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## 2017 6 Nations – 3rd weekend review

The 3rd weekend of the 2017 6 Nations saw home victories in all three matches. Scotland beat Wales at Murrayfield, Ireland beat France in Dublin, and England beat Italy at Twickenham.

## Scotland v Wales

Scotland beat Wales for the first time since 2007, with a well-deserved win that saw them score 20 unanswered points in the 2nd half. Wales went into half-time with a slim 13-9 lead. It could have been more, with Wales missing a chance to go 16-6 up with only a few minutes left of the first half. Instead, Scotland scored a penalty just before half-time and instead of 16-6 it was 13-9.

If ever there was a game of two halves this was it. The second half saw Scotland score 20 unanswered points, including two well deserved tries. It is true that Wales did have their chances to score points in the 2nd half, but Scotland were by far the better team after the break and fully deserved their victory over a largely ineffective Welsh team.

Scotland beat Wales 29-13 in a well-deserved victory, their first win over Wales since 2007.

Whereas the Scottish team took their chances, Wales were unable to come away with points when they were in the Scottish 22. It was a poor performance by Wales, but also a very good one by Scotland. After being uncompetitive for most of the last 10 years, it is good to see Scotland back mixing it with the other countries in the 6 Nations.

## Ireland v France

This was a cracker of a match. For a neutral, it had everything. Ireland went into half-time with a narrow 7-6 lead, and it was not clear well into the second half who was going to win this keenly contested match. In the end the better play of Irish half-backs Connor Murray and Johnny Sexton proved decisive, Ireland emerging victorious 19-9. With first Ireland, and then France, being the last two sides that Wales will face, the performance of both teams in this match does not bode well for Wales’ chances of winning either match, in my opinion.

## England v Italy

I have not seen this match, but am looking forward to seeing the highlights, mainly for the tactics which Italy adopted. By not committing any men to the breakdown, they avoided any rucks forming and thus the off-side line which normally exists did not apply. This allowed them to have men standing between the English backs, completely disrupting the game that England hoped to play. In fact, so effective was this tactic that Italy were 10-5 ahead at half-time. It is, apparently, only in the second half that England got to grips with this novel tactic, and eventually ran away to win 36-15.

## The 4th weekend

The 6 Nations now has another brief hiatus, before the next round of matches in just under two weeks. Ireland come to Cardiff to play Wales on Friday evening (10th March), then on the Saturday Italy take on France before the big game of the weekend, England v Scotland. Not only is this for the Calcutta Cup, but both countries are going for the Triple Crown. It will  be the first time in probably some 20 years that Scotland will go to Twickenham with even the slightest hope of winning.

Based on our performance against Scotland, I cannot see Wales beating Ireland, which leaves us with the real possibility of finishing the 2017 6 Nations with only one win. Quite where this will leave the Welsh team and management is a good question,  but it is clear from our performances this season that Wales have not developed the creativity in attack which is needed to win matches. Scotland showed us how it should be done; all we could do was look on and marvel at how much they have developed and how much we have regressed.

## 2017 6 Nations – 3rd weekend preview

This weekend sees the 2017 6 Nations come out of its hiatus for the 3rd weekend of matches. The weekend kicks off with Scotland v Wales at Murrayfield, followed by Ireland v France in Dublin, and Sunday sees England v Italy at Twickenham. It is an important weekend in the Championships, particularly for Wales, Scotland, Ireland and France. Win this weekend and their records will be 2 wins from 3, but lose and it will be 1 win from 3, a big difference.

This weekend is an important one in the 2017 6 Nations. For Wales, Scotland, Ireland and France it will make the difference between being 2 wins from 3 or 1 win from 3.

## Scotland v Wales

Wales go to Murrayfield having not lost to Scotland since 2007, either home or away. George North comes back into the team that lost to England, after injury kept him out of that game. Otherwise the Welsh team is unchanged, with perhaps the biggest surprise being that Ross Moriarty keeps his starting position, forcing Taulupe Faletau to start on the bench. Given how well Moriarty played against England I think it is the correct call by Rob Howley.

This will be the sternest test Wales have had against Scotland in the last 10 years, and they will need to be at their best to beat a resurgent Scotland. It will be a fascinating encounter, and I am going to be on the edge of my seat watching it.

## Ireland v France

Ireland will be wanting to continue getting their 2017 campaign back on track after their opening loss to Scotland. They beat Italy easily in the 2nd weekend of matches, so if they can beat France their campaign to win the 2017 Championship will still be possible. Lose and they can forget about being crowned champions. France too will be wanting to make it 2 wins from 3, and an away win against Ireland would be a scalp to bolster their confidence of being in contention for the Championship title come the final weekend.

## England v Italy

Sunday’s game seems England play Italy at home. This should be a rout, it is really just a question of how many points England can put on Italy. There has been a lot of debate this Championships as to whether Italy deserve to be part of the 6 Nations, and this match will probably do nothing to quell that debate. I expect a cricket score of 60+ points for England.

## Imaging the Galaxy’s supermassive blackhole – part 1

Last week, I blogged about the theoretical arguments for the Galaxy harbouring a supermassive black hole at its centre, and here I blogged about the observational evidence. The work done by the UCLA and MPE teams, discussed here, has led to a determination that the central black hole has a mass of between 4.4 and 4.5 million solar  masses. I am going to take the upper end  of this range, just for convenience.

An artist’s impression of Sgr A*, showing the central supermassive black hole and the accretion disk which surrounds it.

## The size of the event horizon

In this blog here I showed that the radius of a blackhole’s event horizon can be calculated by using the equation for the escape velocity $v_{esc}$ when that velocity is equal to the speed of light $c$. That is

$v_{esc} = c = \sqrt{ \frac{2GM }{ R } }$

where $M$ is the mass of the blackhole, $G$ is the universal gravitational constant, and $R$ is the size of the object, which in this case is the radius of the event horizon (also known as the Swarzchild radius $R_{s}$). So, we can write

$R_{s} = \frac{ 2GM }{ c^{2} }$

Putting in a mass of 4.5 million solar masses, we find

$R_{s} = 1.33 \times 10^{10} \text{ metres}$

Converting this to AUs, we find the radius of the event horizon is 0.09 AUs, much smaller than the radius of Mercury’s orbit, which is about 0.3 AUs.

At the distance of the Galactic centre, 8 kpc, this would subtend an angle of
$\theta = 6.17 \times 10^{-9} \text{ degrees}$ (remember to double $R_{s}$ to get the diameter of the event horizon). This is the same as

$\boxed{ \theta = 22.22 \text{ micro arc seconds} }$

Converting this to radians, we get

$\theta ( \text{in radians}) = 1.08 \times 10^{-10}$

In fact, we do not need to resolve the event horizon itself, but rather the “shadow” of the event horizon, which is about four times the size, so we need to resolve an angle of

$\theta ( \text{in radians}) \approx 4 \times 10^{-10}$

## The resolution of a telescope

There is a very simple formula for the resolving power of a telescope, it is given by

$\theta( \text{in radians}) = \frac{ 1.22 \lambda }{ D }$

where $D$ is the diameter of the telescope and $\lambda$ is the wavelength of the observation. Let us work out the diameter of a telescope necessary to resolve an object with an angular size of $50 \times 10^{-4} \text{ radians }$ at various wavelengths.

For visible light, assuming $\lambda = 550 \text{ nanometres}$

$D = \frac{ 1.22 \times 550 \times 10^{-9} }{ 4 \times 10^{-10 } }, \boxed{ D = 1.68 \text { km} }$

There is no visible light telescope this large, nor will there ever be. At the moment, visible-light interferometry is still not technically feasible over this kind of a baseline, so imaging the event horizon of the Galaxy’s supermassive blackhole is not currently possible at visible wavelengths.

$D = \frac{ 1.22 \times 21 \times 10^{-2} }{ 4 \times 10^{-10 } }, \boxed{ D = 640,000 \text { km} }$

This is more than the distance to the Moon (which is about 400,000 km away). So, until we have a radio dish in space, we cannot resolve the supermassive blackhole at 21cm either.

For millimetre waves, we have

$D = \frac{ 1.22 \times 1 \times 10^{-3} }{ 4 \times 10^{-10 } }, \boxed{ D = 3,100 \text { km} }$

which is feasible with very long baseline interferometry (VLBI). So, with current technology, imaging the event horizon of the Milky Way’s supermassive blackhole is only feasible at millimetre wavelengths. Millimetre waves lie in a niche between visible light and radio waves. They are long enough that we can do VLBI, but they are short enough that the baseline to image the supermassive black hole’s event horizon is small enough to be possible with telescope on the Earth.

Next week I will talk about a project to do just that!

## Free As A Bird – The Beatles (song)

Today I thought I would share this lesser known song by The Beatles – “Free As A Bird”. It was released in 1996, when The Beatles put together their Anthology series of CDs, book and TV series. By this time, of course, John Lennon had been dead for 16 years. The other Beatles got in touch with Yoko who gave them a recording that Lennon had made of this song. They added their own voices and musical accompaniment to his original track, to create the closest thing possible to a new Beatles song. There are also some lyrics added by Paul McCartney, the parts that he sings on his own, rather than in harmony to Lennon’s voice.

The Beatles single “Free As A Bird” was released in 1996, to coincide with the release of the Anthology CDs, book and TV series

Free as a bird
It’s the next best thing to be
Free as a bird

Home, home and dry
Like a homing bird I’ll fly
As a bird on wings

Whatever happened to
The life that we once knew?
Can we really live without each other?

Where did we lose the touch
That seemed to mean so much?
It always made me feel so…

Free as a bird
Like the next best thing to be
Free as a bird

Home, home and dry
Like a homing bird I’ll fly
As a bird on wings

Whatever happened to
The life that we once knew?
Always made me feel so free

Ah…
Ah…
Ah…

Free as a bird
It’s the next best thing to be
Free as a bird
Free as a bird
Free as a bird
Oooooo

Free…

[Turn out nice again, mother.]

The video to accompany “Free As A Bird” is fascinating. For aficionados of Beatles trivia, there are all kinds of obvious and less obvious references to Beatles songs in the video. See how may you can spot?

Enjoy!

## The February 2017 annular solar eclipse

Some of you may be aware that there is an annular eclipse of the Sun on Sunday 26 February, which is why I am posting this blog a few days before it. Annular eclipses occur when the Moon is a little too far away to block the Sun out entirely, so instead we see a ring of light around the Moon, as this picture below shows. This particular picture was taken during the May 20 2012 annular eclipse

An annular eclipse happens when the Moon is slightly too far away to block out the Sun entirely. This is a picture of the May 20 2012 annular eclipse.

## The Moon’s elliptical orbit about the Earth

The diagram below shows an exaggerated cartoon of the Moon’s orbit about the Earth. The Moon’s orbit is an ellipse, it has an eccentricity of 0.0549 (a perfect circle has an eccentricity of 0). The average distance of the Moon from the Earth (actually, the distance between their centres) is 384,400 kilometres. The point at which it is furthest from the Earth is called the apogee, and is at a distance of 405,400 km. The point at which it is closest is called the perigee, and it is at a distance of 362,600 km.

The Moon orbits the Earth in an ellipse, not a circle. The furthest it is from the Earth in its orbit (the apogee) is at a distance of 405,400 km, the nearest (the perigee) is at a distance of 362,600 km.

## The angular size of the Moon

It is pure coincidence that the Moon is the correct angular size to block out the Sun. The Moon is slightly oblate, but has a mean radius of 1,737 km. With its average distance of 384,400, this means that from the Earth’s surface (the Earth’s mean radius is 6,371 km) the Moon has an angular size on the sky of

$2 \times \tan^{-1} \left( \frac{ (1.737 \times 10^{6}) }{ (3.84 \times 10^{8} - 6.371 \times 10^{6}) } \right) = 2 \times \tan^{-1} (4.59975 \times 10^{-3} )$

$= 2 \times 0.2635 = \boxed{ 0.527 ^{\circ} \text{ or } 31.62 \text{ arc minutes} }$
So, just over half a degree on the sky. But, this of course will vary depending on its distance. When it is at apogee (furthest away), its angular size will be

$\boxed{ \text{ at apogee } 29.93 \text{ arc minutes } }$

and when it is at perigee (closest) it will be

$\boxed{ \text{ at perigee } 33.53 \text{ arc minutes } }$

## The angular size of the Sun

The Sun has an equatorial radius of 695,700 km, and its average distance from us is 149.6 million km (the Astronomical Unit – AU). So, at this average distance the Sun has an angular size of

$2 \times \tan^{-1} \left( \frac{ (6.957 \times 10^{8}) }{ (1.496 \times 10^{11} - 6.371 \times 10^{6} ) } \right) = 2 \times 0.266 = \boxed {0.533^{\circ} }$

Converting this to arc minutes, we get that the angular size of the Sun at its average distance is

$\boxed{ 31.97 \text{ arc minutes} }$

Compare this to the angular size of the Moon at its average distance, which we found to be $31.62 \text{ arc minutes}$.

The angular size of the Sun varies much less than the variation in the angular size of the Moon, at aphelion (when we are furthest) from the Sun, we are at a distance of 152.1 million km, so this gives an angular size of

$\boxed{ \text{ at aphelion } 31.44 \text{ arc minutes } }$

and, at perielion, when the distance to the Sun is 147.095 million km, the angular size of the Sun is

$\boxed{ \text{ at perihelion } 32.52 \text{ arc minutes } }$

## Annular Eclipses

So, from the calculations above one can see that, if the Moon is at or near perigee, its angular size of $33.53 \text{ arc minutes }$ is more than enough to block out the Sun. When the Moon is at its average distance, its angular size is $31.62 \text { arc minutes }$, which is enough to block out the Sun unless we are near perihelion. But, when the Moon is near apogee, its angular size drops to $29.93 \text{ arc minutes }$, and this is not enough to block out the Sun, even if we are at aphelion.

The Earth is currently at perihelion in early January (this year it was on January 4), so the Sun is slightly larger in the sky that it will be in August for the next solar eclipse. This, combined with the Moon being near its apogee, which occurred on February 18, (for a table of the dates of the Moon’s apogees and perigees in 2017 follow this link) means that the solar eclipse on Sunday February 26 is annular, and not total.

## The February 26 2017 Annular Eclipse

Here is a map of the path of the eclipse, it is taken from the wonderful NASA Eclipse website. If you follow this link, you can find interactive maps of all the eclipses from -1999 BC to 3000 AD! If you have about 6 years to waste, this is an ideal place to do it!

The February 26 2017 annular eclipse will start in the southern Pacific ocean, sweep across Chile and Argentina, then across the Atlantic Ocean, before reaching Angola, Zambia and the Democratic Republic of Congo (Congo-Kinshasa)

The eclipse finishes in Africa and, as luck would have it, I am going to be in Namibia on the day of the eclipse. In fact, if you are reading this anytime in the week before the eclipse, I am already there. I am in Namibia for a week as part of Cardiff University’s Phoenix Project, and I will be giving a public lecture at the University of Namibia about the eclipse on Wednesday 22 February. I also hope to give a public lecture to the Namibian Scientific Society on the Friday, and on the Sunday I will be helping University of Namibia astronomers with a public observing session in Windhoek.

The February 20 2017 annular eclipse will finish in Africa, passing through Angola, Zambia and the Democratic Republic of Congo (Congo Kinshasa)

The interactive map to this eclipse, which you can find by following this link, allows you to click on any place and find out the eclipse details for that location. So, for Windhoek, the eclipse begins at 15:09 UT (which will be 17:09 local time), with the maximum of the partial eclipse being at 16:16 UT (18:16 local time), and the eclipse ending at 17:16 UT (19:16 local time). Because Windhoek is to the south of the path which will experience an annular eclipse, it will be a partial eclipse, with a coverage of 69%.

As seen from Windhoek, where I will be for the annular eclipse, the obscuration will be 69%.

So, if you are anywhere Chile, Argentina, in western South Africa, in Namibia, in Angola, or the western parts of Congo-Kinshasa and Congo-Brazzaville, look out for this wonderful astronomical event this coming Sunday. And, remember to follow the safety advise when viewing an eclipse; never look directly at the Sun and only look through a viewing device that has correct filtration. Failure to follow these precautions can result in permanently damaging your eyesight.

## Beyond Here Lies Nothin’ – Bob Dylan (song)

Today I will continue with my series of blogposts of some Bob Dylan songs, in celebration of his winning the 2016 Nobel prize for literature. I am concentrating on songs which are on  his official Vevo channel, as other songs of his which are uploaded to YouTube are almost always swiftly removed.

The song I am sharing today is “Beyond Here Lies Nothin'”, which appears on his 2009 album Together Through Life. According to his website, he first performed this live in July 2009 and the most recent live performance was earlier this year, in July 2016. As of my writing this, he has performed it 398 times!

“Beyond Here Lies Nothin'” appears on Dylan’s 2009 album Together Through Life.

Here are the lyrics to this song, which you can find here on Dylan’s official website.

I love you pretty baby
You’re the only love I’ve ever known
Just as long as you stay with me
The whole world is my throne
Beyond here lies nothin’
Nothin’ we can call our own

I’m movin’ after midnight
Down boulevards of broken cars
Don’t know what to do without it
Without this love that we call ours
Beyond here lies nothin’
Nothin’ but the moon and stars

Down every street there’s a window
And every window made of glass
We’ll keep on lovin’ pretty baby
For as long as love will last
Beyond here lies nothin’
But the mountains of the past

My ship is in the harbor
Listen to me pretty baby
Beyond here lies nothin’
Nothin’ done and nothin’ said

Here is the official video of “Beyond Here Lies Nothin'” from Dylan’s Vevo channel. Enjoy!

## Evidence for the Galaxy’s supermassive black hole

On Tuesday, I blogged about the theoretical work done in the early 1970s by Martin Rees, and others, which proposed that there may be a supermassive black hole at the centre of our Galaxy and most spiral galaxies.

## What about the observational evidence?

In the early 1980s two teams set about observing the orbits of stars near Sgr A*. The two teams, working separately, were at UCLA and The Max Planck Institute For Extra-terrestrial Physics (MPE). The team at UCLA is known as the UCLA Galactic Center Group, the team at the MPE doesn’t have a snazzy name, but their website can be found here. Gradually, over many years, each of the two teams has determined the orbits of several dozens of stars, and hence have been able to use the laws of gravity to determine the mass of the enclosed mass which the stars are orbiting.

Below is an image of Sgr A* taken by the MPE team using the NACO near-infrared camera on the VLT with adaptive optics. The entire image is only 30 arc seconds across.

A combined H, K and L-band near infrared image of the Galactic Centre obtained by the NACO camera on the VLT using adaptive optics. This image is from the MPE website.

Here is a paper, published in 2009, entitled “Monitoring Stellar Orbits Around the Massive Black Hole in the Galactic Center”, published by the MPE group in The Astrophysical Journal. Here is a link to the paper.

This paper, published in The Astrophysical Journal in 2009, is one of several showing overwhelming evidence for a supermassive blackhole at the centre of the Milky Way galaxy.

In this paper, entitled “The Galactic Center massive black hole and nuclear star cluster”, Reinhard Genzel (the director of the MPE) and colleagues summarise the evidence from their studies of their being a supermassive blackhole at the centre of the Milky Way, with a calculated mass of about 4.4 million solar masses. Here is a link to the paper.

In a paper entitled “The Galactic Center massive black hole and nuclear star cluster”, Genzel etal. summarise their finding that the Galaxy harbours a massive black hole with a mass of about 4.4 million solar masses.

The UCLA group published this paper “Measuring Distance and Properties of the Milky Way’s Central Supermassive Black Hole With Stellar Orbits”, in 2008 in The Astrophysical Journal (here is a link to the paper). In it,they calculate the mass of the supermassive black hole to be 4.5 million solar masses, with an error of plus or minus 0.4 million solar masses.

Ghez etal. (2008) find the  mass of the supermassive black hole to be 4.5 million solar masses, slightly higher than the MPE group, but well within the errors of the two groups’ measurements.

The mass of this black hole is about 4.45 million times the mass of the Sun (the two groups calculate different masses, with the UCLA group calculating about 4.5 million solar masses, the MPE group about 4.4 million solar masses). Let us assume it is 4.5 million solar masses, just to round up to the nearest half a million solar masses.

As some of you may know, blackholes are observable in certain ways. They clearly affect the orbit of nearby objects (this is how the UCLA and MPE teams have garnered the evidence for the supermassive blackhole), but also the accretion disk which usually surrounds a blackhole has very hot gas spiralling into the blackhole. This very hot gas emits radiation as a blackbody, so most of it comes out in the X-ray part of the spectrum due to the very high temperature of several millions of Kelvin.

But, a blackbody will also radiate at other wavelengths (see my blog here to remind yourself of the shape of a blackbody curve), so such accretion disks will also radiate visible light, infrared light, and even radio emission. The question then arises, is it possible to observe the accretion disk way in towards the event horizon of the Galaxy’s supermassive blackhole?

I will answer that question next week.

## The Galaxy’s supermassive black hole

There is now overwhelming evidence that our Galaxy harbours a supermassive black hole at its centre. Not only that, but the Hubble Space Telescope has discovered that all spiral galaxies harbour supermassive black holes at their centres, and the mass of that black hole is directly proportional to the mass of the galaxy in which it resides. The reasons for this are still unclear.

The idea of supermassive black holes driving the prodigious energy output at the centre of some galaxies was first proposed by Edwin Salpeter in a 1964 paper. Salpeter is probably better known for his work on the initial mass function of star formation, but in this paper (follow this link to read it), Salpeter proposed that supermassive black holes may be the energy source behind the then newly discovered quasars.

In a 1964 paper, Edwin Salpeter (possibly more famous for his work on the initial mass function of star-formation) was the first to propose supermassive black holes as the energy source of the newly-discovered quasars (or QSOs).

In 1971, Donald Lynden-Bell and Martin Rees wrote an important paper entitled “On quasars, dust and the galactic centre”, (follow this link to the paper). It was the first paper to suggest that our own Galaxy, the Milky Way, may harbour a supermassive black hole at its centre.

The possibility that our Milky Way harboured a supermassive blackhole at its centre was first proposed by Donald Lynden-Bell and Martin Rees in 1971.

Another important paper entitled “Accretion onto Massive Black Holes” was written in 1973 by Pringle, Rees and Pacholczyk (follow this link), who considered the observable effects that matter accreting onto a (super)massive black hole would have.

In a 1973 paper, Pringle, Rees and Packolczyk considered the observable effects of the accretion of matter onto a supermassive black hole.

Pringle etal. draw two main conclusions, the second of which is possibly the more important; that material falling onto a (super)massive black hole will emit a huge amount of radiation.

The two main conclusions of the Pringle etal. (1973) paper.

In a 1974 review article in The Observatory entitled “Black Holes”, Martin Rees further stated the arguments for supermassive black holes at the centres of galaxies.

In a 1974 review article in The Observatory, Martin Rees wrote that “a black hole might lurk in the centres of most galaxies.”. 35-40 years later, he was shown to be correct.

He stated (my highlight)

If we regard quasars as hyperactive galactic nuclei, then a black hole might lurk in the centres of most normal galaxies.

How prescient were these words!

Later in the same year, radio astronomers Bruce Balick and Robert Brown discovered a compact radio source in the constellation Sagittarius. They announced their result in a paper entitled “Intense Sub-Arcsecond Structure in the Galactic Center” (here is a link to the paper).

In 1974, Bruce Balick and Robert Brown used the Very Large Array of radio telescopes in New Mexico to discover a compact radio source at the centre of the Milky Way. We now call this source Sagittarius A*

Using the Very Large Array of radio telescopes in New Mexico, Balick and Brown found a sub-arcsecond radio source at both 2695 MHz (11cm) and 8085 MHz (3.7cm). We now call this source Sagittarius A*, and it is believed to be where the Galaxy’s supermassive black hole resides. Here is their image obtained at 2695 MHz (at right, the image at left is the base-line coverage in the u-v plane, the interferometry plane of the array).

Bruce Balick and Robert Brown discovered a sub-arcsecond radio source in the constellation Sagittarius. We now call this source Sagittarius A*. Their discovery was made at 2695 MHz (which corresponds to a wavelength of 11 centimetres)

Since these discovery observations, Sagittarius A* (or Sgr A* as it is often known) has been observed at many other wavelengths (but not in visible light, the dust extinction is too great). For example, here is a combined infrared and X-ray image.

A composite infrared and X-ray image of Sagittarius A*

And, here are some images taken by the SPIRE camera on the Herschel Space Observatory at 250, 350 and 500 microns.

Images of Sagittarius A* taken by the SPIRE camera on the Herschel Space Telescope. The observations are at (from left to right) 250, 350 and 500 microns.

Later this week I will blog about the observational evidence for this compact object being a supermassive black hole.

## The 2nd weekend of the 2017 6 Nations

The second weekend of the 2017 6 Nations is over, and I think it is fair to say that this year is on course to be one of the most exciting 6 Nations in history. After Ireland running rampant in Italy, England narrowly beat Wales in Cardiff by scoring a converted try in the last 4 minutes, and France narrowly beat Scotland in Paris. It seems that 5 of the 6 nations are pretty evenly matched, so most of the games look like they could be very close.

## Italy v Ireland

Ireland went to Rome seeking to make amends for their loss last week to Scotland. They won comfortably by 63 points to 10, running in 9 tries. Italy were woeful, not to take anything away from Ireland, and it has re-opened the debate as to whether Italy deserve to be in the 6 Nations at all. An idea which has been floating around for a while is that there should be relegation from the 6 Nations, with the second tier European countries like Romania and Georgia in a play-off against the bottom placed country in the 6 Nations table to see who should make up the sixth place the following year.

But, one needs to bear in mind that, when France joined the then 4 Nations to make the 5 Nations (back in 1910) they barely won any matches for well over a decade. Their initial record in the 5 Nations was worse that Italy’s current record in the 6 Nations, but they went on to be one of the best teams in the world. So, in my opinion we should give Italy more time to bring their rugby up to speed, not expel them yet.

## Wales v England

This was, for most people, the big game of the weekend. Wales and England have been playing each other since 1881, and had played each other a total of 129 times prior to this weekend. With the record at 60 wins to England, 57 to Wales and 12 drawn matches, long-term they could not be more evenly matched.

It was a cracking match; at no point were the two sides more than 5 points apart, and the ferocity and competitiveness on display was something to behold. This was the best Welsh performance since beating England in the world cup in September 2014. And, it is a match that Wales could have won and should have won. At half time we were 13-8 up, and between 50 and 75 minutes we were, in my opinion, by far the better side. England looked rattled, and at 74 minutes the score was 16-14 to Wales.

With England threatening the Welsh try line, Dan Biggar made a superb interception and broke up field. Isolated, he kicked ahead and a desparate English defence kicked it into touch, giving Wales an attacking line-out in the English 22. But, we fluffed it, and this for me was the incident which lost us the match. With only 5 minutes on the clock, we could have won the line-out, and played out the remaining 5 minutes in England’s 22, or even gone for a drop goal.

But, instead, we found ourselves back under pressure near our own try line. The ball came back to centre Jonathan Davies, who kicked up-field. But, rather than kicking into touch, he kicked in-field, something the Welsh team had been doing all afternoon. It was a crazy decision; he should have belted it into the stands and given the Welsh forwards as much time as possible to slow the game down and regain their breath for the line-out. Instead, George Ford caught Davies’ in-field kick, passed it to Owen Farrell who spun a long pass to winger Elliot Daly. Daly outclassed a poor Alex Cuthbert to score in the corner, and Farrell slotted the conversion to rub salt into the wound.

We went from being in control and 16-14 ahead to losing 21-16.

Wales lost to England 21-16 in Cardiff, with England snatching a try with only 4 minutes left on the clock to break Welsh hearts.

As I say, in my opinion not only were Wales the better team, but it is a match that we lost rather than England winning it. Poor decision making, particularly in the last 5 minutes, cost us what would have been a memorable win over a very good England team. On the plus side, it was the closest this England team under Eddie Jones has been pushed during his 14 months in charge. And, Wales are finally beginning to develop their game beyond the Warren-ball of the last too many years. I hope we can back it up by winning in Scotland in two weeks’ time.

## France v Scotland

This was another close fought match, with France sneaking it 22-16. Scotland had their chances, and they are a much improved side. They have so many good back-line players that they look dangerous from anywhere on the field, and it is great to see them so competitive again. They are going to be a tough prospect for Wales in what is our next game.

## Round 3 of the 6 Nations

The 6 Nations now takes a slight hiatus; the next round of matches are in a fortnight’s time rather than this coming weekend. On Saturday 25th February, Wales head up to Murrayfield to take on Scotland, with the KO at 14:25. The second match of the day is Ireland at home to France, KO is 16:50. On Sunday 26th, England host Italy at Twickenham, with the KO at 15:00. I will preview these matches closer to the time.

## Wales v England 2017 preview

Later today Wales take on England in the 2017 6 Nations. Unfortunately for Wales, this is only our second game of the championships, and I don’t think we are up to speed yet. Yes, we won convincingly in Italy last Sunday, but our first half display showed many of our weaknesses. England, on the other hand, battled to a win against France, and with only 10 minutes to go they looked like they would lose. So, which side will be better prepared for today’s clash?

Wales have two injury worries to deal with. Both Dan Biggar (outside-half) and George North (wing) suffered injuries against Italy. They have been named in the starting 15, but Rob Howley may find that they fail final fitness tests and do not play. Should Dan Biggar not start, then Sam Davies will start in his place. Davies is a more attacking outside-half than Biggar, and personally I think that he gets the backline moving better. But, if Biggar is fit I would prefer to see him start, with Davies coming on at about 60 minutes as the game opens up for the last quarter.

Both Dan Biggar and George North are doubts for today’s clash with England. They have been named in the starting 15, but both suffered injuries against Italy last Sunday.

The other impact player which Wales will hope to bring on during the last quarter is our outstanding number 8 Taulupe Faletau. He has been injured, so has not played much rugby in the last few months. But, I am sure that he has kept himself fit enough to put in a barn-storming 20 minutes and I’m in no doubt that he will be brought on as impact replacement.

England have only won two of their last seven visits to Cardiff, so their coach Eddie Jones will be doing all he can to tell his players that they should not fear the Millennium Stadium. But, if you have ever been to this cathedral of rugby you will know that it is intimidating. England have opted to leave the roof open, which does reduce the noise in the stadium somewhat. But, with 74,500 supporters, most of them shouting for Wales, do not underestimate how intimidating that will be for England. And, with England unbeaten under Eddie Jones, there is nothing Wales would like more than to be the first country to break this winning streak. Beating England is not the be-all and end-all of Welsh rugby, but it is very sweet when we do beat our larger, noisier neighbour.

The head-to-head between Wales and England is almost tied. They have played each other 129 times. England have won 60, Wales 57, with 12 matches being drawn. Will it be 61 to 57, or 60 to 58 after today? We will know in a few hours……

DERE ‘MLAEN CYMRU!!!! / COME ON WALES!!!!


text:
Preprint ims97-16

E. de Faria and W. de Melo
Rigidity of critical circle mappings I

Abstract: We prove that two $C^r$ critical circle maps with the same rotation number of bounded type are $C^{1+\alpha}$ conjugate for some $\alpha>0$ provided their successive renormalizations converge together at an exponential rate in the $C^0$ sense. The number $\alpha$ depends only on the rate of convergence. We also give examples of $C^\infty$ critical circle maps with the same rotation number that are not $C^{1+\beta}$ conjugate for any $\beta>0$.
View ims97-16 (PDF format)

format: plain


text:
# American Institute of Mathematical Sciences

May  2013, 33(5): 1741-1771. doi: 10.3934/dcds.2013.33.1741

## Persistence of Hölder continuity for non-local integro-differential equations

 1 Department of Mathematics, University of Texas at Austin, 1 University Station, C1200, Austin, TX 78712, United States

Received  December 2011 Revised  April 2012 Published  December 2012

In this paper, we consider non-local integro-differential equations under certain natural assumptions on the kernel, and obtain persistence of Hölder continuity for their solutions. In other words, we prove that a solution stays in $C^\beta$ for all time if its initial data lies in $C^\beta$. This result has an application for a fully non-linear problem, which is used in the field of image processing. In addition, we show Hölder regularity for solutions of drift diffusion equations with supercritical fractional diffusion under the assumption $b\in L^\infty C^{1-\alpha}$ on the divergent-free drift velocity. The proof is in the spirit of [23] where Kiselev and Nazarov established Hölder continuity of the critical surface quasi-geostrophic (SQG) equation.
Citation: Kyudong Choi. Persistence of Hölder continuity for non-local integro-differential equations. Discrete & Continuous Dynamical Systems - A, 2013, 33 (5) : 1741-1771. doi: 10.3934/dcds.2013.33.1741
##### References:

show all references

##### References:
 [1] Sertan Alkan. A new solution method for nonlinear fractional integro-differential equations. Discrete & Continuous Dynamical Systems - S, 2015, 8 (6) : 1065-1077. doi: 10.3934/dcdss.2015.8.1065 [2] Eitan Tadmor, Prashant Athavale. Multiscale image representation using novel integro-differential equations. Inverse Problems & Imaging, 2009, 3 (4) : 693-710. doi: 10.3934/ipi.2009.3.693 [3] Ramasamy Subashini, Chokkalingam Ravichandran, Kasthurisamy Jothimani, Haci Mehmet Baskonus. Existence results of Hilfer integro-differential equations with fractional order. Discrete & Continuous Dynamical Systems - S, 2018, 0 (0) : 911-923. doi: 10.3934/dcdss.2020053 [4] Tonny Paul, A. Anguraj. Existence and uniqueness of nonlinear impulsive integro-differential equations. Discrete & Continuous Dynamical Systems - B, 2006, 6 (5) : 1191-1198. doi: 10.3934/dcdsb.2006.6.1191 [5] Olivier Bonnefon, Jérôme Coville, Jimmy Garnier, Lionel Roques. Inside dynamics of solutions of integro-differential equations. Discrete & Continuous Dynamical Systems - B, 2014, 19 (10) : 3057-3085. doi: 10.3934/dcdsb.2014.19.3057 [6] Eduardo Cuesta. Asymptotic behaviour of the solutions of fractional integro-differential equations and some time discretizations. Conference Publications, 2007, 2007 (Special) : 277-285. doi: 10.3934/proc.2007.2007.277 [7] Tomás Caraballo, P.E. Kloeden. Non-autonomous attractors for integro-differential evolution equations. Discrete & Continuous Dynamical Systems - S, 2009, 2 (1) : 17-36. doi: 10.3934/dcdss.2009.2.17 [8] Yubo Chen, Wan Zhuang. The extreme solutions of PBVP for integro-differential equations with caratheodory functions. Conference Publications, 1998, 1998 (Special) : 160-166. doi: 10.3934/proc.1998.1998.160 [9] Narcisa Apreutesei, Arnaud Ducrot, Vitaly Volpert. Travelling waves for integro-differential equations in population dynamics. Discrete & Continuous Dynamical Systems - B, 2009, 11 (3) : 541-561. doi: 10.3934/dcdsb.2009.11.541 [10] Tianling Jin, Jingang Xiong. Schauder estimates for solutions of linear parabolic integro-differential equations. Discrete & Continuous Dynamical Systems - A, 2015, 35 (12) : 5977-5998. doi: 10.3934/dcds.2015.35.5977 [11] Sebti Kerbal, Yang Jiang. General integro-differential equations and optimal controls on Banach spaces. Journal of Industrial & Management Optimization, 2007, 3 (1) : 119-128. doi: 10.3934/jimo.2007.3.119 [12] Patricio Felmer, Ying Wang. Qualitative properties of positive solutions for mixed integro-differential equations. Discrete & Continuous Dynamical Systems - A, 2019, 39 (1) : 369-393. doi: 10.3934/dcds.2019015 [13] Yi Cao, Jianhua Wu, Lihe Wang. Fundamental solutions of a class of homogeneous integro-differential elliptic equations. Discrete & Continuous Dynamical Systems - A, 2019, 39 (3) : 1237-1256. doi: 10.3934/dcds.2019053 [14] Frederic Abergel, Remi Tachet. A nonlinear partial integro-differential equation from mathematical finance. Discrete & Continuous Dynamical Systems - A, 2010, 27 (3) : 907-917. doi: 10.3934/dcds.2010.27.907 [15] Nestor Guillen, Russell W. Schwab. Neumann homogenization via integro-differential operators. Discrete & Continuous Dynamical Systems - A, 2016, 36 (7) : 3677-3703. doi: 10.3934/dcds.2016.36.3677 [16] Wilhelm Schlag. Spectral theory and nonlinear partial differential equations: A survey. Discrete & Continuous Dynamical Systems - A, 2006, 15 (3) : 703-723. doi: 10.3934/dcds.2006.15.703 [17] Barbara Abraham-Shrauner. Exact solutions of nonlinear partial differential equations. Discrete & Continuous Dynamical Systems - S, 2018, 11 (4) : 577-582. doi: 10.3934/dcdss.2018032 [18] Cyril Imbert, Sylvia Serfaty. Repeated games for non-linear parabolic integro-differential equations and integral curvature flows. Discrete & Continuous Dynamical Systems - A, 2011, 29 (4) : 1517-1552. doi: 10.3934/dcds.2011.29.1517 [19] Luis Silvestre. Hölder continuity for integro-differential parabolic equations with polynomial growth respect to the gradient. Discrete & Continuous Dynamical Systems - A, 2010, 28 (3) : 1069-1081. doi: 10.3934/dcds.2010.28.1069 [20] Xu Chen, Jianping Wan. Integro-differential equations for foreign currency option prices in exponential Lévy models. Discrete & Continuous Dynamical Systems - B, 2007, 8 (3) : 529-537. doi: 10.3934/dcdsb.2007.8.529

2018 Impact Factor: 1.143


text:
## Section: New Results

### Expressive power of models and formalisms for concurrency

Participants : Jesus Aranda, Romain Beauxis, Catuscia Palamidessi, Frank Valencia.

#### On the Expressive Power of Restriction in CCS with Replication

Busi et al. [40] showed that CCS! (CCS with replication instead of recursion) is Turing powerful by providing an encoding of Random Access Machines (RAMs) which preserves and reflects convergence (i.e., the existence of terminating computations). The encoding uses an unbounded number of restrictions arising from having restriction operators under the scope of replication. On the other hand, in [39] they had shown that there is no encoding of RAMs into CCS! which preserves and reflects divergence.

In [23] we have studied the expressive power of restriction and its interplay with replication. We have done this by considering several syntactic variants of CCS! which differ from each other in the use of restriction with respect to replication. We have considered three syntactic variations which do not allow the use of an unbounded number of restrictions: is the fragment of CCS! not allowing restrictions under the scope of a replication. is the restriction-free fragment of CCS! . The third variant is which extends CCS with PhillipÕs priority guards. We have shown that the use of unboundedly many restrictions in CCS! is necessary for obtaining Turing expressiveness in the sense of Busi et al. We have done this by showing that there is no encoding of RAMs into CCS which preserves and reflects convergence. We have also proved that up to failures equivalence, there is no encoding from CCS! into CCS nor from CCS into CCS . As lemmata for the above results we have proved that convergence is decidable for CCS and that language equivalence is decidable for CCS . As corollary it follows that convergence is decidable for restriction-free CCS. Finally, we have shown the expressive power of priorities by providing an encoding of RAMs in CCS : Not only does the encoding preserve and reflect convergence but it also preserves and reflects divergence (the existence of infinite computations). This is to be contrasted with the result of Busi et al. mentioned above.

#### Linearity vs persistence

In his PhD thesis [11] Aranda has presented an expressiveness study of linearity vs persistence in the asynchronous -calculus (A ), a representative process calculus, w.r.t. De Nicola and Hennessy's testing scenario which is sensitive to divergence. The work considers A and three sub-languages of it, each capturing one source of persistence: the persistent-input A -calculus (PI ), the persistent-output A -calculus (PO ) and the persistent A -calculus (P ). It is shown that, under some general conditions related to compositionality of the encoding and preservation of the infinite behaviour, there cannot be an encoding from A into a (semi)-persistent calculus preserving the must testing semantics. It also shown that, unlike for A , convergence and divergence are decidable for PO (and P ). As a consequence there is no encoding preserving and reflecting divergence or convergence from A into PO (and P ). This work confirms informal expressiveness claims in the literature of CCP.

#### Expressiveness of ntcc

In the context of the modeling expressiveness of process calculi, in [31] we have studied the suitability of the ntcc (nondeterministic timed concurrent constraint) calculus for modeling, simulating and analyzing biological systems. In particular, we have explored if it is possible to model membrane systems in ntcc. As the main contribution of this paper, we have proposed a general mechanism for modeling membrane systems in ntcc. The application of this mechanism has been illustrated with a model for the LDL cholesterol degradation pathway using membrane systems defined in ntcc. We have simulated the model of the LDL cholesterol degradation pathway by using ntccSim, a tool to run program specifications in ntcc.

#### Fairness

In [16] we have defined fair computations in the -calculus. We have followed Costa and Stirling's approach for CCS-like languages [44] , [45] but exploited a more natural labeling method of process actions to filter out unfair process executions. The new labeling has allowed us to prove all the significant properties of the original one, such as unicity, persistence and disappearance of labels. It has also turned out that the labeled -calculus is a conservative extension of the standard one. We have contrasted the existing fair testing notions [38] , [48] with those that naturally arise by imposing weak and strong fairness. This comparison provides the expressiveness of the various fair testing-based semantics and emphasizes the discriminating power of the one already proposed in the literature.

#### On the asynchronous nature of the asynchronous -calculus

In [12] and [37] we have addressed the question of what kind of asynchronous communication is exactly modeled by the asynchronous -calculus (a ). To this purpose we have defined a calculus where channels are represented explicitly as special buffer processes. The base language for is the (synchronous) -calculus, except that ordinary processes communicate only via buffers. We have compared this calculus with a , and we have shown that there is a strong correspondence between a and in the case that buffers are bags: there are indeed encodings which map each a process into a strongly asynchronous bisimilar process, and each process into a weakly asynchronous bisimilar a process. In case the buffers are queues or stacks, on the contrary, the correspondence does not hold. We have shown indeed that it is not possible to translate a stack or a queue into a weakly asynchronous bisimilar a process. Actually, for stacks we have shown an even stronger result, namely that they cannot be encoded into weakly (asynchronous) bisimilar processes in a -calculus without mixed choice.

Logo Inria


text:
# ${\boldsymbol m}_{{{\boldsymbol \chi}_{{b2}}{(2P)}}}–{\boldsymbol m}_{{{\boldsymbol \chi}_{{b1}}{(2P)}}}$ INSPIRE search

VALUE (MeV) DOCUMENT ID TECN  COMMENT
$\bf{ 13.10 \pm0.24}$ OUR AVERAGE
$12.3$ $\pm2.6$ $\pm0.6$ 1
 2014 BG
LHCB ${{\mathit p}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \mu}^{+}}{{\mathit \mu}^{-}}{{\mathit X}}$
$13.04$ $\pm0.26$
 2014 M
BABR ${{\mathit \Upsilon}{(3S)}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \gamma}}{{\mathit \mu}^{+}}{{\mathit \mu}^{-}}$
$13.5$ $\pm0.4$ $\pm0.5$ 2
 1992
CSB2 ${{\mathit e}^{+}}$ ${{\mathit e}^{-}}$ $\rightarrow$ ${{\mathit \gamma}}$ X, ${{\mathit \ell}^{+}}{{\mathit \ell}^{-}}{{\mathit \gamma}}{{\mathit \gamma}}$
1  From the ${{\mathit \chi}_{{bj}}{(2P)}}$ $\rightarrow$ ${{\mathit \Upsilon}{(1S)}}{{\mathit \gamma}}$ transition.
2  From the average photon energy for inclusive and exclusive events. Supersedes NARAIN 1991 .
References:
 AAIJ 2014BG
JHEP 1410 088 Measurement of the ${{\mathit \chi}_{{b}}{(3P)}}$ Mass and of the Relative Rate of ${{\mathit \chi}_{{b1}}{(1P)}}$ and ${{\mathit \chi}_{{b2}}{(1P)}}$ Production
 LEES 2014M
PR D90 112010 Bottomonium Spectroscopy and Radiative Transitions Involving the ${{\mathit \chi}_{{bJ}}{(1P,2P)}}$ States at BABAR
 HEINTZ 1992
PR D46 1928 ${\mathit {\mathit b}}{\mathit {\overline{\mathit b}}}$ Spectroscopy from the ${{\mathit \Upsilon}{(3S)}}$ State


text:
# Is every manifold sigma-compact?

Thanks for reading my post. Here is my question. I want to know if every surface is hemicompact, i.e., there is a compact exhaustion. I think that question could be asked for every manifold. I know that every locally compact and sigma compact space is hemicompact. Any manifold is locally compact, and second countable, so I wonder if these imply that the countable basis could be choosen in such way that every open subset in the basis is precompact, and therefore sigma compact. Thanks in advance for your answer.

-
There is a basis formed by standardly embedded balls. –  Mariano Suárez-Alvarez Dec 20 '12 at 17:21
Every metrizable manifold is $\sigma$-compact, and your definition of manifold implies metrizability. The long line is a non-metrizable manifold that is not $\sigma$-compact. –  Brian M. Scott Dec 20 '12 at 17:26
Thank you very much! –  math_failure Dec 31 '12 at 15:38


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# How do you find the axis of symmetry, and the maximum or minimum value of the function y = x^2 - 6x + 4?

Jul 25, 2016

Axis if symmetry: $\textcolor{g r e e n}{x = 3}$
Minimum value: $\textcolor{g r e e n}{y = - 5}$

#### Explanation:

A quadratic equation opens upward if the coefficient of the squared variable is greater than zero (i.e. it has a minimum value).

The minimum value is attained when the slope of the tangent (as given by the derivative) is equal to zero.

That is the minimum occurs when
$\textcolor{w h i t e}{\text{XXX}} \frac{d \left({x}^{2} - 6 x + 4\right)}{\mathrm{dx}} = 2 x - 6 = 0$
or
$\textcolor{w h i t e}{\text{XXX}} x = 3$

The axis of symmetry is a vertical line (when $x$ is the dependent variable) passing though the minimum.
Therefore the axis of symmetry is $\textcolor{g r e e n}{x = 3}$

The value of the function $y = {x}^{2} - 6 x + 4$ at the minimum (i.e. when $x = 3$) is
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{y =} {\left(3\right)}^{2} - 6 \left(3\right) + 4 = 9 - 18 + 4 = \textcolor{g r e e n}{- 5}$


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# Help Center > Badges > Custodian

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text:
## Permutation Group

A finite Group of substitutions of elements for each other. For instance, the order 4 permutation group would rearrange the elements in the order . A Substitution Group of two elements is called a Transposition. Every Substitution Group with elements can be written as a product of transpositions. For example,

Conjugacy Classes of elements which are interchanged are called Cycles (in the above example, the Cycles are ).


text:
nLab information metric

Context

Measure and probability theory

measure theory

probability theory

Contents

Idea

In information geometry, a (Fisher-)information metric is a Riemannian metric on a manifold of probability distributions over some probability space $X$ (the latter often assumed to be finite).

Definition

On a finite probability space $X \in$ Set a positive measure is a function $\rho : X \to \mathbb{R}_+$ and a probability distribution is one such that $\sum_{x \in X} \rho(x) = 1$.

This space is actually a submanifold of $\mathbb{R}_{\geq 0}^{\vert X \vert}$. For $\{\frac{\partial}{\partial x^i}\}$ the canonical basis of tangent vectors on this wedge of Cartesian space, the information metric $g$ is given by

$g(\frac{\partial}{\partial x^i}, \frac{\partial }{\partial x^j})(\rho) = \frac{1}{\rho(x^i)} \delta_{i j} \,.$

References

• L. L. Campbell, An extended Čencov characterization of the information metric Journal: Proc. Amer. Math. Soc. 98 (1986), 135-141. (AMS)

Created on June 17, 2011 17:48:10 by Urs Schreiber (89.204.137.105)


text:
# Discrete Bayes¶

words go here

Copyright 2015 Roger R Labbe Jr.

FilterPy library. http://github.com/rlabbe/filterpy

Supporting book at: https://github.com/rlabbe/Kalman-and-Bayesian-Filters-in-Python

filterpy.discrete_bayes.normalize(pdf)[source]

Normalize distribution pdf in-place so it sums to 1.0.

Returns pdf for convienence, so you can write things like:

>>> kernel = normalize(randn(7))

Parameters: pdf : ndarray discrete distribution that needs to be converted to a pdf. Converted in-place, i.e., this is modified. pdf : ndarray The converted pdf.

filterpy.discrete_bayes.update(likelihood, prior)[source]

Computes the posterior of a discrete random variable given a discrete likelihood and prior. In a typical application the likelihood will be the likelihood of a measurement matching your current environment, and the prior comes from discrete_bayes.predict().

Parameters: likelihood : ndarray, dtype=flaot array of likelihood values prior : ndarray, dtype=flaot prior pdf. posterior : ndarray, dtype=float Returns array representing the posterior.

Examples

# self driving car. Sensor returns values that can be equated to positions
# on the road. A real likelihood compuation would be much more complicated
# than this example.

prior = predict(posterior, velocity, kernel)
posterior = update(likelihood, prior)

filterpy.discrete_bayes.predict(pdf, offset, kernel, mode=u'wrap', cval=0.0)[source]

Performs the discrete Bayes filter prediction step, generating the prior.

pdf is a discrete probability distribution expressing our initial belief.

offset is an integer specifying how much we want to move to the right (negative values means move to the left)

We assume there is some noise in that offset, which we express in kernel. For example, if offset=3 and kernel=[.1, .7., .2], that means we think there is a 70% chance of moving right by 3, a 10% chance of moving 2 spaces, and a 20% chance of moving by 4.

It returns the resulting distribution.

If mode=’wrap’, then the probability distribution is wrapped around the array.

If mode=’constant’, or any other value the pdf is shifted, with cval used to fill in missing elements.

Examples

belief = [.05, .05, .05, .05, .55, .05, .05, .05, .05, .05]
prior = predict(belief, offset=2, kernel=[.1, .8, .1])


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

Whole Number Exponents

Distinguish bases and powers.

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Color Highlighted Text Notes

Vocabulary Language: English

TermDefinition
Base When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Cubed The cube of a number is the number multiplied by itself three times. For example, "two-cubed" = $2^3 = 2 \times 2 \times 2 = 8$.
Power The "power" refers to the value of the exponent. For example, $3^4$ is "three to the fourth power".
Squared Squared is the word used to refer to the exponent 2. For example, $5^2$ could be read as "5 squared". When a number is squared, the number is multiplied by itself.
Whole Numbers The whole numbers are all positive counting numbers and zero. The whole numbers are 0, 1, 2, 3, ...


text:
# calcium fluoride solubility

How to limit population growth in a utopia? You can do the experiment yourself to find out which one is more soluble, since all solubility facts and numbers we know come from experiments. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. What if the P-Value is less than 0.05, but the test statistic is also less than the critical value? Thanks for contributing an answer to Chemistry Stack Exchange! Asking for help, clarification, or responding to other answers. Nevertheless, there is also a major advantage to this. The polarisation effects are way much less due to the additional small size (relative) of the fluorine anion (low polarisability). Calculating the pH of a saturated calcium fluoride solution. Deze pagina is voor het laatst bewerkt op 17 jun 2018 om 17:05. MathJax reference. Low concentrations of fluoride in the oral fluid in the range of 0.1 ppm F − derived from dentifrices or mouth rinses have been shown to reveal a profound effect on the progression of dental caries. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. A few water molecules do not have enough strength to get in between and dissolve it. Unlike sodium fluoride, calcium fluoride is very difficult to dissolve in water. In this case, the main difference between the compounds is the bonds’ strength. Asking for help, clarification, or responding to other answers. Why is silver chloride less soluble than silver nitrate? You may also expect increase in bond length depending on the crystal structure, which may also contribute to the solubility properties of the given compound. Is ground connection in home electrical system really necessary? Is it too late for me to get into competitive chess? How to calculate mass of calcium fluoride that will dissolve in sodium fluoride solution? I have a thought of the crystal structures. Shouldn't some stars behave as black hole? What is the cost of health care in the US? Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. Now when we replace $\ce{F-}$ with larger $\ce{Cl-}$ and $\ce{Br-}$there would be again a change in the limiting radius ratio of the crystal which would enhance dissociation and hence solubility. On the other hand, when the cation size increases, let it be $\ce{Mg^2+}$ or $\ce{Ca^2+}$, they could be large enough to accommodate $\ce{F-}$ ions in their tetrahedral voids making the lattice stable. Is a software open source if its source code is published by its copyright owner but cannot be used without a commercial license? $\ce{NaCl}$ is soluble, but $\ce{CaF2}$ is not. How do we get to know the total mass of an atmosphere? Where should small utility programs store their preferences? Calcium fluoride in the fluorite state is of significant commercial importance as a fluoride source. There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in this reference. What's the implying meaning of "sentence" in "Home is the first sentence"? Chloride is less electronegative, thus $\ce{CaCl2}$ has weaker bonds, then comes $\ce{CaBr2}$ and at last $\ce{CaI2}$, the weakest of them all. MathJax reference. Making statements based on opinion; back them up with references or personal experience. However, I did some research and the Wikipedia article on $\ce{CaI2}$ claims it’s less soluble than $\ce{CaBr2}$, which is not the expected, but it does not provide a source for that information. Why is the concept of injective functions difficult for my students? How do smaller capacitors filter out higher frequencies than larger values? There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in this reference. To learn more, see our tips on writing great answers. The disadvantage of this is that it is difficult to process in oral care products. In this case, the main difference between the compounds is the bonds’ strength. Calciumfluoride is het calciumzout van waterstoffluoride, met als brutoformule CaF2. It only takes a minute to sign up. Looking for instructions for Nanoblock Synthesizer (NBC_038).

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# Proof for non-positive semi-definite covariance matrix estimator

It is well known that the standard estimator of the covariance matrix can lose the property of being positive-semidefinite if the number of variables (e.g. number of stocks) exceeds the number of observations (e.g. trading days). I think the matrix can become singular. I have a clear idea why (inspired by the geometry of the problem) but does anybody have a short but rigorous proof for this fact?

-
Me and a colleague tried to prove along the following lines:Let $X = (X_t^j)$ where $t = 1,\ldots,T$ and $j=1,\ldots,N$ and $T$ denotes the number of observations and $N$ the number of assets. Assuming that $T<N$ and that the mean returns are subtracted we have $$\hat{\Sigma} = \frac{1}{T-1} X^T X$$ and the claim follows as $$rank(\hat{\Sigma}) \le \min(rank(X),rank(X^T)) \le T$$ which implies $rank(\hat{\Sigma}) \le T < N$. Thus $\hat{\Sigma}$ is singular. Something more intuitive but still rigorous would be interesting. –  Richard Jun 29 '12 at 15:54
This isn't guaranteed to generate a positive semi-definite matrix –  pyCthon Jul 2 '12 at 2:17
This is what I am saying. But an easier proof for it, that's the question. In fact the matrix is singular for sure (as its rank is too little). –  Richard Jul 3 '12 at 20:09

The standard estimator of the covariance matrix is: $$\widehat{ \mathrm{cov}}(X) = \frac 1 {n-1} \sum_{i=1}^n (X_i-\bar X)(X_i-\bar X)^T,$$ where $X_i$ is the column vector containing the $i$th observation of all the observables. Each summand is an outer product of a vector with itself, i.e., a square matrix having rank at most one. Therefore $$\mathrm{rk\;}\widehat{\mathrm{cov}}(X) \le n$$ and the matrix not only can be but is always singular if $$n \lt \dim X,$$ i.e., if the number of observations is less than the number of variables.
Edit: regarding positive-semidefiniteness: $\widehat{ \mathrm{cov}}(X)$ is always positive-semidefinite because it is Gramian, even if its rank is not full. It loses the property of being positive-definite if and only if it is singular.


text:
What does a vertical bar mean in this context?

I'm doing a few exercises regarding Schnorr's identification scheme. I have the exercise starting off like this, with the values defined:

Let $$p = 311$$ and $$r = 31\ |\ (p - 1)$$. Let $$g = 169$$, which has order $$r$$.

I just really can't figure out what the vertical bar means here?

Sometimes in discrete maths, a vertical bar means absolute value, sometimes two them are cardinalty? Programming would suggest that it means a logical or?

One place I saw something indicating that it might be xor, but I really have no idea.

• Do you have more context? $|$ can also mean divides in that 31 divides 310, but it doesn't make sense to assign that to $r$ May 18 at 15:57
• @AmanGrewal This is apparently a widely used problem, see books.google.com/… for one example. May 18 at 16:52

The meaning of that $$\ \vert\$$ in this context is divides (as in evenly divides, or is a divisor of), and that's a standard usage of this sign. The quote should be read as:

let $$p=311$$ and $$r=31$$, which divides $$(p-1)$$. Let $$g=169$$

In other words: $$r$$ is a divisor of $$p-1$$. Or, exists integer $$q$$ with $$r\times q=p-1$$. Or, $$((p-1)\bmod r)=0$$, also writable as $$p-1\bmod r=0$$ or $$p-1\equiv0\pmod r$$ or $$p\equiv1\pmod r$$. In many common programing languages, (p-1)%r == 0. That's because $$31$$ (evenly) divides $$311-1$$, since $$31\times10=310$$.

That was correctly guessed by Aman Grewal in comment, but as noted, proximity with the assignment makes the notation confusing. Elision of the implied which is something I would try to avoid.

The end of the sentence says «Let $$g=169$$, which has order $$r$$». Does that mean that $$g$$ is really 169%31?

No. The term order is used in its meaning in group theory. In this context, it means that when we repeatedly multiply $$1$$ by $$g$$, reducing modulo $$p$$ after each multiplication, we'll first get back to $$1$$ after performing $$r$$ multiplications. That's related to $$r\,\vert\,(p-1)$$, because the order of any element in a finite group is a divisor of the order of the group, that is the number of elements in the group. Here the group is the multiplicative group modulo $$p$$, noted $$\mathbb Z_p^*$$ or $$(\mathbb Z/p\mathbb Z)^\times$$, which has $$p-1$$ elements since $$p$$ is prime. The powers of $$g$$ form a subgroup of order $$r$$, called a Schnorr group.

• Thanks, One more thing that I don't understand. The end of the sentence says Let G=169, which has order r does that mean that G is really 169%31? May 18 at 17:37
• thanks a lot again. A quick addition though,. I have just received an answe that the actual value of r is actually 8? How does this make sense? May 18 at 18:04
• @Garsty100: Asking Wolfram order of 169 modulo 311 yields 31, thus the quote is correct. I recommend studying some group theory. That serves in many fields (including the non-mathematical sense of that).
– fgrieu
May 18 at 18:42
• “Let… $r$ be 31, which divides $(p-1)$…” May 19 at 18:48


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# Three numbers

We have three different non-zero digits. We will create all 3 digits numbers from them to use all 3 figures in each number. We add all the created numbers, and we get the sum of 1554. What were the numbers?

Correct result:

a =  1
b =  2
c =  4

#### Solution:

$b=2$
$c=4$

We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!

Tips to related online calculators
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How many different three-digit numbers divisible by five can we create from the digits 2, 4, 5? The digits can be repeated in the created number.
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How to keep page numbering below text?

I have started writing my report using LaTeX. I have hid the page numbers for the first pages, table of contents included. However, after table of contents page numbers seems to be above the lowest point that the content reaches. I am using

\documentclass[a4paper,12pt]{article}
\usepackage{fullpage}
\title{\bf Title\vspace{2cm}}
\author{\emph{Author:}\\author\vspace{0.5cm} \\ \emph{Supervisor:}\\supervisor\vspace{2cm}}
\date{\today}
\begin{document}
\maketitle
\thispagestyle{empty}
\newpage
\begin{abstract}
\end{abstract}
\thispagestyle{empty}
\newpage
\renewcommand{\abstractname}{Acknowledgements}
\begin{abstract}
\end{abstract}
\thispagestyle{empty}
\newpage
\thispagestyle{empty}
\tableofcontents
\thispagestyle{empty} \newpage
\setcounter{page}{1}

and resetting counter after ToC like this :\setcounter{page}{1}
What do i have to change in order the page numbers to appear below my text?

-
Please add to your question a minimal working example illustrating the problem mentioned. –  Gonzalo Medina Mar 8 '12 at 1:10
The code you just posted doesn't reproduce the problem you mentioned; please add a proper minimal example. –  Gonzalo Medina Mar 8 '12 at 1:44

I agree with recluze regarding the selection of document class. You should not be using article class for a report.

However, if you want the page numbers to be uniformly placed at a particular height you can use geometry package (footskip=.25in) along with fancyhdr.

\documentclass[12pt,oneside]{book}
\usepackage{fancyhdr}%<-------------to control headers and footers
\usepackage[a4paper,margin=1in,showframe,footskip=.25in]{geometry}
\usepackage{lipsum}
\title{Title of the report}
\begin{document}
\maketitle
\fancyhf{}
\fancyfoot[C]{\thepage} %<----to get page number below your text
\pagestyle{fancy} %<-------the page style itself
\frontmatter
\tableofcontents
\mainmatter
\chapter{First chapter}
\section{One}
\lipsum[1-3]
\section{Two}
\lipsum[1-13]
\chapter{Second chapter}
\section{One}
\lipsum[1-3]
\section{Two}
\lipsum[1-13]
\backmatter
\end{document}

You may remove showframe from geometry options in the final version.

-
To write a report, you should use the report document class (or memoir if you want better output. I prefer it myself.) Use the memoir manual to find out how to separate frontmatter, mainmatter and backmatter. That should solve all the numbering issues.
If you want to hide the TOC from the TOC itself, use \section*{Table of Contents} (or \chapter*{Table of Contents} if you're using report or memoir). Then, post any follow up questions.


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# Kuranishi Family as Local Moduli space

Let $X$ be a compact complex manifold. People always refer the Kuranishi family of $X$ as the "local moduli space" of the moduli space of complex structures on $X$.

I am confused with this terminology. When I look at the 2-torus case $T^2$. It is known that the moduli space of complex structure on $T^2$ is given by the quotient $$\mathbb{H}/SL(2,\mathbb{Z}),$$ where $\mathbb{H}$ is the upper-half space and $SL(2,\mathbb{Z})$ acts on it by Mobius transform. However, the Kuranishi space of $T^2$ is simply given by the unit disk (or $\mathbb{H}$), because the Maurer-Cartan element is given by $$td\bar{z}\otimes\frac{\partial}{\partial z}.$$ Hence for different $t$'s, the corresponding elliptic curves can be isomorphic. So, how should I interpret the Kuranishi family as the "local moduli space"?

Thank you!

• I'm sorry I can't help much with this question (I don't know some of the concepts you're using), but is it possible that the Kuranishi family of $X$ is actually the Teichmuller space of $X$? The moduli space is a quotient of the Teichmuller space. For a little information on this, see this delightful paper. – Myridium Oct 17 '16 at 7:29
• Never mind. At least your comment leads me to read the following paper arxiv.org/pdf/1106.1368.pdf. Thank you anyway! – Marco Oct 17 '16 at 7:45


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# Coxeter–Dynkin diagram

(Redirected from Coxeter diagram)
Coxeter–Dynkin diagrams for the fundamental finite Coxeter groups
Coxeter–Dynkin diagrams for the fundamental affine Coxeter groups

In geometry, a Coxeter–Dynkin diagram (or Coxeter diagram, Coxeter graph) is a graph with numerically labeled edges (called branches) representing the spatial relations between a collection of mirrors (or reflecting hyperplanes). It describes a kaleidoscopic construction: each graph "node" represents a mirror (domain facet) and the label attached to a branch encodes the dihedral angle order between two mirrors (on a domain ridge). An unlabeled branch implicitly represents order-3.

Each diagram represents a Coxeter group, and Coxeter groups are classified by their associated diagrams.

Dynkin diagrams are closely related objects, which differ from Coxeter diagrams in two respects: firstly, branches labeled "4" or greater are directed, while Coxeter diagrams are undirected; secondly, Dynkin diagrams must satisfy an additional (crystallographic) restriction, namely that the only allowed branch labels are 2, 3, 4, and 6. Dynkin diagrams correspond to and are used to classify root systems and therefore semisimple Lie algebras.[1]

## Description

Branches of a Coxeter–Dynkin diagram are labeled with a rational number p, representing a dihedral angle of 180°/p. When p = 2 the angle is 90° and the mirrors have no interaction, so the branch can be omitted from the diagram. If a branch is unlabeled, it is assumed to have p = 3, representing an angle of 60°. Two parallel mirrors have a branch marked with "∞". In principle, n mirrors can be represented by a complete graph in which all n(n − 1) / 2 branches are drawn. In practice, nearly all interesting configurations of mirrors include a number of right angles, so the corresponding branches are omitted.

Diagrams can be labeled by their graph structure. The first forms studied by Ludwig Schläfli are the orthoschemes which have linear graphs that generate regular polytopes and regular honeycombs. Plagioschemes are simplices represented by branching graphs, and cycloschemes are simplices represented by cyclic graphs.

## Schläfli matrix

Every Coxeter diagram has a corresponding Schläfli matrix (so named after Ludwig Schläfli), with matrix elements ai,j = aj,i = −2cos (π / p) where p is the branch order between the pairs of mirrors. As a matrix of cosines, it is also called a Gramian matrix after Jørgen Pedersen Gram. All Coxeter group Schläfli matrices are symmetric because their root vectors are normalized. It is related closely to the Cartan matrix, used in the similar but directed graph Dynkin diagrams in the limited cases of p = 2,3,4, and 6, which are NOT symmetric in general.

The determinant of the Schläfli matrix, called the Schläflian, and its sign determines whether the group is finite (positive), affine (zero), indefinite (negative). This rule is called Schläfli's Criterion.[2]

The eigenvalues of the Schläfli matrix determines whether a Coxeter group is of finite type (all positive), affine type (all non-negative, at least one is zero), or indefinite type (otherwise). The indefinite type is sometimes further subdivided, e.g. into hyperbolic and other Coxeter groups. However, there are multiple non-equivalent definitions for hyperbolic Coxeter groups. We use the following definition: A Coxeter group with connected diagram is hyperbolic if it is neither of finite nor affine type, but every proper connected subdiagram is of finite or affine type. A hyperbolic Coxeter group is compact if all subgroups are finite (i.e. have positive determinants), and paracompact if all its subgroups are finite or affine (i.e. have nonnegative determinants).

Finite and affine groups are also called elliptical and parabolic respectively. Hyperbolic groups are also called Lannér, after F. Lannér who enumerated the compact hyperbolic groups in 1950,[3] and Koszul (or quasi-Lannér) for the paracompact groups.

### Rank 2 Coxeter groups

For rank 2, the type of a Coxeter group is fully determined by the determinant of the Schläfli matrix, as it is simply the product of the eigenvalues: Finite type (positive determinant), affine type (zero determinant) or hyperbolic (negative determinant). Coxeter uses an equivalent bracket notation which lists sequences of branch orders as a substitute for the node-branch graphic diagrams. Rational solutions [p/q],      , also exist, with gcd(p,q)=1, which define overlapping fundamental domains. For example, 3/2, 4/3, 5/2, 5/3, 5/4. and 6/5.

Type Finite Affine Hyperbolic
Geometry         ...
Coxeter
[ ]

[2]

[3]

[4]

[p]

[∞]

[∞]

[iπ/λ]
Order 2 4 6 8 2p
Mirror lines are colored to correspond to Coxeter diagram nodes.
Fundamental domains are alternately colored.

### Geometric visualizations

The Coxeter–Dynkin diagram can be seen as a graphic description of the fundamental domain of mirrors. A mirror represents a hyperplane within a given dimensional spherical or Euclidean or hyperbolic space. (In 2D spaces, a mirror is a line, and in 3D a mirror is a plane).

These visualizations show the fundamental domains for 2D and 3D Euclidean groups, and 2D spherical groups. For each the Coxeter diagram can be deduced by identifying the hyperplane mirrors and labelling their connectivity, ignoring 90-degree dihedral angles (order 2).

 Coxeter groups in the Euclidean plane with equivalent diagrams. Reflections are labeled as graph nodes R1, R2, etc. and are colored by their reflection order. Reflections at 90 degrees are inactive and therefore suppressed from the diagram. Parallel mirrors are connected by an ∞ labeled branch. The prismatic group ${\displaystyle {\tilde {I}}_{1}}$ x${\displaystyle {\tilde {I}}_{1}}$  is shown as a doubling of the ${\displaystyle {\tilde {C}}_{2}}$ , but can also be created as rectangular domains from doubling the ${\displaystyle {\tilde {G}}_{2}}$  triangles. The ${\displaystyle {\tilde {A}}_{2}}$  is a doubling of the ${\displaystyle {\tilde {G}}_{2}}$  triangle. Many Coxeter groups in the hyperbolic plane can be extended from the Euclidean cases as a series of hyperbolic solutions. Coxeter groups in 3-space with diagrams. Mirrors (triangle faces) are labeled by opposite vertex 0..3. Branches are colored by their reflection order.${\displaystyle {\tilde {C}}_{3}}$  fills 1/48 of the cube. ${\displaystyle {\tilde {B}}_{3}}$  fills 1/24 of the cube. ${\displaystyle {\tilde {A}}_{3}}$  fills 1/12 of the cube. Coxeter groups in the sphere with equivalent diagrams. One fundamental domain is outlined in yellow. Domain vertices (and graph branches) are colored by their reflection order.

## Finite Coxeter groups

See also polytope families for a table of end-node uniform polytopes associated with these groups.
• Three different symbols are given for the same groups – as a letter/number, as a bracketed set of numbers, and as the Coxeter diagram.
• The bifurcated Dn groups is half or alternated version of the regular Cn groups.
• The bifurcated Dn and En groups are also labeled by a superscript form [3a,b,c] where a,b,c are the numbers of segments in each of the three branches.
Connected finite Coxeter-Dynkin diagrams (ranks 1 to 9)
Rank Simple Lie groups Exceptional Lie groups
${\displaystyle {A}_{1+}}$  ${\displaystyle {B}_{2+}}$  ${\displaystyle {D}_{2+}}$  ${\displaystyle {E}_{3-8}}$  ${\displaystyle {F}_{3-4}}$  ${\displaystyle {G}_{2}}$  ${\displaystyle {H}_{2-4}}$  ${\displaystyle {I}_{2}(p)}$
1 A1=[ ]

2 A2=[3]

B2=[4]

D2=A1A1

G2=[6]

H2=[5]

I2[p]

3 A3=[32]

B3=[3,4]

D3=A3

E3=A2A1

F3=B3

H3

4 A4=[33]

B4=[32,4]

D4=[31,1,1]

E4=A4

F4

H4

5 A5=[34]

B5=[33,4]

D5=[32,1,1]

E5=D5

6 A6=[35]

B6=[34,4]

D6=[33,1,1]

E6=[32,2,1]

7 A7=[36]

B7=[35,4]

D7=[34,1,1]

E7=[33,2,1]

8 A8=[37]

B8=[36,4]

D8=[35,1,1]

E8=[34,2,1]

9 A9=[38]

B9=[37,4]

D9=[36,1,1]

10+ .. .. .. ..

## Application with uniform polytopes

 In constructing uniform polytopes, nodes are marked as active by a ring if a generator point is off the mirror, creating a new edge between a generator point and its mirror image. An unringed node represents an inactive mirror that generates no new points. Two orthogonal mirrors can be used to generate a square,    , seen here with a red generator point and 3 virtual copies across the mirrors. The generator has to be off both mirrors in this orthogonal case to generate an interior. The ring markup presumes active rings have generators equal distance from all mirrors, while a rectangle can also represent a nonuniform solution.

Coxeter–Dynkin diagrams can explicitly enumerate nearly all classes of uniform polytope and uniform tessellations. Every uniform polytope with pure reflective symmetry (all but a few special cases have pure reflectional symmetry) can be represented by a Coxeter–Dynkin diagram with permutations of markups. Each uniform polytope can be generated using such mirrors and a single generator point: mirror images create new points as reflections, then polytope edges can be defined between points and a mirror image point. Faces can be constructed by cycles of edges created, etc. To specify the generating vertex, one or more nodes are marked with rings, meaning that the vertex is not on the mirror(s) represented by the ringed node(s). (If two or more mirrors are marked, the vertex is equidistant from them.) A mirror is active (creates reflections) only with respect to points not on it. A diagram needs at least one active node to represent a polytope. An unconnected diagram (subgroups separated by order-2 branches, or orthogonal mirrors) requires at least one active node in each subgraph.

All regular polytopes, represented by Schläfli symbol {p, q, r, ...}, can have their fundamental domains represented by a set of n mirrors with a related Coxeter–Dynkin diagram of a line of nodes and branches labeled by p, q, r, ..., with the first node ringed.

Uniform polytopes with one ring correspond to generator points at the corners of the fundamental domain simplex. Two rings correspond to the edges of simplex and have a degree of freedom, with only the midpoint as the uniform solution for equal edge lengths. In general k-ring generator points are on (k-1)-faces of the simplex, and if all the nodes are ringed, the generator point is in the interior of the simplex.

A secondary markup conveys a special case nonreflectional symmetry uniform polytopes. These cases exist as alternations of reflective symmetry polytopes. This markup removes the central dot of a ringed node, called a hole (circles with nodes removed), to imply alternate nodes deleted. The resulting polytope will have a subsymmetry of the original Coxeter group. A truncated alternation is called a snub.

• A single node represents a single mirror. This is called group A1. If ringed this creates a line segment perpendicular to the mirror, represented as {}.
• Two unattached nodes represent two perpendicular mirrors. If both nodes are ringed, a rectangle can be created, or a square if the point is at equal distance from both mirrors.
• Two nodes attached by an order-n branch can create an n-gon if the point is on one mirror, and a 2n-gon if the point is off both mirrors. This forms the I1(n) group.
• Two parallel mirrors can represent an infinite polygon I1(∞) group, also called Ĩ1.
• Three mirrors in a triangle form images seen in a traditional kaleidoscope and can be represented by three nodes connected in a triangle. Repeating examples will have branches labeled as (3 3 3), (2 4 4), (2 3 6), although the last two can be drawn as a line (with the 2 branches ignored). These will generate uniform tilings.
• Three mirrors can generate uniform polyhedra; including rational numbers gives the set of Schwarz triangles.
• Three mirrors with one perpendicular to the other two can form the uniform prisms.
 There are 7 reflective uniform constructions within a general triangle, based on 7 topological generator positions within the fundamental domain. Every active mirror generates an edge, with two active mirrors have generators on the domain sides and three active mirrors has the generator in the interior. One or two degrees of freedom can be solved for a unique position for equal edge lengths of the resulting polyhedron or tiling. Example 7 generators on octahedral symmetry, fundamental domain triangle (4 3 2), with 8th snub generation as an alternation

The duals of the uniform polytopes are sometimes marked up with a perpendicular slash replacing ringed nodes, and a slash-hole for hole nodes of the snubs. For example,     represents a rectangle (as two active orthogonal mirrors), and     represents its dual polygon, the rhombus.

### Example polyhedra and tilings

For example, the B3 Coxeter group has a diagram:      . This is also called octahedral symmetry.

There are 7 convex uniform polyhedra that can be constructed from this symmetry group and 3 from its alternation subsymmetries, each with a uniquely marked up Coxeter–Dynkin diagram. The Wythoff symbol represents a special case of the Coxeter diagram for rank 3 graphs, with all 3 branch orders named, rather than suppressing the order 2 branches. The Wythoff symbol is able to handle the snub form, but not general alternations without all nodes ringed.

The same constructions can be made on disjointed (orthogonal) Coxeter groups like the uniform prisms, and can be seen more clearly as tilings of dihedrons and hosohedrons on the sphere, like this [6]×[] or [6,2] family:

In comparison, the [6,3],       family produces a parallel set of 7 uniform tilings of the Euclidean plane, and their dual tilings. There are again 3 alternations and some half symmetry version.

In the hyperbolic plane [7,3],       family produces a parallel set of uniform tilings, and their dual tilings. There is only 1 alternation (snub) since all branch orders are odd. Many other hyperbolic families of uniform tilings can be seen at uniform tilings in hyperbolic plane.

## Affine Coxeter groups

Families of convex uniform Euclidean tessellations are defined by the affine Coxeter groups. These groups are identical to the finite groups with the inclusion of one added node. In letter names they are given the same letter with a "~" above the letter. The index refers to the finite group, so the rank is the index plus 1. (Ernst Witt symbols for the affine groups are given as also)

1. ${\displaystyle {\tilde {A}}_{n-1}}$ : diagrams of this type are cycles. (Also Pn)
2. ${\displaystyle {\tilde {C}}_{n-1}}$  is associated with the hypercube regular tessellation {4, 3, ...., 4} family. (Also Rn)
3. ${\displaystyle {\tilde {B}}_{n-1}}$  related to C by one removed mirror. (Also Sn)
4. ${\displaystyle {\tilde {D}}_{n-1}}$  related to C by two removed mirrors. (Also Qn)
5. ${\displaystyle {\tilde {E}}_{6}}$ , ${\displaystyle {\tilde {E}}_{7}}$ , ${\displaystyle {\tilde {E}}_{8}}$ . (Also T7, T8, T9)
6. ${\displaystyle {\tilde {F}}_{4}}$  forms the {3,4,3,3} regular tessellation. (Also U5)
7. ${\displaystyle {\tilde {G}}_{2}}$  forms 30-60-90 triangle fundamental domains. (Also V3)
8. ${\displaystyle {\tilde {I}}_{1}}$  is two parallel mirrors. ( = ${\displaystyle {\tilde {A}}_{1}}$  = ${\displaystyle {\tilde {C}}_{1}}$ ) (Also W2)

Composite groups can also be defined as orthogonal projects. The most common use ${\displaystyle {\tilde {A}}_{1}}$ , like ${\displaystyle {\tilde {A}}_{1}^{2}}$ ,         represents square or rectangular checker board domains in the Euclidean plane. And ${\displaystyle {\tilde {A}}_{1}{\tilde {G}}_{2}}$          represents triangular prism fundamental domains in Euclidean 3-space.

Affine Coxeter graphs up to (2 to 10 nodes)
Rank ${\displaystyle {\tilde {A}}_{1+}}$  (P2+) ${\displaystyle {\tilde {B}}_{3+}}$  (S4+) ${\displaystyle {\tilde {C}}_{1+}}$  (R2+) ${\displaystyle {\tilde {D}}_{4+}}$  (Q5+) ${\displaystyle {\tilde {E}}_{n}}$  (Tn+1) / ${\displaystyle {\tilde {F}}_{4}}$  (U5) / ${\displaystyle {\tilde {G}}_{2}}$  (V3)
2 ${\displaystyle {\tilde {A}}_{1}}$ =[∞]

${\displaystyle {\tilde {C}}_{1}}$ =[∞]

3 ${\displaystyle {\tilde {A}}_{2}}$ =[3[3]]
*
${\displaystyle {\tilde {C}}_{2}}$ =[4,4]
*
${\displaystyle {\tilde {G}}_{2}}$ =[6,3]
*
4 ${\displaystyle {\tilde {A}}_{3}}$ =[3[4]]
*
${\displaystyle {\tilde {B}}_{3}}$ =[4,31,1]
*
${\displaystyle {\tilde {C}}_{3}}$ =[4,3,4]
*
${\displaystyle {\tilde {D}}_{3}}$ =[31,1,3−1,31,1]
= ${\displaystyle {\tilde {A}}_{3}}$
5 ${\displaystyle {\tilde {A}}_{4}}$ =[3[5]]
*
${\displaystyle {\tilde {B}}_{4}}$ =[4,3,31,1]
*
${\displaystyle {\tilde {C}}_{4}}$ =[4,32,4]
*
${\displaystyle {\tilde {D}}_{4}}$ =[31,1,1,1]
*
${\displaystyle {\tilde {F}}_{4}}$ =[3,4,3,3]
*
6 ${\displaystyle {\tilde {A}}_{5}}$ =[3[6]]
*
${\displaystyle {\tilde {B}}_{5}}$ =[4,32,31,1]
*
${\displaystyle {\tilde {C}}_{5}}$ =[4,33,4]
*
${\displaystyle {\tilde {D}}_{5}}$ =[31,1,3,31,1]
*

7 ${\displaystyle {\tilde {A}}_{6}}$ =[3[7]]
*
${\displaystyle {\tilde {B}}_{6}}$ =[4,33,31,1]

${\displaystyle {\tilde {C}}_{6}}$ =[4,34,4]

${\displaystyle {\tilde {D}}_{6}}$ =[31,1,32,31,1]

${\displaystyle {\tilde {E}}_{6}}$ =[32,2,2]

8 ${\displaystyle {\tilde {A}}_{7}}$ =[3[8]]
*
${\displaystyle {\tilde {B}}_{7}}$ =[4,34,31,1]
*
${\displaystyle {\tilde {C}}_{7}}$ =[4,35,4]

${\displaystyle {\tilde {D}}_{7}}$ =[31,1,33,31,1]
*
${\displaystyle {\tilde {E}}_{7}}$ =[33,3,1]
*
9 ${\displaystyle {\tilde {A}}_{8}}$ =[3[9]]
*
${\displaystyle {\tilde {B}}_{8}}$ =[4,35,31,1]

${\displaystyle {\tilde {C}}_{8}}$ =[4,36,4]

${\displaystyle {\tilde {D}}_{8}}$ =[31,1,34,31,1]

${\displaystyle {\tilde {E}}_{8}}$ =[35,2,1]
*
10 ${\displaystyle {\tilde {A}}_{9}}$ =[3[10]]
*
${\displaystyle {\tilde {B}}_{9}}$ =[4,36,31,1]

${\displaystyle {\tilde {C}}_{9}}$ =[4,37,4]

${\displaystyle {\tilde {D}}_{9}}$ =[31,1,35,31,1]

11 ... ... ... ...

## Hyperbolic Coxeter groups

There are many infinite hyperbolic Coxeter groups. Hyperbolic groups are categorized as compact or not, with compact groups having bounded fundamental domains. Compact simplex hyperbolic groups (Lannér simplices) exist as rank 3 to 5. Paracompact simplex groups (Koszul simplices) exist up to rank 10. Hypercompact (Vinberg polytopes) groups have been explored but not been fully determined. In 2006, Allcock proved that there are infinitely many compact Vinberg polytopes for dimension up to 6, and infinitely many finite-volume Vinberg polytopes for dimension up to 19,[4] so a complete enumeration is not possible. All of these fundamental reflective domains, both simplices and nonsimplices, are often called Coxeter polytopes or sometimes less accurately Coxeter polyhedra.

### Hyperbolic groups in H2

Poincaré disk model of fundamental domain triangles
Example right triangles [p,q]

[3,7]

[3,8]

[3,9]

[3,∞]

[4,5]

[4,6]

[4,7]

[4,8]

[∞,4]

[5,5]

[5,6]

[5,7]

[6,6]

[∞,∞]
Example general triangles [(p,q,r)]

[(3,3,4)]

[(3,3,5)]

[(3,3,6)]

[(3,3,7)]

[(3,3,∞)]

[(3,4,4)]

[(3,6,6)]

[(3,∞,∞)]

[(6,6,6)]

[(∞,∞,∞)]

Two-dimensional hyperbolic triangle groups exist as rank 3 Coxeter diagrams, defined by triangle (p q r) for:

${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}+{\frac {1}{r}}<1.}$

There are infinitely many compact triangular hyperbolic Coxeter groups, including linear and triangle graphs. The linear graphs exist for right triangles (with r=2).[5]

Compact hyperbolic Coxeter groups
Linear Cyclic
[p,q],      :
2(p+q)<pq

...

...

...

∞ [(p,q,r)],  : p+q+r>9
 ...

Paracompact Coxeter groups of rank 3 exist as limits to the compact ones.

Linear graphs Cyclic graphs
• [p,∞]
• [∞,∞]
• [(p,q,∞)]
• [(p,∞,∞)]
• [(∞,∞,∞)]

#### Arithmetic triangle group

The hyperbolic triangle groups that are also arithmetic groups form a finite subset. By computer search the complete list was determined by Kisao Takeuchi in his 1977 paper Arithmetic triangle groups.[6] There are 85 total, 76 compact and 9 paracompact.

Right triangles (p q 2) General triangles (p q r)
Compact groups: (76)
,      ,      ,      ,      ,      ,      ,      ,      ,       ,
,      ,      ,      ,      ,      ,
,      ,      ,      ,      ,
,      ,      ,      ,
,      ,      ,      ,      ,       ,       ,

Paracompact right triangles: (4)

,      ,      ,
General triangles: (39)
,        ,        ,        ,        ,        ,        ,
,        ,        ,        ,        ,        ,        ,         ,         ,
,        ,        ,        ,        ,        ,        ,
,        ,        ,
,        ,          ,        ,        ,        ,


text:
# Increase in entropy

1. Oct 1, 2008

### LocationX

A sealed and thermally insulated container of total volume V is divided into two equal volumes by an impermeable wall. The left half of the container is initially occupied by n moles of an ideal
gas at temperature T. Which of the following gives the change in entropy of the system when
the wall is suddenly removed and the gas expands to fill the entire volume?

thermo is a very weak subject for me... any help is appreciated

I know that entropy is s=k*g where g is the number of accessible states, but i'm not sure where to go from here

2. Oct 2, 2008

### kreil

$$\Delta S(T,P)=\frac{3}{2}Nk ln( \frac{T_2}{T_1} ) + Nk ln( \frac{V_2}{V_1} )$$

This is the state equation for the entropy in terms of temperature and volume. Since the container is thermally insulated, T2=T1 and the first term drops out since ln(1)=0. Since V2 > V1, the second term is positive and the entropy of the system will INCREASE.

3. Oct 2, 2008

### LocationX

where did you get this equation from? Is this a standard equation?

4. Oct 2, 2008


text:
SCIENTIA SINICA Informationis, Volume 47 , Issue 4 : 482-491(2017) https://doi.org/10.1360/N112016-00198

## Generality of the multi-solution phenomenon in the P3P problem

• AcceptedNov 24, 2016
• PublishedFeb 13, 2017
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### Abstract

The P3P problem is a significant single-view-based pose estimation method, and its multi-solution phenomenon has been the focus of investigation since its inception. In this work, we demonstrate from a geometrical point of view that when the distance from the camera's optical center to the three control points is relatively large, the corresponding P3P problem cannot have a unique solution in general; alternatively, or it should have at least two solutions satisfying the three original constraints. A further contribution of this study is the finding that the intersection of the three 3D pumpkin surfaces in the P3P problem can be generally deduced from the intersection of the three 2D intersecting curves of the three pumpkin surfaces with the control point plane, which provides a new way to investigate the multi-solution phenomenon. Our results may be significant for both P3P theorists and practitioners.

Appendix

Citations

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