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# Luminosity of Neowise

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So, I have been trying to figure out the bolometric luminosity of the comet C/2020 F3 (Neowise) on the 22nd of June, 2020. At that time, Neowise was $$1.4$$ AU away from Earth and had an apparent magnitude of $$m=3$$. By using the formula $$m=-2.5logBigl(frac{F}{F_O}Bigl)$$ and $$F=frac{L}{4pi D^2}$$ where $$F_0$$ is the flux of Vega, $$F$$ is the flux of the comet, $$D$$ is the distance from Earth to the comet. I found that the bolometric luminosity, $$L$$ ,of the comet Neowise was around $$7.45cdot10^{14}$$ watts when it was $$1.4$$ AU away from Earth. However, on the 22nd of June, Neowise was also $$0.4$$ AU away from the Sun. I have tried to use the flux from the Sun at this distance to determine the luminosity of Neowise again. I have found the flux from the Sun at that distance to be $$8488.26$$ $$Wm^{-2}$$ using the above formula for flux, but now using $$D=0.4 AU$$. After that I thought of calculating the luminosity by using flux from the sun times the area of Neowise. Assuming the nucleus of Neowise has a diameter of $$5 km$$, and assuming that the comet is fully illuminated,

should I assume that the nucleus of Neowise is spherical or circular? And why should the 2 luminosities be different?

I think I get it - you are trying to calculate a luminosity based on some assumption about how much light is reflected/scattered from the Sun towards the Earth - hence the relevance of the flux from the Sun at the comet.

I think it's very hard to adopt this approach. There is uncertainty about what is the appropriate geometry to use for the comet and how the reflection/scattering should be treated. It is doubtful that assuming a comet diameter of 5 km will be appropriate because I would have thought most of the light we see is scattered from a coma around the comet, not from the surface of the comet itself.

## Comet NEOWISE could be last comet visible to Earth, warn researchers

### Comet NEOWISE can be seen with 'naked eye' says the RAS

When you subscribe we will use the information you provide to send you these newsletters. Sometimes they'll include recommendations for other related newsletters or services we offer. Our Privacy Notice explains more about how we use your data, and your rights. You can unsubscribe at any time.

Comet NEOWISE is currently visible in the night's sky and will continue to be for the coming weeks as it voyages through the solar system. But it could be the last one many of us will be able to see, according to a dire warning from scientists. Light pollution is increasingly making it difficult for astronomers and amateur stargazers alike, as artificial lighting is constantly on the increase, a team of researchers say.

### Related articles

According to the Natural History museum, light pollution caused by artificial lighting is increasing by an average of six percent a year.

And as things get lighter here on Earth, the sky at night seemingly gets darker.

Gareth Dorrian, post doctoral research fellow in Space Science at the University of Birmingham, and Ian Whittaker, senior lecturer in physics from Nottingham Trent University, said everyone should take advantage of the current comet, the first to be visible from Earth since the 90s, as it could be our last.

The pair wrote in The Conversation: "So comet NEOWISE will only be seen for a few weeks near Earth while it is near perihelion (its closest approach to the Sun).

Comet NEOWISE could be last comet visible to Earth, warn researchers (Image: GETTY)

Spot Comet NEOWSISE among the light pollution (Image: GETTY)

"It will then spend thousands of years moving slowly near the other end of its orbit.

"It&rsquos aphelion (farthest point) is estimated at 630 astronomical units (AU), with one AU being the distance between the Earth and the Sun.

"With the constant increase of light pollution in the night sky the observation of comets with the naked eye is becoming much rarer.

"For now, though, NEOWISE presents a fantastic opportunity for millions of people to see a night sky phenomenon which typically only presents itself perhaps once in a decade or more. Enjoy the view!"

Light pollution is on the rise (Image: GETTY)

### Related articles

The comet will reach peak luminosity from Earth's perspective on July 22, before it voyages nearer to the Sun and swings back round - by which time Earth will be far away from its orbit.

To view it now, Mr Dorrian and Mr Whittaker said: "Even though NEOWISE is very distant from Earth, with its closest approach on July 22 being almost as far away as Mars, it is still visible in the night sky to the naked eye &ndash hovering near the northern horizon.

"The comet is estimated to currently be at magnitude 1.4 &ndash a measure of brightness astronomers use, with smaller numbers denoting brighter objects.

"Venus, which is the brightest planetary object in the sky, is about -4.

Comets, asteroids and meteors (Image: EXPRESS)

### Trending

"Comet Hale-Bopp reached a maximum magnitude of 0 in 1997 due to its exceptionally large size, while comet McNaught was visible from the southern hemisphere with a maximum magnitude of -5.5.

"NEOWISE may get brighter over the next week, but which level of brightness it reaches will depend primarily on how much material erupts from its surface rather than the distance from the Earth.

"This material consists of highly reflective water ice particles from the nucleus of the comet erupting outwards, shining when they catch the sunlight."

## Contents

When not qualified, the term "luminosity" means bolometric luminosity, which is measured either in the SI units, watts, or in terms of solar luminosities ( L ). A bolometer is the instrument used to measure radiant energy over a wide band by absorption and measurement of heating. A star also radiates neutrinos, which carry off some energy (about 2% in the case of our Sun), contributing to the star's total luminosity. [5] The IAU has defined a nominal solar luminosity of 3.828 × 10 26 W to promote publication of consistent and comparable values in units of the solar luminosity. [6]

While bolometers do exist, they cannot be used to measure even the apparent brightness of a star because they are insufficiently sensitive across the electromagnetic spectrum and because most wavelengths do not reach the surface of the Earth. In practice bolometric magnitudes are measured by taking measurements at certain wavelengths and constructing a model of the total spectrum that is most likely to match those measurements. In some cases, the process of estimation is extreme, with luminosities being calculated when less than 1% of the energy output is observed, for example with a hot Wolf-Rayet star observed only in the infrared. Bolometric luminosities can also be calculated using a bolometric correction to a luminosity in a particular passband. [7] [8]

The term luminosity is also used in relation to particular passbands such as a visual luminosity of K-band luminosity. [9] These are not generally luminosities in the strict sense of an absolute measure of radiated power, but absolute magnitudes defined for a given filter in a photometric system. Several different photometric systems exist. Some such as the UBV or Johnson system are defined against photometric standard stars, while others such as the AB system are defined in terms of a spectral flux density. [10]

A star's luminosity can be determined from two stellar characteristics: size and effective temperature. [11] The former is typically represented in terms of solar radii, R, while the latter is represented in kelvins, but in most cases neither can be measured directly. To determine a star's radius, two other metrics are needed: the star's angular diameter and its distance from Earth. Both can be measured with great accuracy in certain cases, with cool supergiants often having large angular diameters, and some cool evolved stars having masers in their atmospheres that can be used to measure the parallax using VLBI. However, for most stars the angular diameter or parallax, or both, are far below our ability to measure with any certainty. Since the effective temperature is merely a number that represents the temperature of a black body that would reproduce the luminosity, it obviously cannot be measured directly, but it can be estimated from the spectrum.

An alternative way to measure stellar luminosity is to measure the star's apparent brightness and distance. A third component needed to derive the luminosity is the degree of interstellar extinction that is present, a condition that usually arises because of gas and dust present in the interstellar medium (ISM), the Earth's atmosphere, and circumstellar matter. Consequently, one of astronomy's central challenges in determining a star's luminosity is to derive accurate measurements for each of these components, without which an accurate luminosity figure remains elusive. [12] Extinction can only be measured directly if the actual and observed luminosities are both known, but it can be estimated from the observed colour of a star, using models of the expected level of reddening from the interstellar medium.

In the current system of stellar classification, stars are grouped according to temperature, with the massive, very young and energetic Class O stars boasting temperatures in excess of 30,000 K while the less massive, typically older Class M stars exhibit temperatures less than 3,500 K. Because luminosity is proportional to temperature to the fourth power, the large variation in stellar temperatures produces an even vaster variation in stellar luminosity. [13] Because the luminosity depends on a high power of the stellar mass, high mass luminous stars have much shorter lifetimes. The most luminous stars are always young stars, no more than a few million years for the most extreme. In the Hertzsprung–Russell diagram, the x-axis represents temperature or spectral type while the y-axis represents luminosity or magnitude. The vast majority of stars are found along the main sequence with blue Class O stars found at the top left of the chart while red Class M stars fall to the bottom right. Certain stars like Deneb and Betelgeuse are found above and to the right of the main sequence, more luminous or cooler than their equivalents on the main sequence. Increased luminosity at the same temperature, or alternatively cooler temperature at the same luminosity, indicates that these stars are larger than those on the main sequence and they are called giants or supergiants.

Blue and white supergiants are high luminosity stars somewhat cooler than the most luminous main sequence stars. A star like Deneb, for example, has a luminosity around 200,000 L, a spectral type of A2, and an effective temperature around 8,500 K, meaning it has a radius around 203 R (1.41 × 10 11 m). For comparison, the red supergiant Betelgeuse has a luminosity around 100,000 L, a spectral type of M2, and a temperature around 3,500 K, meaning its radius is about 1,000 R (7.0 × 10 11 m). Red supergiants are the largest type of star, but the most luminous are much smaller and hotter, with temperatures up to 50,000 K and more and luminosities of several million L, meaning their radii are just a few tens of R. For example, R136a1 has a temperature over 46,000 K and a luminosity of more than 6,100,000 L [14] (mostly in the UV), it is only 39 R (2.7 × 10 10 m).

The luminosity of a radio source is measured in W Hz −1 , to avoid having to specify a bandwidth over which it is measured. The observed strength, or flux density, of a radio source is measured in Jansky where 1 Jy = 10 −26 W m −2 Hz −1 .

For example, consider a 10 W transmitter at a distance of 1 million metres, radiating over a bandwidth of 1 MHz. By the time that power has reached the observer, the power is spread over the surface of a sphere with area 4 π r 2 or about 1.26×10 13 m 2 , so its flux density is 10 / 10 6 / 1.26×10 13 W m −2 Hz −1 = 10 8 Jy .

More generally, for sources at cosmological distances, a k-correction must be made for the spectral index α of the source, and a relativistic correction must be made for the fact that the frequency scale in the emitted rest frame is different from that in the observer's rest frame. So the full expression for radio luminosity, assuming isotropic emission, is

For example, consider a 1 Jy signal from a radio source at a redshift of 1, at a frequency of 1.4 GHz. Ned Wright's cosmology calculator calculates a luminosity distance for a redshift of 1 to be 6701 Mpc = 2×10 26 m giving a radio luminosity of 10 −26 × 4 π (2×10 26 ) 2 / (1 + 1) (1 + 2) = 6×10 26 W Hz −1 .

To calculate the total radio power, this luminosity must be integrated over the bandwidth of the emission. A common assumption is to set the bandwidth to the observing frequency, which effectively assumes the power radiated has uniform intensity from zero frequency up to the observing frequency. In the case above, the total power is 4×10 27 × 1.4×10 9 = 5.7×10 36 W . This is sometimes expressed in terms of the total (i.e. integrated over all wavelengths) luminosity of the Sun which is 3.86×10 26 W , giving a radio power of 1.5×10 10 L .

The Stefan–Boltzmann equation applied to a black body gives the value for luminosity for a black body, an idealized object which is perfectly opaque and non-reflecting: [11]

where A is the surface area, T is the temperature (in kelvins) and σ is the Stefan–Boltzmann constant, with a value of 5.670 374 419 . × 10 −8 W⋅m −2 ⋅K −4 . [16]

Imagine a point source of light of luminosity L that radiates equally in all directions. A hollow sphere centered on the point would have its entire interior surface illuminated. As the radius increases, the surface area will also increase, and the constant luminosity has more surface area to illuminate, leading to a decrease in observed brightness.

The surface area of a sphere with radius r is A = 4 π r 2 > , so for stars and other point sources of light:

For stars on the main sequence, luminosity is also related to mass approximately as below:

If we define M as the mass of the star in terms of solar masses, the above relationship can be simplified as follows:

Luminosity is an intrinsic measurable property of a star independent of distance. The concept of magnitude, on the other hand, incorporates distance. The apparent magnitude is a measure of the diminishing flux of light as a result of distance according to the inverse-square law. [17] The Pogson logarithmic scale is used to measure both apparent and absolute magnitudes, the latter corresponding to the brightness of a star or other celestial body as seen if it would be located at an interstellar distance of 10 parsecs (3.1 × 10 17 metres). In addition to this brightness decrease from increased distance, there is an extra decrease of brightness due to extinction from intervening interstellar dust. [18]

By measuring the width of certain absorption lines in the stellar spectrum, it is often possible to assign a certain luminosity class to a star without knowing its distance. Thus a fair measure of its absolute magnitude can be determined without knowing its distance nor the interstellar extinction.

In measuring star brightnesses, absolute magnitude, apparent magnitude, and distance are interrelated parameters—if two are known, the third can be determined. Since the Sun's luminosity is the standard, comparing these parameters with the Sun's apparent magnitude and distance is the easiest way to remember how to convert between them, although officially, zero point values are defined by the IAU.

The magnitude of a star, a unitless measure, is a logarithmic scale of observed visible brightness. The apparent magnitude is the observed visible brightness from Earth which depends on the distance of the object. The absolute magnitude is the apparent magnitude at a distance of 10 pc (3.1 × 10 17 m), therefore the bolometric absolute magnitude is a logarithmic measure of the bolometric luminosity.

The difference in bolometric magnitude between two objects is related to their luminosity ratio according to:

The zero point of the absolute magnitude scale is actually defined as a fixed luminosity of 3.0128 × 10 28 W . Therefore, the absolute magnitude can be calculated from a luminosity in watts:

where L0 is the zero point luminosity 3.0128 × 10 28 W

and the luminosity in watts can be calculated from an absolute magnitude (although absolute magnitudes are often not measured relative to an absolute flux):

## How To Find Clear Skies

The easiest way to make sure you aren’t “skunked” by clouds is, of course, to simply check the weather! Indeed, there are tons of apps available that can help you do this, but as a landscape and nightscape photographer, I have found a couple that are the most helpful for not just predicting perfectly clear skies, but also things like predicting the perfect sunset!

The first app I always check is called MeteoBlue, it is unique because it gathers forecasts from as many different actual meteorological (weather) stations as possible, and presents them all to you. In addition to that, it actually shows a very useful graphic of cloud cover and elevation, so, for example, you can predict if there will be low-lying fog, or high-altitude clouds. (The latter of which are what often make for epic sunsets and sunrises, by the way!)

## Is it possible to see the tail of neowise without binoculars/telescope?

Im looking up at the sky right now at the big dipper. All the bright dots pretty much look the same. I don't know if one of those dots is neowise. If it helps I'm in the bay area.

You’re looking for something that kind of looks like a smudge

Having seen it helps now its its dimmer. But like the other dude said you're looking for a faint smudge in the sky. Almost like a dim white line.

It's about midway between the big dipper and the horizon. But it has lost a lot of luminosity, so it's barely visible with the naked eye.

Maybe but doubtful. It's dimming quickly now. Use binoculars.

It was very clearly visible a few weeks ago. Right now (lat 48 deg N) it was just a tiny smudge with a faint streak above it. Possible to see if you knew where to look and what to look for but way dimmer than before.

## Comet NEOWISE

Great shot. Very nice work capturing the nucleus. Probably the best shot I've seen in that regard. Well done indeed.

Seeing it in person has been great. Went out the last two nights at 3:30 am.

What a sky. Jupiter, Saturn, Mars, Venus, a waning moon and a comet!

Yes, the past week has been amazing!

We can see all of that right now? Sorry I'm new to the community

I’m jealous! I’ve seen all the planets and moon and even Jupiter’s moons and saturns rings for the first time with my eye (telescope and not a picture in a book or online). Hoping to see the comet before it’s gone.

Nikon 70-200 lens at 135mm

Stacking: Deep Sky Stacker

Photoshop: Levels, Curves, Camera Raw

Follow my Instagram for more photos like these: hk.astro

I wish I could catch it but I think too low in horizon, living in suburbs sucks

Same here! I traveled to a park about 20-30 minutes away to get this (well not a park a soccer field inside a commercial space)

Look for a reservoir, a lake, a flood zone. I saw it at our local flood reservoir.

Same case here. My eyes are literally watering now as I've been awake for over an hour now by the time of writing this comment. It's too damn low and the sunrise is inbound in 33 minutes from now. I will try later this evening to get the best shot. Meanwhile please tell me whether I'll ever be able to view it with naked eyes or do I need a telescope ASAP from the data - screenshots I took from my Mob. Observatory Pro 3 app?

## Luminosity of Neowise - Astronomy

The comet C/2020 F3 (NEOWISE) was discovered in March 2020 by a space probe named NEOWISE. On 03 July 2020 Neowise passed the point of its orbit closest to the sun, the so-called perihelion. Since then it has been moving away from it again. On 23 July the comet came within 103 million kilometers of our Earth. Especially the first image of the comet shows that it has formed a second bluish tail in addition to the whitish comet tail. Before a tail can develop at all, the comet must come relatively close to the sun. Actually, a comet like NEOWISE spends most of its existence in the icy cold of a great distance from the sun beyond the orbit of Neptune. The C in its name refers to this circumstance. The C receive non-periodic comets with an orbital period of over 200 years. For comparison: The Neptune year lasts 165 earth years.

Comets are dirty snowballs. When a comet - for whatever reason - crashes towards the sun, its icy components sublimate: Ice suddenly turns into gas that carries away dust particles. At about the distance of Jupiter, a cloud of gas and dust - called a coma - forms around the comet nucleus, which is actually only a few kilometers in size. When the solar wind and solar radiation hit this coma, the particles are entrained. A tail is formed. By the way: Ice here means not only water ice, but also frozen ammonia, methane, carbon monoxide, carbon dioxide, methanol, ethane, hydrogen sulfide and other substances that are not solid substances at the distance of our earth from the sun. There are two different mechanisms for the formation of the tail: The pressure of solar radiation (light pressure) slowly sets the dust particles in motion and forms the dust tail. The inert mass of the dust particles also gives them an impulse in the direction of the original orbit of the comet. Therefore the dust tail is curved.

## Comet NEOWISE could be last comet visible to Earth, warn researchers

### Comet NEOWISE can be seen with 'naked eye' says the RAS

When you subscribe we will use the information you provide to send you these newsletters. Sometimes they'll include recommendations for other related newsletters or services we offer. Our Privacy Notice explains more about how we use your data, and your rights. You can unsubscribe at any time.

Comet NEOWISE is currently visible in the night's sky and will continue to be for the coming weeks as it voyages through the solar system. But it could be the last one many of us will be able to see, according to a dire warning from scientists. Light pollution is increasingly making it difficult for astronomers and amateur stargazers alike, as artificial lighting is constantly on the increase, a team of researchers say.

### Related articles

According to the Natural History museum, light pollution caused by artificial lighting is increasing by an average of six percent a year.

And as things get lighter here on Earth, the sky at night seemingly gets darker.

Gareth Dorrian, post doctoral research fellow in Space Science at the University of Birmingham, and Ian Whittaker, senior lecturer in physics from Nottingham Trent University, said everyone should take advantage of the current comet, the first to be visible from Earth since the 90s, as it could be our last.

The pair wrote in The Conversation: "So comet NEOWISE will only be seen for a few weeks near Earth while it is near perihelion (its closest approach to the Sun).

Comet NEOWISE could be last comet visible to Earth, warn researchers (Image: GETTY)

Spot Comet NEOWSISE among the light pollution (Image: GETTY)

"It will then spend thousands of years moving slowly near the other end of its orbit.

"It&rsquos aphelion (farthest point) is estimated at 630 astronomical units (AU), with one AU being the distance between the Earth and the Sun.

"With the constant increase of light pollution in the night sky the observation of comets with the naked eye is becoming much rarer.

"For now, though, NEOWISE presents a fantastic opportunity for millions of people to see a night sky phenomenon which typically only presents itself perhaps once in a decade or more. Enjoy the view!"

Light pollution is on the rise (Image: GETTY)

### Related articles

The comet will reach peak luminosity from Earth's perspective on July 22, before it voyages nearer to the Sun and swings back round - by which time Earth will be far away from its orbit.

To view it now, Mr Dorrian and Mr Whittaker said: "Even though NEOWISE is very distant from Earth, with its closest approach on July 22 being almost as far away as Mars, it is still visible in the night sky to the naked eye &ndash hovering near the northern horizon.

"The comet is estimated to currently be at magnitude 1.4 &ndash a measure of brightness astronomers use, with smaller numbers denoting brighter objects.

"Venus, which is the brightest planetary object in the sky, is about -4.

Comets, asteroids and meteors (Image: EXPRESS)

### Trending

"Comet Hale-Bopp reached a maximum magnitude of 0 in 1997 due to its exceptionally large size, while comet McNaught was visible from the southern hemisphere with a maximum magnitude of -5.5.

"NEOWISE may get brighter over the next week, but which level of brightness it reaches will depend primarily on how much material erupts from its surface rather than the distance from the Earth.

"This material consists of highly reflective water ice particles from the nucleus of the comet erupting outwards, shining when they catch the sunlight."

## Enjoy the view on the comet Neowise from the ISS

I believe your social media timeline is covered in photos of the comet Neowise. Perhaps you’ve taken some photos yourself, too. But here’s something a little bit different. The astronauts on board the International Space Station (ISS) have captured photos of the comet from some 254 miles above the Earth. And then, Seán Doran turned them into this beautiful 7-minute video.

Seán is a visual artist and he’s interested in science and astronomy. We have seen painting-like photos of Jupiter he has processed, as well as an animation of Jupiter that he created. To make the video of Neowise from the ISS, he used NASA’s timelapse photography from 5 July 2020. He repaired and processed the material, and converted it to the real-time video you can see above.

I tried finding the comet this weekend in the countryside. I was without my camera and without luck – because I couldn’t see it with the naked eye. I plan to try my luck again tomorrow night. I came back to my flat for the camera, I’m traveling again tomorrow, and this video has made even more determined to find it and add one more photo of Neowise to people’s timeline. :)

## NEOWISE from Arizona

This was taken on Wednesday night (7/22) from Dateland, AZ.

That was the closest clear skies I could find that night..

Amazing capture. Imagine a time lapse from the front of the comet, all the things it would see, all the distance it travels. Would be mind blowing!

Still no views here in North India :(

Nice picture! Great capture of the ion tail.

Thanks! This was on 7/24 around 10pm..

bro imagine not having light pollution

When is a good time to see it

Last week. It's all but invisible to the naked eye now.

I'm a bot, bleep, bloop. Someone has linked to this thread from another place on reddit:

If you follow any of the above links, please respect the rules of reddit and don't vote in the other threads. ( Info / ^Contact)


text:
#### 题目列表

Those who criticized Coco Chanels later clothing designs misjudged as _____ the style that generations of women to come would regard as the epitome of high fashion.
Since receiving a promotion to departmental chair, Brookstone has been even more prone to _____ against the universitys administration, and consequently has lost several professional allies.
Sergeis belief in astrology, a pseudoscience whose practitioners provide results than can never be conclusively proven or falsified, left him vulnerable to _____.
The _____ international aid agencies have toward selecting a fresh cause to champion approximately every five years is indicative of their desire to avoid apathy due to overexposure and, instead, continue to stimulate donor interest.
Discussions about the use of high-fructose corn syrup as a sweetener lead to _____ among nutritionists, and the use of aspartame, which is also common, is just as controversial.
Further recognition of the destructive effects of targeting fast food advertising at young children may lead to more efforts to _____ such tactics: there is already clear evidence that the necessary prohibitive regulations have widespread support.
Even though the judge personally found the law _____, his moral objection did not provide a legal basis on which to rule the law unconstitutional.
Completely impenetrable to the layperson, the ancient text was _____ even to experts in the field.
The feudalism practiced by Carolingian rulers was still in its _____ stage; only later did features such as subinfeudation and the consequent necessity of designating a liege lord lead to the fully-developed system familiar to students of the High Middle Ages.
A key element of The Smiths recognizable sound came from the tension between Morrisseys _____ lyrics and the cheerful, almost bouncy music composed by Johnny Marr.
The bride was mortified to learn that her dress-which had appeared delicate, even _____, in the artificial light of the boutique-was nearly transparent in the bright sunlight of her outdoor wedding.
The publics fascination with celebrities coupled with the innovations of the electronic age may inspire a new cadre of amateur "paparazzi": there are certainly indications that such a trend is _____.
The doctors real mistake, from the perspective of his _____ professional friends who quickly jilted him, was not that his choice of treatment was inappropriate, but rather that it was viscerally objectionable to the medical establishment.
By reordering the sentences in the problem and adding distracting figures, the professor successfully _____ a previously easy exam question and made it almost impossible to solve.
Because she always had the correct answers to lifes difficult dilemmas, my grandmother was sought after for her _____ by family members and neighbors.
The _____ road, made what seemed like a short trip on a map much longer in reality; it twisted its way through mountains to get from one valley to another.
Classmates who had pre-judged Lucy as unaware were surprised when she made the _____ observation that their professors missing coffee mug indicated that he had left for the day.
Madelines guests all agreed that had it not been for the terrible weather, her wedding day, complete with white dress and three-tiered cake, would have been _____.
The Shakespeare scholar argued that in all of the playwright`s 37 works, he had never written the part of a _____ character, only relying on vibrant and colorful individuals to propel his stories forward.
The intricate, complex photographic process of the daguerreotype contributed to the rapid development of numerous related processes, a _____ that included tintypes and calotypes.

25000 +道题目

121本备考书籍

format: plain


text:
## Tuesday, 2 December 2014

### Week 12: A Fun Induction Problem

#### Introducing a Fellow Slogger:

Please note that this is not a problem solving episode (see Week 9).  This final week, the instructors introduce simple induction.  A fellow slogger gives a nice induction post under Week 12 here.

You can also find my comments to the following slog under Week 3:

What is really cool about the induction post is that the slogger talks about how induction is used in a proof of the following statement: the determinant of any $n \times n$ identity matrix is $1$.  In fact, I am so excited that a fellow slogger uses CSC165 concepts for greater and better things that I feel compelled to follow the slogger's footsteps.  I will also use CSC165 concepts as a guide for a problem that is outside the scope of this course:

#### The Fun Problem:

Problem:  Is the following statement true: the determinant of any upper triangular matrix is
the product of its diagonal entries?

The truth is I have no clue.  In my Problem Solving Episode under Week 9 (under the heading Step 1: Understand the Problem), I work out examples of an idea that I do not fully understand to gain intuition.  We should work out examples for this problem too.

First we know that an upper triangular matrix is a $n \times n$ matrix $A$ that is defined by:

$A_{ij} = \begin{cases} a_{ij} & i \leq j \\ 0 & i > j \end{cases}$.

Then the following matrices satisfy the definition nicely:

$\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}, \begin{bmatrix} 4 & 5 & 6 \\ 0 & 7 & 8 \\ 0 & 0 & 9 \end{bmatrix}$.

Secondly, let $A$ be a $n \times n$ matrix and $F$ be a field.   Then the determinant is a function $det: M_{n \times n}(F) \to F$ that is defined by:

$det(A) = \begin{cases} A_{ij} & n = 1 \\ \sum\limits_{j= 1}^n (-1)^{i + j} A_{ij} det(\bar{A_{ij}}) & n \geq 2 \end{cases}$.

There is some fancy notation here but it is not difficult to understand.  For simplicity sakes, think of $F$ as the set of real numbers.  Also, the strange bar on top of $A_{ij}$ just means $A$ with both row $i$ and column $j$ removed.  We see that the determinant is recursively defined.  This may be useful later.

To find the determinant of the above matrices, use the definition.  For the $2 \times 2$ matrix, we have:

$det(\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}) = (-1)^{1 + 1}(1)(3) + (-1)^{1 + 2}(0) = 3$.

For the $3 \times 3$ matrix, we have:

$det(\begin{bmatrix} 4 & 5 & 6 \\ 0 & 7 & 8 \\ 0 & 0 & 9\end{bmatrix}) =$

$(-1)^{1 + 1}(4)det(\begin{bmatrix} 7 & 8 \\ 0 & 9 \end{bmatrix}) + (-1)^{1 + 2}(5)det(\begin{bmatrix} 0 & 8 \\ 0 & 9 \end{bmatrix}) + (-1)^{1 + 3}(6)det(\begin{bmatrix} 0 & 7 \\ 0 & 0 \end{bmatrix}) =$

$(4)(-1)^{1 + 1}(7)(8) + (-1)(5)(-1)^{1 + 1}(0)(9) + (5)(-1)^{1 + 1}(0)(0) =$

$(4)(7)(8) + 0 + 0 = 224$.

It seems like the statement is true and that we should try proving it.  Something interesting in the calculation is that the determinant of a matrix with a column of $0$s is $0$.  Again, this may be useful later.

#### The Attempt at a Solution:

How should we prove the statement?  A direct proof comes to mind but remember that the determinant is recursively defined.  This makes it difficult to work with unless we know the determinant of a previous recursive case.  Proof by simple induction sounds attractive here.  The induction hypothesis may give us the determinant of a previous recursive case.  Let us try it.

Since a $0 \times 0$ matrix is undefined, then we show that $\forall n \in N, n \geq 1 \implies$ the determinant of an upper triangular $n \times n$ matrix is the product of its diagonal entries.  By simple induction on $n$, we show that this statement is true.  The CSC165 proof structure for induction is our guide.

For the base case, assume that $n = 1$.  Let $A$ be a $1 \times 1$ upper triangular matrix.  By the definition of the determinant, then $det(A) = A_{ij}$.  Since $A_{ij}$ is the only diagonal entry, then statement is true for $n = 1$.

For the induction step, assume that $n \in N, n \geq 1$ and that the statement is true for $n$.  That will be our induction hypothesis.  We need to show that the statement is true for $n + 1$.  Let $A$ be a $(n + 1) \times (n + 1)$ upper triangular matrix.   By the definition of the determinant, we have that

$det(A) = \sum\limits_{j = 1}^{n + 1} (-1)^{1 + j}A_{1j} det(\bar{A_{1j}})$.

Remember this observation: the determinant of a matrix with a column of $0$s is $0$?  Turns out, that is because a square matrix with a column of $0$s is not invertible and some fancy theorem tells us that the determinant of such a matrix is $0$.

Luckily, we make the following observation:

$det(A) = \sum\limits_{j = 1}^{n + 1} (-1)^{1 + j} A_{1j} det(\bar{A_{1j}}) = (-1) ^{1 + 1}A_{11} det(\bar{A_{11}}) + \sum\limits_{j = 2}^{n + 1} (-1)^{1 + j}A_{1j} det(\bar{A_{1j}})$.

Notice that the matrix $\bar{A_{1j}}$ is $A$ without row $1$ and without a column $j \geq 2$.  Minus the first element, column $1$ of $A$ is a column of $0$s  So the matrix $\bar{A_{1j}}$ also has a column of $0$s.  Thus, $det(\bar{A_{1j}}) = 0$  and

$det(A) = (-1) ^{1 + 1}A_{11} det(\bar{A_{11}}) + \sum\limits_{j = 2}^{n + 1} (-1)^{1 + j}A_{1j} det(\bar{A_{1j}}) = (-1) ^{1 + 1}A_{11} det(\bar{A_{11}})$.

The matrix $\bar{A_{11}}$ has both row $1$ and column $1$ removed but it is still an upper triangular $n \times n$ matrix.  By the induction hypothesis, then $det(\bar{A_{11}})$ is the product of its diagonal entries.  Notice that the diagonal entries of $det(\bar{A_{11}})$  are the diagonal entries of $A$ without $A_{11}$.  Since

$det(A) = (-1) ^{1 + 1}A_{11} det(\bar{A_{11}}) = A_{11} det(\bar{A_{11}})$,

then we have $det(A)$ as the product of its diagonal entries.  Since $A$ is a $(n + 1) \times (n + 1)$ upper triangular matrix, then the statement is true for $n + 1$.  By simple induction, then $\forall n \in N, n \geq 1 \implies$ the statement is true for $n$.  So that is my attempt is at this fun problem.

## Wednesday, 26 November 2014

### Week 11: Harder Time Complexity Problems

There is no class this week so for fun, we go over a few harder time complexity problems.  Consider the sequences $f(n) = n^4 - 165n^3 - 240n^2 - 265n - 373$ and  $g(n) = 157n^3 + 257n^2 + 337n + 457$.

Problem 1: Show that $f \notin O(g)$ without using any limit definitions.

Problem 2:  Show that $f \in \Omega (g)$ without using any limit definitions.

#### Problem 1 Tips:

The definition that we have is $f \in O(g)$, which is defined as:

$\exists c \in R^+, \exists B \in N, ( \forall n \in N, n \geq B \implies f(n) \leq g(n))$.

Thus, its negation is:

$\forall c \in R^+, \forall B \in N, ( \exists n \in N, n \geq B \wedge f(n) > g(n))$.

We must choose $n$.  Using the recommended strategies of my instructors, we get:

$n^4 - 165n^3 - 240n^2 - 265n - 373 \geq$

$n^4 - 1043n^3$

$???$

$c(1208n^3) \geq$

$c(157n^3 + 257n^2 + 337n + 457)$.

The $???$ part indicates the area where the strategies of my instructors do not cover.  So I think hard here (like really hard) and I come up with this.  All we must do is find a $n$ that satisfies the inequality $n^4 - 1043n^3 > c(1208n^3)$.  We have:

$n^4 - 1043n^3 > c(1208n^3)$

Move $-1043n^3$ on the left side to the right and factor out $n^3$ from the resulting right:

$n^4 > n^3(c1208 + 1043)$

It looks like we will choose $n > 0$.  That way, we can divide both sides by $n^3$:
$n > c1208 + 1043$.

Our choice of $n$ is then $max(\lceil c1208 + 1043 \rceil + 1, B)$.

#### Problem 2 Tips:

Likewise, we show that $f \in \Omega (g)$, which is defined as:

$\exists c \in R^+, \exists B \in N, ( \forall n \in N, n \geq B \implies f(n) \geq g(n))$.

We must choose $c, B$.  We have a similar inequality to problem 1:

$n^4 - 1043n^3 \geq 1208n^3$

Again, we choose $B >0$ so that we can divide both sides by $n^3$.  Then add both sides by $1043$ to yield:

$n \geq 2051$.

Of course, to choose $B = 2051$, we must choose $c = 1$.  Otherwise, the first step is irreversible.

## Tuesday, 18 November 2014

### Week 10: Computability and Confusion

Computability is the first topic in CSC165 that is new to me.  It is both interesting and somewhat confusing.  Unfortunately, any confusion is undesirable at this point in the course.  After thinking about my confusion, I realize that the halting problem is the culprit.  Why is this?

#### Elementary Definitions and Theorems:

First, the stuff that makes sense.  Consider a set of inputs $I$ and a set of outputs $O$.  Computability begins with three relatively simple definitions and a single theorem.

Definition 1:  Let $f: I \to O$ be a function.  Then $f$ is well-defined $\iff$ for each $x \in I$, there is a unique $y \in O$ such that $f(x) = y$.

For remaining definitions and theorems, let $g, h: I \to O$ be well-defined functions.

Definition 2:  Then $g$ is computable $\iff$ there is a program $P$ such that $\forall x \in I, g(x) = P(x)$.

Definition 3:  Then $g$ reduces to $h \iff g$ can be implemented with $h$.  Below is what $g, h$ look like in Python code if $g$ reduces to $h$.  Let $P$ be the implementation of $g$ and $Q$ be the implementation of $h$.

def P():
"""An implementation of the function f."""
Q():

Theorem 1:  Let $g$ reduce to $h$.  Then $h$ is computable $\implies g$ is computable.

#### The Confusing Part:

Now, the stuff that is not so clear.  The function halt(f, i) tells us if a function $f$ with an input $i$ halts.  The halting function is not computable.  The proof of this statement starts simple enough:  Assume to the contrary that there is an implementation of halt(f, i) that works for all functions and inputs.  Then consider the function what_is_love(f):

def what_is_love(f):

if halt(f, f):
while True:
print(“Look to the skies.”)
else:
print(“Nobody knows.”)

The “while True” statement is the incomprehensible part of the proof.  I ask in lecture why it is there.  The instructor lets a classmate several rows ahead of me answer my question.  The explanation takes nearly half a minute.  I did not get any of it.  Apparently, the instructor did not have any problems with the explanation nor did my other classmates.

#### The Comprehending Part:

It takes several days for me to understand why the “while True” statement is there.  Eventually, I look at what we are proving.  We assume that there is an implementation of halt(f, i) that works for all functions and inputs.  To disprove, we only have to show that one function and input does not work for halt(f, i).  This is a classic existential proof.  We are free to construct the function what_is_love(f) however we want to so long as it disproves the computability of halt(f, i).

The “while True” statement guarantees that what_is_love(what_is_love) goes into an infinite loop if halt(what_is_love, what_is_love) returns True.  Likewise, what_is_love(what_is_love) halts if halt(what_is_love, what_is_love) returns False.  This is the contradiction and it disproves the computability of halt(f, i).  I wonder why that classmate who answered my question did not explain it like this....

## Wednesday, 12 November 2014

### Week 9: Problem Solving Episode: The Stockbroker's Algorithm

#### Context

So you graduate from UofT and land your dream job.  You work a few years and realize there is this thing called inflation.  So you get a brokerage firm to invest your savings.

To guide its investment decisions, the firm uses the Stockbroker’s algorithm, which inputs a list of real numbers and outputs the max sum over a slice of the list.

Unfortunately, the firm uses the following implementation of the algorithm, which is called John Smith’s Sum as nobody knows the name of the employee who wrote it:

def john_smith_sum(L: list) -> int:
"""
Return the maximum sum over a slice of L in cubic time.
"""

max = front_index = 0
n = len(L)

while front_index < n:
end_index = front_index

while end_index <= n:
middle_index = front_index
sum = 0

while middle_index < end_index:
sum += L[middle_index]
middle_index += 1

if sum > max:
max = sum

end_index += 1

front_index += 1

return max

The runtime of john_smith_sum is cubic.  This means that it is agonizingly slow for large inputs and that you will miss out on great investments.  So you decide to do the following:

Main Problem:  Come up with an implementation of the Stockbroker’s algorithm with a linear runtime.

You will then give it to the firm because you are very generous person.  First things first; how will you come up with an implementation?

#### Step 1:  Understand the Problem

Polya’s How to Solve It recommends understanding the problem first.  We have to meet two goals: our implementation should be correct and linear in its runtime.  Let us answer the following to see if we know what these goals mean:

(1) What does an iterative algorithm with a linear runtime look like?
(2) What is the correct output of the Stockbroker’s algorithm for different inputs?

For (1), a single loop and a monotone loop index should suffice.  For (2), work out examples:  Consider the list $[3, 1, 4]$.  The max sum is $8$ so that is what our implementation should return.  How about $[1, -2, 3, -4, 5, -6]$?  Each negative number makes all the preceding partial sums negative so the max sum is $5$ here.  What about $[3, -1, 4, -1, 5]$?.  We may be tempted to look only at positive numbers but we see that adding $-1$ after $3$ and $4$ still keeps the partial sum positive.  Hence, the max sum is $10$.

#### Step 2: Devise a Plan

“No plan survives contact with the enemy” – The Desert Fox

There are a couple of approaches to this problem.  One is to assume an arbitrary list of real numbers as the input.  We write a single implementation for that input.  However, this is a little hard as we have to juggle negative and non-negative numbers simultaneously.

A second approach is to implement by cases through the lists in (1).  Then combine them into a single implementation.  However, we are lazy and we do not want to write multiple unrelated implementations.

Check out this approach.  We write a single implementation that works for a simple input and then extend it to all inputs.  This may be the best approach as there appears to be only a few cases.  Hopefully, this assumption is correct.  Then this approach is our plan.

#### Step 3: Carry out the Plan

This is the part where we execute the plan so try not to lose your head.  Of the three lists in (1), the easiest one to write an implementation for is $[3, 1, 4]$ as all the elements are non-negative.  By delegating a variable to store the max sum, then we have version one of our implementation:

My Sum Version 1:

def my_sum(L: list) -> int:
"""
Return the maximum sum over a slice of L in linear time.
"""

n = len(L)
sum = max = index = 0

while index < n:
element = L[index]
sum += element

if max < sum:
max = sum

index += 1

return max

This version features a single loop with a monotone increasing loop index so we satisfy the linear runtime component.  Unfortunately, this version only works if all the elements are positive.  It falters for lists of the following nature: $[1, -2, 3, -4, 5, -6]$.  So we break down each iteration into two cases: one for negative and one for nonnegative.  This leads to version two of our implementation:

My Sum Version 2:

def my_sum(L: list) -> int:
"""
Return the maximum sum over a slice of L in linear time.
"""

n = len(L)
sum = max = index = 0

while index < n:
element = L[index]

if element >= 0:
sum += element

else:
sum = 0

if max < sum:
max = sum

index += 1

return max

This version resets the sum once it iterates over a negative number.  The linear runtime is still preserved.  However, consider the following list: $[3, -1, 4, -1, 5]$.  The element $-1$ is not enough to reduce the sum to below zero.  Then we should not reset the sum unless it is below zero.  This also allows us to eliminate some code: the negative and non-negative checks and the memoization step where L[index] is assigned to the variable element:

My Sum Version 3:

def my_sum(L: list) -> int:
"""
Return the maximum sum over a slice of L in linear time.
"""

n = len(L)
sum = max = index = 0

while index < n:
sum += L[index]

if sum < 0:
sum = 0

if max < sum:
max = sum

index += 1

return max

#### Step 4: Look Back

Version three is our final version but our obsessive tendencies force us to double check its correctness and runtime.  Consider the test code below.  It checks that my_sum and john_smith_sum have matching outputs for each input.

Verify this by running the code from the link on Python.  It contains the code for john_smith sum, the final version of my_sum and the test cases:

Test Cases:

if __name__ == "__main__":
L = [[3, 1, 4], [3, -1, 4, 1], [3, -1, 4, -1, 5], [3, 1, -4, 1, -5, -9],
[-3, -1, -4], [1, -2, 3, -4, 5, -6], [3, -5, 7, 1, -2, 0, 3, -2, 1]]

for x in L:
assert john_smith_sum(x) == my_sum(x)

Likewise for the runtime, the loop executes $7$ steps for each iteration in the worst case so that is $7n$ steps in total.  Add the steps outside the loop plus the loop guard and we have $7n + 4$ as the runtime, which is linear.  Hence, we have a credible implementation of the Stockbroker's algorithm.

## Tuesday, 4 November 2014

### Week 8: Time Complexity

The post in Week 7 talks about algorithms being linear, quadratic or cubic.  Such descriptions are neither precise nor fancy enough keep computer scientists satisfied.  So we have this thing called the time complexity of an algorithm, which describes the growth in runtime over the growth in input size.

#### Here Comes the Notation!

Time complexity is described in Big O.  It looks something like this.  Let $f: N \to R^{\geq 0}$ be a sequence that describes the runtime of an algorithm $P$ and $g: N \to R^{\geq 0}$ be a sequence that describes an arbitrary runtime (i.e. linear, quadratic, cubic, etc).  If we say that $f \in O(g)$, then we mean that:

$\exists c \in R^+, \exists B \in N, (\forall n \in N, n \geq B \implies f(n) \leq cg(n))$

Intuitively, what these little symbols say is that as the input becomes arbitrarily large, then the runtime of $P$ performs better than or as well as $g$.  Likewise, if we say that $f \in \Omega (g)$, then we mean that:

$\exists c \in R^+, \exists B \in N, (\forall n \in N, n \geq B \implies f(n) \geq cg(n))$

Again, this is just a fancy way of saying that as the input becomes arbitrarily large, then the runtime of $P$ performs worse than or as well as $g$.

#### A Fun Problem

Let $f: N \to R^{\geq 0}$ be a polynomial sequence of degree $m$ and $g: N \to R^{\geq 0}$ be a polynomial sequence of degree $n$ with $m, n \in R$.  Find all $m, n$ such that $f \in O(g)$ and $(f \times g) \in \Omega(f)$.

## Monday, 27 October 2014

### Week 7: Nice Mathematical Toys in Algorithm Analysis

Constant, linear, quadratic, logarithmic or linearithmic are terms that vaguely describe the worst case runtime of algorithms in CSC148 but they are not precise enough in CSC165.  We need the exact number of steps.  This calls for some nice mathematical toys, which we introduce with the following algorithm:

def mystery_sort(L: list):
“””
Sort the elements of L in mysterious order.
“””

n = len(L)
outer_index  = 1

while outer_index < n:
value = L[outer_index]
inner_index = outer_index

while inner_index > 0 and A[inner_index – 1] < value:
A[inner_index] = A[inner_index – 1]
inner_index = inner_index – 1

A[inner_index] = value
outer_index = outer_index + 1

Since mystery_sort has a nested loop with monotone loop indexes, then its worst-case runtime is quadratic.  What is less obvious is the exact number of steps that it takes for mystery_sort to run in the worst case.   Our procedure to count mystery_sort is as follows:

(1) Assume that each line represents one step to make the analysis easier.
(2) Count from the inner loop to the outer loop.
(3) Add up the steps that are outside the outer loop.

So assume (1).  Then do (2) and (3).

#### Inner Loop

To count the steps of the inner loop, our first mathematical toy comes into play: abstraction.  It helps us determine the number of times that the inner loop executes.  Observe that the inner loop depends on the outer_index.  However, the outer_index is not concrete since it changes each time the outer loop executes.  So we must think abstractly; use an arbitrary integer $i$ to represent the outer_index.  Then $i$ becomes the number of times that the inner loop executes.  Now the easy part: the inner loop performs three steps for each iteration.  Since the inner loop executes $i$ times, then the number of steps of the inner loop is $3i$.

#### Outer Loop

In addition to the number of steps of the inner loop, we have the following steps inside each outer loop:

(1) while outer_index < n
(2) value = L[outer_index]
(3) inner_index = outer_index
(4) loop guard of the inner loop
(5) A[inner_index] = value
(6) outer_index = outer_index + 1

This is $6$ steps in total.  Add it to the steps of the inner loop and you see that each iteration of the outer loop executes $3i + 5$ steps.  We still have to add up the steps for each iteration so here our second mathematical toy comes into play: summation.  Since the outer loop runs $n - 1$ times, then the total steps of the outer loop is the following summation:

$\sum_{i = 1}^{n - 1} (3i + 6)$.

To find the sum of this series, our mathematical final toy comes and it is Euler’s trick.  This is a technique for finding the sum of an arithmetic series.  First decompose the summation:

$\sum_{i = 1}^{n - 1} (3i + 6) = 3 \sum_{i = 1}^{n - 1} i + \sum_{i = 1}^{n - 1} 6$

Then perform the following calculation:

$\sum_{i = 1}^{n - 1} i = 1 + 2 + 3 + … + (n - 3) + (n - 2) + (n - 1) =$
$\frac{1}{2}(1 + 2 + 3 + … + (n - 3) + (n - 2) + (n - 1) +$
$(n - 1) + (n - 2) + (n - 3) + … + 3 + 2 + 1) =$
$\frac{n(n - 1)}{2} = \frac{n^2 - n}{2}$

Substitute back into the series and we get the number of steps that outer loop executes:

$\frac{3n^2 - 3n + 12n - 12}{2} =$
$\frac{3n^2 - 9n - 12}{2}$

#### Outside Both Loops

There are still the following steps to consider:

(1) n = len(L)
(2) outer_index = 1
(3) loop guard of the outer loop

This is $3$ steps so the total number of steps that the algorithm performs is:

$\frac{3n^2 - 9n - 6}{2}$

## Monday, 20 October 2014

### Week 6: $\epsilon - \delta$ Definition of Limit

As someone who has no prior experience with the $\epsilon - \delta$ definition of limit, I will teach myself about them with the help of my favourite math textbook and some creativity.  This is what I gleaned so far.

In high school, we calculate limits.  In university, we like to think that we are sophisticated thinkers. Knowing how to calculate is not enough for us.  We must know why our calculation is correct.  Then we need a definition to work with.  This is where the the $\epsilon - \delta$ definition of limit comes in:

Let $f:X \to R$ be a function, $L \in R$, $a \in X$ and $X \subseteq R$.  Then

$\lim_{x \to a} f(x) = L \iff$
$\forall \epsilon \in R^+, \exists \delta \in R^+, (\forall x \in R, |x - a| < \delta \implies |f(x) - L| < \epsilon).$

After digesting this definition, we can prove or disprove statements about the limits of functions.
For example, consider the following statement:

$\lim_{x \to 2} x^2 = 4$

By substituting two into $x$, we discover that this statement should be true.  To show that our hunch is correct, we plug in $a = 2$ and $L = 4$ into the $\epsilon - \delta$ definition and prove it:

$\forall \epsilon \in R^+, \exists \delta \in R^+, (\forall x \in R, |x - 2| < \delta \implies |x^2 – 4| < \epsilon)$

Our value for $\epsilon$ will be arbitrary but our value for $\delta$ must be concrete since $\epsilon$ has a $\forall$ quantifier but $\delta$ has a $\exists$ quantifier.  The hard part is choosing $\delta$.  How do we do it?

#### Where to Start

Think of $\epsilon$ as a distance from $L$ on the vertical axis and $\delta$ as a distance from $2$ on the horizontal axis.  Once the distance of $x$ from $2$ is less than $\delta$ on the horizontal, then the $\epsilon – \delta$ definition of limit guarantees that the distance of $x^2$ from $4$ is less than $\epsilon$ on the vertical.  Hence, our choice of $\delta$ must satisfy the definition for all $\epsilon$.  To achieve this, we express $\delta$ in terms of $\epsilon$, which is done by working backwards from the definition.

#### How to Work Backwards

Since the definition is an implication, then working backwards begins with the consequent:

$|x^2 - 4| < \epsilon$

Since we operate in $R$, then we can manipulate the left side of the inequality:

$|x + 2||x - 2| = |x^2 - 4| < \epsilon$

We see that the $|x - 2|$ in $|x + 2||x - 2| < \epsilon$ will be useful in getting the antecedent of the definition, which is $|x - 2| < \delta$.  However, there is still a bit of baggage to unload.

#### The Creative Step

Since we choose our $\delta$, then we can give it an upper bound.  For simplicity, we choose $1$ as an upper bound for $\delta$.  So if $|x - 2| < d$, then $|x - 2| < 1$.

From $|x - 2| < 1$, the absolute value definition gives $-1 < x - 2 < 1$.  By adding $4$ to all sides of the inequality, then we have $3 < x + 2 < 5$.  Use the absolute value definition again, which gives us $|x + 2| < 5$.

Now we can unload $|x + 2|$ from $|x + 2||x - 2| < \epsilon$.  Since $|x + 2| < 5$, then $|x + 2||x - 2| < 5|x - 2|$.  Then $5|x - 2| < e$ is equivalent to $|x - 2| < \frac{\epsilon}{5}$.

#### Endgame

We can now chose $\delta$.  If $|x - 2| < \frac{\epsilon}{5}$ is true, then its equivalent statement $5|x - 2| < e$ guarantees that $|x - 2||x + 2| = |x^2 - 4| < \epsilon$ since $|x - 2| < 5$.  Yet we chose $1$ as an upper bound for $\delta$.  So if $\frac{\epsilon}{5}>1$, then $\delta = 1$.  Hence our choice of $\delta$ is $min(1, \frac{\epsilon}{5})$.  The proof follows naturally from the scratch work so it will be omitted.


text:
## BSC approach at each strength? [Dissolution / BCS / IVIVC]

» If your 3 strengths have a different composition, »

Thank you dear Ohlbe.
But I'm assuming proportionally the same qualitative composition for each strength.


text:
# Let $V = \mathbb{R}^2$, show that $V$ is a vector space

I am new to the concepts of vector spaces (as far as I remember), and I have some difficulties in understanding how can I show, in practice, if a set is a vector space or not.

I have an exercise in my last linear algebra assignment, which is the following:

Let $V = \mathbb{R}^2$, with addition $\oplus$ defined by:

$$\left(\begin{matrix} x \\ y \end{matrix}\right) \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) = \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right)$$

and scalar multiplication $\odot$ defined by:

$$\alpha \odot \left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} \alpha x - 3\alpha +3\\ \alpha y + \alpha - 1 \end{matrix}\right)$$

Show that $V$ is a vector space by showing that:

a) $V$ is closed with respect to $\oplus$ and $\odot$ and i.e.

• $\forall$ v, w $\in V \rightarrow$ v $\oplus$ w $\in V$.

• $\forall \alpha \in \mathbb{R}$, $\forall$ v $\in V \rightarrow \alpha \odot$ v $\in V$.

b) All $8$ vector spaces axioms hold. The vector $\left(\begin{matrix} 3 \\ -1 \end{matrix}\right)$ acts as a zero vector.

This can be all interesting if you have at least an idea on how to do it, and maybe you can do it, but this is not exactly my case.

I know that to show a set is a vector space I have to show it's closed under addition and under scalar multiplication (nice!). I know, from my notes, what these axioms are.

But in this case, how can I show that:

$$\left(\begin{matrix} x \\ y \end{matrix}\right) \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) = \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right)$$

is closed under addition, if we don't know the values of $x$, etc?

The same for all other options (show that is closed under scalar multiplication, and show that the axioms hold).

In general, how would you prove that a set like this is a vector space? Can I see an example? I am really not understanding what I have to prove, and the examples I have tried to see for me are like Chinese.

• Just a comment: to say that a set $V$ is or is not a vector space is actually wrong. In addition to the set you always need a field $F$ and two well-defined binary operations $\oplus : V \times V \to V$ and $\odot : F \times V \to V$. Without these, you cannot prove that the vector space axioms hold. Sep 24, 2015 at 14:34

Although you may not know what the values of your vector is, you do know what vectors of the set look like. In this case $\mathbb{R}^2 = \{ \begin{pmatrix} a \\ b \end{pmatrix} \colon b, a \in \mathbb{R} \}$.

Given $\begin{pmatrix} x \\ y \end{pmatrix}$ and $\begin{pmatrix} z \\ w \end{pmatrix}$, is the vector $\begin{pmatrix} x + w - 3 \\ y + z + 1 \end{pmatrix}$ in $\mathbb{R}^2$ ?

• Well, you're right! The vector $\begin{pmatrix} x + w - 3 \\ y + z + 1 \end{pmatrix}$ is still in $\mathbb{R}^2$ regardless of the choice of $x, y, z, w \in \mathbb{R}$, so the operation is indeed closed. The argument for multiplication follows similarly. To check that $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ is the zero vector, pick up an arbitrary element $\begin{pmatrix} x \\ y \end{pmatrix}$ in $\mathbb{R}^2$ and show that $\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$. Mar 15, 2015 at 16:47

To prove that $\mathbb{R}^2$ is closed under $\oplus$, you need to prove that ${x+w-3\choose y+z+1}\in \mathbb{R}^2$, regardless of the values of $x,w,y,z$. Fortunately, this is easy, since $\mathbb{R}$ is itself closed under addition, so the sum of the three real numbers $x,w,-3$ is again a real number. Similarly, the sum of the three real numbers $y,z,1$ is again a real number. Thus ${x+w-3\choose y+z+1}$ is an ordered pair of real numbers, which is the definition of $\mathbb{R}^2$.

Proving closure under scalar multiplication is similar.

The hard parts are the vector space axioms; you should try them yourself rather than having people work through them all for you.

In the equation $$\left(\begin{matrix} x \\ y \end{matrix}\right) \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) = \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right)$$ the right side is a vector in $\mathbb{R}^2$ and so it is closed under your specific type of addition. This is just regular addition with the specific modified zero vector added in as well.

The same goes for your modified version of scalar multiplication. $\alpha\odot\mathbf{v}=\alpha\mathbf{v}+(\alpha-1)\mathbf{v_0}$ where $\mathbf{v}_0$ is your modified zero vector. This still results in a vector in your space.

Checking the other 8 requirements should be fairly straightforward once you get past the notation.

For the first axiom, you have to prove that $$\left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right) \in V = \mathbb{R}^2$$ which is true for $x,y,z \in \mathbb{R}$, since you original vectors were in $\mathbb{R}^2$ you have $x,y,z \in \mathbb{R}$ and the result follows.

Similarly for the clousure with respect to scalar product, $$\alpha \odot \left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} \alpha x - 3\alpha +3\\ \alpha y + \alpha - 1 \end{matrix}\right) \in \mathbb{R}^2$$ if $\alpha, x, y \in \mathbb{R}$, which is true for the same argument above.

The rest are a little easier because you can use the propierties of $\mathbb{R}$ after you reach a certain place, like for example commutativity:

$$\left(\begin{matrix} x \\ y \end{matrix}\right) \oplus\left(\begin{matrix} w \\ z \end{matrix}\right) = \left(\begin{matrix} x + w -3\\ y + z + 1 \end{matrix}\right) = \left(\begin{matrix} w + x -3\\ z + y + 1 \end{matrix}\right) = \left(\begin{matrix} w \\ z \end{matrix}\right) \oplus\left(\begin{matrix} x \\ y \end{matrix}\right)$$

Note that in the middle equality we used the conmutativity of $\mathbb{R}$ to interchange $w,x$ and $y,z$.

• Yes, buy you'd have to do it with the operation $\oplus$ defined in you question. Mar 15, 2015 at 17:14
• You have to follow the parenthesis, first you have to do $\left(\begin{matrix} w \\ z \end{matrix}\right) + \left(\begin{matrix} a \\ b \end{matrix}\right)$, that gives you one vector (that will look a little weird but is only one vector), then to that vector you have to sum $\left(\begin{matrix} x \\ y \end{matrix}\right)$ Mar 15, 2015 at 17:29


text:
# DX11 Urho3D Graphics abstraction

## Recommended Posts

Hello,

I've stumbled upon Urho3D engine and found that it has a really nice and easy to read code structure.
I think the graphics abstraction looks really interesting and I like the idea of how it defers pipeline state changes until just before the draw call to resolve redundant state changes.
This is done by saving the state changes (blendEnabled/SRV changes/RTV changes) in member variables and just before the draw, apply the actual state changes using the graphics context.
It looks something like this (pseudo):

void PrepareDraw()
{
if(renderTargetsDirty)
{
pD3D11DeviceContext->OMSetRenderTarget(mCurrentRenderTargets);
renderTargetsDirty = false
}
if(texturesDirty)
{
texturesDirty = false
}
....
//Some more state changes
}

This all looked like a great design at first but I've found that there is one big issue with this which I don't really understand how it is solved in their case and how I would tackle it.
I'll explain it by example, imagine I have two rendertargets: my backbuffer RT and an offscreen RT.

Say I want to render my backbuffer to the offscreen RT and then back to the backbuffer (Just for the sake of the example).
You would do something like this:

//Render to the offscreen RT
pGraphics->SetRenderTarget(pOffscreenRT->GetRTV());
pGraphics->SetTexture(diffuseSlot, pDefaultRT->GetSRV())

pGraphics->SetTexture(diffuseSlot, nullptr); //Remove the default RT from input

//Render to the default (screen) RT
pGraphics->SetRenderTarget(nullptr); //Default RT
pGraphics->SetTexture(diffuseSlot, pOffscreenRT->GetSRV())
pGraphics->DrawQuad();

The problem here is that the second time the application loop comes around, the offscreen rendertarget is still bound as input ShaderResourceView when it gets set as a RenderTargetView because in Urho3D, the state of the RenderTargetView will always be changed before the ShaderResourceViews (see top code snippet) even when I set the SRV to nullptr before using it as a RTV like above causing errors because a resource can't be bound to both input and rendertarget.

What is usually the solution to this?

Thanks!

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Posted (edited)

I deal with a similar issue in my graphics library. My API is not exactly like what you've posted (it's closer to Vulkan/D3D12 with multiple resources being bound as a single set), but there's a similar problem that needs to be tackled in the Direct3D11 backend.

In my case, I just keep track of all SRVs and UAVs that are currently bound, and check for invalid combinations when a new SRV or UAV is bound. For example, if you try to bind an SRV for Texture A to the context, then I'll check if Texture A has any bound UAV's. If it does, then those are removed. Then, the SRV is bound to the context and added to the map that tracks SRV state. Same process when a UAV is bound.

A bit hand-wavey (I could elaborate if it doesn't make sense), but overall it's not a very complicated system. Urho3D might be doing something more clever than I am.

Edited by mellinoe

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That sounds pretty straight forward and I've thought of doing that as well.
Although:

7 hours ago, mellinoe said:

If it does, then those are removed

The downside of this, is that the graphics system would be doing "magic" just to make it work behind the back of the user while it might not have been intended behavior. It wouldn't point out when the user made an actual mistake, it would just solve it by unbinding the problematic resource while the user might not realize that.

I wonder if there's a cleaner solution for this.

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Posted (edited)
45 minutes ago, simco50 said:

The downside of this, is that the graphics system would be doing "magic" just to make it work behind the back of the user while it might not have been intended behavior.

I just treat this as a quirk of the Direct3D11 backend's state tracking behavior. It's not legal for users to bind incompatible resources to the pipeline through my library's regular API, and they would get a descriptive error message if they did that.

Nevertheless the D3D11 context might get into a state where there are resource conflicts after a sequence of legal operations, because unused resources aren't removed. I could probably aggressively purge those unused resources, but it would be unnecessary most of the time.

EDIT: To clarify, in response to this:

Quote

It wouldn't point out when the user made an actual mistake, it would just solve it by unbinding the problematic resource while the user might not realize that.

I'm only talking about the cases where the user hasn't made a mistake. If they try to actually do something illegal (e.g. bind a Texture as both read and read-write), then you should catch that separately and give them a descriptive error.

Edited by mellinoe

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That sounds like the right thing to do.
What I do now, is before a conflict, flush the staged states (submit the state changes to the context with PrepareDraw() in my case)
That works but I wonder if there's a smarter solution for this.

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I'm no longer actively involved in the development of this library, but thought I'd chime in..

When a rendertarget is set such that it's still also being bound as a SRV, what should happen is that the runtime cleans up the still bound SRV, and if you have the debug device flag enabled, you see this as output log spam.

(D3D11 WARNING: ID3D11DeviceContext::OMSetRenderTargets[AndUnorderedAccessViews]: Forcing VS shader resource slot 0 to NULL)

That's why the D3D11 Graphics::PrepareDraw() in Urho3D sets rendertargets first, then textures.

The situation also easily happens also in Urho3D's own rendering, when pingponged postprocesses are involved.

In case you see an actual rendering bug resulting from this (ie. black screen) please raise it in Urho3D's issue tracker.

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Setting render targets is very rare compared to setting textures, so I take the heavy-handed approach of unbinding all SRV's when setting new render-targets (and marking the SRV bindings as dirty so the next draw will rebind any that are required). In non-shipping builds, I also implement my own version of the D3D runtime warnings -- when binding a SRV, I check to see if it's already bound as a RTV/DSV/UAV, and assert if so.

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That's why the D3D11 Graphics::PrepareDraw() in Urho3D sets rendertargets first, then textures.

Yeah, that's what I saw. I actually had no idea DirectX solves it by itself and the warning doesn't really break anything.

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# Diskite:Modèl Lhermite yo

Sauter à la navigation Sauter à la recherche

Diskite atik sa a an anglè e ede amelyore

Lhermite's models are interesting ways to synthesize some objects that are apparently scattered.

## Prime numbers and the model of three arrows

${\displaystyle P_{n}=\sum _{i=1}^{2^{2^{n}}}\left(\left\lfloor {\frac {1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}{n+1}}\right\rfloor \times {\left\lfloor {\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}}\right\rfloor }\times {i}\times {\left({1-\left\lfloor {\frac {\left\lfloor ({\frac {\left(i!\right)^{2}}{i^{3}}}\right\rfloor }{\frac {\left(i!\right)^{2}}{i^{3}}}}\right\rfloor }\right)}\right)}$
${\displaystyle P_{n}=\sum _{i=1}^{2^{n}}\left(\left\lfloor {\frac {1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}{n+1}}\right\rfloor \times {\left\lfloor {\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}}\right\rfloor }\times {i}\times {\left({1-\left\lfloor {\frac {\left\lfloor ({\frac {\left(i!\right)^{2}}{i^{3}}}\right\rfloor }{\frac {\left(i!\right)^{2}}{i^{3}}}}\right\rfloor }\right)}\right)}$
${\displaystyle P_{n}=\sum _{i=1}^{1+n!}\left(\left\lfloor {\frac {1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}{n+1}}\right\rfloor \times {\left\lfloor {\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left\lfloor {\frac {\left\lfloor {\frac {\left(m!\right)^{2}}{m^{3}}}\right\rfloor }{\frac {\left(m!\right)^{2}}{m^{3}}}}\right\rfloor \right)}}}\right\rfloor }\times {i}\times {\left({1-\left\lfloor {\frac {\left\lfloor ({\frac {\left(i!\right)^{2}}{i^{3}}}\right\rfloor }{\frac {\left(i!\right)^{2}}{i^{3}}}}\right\rfloor }\right)}\right)}$
${\displaystyle P_{n}=\sum _{i=1}^{2^{n}}\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{n+1}}\right]\times {\left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]}\times {i}\times {\left({1-\left[{\frac {\left[({\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]}\right)}\right)}$
${\displaystyle P_{n}=\sum _{i=1}^{2^{2^{n}}}\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{n+1}}\right]\times {\left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]}\times {i}\times {\left({1-\left[{\frac {\left[({\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]}\right)}\right)}$

## Red balls and blue balls and prime numbers

${\displaystyle P_{\left(\left(1-\left[{\frac {\left[{\frac {\left(n!\right)^{2}}{n^{3}}}\right]}{\frac {\left(n!\right)^{2}}{n^{3}}}}\right]\right)\times \left(\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}-i\right)+i\right)}=\left(P_{i}-n\right)\times \left[{\frac {\left[{\frac {\left(n!\right)^{2}}{n^{3}}}\right]}{\frac {\left(n!\right)^{2}}{n^{3}}}}\right]+n}$

## Prime numbers and the model of three arrows according to Wilson's theorem

${\displaystyle \forall n\in \mathbb {N'} }$
${\displaystyle (n-1)!\equiv \ -1{\pmod {n}}\Leftrightarrow n\in \mathbb {P} }$

in the same way, it is advanced that

${\displaystyle \forall n\in \mathbb {N'} }$
${\displaystyle \left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]=1\Leftrightarrow n\in \mathbb {P} }$

It's very evident that

${\displaystyle \forall n\in \mathbb {N'} }$
${\displaystyle \left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]=0\Leftrightarrow n\notin \mathbb {P} }$

Therefore, according to Lhermite's models and Wilson's theorem, there are two evident theorems :

${\displaystyle \forall n\in \mathbb {N^{*}} }$
${\displaystyle \left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]-\left[{\frac {1}{n}}\right]=1\Leftrightarrow n\in \mathbb {P} }$
${\displaystyle \forall n\in \mathbb {N^{*}} }$
${\displaystyle \left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]-\left[{\frac {1}{n}}\right]=0\Leftrightarrow n\notin \mathbb {P} }$

Therefore the following relation becomes true :

${\displaystyle \forall n\in \mathbb {N^{*}} }$
${\displaystyle \left[{\frac {\left[{\frac {\left(n-1\right)!+1}{n}}\right]}{\frac {\left(n-1\right)!+1}{n}}}\right]-\left[{\frac {1}{n}}\right]=1-\left[{\frac {\left[{\frac {\left(n!\right)^{2}}{n^{3}}}\right]}{\frac {\left(n!\right)^{2}}{n^{3}}}}\right]}$

Let's choose one of the formulas that are indicated in the first section :

${\displaystyle P_{n}=\sum _{i=1}^{2^{2^{n}}}\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{n+1}}\right]\times {\left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]}\times {i}\times {\left({1-\left[{\frac {\left[({\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]}\right)}\right)}$

let's replace

${\displaystyle 1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]by\left[{\frac {\left[{\frac {\left(m-1\right)!+1}{m}}\right]}{\frac {\left(m-1\right)!+1}{m}}}\right]-\left[{\frac {1}{m}}\right]}$

and

${\displaystyle 1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]by\left[{\frac {\left[{\frac {\left(i-1\right)!+1}{i}}\right]}{\frac {\left(i-1\right)!+1}{i}}}\right]-\left[{\frac {1}{i}}\right]}$

Therefore an equivalent expression is :

${\displaystyle P_{n}=\sum _{i=1}^{2^{2^{n}}}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(\left[{\frac {\left[{\frac {\left(m-1\right)!+1}{m}}\right]}{\frac {\left(m-1\right)!+1}{m}}}\right]-\left[{\frac {1}{m}}\right]\right)}}{n+1}}\right]\times \left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(\left[{\frac {\left[{\frac {\left(m-1\right)!+1}{m}}\right]}{\frac {\left(m-1\right)!+1}{m}}}\right]-\left[{\frac {1}{m}}\right]\right)}}}\right]\times i\times \left(\left[{\frac {\left[{\frac {\left(i-1\right)!+1}{i}}\right]}{\frac {\left(i-1\right)!+1}{i}}}\right]-\left[{\frac {1}{i}}\right]\right)\right)}}$

## Function Ω according to Lhermite's models

${\displaystyle \Omega \left(n\right)=\sum _{j=1}^{n}\left({\sum _{i=1}^{n}{\left({{\left[{\frac {\left[{\frac {n}{i^{j}}}\right]}{\left({\frac {n}{i^{j}}}\right)}}\right]}\times \left(1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)}\right)}}\right)}$

## The Liouville function and Lhermite's models

${\displaystyle \lambda \left(n\right)=\left(-1\right)^{\left(\sum _{j=1}^{n}\left({\sum _{i=1}^{n}{\left({{\left[{\frac {\left[{\frac {n}{i^{j}}}\right]}{\left({\frac {n}{i^{j}}}\right)}}\right]}\times \left(1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)}\right)}}\right)\right)}}$

## Three Arrows or Jonatan's Arrows

Let's explain Jonatan's arrows for any context. This model has a certain relation with Boole's works and Leibniz work's.


text:
## The bounded proper forcing axiom.(English)Zbl 0819.03042

Summary: The bounded proper forcing axiom BPFA is the statement that for any family of $$\aleph_ 1$$ many maximal antichains of a proper forcing notion, each of size $$\aleph_ 1$$, there is a directed set meeting all these antichains. A regular cardinal $$\kappa$$ is called $$\Sigma_ 1$$- reflecting, if for any regular cardinal $$\chi$$, for all formulas $$\varphi$$, “$$H (\chi) \vDash \text{} \varphi \text{' }$$” implies “$$\exists \delta< \kappa$$, $$H(\delta) \vDash \text{} \varphi \text{' }$$”. We investigate several algebraic consequences of BPFA, and we show that the consistency strength of the bounded proper forcing axiom is exactly the existence of a $$\Sigma_ 1$$-reflecting cardinal (which is less than the existence of a Mahlo cardinal). We also show that the question of the existence of isomorphisms between two structures can be reduced to the question of rigidity of a structure.

### MSC:

 03E35 Consistency and independence results 03C55 Set-theoretic model theory 03E55 Large cardinals 03C50 Models with special properties (saturated, rigid, etc.)
Full Text:

### References:

 [1] DOI: 10.1016/0003-4843(72)90017-4 · Zbl 0255.02069 [2] DOI: 10.1305/ndjfl/1093634482 · Zbl 0787.03041 [3] Surveys in set theory 87 pp 1– (1983) · Zbl 0511.00004 [4] Handbook of set-theoretic topology pp 913– (1984) [5] Israel Journal of Mathematics 25 (1976) [6] Proper and improper forcing · Zbl 0889.03041 [7] Proper forcing 940 (1982) [8] DOI: 10.1016/0003-4843(78)90009-8 · Zbl 0383.03019 [9] Axiomatic set theory pp 209– (1984)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# MECH 221 FLUID MECHANICS (Fall 06/07) Chapter 7: INVISCID FLOWS

## Presentation on theme: "MECH 221 FLUID MECHANICS (Fall 06/07) Chapter 7: INVISCID FLOWS"— Presentation transcript:

MECH 221 FLUID MECHANICS (Fall 06/07) Chapter 7: INVISCID FLOWS
Instructor: Professor C. T. HSU

7.1 Inviscid Flow Inviscid flow implies that the viscous effect is negligible. This occurs in the flow domain away from a solid boundary outside the boundary layer at Re. The flows are governed by Euler Equations where , v, and p can be functions of r and t .

7.1 Inviscid Flow On the other hand, if flows are steady but compressible, the governing equation becomes where  can be a function of r For compressible flows, the state equation is needed; then, we will require the equation for temperature T also.

7.1 Inviscid Flow Compressible inviscid flows usually belong to the scope of aerodynamics of high speed flight of aircraft. Here we consider only incompressible inviscid flows. For incompressible flow, the governing equations reduce to where  = constant.

7.1 Inviscid Flow For steady incompressible flow, the governing eqt reduce further to where  = constant. The equation of motion can be rewrited into Take the scalar products with dr and integrate from a reference at  along an arbitrary streamline =C , leads to since

7.1 Inviscid Flow If the constant (total energy per unit mass) is the same for all streamlines, the path of the integral can be arbitrary, and in the flow domain except inside boundary layers. Finally, the governing equations for inviscid, irrotational steady flow are Since is the vorticity , flows with are called irrotational flows.

7.1 Inviscid Flow Note that the velocity and pressure fields are decoupled. Hence, we can solve the velocity field from the continuity and vorticity equations. Then the pressure field is determined by Bernoulli equation. A velocity potential  exists for irrotational flow, such that, and irrotationality is automatically satisfied.

7.1 Inviscid Flow The continuity equation becomes which is also known as the Laplace equation. Every potential satisfy this equation. Flows with the existence of potential functions satisfying the Laplace equation are called potential flow.

7.1 Inviscid Flow The linearity of the governing equation for the flow fields implies that different potential flows can be superposed. If 1 and 2 are two potential flows, the sum =(1+2) also constitutes a potential flow. We have However, the pressure cannot be superposed due to the nonlinearity in the Bernoulli equation, i.e.

7.2 2D Potential Flows If restricted to steady two dimensional potential flow, then the governing equations become E.g. potential flow past a circular cylinder with D/L <<1 is a 2D potential flow near the middle of the cylinder, where both w component and /z0. U L y x z D

7.2 2D Potential Flows The 2-D velocity potential function gives and then the continuity equation becomes The pressure distribution can be determined by the Bernoulli equation, where p is the dynamic pressure

7.2 2D Potential Flows For 2D potential flows, a stream function (x,y) can also be defined together with (x,y). In Cartisian coordinates, where continuity equation is automatically satisfied, and irrotationality leads to the Laplace equation, Both Laplace equations are satisfied for a 2D potential flow

7.2 Two-Dimensional Potential Flows
For two-dimensional flows, become: In a Cartesian coordinate system In a Cylindrical coordinate system and and

7.2 Two-Dimensional Potential Flows
Therefore, there exists a stream function such that in the Cartesian coordinate system and in the cylindrical coordinate system. The transformation between the two coordinate systems

7.2 Two-Dimensional Potential Flows
The potential function and the stream function are conjugate pair of an analytical function in complex variable analysis. The conditions: These are the Cauchy-Riemann conditions. The analytical property implies that the constant potential line and the constant streamline are orthogonal, i.e., and to imply that

7.3 Simple 2-D Potential Flows
Uniform Flow Stagnation Flow Source (Sink) Free Vortex

7.3.1 Uniform Flow For a uniform flow given by , we have Therefore,
Where the arbitrary integration constants are taken to be zero at the origin. and and

7.3.1 Uniform Flow This is a simple uniform flow along a single direction.

7.3.2 Stagnation Flow For a stagnation flow, Hence, Therefore,

7.3.2 Stagnation Flow The flow an incoming far field flow which is perpendicular to the wall, and then turn its direction near the wall The origin is the stagnation point of the flow. The velocity is zero there. x y

7.3.3 Source (Sink) Consider a line source at the origin along the z-direction. The fluid flows radially outward from (or inward toward) the origin. If m denotes the flowrate per unit length, we have (source if m is positive and sink if negative). Therefore,

7.3.3 Source (Sink) The integration leads to and
Where again the arbitrary integration constants are taken to be zero at and

7.3.3 Source (Sink) A pure radial flow either away from source or into a sink A +ve m indicates a source, and –ve m indicates a sink The magnitude of the flow decrease as 1/r z direction = into the paper. (change graphics)

7.3.4 Free Vortex Consider the flow circulating around the origin with a constant circulation . We have: where fluid moves counter clockwise if is positive and clockwise if negative. Therefore,

7.3.4 Free Vortex The integration leads to where again the arbitrary integration constants are taken to be zero at and

7.3.4 Free Vortex The potential represents a flow swirling around origin with a constant circulation . The magnitude of the flow decrease as 1/r.

7.4. Superposition of 2-D Potential Flows
Because the potential and stream functions satisfy the linear Laplace equation, the superposition of two potential flow is also a potential flow. From this, it is possible to construct potential flows of more complex geometry. Source and Sink Doublet Source in Uniform Stream 2-D Rankine Ovals Flows Around a Circular Cylinder

7.4.1 Source and Sink Consider a source of m at (-a, 0) and a sink of m at (a, 0) For a point P with polar coordinate of (r, ). If the polar coordinate from (-a,0) to P is and from (a, 0) to P is Then the stream function and potential function obtained by superposition are given by:

7.4.1 Source and Sink

7.4.1 Source and Sink Hence, Since We have

7.4.1 Source and Sink We have By Therefore,

7.4.1 Source and Sink The velocity component are:

7.4.1 Source and Sink

7.4.2 Doublet The doublet occurs when a source and a sink of the same strength are collocated the same location, say at the origin. This can be obtained by placing a source at (-a,0) and a sink of equal strength at (a,0) and then letting a  0, and m , with ma keeping constant, say 2am=M

7.4.2 Doublet For source of m at (-a,0) and sink of m at (a,0)
Under these limiting conditions of a0, m , we have

7.4.2 Doublet Therefore, as a0 and m with 2am=M
The corresponding velocity components are

7.4.2 Doublet

7.4.3 Source in Uniform Stream
Assuming the uniform flow U is in x-direction and the source of m at(0,0), the velocity potential and stream function of the superposed potential flow become:

7.4.3 Source in Uniform Stream

7.4.3 Source in Uniform Stream
The velocity components are: A stagnation point occurs at Therefore, the streamline passing through the stagnation point when The maximum height of the curve is

7.4.3 Source in Uniform Stream
For underground flows in an aquifer of constant thickness, the flow through porous media are potential flows. An injection well at the origin than act as a point source and the underground flow can be regarded as a uniform flow.

D Rankine Ovals The 2D Rankine ovals are the results of the superposition of equal strength sink and source at x=a and –a with a uniform flow in x-direction. Hence,

D Rankine Ovals Equivalently,

7.4.4 2-D Rankine Ovals The stagnation points occur at
where with corresponding

7.4.4 2-D Rankine Ovals The maximum height of the Rankine oval is
located at when ,i.e., which can only be solved numerically.

D Rankine Ovals rs ro

7.4.5 Flows Around a Circular Cylinder
Steady Cylinder Rotating Cylinder Lift Force

Steady Cylinder Flow around a steady circular cylinder is the limiting case of a Rankine oval when a0. This becomes the superposition of a uniform parallel flow with a doublet in x-direction. Under this limit and with M=2a. m=constant, is the radius of the cylinder.

Steady Cylinder The stream function and velocity potential become: The radial and circumferential velocities are:

Rotating Cylinder The potential flows for a rotating cylinder is the free vortex flow given in section Therefore, the potential flow of a uniform parallel flow past a rotating cylinder at high Reynolds number is the superposition of a uniform parallel flow, a doublet and free vortex. Hence, the stream function and the velocity potential are given by

Rotating Cylinder The radial and circumferential velocities are given by

Rotating Cylinder The stagnation points occur at From

Rotating Cylinder

Rotating Cylinder

7.4.5.2 Rotating Cylinder The stagnation points occur at Case 1:

Rotating Cylinder Case 1:

7.4.5.2 Rotating Cylinder Case 2:
The two stagnation points merge to one at cylinder surface where

7.4.5.2 Rotating Cylinder Case 3:
The stagnation point occurs outside the cylinder when where The condition of leads to Therefore, as , we have

Rotating Cylinder Case 3:

Lift Force The force per unit length of cylinder due to pressure on the cylinder surface can be obtained by integrating the surface pressure around the cylinder. The tangential velocity along the cylinder surface is obtained by letting r=ro,

Lift Force The surface pressure as obtained from Bernoulli equation is where is the pressure at far away from the cylinder.

Lift Force Hence, The force due to pressure in x and y directions are then obtained by

Lift Force The development of the lift on rotating bodies is called the Magnus effect. It is clear that the lift force is due to the development of circulation around the body. An airfoil without rotation can develop a circulation around the airfoil when Kutta condition is satisfied at the rear tip of the air foil. Therefore, The tangential velocity along the cylinder surface is obtained by letting r=ro: This forms the base of aerodynamic theory of airplane.

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# How to quantify the thermal advantages of a white thermos?

My understanding is that you could expect a white thermos to hold heat better than a black one as it doesn't lose heat through radiation as much. Is there a way to calculate the difference in heat retention?

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It is the emissivity at the IR (infrared) wavelengths that determines whether the heat is radiated away or not. Having a white color (low absorption=low emissivity) in the visible does not mean that it is low emissivity in the mid-IR (3-6 microns) and slightly longer wavelengths, out to about 20 microns, where most of the thermal radiation will occur at from temperatures around 300-350 Kelvin. (Wien's law and/or the Planck function tells you the wavelengths of interest). For maximum containment of the heat, if the outer surface becomes warm, the outer surface needs to have low emissivity in the infrared, and it will thereby will radiate less. ## \\ ## For thermos bottle design, it is really more optimal to contain the heat in the interior, by using glass surfaces whose outer coating is metallic. Some metals do have low emissivity throughout the infrared. These surfaces also work for keeping the inside cool in the case of cold beverages, by reflecting away any incident (radiated) thermal energy that comes from the outside. ## \\ ## Oftentimes, the design in thermos bottles is a vacuum between two glass layers, I believe with metallic coating on the glass surfaces that make up the vacuum cavity. That way, heat conduction is minimized by the vacuum, and radiative transfer is also minimized. ## \\ ## Meanwhile, even if a thermos bottle uses a high quality solid insulator instead of a vacuum type design, the emissivity of the outer surface is normally of a lower priority for ensuring good containment of the heat.

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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeOct 26th 2009

Mike Stay kindly added the standard QM story to path integral.

I changed the section titles a bit and added the reference to the Baer-Pfaeffle article on the QM path integral. Probably the best reference there is on this matter.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeMar 16th 2010

added more references to path integral. Also added an Idea-section

much more needs to be added here, eventually.

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeMar 16th 2010

In general one needs to have integration on paths in phase space, only for special Lagrangeans it reduces to integration only over paths in configuration space. Momenta integration can not be in general skipped over.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJul 11th 2013
• (edited Jul 11th 2013)

I have added to path integral a bunch of on the formalization in quantum mechanics by Wiener integration over stochastic processes.

Then I started a new subsection Realization – As an integral against the Wiener measure with a (very) brief indication of that…

…ending in an observation on what the passage from gauge coupling action functionals to the quantum propagator integral kernel looks like when viewed in higher geometric prequantum theory as an operation on correspondences in the slice (infinity,1)-topos over the moduli of circle bundles and of vector bundles, respectively.

• CommentRowNumber5.
• CommentAuthorzskoda
• CommentTimeJul 11th 2013
• (edited Jul 11th 2013)

Wiener integral needs to be analytically continued to become a QM path integral, Wiener measure certainly does not exist after the Wick rotation…I do not detect this basic idea in the entry…

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeJul 11th 2013

Right, I have added a “Eudlidean”-qualifier for the moment.

• CommentRowNumber7.
• CommentAuthorzskoda
• CommentTimeJul 12th 2013

Added reference

• Theo Johnson-Freyd, The formal path integral and quantum mechanics, J. Math. Phys. 51, 122103 (2010) arxiv/1004.4305, doi; On the coordinate (in)dependence of the formal path integral, arxiv/1003.5730
• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeJul 13th 2013

added a remark stating that the expression “holonomy integrated against the Wiener measure” is precisely what appears in the worldline formalism for computing QFT scattering amplitudes.

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Statistikk og Simulering

# Økt 14-16. Predator/Prey Modelling

## Statistikk og Simulering

### 15. Økt 14-16. Predator/Prey Modelling

Oppgåve 15.1 Implement a predator-prey simulator with visualisation for an eco-system with species of your choice. You should use the rabbit and fox example from Barnes and Kolling as a starting point, but also add a few new features of your own, in order to make the model more realistic. Such additions could include food for the prey species (e.g. grass), more detailled movement patterns, or other things. You can take inspiration from Project 4 (wolf/sheep/grass) of Shiflet & Shiflet, page 512 (see attachment).

Use the similator to predict (estimate) average life span of each of the species involved. Optionally, you can extract other statistical data for analysis as well. Work in a group of three (or two if numbers don’t fit).

The project consists of three phases, with an oral presentation of current results at the end of each phase. The three phases have different focus:

1.
Modelling
2.
Implementation
3.
Analysis

Note that the modelling stage gives you a lot of options. Modelling is always the exercise of selecting key features to be captured and less essential detail to be ommitted. In your group, you need to discuss what species to model and which features and properties to prioritise.


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Pie Shop

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Points: 17 (partial)
Time limit: 1.0s
Memory limit: 64M

Author:
Problem type
Maniacal Midsummer Marathon 2014 by AL, TL, JJ

Dorian loves pies, and he especially loves his local pie shop - that's why he's bought an all-you-can-eat pie dinner! The pies have been arranged into a grid with rows and columns . The pie in row and column has a tastiness value . Dorian is a messy eater, and whenever he eats a pie, he destroys all pies adjacent to it. He would like to maximize the sum of the tastiness values of the pies he eats, and also figure out how many ways he can achieve this sum.

Two pies respectively located at and , are adjacent if they are exactly unit of distance apart; that is, .

Input Specification

The first line contains two space-separated integers and .
The next lines each contain space-separated integers representing the tastiness values of the pies.

Output Specification

A single line containing two space-separated integers: the maximum possible sum and the number of ways to reach this sum. Since the number of ways may be very large, you should give the value modulo ().

Sample Input 1

3 4
1 2 3 1
5 7 2 1
1 4 1 0

Sample Output 1

14 3

Sample Input 2

3 3
1 1 0
0 0 1
1 1 0

Sample Output 2

3 5

Scoring

If your output is incorrectly formatted, then your program will score points.
Otherwise, if you correctly determine the first number but not the second, then your program will score of points for the test case.
If both numbers are correct, then you will receive of points for the test case.