text:
# How can KCl be enriched in 40K at home?

Tags:
1. Feb 11, 2019

### ORF

Hello,

I have pure KCl, but its natural radioactivity is low. I would like to increase it at least by a factor of 10. Is there any simple way (diy) for enriching the natural KCl in 40K?

Alternatively, is there a compound with higher concentration of potassium (in volume) than KCl, and that you can have at home?

Thank you for your time.

Regards,
Alvaro.

2. Feb 11, 2019

### phyzguy

I think the answer is no to both questions. What exactly are you trying to do? Radioactive materials are strictly regulated, for good reasons.

3. Feb 11, 2019

### .Scott

I am not sure what the rules and regulations are.
Here is a site which will could sell you the material - and perhaps advise you on what would be required for the handling, reporting, and other regulatory requirements:

Realistically, the only way you're going to get a 10-fold increase in the K40 concentration is to have someone do it for you.

4. Feb 11, 2019

### .Scott

A quick look at the hazard you're considering:

If you procured 20 grams of KCl with K40 enriched 10-fold, it would include about 9 grams of Potassium. It would also include about 610 Becquerels of K40 - thus about 610 radioactive decays per second.

Of course, ingesting 20 grams of KCl in a shot presents some risk just because of the chemistry.
But perhaps the best way to assess the radiation health risk it to determine its "Banana Equivalent Dose": https://en.wikipedia.org/wiki/Banana_equivalent_dose

The BED for 9 grams of Potassium is: 9000mg/422mg = 21 BEDs.
2.1 microsieverts of exposure
roughly 21% of the typical daily radiation exposure.
committed dose equivalent over 50 years: 1.7 microsieverts

In contrast, a dental xray will expose you to between 5 and 50 microsieverts.

5. Feb 11, 2019

Staff Emeritus
40K is not that radioactive. Nevertheless, if you try and buy some, you're going to have to deal with all the regulations.

I don't think you can do this with chemistry. 40K is chemically identical to other isotopes, and even if there were a process, you would need to separate it from both 39K, which is lighter, and 41K, which is heavier.

If you buy it, I expect a factor of 10 enrichment would cost \$5,000-10,000 per gram.

6. Feb 11, 2019

### Ygggdrasil

If you do find a DIY method for isotope enrichment at home, I'm sure there are plenty of people in Iran and other countries who would be interested in hearing about it.

7. Feb 11, 2019

### TeethWhitener

In principle, you could separate out the isotopes by their reaction rates in some appropriately chosen reaction:
https://en.m.wikipedia.org/wiki/Kinetic_isotope_effect

In practice, this will be very difficult for potassium, for two reasons:
1) The kinetic isotope effect tends to be a function of the mass ratio between isotopes, as it is fundamentally a result of different zero point vibrational energies. So it works well for separating hydrogen/deuterium, but not so much for heavier atoms.

2) Potassium compounds tend to be either really unreactive (like KCl) or really reactive (like potassium cyclopentadienide). I suppose you may be able to figure out some really clever salt metathesis, but it would probably be really inefficient.

8. Feb 11, 2019

### TeethWhitener

To answer your second question, KF has a slightly higher volume fraction of potassium than KCl. It’s not particularly safe, though. KH has a much higher volume fraction of potassium, but it’s even less safe. At that point, you might as well be working with potassium metal directly.

9. Feb 12, 2019

### DrDu

I wonder whether some enrichment can be obtained by ion exchange chromatography.

10. Feb 12, 2019

### TeethWhitener

In principle, as long as there is a contribution from interatomic vibrations and the zero point energy differences of the reactant states are different from the zero point energy differences of the transition state resonances, any chemical interaction would work. For instance, pure D2O boils at a slightly higher temperature than pure H2O, and in fact, distillation is used industrially to enrich heavy water.

Would it work for potassium? I doubt it would be very efficient, based on reason #1 from post #7.

11. Feb 12, 2019

### DrDu

12. Feb 12, 2019

Staff Emeritus
Again, you can't do this with chemistry. If you use any of the various techniques to favor heavy nuclei (the bulk of potassium is 39K), you will build up the 40K component, but you will also build up the 41K component even faster. The best you can possibly do is a factor of 2.

This assumes your reagents are isotopically pure. If you are using anything water-based, you will quickly discover the process - whatever it is - is doing a much better job of getting the deuterium out of the water (and much, much better at getting the 18O out of the water) than the 40K out of the potassium.

Of course, you can always use a mass spec. That's essentially how they make it commercially. A good rule of thumb is that the cheapest isotopes you can get by this process run a few thousand per gram.

13. Feb 12, 2019

### TeethWhitener

There might be a clever way to integrate or hybridize this with a recycling HPLC to get a functionally unlimited column length. We used to use one to separate out C60 from He@C60. It takes forever, wastes a bunch of solvent, and is extremely energy intensive and inefficient. Existing isotope separation methods are probably cheaper. Nonetheless, I don't think it's out of the realm of possibility. But definitely not a DIY at home project.

Or with a factor of 1.007 enrichment on each pass, all you'd need to do is run the column $\frac{\ln 10}{\ln 1.007} \approx 330$ times to get a factor of 10 enrichment.

Edit: just looked at the paper: it's looking specifically at 41K/39K ratios. It doesn't say what the 40K enrichment is. So just run the column a few more thousand times and take the middle fraction.

14. Feb 12, 2019

### phyzguy

How could this work, given Vanadium50's points in post #12?

15. Feb 12, 2019

### TeethWhitener

The point is that the three isotopes all have different retention times (different partition coefficients, if you want to think of it that way). So given a long enough column, the three isotopes would separate into different fractions.

Edit: just so we're clear. It's completely impractical. The effective column length would probably have to be hundreds of kilometers. Even to separate out 41K would take a column length of 330×4 = 1320 meters.

16. Feb 12, 2019

### DrDu

most methods for isotopic enrichment are either impractical or absurdly energy intensive.

17. Feb 12, 2019

Staff Emeritus
If you need something isotopically pure, it's best if you can use an element that is already a single isotope. Thta's why the LHC's lead is much more expensive than RHIC's gold: gold is isotopically pure when you dig it out of the ground. That's why aluminum good, cobalt good, but tin? Very very bad.

18. Feb 12, 2019

Staff Emeritus
I have been so focused on the potassium, I forgot about the chlorine.

You can't use KCl. There's a thousand times as much 37Cl as 40K. You'll need another compound or you will find yourself separating the chlorine and not the potassium.

19. Feb 12, 2019

### TeethWhitener

The relevant interaction in ion exchange chromatography is electrostatic, where the analyte of interest and the stationary phase have opposite charges. So in the case of (e.g.) cation exchange chromatography, separation only happens between positively charged species:
https://en.m.wikipedia.org/wiki/Ion_chromatography


text:
Solution to Dummy Packet Assembler (Task from r/dailyprogrammer)

I tried to solve the Packet Assembler task posted at r/dailyprogrammer.

The Question is this:

When a message is transmitted over the internet, it is split into multiple packets, each packet is transferred individually, and the packets are reassembled into the original message by the receiver. Because the internet exists in the real world, and because the real world can be messy, packets do not always arrive in the order in which they are sent. For today's challenge, your program must collect packets from stdin, assemble them in the correct order, and print the completed messages to stdout.

The point of reading from stdin is to simulate incoming packets. For the purposes of this challenge, assume there is a potentially unlimited number of packets. Your program should not depend on knowing how many packets there are in total. Simply sorting the input in its entirety would technically work, but defeats the purpose of this exercise.

Input Format is this:

Each line of input represents a single packet. Each line will be formatted as X Y Z some_text, where X Y and Z are positive integer and some_text is an arbitrary string. X represents the message ID (ie which message this packet is a part of). Y represents the packet ID (ie the index of this packet in the message) (packets are zero-indexed, so the first packet in a message will have Y=0, the last packet in a message will have Y=Z-1). Z represents the total number of packets in the message.

Here is the Python3 solution I did

from re import split

packets[p_id] = {}
packets[p_id]["len"] = len
packets[p_id]["packets"] = {}

return packets

def add_packet(packets, p_id, p_num, p_len, msg):
if p_id not in packets.keys():

if p_num > packets[p_id]["len"]:
return packets

packets[p_id]["packets"][p_num] = msg

return packets

def main():
packets = {}
while True:
try:
recv = split(r'\s+', input())

int(recv[2]), " ".join(recv[3:]))
except EOFError:
break

for p_id in sorted(packets):
for packet in range(packets[p_id]["len"]):
print("%d %d %d %s" %(p_id, packet, packets[p_id]["len"], packets[p_id]["packets"][packet]))

if __name__ == '__main__':
main()

Please rate this solution and suggest some improvements that can be made?

• Rate this solution is rather subjective (reviewers don't rate code on this site), but you'll definitely receive feedback and improvements on any/all aspects of the code. – Mathieu Guindon Oct 13 '17 at 13:38
• @Mat'sMug I mean just reviewing it. – Jøê Grèéñ Oct 19 '17 at 14:47

I think this code is very straightforward and easy to read. Good job!

Some nitpicks are,

You have a few PEP8 violations none all that concerning.

1. Should add a second white line between imports and your functions
2. Line too long, at your print statement
3. Continuation line under-indented for visual indent
4. printf style formatting is recommended against

Code suggestions

I would use a namedtuple instead of a dictionary, since it is much cleaner looking and easier to use. When to use a namedtuple!

Using the above suggestions and adhering PEP8 I get the following.

from re import split
from collections import namedtuple

def get_messages():
Message = namedtuple('Message', 'idx num length message')
messages = []

while True:
try:
recv = split(r'\s+', input())
messages.append(Message(recv[0], recv[1], recv[2], ' '.join(recv[3:])))
except EOFError:
return messages

def print_messages(messages):
for message in messages:
print("{}  {} {} {}"
.format(message.idx, message.num, message.length, message.message))

def main():
messages = sorted(get_messages(), key=lambda x: (x.idx, x.num))
print_messages(messages)

if __name__ == '__main__':
main()

• Thanks! TBH i am not too familiar with NamedTuples, that's why I don't use them! Thanks for PEP8s too! – Jøê Grèéñ Oct 19 '17 at 14:52

format: plain


text:
# 6.3 Geometry  (Page 5/7)

 Page 5 / 7
$\begin{array}{ccc}\hfill PS& =& \sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}\hfill \\ & =& \sqrt{{\left(0-2\right)}^{2}+{\left(-0.5-1\right)}^{2}}\hfill \\ & =& \sqrt{{\left(-2\right)}^{2}+{\left(-1.5\right)}^{2}}\hfill \\ & =& \sqrt{4+2.25}\hfill \\ & =& \sqrt{6.25}\hfill \end{array}$

and

$\begin{array}{ccc}\hfill QS& =& \sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}\hfill \\ & =& \sqrt{{\left(0-\left(-2\right)\right)}^{2}+{\left(-0.5-\left(-2\right)\right)}^{2}}\hfill \\ & =& \sqrt{{\left(0+2\right)\right)}^{2}{+\left(-0.5+2\right)\right)}^{2}}\hfill \\ & =& \sqrt{{\left(2\right)\right)}^{2}{+\left(-1.5\right)\right)}^{2}}\hfill \\ & =& \sqrt{4+2.25}\hfill \\ & =& \sqrt{6.25}\hfill \end{array}$

It can be seen that $PS=QS$ as expected.

The following video provides a summary of the midpoint of a line.

## Co-ordinate geometry

1. In the diagram given the vertices of a quadrilateral are F(2;0), G(1;5), H(3;7) and I(7;2).
1. What are the lengths of the opposite sides of FGHI?
2. Are the opposite sides of FGHI parallel?
3. Do the diagonals of FGHI bisect each other?
2. A quadrialteral ABCD with vertices A(3;2), B(1;7), C(4;5) and D(1;3) is given.
2. Find the lengths of the sides of the quadrilateral.
3. ABCD is a quadrilateral with verticies A(0;3), B(4;3), C(5;-1) and D(-1;-1).
1. Show that:
2. AB $\parallel$ DC
2. What name would you give to ABCD?
3. Show that the diagonals AC and BD do not bisect each other.
4. P, Q, R and S are the points (-2;0), (2;3), (5;3), (-3;-3) respectively.
1. Show that:
1. SR = 2PQ
2. SR $\parallel$ PQ
2. Calculate:
1. PS
2. QR
5. EFGH is a parallelogram with verticies E(-1;2), F(-2;-1) and G(2;0). Find the co-ordinates of H by using the fact that the diagonals of a parallelogram bisect each other.

## Transformations

In this section you will learn about how the co-ordinates of a point change when the point is moved horizontally and vertically on the Cartesian plane. You will also learn about what happens to the co-ordinates of a point when it is reflected on the $x$ -axis, $y$ -axis and the line $y=x$ .

## Translation of a point

When something is moved in a straight line, we say that it is translated . What happens to the co-ordinates of a point that is translated horizontally or vertically?

## Discussion : translation of a point vertically

Complete the table, by filling in the co-ordinates of the points shown in the figure.

 Point $x$ co-ordinate $y$ co-ordinate A B C D E F G

What do you notice about the $x$ co-ordinates? What do you notice about the $y$ co-ordinates? What would happen to the co-ordinates of point A, if it was moved to the position of point G?

When a point is moved vertically up or down on the Cartesian plane, the $x$ co-ordinate of the point remains the same, but the $y$ co-ordinate changes by the amount that the point was moved up or down.

For example, in [link] Point A is moved 4 units upwards to the position marked by G. The new $x$ co-ordinate of point A is the same ( $x$ =1), but the new $y$ co-ordinate is shifted in the positive $y$ direction 4 units and becomes $y$ =-2 +4 =2. The new co-ordinates of point A are therefore G(1;2). Similarly, for point B that is moved downwards by 5 units, the $x$ co-ordinate is the same ( $x=-2,5$ ), but the $y$ co-ordinate is shifted in the negative $y$ -direction by 5 units. The new $y$ co-ordinate is therefore $y$ =2,5 -5 =-2,5.

anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers!


text:
×

What is the remainder when 501 * 502 * 503 is divided by 25?

Note by Pramita Kastha
2 years, 1 month ago

Sort by:

$$501 \times502 \times503 = (500+1)(500+2)(500+3) =500(500+2)(500+3) + (500(500+3)) +2(500) + 6$$ .....when dividing by $$25$$ the only term left out is $$6$$ the remaining are evenly divisible by $$25$$ ..therefore the remainder is $$\boxed{6}$$.

- 2 years, 1 month ago

6

- 1 year, 2 months ago

(500+1)X(500+2)X(500+3)=Multiple of 100+6 i.e. last 2 digits are 06. Therefore remainder when divided by 25 is 6.

- 1 year, 8 months ago

6

- 2 years, 1 month ago

6

- 2 years, 1 month ago

6

- 2 years, 1 month ago

Check out Modular Arithmetic - Multiplication.

Staff - 2 years, 1 month ago


text:
# Why does milk clot while making yogurt?

Why does milk clot when it becomes yogurt? Is it only because of the decrease in the $\mathrm{pH}$ of milk? If it does so, what is happening when I increase the $\mathrm{pH}$ level? And by decreasing the $\mathrm{pH}$ level can I clot any type of liquid like milk?

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. – Martin - マーチン May 11 '16 at 8:50
The clotting that you see is caused by denatured proteins. You can denature proteins physically and chemically. You can use acids or bases (chemical). Also, you can heat it up shake the protein (physical). The $\mathrm{pH}$ of yogurt is different from milk because of fermentation. Fermentation causes production of lactic acid in milk. Which is responsible for the change in $\mathrm{pH}$ (yogurt $\mathrm{pH}$ 4.0 vs milk pH 6.5).


text:
Lemma 13.31.3. Let $\mathcal{A}$ be an abelian category. Let $(K, L, M, f, g, h)$ be a distinguished triangle of $K(\mathcal{A})$. If two out of $K$, $L$, $M$ are K-injective complexes, then the third is too.

Proof. Follows from the definition, Lemma 13.4.2, and the fact that $K(\mathcal{A})$ is a triangulated category (Proposition 13.10.3). $\square$

There are also:

• 5 comment(s) on Section 13.31: K-injective complexes

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# How To Calculate Shear Strain: Process,Formula,Example And Exhaustive FACTS

This article explains how to calculate shear strain. Shear strain is the ratio of change in dimensions due to shear force to the original dimensions of the work piece.

The change in dimensions is in the form of angular displacement. The shear force acts parallel to the plane of cross section. The plane shifts by some amount which is the linear displacement. The angle subtended by this displacement gives the angular displacement.

## What is shear strain?

As discussed in above section, shear strain is the ratio of change in dimensions due to shear force to the original dimensions of the work piece. It tells us about the deformation in form of percentage value.

The work piece deforms in the plane parallel to the cross section of the work piece. It appears as if a layer of the work piece is moving.

Mathematically, shear strain can be given as-

where,

gamma is shear strain

L is the original length of the work piece.

Image credits: Bob ClemintimeUniaxial Shear StrainCC BY-SA 4.0

## What is engineering strain?

Engineering strain is a dimensionless number that represents the ratio of deformation of the work piece to its original dimension. It is directionally proportional to the amount of elongation along the length.

Formula given in above section represents engineering strain. When the type of strain is not mentioned, it is understood that engineering strain is being referred.

## What is true strain?

True strain is the natural log of final length of work piece to the original length of the work piece. It shows the instantaneous strain being developed in the work piece.

The strain developed in the work piece at a specific moment is called the true strain. It gives better representation of the material will behave under stress.

Mathematically, true strain can be given as-

∈ = ln x (A0/A1)

where,

eta represents the true strain

## How to calculate shear strain?

Shear strain is the ratio of deformation length of the work piece to the original length of the work piece.

Following steps are followed to calculate shear strain-

• Measure the original length of the work piece.
• Apply shear stress on the material.
• Measure the angle of deformation.
• tan of angle of deformation gives the shear strain.

## How to calculate shear strain rate?

Strain rate can be defined as the rate of change of strain with respect to time. Hence it is a function of time. It is the first differential of shear strain with respect to time.

Shear strain rate can be calculated by the formula given below-

γ = dγ/dt

where the dot on gamma represents shear strain rate

## Torsional stress

A cylindrical bar experiences shear stress on its surface when subjected under torsional stress. Torsional stress is the stress which tends to twist the plane of cylindrical bar by some angle.

Torsional stress can be given by the relation given below. The torsional formula can be used to find angle of twist, shear stress, shear strain torsional moment etc if adequate data is provided.

## How to calculate shear strain in torsion?

The formula discussed in above section can be used to find shear strain when the work piece is subjected to torsional stress.

The formula for shear strain is discussed in above sections.

From the torsional formula, shear strain comes out to be-

γ = Rθ/L

where,

theta is the angle of twist

L is the length of the work piece

## How to calculate shear strain from shear stress?

Shear modulus of rigidity is the ratio of shear stress to shear strain. To find the shear strain from shear stress, we divide shear stress by the shear modulus. We get,

where,

G is the modulus of rigidity

## How to find angle of shear strain?

The angle of shear strain is the angle by which the work piece is deformed.

The angle of shear strain can be found using simple trigonometry. Angle of shear strain is the tan of angle subtended by the displacement, X and the original length, L. The adjacent side of the triangle formed acts as the original length of the work piece and the displacement acts as the opposite side in the triangle.

The angle of shear can be found using the relation given below

where,

Phi is the angle of shear strain

## How to find average shear strain?

The concept of shear strain comes into play when the work piece experiences strain from multiple directions.

Lets consider shear strain occuring at two ends making angle of shear strain alpha and beta respectively.

The average shear strain will become-

γ = α + β

## How to find angular shear strain?

Angular shear strain is discussed in above section. The shear strain that describes the change in angle between two lines which were initially perpendicular is called as angular shear strain.

Mathematically, it can be shown as

γ = tanα

where,

alpha is the angular shear strain

## Shear stress in rectangular beam

A rectangular beam is a beam whose cross section is rectangular or square. The amount of shear stress experienced by different points of cross section are different.

This happens because the distance of the point from the application of force is variable for different points. So different amount of force acting on different points lead to variable shear stress.

The shear stress is maximum at the centroid and minimum at the ends of the cross section. Shear stress is parabolically distributed across the rectangular cross section.

## What is bending stress?

As the name suggests, this stress causes bending in work piece when a load is applied.

Bending stress is the stress which acts normally on the work piece and causes bending along the longitudinal axis of the work piece.

## What is bending stress formula?

Similar to torsional formula. Bending formula also exists that makes it easy for us to find the bending stress. The formula for bending stress is given below

σb = My/I

## What is polar moment of inertia?

Polar moment of inertia or second moment of inertia is a quantity used to explain the resistance to deformation caused by torsional stress.

Mathematically, it can be given as-

where,

J is the polar moment of inertia

## What is section modulus?

Section modulus is used while designing flexural members of a structure or beams. It can be defined as the ratio of moment of inertia of the beam’s cross section about the neutral axis to the distance of extreme fiber from the center.

Mathematically, it can be shown as-

Z = I/ymax

where,

I is the moment of inertia

Z is the section of modulus

Scroll to Top


text:
Opened 7 years ago

Closed 7 years ago

timestamp out of range for platform time_t error with Trac 0.12 (suggested patch is now attached)

Reported by: Owned by: eric.portelance@… Joachim Hoessler normal EstimationToolsPlugin normal scottj@… 0.12

Description

After installing Trac 0.12 we are getting a "timestamp out of range for platform time_t" error.

comment:1 Changed 7 years ago by Odd Simon Simonsen

Traceback, please. Use Trac debug logging to get the full error from the server. Anything else needed to reproduce this (like macro or macro options)?

Changed 7 years ago by Björn Harrtell

use new date format conversion method as req 0.12+

comment:2 Changed 7 years ago by bjorn

I think he is hitting a bug that I think I have fixed in attached patch. The problem is that 0.12 uses new timestamp format:

Not sure if I've messed up UTM zone consideration though.

comment:3 Changed 7 years ago by anonymous

Summary: timestamp out of range for platform time_t error with Trac 0.12 → timestamp out of range for platform time_t error with Trac 0.12 (suggested patch is now attached)

Changed 7 years ago by Odd Simon Simonsen

Updated patch that should support both 0.11 and 0.12.

comment:4 Changed 7 years ago by Odd Simon Simonsen

Thanks for the patch. I've updated it, and made it work across Trac versions to avoid branching just for this. Please try attachment:t7355-to_utimestamp_compat-r5360.diff

Untested on 0.12 by me, so please report back after testing.

comment:5 Changed 7 years ago by Joachim Hoessler

Status: new → assigned

If someone could verify that it works on 0.12, I will apply osimons patch on trunk.

comment:6 Changed 7 years ago by Björn Harrtell

Verified as working on 0.12, nice work making it compatible.

comment:7 Changed 7 years ago by Joachim Hoessler

Resolution: → fixed assigned → closed

(In [8296]) applied patch by bjorn@… and osimons to support trac 0.12. Fixes #7355.

comment:8 Changed 7 years ago by scottj@…

I'm still getting this error with BurndownChart, after pulling from trunk and building r9029. Do I need to do anything special to reconfigure the plugin after installation?

comment:10 follow-up:  11 Changed 6 years ago by anonymous

fast fix:

echo "update wiki set time= substr(time, 0, length(time)-6);" | sqlite3 /var/trac/db/trac.db

comment:11 in reply to:  10 Changed 6 years ago by Odd Simon Simonsen

fast fix:

echo "update wiki set time= substr(time, 0, length(time)-6);" | sqlite3 /var/trac/db/trac.db

This is not a fast fix for anything. It is a super-fast way of making your Trac 0.12+ wiki think it is ~42 years back in time... Don't run it!

Speaking of, how can changing timestamps in wiki table ever affect the ticket timestamps that the EstimationToolsPlugin uses to render ticket graphs and stats???

Modify Ticket

Change Properties


text:
### PRESENT

Presents a list of images to CCDPACK

#### Description:

This routine enters reduction information into the CCDPACK extensions of a list of images. This information is required if an automated reduction schedule is to be produced using SCHEDULE. Before using this routine you should set up the CCDPACK global parameters, describing the CCD characteristics, using the CCDSETUP application.

If the input images have not already been categorised then this routine performs this task for the "frame types" BIAS, TARGET, DARK, FLASH, FLAT, MASTER_BIAS, MASTER_FLAT, MASTER_DARK and MASTER_FLASH (these are input as different groups of images).

Missing exposure times for DARK and FLASH counts can be entered as can filter types.

This routine can also be used to check that a list of images have the minimum amount of information in their CCDPACK extensions to allow an automated scheduling.

#### Usage:

present modify=? simple=? in=? bias=? target=? dark=? flash=? flat=? ftype=? filter=? darktime=? flashtime=?

#### Parameters:

##### ADC = _DOUBLE (Read)
The Analogue-to-Digital conversion factor. CCD readout values are usually given in Analogue-to-Digital Units (ADUs). The ADC factor is the value which converts ADUs back to the number of electrons which were present in each pixel in the CCD after the integration had finished. This value is required to allow proper estimates of the inherent noise associated with each readout value. CCDPACK makes these estimates and stores them in the variance component of the final images. Not supplying a value for this parameter (if prompted) may be a valid response if variances are not to be generated.

This parameter normally accesses the value of the related CCDPACK global association. This behaviour can only be superceded if ADC=value is used on the command-line or if a prompt is forced (using the PROMPT keyword). The value of this parameter will be entered into the extension of the input images only if MODIFY is TRUE or the related extension item does not exist. [!]

##### ADDDARK = _LOGICAL (Read)
Whether or not to prompt for a dark exposure time for the input images which require one. [Dynamic default, TRUE if dark count frames are given, FALSE otherwise]
##### ADDFLASH = _LOGICAL (Read)
Whether or not to prompt for a pre-flash exposure time for the input images which require one. [Dynamic default, TRUE if pre-flash frames are given, FALSE otherwise]
##### BIAS = LITERAL (Read)
A list of the names of the images which contain the raw bias data. These are the images which are to used to produce a "master" bias image. On exit these images will have their FTYPE extension item set to the value "BIAS". [!]
##### BIASVALUE = _DOUBLE (Read)
If no raw bias frames exist and the data does not have any bias strips, then the only way to remove the bias contribution is to subtract a constant. If your data has already had its bias contribution subtracted and you want to process it using CCDPACK (so that you can generate variances for instance) then set this value to zero. This parameter defaults to ! and is not prompted for so the only way that a value can be supplied is on the command-line or by using the PROMPT keyword. [!]
##### BOUNDS( 2 or 4 ) = _INTEGER (Read)
The bounds of the detector bias strips (if any exist). The bounds (if given) should be in pixel indices and be given in pairs up to a limit of 2. The sense of the bounds is along the readout direction. For example, 2,16,400,416 means that the bias strips are located between pixels 2 to 16 and 400 to 416 inclusive along the readout direction. The bias strips are used to either offset the master bias image or as an estimate of the bias which is to be interpolated across the image in some way (see DEBIAS). Not supplying values for this parameter may be a valid response if the bias frame is to be directly subtracted from the data without offsetting or if a single constant is to be used as the bias value for the whole image.

This parameter normally accesses the value of the related CCDPACK global association. This behaviour can only be superceded if BOUNDS=[value,...] is used on the command-line or if a prompt is forced (using the PROMPT keyword). The value of this parameter will be entered into the extension of the input images only if MODIFY is TRUE or the related extension item does not exist. [!]

##### DARK = LITERAL (Read)
A list of the names of the images which contain the raw dark count data. These are the images which are to used to produce a "master" dark counts image. On exit these images will have their FTYPE extension item set to the value "DARK". [!]
##### DARKTIME = _DOUBLE (Read)
The time for which the data in the current image collected dark count electrons. The dark count is basically charge which accumulates in the detector pixels due to thermal noise. The effect of dark current is to produce an additive quantity to the electron count in each pixel. Most modern devices only produce a few ADU (or less) counts per pixel per hour and so this effect can generally be ignored. This, however, is not the case for Infra-Red detectors.

The value given does not need to be a number of seconds or minutes and can be ratio of some kind, as long as it is consistently used for all images (so if all your images have the same darktime then the value 1 could be used). Images which have no dark count should be given a DARKTIME of 0. This parameter is only used if ADDDARK is TRUE. [!]

##### DEFERRED = _DOUBLE (Read)
The deferred charge value. Often known as the "fat" or "skinny" zero (just for confusion). This is actually the charge which is not transferred from a CCD pixel when the device is read out. Usually this is zero or negligible and is only included for completeness and for processing very old data.

This parameter normally accesses the value of the related CCDPACK global association. This behaviour can only be superceded if DEFERRED=value is used on the command-line or if a prompt is forced (using the PROMPT keyword). The value of this parameter will be entered into the extension of the input images only if MODIFY is TRUE or the related extension item does not exist. [!]

##### DIRECTION = LITERAL (Read)
The readout direction of the detector. This may take the values X or Y. A value of X indicates that the readout direction is along the first (horizontal) direction, an Y indicates that the readout direction is along the direction perpendicular to the X axis.

This parameter normally accesses the value of the related CCDPACK global association. This behaviour can only be superceded if DIRECTION=value is used on the command-line or if a prompt is forced (using the PROMPT keyword). The value of this parameter will be entered into the extension of the input images only if MODIFY is TRUE or the related extension item does not exist. [!]

##### EXTENT( 4 ) = _INTEGER (Read)
The extent of the useful detector area in pixel indices. The extent is defined as a range in X values and a range in Y values (XMIN, XMAX, YMIN, YMAX). These define a section of an image (see SUN/33). Any parts of the detector surface area outside of this region will not be present in the final output. This is useful for excluding bias strips, badly vignetted parts etc.

This parameter normally accesses the value of the related CCDPACK global association. This behaviour can only be superceded if EXTENT=[XMIN, XMAX, YMIN, YMAX] is used on the command-line or if a prompt is forced (using the PROMPT keyword). The value of this parameter will be entered into the extension of the input images only if MODIFY is TRUE or the related extension item does not exist. [!]

##### FILTER = LITERAL (Read)
The filter name associated with the current image. The filter name is stored in the extension item FILTER and is used when determining which flatfields should be used for which data. images with a frame type which is independent of the filter will not use this parameter. The filter type is a case sensitive string. [Current value]
##### FLASH = LITERAL (Read)
A list of the names of the images which contain the raw pre-flash correction data. These are the images which are to used to produce a "master" pre-flash correction image. On exit these images will have their FTYPE extension item set to the value "FLASH". [!]
##### FLASHTIME = _DOUBLE (Read)
The time for which the data in the current image was exposed to pre-flash.

The value given does not need to be a number of seconds or minutes and can be ratio of some kind, as long as it is consistently used for all images (so if all your images have the same darktime then the value 1 could be used). Images which have no pre-flash should be given a FLASHTIME of 0. This parameter is only used if ADDFLASH is TRUE. [!]

##### FLAT = LITERAL (Read)
A list of the names of the images which contain the raw flatfield data. These are the images which are to used to produce "master" flatfields (one for each filter type). On exit these images will have their FTYPE extension item set to the value "FLAT". [!]
##### FTYPE = LITERAL (Read)
The "frame" type of the current image. Each image is processed in turn and if SIMPLE is TRUE and a frame type extension item does not exist then this parameter will be used to prompt for a value. A prompt will also be made if SIMPLE is TRUE and MODIFY is TRUE regardless of whether the item already exists or not. If SIMPLE is FALSE then this parameter will not be used. [Current value]
##### IN = LITERAL (Read)
A list of the names of the images which contain the raw CCD data. Images entered using this parameter must already have the correct "frame type" information (extension item FTYPE) entered into their CCDPACK extensions. This parameter is only used if SIMPLE is TRUE.

The image names should be separated by commas and may include wildcards.

##### LOGFILE = FILENAME (Read)
Name of the CCDPACK logfile. If a null (!) value is given for this parameter then no logfile will be written, regardless of the value of the LOGTO parameter.

If the logging system has been initialised using CCDSETUP, then the value specified there will be used. Otherwise, the default is "CCDPACK.LOG". [CCDPACK.LOG]

##### LOGTO = LITERAL (Read)
Every CCDPACK application has the ability to log its output for future reference as well as for display on the terminal. This parameter controls this process, and may be set to any unique abbreviation of the following:
• TERMINAL – Send output to the terminal only

• LOGFILE – Send output to the logfile only (see the LOGFILE parameter)

• BOTH – Send output to both the terminal and the logfile

• NEITHER – Produce no output at all

If the logging system has been initialised using CCDSETUP then the value specified there will be used. Otherwise, the default is "BOTH". [BOTH]

##### MASTERBIAS = LITERAL (Read)
The name of a master bias frame. If this has been created by CCDPACK then there is no need to present it. This parameter is designed for the import of frames created by other packages. [!]
##### MASTERDARK = LITERAL (Read)
The name of a master dark counts frame. If this has been created by CCDPACK then there is no need to present it (unless for some reason it has been assigned the wrong frame type). This parameter is designed for the import of frames created by other packages.
##### MASTERFLASH = LITERAL (Read)
The name of a master pre-flash frame. If this has been created by CCDPACK then there is no need to present it (unless for some reason it has been assigned the wrong frame type). This parameter is designed for the import of frames created by other packages. [!]
##### MASTERFLAT = LITERAL (Read)
The names of a set of master flatfield frames (one for each filter type used). If these have been created by CCDPACK then there is no need to present them (unless for some reason they have been assigned the wrong frame type or filter). This parameter is designed for the import of frames created by other packages (such as those that specifically process spectral data). [!]
##### MASTERS = _LOGICAL (Read)
If this parameter is TRUE then prompts will be made for all the master calibration types (MASTERBIAS, MASTERDARK, MASTERFLAT and MASTERFLASH). [FALSE]
##### MODIFY = _LOGICAL (Read)
If the input images already contain information in their CCDPACK extensions, then this parameter controls whether this information will be overwritten (if a new value exists) or not. [TRUE]
##### MULTIENTRY = _LOGICAL (Read)
Whether or not the names of the input images, their frame types, filters and related exposure factors are all given in response to the IN parameter (SIMPLE must be TRUE). If this option is selected then the parameters FTYPE, FILTER, DARKTIME and FLASHTIME will be set up with these values as defaults. If MODIFY is TRUE then you will be given an opportunity to modify them, otherwise these values will be entered into the image CCDPACK extensions.

The input record format is five fields separated by commas. These are:

• 1 Image name

• 2 Frame type

• 3 Filter name

• 4 Dark exposure time

• 5 Flash exposure time

The latter three fields can be specified as "!" in which case they are not set (they may not be relevant). Multiple records can be entered and can be read in from a text file. So for instance if the file "XREDUCE.NDFS" had the following as its contents:

DATA1,target,!,!,!
DATA2,target,!,!,!
DATA3,target,!,!,!
FF1,flat,!,!,!
FF2,flat,!,!,!
FF3,flat,!,!,!
BIAS1,bias,!,!,!
BIAS2,bias,!,!,!
BIAS3,bias,!,!,!

Then it would be invoked using parameters

• SIMPLE MULTIENTRY IN=^XREDUCE.NDFS

This parameter is intended as an aid when using this program non-interactively (i.e. from scripts) and shouldn’t normally be used, hence its default is FALSE and this can only be overridden by assignment on the command line or in response to a forced prompt. [FALSE]

##### NAMELIST = LITERAL (Read)
The name of a file to contain a listing of the name of the input images. This is intended to be of use when using these same names with other applications (such as SCHEDULE). [!]
##### ONEDARKTIME = _LOGICAL (Read)
If the input data have the same dark count exposure time then this parameter may be set to inhibit repeated prompting for an exposure for every frame. This parameter is of particular use when running from scripts. [FALSE]
##### ONEFILTER = _LOGICAL (Read)
If the input data have only one filter type then this parameter may be set to inhibit repeated prompting for a filter name for every frame (that is filter dependent). This parameter is of particular use when running from scripts. [FALSE]
##### ONEFLASHTIME = _LOGICAL (Read)
If the input data have the same pre-flash exposure time then this parameter may be set to inhibit repeated prompting for an exposure for every frame. This parameter is of particular use when running from scripts. [FALSE]
##### RNOISE = _DOUBLE (Read)
The readout noise of the detector (in ADUs). Usually the readout noise of a detector is estimated by the observatory at which the data was taken and this is the value which should be supplied. Not supplying a value for this parameter may be a valid response if variances are not to be generated.

This parameter normally accesses the value of the related CCDPACK global association (which is the readout noise value). This behaviour can only be superceded if RNOISE=value is used on the command-line or if a prompt is forced (using the PROMPT keyword). The value of this parameter will be entered into the extension of the input images only if MODIFY is TRUE or the related extension item does not exist. [!]

##### SATURATION = _DOUBLE (Read)
The saturation value of the detector pixels (in ADUs).

This parameter normally accesses the value of the related CCDPACK global association. This behaviour can only be superceded if SATURATION=value is used on the command-line or if a prompt is forced (using the PROMPT keyword). The value of this parameter will be entered into the extension of the input images only if MODIFY is TRUE or the related extension item does not exist. [!]

##### SIMPLE = _LOGICAL (Read)
Whether or not the input images already contain "frame type" (extension item FTYPE) information in their CCDPACK extensions or not. Usually images to be presented to CCDPACK do not contain this information, unless it has been imported from FITS information using IMPORT, or the images have already been presented and this pass is to modify existing extension items. [FALSE]
##### TARGET = LITERAL (Read)
A list of the names of the images which contain the "target" data. These are the images which contain the images or spectra etc. On exit these images will have their FTYPE extension item set to the value "TARGET". [!]
##### ZEROED = _LOGICAL (Read)
If a master bias frame is given, then this parameter indicates whether or not it has a mean value of zero. If SIMPLE and MULTIENTRY are TRUE then this value (TRUE or FALSE) can be entered as the fourth field to the IN parameter. [FALSE]

#### Examples:

present simple in=’$\ast$’ modify
In this example PRESENT processes all the images in the current directory. The images should already have a valid frame type (such as TARGET, FLAT etc.). The any existing global variables describing the detector are accessed and written into the image extension overwriting any values which already exist.
present simple=false bias=’bias$\ast$’ target=’data$\ast$’ dark=! flash=! flat=’ff$\ast$’
In this example the input images are organised into their respective frame types using the specially designed input parameters. On exit the output images will have the correct frame types entered into their CCDPACK extensions (provided MODIFY is TRUE).
present modify=false simple=true in=’$\ast$’
In this example all the images in the current directory are accessed. If any required extension or global associated items are missing then they will be entered into the image extension. If all extension items are present then a listing of their values will be made.
present masters simple=false masterflat=2dspectraff
In this example a master flatfield is imported to be used in an automated reduction of spectral data.


text:
How to use Control Chart Constants? | A step by step Guide

# How to use Control Chart Constants? | A step by step Guide

Last updated on 12th Jul 2020, Blog, General

Kailash (Data Entry Executive )

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In statistical process control (SPC) charting, we use the A2 and E2 constants to calculate control limits for an Average (X-bar chart) and Individuals charts. But where do the A2 and E2 constants come from? Let’s look at the following example, for an X-bar chart, that will explain how we derive the A2 constant.

The Generalize Control Limit Equation For Variable Charting

The expression used to compute the control limits for an X-bar chart is:

In this expression parameters μ, σ, and n represent the mean, standard deviation, and sample size. The expression, σ/√n is also called the standard error of the mean.

Control Chart Constants – X-Bar Chart

When using an X-bar Chart we collect several consecutive samples of size, n, to form a homogeneous subgroup and compute a subgroup average. Once we have enough subgroups, say 30 or more, we can estimate the population average. To do so, we compute the average of the subgroup averages. We call this estimate of the mean X-double bar. The modified expression appears below.

Next we need to estimate the standard error of the mean. Recall in a earlier post (Estimating Gage Repeatability Using Range Statistics), I showed you how to estimate the standard deviation using the average range from the following expression.

Estimating The Standard Deviation Using The Average Range

To estimate the standard deviation we compute the range for each subgroup. Recall the range is the difference between the smallest from the largest value. Once we have the range for each subgroup we then calculate the average range and divide by the d2 constant. To learn more about the d2 constant read the following post (Range Statistics and the d2 Constant).

Control Limit Equation For X-Bar Chart

We can now substitute equation (3) into equation (2) to get equation (4) as shown below.

We now have the final equation to compute the control limits for the X-bar Chart based on the average range (R-bar). Take special notice of the expression 3/d2√n. This is the A2 constant.

The A2 constant is a function of the sample size n. Once we know the sample size, n, we can find the value for d2 and compute the value for A2.

Control Chart Constants For A2 At n=5, n=7

Let’s assume that we want to build control limits using a sample size of n=5. In this case the d2 constant is d2=2.326. Substituting these values into equation (5) we have:

Let’s assume that we want to build control limits using a sample size of n=7. In this case the d2 constant is 2.704. Substituting these values into equation (5) we have:

Control Chart Constants Depend On d2

The X-Bar chart and Individuals chart both use A2 and E2 constants to compute their upper and lower control limits. In both cases we need the d2 constant. Without it we cannot estimate the control limits using equation (4). To learn more about the d2 constant and how you can derive the d2 constant read the following post Range Statistics and d2 Constant – How to Calculate Standard Deviation.

Control Chart Constants For A2 At n=2 Thru n=7

In Table 1, shown are the d2 and A2 constants for various samples sizes, n=2 through n=7. We can use these d2 and A2 values to calculate the control limits for the X-Bar Chart.

Control Chart Constants – Individuals Chart

Let’s apply this new-found knowledge to derive the E2 constants used to compute the control limits for an Individuals Chart. When using an Individuals Chart the subgroup sample size is n=1. In this case, we can change equation (4) and use the following expression shown in equation (6).

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Since n=1, notice that the sample size, n, does not appear in equation (6). Take special notice of the expression 3/d2. This is the E2 constant.

Control Chart Constants – E2

Because d2 is a function of the Average Moving Range (MR-Bar), we often compute MR-Bar based on a Moving Range of MR=2. For example, the first moving range (MR1) is the absolute value of the difference between the 1st and 2nd values. Likewise, the second moving range (MR2) is the absolute value of the difference between the 2nd and 3rd values and so on.

We can also compute MR-Bar based on a Moving Range of MR=3. In this case, the first moving range (MR1) is the absolute value of the difference between the 1st and 3rd values. Likewise, the second moving range (MR2) is the absolute value of the difference between the 2nd and 4th values and so on.

Control Chart Constants – Individuals Chart

Let’s assume that we want to build control limits using a Moving Range=2. In this case the d2 constant is d2=1.1.128. Notice this is the same d2 constant used for a subgroup size of n=2. Substituting this value into equation (7) we have:

Let’s assume that we want to build control limits using a Moving Range span of 3 values. In this case, we use the d2 constant for a sample size of n=3 which is 1.693. Notice this d2 value is the same used for a subgroup size of n=3 for an Xbar chart. Substituting this value into equation (7) we have:

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Control Chart Constants for E2 at MR=2 thru MR=5

In Table 2, shown are the d2 and E2 constants for various Moving Ranges, n=2 through n=7. We can use these d2 and E2 values to calculate the control limits for the Individuals Chart.

This post on Control Chart Constants is a subset of the broader topic of Statistical Process Control Charting.

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text:
# $\lVert Tf\rVert_{L^{p}}\le \gamma\lVert f\rVert_{L^{p}}$

For $s,t\geq0,$ let $K(s,t)$ satisfy $K\geq0$ and $K(\lambda s,\lambda t)=\lambda^{-1}K(s,t)$ for every $\lambda>0$ and suppose that $\int_{0}^{\infty}t^{\frac{-1}{p}}K(1,t)dt=\gamma<\infty$ for some $p\in[1,\infty]$.

Show that if $$(Tf)(s)=\int_{0}^{\infty}f(t)K(s,t)dt~~~,(f\ge 0)$$ then $$\lVert Tf\rVert_{L^{p}}\le \gamma\lVert f\rVert_{L^{p}}$$

Here's my working :

Now we may assume $\lVert f\rVert_{L^{p}}<\infty$ for $p\in[1,\infty],$otherwise our conclusion holds trivially.

Case i) : $1\le p<\infty$

Note that for $s>0$ , we have

$$Tf(s)=\int_{0}^{\infty}f(t)K(s,t)~dt=\int_{0}^{\infty}f(t)s^{-1}K(1,\frac{t}{s})~dt=\int_{0}^{\infty}f(ts)K(1,t)~dt$$ Then, one has

\begin{align} \lVert Tf\rVert_{L^{p}}&=\bigg\lVert \int_{0}^{\infty}f(ts)K(1,t)~dt\bigg\lVert_{L^{p}(\mathbb{R_{>0}}~~,~ds)}\\ &\le\int_{0}^{\infty}\lVert f(ts)K(1,t)\rVert_{L^{p}(\mathbb{R_{>0}}~~,~ds)}~dt\\ &=\int_{0}^{\infty} K(1,t)\lVert f(ts) \rVert_{L^{p}(\mathbb{R_{>0}}~~,~ds)}~dt\\ &=\int_{0}^{\infty} K(1,t) t^{\frac{-1}{p}}\lVert f \rVert_{L^{p}}~dt\\ &=\lVert f \rVert_{L^{p}}\int_{0}^{\infty}t^{\frac{-1}{p}}K(1,t)dt\\ &=\gamma\lVert f \rVert_{L^{p}}\ \end{align}

,where the first inequality holds by Minkowski's integral inequality and $\mathbb{R_{>0}}=\{x>0 :x\in\mathbb{R}\}$.

Case ii) $p=\infty,$ for almost everywhere $s\in (0,\infty)$ one has,

$$|Tf(s)|=\bigg|\int_{0}^{\infty} f(ts)K(1,t)dt\bigg|\le \lVert f \rVert_{L^{\infty}} \bigg|\int_{0}^{\infty}K(1,t)\bigg|=\gamma\lVert f \rVert_{L^{\infty}}~.$$

Henceforth, $$\lVert Tf\rVert_{L^{\infty}}\le \gamma\lVert f\rVert_{L^{\infty}}~.$$

Is there anyone checking my attempt for validity ? any advice or comment will be appreciated. Thanks for considering my request.

• The words "hence" and "henceforth" mean different things. You meant "hence".... – David C. Ullrich Dec 28 '17 at 19:01

For the last one, you should write \begin{align*} \left|\int_{0}^{\infty}f(ts)K(1,t)dt\right|\leq\int_{0}^{\infty}|f(ts)|\cdot|K(1,t)|dt\leq\|f\|_{L^{\infty}}\int_{0}^{\infty}K(1,t)dt=\gamma\|f\|_{L^{\infty}} \end{align*} since $K\geq 0$.


text:
# Flat limit (of twisted cubic) contained in surfaces

Let $$H$$ denote the irreducible component of $$\text{Hilb}^{3t+1}\mathbb{P}^3$$ whose general member corresponds to a non-singular twisted cubic. Let $$C$$ be a subscheme lying in the boundary of $$H$$ and assume it lies in a surface $$S \subseteq \mathbb{P}^3$$.

Then why is it possible that we can find families $$C_R, S_R \subseteq \mathbb{P}^3_R$$ over a DVR $$R$$ with fraction field $$K$$ such that

1) $$C_R \subseteq S_R$$

2) The generic fiber $$C_K$$ is a non-singular twisted cubic

3) $$C \subseteq S$$ are the closed fibers of the family.

The authors in "Hilbert Scheme Compactification of the Space of Twisted Cubics": https://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/Hilbertscheme.pdf make the claim on page 4 (pg 763), line 7 of the proof. Although they are studying embedded points, this claim seems to be something more general about flat limits?

More generally, is it true that if something on the boundary of my component in a Hilbert scheme lied in a hypersurface, then I could find a family over a DVR like above?

• If I understand correctly the question (not sure), this is the valuative criterion of specialisation, see EGA II, Proposition 7.1.4. This tells you that there exists a DVR $R$ and a morphism $\operatorname{Spec}(R)\rightarrow H$ mapping the generic point to the generic point of $H$, and the special point to $[C]\in H$. Then you just pull back the universal subscheme over $H$ to $\operatorname{Spec}(R)$. – abx Dec 1 '18 at 5:18
• Hi thanks for the response. I understand how to lift the curve and surface separately, but my confusion is how could we lift them together i.e. we have a containment in the generic fiber, $C_K \subseteq S_K$. This seems to imply that there's a non-singular twisted cubic that lies in a surface of degree $d$, if some degenerated twisted cubic lies in a surface of degree $d$. – Randomness Dec 1 '18 at 5:40
• I think this would follow from knowing that the Hilbert function is constant in the family of degenerations of twisted cubic. Do you know if this holds? – ulrich Dec 1 '18 at 5:45
• Hi. I don't think the Hilbert function is constant as they are proving that the degenerate curve has an extra cubic in its ideal (twisted cubic is cut out by three quadrics). I also think they use this family of surfaces to show that the degenerate curve doesn't lie in a plane. – Randomness Dec 1 '18 at 7:11

As @abx and @ulrich explain in the comments, the original question is equivalent to a question about constancy of Hilbert functions for the universal family restricted over the irreducible component $$H$$. The Hilbert function is constant on $$H$$, as I explain below. I believe the OP is confused because, at this point in the article of Piene and Schlessinger, they have not proved constancy of the Hilbert function. Nor do they need this. For instance, Piene and Schlessinger point out that for the purposes of their proof, it is fine to replace a hyperplane that contains the curve by a quadric hypersurface that contains the curve (in fact, as follows from constancy of the Hilbert function, there is no hyperplane containing a curve parameterized by $$H$$). So my advice to the OP for reading the article is: just read the remainder of the proof and then come back to this issue after a complete read-through.

Anyway, the Hilbert function is constant. Denote by $$p(t)$$ the Hilbert polynomial $$p(t)=3t+1$$. On the Hilbert scheme $$\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$$, the natural action of $$\textbf{PGL}_{4,k}$$ on $$\mathbb{P}^3_k$$ induces an action on $$\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$$. Denote by $$H_0$$ the unique open orbit. Denote by $$H$$ the closure of $$H_0$$ in $$\text{Hilb}^{p(t)}_{\mathbb{P}^3_k/k}$$. Denote the restriction of the universal closed subscheme over $$H$$ by $$Z_H\subset H\times_{\text{Spec}\ k}\mathbb{P}^3_k.$$

Claim. Every geometric fiber $$Z_t$$ of the projection $$Z_H\to H$$ has Hilbert function, $$h_{Z_t}:\mathbb{Z}_{\geq 0} \to \mathbb{Z}_{\geq 0}, \ \ h_{Z_t}(d) := h^0(\mathbb{P}^3_k,\mathcal{O}(d)) - h^0(\mathbb{P}^3_k,\mathcal{I}_{Z_t}(d)),$$ equal to $$h_{Z_t}(d) = 3d+1$$.

Proof. Probably the fastest way to prove this is to use the stratification of $$H$$ according to the "type" of $$Z_t$$. Some version of this is contained in Joe Harris's monograph.

MR0685427 (84g:14024)
Harris, Joe
Curves in projective space.
With the collaboration of David Eisenbud.
Séminaire de Mathématiques Supérieures, 85.
Presses de l'Université de Montréal, Montreal, Que., 1982.
138 pp. ISBN: 2-7606-0603-1

I have a vague recollection that there is a small mistake in Harris's description of the orbit decomposition. I am more familiar with the honors thesis of Yoon-Ho Alex Lee. The most relevant result is Figure 4.4, p. 47.

Yoon-Ho Alex Lee
The Hilbert scheme of curves in $$\mathbb{P}^3$$
https://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/ALee_Hilbertschemes.pdf

Since dimensions of cohomology groups are upper semicontinuous, it suffices to prove that $$h_{Z_t}(d)$$ equals $$3d+1$$ for $$Z_t$$ in the two "deepest" strata, XVI and XVII in Lee's notation. With respect to homogeneous coordinates $$[ s,t,u,v ]$$, the ideal for XVI is $$\langle u^2,ut,uv,v^3\rangle$$, and the ideal for XVII is $$\langle u^2,uv,v^2 \rangle$$. It is straightforward in each of these cases to compute that $$h_{Z_t}(d)$$ equals $$3d+1$$. QED


text:
# Galaxy lensing and dark matter

When we observe galaxy lensing of background objects taking place, is it possible to estimate the total mass of the galaxy carrying out the lensing and from that confirm that dark matter is needed to be present to provide sufficient mass to bend the light by the amount observed? Is there agreement?

http://www.space.com/28489-dark-matter-light-halos.html

This is just another article which I thought interesting:
http://phys.org/news/2015-02-big-quantum-equation-universe.html

Chalnoth
The evidence comes from galaxy clusters rather than individual galaxies. This is because most of the normal matter mass in galaxy clusters is visible in the x-ray band: it exists as a diffuse, high-temperature gas throughout the cluster. The reason why the temperature is in the X-ray range is because of the depth of the gravitational potential well. You can figure out the mass of the galaxy cluster from either the temperature of the gas, or from gravitational lensing. And the luminosity of the gas gives you how much gas there is, so it's easy to see that there isn't near enough mass in the cluster gas to produce the deep gravitational potential well of the galaxy cluster.

To get a really stark picture, it's sufficient to look at collisions between galaxy clusters:
http://www.preposterousuniverse.com/blog/2006/08/21/dark-matter-exists/

In general, yes, there is quite good agreement that the cosmological data as a whole strongly supports the existence of dark matter. There are still a few holdouts that are trying to produce alternative gravity models that do not require dark matter, but their attempts have become ever more feeble, and seem highly unlikely to ever succeed.

Tanelorn
phinds
Gold Member
2021 Award
I'm not aware that there has ever been any theory that says that dark matter exists only at the galactic rim. What HAS been said that because of the yo-yo motion of dark matter through the middle of galaxies and back out the other side, and the way this motion works, most dark matter spends most of its time in the halo.

Chalnoth
I'm not aware that there has ever been any theory that says that dark matter exists only at the galactic rim. What HAS been said that because of the yo-yo motion of dark matter through the middle of galaxies and back out the other side, and the way this motion works, most dark matter spends most of its time in the halo.
Indeed. The largest density is near the center of the galaxy.

I think the problem is that the term "halo" misleadingly implies that dark matter surrounds the galaxy. A better visualization is that a galaxy is made up of a large cloud of dark matter, and in the center of that cloud is a small but relatively dense bunch of normal matter (the visible part of the galaxy).

phinds
Gold Member
2021 Award
Indeed. The largest density is near the center of the galaxy.

I think the problem is that the term "halo" misleadingly implies that dark matter surrounds the galaxy. A better visualization is that a galaxy is made up of a large cloud of dark matter, and in the center of that cloud is a small but relatively dense bunch of normal matter (the visible part of the galaxy).
Wait, do you mean the largest density of DARK matter is in the center? I though the halo was slightly more dense than the center in terms of dark matter only. I've got that wrong?

Chalnoth
Yup. One of the more commonly-used density profiles is the NFW profile:
https://en.wikipedia.org/wiki/Navarro–Frenk–White_profile

$$\rho(r) = {\rho_0 \over {r \over R_s}\left(1 + {r \over R_s}\right)^2}$$

It's not quite correct, but neither is it horribly wrong. If I recall correctly, one of the unsolved problems with dark matter simulations is that they predict a dark matter density profile that is much denser at the galactic center than observations suggest. But either way, yes, the density continues to increase. The normal matter has such a large impact because it is more dense.

phinds
Gold Member
2021 Award
OK, thanks for that correction.

Chronos
Gold Member
The density profile of DM, from halo to core, is a valuable clue about its collisional cross section.

phinds


text:
# Extending R with Intel MKL

Published: 02/20/2013

Last Updated: 11/01/2017

Introduction

The Using Intel MKL with R article discusses building R with the Intel® Math Kernel Library (Intel® MKL) BLAS and LAPACK to improve the performance of those parts of R that rely on matrix computations.  Since the BLAS and LAPACK routines are a long-standing set of functions with a standard API, the R build scripts have a built-in facility to use these. Setting R to use Intel MKL can provide significant performance improvement on parts of R, but there are other functions in Intel MKL that can provide great performance as well. This article will give you a quick tutorial on how to extend R by writing a wrapper for a function in Intel MKL and call it from your R script.

The wrapper

Let's begin by selecting a simple function in Intel MKL that is not a part of the BLAS or LAPACK. For this exercise we'll choose the vdPow function which takes two arrays, x and y, and creates a new array z whose nth element is x[n] to the y[n] power.

#include "R.h" #include "Rdefines.h" #include "Rinternals.h" #include "mkl.h" SEXP mkl_vdpow(SEXP N, SEXP X, SEXP Y) {   SEXP Z;   int *n;   int nn;   double *x, *y, *z;     /* Convert input arguments to C types */   PROTECT(N = AS_INTEGER(N));   PROTECT(X = AS_NUMERIC(X));   PROTECT(Y = AS_NUMERIC(Y));     n = INTEGER_POINTER(N);   x = NUMERIC_POINTER(X);   y = NUMERIC_POINTER(Y);     /* Allocate memory to store results */   nn = *n;   PROTECT(Z = NEW_NUMERIC(nn));   z = NUMERIC_POINTER(Z);     vdPow(nn, x, y, z);     UNPROTECT(4);     return Z; }

Building the wrapper

On Windows this is easily done by obtaining the Windows package and Rtools from CRAN. I downloaded R 2.15.2 and Rtools version 3.0.0.1927 and then created a file to set my environment as follows:

SET R_TOOLS_ROOT="C:\Program Files\R\Rtools"

SET MINGW64_ROOT=%R_TOOLS_ROOT%\gcc-4.6.3

SET MINGW64_CMPLR_ROOT=%MINGW64_ROOT%\i686-w64-mingw32

SET PATH=%MINGW64_ROOT%\bin;%MINGW64_CMPLR_ROOT%\bin;%R_TOOLS_ROOT%\bin;%MINGW64_ROOT%\libexec\gcc\i686-w64-mingw32\4.6.3;%PATH%

SET LIB=%MINGW64_CMPLR_ROOT%\lib64;%MINGW64_CMPLR_ROOT%\lib;%LIB%

I then build using the following commands:

SET R_HOME="C:\Program Files\R\R-2.15.2"

SET MKL_ROOT="C:\Program Files (x86)\Intel\Composer XE 2013"

c:\"program files"\r\rtools\gcc-4.6.3\bin\gcc -c -O2 -Wall  -std=gnu99 -I%R_HOME%\include -I%MKL_ROOT%\mkl\include pow_wrp.c -o pow_wrp.o

c:\"program files"\r\rtools\gcc-4.6.3\bin\gcc -shared -Wl,-soname,pow_wrp.so -o pow_wrp.so pow_wrp.o %R_HOME%\bin\i386R.dll %MKL_ROOT%\mkl\lib\ia32\mkl_rt.lib %MKL_ROOT%\compiler\lib\intel64\libiomp5md.lib

Calling the function from R

Here then is a simple script that calls the function we've just built

dyn.load("c:/full_path_or_directory_on_path/pow_wrp.so")

mkl_pow <- function(n, x, y) .External("mkl_vdpow", n, x, y)

n <- 1000

x <- runif(n, min=2, max=10)

y <- runif(n, min=-2, max=-1)

z <- mkl_pow(n, x, y)

Performance

Let's see what sort of performance difference the use of Intel MKL can make by comparing this new power function with the simple means available within R. To do the similar operation in R we could use a loop to perform each scalar operation. The following is the result of running the interactively in the 32-bit RGui that comes with the R-2.15.2 pre-built Windows package on my laptop (Intel MKL 11.0 update 2, R-2.15.2, 2-core Intel® Core™ i5-540M processor (2.53GHz) 4GB RAM, Microsoft Windows 7):

> dyn.load("c:/a/collateral/kb/r/files/pow_wrp.so")

> mkl_pow <- function(n, x, y) .Call("mkl_vdpow", n, x, y)

> n <- 1000000

> x <- runif(n, min=2, max=10)

> y <- runif(n, min=-2, max=-1)

> start <- proc.time()

> z <- mkl_pow(n, x, y)

> end1 <- proc.time() - start

> end1

user system elapsed

0.03 0.02 0.05

> i <- n

> start <- proc.time()

> repeat{ z[i] <- x[i]^y[i] i <- i - 1 if (i==0) break() }

> end2 <- proc.time() - start

> end2

user system elapsed

5.49 0.03 5.17

For this very large, size the Intel MKL function makes a large difference. The loop approach takes 5.49 seconds while the Intel MKL function takes 0.03 seconds. For smaller problems the performance will be less, but for systems with higher core-counts, compute intensive functions can show even greater speed-up.

Given the potential for large performance improvements, it may be worthwhile finding out the functions available in Intel MKL

#### Product and Performance Information

1

Performance varies by use, configuration and other factors. Learn more at www.Intel.com/PerformanceIndex.


text:
# A particle (m=5 kg) hits another identical particle. If v^0= 10m/s= initial speed. What are both...

###### Question:

A particle (m=5 kg) hits another identical particle. If v^0= 10m/s= initial speed. What are both particles final speeds if: A) They stick together?

B) The collisio is elastic?

C) The incoming particle bounces backwards at vf=1o m/s? (vxf=-10 m/s)

#### Similar Solved Questions

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text:
# How do use the discriminant to find all values of b for which the equation x^2-bx+4=0 has one real root?

Jul 6, 2017

$b = \pm 4$

#### Explanation:

The discriminant of a quadratic is ${b}^{2} - 4 a c$.

In this case, a=1, b=?, c=-4.

Equal roots means ${b}^{2} - 4 a c = 0$

b^2-4(1*4=b^2-16=0

${b}^{2} = 16$

$b = \pm 4$

Jul 6, 2017

$b = \pm 4.$

#### Explanation:

For the Qudr. Eqn. $a {x}^{2} + b x + c = 0 ,$ to have only One Root, or, to

have Two Identical Roots,

$\text{The Dicriminant } \Delta = {b}^{2} - 4 a c = 0.$

In our case, ${\left(- b\right)}^{2} - 4 \left(1\right) \left(4\right) = {b}^{2} - 16 = 0 \Rightarrow b = \pm 4.$

Incidentally, if

$b = 4 , \text{ the root is, "2; &, if b=-4," the root is } - 2.$

Jul 6, 2017

In $y = a {x}^{2} + b x + c$ we have:

$b = 2 \sqrt{a c} = \pm 4$

#### Explanation:

Given the standard form $y = a {x}^{2} + b x + c$

where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

For there to only be 1 real root we need the form

$x = \frac{- b}{2 a} \pm 0$

This means that

$\frac{\sqrt{{b}^{2} - 4 a c}}{2 a} = 0$

Multiply both sides by $2 a$

$\sqrt{{b}^{2} - 4 a c} = 0$

square root both sides

${b}^{2} - 4 a c = 0$

$b = \sqrt{4 a c} = 2 \sqrt{a c}$

But $a = 1 \mathmr{and} c = 4$ giving

$b = 2 \sqrt{1 \times 4} = \pm 4$


text:
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text:
# Conditional convergence of a power series on the $z \in \mathbb{C}$ for which $|z|=\rho$ [duplicate]

Consider the following complex power series $$\sum_{n \geq 1} \frac{z^n}{n} \,\,\,\,\,\,\, z \in \mathbb{C}$$

It surely converges conditionally for $z=-1$ (for alternating series test) and for $z=1$ it diverges (it is the harmonic series).

My question is: how can one show that the power series converges conditionally for any $z \in \mathbb{C}$ such that $|z|=1$ (except for $z=1$)?

## marked as duplicate by Martin R, Daniel Fischer♦ complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 12 '16 at 19:30

We can use the Dirichlet's test to prove the desired result: with $a_n=\frac1n$ and $b_n=e^{in\theta}$ we verify that
• $(a_n)$ is decreasing to 0
• $\sum_{n=1}^N b_n$ is bounded
so the series $\sum a_nb_n$ is convergent.


text:
## Friday, December 18, 2015

### Codes and Coding

as my master's thesis was about Linear Codes and Applications in Cryptography I'd like to provide some content of it in a nutshell and to give some recommendations of software that I used in the course of writing it.

 multiple scientific fields are involved (created using TikZ)
Modern cryptography covers among others the three areas privacy, authentication and integrity of messages. The aim of coding theory on the other hand described by one word is reliability - adding redundancy to messages in order to detect transmission errors. Both fields cover seemingly different areas in the field of digital information processing.

The motivation of my thesis was to showcase the combination of multiple scientific fields within coding theory and modern cryptography and to explore how they interact in the domain mentioned above and why they are a good choice.

In the center of attention stood the McEliece public-key cryptosystem, it's variation (Niederreiter's cryptosystem) and generalizations to other code classes after those code classes of interest were introduced and discussed.

The McEliece public-key cryptosystem was published in 1978, is based on disguising a special code as a general code and it is of interest back then and today because it seems to stand solid in the so-called post-quantum era, when a powerful quantum computer is available to attackers. More than 35 years ago the original code parameters: codeword length $n = 1024$, code dimension (or message bits) $k = 524$ and error-correcting capability $t = 50$ were proposed for the binary code underlying the McEliece public-key cryptosystem for $2^64$ bit security. To achieve the a reasonable security level today they have been adjusted to $n = 6960, k = 5413, t = 119$ because of attacks that were explored since then.

Given the number $t$ of erroneous positions and a general code $C$, the task of decoding $y = c + e$ is to find a codeword $c \in C$ of Hamming weight or number of non-zero entries: $w(y-c) = w(e) = t$. In the complexity theoretic literature the result is that this problem is NP-complete - there is no algorithm known that fulfills this task in polynomial time depending on the input size (length of $y$ respectively $c$).

The choice to use a Goppa codes as private key, which is multiplied by permutation matrices to obtain the public key looking as if it was a general code, was due to performance reasons when decoding messages thanks to Patterson's algorithm.

Interestingly many attacks are not applicable to the class of binary Goppa codes that McEliece's scheme was based on but on similar schemes and generalizations allowing other alphabets or other structures - binary Goppa codes were seemingly proposed with foresight.

In spite of some advantages and the - so far - quantum computer resistant attacks the biggest drawback of public-key cryptosystems based on codes are rather long public keys compared to other schemes. If one adds a security margin, under the assumption of a working quantum computer performing the general attack through the application of Grover's algorithm, the public key size is as big as 1 MB.

In the thesis I implemented the fast algorithm (by N. J. Patterson) for algebraically decoding binary Goppa codes in Sage and demonstrated the full functionality of the system in yet another chapter giving an example. Since I was writing it all up in $\LaTeX$, using sagetex even allowed the computation and insertion of the results into the $\verb!.tex!$ file one the fly when the document is being compiled!

My recommendation (to my fellows) for researching, analyzing, developing & testing of algorithms in a structured manner with the ability to create nice plots and documents rather easily is to give Sage or Spyder a try. Even though it might not be a necessity in your research to actually code something, playing around with these tools already available is instructive and worth a try.

Sage is available online or offline via installation and it is a good choice if you need access to i.e. number theory. Maybe the aforementioned Spyder is preferable if a somewhat more automated handling even closer to the programming language Python of the implemented primitive at hand is needed in your case.

 Autumnal Bochum around Oct 31
As usual, a little plurivalent:
Happy Coding and Happy Halloween - because $Dec(25) = Oct(31)$, right?!


text:
# All Questions

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text:
Here, we only show the full-dimensional entanglement polytopes, corresponding to genuinely four-qubit entangled quantum states in $$\mathbb C^2 \otimes \mathbb C^2 \otimes \mathbb C^2 \otimes \mathbb C^2$$.

$$\lambda_1^{\max}$$ = 0.75
No. Class Description Vertices Faces Perms.
1 $$L_{0_{7\oplus \bar{1}}}$$ Origin (0.5,0.5,0.5,0.5) replaced by (0.75,0.5,0.5,0.5). 12 13 4
2 $$L_{0_{5\oplus 3}}$$ Vertices (0.5,0.5,0.5,0.5) and (1,0.5,0.5,0.5) missing. 10 13 4
3 $$L_{a_4}$$ ($$a=0$$) Vertices (0.5,1,0.5,0.5), (0.5,0.5,0.5,1), (0.5,0.5,0.5,0.5) missing. 9 14 6
4 $$L_{\text{ab}_3}$$ ($$b=-a$$) All three-partite entangled vertices missing. 8 16 1
5 $$L_{\text{ab}_3}$$ ($$a=b=0$$) W-state. All three- and four-partite entangled vertices missing. 7 9 1
6 $$L_{a_2b_2}$$ ($$b = -a$$) Vertices (0.5,1,0.5,0.5), (0.5,0.5,0.5,1) missing. 10 14 6
7 $$G_{\text{abcd}}$$ (generic) Generic polytope. 12 12 1


text:
# corrections

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