oct 29 '16 at 18:23
• Haha, well it doesn't, surprisingly! – user91670 Oct 29 '16 at 18:57

• @user91670: This (random sampling) is an extremely common method, that is used by most raytracers to achieve antialiasing. ($N^2$ random samples produce better (psychovisual) results than $N \times N$ regular samples in a grid.) One very nice speedup is that if you know which surfaces/edges are involved in the calculation for the four points at $(x,y)$, $(x+1,y)$, $(x,y+1)$, $(x+1,y+1)$, it is enough to consider those surfaces/edges in the random samples. – Nominal Animal Oct 29 '16 at 19:08


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# Elliptic coordinates: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Encyclopedia

Elliptic coordinate system

Elliptic coordinates are a two-dimensional orthogonal coordinate system in which the coordinate lines are confocal ellipses and hyperbolae. The two foci F1 and F2 are generally taken to be fixed at a and + a, respectively, on the x-axis of the Cartesian coordinate system.

## Basic definition

The most common definition of elliptic coordinates (μ,ν) is

$x = a \ \cosh \mu \ \cos \nu$
$y = a \ \sinh \mu \ \sin \nu$

where μ is a nonnegative real number and $\nu \in [0, 2\pi).$

On the complex plane, an equivalent relationship is

$x + iy = a \ \cosh(\mu + i\nu)$

These definitions correspond to ellipses and hyperbolae. The trigonometric identity

$\frac{x^{2}}{a^{2} \cosh^{2} \mu} + \frac{y^{2}}{a^{2} \sinh^{2} \mu} = \cos^{2} \nu + \sin^{2} \nu = 1$

shows that curves of constant μ form ellipses, whereas the hyperbolic trigonometric identity

$\frac{x^{2}}{a^{2} \cos^{2} \nu} - \frac{y^{2}}{a^{2} \sin^{2} \nu} = \cosh^{2} \mu - \sinh^{2} \mu = 1$

shows that curves of constant ν form hyperbolae.

## Scale factors

The scale factors for the elliptic coordinates (μ,ν) are equal

$h_{\mu} = h_{\nu} = a\sqrt{\sinh^{2}\mu + \sin^{2}\nu}.$

To simplify the computation of the scale factors, double angle identities can be used to express them equivalently as

$h_{\mu} = h_{\nu} = a\sqrt{\frac{1}{2} (\cosh2\mu - \cos2\nu}).$

Consequently, an infinitesimal element of area equals

$dA = a^{2} \left( \sinh^{2}\mu + \sin^{2}\nu \right) d\mu d\nu$

and the Laplacian equals

$\nabla^{2} \Phi = \frac{1}{a^{2} \left( \sinh^{2}\mu + \sin^{2}\nu \right)} \left( \frac{\partial^{2} \Phi}{\partial \mu^{2}} + \frac{\partial^{2} \Phi}{\partial \nu^{2}} \right).$

Other differential operators such as $\nabla \cdot \mathbf{F}$ and $\nabla \times \mathbf{F}$ can be expressed in the coordinates (μ,ν) by substituting the scale factors into the general formulae found in orthogonal coordinates.

## Alternative definition

An alternative and geometrically intuitive set of elliptic coordinates (σ,τ) are sometimes used, where σ = coshμ and τ = cosν. Hence, the curves of constant σ are ellipses, whereas the curves of constant τ are hyperbolae. The coordinate τ must belong to the interval [-1, 1], whereas the σ coordinate must be greater than or equal to one.

The coordinates (σ,τ) have a simple relation to the distances to the foci F1 and F2. For any point in the plane, the sum d1 + d2 of its distances to the foci equals 2aσ, whereas their difference d1d2 equals 2aτ. Thus, the distance to F1 is a(σ + τ), whereas the distance to F2 is a(σ − τ). (Recall that F1 and F2 are located at x = − a and x = + a, respectively.)

A drawback of these coordinates is that they do not have a 1-to-1 transformation to the Cartesian coordinates

$x = a \left. \sigma \right. \tau$
$y^{2} = a^{2} \left( \sigma^{2} - 1 \right) \left(1 - \tau^{2} \right).$

## Alternative scale factors

The scale factors for the alternative elliptic coordinates (σ,τ) are

$h_{\sigma} = a\sqrt{\frac{\sigma^{2} - \tau^{2}}{\sigma^{2} - 1}}$
$h_{\tau} = a\sqrt{\frac{\sigma^{2} - \tau^{2}}{1 - \tau^{2}}}.$

Hence, the infinitesimal area element becomes

$dA = a^{2} \frac{\sigma^{2} - \tau^{2}}{\sqrt{\left( \sigma^{2} - 1 \right) \left( 1 - \tau^{2} \right)}} d\sigma d\tau$

and the Laplacian equals

$\nabla^{2} \Phi = \frac{1}{a^{2} \left( \sigma^{2} - \tau^{2} \right) } \left[ \sqrt{\sigma^{2} - 1} \frac{\partial}{\partial \sigma} \left( \sqrt{\sigma^{2} - 1} \frac{\partial \Phi}{\partial \sigma} \right) + \sqrt{1 - \tau^{2}} \frac{\partial}{\partial \tau} \left( \sqrt{1 - \tau^{2}} \frac{\partial \Phi}{\partial \tau} \right) \right].$

Other differential operators such as $\nabla \cdot \mathbf{F}$ and $\nabla \times \mathbf{F}$ can be expressed in the coordinates (σ,τ) by substituting the scale factors into the general formulae found in orthogonal coordinates.

## Extrapolation to higher dimensions

Elliptic coordinates form the basis for several sets of three-dimensional orthogonal coordinates. The elliptic cylindrical coordinates are produced by projecting in the z-direction. The prolate spheroidal coordinates are produced by rotating the elliptic coordinates about the x-axis, i.e., the axis connecting the foci, whereas the oblate spheroidal coordinates are produced by rotating the elliptic coordinates about the y-axis, i.e., the axis separating the foci.

## Applications

The classic applications of elliptic coordinates are in solving partial differential equations, e.g., Laplace's equation or the Helmholtz equation, for which elliptic coordinates allow a separation of variables. A typical example would be the electric field surrounding a flat conducting plate of width 2a.

The geometric properties of elliptic coordinates can also be useful. A typical example might involve an integration over all pairs of vectors $\mathbf{p}$ and $\mathbf{q}$ that sum to a fixed vector $\mathbf{r} = \mathbf{p} + \mathbf{q}$, where the integrand was a function of the vector lengths $\left| \mathbf{p} \right|$ and $\left| \mathbf{q} \right|$. (In such a case, one would position $\mathbf{r}$ between the two foci and aligned with the x-axis, i.e., $\mathbf{r} = 2a \mathbf{\hat{x}}$.) For concreteness, $\mathbf{r}$, $\mathbf{p}$ and $\mathbf{q}$ could represent the momenta of a particle and its decomposition products, respectively, and the integrand might involve the kinetic energies of the products (which are proportional to the squared lengths of the momenta).

• Three dimensional orthogonal coordinate systems

## References

• Korn GA and Korn TM. (1961) Mathematical Handbook for Scientists and Engineers, McGraw-Hill.


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# How do you solve  |x – 4| > |3x – 1|?

Sep 13, 2017

$x \in \mathbb{R}$

#### Explanation:

When there is absolute value, we have to take two options the first where the first is positive the next is negative.
$| a + b |$
$= a + b$
or $= - \left(a + b\right) = - a - b$
|x–4|>|3x–1|

1. $= x - 4 > | 3 x - 1 |$
again doing the same with the left hand side:
$x - 4 > | 3 x - 1 |$

1a. $= x - 4 > 3 x - 1$

Subtracting x from both sides of the equation to get the unknown's terms to one side of the equation:
$\cancel{x} \cancel{- x} - 4 > 3 x - 1 - x$
$- 4 > 2 x - 1$

Adding 1 to both sides of the equation to isolate the unknown's term on the left hand side:
$- 4 + 1 > 2 x \cancel{- 1} \cancel{+ 1}$
$- 3 > 2 x$

Dividing both sides of the equation by 2 to isolate x:
$- \frac{3}{2} > \frac{\cancel{2} x}{\cancel{2}}$

1a. color(red)(x<-3/2

and

1b. $x - 4 > - \left(3 x - 1\right)$
$= x - 4 > - 3 x + 1$
Doing same as before to isolate x:
$\cancel{x} - 4 \cancel{- x} > - 3 x + 1 - x$
$- 4 > - 4 x$
$\frac{\cancel{- 4}}{\cancel{- 4}} > \frac{\cancel{- 4} x}{\cancel{- 4}}$
$1 > x$
1b. color(red)(x<1

and

1. $= - \left(x - 4\right) > | 3 x - 1 |$
$= - x + 4 > | 3 x - 1 |$

2a.$= - x + 4 > 3 x - 1$

simplifying: (adding 1 and subtracting x from both sides of the equation and then dividing by 4):
$\cancel{- x} \cancel{+ x} + 4 + 1 > 3 x \cancel{- 1} + x \cancel{+ 1}$
$4 + 1 > 3 x + x$
$5 > 4 x$

$\frac{5}{4} > \frac{\cancel{4} x}{\cancel{4}}$
$\frac{5}{4} > x$
2a.color(red)(x<5/4

and

2b.$= - x + 4 > - 1 \left(3 x - 1\right)$
$= - x + 4 > - 3 x + 1$

Simplifying by adding 3x to, subtracting 4 from both sides and then dividing by 2:
$- x \cancel{+ 4} + 3 x \cancel{- 4} > \cancel{- 3 x} + 1 \cancel{+ 3 x} - 4$
$- x + 3 x > 1 - 4$
$2 x \succ 3$
$\frac{\cancel{2} x}{\cancel{2}} > - \frac{3}{2}$
2b. $\textcolor{red}{x > - \frac{3}{2}}$

Simplifying 1a , 2a ,and 2b :
$x < - \frac{3}{2}$
$x < 1$
$x < \frac{5}{4} \rightarrow x < 1.25$
$\rightarrow x < - \frac{3}{2} < 1 < \frac{5}{4}$
rarr color(blue)(x<5/4

and

From 2a:
color(blue)(x> -3/2

since x is smaller than $\frac{5}{4}$ and bigger than $- \frac{3}{2}$ then x defined for any number.


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# How to read a Q sphere representation?

I'm trying to understand the Q-sphere representation of a 3-qubit system. I get that the 3-qubits are in a superposition of 2 different states. The first qubit (rightmost) is in a superposition of 0 and 1 with a phase difference of pi. And the other two qubits are in ground states. I hope I got this right till here.

What does the location of the point on the 1st ring represent? In the documentation, it's mentioned that the latitude is the Hamming distance from zero state. What does that indicate?

We first need to understand what do we mean by Hamming Distance, simply put, for two strings of equal length, it is the number of different symbols in corresponding locations. For example: since we're exclusively dealing with binary strings, let's say we are calculating Hamming distance between 001 and 000. Since only the right-most values differ, we get a hamming distance of $$1$$.
Another example: 101 and 010 has a hamming distance of $$3$$ since they are different at each position. Hamming distances are widely used in Coding Theory for Error detection and can also be used in Machine Learning as an alternative measure to Euclidean distance. In both the applications it is used to objectively calculate how separate two values are, more the hamming distance more dissimilar they are considered to be.
Here, hamming distance has the same purpose, to calculate how dissimilar the states are. As you move farther away from $$|000\rangle$$ (going towards $$|111\rangle$$) the bigger will be the difference between the states and hence the bigger will be the Hamming distance. Going with the same logic, $$|000\rangle$$ and $$|111\rangle$$ have a hamming distance of 3 hence signifying that they are very different (or better said, completely opposite since hamming distance = string length).


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 Sciencemadness Discussion Board » Fundamentals » Reagents and Apparatus Acquisition » Cheap Potassium Iodide Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Detritus   » Test Forum

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Author: Subject: Cheap Potassium Iodide
Deathunter88
National Hazard

Posts: 356
Registered: 20-2-2015
Location: Beijing, China
Member Is Offline

Mood: No Mood

What's the average price of KI in the USA? I fail to see how $6 for a mere 100g is cheap. Here in China 500g of reagent grade KI is only$10 or so, and even a brand name is $20 for 500g, so I don't really see the incentive to sell fake KI, especially at such a high price. Mr. Rogers Hazard to Others Posts: 146 Registered: 30-10-2017 Location: Ammonia Avenue Member Is Offline Mood: No Mood I bought 300g from the OP's link and, as someone stated, it was slightly wet but it's definitely KI. It was shipped in a leaking plastic bag inside an envelope and the customs form said it was a necklace. I think it's "fine" for the price. [Edited on 27-11-2018 by Mr. Rogers] Mr. Rogers Hazard to Others Posts: 146 Registered: 30-10-2017 Location: Ammonia Avenue Member Is Offline Mood: No Mood  Quote: Originally posted by Deathunter88 What's the average price of KI in the USA? I fail to see how$6 for a mere 100g is cheap. Here in China 500g of reagent grade KI is only $10 or so, and even a brand name is$20 for 500g, so I don't really see the incentive to sell fake KI, especially at such a high price.

It's much more expensive in the US - probably 4-5x as much. Dunno why.
woelen

Posts: 6631
Registered: 20-8-2005
Location: Netherlands
Member Is Offline

Mood: interested

Average price where I live (EU) is appr. EUR 20 per 100 gram, so $6 (appr. EUR 5) per 100 gram is very cheap for us. Prices in the USA will not be very different. The art of wondering makes life worth living... Want to wonder? Look at http://www.oelen.net/science Ubya National Hazard Posts: 383 Registered: 23-11-2017 Location: Rome-Italy Member Is Offline Mood: I'm a maddo scientisto!!!  Quote: Originally posted by Deathunter88 What's the average price of KI in the USA? I fail to see how$6 for a mere 100g is cheap. Here in China 500g of reagent grade KI is only $10 or so, and even a brand name is$20 for 500g, so I don't really see the incentive to sell fake KI, especially at such a high price.

if you can find a source that would sell internationally KI for that price we would be really happy

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feel free to correct my grammar, or any mistakes i make
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DavidJR
International Hazard

Posts: 537
Registered: 1-1-2018
Location: Scotland
Member Is Offline

Mood: Anxious

 Quote: Originally posted by Deathunter88 What's the average price of KI in the USA? I fail to see how $6 for a mere 100g is cheap. Here in China 500g of reagent grade KI is only$10 or so, and even a brand name is $20 for 500g, so I don't really see the incentive to sell fake KI, especially at such a high price. I just looked at one of my suppliers in the UK and they have KI for £11.25/100g and NaI for £26/100g.$6 US is about £4.70 at the moment. So yes, if it wasn't a scam, this would be very cheap for me.
S.C. Wack
bibliomaster

Posts: 1895
Registered: 7-5-2004
Location: Cornworld, Central USA
Member Is Offline

Mood: Enhanced

 Quote: Originally posted by Deathunter88 What's the average price of KI in the USA? I fail to see how $6 for a mere 100g is cheap. Here in China 500g of reagent grade KI is only$10 or so, and even a brand name is $20 for 500g, so I don't really see the incentive to sell fake KI, especially at such a high price. A price beyond competitive on amazon is$85/kg...best is $56/kg, for 5, which makes me wish I needed that much, that's insanely low. These are just repackers of probably Chinese goods. There are sellers of I, 50 g. for$10; technically, to import legally one needs a license from the DEA, which they won't provide.

Someone who has been pricing chemicals for a time might be suspicious of cheap imported iodide, and would want to test it well. Does China control some vast natural source of iodide? I suspect that the price from a non-Chinese source is higher than the amounts given above.

[Edited on 28-11-2018 by S.C. Wack]

"You're going to be all right, kid...Everything's under control." Yossarian, to Snowden
Sulaiman
International Hazard

Posts: 2229
Registered: 8-2-2015
Location: Walsall, England
Member Is Offline

With the recently purchased potassium iodide I tried to precipitate the iodine using hydrogen peroxide and got an indication of low iodine content.

Today I made up a 500ml solution of what should be a 1.0M solution of the suspect potassium iodide,
(which I bought via eBay seller 94racing-au, the cheapest at the time)
and bubbled chlorine gas through 100ml of the solution
to replace the iodide with chloride and precipitate iodine.

36% HCl in a 3ml disposable pipette, TCCA in a side-arm test tube.

The crop of wet iodine on a 90mm diameter filter paper.

Ideally there should be 1/10th mole of iodine = 12.7 g
A little was lost to physical handling, and there should be a little Iodine in the filtrate/solution.
I guess that potassium chloride may increase the solubility of iodine in a manner similar to potassium iodide ?
Could anyone please tell me, or point to, or calculate,
the solubility of iodine in 1.0M Potassium Chloride ?
especially in the 10 to 20 Celcius range.

Result : 8.05g dry iodine

The filtrate has a strong yellow tint and the tissues used beneath the filter paper absorbed a lot of iodine giving a strong blue tint,
so I still suspect that KCl increases the solubilit of iodine in water..
Although the yield was only 63% I suspect that the KI was pure.
now I need to do a titration to be sure.

[Edited on 6-12-2018 by Sulaiman]

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DavidJR
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I just got a refund for my fake stuff.
Ubya
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 Quote: Originally posted by Sulaiman I guess that potassium chloride may increase the solubility of iodine in a manner similar to potassium iodide ?

iodine is solubile in iodide solution because it forms triiodide, which is soluble

---------------------------------------------------------------------
feel free to correct my grammar, or any mistakes i make
---------------------------------------------------------------------
S.C. Wack
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 Quote: Originally posted by Sulaiman I guess that potassium chloride may increase the solubility of iodine in a manner similar to potassium iodide ? Could anyone please tell me, or point to, or calculate, the solubility of iodine in 1.0M Potassium Chloride ? especially in the 10 to 20 Celcius range.

You have the iodine, KCl is hopefully not a challenge...

So I looked in Mellor and he gives the exact answer actually and it's a good thing, because CA abstracts the same article but doesn't provide details...just like the publisher of Gmelin's didn't provide a pdf of any I volume for sale last time I looked...if you believe 1920 Gazz It, KCl accounts for less than 1/10 gram of the missing I.

[Edited on 6-12-2018 by S.C. Wack]

"You're going to be all right, kid...Everything's under control." Yossarian, to Snowden
Sulaiman
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Actually I would have to buy or make some KCl so the data is useful, thanks.
I dowloaded Mellor years ago but forgot the contents

After tidying up I dropped a pinch of TCCA into the filtrate,
I'll check for any precipitate tomorrow.

To be sure I'll boil dry the filtrate to dryness then heat the residue to see how much iodine vapour is produced.

I just realised that the iodine may have precipitated in particles too small to be caught by the medium filter paper that I used,
which now seems more likely than dissolution carrying the iodine into the tissues used beneath the paper for drying the iodine,
but if true then the light yellow colour of the filtrate should be much darker.
I'm confused.

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S.C. Wack
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The thinking was neutralizing HCl and K (bi)carbonate or hydroxide exactly should be a low hurdle.

IIRC H2O2/HCl and KI is popular enough that there were calculators for the different parameters, for meth cooks, who seemed satisfied. Perhaps with Cl as a gas, bad things are more likely to happen?

"You're going to be all right, kid...Everything's under control." Yossarian, to Snowden
MrHomeScientist
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I know for bromine it can form interhalogens with chlorine; that's why I avoided Cl when I made my Br. I wonder if something similar can happen with iodine.
UC235
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 Quote: Originally posted by MrHomeScientist I know for bromine it can form interhalogens with chlorine; that's why I avoided Cl when I made my Br. I wonder if something similar can happen with iodine.

ICl and ICl3 are well known. Bubbling Cl2 through iodide solution is how woelen made periodate.
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### glDrawPixels problem

• Email
• Follow

I've a problem with the glDrawPixels openGL primitive....
I'm not able to draw at screen a figure described into a GLubyte
dataCulling[480][640] matrix...
This is a resuming of my program:

char valid;
float position[4];
GLubyte GLubyte dataCulling[480][640];
.....
operation with dataCulling....
at the end of operation I've a matrix dataCulling composed only by GLubyte
value 0 or 127 or 255 ....
.....
glRasterPos2i(0,0);
glGetFloatv(GL_CURRENT_RASTER_POSITION,&position[0]);
glGetBooleanv(GL_CURRENT_RASTER_POSITION_VALID,&valid);
glDrawPixels(640,480,GL_GREEN,GL_UNSIGNED_BYTE,dataCulling);
glFlush(); <====== DEBUG STOP

I've made the debugging and I've find out that:
position => { 0,480,0.5,0 }
valid => 0
and the screen remain all black....but in the dataCulling matrix ( I've
check it ) there are several 127 or 255 values....not only 0......

now......what I've to do ?!?
How can I resolve my problem ?!?
I want only see my green figure ( described into dataCulling) in the
screen...

 0

See related articles to this posting

> position => { 0,480,0.5,0 }
> valid => 0

Well, valid must be TRUE, otherwise "the current raster position become
indeterminate".

However, I founded a strange behaviour. If I specify GL_GREEN in
glDrawPixels nothing appears to the screen (even if
GL_CURRENT_RASTER_POSITION_VALID is true). With GL_LUMINANCE ... I get the
correct result. Why? ... I'm upset..

Manuel

 0

> Well, valid must be TRUE, otherwise "the current raster position become
> indeterminate".

but, in my case.....why the raster position validity is FALSE ?!

> However, I founded a strange behaviour. If I specify GL_GREEN in
> glDrawPixels nothing appears to the screen (even if
> GL_CURRENT_RASTER_POSITION_VALID is true).
> With GL_LUMINANCE ... I get the
> correct result. Why? ... I'm upset..

I think is the same problem of mine..

 0

> Maybe your graphics driver is rubbish. Many gaming-card
> drivers will fail when you try to do "weird things".

....I've a geForce 420 go ( Mobile computer ) with 16 MB of ram....
and drivers are latest detonator of nVIDIA site....

 0

> > Maybe your graphics driver is rubbish. Many gaming-card
> > drivers will fail when you try to do "weird things".
>
> ...I've a geForce 420 go ( Mobile computer ) with 16 MB of ram....
> and drivers are latest detonator of nVIDIA site....

But Nvidia's mobile chipsets don't use the Dets from Nvidia's website, you're supposed to get

"NOTE: GeForce2 Go and GeForce4 Go mobile processors are not supported in this driver. Please
contact the notebook's manufacturer for graphics drivers for your notebook PC."

Perhaps if you installed the Dets you've got incorrect drivers installed?

Andrew

 0

>  glDrawPixels(width,height,GL_RED,GL_UNSIGNED_BYTE,dataCulling);
>
> valida=1
> position= { 0  0  0.5  1 }
>

Ah, maybe I forgot one thing, before glMatrixMode(GL_PROJECTION) you should
But, It seems you're already doing this (you got valid=1).

You can try using GL_LUMINANCE (with glColorMask() if necessary). On my
computer that seem to work (while GL_GREEN,GL_RED,GL_BLUE are evils..).

Last try: use GL_RGB and fill your dataCulling[] with 3 components. This
MUST works.. well,I hope! I've tried this on a GeForce3 and on a 9700PRO and
it works.

However, glDrawPixels is really a sloow function...

Manuel

(PS : sorry for my poor english... I'm italian)

 0

zaero wrote:
>
> > However I think the correct way to code is:
> >
> >  ........
>
> I've used your precious help...an I've try with the follow code...
>
> glMatrixMode(GL_PROJECTION);
>  glPushMatrix();
>  glOrtho(0,width,0,height,-1.0f,1.0f);

glMatrixMode(GL_MODELVIEW);
glPushMatrix();

>  glRasterPos2f(0,0);
>  glGetFloatv(GL_CURRENT_RASTER_POSITION,position);
>  glGetBooleanv(GL_CURRENT_RASTER_POSITION_VALID,&valida);
>  glDrawPixels(width,height,GL_RED,GL_UNSIGNED_BYTE,dataCulling);
>  glFlush();

glPopMatrix();

>  glMatrixMode(GL_PROJECTION);
>  glPopMatrix(); <=======================STOP DEBUGGING
>
> Debugging says that:
>
> valida=1
> position= { 0  0  0.5  1 }
>
> BUT    I dont see anything at screen.......why ?!?!? Why glDrawPixels
> doesn't work ?!?
> ====
>

Maybe these inserts might help? Dunno what the rest of your code is
doin'

--
Rob Fletcher, University of York, UK
[Spamtrap - Remove the "y" to reply]

 0

> "NOTE: GeForce2 Go and GeForce4 Go mobile processors are not supported in
> contact the notebook's manufacturer for graphics drivers for your notebook
PC."
>
> Perhaps if you installed the Dets you've got incorrect drivers installed?

.....well....I've installed detonator and....so far I haven't had any
problem.....
but.....I've search to download specific driver for my sony vaio geforce 420
go and I've found out anything...
Do You know where I could find out them ?!?

 0

> > "NOTE: GeForce2 Go and GeForce4 Go mobile processors are not supported in
> > contact the notebook's manufacturer for graphics drivers for your notebook
> > PC."
> >
> > Perhaps if you installed the Dets you've got incorrect drivers installed?
>
> ....well....I've installed detonator and....so far I haven't had any
> problem.....
> but.....I've search to download specific driver for my sony vaio geforce 420
> go and I've found out anything...
> Do You know where I could find out them ?!?

Well somewhere on Sony's website I presume... can't get very far without knowing your model number

http://ciscdb.sel.sony.com/cgi-bin/select-p-n.pl

Andrew

 0

> Ah, maybe I forgot one thing, before glMatrixMode(GL_PROJECTION) you
should
> loadIdentity() to reset the modelview_matrix.
> But, It seems you're already doing this (you got valid=1).

> You can try using GL_LUMINANCE (with glColorMask() if necessary). On my
> computer that seem to work (while GL_GREEN,GL_RED,GL_BLUE are evils..).

I've used glDrawPixels with GL_LUMINANCE but....
my screen is still all black !!!!

> Last try: use GL_RGB and fill your dataCulling[] with 3 components. This
> MUST works.. well,I hope! I've tried this on a GeForce3 and on a 9700PRO
and
> it works.

....still to try.....

> However, glDrawPixels is really a sloow function...

I know that but....how can I draw at screen an image from a matrix in an
other way ?!?!?

> (PS : sorry for my poor english... I'm italian)

You undervalue yourself !!!

 0

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text:
# Carleman matrix

In mathematics, a Carleman matrix is a matrix used to convert function composition into matrix multiplication. It is often used in iteration theory to find the continuous iteration of functions which cannot be iterated by pattern recognition alone. Other uses of Carleman matrices occur in the theory of probability generating functions, and Markov chains.

## Definition

The Carleman matrix of a function ${\displaystyle f(x)}$ is defined as:

${\displaystyle M[f]_{jk}={\frac {1}{k!}}\left[{\frac {d^{k}}{dx^{k}}}(f(x))^{j}\right]_{x=0}~,}$

so as to satisfy the (Taylor series) equation:

${\displaystyle (f(x))^{j}=\sum _{k=0}^{\infty }M[f]_{jk}x^{k}.}$

For instance, the computation of ${\displaystyle f(x)}$ by

${\displaystyle f(x)=\sum _{k=0}^{\infty }M[f]_{1,k}x^{k}.~}$

simply amounts to the dot-product of row 1 of ${\displaystyle M[f]}$ with a column vector ${\displaystyle \left[1,x,x^{2},x^{3},...\right]^{\tau }}$.

The entries of ${\displaystyle M[f]}$ in the next row give the 2nd power of ${\displaystyle f(x)}$:

${\displaystyle f(x)^{2}=\sum _{k=0}^{\infty }M[f]_{2,k}x^{k}~,}$

and also, in order to have the zero'th power of ${\displaystyle f(x)}$ in ${\displaystyle M[f]}$, we aadopt the row 0 containing zeros everywhere except the first position, such that

${\displaystyle f(x)^{0}=1=\sum _{k=0}^{\infty }M[f]_{0,k}x^{k}=1+\sum _{k=1}^{\infty }0*x^{k}~.}$

Thus, the dot product of ${\displaystyle M[f]}$ with the column vector ${\displaystyle \left[1,x,x^{2},...\right]^{\tau }}$ yields the column vector ${\displaystyle \left[1,f(x),f(x)^{2},...\right]^{\tau }}$

${\displaystyle M[f]*\left[1,x,x^{2},x^{3},...\right]^{\tau }=\left[1,f(x),(f(x))^{2},(f(x))^{3},...\right]^{\tau }.}$

## Bell matrix

The Bell matrix of a function ${\displaystyle f(x)}$ is defined as

${\displaystyle B[f]_{jk}={\frac {1}{j!}}\left[{\frac {d^{j}}{dx^{j}}}(f(x))^{k}\right]_{x=0}~,}$

so as to satisfy the equation

${\displaystyle (f(x))^{k}=\sum _{j=0}^{\infty }B[f]_{jk}x^{j}~,}$

so it is the transpose of the above Carleman matrix.

## Jabotinsky matrix

Eri Jabotinsky developed that concept of matrices 1947 for the purpose of representation of convolutions of polynomials. Several authors refer to the Bell matrices as "Jabotinsky matrix" since (D. Knuth 1992, W.D. Lang 2000), and possibly this shall grow to a more canonical name.

## Generalization

A generalization of the Carleman matrix of a function can be defined around any point, such as:

${\displaystyle M[f]_{x_{0}}=M_{x}[x-x_{0}]M[f]M_{x}[x+x_{0}]}$

or ${\displaystyle M[f]_{x_{0}}=M[g]}$ where ${\displaystyle g(x)=f(x+x_{0})-x_{0}}$. This allows the matrix power to be related as:

${\displaystyle (M[f]_{x_{0}})^{n}=M_{x}[x-x_{0}]M[f]^{n}M_{x}[x+x_{0}]}$

## Matrix properties

These matrices satisfy the fundamental relationships:

which makes the Carleman matrix M a (direct) representation of ${\displaystyle f(x)}$, and the Bell matrix B an anti-representation of ${\displaystyle f(x)}$. Here the term ${\displaystyle f\circ g}$ denotes the composition of functions ${\displaystyle f(g(x))}$.

Other properties include:

## Examples

The Carleman matrix of a constant is:

${\displaystyle M[a]=\left({\begin{array}{cccc}1&0&0&\cdots \\a&0&0&\cdots \\a^{2}&0&0&\cdots \\\vdots &\vdots &\vdots &\ddots \end{array}}\right)}$

The Carleman matrix of the identity function is:

${\displaystyle M_{x}[x]=\left({\begin{array}{cccc}1&0&0&\cdots \\0&1&0&\cdots \\0&0&1&\cdots \\\vdots &\vdots &\vdots &\ddots \end{array}}\right)}$

The Carleman matrix of a constant addition is:

${\displaystyle M_{x}[a+x]=\left({\begin{array}{cccc}1&0&0&\cdots \\a&1&0&\cdots \\a^{2}&2a&1&\cdots \\\vdots &\vdots &\vdots &\ddots \end{array}}\right)}$

The Carleman matrix of a constant multiple is:

${\displaystyle M_{x}[cx]=\left({\begin{array}{cccc}1&0&0&\cdots \\0&c&0&\cdots \\0&0&c^{2}&\cdots \\\vdots &\vdots &\vdots &\ddots \end{array}}\right)}$

The Carleman matrix of a linear function is:

${\displaystyle M_{x}[a+cx]=\left({\begin{array}{cccc}1&0&0&\cdots \\a&c&0&\cdots \\a^{2}&2ac&c^{2}&\cdots \\\vdots &\vdots &\vdots &\ddots \end{array}}\right)}$

The Carleman matrix of a function ${\displaystyle f(x)=\sum _{k=1}^{\infty }f_{k}x^{k}}$ is:

${\displaystyle M[f]=\left({\begin{array}{cccc}1&0&0&\cdots \\0&f_{1}&f_{2}&\cdots \\0&0&f_{1}^{2}&\cdots \\\vdots &\vdots &\vdots &\ddots \end{array}}\right)}$

The Carleman matrix of a function ${\displaystyle f(x)=\sum _{k=0}^{\infty }f_{k}x^{k}}$ is:

${\displaystyle M[f]=\left({\begin{array}{cccc}1&0&0&\cdots \\f_{0}&f_{1}&f_{2}&\cdots \\f_{0}^{2}&2f_{0}f_{1}&f_{1}^{2}&\cdots \\\vdots &\vdots &\vdots &\ddots \end{array}}\right)}$


text:
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# CEA-jiangming / DecGMCA Public

Joint Multichannel Deconvolution and Blind Source Separation algorithm

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# DecGMCA

DecGMCA (Deconvolution Generalized Morphological Component Analysis) is a sparsity-based algorithm aiming at solving joint multichannel Deconvolution and Blind Source Separation (DBSS) problem.

For more details, please refer to the paper Joint Multichannel Deconvolution and Blind Source Separation (https://arxiv.org/abs/1703.02650)

## Introduction

Considering a multichannel DBSS problem:

Y = H * ( AS ) + N

Y: observation of size Nc by Np for an Nc-channel and Np-pixel imaging system.

H: a linear convolution kernel of size Nc by Np (e.g. downsampling matrix, psf), often ill-conditioned in practice.

A: unknown mixing matrix of size Nc by Ns, representing Ns sources are blindly linearly mixed (entities of row i represents the weighted contribution of sources at the given channel number i).

S: sources of size Ns by Np, representing Ns sources of Np pixels (sources are aligned to row vectors).

N: additional noise of size Nc by Np, supposed to be Gaussian.

This problem can be conveniently written in Fourier space, which has a lot of interests in Fourier imaging systems such as radio interferometry, MRI, etc. In Fourier space, noticing the convolution is transformed to product and A is unchanged as its entities are actually scalar factors applied to sources, the problem is written as

\hat{Y} = \hat{H} ( A \hat{S} ) + \hat{N}

The sources are sparse in a given dictionary $\Phi$. Thus, the yielding optimization problem is written as follows:

min_{A,S} ||\hat{Y} - \hat{H} A \hat{S}||_2^2 + \lambda ||S \Phi||_p, where p-norm is 0-norm or 1-norm.

## Optimization

Challenges:

1. As the above optimization is non-convex, only critical point can be expected.
2. Convolution kernel is ill-conditioned, leading to the unstability of the solution.

The main idea of solution is based on alternating minimization but we do not directly apply alternating proximity-based algorithms due to its computational demanding. Our DecGMCA employs an alternating projected least-squares procedure to approach the critical point plus a proximity-based procedure to finally refine the solution. Thus, DecGMCA is structured as:

• Intitialization
• Alternating projected least-squares
• Refinement step

### Initialization

A simple initialization (of matrix A) can be realized by randomization. One can also have more accurate initialization which depends on the form of the convolution kernel H:

• If H is a downsampling matrix, we apply several iterations of matrix completion scheme (e.g. SVT algorithm). Then the initialization is acheived by selecting Ns eigonvectors of left singular matrix after the application of singular value decomposition (svd) on the completed data.
• If H is a convolution kernel (not a downsampling matrix), we only keep Ns eigonvectors of left singular matrix as the initialization after svd on the completed data.

### Alternating projected least-squares

The procedure is based on the alternating update of one variable with respect to the other.

#### Update S with respect to A

This update can be divided into two steps: approximation of S via least-squares and sparsity thresholding.

As for the approximation of S via least-squares, due to the ill-conditioned kernel H, a regularization parameter $\epsilon$ is involved to stablize the deconvolution. This parameter acts as a Tikhonov parameter: If $\epsilon$ is large, the system will be more regularized but the solution is less accurate. Conversely, if $\epsilon$ is small, the system will be less regularized but the solution is more accurate. The second step is sparsity thresholding to have a clean estimate of S. This step is realized by hard-thresholding with threshold $\lambda$. (Although soft-thresholding for l1-norm has more beautiful mathematical convergence proof, we argue that hard-thresholding has no biais effect and has empirical convergence according to our experiments). The consideration of $\epsilon$ and $\lambda$ is extensively studied in the paper.

#### Update A with respect to S

This update is just a simple least-squares. One should notice that columns of A should be l2 normalized.

### Refinement step

The alternating projected least-squares is efficient but does not necessarily ensure the optimal solution. This is owing to the fact that the projected least-squares has algorithmic biais compared to the exact projection realized by proximal algorithms. Thus, in order to refine the solution (often S), we resolve the problem of updating S with respect to A by using proximal algorithms such as Forward-Backward, Condat-Vu primal dual, etc.

## Python DecGMCA package

### Prerequisites

#### Basic python environment

This package has been tested with python 2.7, some python libraries are required to ensure a correct working:

• numpy
• scipy
• matplotlib
• astropy The above libraries are accessible in macport, pip or other package manager systems. For instance, via macport:
port install some-package-name

or via pip:

pip install some-package-name

One may need root permission for the above operations.

#### Accelaration of codes? Interface python with C++ and paralization

For large-scale data, one may be not satisfied python (python can be up to 50 times slower than C/C++). In this DecGMCA package, we have an option to interface python with C++ and paralize the codes. The following packages are required:

• GCC (tested with GCC 4.9)
• CMake (tested with v3.9)
• Boost (tested with v1.58)
• Cfitsio
• OMP (tested with GCC 4.9) These packages are easily installed via apt-get (Linux), homebrew (MacOS) or directly from the website. For instance, via apt-get
apt-get install some-package-name

or via homebrew

brew install some-package-name

#### Problem with MacOS?

The default C compiler of MacOS is Clang. One should set GCC as the default C compiler and compile all dependencies (Boost, Cfitsio, etc.)

Among all packages, the Boost package is the most troublesome on MacOS. Here is a rapid solution:

tar -xzvf boost_1_58_0.tar.gz

• Inside the boost directory run the bootstrap.sh script specifying a path to where the libraries are to be installed as follows:
./bootstrap.sh --prefix=/opt/local/ --with-toolset=gcc

• Run the b2 script as follows:
./b2 install

## Installing

If all of the prerequisites are installed properly, one only needs to download the whole repository on his local machine.

The package includes:

• pyDecGMCA: main DecGMCA package
• mathTools.py: some useful mathematical operations
• pyUtils.py: useful routines such as two stages of update
• algoDecG.py: python DecGMCA algorithm
• pyProx.py: python proximal algorithms (used for the refinement step)
• boost_Prox.py: partially accelerated proximal algorithms
• boost_algoDecG.py: accelerated DecGMCA algorithm
• boost_utils.py: accelerated routines such as two stages of update. One needs to compile DecGMCA_utils previously.
• pyWavelet: python wavelet tools
• waveTools.py: some operations for wavelet coefficients
• wav1d.py: 1D wavelet
• wav2d.py: 2D wavelet
• starlet_utils.py: accelerated starlet transform. One needs to compile pystarlet previously.
• simulationsTools: used to generate sources for tests
• MakeExperiment.py
• simu_CS_deconv: used for running tests
• param.py: global parameters for tests
• test_CS.py: test script for compressed sensing test
• test_deconv.py: test script for deconvolution test
• evaluation: used to evaluate results, such as criteria A and S.
• evaluation.py

The following packages are needed to interface python with C++

• DecGMCA_utils
• pystarlet

Instructions for compilation (e.g. DecGMCA_utils):

• Inside the DecGMCA_utils directory create a build dossier
mkdir build

• Inside the build directory and run
cmake ..
make

Then a shared object decG.so is created. Please move this object to the directory pyDecGMCA. The same compilation goes with pystarlet as well. After the compilation, a shared object starlet.so should be moved to the directory pyWavelet.

## Execution

Given correct parameters, one only needs to run the function DecGMCA located in pyDecGMCA.

### Example

Assume noised multichannel 1D data (in Fourier space, Fourier plane option (FTPlane) should be set True) is V_N, linear operator is Mask which downsamples the data points, the number of sources is 5, the size of each source is 1 by 4096. The number of iterations is set 500, the initial epsilon for regularization is 10^{-1} and the final epsilon is 10^{-4}, the wavelet option is True with starlet wavelet and 4 scales of decomposition, the thresholding strategy is 2 with threshold 3. To avoid that the low frequency data affect the sources separation quality, a high-pass filter is applied before the update of mixing matrix. The cut-off frequency of this high-pass filter is set 1./16 without logistic smoothness (False). Both of the positivity constraints are False. The refinement step (postProc) is True with max iterations 50 and parameter (Ksig) 3.

(S_est,A_est) = DecGMCA(V_N,Mask,5,1,4096,500,1e-1,1e-4,Ndim=1,wavelet=True,scale=4,mask=True,deconv=False,wname='starlet',thresStrtg=2,FTPlane=True,fc=1./16,logistic=False,postProc=True,postProcImax=50,Kend=3.0,Ksig=3.0,positivityS=False,positivityA=False)

### Run simulations

This package consists of reproducible simulations presented in the paper. One needs to enter the directory simu_CS_deconv and run corresponding simulation scripts.

#### Multichannel compressed sensing and blind source separation simulation

This simulation corresponds to the case where H is a downsampling matrix.

In the paper, we studied the performance of DecGMCA in terms of the sub-sampling effect, the number of sources, the SNR. One can change these parameters in the file param.py. To run the simulation, only need to run the script test_CS.py.

• Remarks of parameters (in param.py):

• pcArr: array of different ratios of present data (used for compressed sensing simulation)
• ratioArr: array of different resolution ratios (used for deconvolution simulation)
• nArr: array of number of sources
• dbArr: array of different SNRs
• bdArr: array of number of bands (channels)
• numTests: number of Monte-Carlo tests
• Comparison with other methods The DecGMCA method is compared with other methods:

• Matrix completion + BSS (GMCA): controled by the option MC_GMCA_flag in the script.

The results (mixing matrix A and source matrix S) of the simulation will be all saved in the same directory. In order to evaluate the quality of the results, one needs to go to the directory evaluation and run the script script_CS.py.

#### Multichannel deconvolution and blind source separation simulation

This simulation corresponds to the case where H is an ill-conditioned linear kernel (e.g. PSF).

The design of this simulation has the same structure as the above one. In this part, we studied the performance of DecGMCA in terms of the resolution ratio (corresponding to the condition number of H), the number of sources, the SNR. Similarly, one can play with these parameters in the same file param.py. To run the simulation, only need to run the script test_deconv.py.

• Comparison with other methods The DecGMCA method is compared with other methods:
• BSS alone (GMCA): controled by the option GMCA_flag in the script.
• Sequential deconvolution (ForWaRD) and BSS (GMCA): controled by the option ForWaRD_GMCA_flag in the script.

The results (mixing matrix A and source matrix S) of the simulation will be all saved in the same directory. In order to evaluate the quality of the results, one needs to go to the directory evaluation and run the script script_kernel.py.

## Acknowledgments

This work is supported by the CEA DRF impulsion project COSMIC and the European Community through the grants PHySIS (contract no. 60174), DEDALE (contract no. 665044) and LENA (contract no. 678282) within the H2020 Framework Programe.

Joint Multichannel Deconvolution and Blind Source Separation algorithm

## Releases

No releases published

## Packages 0

No packages published


text:
# Can pair production happen in vacuum?

1. Apr 19, 2005

### skpang82

Hi.

Can pair production happen in vacuum?

I had this question for a test. And the answer is: no, it cannot because it would violate the conservation of momentum, since you can't have a momentum zero photon.

But i seem to recall that pair production happen in space (?). Due to Heisenberg's uncertainty relation, there is a constant energy flux, which allows for pair production. The pairs are produced, and they annihilate each other again within a certain time frame. And if it happens near a black hole, one of the pair may pass beyond the event horizon of the black hole, thus making it appear as if the black hole is radiating. I think i got this info quite some time ago when i was reading on Stephen Hawking. Not sure if i remember it correctly.

2. Apr 19, 2005

### marlon

The answer is YES. Indeed conservation of both energy and momentum will be violated but during some short period of time, that is actually allowed by the HUP. Ofcourse we are talking about virtual particles here...Check out :
https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=2 [Broken]

Scroll down to the 'what are virtual particles' entry

marlon

Last edited by a moderator: May 2, 2017
3. Apr 19, 2005

### dextercioby

Why are you treating differently "space" and "vacuum"...?In QFT,"vacuum" has another description,at least in the interacting theories...QED is a good example.

Free field theories have the same notion of vacuum just like the "space" in SR and Class.Field Theories.The fields are quantized,yet the "vacuum state" is seen as the state of zero particles and defined through

$$\hat{a}|0\rangle=|0\rangle$$

,where $$\hat{a}$$ is a generic notation for an annihilation operator...

In interacting field theories,the notion of vacuum is different.It can accomodate particles.

Daniel.

4. Apr 19, 2005

### ZapperZ

Staff Emeritus
You are confusing between "pair production" and "virtual particles".

Pair production produces a particle-antiparticle pair from high energy photons. These particles are REAL, they can interact, and they don't need to die before the rest of the universe detect them. Their lifetimes are not limited by the HUP. And yes, they need to obey both conservation of momentum and energy.

Virtual particles, on the other hand, can appear out of nowhere. When this happens, they temporarily violate conservation of energy (or energy+mass, to be exact), but not conservation of momentum (they exist only via an interaction). However, their lifetimes and other properties are confined within what are allowed by the HUP.

Do not confuse the two...

Zz.

5. Apr 19, 2005

### EL

Must say I'm a little sceptic when it comes to speaking about virtual particles as if they truely exist (which I'm not sure if you are doing or not). If something only can exist virtually, how can we say it is there?

6. Apr 19, 2005

### marlon

I suggest you look up the Casimir force or Casimir effect...Try Wikipedia.

ps : a general remark : i have been reading that only energy conservation is violated when virtual particles arise. This is only partially correct : also momentum conservation can be violated during the existence (ie the lifetime) of virtual particles. A particle can emit a virtual particle that is more heavy then the actual emitting particle. In calculating the propagator's amplitude one integrates over all possible internal momenta. Ofcourse eventually momentum conservation must be respected between the initial and final state. In interactions, the virtual particles will carry a definite momentum-value (so they are everywhere because of the HUP). One value will respect the momentum conservation law, however other values are also present. Besides, in QFT one can prove that the amplitude of the propagator lowers if virtual particles are more and more in violation with energy conservation laws.Besides it is also wrong to claim that virtual particles exist ONLY because of some interaction. This is certainly not enough. Other then that , they exist because of the non-zero-groundstate value (non-zero zero point energy) and MAINLY because of the HUP. However, we need to make sure that we are not talking about semantics here. It is only when you compare the very initial state with the very final state that both energy and momentum MUST be conserved...

regards
marlon

Last edited: Apr 19, 2005
7. Apr 19, 2005

### EL

Dear Marlon. I'm well aware of the Casimir effect. What I wanted to discuss was mearly the semantics.
I mean, virtual particles are not what we usually call particles, but only abstract interpretations of mathematical expressions arising in QFT.
Therefore, if someone asks wheter som "particles" are created, it sounds a little strange to answer: YES, "virtual particles". But of course, as long as everyone knows what it ment by a virtual particle it is ok. However it may be confusing for anyone not so familiar with it.

8. Apr 19, 2005

### marlon

Ahh so you are not argiung that virtual particles are indeed REAL...

Sorry but then i misinterpreted your words...again apologies

regards
marlon

9. Apr 19, 2005

### ZapperZ

Staff Emeritus
Here's the deal...

If virtual particles picture is valid, then there are a number of consequences. One of them being higher order interactions such as those we obtain when we do perturbation expansion. We see this often via the self-energy interactions such as in QED.

Guess what? This is how QED can make the most accurate prediction of the electron gyromagnetic, something that classical field theory could not. Higher order corrections based on quantum field theory picture produce a more accurate value of a physical quantity that we measure.

Now we can argue back and forth on the philosophical definition if something is "real". However as an experimentalist, if I see a theoretical idea in which the consequences of that idea actually predicts with very high accuracy of a measured quantity, I'd say there's something to that theory. Such a description is better and more accurate. In the end, that's as best as we can say.

Zz.

10. Apr 20, 2005

### EL

Just to make myself more clear. I'm really quite familiar with QED, and I'm absolutely NOT questioning the theory.
What I'm questioning is the use of the term "virtual particles" as if they really are "particles". We just choose to call parts of the mathematical expressions arising in QED for "virtual particles" since it makes it easier for us to keep track on what the end results, i.e. what can be measured, will be.

11. Apr 20, 2005

### dextercioby

Okay,let us call them "virtual particles" and u call'em "internal lines"...

Daniel.

12. Apr 20, 2005

### EL

I call the "internal lines" for "virtual particles" all the time. It's not about that. Just wanted to cleraify for "skpang82", who is maybe not so familiar with the concept.
I know you guys are quite fast to debunk (and you do it really well!), but in this case you have misinterpreted my intentions.

Last edited: Apr 20, 2005
13. Apr 20, 2005

### ZapperZ

Staff Emeritus
But think of it this way - if we consider them as real particles, we would not have attached the word "virtual" to them. We'd just call them.... er..... PARTICLES!

The fact that we don't, and call it by an additional qualifier (virtual), implies that they have additional limitations to them. We can go on with this forever and include other "particles" such as phonons, magnons, spinons, holons, polaritons, etc. etc.... Some of these do not even have the world virtual attached to them, yet, we still loosely call them "particles" in the sense that we treat them the same way as any exchange particle in QFT.

Zz.

14. Apr 20, 2005

### EL

Yes, yes, yes, yes, yes. I agree, I understand.
You are reading something between my lines which was not ment to be there.

15. Apr 22, 2005

### Sterj

I read this thread and it seems to be interesting (not the part of pair creaton where all particles are real but the part with virtual particles).

I think at something: Consider a system that is in vacuum state. The system contains a nuclei and a electromagnetic field in ground state (actually one oscillation).

To get an electron (that is in a negative energy state) go in a positive energy state we need a ground state oscillation as follows (w):
w*h/(4pi)=2*m(electron)*c*c. So this oscillation needs many energy. If the oscillation "decides " to create an e-e+ pair it needs that energy.

right?


text:
# cokernel and kernel of adjoint operator

Let $L$ be a linear operator from Banach space $X$ to $Y$. Is the dimension of the kernel of the adjoint of $L$ the same as the dimension of the cokernel? The cokernel is $Y/(Im L)$.

Also, is the index of operators for which this quantity is defined on, same some subspace of the bounded linear operators from $X$ to $Y$, continuous on this set? This is the case for Fredholm operators, but I was wondering if it was true in more general settings.

-
In the "Encyclopedic Dictionary of Mathematics" it says that this is true if the range is closed and the domain is dense. But it says this under the heading of Fredholm operators, so I don't know if this is true without the Fredholm hypothesis. –  Q-the-curvature Nov 17 '11 at 3:27
So, you want unbounded operators? You should say so, there's nothing in your question that hints at that. Be that as it may, it is completely wrong if the range isn't closed, even for bounded operators: In fact for bounded operators you have $\ker{L^\ast} = (\operatorname{im}{L})^\perp$, so in particular if the range is dense you have that $\ker{L^\ast} = 0$. If the range is dense the image must have infinite co-dimension, so the cokernel of $L$ is infinite-dimensional. On the other hand, the best you can hope for in the second question is upper semi-continuity: perturb the identity with ... –  t.b. Nov 17 '11 at 5:36
... the orthogonal projection $P$ onto a subspace of a Hilbert space, so consider $I + tP$. For real $t \neq -1$ this operator is self-adjoint and invertible, so has zero cokernel, while for $t = -1$ you get a cokernel isomorphic to the range of $P$ which can be of arbitrary dimension. –  t.b. Nov 17 '11 at 5:38


text:
# Moving Walkways

The speed of a moving walkway is
typically about 2.5 feet per second. Walking on such a
moving walkway, it takes Karen a total of 40 seconds to
travel 50 feet with the movement of the walkway and then
back again against the movement of the walkway.What is
Karen’s normal walking speed?

Tags:

### I think this problem is just

I think this problem is just similar to boat and river, that boat going upstream and boat going downstream problems. When Karen walks with the movement of the walkway, it is similar to boat going downstream and with Karen in the direction against the walkway, it is similar to boat going upstream.

Fundamentally, this problem is about motion with constant speed with formula $s = vt$. For speed $v$, use the net speed. For more details details, check the motion in a current of water or air.

Let $x$ = Karen’s normal walking speed
$t_\text{50' with the walkway} + t_\text{50' against the walkway} = 40 \text{ sec}$

$\dfrac{50}{x + 2.5} + \dfrac{50}{x - 2.5} = 40$

$x = 4.045 \text{ ft/sec}$

Important:
Tell Karen not to walk against the walkway. For her safety and for the safety of other people using the walkway, she must alight first then look for another walkway in opposite direction for her to go back. :)

• Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and $...$ for displayed mathematics, and $...$ and $...$ for in-line mathematics.


text:
Chi-squared test

Jump to: navigation, search

A chi-squared test, also referred to as chi-square test or $χw²$ test, is any statistical hypothesis test in which the sampling distribution of the test statistic is a chi-squared distribution when the null hypothesis is true. Also considered a chi-squared test is a test in which this is asymptotically true, meaning that the sampling distribution (if the null hypothesis is true) can be made to approximate a chi-squared distribution as closely as desired by making the sample size large enough.

Some examples of chi-squared tests where the chi-squared distribution is only approximately valid:

One case where the distribution of the test statistic is an exact chi-squared distribution is the test that the variance of a normally distributed population has a given value based on a sample variance. Such a test is uncommon in practice because values of variances to test against are seldom known exactly.

Chi-squared test for variance in a normal population

If a sample of size n is taken from a population having a normal distribution, then there is a result (see distribution of the sample variance) which allows a test to be made of whether the variance of the population has a pre-determined value. For example, a manufacturing process might have been in stable condition for a long period, allowing a value for the variance to be determined essentially without error. Suppose that a variant of the process is being tested, giving rise to a small sample of n product items whose variation is to be tested. The test statistic T in this instance could be set to be the sum of squares about the sample mean, divided by the nominal value for the variance (i.e. the value to be tested as holding). Then T has a chi-squared distribution with n − 1 degrees of freedom. For example if the sample size is 21, the acceptance region for T for a significance level of 5% is the interval 9.59 to 34.17.

References

• Corder, G.W., Foreman, D.I. (2009). Nonparametric Statistics for Non-Statisticians: A Step-by-Step Approach Wiley, ISBN 978-0-470-45461-9
• Greenwood, P.E., Nikulin, M.S. (1996) A guide to chi-squared testing. Wiley, New York. ISBN 0-471-55779-X
• Nikulin, M.S. (1973). "Chi-squared test for normality". In: Proceedings of the International Vilnius Conference on Probability Theory and Mathematical Statistics, v.2, pp. 119–122.
• Bagdonavicius, V., Nikulin, M.S. (2011) "Chi-square goodness-of-fit test for right censored data". The International Journal of Applied Mathematics and Statistics, p. 30-50.[full citation needed]


text:
Computing the index in a structured way

I want to map the various combinations to an unique index: For a given $n$ and $r$, we would have $\binom{n}{r}$ arrangement for values:$[0,\dots,n)$:

Ex: For n = 6, r = 3
[012,
013,
014,
015,
...,
345]
So, for any given arrangement [012] the index would 0 as its the very first element.
Similarly, the index for arrangement for [015] should be 3. Trying to come up a formula to determine the correct index for a given arrangement.

• – D.W. Feb 12 '16 at 18:22
• @D.W. Now the question has become unanswerable. Why not convert your comment to an answer? – Yuval Filmus Feb 12 '16 at 18:49
• @YuvalFilmus, I don't understand why the question has become unanswerable. I don't have time right now to write up a full answer, but anyone else who would like to is welcome to do so. – D.W. Feb 12 '16 at 23:41


text:
(G,X)-manifold

In mathematics, the concept of a (G, X)-manifold generalizes several different kinds of manifolds, including Riemannian manifolds, affine manifolds, piecewise linear manifolds, etc.

Definition

Let G be a group acting on a manifold X via diffeomorphisms—i.e. for each $g\in G$, the map $x\mapsto gx$ from X to itself is a diffeomorphism. A manifold M which satisfies the following conditions is called a (G, X)-manifold:[1]

1. There exists an open cover $\{U_\alpha\}$ of M and a family $\{\varphi_\alpha\colon U_\alpha \to V_\alpha\}$ of diffeomorphisms taking each Uα to an open subset Vα of X.
2. For each Uα, Uβ with nonempty intersection, there exists a $g\in G$ such that $gx=\phi_\alpha\circ\phi_\beta^{-1}(x)$ for all $x\in V_\alpha\cap V_\beta$. In other words, viewing the elements of G as diffeomorphisms, each transition map $\phi_\alpha\circ\phi_\beta^{-1}\colon V_\alpha\cap V_\beta\to V_\alpha\cap V_\beta$ is the restriction of an element of G to $V_\alpha\cap V_\beta$.

Citations

1. ^ Thurston 2002, p. 27


text:
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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJan 31st 2012
• (edited Jan 31st 2012)

just for completeness, I have created an entry almost connected topological group.

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeFeb 2nd 2012
• (edited Feb 2nd 2012)

It says the “quotient topological space $G/G_0$” and it means the quotient topological group. I know, you wanted to say the underlying space of the quotient topological group, but if one talks about quotient topological space then this means that only $G_0$ is contracted to a point, while the rest stays the same. In other words, the notation $G/G_0$ in topological category is much bigger than the underlying space of $G/G_0$ in the category of topological groups. The quotient in the category of topological spaces is different than the underlying topological space of a quotient in the category of topological groups. I corrected the statement.

But a real question is if the entry assumes Hausdorfness. I wonder how one can have compact quotient of topological groups by a connected component if not finite, as if something is an accumulation point, by the homogeneity, everything would be an accumulation point. Example ?

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 2nd 2012

Zoran, as an example, consider the $p$-adic integers under addition. This is totally disconnected, so the connected component $G_0$ is just the identity element. Therefore $G/G_0 \cong G$, and $G$ is compact Hausdorff, so this is an example.

• CommentRowNumber4.
• CommentAuthorzskoda
• CommentTimeFeb 2nd 2012

Thanks.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2012

This example is mentioned already at maximal compact subgroup. But it would be nice if somebody found the time to add it also to almost connected topological group, highlighting the point just discussed.

(I guess I could do it later. But I won’t protest if somebody does it before me ;-)

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 2nd 2012

The example if $p$-adic integers isn’t in either article. And I can’t decide where to put it in either article.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2012
• (edited Feb 2nd 2012)

And I can’t decide where to put it in either article.

Maybe we should splitt off an entry p-adic integer from p-adic number anyway. The non almost-connectedness would naturally be discussed there and then we could point to that from the other two entries.


text:
# blog

Vanessa McHale's blog (3)
• ## Names in Haskell Compilers

One common oversight in Haskell compilers is failing to intern identifiers using Ints and failing to prefer IntMaps and IntSets. The PureScript compiler, for instance, uses Maps as of writing.

• ## Finding Cruft in Patches With Jacinda

Suppose we want to clean up all TODOs in a branch before merging. We can check for TODOs introduced with

• ## Generating Other-extensions for Cabal Files

One would like to be able to find all language extensions in a given Haskell project in order to populate the other-extensions field of the .cabal file (cabal-install uses this for dependency resolution).

• ## Finding Haskell Libraries Used in an Executable

We can find all (Haskell) libraries linked against the final executable with:

• ## Transparent Programming

J and APL support (and encourage) a certain form of programming without error handling or library code reuse. The alternative wisdom goes against typical programming but it works together.

|


text:
Pełnotekstowe zasoby PLDML oraz innych baz dziedzinowych są już dostępne w nowej Bibliotece Nauki.
Zapraszamy na https://bibliotekanauki.pl

PL EN

Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
• # Artykuł - szczegóły

## Studia Mathematica

2014 | 220 | 2 | 119-140

## A new characteristic property of Mittag-Leffler functions and fractional cosine functions

EN

### Abstrakty

EN
A new characteristic property of the Mittag-Leffler function $E_{α}(at^{α})$ with 1 < α < 2 is deduced. Motivated by this property, a new notion, named α-order cosine function, is developed. It is proved that an α-order cosine function is associated with a solution operator of an α-order abstract Cauchy problem. Consequently, an α-order abstract Cauchy problem is well-posed if and only if its coefficient operator generates a unique α-order cosine function.

119-140

wydano
2014

### Twórcy

autor
• School of Mathematics and Statistics, Xi'an Jiaotong University, Xi'an 710049, China
autor
• School of Mathematics and Statistics, Xi'an Jiaotong University, Xi'an 710049, China
autor
• School of Mathematics and Statistics, Xi'an Jiaotong University, Xi'an 710049, China


text:
Charge conjugation matrix in baryon current

In his paper Calculation of baryon masses in quantum chromodynamics (ScienceDirect), B.L. Ioffe considers currents describing baryons. In equation (13) he gives an interpolating current for the isobar $\Delta^{++}$, $$\eta_\mu (x) = \left( u^a(x) C \gamma_\mu u^b(x) \right) u^c(x) \varepsilon^{abc},$$ where $u^a(x)$ is an up quark field of colour $a$ and $C$ is the "charge conjugation matrix". This current has the proper isospin of $3/2$ (three up quarks) and has the correct spin (carries one Lorentz index and implicitly one spinor index at $u^c$).

How does $C$ come into this equation though? If I understand the expression correctly, $u^a(x)$ is implicitly transposed so that the expression in the big parentheses is just a number in spinor space. So I do not think that $u^a(x) C$ is some other way of writing $\bar u^a(x)$ – in which case the current would consist of two quarks and one antiquark. What is $C$ doing then?

Furthermore, how is this $C$ related to the charge conjugation matrix (say $\tilde C$ to distinguish) introduced e.g. in Peskin/Schroeder? $\tilde C$ exchanges particles with antiparticles by $\tilde C a_{\mathbf p}^s \tilde C = b_{\mathbf p}^s$ and $\tilde C b_{\mathbf p}^s \tilde C = a_{\mathbf p}^s$ resulting in e.g. $$\tilde C \psi \tilde C = - i \left(\bar\psi \gamma^0\gamma^2\right)^T$$ and seems to always appear in pairs.