text:
## Definitions

• Webster's Revised Unabridged Dictionary
• ***

## Quotations

• Will Rogers
“The worst thing that happens to you may be the best thing for you if you don't let it get the best of you.”
• Thomas Carlyle
“After all manner of professors have done their best for us, the place we are to get knowledge is in books. The true university of these days is a collection of books.”
• Don G. Mitchell
“Determine to become one of the best. Sufficient money will almost automatically follow if you get to be one of the best in your chosen field, whatever it is.”
• Gabriel Riqueti Mirabeau
“If honesty did not exist, we ought to invent it as the best means of getting rich.”
• Richard Saunders
“The lead dog gets the best view. The rest of the dogs view is butt ugly. Of course, the lead dog is also the first to fall into the ravine.”
• Henry Ford
“If you think of standardization as the best that you know today, but which is to be improved tomorrow; you get somewhere.”

## Usage

### In literature:

So it is best to get behind and a little lower than the tail of the enemy plane.
"Georges Guynemer" by Henry Bordeaux
He would have to make the best of it, dispose of it for what he could get.
"The Hidden Places" by Bertrand W. Sinclair
Most of us are willing to spend this time and labor if we get results and to get the best results with peonies we must have good varieties.
"Trees, Fruits and Flowers of Minnesota, 1916" by Various
The best judgment of the most skilled physicians is the least that the victim owes himself in his effort to get well.
"The Third Great Plague" by John H. Stokes
Had we not better make the best of it and get back to Morlaix?
"The Argosy" by Various
Parents therefore should see to it that their children get the kind of education they need, the kind that will serve them best in after life.
"Explanation of Catholic Morals" by John H. Stapleton
The best teaching is that which leads the child to get the clearest apprehension of what is taught.
"The Science of Human Nature" by William Henry Pyle
I shall be glad of this opportunity to get a few dollars, and then make the best of my way to Sakkatou.
"Narrative of a Mission to Central Africa Performed in the Years 1850-51, Volume 2" by James Richardson
If they make a break it'll be up to us to get the best of the trouble.
"Raw Gold" by Bertrand W. Sinclair
"The Swindler and Other Stories" by Ethel M. Dell
Now we have got to talk over the best plan of getting him out.
"The Tiger of Mysore" by G. A. Henty
Then, if we are getting the best of it, the rascal will join us, for the sake of the advantages which he expects to gain.
"With Clive in India" by G. A. Henty
"The Four Feathers" by A. E. W. Mason
I think I will dispose of this lot to Talbot; he is the best paymaster, and the first dark night I will get them away from here.
"The Brother Clerks" by Xariffa
Tinker set the car going, and began to try his hardest to get her best speed out of her.
"The Admirable Tinker" by Edgar Jepson
After that, I'd just have to call a couple of the big restaurants in Venusport and get the best price.
"The Revolt on Venus" by Carey Rockwell
"Stories That Words Tell Us" by Elizabeth O'Neill
Of the best version of "Peace Egg" which I have seen performed, I have as yet quite vainly endeavoured to get any part transcribed.
"The Peace Egg and Other tales" by Juliana Horatia Ewing
But he stuck to his post, knowing full well that, if he let up, the fire would soon get the best of them.
"Young Captain Jack" by Horatio Alger and Arthur M. Winfield
I intend to do my best to get Anne away this very night, if it upsets the entire town of Oakdale.
"Grace Harlowe's Plebe Year at High School" by Jessie Graham Flower
***

### In poetry:

It is right to get up and do,
But why outstrip the rest?
Why should one of the many be one of the few?
Why should you think to be best?
"Willie's Question" by George MacDonald
As John ne'er prosper'd, all allow,
In getting heirs, to kill a cow;
It follows, the two first, we see,
Were the best work-men of the three.
"The Wager" by William Hutton
He's going to send some goods to the World's Fair,
And I hope of patronage he will get the biggest share;
Because his Tweed cloth is the best I ever did see,
In the year of our Lord eighteen hundred and ninety-three.
"Lines in Praise of Mr. J. Graham Henderson, Hawick" by William Topaz McGonagall
Sallying forth, we fought at several points—but in each the luck was
against us;
Our foe advancing, steadily getting the best of it, push'd us back to
the works on this hill;
Till we turn'd, menacing, here, and then he left us.
"The Centerarian's Story" by Walt Whitman

### In news:

Carsonified is brilliant in putting together these online conferences, we get to watch some of the best speakers at the convenience of our space and theirs.
I finally get to write about my one true love of the year, the book I will champion against all others as the be all, end all best book of the year.
0Steve and Teresa LaFountain of Kalamazoo are trying their best to bring back the Fenton Cinema but the couple needs your help with funding in order to get the project going.
Well, it'll be a while before we get to see the film that was voted best picture of 2012 by the New York Film Critics Circle on Monday.
Saddled with financial woes and struggling through life in a dangerous neighborhood, washed-up actress Diane (Peters) refuses to let her depression get the best of her.
"When it was all over with, we got through cooking that was the best part of it when you tell everyone to come and get it," said Ricky Hicks.
In many ways, early autumn is the best time of year to get to the beach.
Calhoun County residents don't have to travel far to get one of the best cupcakes in the state.
RENTON – One of the best run defenses in the league the first half of the season, the Seattle Seahawks may have to get some extra work in on tackling dummies this week.
Not long ago, the city of Folsom was best known for its state prison – and as a place to drive through to get to Folsom Lake.
Since Everyone's Getting Engaged This Season, I Pulled Out Some Of The Best Wedding Fails For You To Enjoy .
The south end of Abercorn is crowded on the best of days, and it's going to get worse.
They have so much energy, and I am constantly in search of the best way to get their creative juices flowing.
And Myers will get to battle agains some of the best junior players in the world as Kelowna looks to win its second Memorial Cup in six years.
***

### In science:

Maps were made using both uniform uv weighting (to get the highest resolution possible, at the cost of noise level) and natural uv weighting (to get the best signal-to-noise ratio possible, at the cost of resolution).
CLASS B1152+199 and B1359+154: Two New Gravitational Lens Systems Discovered in the Cosmic Lens All-Sky Survey
Of all the OC’s we have studied photometrically, four have the metallicity measured by high dispersion spectroscopy, the best technique to get reliable results, with uncertainties of less than about 0.1 dex in [Fe/H] (see Gratton 1999 for a recent review).
Old Open Clusters as Tracers of Galactic Evolution
We have applied the plane-parallel shock model of Michael et al. (2001) with exactly the same best ﬁt parameters to the XMM-Newton data but we failed to get an acceptable ﬁt.
X-rays from Supernova Remnants
For a given detector mass, one gets the best sensitivity for seeing a non-zero value of the muon to electron neutrino transition probability by placing an experiment near or slightly below the ﬁrst oscillation maximum (1.27∆m2L/E ≈ π/2).
Summary of Working Group 2
It would be desirable to deﬁne a criterion of closeness with quantum predictions and, after that, to optimize the functions ρ and Pj in order to get the best LHV model of the said family, but this program will be left for the future.
Optical tests of Bell's inequalities not resting upon the absurd fair sampling assumption
Its fudge aspects are revealed in the different best-ﬁt alphas you get for different choices of how to normalize composition to the changing one of the sun (Ferraro et al. 2006).
Astrophysics in 2006
In any case, the best thing to do in order to get an expression for the tension of the strings, including this 1-loop correction, is to make a numerical model, ﬁnd the string-solution and integrate the energy-density to get a corrected tension.
D-term inflation after spontaneous symmetry breaking
If Rτ, S is measured with a 1% precision, the resulting Vus uncertainty will get reduced to around 0.6%, i.e. ±0.0013, making τ decay the best source of information about Vus .
QCD Description of Hadronic Tau Decays
Ratio of data to a ﬁt to a power law function for preliminary central Au+Au spectra from STAR after artiﬁcial scaling to get the best agreement between species.
What's Interesting About Strangeness Production? - An Overview of Recent Results
If our ordering of the variables is right, the best m belongs to M1 and we get an analogue of (5.7).
The Brouwer Lecture 2005: Statistical estimation with model selection
In the next step, we will try to get best ﬁtting of these data using regression analysis.
Statistical Analysis of Quasar Data and Hubble Law
The above scaling factor was determined in an attempt to get the best possible agreement to the ARPES quasiparticle self energy Σ1 (ω ) without changing the shape of I 2χ(ω )tr .
Finite band inversion of ARPES in Bi$_2$Sr$_2$CaCu$_2$O$_{8+\delta}$ in comparison with optics
Nevertheless, to get H s estimates and the best nonlinear results as possible, it is important to have the special structure of the two last terms in (23).
Geometric optics and boundary layers for Nonlinear Schrodinger equations
The conditional probability that each task (period) gets at least one edge given that K edges exist a priori and have been assigned in the best possible way (an edge each to K periods) is still dependent on the total number of edges, M, which is greater than n.
Sharp utilization thresholds for some real-time scheduling problems
Proof of Theorem 1.6: If we apply Theorem 1.5 to almost Mathieu operator directly, we can not get the best estimate, since by the local embedding theorem, we lose some analyticity, in order to get ﬁner estimate, we need the following: Theorem 6.1.
Embedding of Analytic Quasi-Periodic Cocycles into Analytic Quasi-Periodic Linear Systems and its Applications
***


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# Time lost or gained by clock

1. Dec 9, 2016

### Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I usually have trouble with these clock problems . In this problem I can certainly rule out option d) . I considered a simple situation in which initially the clock is at 12' o clock . After 65 minutes , the minute hand would be at 1'o clock mark wheras the hour hand would be somewhere between 1 and 2 . This means when the two hands meet the actual time elapsed should be more than 65 minutes .

Now the problem is how much time the clock has gained . Consider the hour hand . It covers 5 minute units in 60 minutes time . So, the time gained should be less than (5/60) times 86400 seconds . I am sure I am thinking incorrectly .

Thanks .

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2. Dec 9, 2016

### Cutter Ketch

Yes, it's easy to get lost in this one. I think there must be more than one way to think about it, too. My approach was to think about how many times the minute hand catches up to the hour hand in one "12 hour" period.

3. Dec 9, 2016

### Vibhor

Hello ,

Did you get one of the answers ? Could you elaborate your approach.

4. Dec 9, 2016

### TomHart

I think the first thing should be to write an equation for the angle of rotation for each of the clock hands as a function of time (seconds or minutes). Then it seems to make sense to me to start at 1:00. At 1:00, write an equation for how much the minute hand moves relative to how much the hour hand moves until they align.

After I started writing, but before I posted this message, I read @Cutter Ketch approach. I would recommend using his approach. The math is much simpler. :)

5. Dec 9, 2016

### Vibhor

The minute hand catches up with hour hand 11 times i.e after every 43200/11 = 3927 seconds . 65 minutes = 3900 seconds . This means the clock gains 27 seconds per hour .Or in other words 648 seconds in a day .So the answer should be option b) .

Do you agree ?

6. Dec 9, 2016

### TomHart

43200 is the number of seconds in 12 hours. So dividing that number by 11 gives the number of seconds in 12/11 of an hour, or 3927.2727 seconds. And (65)(60) = 3900 is the measured number of seconds (measured by a perfect clock) in 12/11 of an hour. So when you subtract those two numbers (3927.27 - 3900), the result does not give the clock error per hour. It gives the clock error per 12/11 of an hour. I hope I said that right.

Edit: Actually I think this problem is easier to work using minutes.

7. Dec 9, 2016

### Cutter Ketch

Almost correct. (and in multiple choice questions almost can be good enough). However, still wrong in two ways. First, it gets there 27 seconds early, but a true 65 minutes hasn't passed yet. After a true 65 minutes the margin will be slightly larger. Also, it's every 65 minutes, not every hour, so you scaled badly. Fix those things and your approach is correct.

I had a slightly different approach. The hands meet 11 times per 12 hours as you note. So if the hands meet in 65 minutes, then the minute hand covered 1 hour in 11 * 65 / 12 minutes. Divide that into 24 hours and that is how many hours the false clock shows in 24 true hours.

8. Dec 9, 2016

### Cutter Ketch

I don't know why I felt better working in hours. Your approach can quickly produce the right answer. You need the ratio false clock per true clock. You have that the false clock showed 3927 when only 3900 had truly passed. Take the ratio and multiply by 24 hours. (But don't use the rounded value of 3927. Keep all the digits)

9. Dec 9, 2016

### Vibhor

Another approach -

Hour hand meets minutes hand after every 65 minutes . This means actual time elapsed when the two hands meet for the 11th time is 65 × 22 × 60 = 85800 seconds . But time elapsed as per the clock is 86400 seconds .

The difference between the two is 600 seconds i.e option C) .

Do you agree/disagree with this line of reasoning ?

10. Dec 9, 2016

### Cutter Ketch

Everything except "option c". As I hinted, your last try although slightly flawed got the right multiple choice answer. Here as in the previous case you've lost track of which is true and which is false. 600 seconds is how far ahead the false clock is when IT says 24 hours have passed. However it is fast, so 24 hours haven't actually passed yet. By the time 24 hours have truly passed it will be even a little further ahead.

To get exactly how far ahead, you need the ratio of false time / true time for any interval you know. Multiply that by 24 true hours and you get how far ahead false time is after 24 true hours.

11. Dec 9, 2016

### Vibhor

I guess this is perhaps the exact reasoning problem setter had in his mind . This also looks to be the fastest way to arrive at the answer .

Impressive analysis Cutter Ketch !

As I said in OP I mess up big time in clock problems .

12. Dec 9, 2016

### Vibhor

I have not been able to understand your line of reasoning . Sorry .

Is it similar to Cutter Ketch's reasoning ?

13. Dec 9, 2016

### Cutter Ketch

Well, maybe it's just me, but I thought this was tricky and easy to get turned around. I certainly screwed up a few times.

14. Dec 9, 2016

### TomHart

As I read post #10 by @Cutter Ketch I realized I had made the same mistake - that I had not calculated how far off the clock was in one day; I had calculated how far off the clock was in what the errant clock reported to be one day - which actually was only 23 hours and 50 minutes. That is why I came up with the wrong answer of 600 seconds. The clock reported that 24 hours had elapsed, when in reality, only 23 hours and 50 minutes had elapsed. So as Cutter pointed out, in the additional 10 minutes, more error is added to the 600 seconds.
The ratio of reported time to actual time is 1440/1430.
So for an actual time of 1440 minutes, the reported time would be (1440)(1440/1430) = 1450.07 minutes.

15. Dec 9, 2016

### Vibhor

@haruspex , would you like to add/point out something in this discussion .

16. Dec 9, 2016

### haruspex

The least head-hurting way might be to figure out how slow the fast clock thinks the ideal clock is, then invert.
According to the fast clock, in 24 hours its hands meet up 22 times, so the ideal clock measures 65*22 minutes.
According to the fast clock, in 24 * 24*60/(65*22) hours the ideal clock measures 65*22 * 24*60/(65*22) minutes = 24 hours.
According to the fast clock, in 24 hours and 10.07 minutes the ideal clock measures 24 hours.

17. Dec 10, 2016

### Vibhor

From post#7
Would you like to review this ? I think true 65 minutes have passed .

$\frac{27.2727}{(65 × 60)} × 86400 = 604.2 sec$

Last edited: Dec 10, 2016
18. Dec 10, 2016

### Cutter Ketch

That's it!

19. Dec 10, 2016

### Vibhor

OK.

But you said in post#7 that when the clock was ahead by 27 seconds , true 65 minutes haven't passed .

I think true 65 minutes had passed .The previous post showed how 65 minutes gives correct answer .

20. Dec 10, 2016

### TomHart

Here's what I got:
When the clock is ahead by 27 seconds, the true time is 64 minutes and 21 seconds.
When the clock is ahead by 27.272727272 seconds, the true time is 65 minutes and 0 seconds.

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# Individual Stocks, Mutual Funds, Index Funds and Exchange Traded Funds (ETFs)

February 09, 2018

Investors are usually faced with a wide variety of options when deciding where to put their money. Each has its own pros and cons. In addition to bonds, derivatives, foreign currencies, and commodities, the investor can choose from among the following options.

# Individual Stocks

Individual stocks are a good option if you are willing to take on higher risk for a higher potential return. They offer you two ways to make gains: by receiving dividends on each share and by an increase in the value of the stock (capital gain).

To make a decision about buying a stock, investors should look into promising companies in promising sectors. Sectors with a high level of anticipated growth represent better investment choices. However, the investor should also choose highly competitive companies within those sectors since the performance of different companies can vary.

There are many considerations that should be taken into account when building a portfolio.

1. The investor should begin with determined financial objectives. The main objective is usually the return on investment.
2. The investor should choose investments that are suitable for his or her tolerable level of risk. This is extremely important.
3. The investor should consider how the composition of stocks in his portfolio will affect his tax payments.
4. The investor should choose an appropriate strategy for managing his portfolio.

The strategy can fall somewhere in between extreme diversification and extreme concentration. The most prudent choice will be somewhere in between. Extreme diversification will limit gains whereas excessive concentration will increase risk.

# Mutual Funds

Mutual funds are pools of investments. Investors gather their funds and provide the capital for the fund. The capital is then managed by investment professionals. The fund invests in either stocks or bonds or both. The fund makes a profit in two ways, the appreciation in the value of the stocks it holds (in case it sells them at a higher price), and from the dividends on stocks and interest on bonds it holds. Usually, earnings are distributed on an annual basis.

1. Professionals manage the fund
2. Economies of scale which allow for higher returns
3. Reduced risk due to diversification

1. The fees of professional management
2. The reduced returns due to diversification
3. The need for large amounts of cash to be liquid, which would reduce invested capital

The history of mutual funds goes back to 1700 in Europe when a Dutch merchant invited investors to share their capitals in one pool to be able to achieve their financial goals with more success. In England, the practice appeared under the name of “investment pooling” in 1800, and it moved to the US from there.

## Mutual Fund Fees and Expenses

There are many types of costs when dealing with a mutual fund, and the vary based on the fund.

Loads are fees paid to brokers who invest on your behalf in the fund. If you invest directly in the fund you do not need to pay those fees.

Transaction Fees

Transaction fees are charged when you conduct transactions such as buying or selling a share. These fees are paid to the fund itself and not to the broker. Usually, they are called purchase fees and redemption fees.

Expense Ratio

Mutual funds are like companies, and they have operating expenses. Those expenses are usually expressed in what is called the expense ratio, or the management expense ratio. These operating expenses typically constitute of hiring costs, distribution and service fees, account fees, and other expenses. Those expenses can range from 0.25% to 2%.

Demo and Real Mutual Fund Expenses

A demo account is an unreal account that an investor can open with virtual money deposited. The investors uses this virtual money to trade in real market conditions to test his performance.

A real account is a live account with real money deposited and with real profit or loss realized.

The expenses of acquiring a share in a mutual fund are not always transparent. They also can vary from the demo account and the real account with the mutual fund. For this reason, investors need to consult with the fund on the expenses charged beforehand and to use a calculator provided by some mutual funds to calculate the expenses and avoid lowering their return.

Exchange traded funds (ETFs) are funds whose securities can be bought and sold on a stock exchange. They track the performance of another asset or assets such as an index, or a basket of assets such as foreign currencies for example. This tracking feature is similar to index funds.

The assets they hold depends on the type of the ETF. If it is a commodity ETF, then the ETF’s performance is similar to that of the commodity.

The price of the share of the ETF usually fluctuates based on the fluctuations in the index it tracks, although the movements may not be exactly precise.

The fund usually purchases many diversified assets and then divides the ownership of those assets into shares. This way, the ETF works like a publicly listed company who has many assets and where investors hold shares. The only difference is that in the ETFs the investors have no direct claim on those assets, although they may receive a residual value in case the fund is liquidated. Also, the holders of the ETF shares are entitled to a proportion of the earnings in terms of dividends and interest (if applicable).

An example of a successful ETF is Vanguard 500 ETF, and its total return in 2017 was 20.2%. It tracks the performance of the S&P 500 Index. This ETF is considered among the cheapest ETF options and it is also among the largest.

ETFs are versatile in nature and investors have many options to choose from depending on their strategy. They can choose from among the following types of ETFs:

• Foreign market index ETFs: those invest purchase assets in foreign markets such as emerging markets.
• Foreign currency ETFs which track a currency or a basket of currencies, since they hold a foreign currency or a basket of foreign currencies
• Sector and Industry ETFs: they hold assets in specific sectors such as Information technology or renewable energies or others.
• Commodity ETFs: they hold commodities such as gold or silver, or others.
• Derivative ETFs: they hold derivative instruments such as future contracts or others.
• Style ETFs: they hold assets based on market capitalization such as large-cap, medium-cap, or small cap.
• Bond ETFs: they hold debt instruments
• Inverse ETFs which are good when the market is falling: those ETFs usually invest selling stocks that are declining or expected to decline.
• Leveraged ETFs: these ETFs track an asset or a basket of assets but they magnify their tracking by using debt instruments.
• Dividend ETFs: they invest in stocks with high dividends
• Innovative ETFs: they hold shares of innovative companies

## Exchange Traded Funds vs. Mutual Funds

Exchange traded funds are similar to mutual funds in that they pool investments together and invest the money in a single portfolio. The investors who hold the fund’s shares receive earnings in terms of dividends, interest payments, and capital gains if applicable.

The difference between mutual funds and ETFs is that the shares of the ETF are traded on exchanges where they can be bought or sold. Thus, ETF’s have higher daily liquidity and lower fees than mutual funds.

## Using the Style Box

The style box is a tool developed by Morningstar company, and it helps investors understand quickly the investment style of the ETF they are considering. The two considerations are the size and the types of investments. The box is shown below with nine categories of ETFs appear as in the below boxes. This box is for equities. A similar box can be provided for bonds where instead of the style, the credit rating is considered.


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# Given an implicit 3D plane, how do I find the orthogonal projection matrix - which projects any point - onto this plane?

The plane is given by the equation $Ax+By+Cz+d = 0$. Can you tell me how can I figure out the 4x4 matrix which orthogonally projects any point given by homogeneous coordinates onto this plane?

I am using homogeneous coordinates, so a point is given by 4 values: $\vec{Q} = [xh,yh,zh,w]$, which in 3D space is $[xh/w, yh/w, zh/w]$.

• Wouldn't that be a $3 \times 3$ matrix? – Joachim Jan 21 '14 at 10:35
• I am using homogeneous coordinates, so a point in the 3D space is represented by 4 values instead of 3. Transforming these [xh,yh,zh,w] vectors, I have to use 4x4 matrices. – Kalevi Jan 21 '14 at 10:37
• But the equation for your plane is in affine (i.e. not homogeneous) coordinates! – Joachim Jan 21 '14 at 10:39
• Yes. But I can't see why this is a problem. Both the point and the plane are in the 3D space. [xh,yh,zh,w] in 3D space is [xh/w, yh/w, zh/w]. – Kalevi Jan 21 '14 at 10:41
• There's nothing wrong with it, per se. However, when posting a question it is best to include all information and to be as clear as possible, so you get the best answer the fastest.. =) – Joachim Jan 21 '14 at 10:48

I'll start by looking about how you can describe the projection of an arbitrary point $Q$ onto your plane, then derive a matrix notation from that.

## Projecting a single point

The point $Q$ and its projection form a line which is perpendicular to the plane. All these perpendicular lines meet in a point $F=(A,B,C,0)^T$, which is the point at infinity orthogonal to your plane. A generic point on the line connecting $Q$ and $F$ can be described as a linear combination $Q+\lambda F$. (This inhomogenous formulation using a single parameter excludes $F$ itself, which is all right since $F$ will not be the projected point in any case.) Now you are looking for the point where this line connecting $F$ and $Q$ intersects the plane, i.e. $\left<Q+\lambda F,h\right>=0$, where $h=(A,B,C,D)^T$ is the vector describing your plane. This you can solve for $\lambda$:

$$\left<Q+\lambda F,h\right>=\left<Q,h\right> + \lambda\,\left<F,h\right>=0\\ \lambda=-\frac{\left<Q,h\right>}{\left<F,h\right>}$$

To avoid the division, you can also use a multiple of that point:

$$Q+\lambda F\;\sim\; \left<F,h\right>\,Q - \left<Q,h\right>F$$

This is the projection of a generic point $Q$.

## Finding the matrix

If you assume the coordinates of $Q$ to be variables, then the coordinates of the result will be linear in these variables, so you can interpret the whole operation as a linear map and therefore write it as a matrix. You can also find the formula of that matrix like this:

$$\left<F,h\right>\,Q - \left<Q,h\right>F = \left<F,h\right>\mathbb 1\cdot Q - Fh^T\cdot Q = \left(\left<F,h\right>\mathbb 1 - Fh^T\right)Q \\ = \begin{pmatrix} B^{2} + C^{2} & -A B & -A C & -A D \\ -A B & A^{2} + C^{2} & -B C & -B D \\ -A C & -B C & A^{2} + B^{2} & -C D \\ 0 & 0 & 0 & A^{2} + B^{2} + C^{2} \end{pmatrix}\cdot Q$$

If anyone reading this post should be interested in central instead of orthogonal projection, simply use that center of projection as $F$, and the above formula will help you compute the matrix for that as well.


text:
When I checked Wikipedia about Marginal likelihood, I find an equality of the form: $$p(a\mid b)=\int p(a\mid c)\cdot p(c\mid b)\,dc$$ and it's valid because the parameter c is marginalized/integrated out. However, when I tried proving its validity by actually performing integration and sum&product rules of probability, I failed... And I just can't figure out how that marginalization really works.
basically it is not necessarily true unless $p(a|c) = p(a|c,b)$. –  Memming Sep 1 '13 at 18:29


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# Reduced Row Echelon Form in $\Bbb Z/3\Bbb Z$?

I'm trying to understand the best way to approach this problem. Short of writing every combination of matrices, I'm wondering if anyone can help me learn how to solve this problem.

How many $3\times 2$ matrices with entries in $\mathbb{Z}/3\mathbb{Z}$ are in reduced row echelon form?

• So it is clear, $\mathbb{Z}/3\mathbb{Z}$ is another notation for $\mathbb{Z}_3$ (in the context of the OP's question) – apnorton Oct 6 '14 at 19:36
• As a first step, consider where the leading ones might be located (and what these imply about the rest of the matrix entries if in reduced row-echelon form). – hardmath Oct 6 '14 at 19:36
• @anorton: I'm not trying to be picky, just reflecting that it should be a field. The notation police got after me for using $\mathbb{Z}_p$ recently, saying it was wrong because that meant $p$-adic arithmetic! – hardmath Oct 6 '14 at 19:38
• @hardmath If I understand you correctly, you're referring to the leading 1's in the first column, since it's reduced there can only be one row with a 1 in the first column? – Russ Oct 6 '14 at 19:40
• @hardmath I know. I wrote that comment primarily for the OP's benefit, so they know that you didn't (really) change the content of their question, but rather just changed the formatting. :) – apnorton Oct 6 '14 at 19:40

## 1 Answer

In a matrix of size $m\times n$ (over any field), the rank is at most the minimum of $m,n$. In the present case then, a $3\times 2$ matrix has rank at most two, and thus a reduced row-echelon form can have at most two leading ones.

Some of the cases then turn out to be quite simple. If there are no leading ones, it must be because all the entries are zero, so there is only one matrix like that.

At the other extreme of having two leading ones, they must appear staggered in different columns, and because all the remaining entries in the two columns are zeros (because of reduced row-echelon form), there is only one matrix like that.

This leaves the slightly more interesting cases with one leading one. Which column is it in? If it is in the second column, that can only be because the entries in the first column were already zeros. And since the leading one in the second column clears out all the other entries there, this again is a case of only one matrix like that.

Finally suppose the first column has a leading one and the second column does not. A picture here is worth a thousand words:

$$\begin{bmatrix} 1 & x \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$$

The other entry in the first row, following its leading one, that can be anything! Well, anything in the field $\mathbb{Z}/3\mathbb{Z}$ is a short list of things: three possibilities.

Altogether there are, as Russ pointed out in a now removed Comment, six possibilities.


text:
# Maximum induced emf with 2 Solenoids

1. Nov 29, 2009

### Fazza3_uae

1. The problem statement, all variables and given/known data

A 3 m long large coil with a radius of 15.7 cm and 130 turns surrounds a 7.2 m long solenoid with a radius of 5.7 cm and 4700 turns. The current in the solenoid changes as $$I=I_0 sin (2\pi f t)$$ where I0= 30 A and f=60 Hz.Inside solenoid has 4700 turns and outside coil has 130 turns. The equation for the emf is $$E= E_0 sin \omega*t$$ .There is also a resistor on the smaller coil that is 34 ohms. Find the maximum induced emf in the large coil. Answer in units of V.

http://img194.imageshack.us/img194/5624/87290481.jpg [Broken]

2. Relevant equations

$$emf = -N\frac{d\phi}{dt}$$

Magnetic field of solenoid

$\mu_0IN/L$

3. The attempt at a solution

I will show my work done till now.If there is any mistake i made or step i forgot please tell me.

First i have to find the magnetic field of the inner solenoid.

Then i have to calculate the flux of the inner solenoid.

After that i have to use the magnetic field of inner solenoid to find the flux of outer solenoid.

Now i have to find the change in flux but the question here is , Do i have to subtract the inner from the outer or the opposite ???

Is is $$\phi$$Outer - $$\phi$$Inner ???

Or $$\phi$$Inner - $$\phi$$Outer ???

Now i know that i have to derive the flux change with respect to time but another question is here , shall i subtract the two fluxe first then derive the change or shall i derive them before subtraction ??? ( I think that i will get the same answer whatever choice i select )

After i get the answer ( which i hope to get it very soon ) i will just multiply the whole thing i mean the time rate of change of that flux with the number of turns for the outer solenoid .

Please check my steps & tell me if i did something wrong.

Last edited by a moderator: May 4, 2017
2. Nov 30, 2009

### Fazza3_uae

Please any one who can point me to the right direction ...

3. Nov 30, 2009

### vishal007win

there is no need to substract inner or outer flux....
after u found out the flux developed by inner solenoid,
find the flux associated with the outer solenoid and then the time derivative will directly yield the induced emf in large coil...

emf= -N d(flux)/dt
the flux is one associated with the coil... no need of substraction ...

4. Nov 30, 2009

### Fazza3_uae

thanks for response

hmmm.. ok there is no need to substract one from another,

how do i find the flux associated with the outer one ?

Do i have to use the inner magnetic field to find the outer flux ??

Or

Do i have to use the inner flux to find the outer flux and how ??

Last edited: Nov 30, 2009


text:
I would like to incorporate a link into a picture using \href. The links are stored in text files. The .tex file should be able to read form those text files (i.e. 1.text) and generate that clickable image.

...but I still have a problem:

% !TeX program = lualatex
\documentclass{article}%
\usepackage{graphicx}

\usepackage{verbatim}% for the input

%To generate a text file for demonstration
\begin{filecontents*}{1.txt}
\end{filecontents*}

\usepackage{hyperref,catchfile}
\newcommand\urlFromFile[1]{%
\CatchFileDef\myurl{#1}{\catcode\#=12\catcode\%=12}%
\expandafter\url\expandafter{\myurl}}

\begin{document}
\href{\urlFromFile{1.txt}}{\includegraphics[width=\linewidth]{example-image}}
\end{document}
%

I get:

line 16: TeX capacity exceeded, sorry [parameter stack size=10000]. \href{\urlFromFile{1.txt}}

\documentclass{article}%
\usepackage{graphicx}

\usepackage{verbatim}% for the input

%To generate a text file for demonstration
\begin{filecontents*}{1.txt}
\end{filecontents*}

\usepackage{hyperref,catchfile}
\newcommand\hrefFromFile[1]{%
\CatchFileDef\myurl{#1}{\catcode\#=12\catcode\%=12\endlinechar=-1}%
\expandafter\href\expandafter{\myurl}}

\begin{document}
\hrefFromFile{1.txt}{\includegraphics[width=\linewidth]{example-image}}
\end{document}

• Thanks a lot ! I noticed that there is an issue with the link. The link looks like "youtube.com%20" instead of "youtube.com". Do you have any ideas how to solve that ? – james Feb 9 '20 at 20:19
• I see no problem when I compile the example. – Ulrike Fischer Feb 9 '20 at 20:46
• Do you get redirected to youtube when you click on the image ? I don´t. – james Feb 10 '20 at 19:12
• I think the problem is that catchfile somwhoe adds a new line to the url and this then gets converted into %20, see stackoverflow.com/questions/1634271/… – james Feb 10 '20 at 19:22
• – james Feb 10 '20 at 19:26


text:
Volume 334 - The 36th Annual International Symposium on Lattice Field Theory (LATTICE2018) - Hadron Spectroscopy and Interactions
Scattering length from BS wave function inside the interaction range
Y. Namekawa,* T. Yamazaki
*corresponding author
Full text: pdf
Published on: 2019 May 29
Abstract
We evaluate $I=2$ two-pion scattering length through the scattering amplitude obtained by the Bethe-Salpeter wave function inside the interaction range.
The scattering length is computed with $m_\pi = 0.52-0.86$ GeV in the quenched lattice QCD.
Furthermore, the half-off-shell amplitude is calculated, from which the effective range is extracted.
Our results are compared with those by the conventional finite size method and by chiral perturbation theory to confirm consistency.
DOI: https://doi.org/10.22323/1.334.0078
Open Access


text:
Results (displaying all 28 matches)

Galois conjugate representations are grouped into single lines.
Label Dimension Conductor Defining polynomial of Artin field $G$ Ind $\chi(c)$
6.16711744.7t6.a.a $6$ $2^{6} \cdot 7^{2} \cdot 73^{2}$ x7 - 2x6 + 2x4 - 2x3 - 2x2 + 2x + 2 $A_7$ $1$ $2$
6.50808384.7t6.a.a $6$ $2^{6} \cdot 3^{8} \cdot 11^{2}$ x7 - 2x6 + 2x + 2 $A_7$ $1$ $2$
6.112021056.7t6.a.a $6$ $2^{6} \cdot 3^{6} \cdot 7^{4}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $1$ $2$
6.988410721.7t6.a.a $6$ $149^{2} \cdot 211^{2}$ x7 - 2x6 - 7x5 + 11x4 + 16x3 - 14x2 - 11x + 2 $A_7$ $1$ $6$
10.14117306610774528.70.a.a 10.14117306610774528.70.a.b $10$ $2^{9} \cdot 3^{14} \cdot 7^{8}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $0$ $-2$
10.351405582803624448.70.a.a 10.351405582803624448.70.a.b $10$ $2^{9} \cdot 3^{18} \cdot 11^{6}$ x7 - 2x6 + 2x + 2 $A_7$ $0$ $-2$
10.9115812039001375232.70.a.a 10.9115812039001375232.70.a.b $10$ $2^{9} \cdot 7^{6} \cdot 73^{6}$ x7 - 2x6 + 2x4 - 2x3 - 2x2 + 2x + 2 $A_7$ $0$ $-2$
10.965633540591074128803235361.70.a.a 10.965633540591074128803235361.70.a.b $10$ $149^{6} \cdot 211^{6}$ x7 - 2x6 - 7x5 + 11x4 + 16x3 - 14x2 - 11x + 2 $A_7$ $0$ $10$
14.21964383255760327839744.15t47.a.a $14$ $2^{12} \cdot 3^{18} \cdot 7^{12}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $1$ $2$
14.21964383255760327839744.21t33.a.a $14$ $2^{12} \cdot 3^{18} \cdot 7^{12}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $1$ $2$
14.247977080428199271469056.21t33.a.a $14$ $2^{12} \cdot 3^{24} \cdot 11^{8}$ x7 - 2x6 + 2x + 2 $A_7$ $1$ $2$
14.388625298846706628890624.21t33.a.a $14$ $2^{12} \cdot 7^{6} \cdot 73^{8}$ x7 - 2x6 + 2x4 - 2x3 - 2x2 + 2x + 2 $A_7$ $1$ $2$
14.2070984217554099625358135296.15t47.a.a $14$ $2^{12} \cdot 7^{6} \cdot 73^{10}$ x7 - 2x6 + 2x4 - 2x3 - 2x2 + 2x + 2 $A_7$ $1$ $2$
14.2430423365276781059668217856.15t47.a.a $14$ $2^{12} \cdot 3^{28} \cdot 11^{10}$ x7 - 2x6 + 2x + 2 $A_7$ $1$ $2$
14.954442544077406345814852730298705281.21t33.a.a $14$ $149^{8} \cdot 211^{8}$ x7 - 2x6 - 7x5 + 11x4 + 16x3 - 14x2 - 11x + 2 $A_7$ $1$ $14$
14.943381243144623486076833919663361832159717601.15t47.a.a $14$ $149^{10} \cdot 211^{10}$ x7 - 2x6 - 7x5 + 11x4 + 16x3 - 14x2 - 11x + 2 $A_7$ $1$ $14$
15.197679449301842950557696.42t294.a.a $15$ $2^{12} \cdot 3^{20} \cdot 7^{12}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $1$ $-1$
15.2231793723853793443221504.42t294.a.a $15$ $2^{12} \cdot 3^{26} \cdot 11^{8}$ x7 - 2x6 + 2x + 2 $A_7$ $1$ $-1$
15.19042639643488624815640576.42t294.a.a $15$ $2^{12} \cdot 7^{8} \cdot 73^{8}$ x7 - 2x6 + 2x4 - 2x3 - 2x2 + 2x + 2 $A_7$ $1$ $-1$
15.954442544077406345814852730298705281.42t294.a.a $15$ $149^{8} \cdot 211^{8}$ x7 - 2x6 - 7x5 + 11x4 + 16x3 - 14x2 - 11x + 2 $A_7$ $1$ $15$
21.1793685113483563715477940482605056.42t299.a.a $21$ $2^{18} \cdot 3^{30} \cdot 7^{16}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $1$ $1$
21.11862004710203242703974199567822205967663104.42t299.a.a $21$ $2^{18} \cdot 3^{44} \cdot 11^{16}$ x7 - 2x6 + 2x + 2 $A_7$ $1$ $1$
21.48159956283065062710236109126029614956937216.42t299.a.a $21$ $2^{18} \cdot 7^{10} \cdot 73^{16}$ x7 - 2x6 + 2x4 - 2x3 - 2x2 + 2x + 2 $A_7$ $1$ $1$
21.910960569944951755248535328155692865403121768144070417465759104897288961.42t299.a.a $21$ $149^{16} \cdot 211^{16}$ x7 - 2x6 - 7x5 + 11x4 + 16x3 - 14x2 - 11x + 2 $A_7$ $1$ $21$
35.354574685454344552857366707698313319755263283342109310976.70.a.a $35$ $2^{30} \cdot 3^{50} \cdot 7^{28}$ x7 - 3x6 + 3x5 + 3x4 - 9x3 + 3x2 + x - 3 $A_7$ $1$ $-1$
35.2941505296061748107334365673192725862524976759105923348765368909824.70.a.a $35$ $2^{30} \cdot 3^{68} \cdot 11^{24}$ x7 - 2x6 + 2x + 2 $A_7$ $1$ $-1$
35.917092692744573842474294019589138832009696605396024114765270054076416.70.a.a $35$ $2^{30} \cdot 7^{18} \cdot 73^{24}$ x7 - 2x6 + 2x4 - 2x3 - 2x2 + 2x + 2 $A_7$ $1$ $-1$
35.869459523932463822137531685068026633401800710814706501873560849812537388615432413560666705776366705233703041.70.a.a $35$ $149^{24} \cdot 211^{24}$ x7 - 2x6 - 7x5 + 11x4 + 16x3 - 14x2 - 11x + 2 $A_7$ $1$ $35$


text:
# NCERT Notes: Geography- Some major minerals and their characteristics

Subject: Geography
Category: The Fundamentals of Physical Geography
Topic: Some major minerals and their characteristics

NCERT notes on important topics for the UPSC civil services exam. These notes will also be useful for other competitive exams like banking PO, SSC, state civil services exams and so on. This article talks about some major minerals and their characteristics.

Major minerals and their characteristics (UPSC Notes):-

Minerals
• The earth is composed of various kinds of elements.
• A mineral is a naturally occurring organic and inorganic substance.
• It has an orderly atomic structure and a definite chemical composition as well as physical properties.
• It is composed of two or more elements.
• Minerals like sulphur, copper, silver, gold, graphite, etc. are single element minerals.
Feldspar
• Oxygen and Silicon are common elements in all types of feldspar.
• Sodium, potassium, calcium, aluminium, etc. are found in specific feldspar variety.
• It is the single most abundant mineral group on Earth and half of the crust of the earth is composed of feldspar.
• It has light cream to salmon pink colour.
• It is used for making ceramics and glass.

Quartz
• It is one of the most important components of granite and sand.
• This hard mineral consists of silica and is virtually insoluble in water.
• This white or colourless mineral is the most important component of granite.
Pyroxene
• Pyroxene comprises of aluminium, calcium, iron, magnesium, and silica.
• It forms 10 per cent of the crust of the earth.
• It is usually found in meteorites and occur in green or black colour.
Amphibole
• Calcium, iron, aluminium, silica, magnesium are the main elements of amphiboles.
• They form 7 % of the crust of the earth.
• It is found in green or black colour.
• It is mainly used in the asbestos industry.
• The other form of amphiboles is Hornblende.
Mica
• Mica comprises of potassium, aluminium, magnesium, iron, silica, etc.
• It forms 4 % of the crust of the earth. It is generally found in metamorphic and igneous rocks.
• Mica is used in electrical instruments.
Olivine
• Magnesium, iron and silica are the main elements of olivine.
• It is used in jewellery.
• It usually occurs in a greenish crystal form and is found in basaltic rocks.

Major minerals and their characteristics (UPSC Notes):-

Tune in to byjus.com/free-ias-prep for more strategy articles and news on UPSC Civil Services IAS Exam.


text:
## Notre Dame Journal of Formal Logic

### The Fact Semantics for Ramified Type Theory and the Axiom of Reducibility

Edwin D. Mares

#### Abstract

This paper uses an atomistic ontology of universals, individuals, and facts to provide a semantics for ramified type theory. It is shown that with some natural constraints on the sort of universals and facts admitted into a model, the axiom of reducibility is made valid.

#### Article information

Source
Notre Dame J. Formal Logic Volume 48, Number 2 (2007), 237-251.

Dates
First available: 16 May 2007

http://projecteuclid.org/euclid.ndjfl/1179323266

Digital Object Identifier
doi:10.1305/ndjfl/1179323266

Mathematical Reviews number (MathSciNet)
MR2306395

Zentralblatt MATH identifier
1137.03004

#### Citation

Mares, Edwin D. The Fact Semantics for Ramified Type Theory and the Axiom of Reducibility. Notre Dame Journal of Formal Logic 48 (2007), no. 2, 237--251. doi:10.1305/ndjfl/1179323266. http://projecteuclid.org/euclid.ndjfl/1179323266.

#### References

• [1] Anderson, C. A., "Russellian intensional logic", pp. 69--103 in Themes from Kaplan, edited by J. Almog, J. Perry, and H. Wettstein, Oxford University Press, Oxford, 1989.
• [2] Church, A., "Comparison of Russell's resolution of the semantical antinomies with that of Tarski", The Journal of Symbolic Logic, vol. 41 (1976), pp. 747--60. Reprinted in R. L. Martin, editor, Recent Essays on Truth and the Liar Paradox, Oxford University Press, Oxford,1984, pp. 289--306.
• [3] Hazen, A. P., and J. M. Davoren, "Russell's 1925 logic", Australasian Journal of Philosophy, vol. 78 (2000), pp. 534--56.
• [4] Hazen, A. P., "Predicative logics", pp. 331--408 in Handbook of Philosophical Logic. Vol. 1. Elements of Classical Logic, 2d edition, edited by D. M. Gabbay and F. Guenthner, D. Reidel, Dordrecht, 1983.
• [5] Jeffrey, R., Formal Logic: Its Scope and Limits, McGraw Hill, New York, 1967.
• [6] Kaplan, D., "The ramified theory of types", unpublished manuscript.
• [7] Landini, G., Russell's Hidden Substitutional Theory, Oxford University Press, New York, 1998.
• [8] Leblanc, H., "That Principia Mathematica, first edition, has a predicative interpretation after all", Journal of Philosophical Logic, vol. 4 (1975), pp. 67--70. Reprinted in H. Leblanc, Existence, Truth, and Provability, SUNY Press, Albany, 1982, pp. 236--39.
• [9] Leblanc, H., and G. Weaver, "Truth-functionality and the ramified theory of types", pp. 148--67 in Truth, Syntax and Modality (Proceedings of the Conference on Alternative Semantics, Temple University, Philadelphia, 1970), vol. 68 of Studies in Logic and the Foundations of Mathematics, North-Holland, Amsterdam, 1973.
• [10] Lewis, D., On the Plurality of Worlds, Blackwell, Oxford, 1986.
• [11] Linsky, B., Russell's Metaphysical Logic, vol. 101 of CSLI Lecture Notes, CSLI Publications, Stanford, 1999.
• [12] Russell, B., The Problems of Philosophy, Oxford University Press, Oxford, 1959.
• [13] Russell, B., An Inquiry into Meaning and Truth, George Allen and Unwin, London, 1980.
• [14] Russell, B., Theory of knowledge: The 1913 manuscript,'' pp. 1--178 in The Collected Papers of Bertrand Russell. Vol. 7, edited by E. R. Eames, George Allen and Unwin, London, 1984.
• [15] Russell, B., The Philosophy of Logical Atomism, Open Court, La Salle, 1985.
• [16] Urquhart, A., "The theory of types", pp. 286--309 in The Cambridge Companion to Bertrand Russell, edited by N. Griffin, Cambridge Companions to Philosophy, Cambridge University Press, Cambridge, 2003.
• [17] Whitehead, A. N., and B. Russell, Principia Mathematica, 2d edition, Cambridge University Press, Cambridge, 1925.


text:
# Are there more types of critical points beyond maxima/minima/saddle points for higher dimensions?

I had a course on single variable calculus and at that point, we had minima and maxima. Now on several variables calculus, there is maxima, minima and saddle points. Certain books of several variables calculus point only those three types of critical points but there is actually no guarantee that these are the only ones - other types could be more advanced and/or irrelevant for the study of calculus, perhaps?

• So are there more kinds of critical points for higher dimensional calculus?

• Supposing there are only these three, how can I know for sure that there are only these three? Supposing that there are more than three, how can I know that there are more than three?

Let $f : \mathbb{R}^n \to \mathbb{R}$ be a smooth function. A point $p \in \mathbb{R}^n$ is called a critical point if $$\frac{\partial f}{\partial x_1}(p) = \dots = \frac{\partial f}{\partial x_n}(p) = 0.$$

A critical point $p$ is called non-degenerate if the Hessian matrix at $p$

$$\left[\frac{\partial^2 f}{\partial x_i\partial x_j}(p)\right]$$

is non-singular (i.e. invertible).

Morse Lemma: Let $p$ be a non-degenerate critical point of $f : \mathbb{R}^n \to \mathbb{R}$. Then there is an neighbourhood $U$ of $p$ and coordinates $(y_1, \dots, y_n)$ on $U$ such that $y_1(p) = \dots = y_n(p) = 0$ and for every $q \in U$, we have $$f(q) = f(p) - y_1(q)^2 - \dots - y_k(q)^2 + y_{k+1}(q)^2 + \dots + y_n(q)^2.$$

The number $k$ is called the index of $f$ at $p$. It is a well-defined quantity (i.e. you can't get a different value of the index by using a different coordinate system). In fact, $k$ is equal to the number of negative eigenvalues of the Hessian matrix at $p$. Note that $0 \leq k \leq n$.

Non-degenerate critical points in $\mathbb{R}^n$ are classified by their index. Indeed, if $n = 2$, there are three possibilities for the index: $0$, $1$, or $2$. These correspond to a local minimum, saddle point, and local maximum respectively.

Index $0$: For all $q \in U$, we have $$f(q) = f(p) + y_1(q)^2 + y_2(q)^2 \geq f(p),$$ so we see that $p$ is a local minimum.

Index $1$: For all $q \in U$, we have $f(q) = f(p) - y_1(q)^2 + y_2(q)^2$.

Along the curve $y_1 = 0$, $$f(q) = f(p) + y_2(q)^2 \geq f(p),$$ so $p$ is a local minimum along $y_1 = 0$.

Along the curve $y_2 = 0$, $$f(q) = f(p) - y_1(q)^2 \leq f(p),$$ so $p$ is a local maximum along $y_2 = 0$.

Therefore, $p$ is a saddle point.

Index $2$: For all $q \in U$, we have $$f(q) = f(p) - y_1(q)^2 - y_2(q)^2 \leq f(p),$$ so we see that $p$ is a local maximum.

It is also worth considering the case $n = 1$ from this point of view. All non-degenerate critical points are either local minima (index $0$) or local maxima (index $1$), and zeroes of the derivative which are neither, sometimes called stationary points of inflection, are degenerate critical points.

As in the specific cases $n = 1, 2$, a critical point with index $0$ is a local minimum and a critical point with index $n$ is a local maximum. For $0 < k < n$, you can intuitively think of the critical point as having $k$ independent directions along which $p$ is a local maximum, and $n - k$ independent directions along which $p$ is a local minimum.

The study of non-degenerate critical points can also take place on a smooth manifold. The relation between such points and the topology of the manifold is known as Morse Theory.

That depends on how you classify them, that is what criterion you have for a critical point being of different type.

When you state that there are only these three in two dimensions you must be inclusive in what you consider saddle point. Basically what they do is to define the saddle point as a critical point which is not a local maximum or minimum - then of course you can never have a fourth type.

You can have an somewhat more detailed classification, by what relative values it takes in around the critical point. You have then strict minimum (only takes larger values), nonstrict minimum (takes both larger and equal values, but no lower), saddle point (take both larger and lower values), nonstrict maximum, strict maximum.

If you on the other hand imagine a saddle point looking like a saddle (but then the observation that there are only those three is basically incorrect) then you get into the question about regions around the critical point where the function takes larger and lower values and here is a slight difference in how these regions may be formed (assuming the function is continuous there):

In two dimensions you will around the critical point have directions in which the value increases and directions where it decreases (and directions inbetween where the value is constant). But you are not limited in having just two regions where it decreases and two where it increases (that is being a saddle point in the normal sense - having the shape of a saddle), you can also have situations where the function increases in on general direction and decreases in another (fx f(x,y)=x^3 at origin).

In some sense what we have in two dimensions that differs from higher is that in two dimensions we can only have one or even number of such regions. This is because if we traverse all directions we may start with non-increasing and then if we leave that we will reach a region where it's non-decreasing and so on alternating until we get back.

In higher dimensions this no longer applies - for example $f(x,y,z) = x^2+y^2-z^2$ has a critical point at origin where you have increasing behavior when $x^2+y^2 > z^2$ (one region), but decreasing when $z^2 > x^2+y^2$ (two non-connected regions).


text:
1

IIT-JEE 1994

If we consider only the principle values of the inverse trigonometric functions then the value of
$$\tan \left( {{{\cos }^{ - 1}}{1 \over {5\sqrt 2 }} - {{\sin }^{ - 1}}{4 \over {\sqrt {17} }}} \right)$$ is
A
$${{\sqrt {29} } \over 3}$$
B
$${{29} \over 3}$$
C
$${{\sqrt 3 } \over {29}}$$
D
$${3 \over {29}}$$
2

IIT-JEE 1986

The principal value of $${\sin ^{ - 1}}\left( {\sin {{2\pi } \over 3}} \right)$$ is
A
$${ - {{2\pi } \over 3}}$$
B
$${{{2\pi } \over 3}}$$
C
$${{{4\pi } \over 3}}$$
D
none
3

IIT-JEE 1983

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {{4 \over 5}} \right) + {{\tan }^{ - 1}}\left( {{2 \over 3}} \right)} \right]$$ is
A
$${{6 \over 17}}$$
B
$${{7 \over 16}}$$
C
$${{16 \over 7}}$$
D
none

Joint Entrance Examination

JEE Main JEE Advanced WB JEE


text:
# The Exposure Triangle Part 3 – Aperture

In this Third post on the exposure triangle, I will be covering aperture, which is in my mind the most interesting part of the exposure triangle. It has two primary affects, depth of field and exposure speed. Like shutter speed aperture works the same for digital and film cameras so will be making no distinctions in this post.

#### Aperture

Aperture is the measurement of the size of the hole in your lens that light travels through to fall on your image sensor.  It is measured with f numbers where a low number means you have a large aperture where more light is allowed onto the sensor.  A high f number means you have a small aperture where less light is allowed onto the sensor. Each of the aperture numbers on your camera is called a stop, and when you change from one full stop to the next or previous full stop you’re doubling or halving the amount of light that will fall on your image sensor. This will double or halve the amount of time you need to properly expose your image. I specify full stop because some lenses will allow you to change the aperture by less than a full stop allowing for a finer control. I will explain this in a bit more detail in the technical aside below but, the f stop scale does not go 1, 2 ,3, and so, full stops actually go 1, 1.4, 2, 2.8, etc.

A small f number also means you will have a shallow depth of field and a large f numbers means you will have a wider depth of field. depth of field is basically how much of your picture is in focus. More specifically it is the width of the plain that is in focus around your primary point of focus.  It may seem like having your photo in focus is something that you would always want, but this is not necessarily true. Most portrait photography actually has a shallow depth of field so that the subject of the photo is in focus but the background is blurry and does not distract from the subject.  Similarly with landscape photography you generally want a wider depth of field so that all of the photo is in focus.  something to consider with depth of field is that aperture is not the only factor in determining its width.  The factors are focal length in mm, longer focal lengths will have a shallower depth of field and distance to the subject where the closer you are to your subject the shallower the depth of field will be.  Depth of field examples are shown below, they were shot at 35mm about 1 foot away and from left to right they used f/ 1.8, f/ 11, and f/ 22

Looking at the photos above it’s easy to see that some parts of the image are clearly in focus and some are clearly out of focus.  There is no one single point of transition from sharp to unsharp so this area of transition is called the “area of confusion”. The region that is beyond the area of confusion and out of focus is called bokeh. Bokeh, pronounced boh-kə, is basically the aesthetic quality of the out of focus parts of the image. The variations in lens aberrations and aperture shape can cause the bokeh to be either pleasing to the eye or distracting, this a a very subjective topic so I will leave it to you to decide what bokeh counts as distracting.  Bokeh is most often thought of when you have distance sources of light in your image because they often appear as perfectly spherical and slightly fuzzy points.  With some creativity, you can make bokeh shapes. For now it’s just a fun little bit of possibility, but I’ll do a follow up post on creating bokeh shapes and why they are formed at a later date.  Classic bokeh balls and bokeh shapes are shown below.

You can usually change your aperture setting by just rotating one of the selection dials on your camera if you are in a mode that allows you to change aperture, but check your manual for details.

#### Technical Details

For those of you still here, welcome to my first the technical section, if you have any additional questions or comments after reading this section feel free to ask below and will do my best to answer.

As discussed earlier the aperture of your lens is measured with f numbers. The f number is the ratio of the focal length of the lens to the effective aperture diameter. This means that the widest aperture that a lens can achieve is dependent on the lens’ length and maximum width. It also explains why quick lenses which have a wide maximum f stop, around 1.4 or 2.8, tend to either be short or very wide at the end of the lens when compared to a similar length lens with a higher maximum f stop. One full stop changes the aperture number by √2 (~1.41), as mentioned above this doubles or halves the amount of light that falls on the image sensor.  This makes perfect sense when you look at what you’re doing mathematically, if you change the diameter of a circle by multiplying or dividing by √2 you’re also effectively doubling or halving the area of that circle as shown by the equations below.

$Area=\pi*r^2 = \frac{\pi*Diameter^2}{4}$

$\frac{\pi*1^2}{4}=0.78$

$\frac{\pi*(1*1.41)^2}{4}=1.56$

Unfortunately the technical details behind depth of field, area of confusion, and bokeh are too vast to cover in this technical aside, I will likely be devoting an entire post on the topic at a later date.

#### End Technical Details

Tune in next post for the exciting finale on the exposure triangle where I will bringing it all together and talking about how I generally choose settings and what to think about when you’re composing a shot.

Peter

## 6 thoughts on “The Exposure Triangle Part 3 – Aperture”

1. Pingback: Metering | petersouthlandphotography


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Astronomy 2e

# Summary

### 25.1The Architecture of the Galaxy

The Milky Way Galaxy consists of a thin disk containing dust, gas, and young and old stars; a spherical halo containing populations of very old stars, including RR Lyrae variable stars and globular star clusters; a thick, more diffuse disk with stars that have properties intermediate between those in the thin disk and the halo; a peanut-shaped nuclear bulge of mostly old stars around the center; and a supermassive black hole at the very center. The Sun is located roughly halfway out of the Milky Way, about 26,000 light-years from the center.

### 25.2Spiral Structure

The gaseous distribution in the Galaxy’s disk has two main spiral arms that emerge from the ends of the central bar, along with several fainter arms and short spurs; the Sun is located in one of those spurs. Measurements show that the Galaxy does not rotate as a solid body, but instead its stars and gas follow differential rotation, such that the material closer to the galactic center completes its orbit more quickly. Observations show that galaxies like the Milky Way take several billion years after they began to form to develop spiral structure.

### 25.3The Mass of the Galaxy

The Sun revolves completely around the galactic center in about 225 million years (a galactic year). The mass of the Galaxy can be determined by measuring the orbital velocities of stars and interstellar matter. The total mass of the Galaxy is about $2×10122×1012$ MSun. As much as 95% of this mass consists of dark matter that emits no electromagnetic radiation and can be detected only because of the gravitational force it exerts on visible stars and interstellar matter. This dark matter is located mostly in the Galaxy’s halo; its nature is not well understood at present.

### 25.4The Center of the Galaxy

A supermassive black hole is located at the center of the Galaxy. Measurements of the velocities of stars located within a few light-days of the center show that the mass inside their orbits around the center is about 4.6 million MSun. Radio observations show that this mass is concentrated in a volume with a diameter similar to that of Mercury’s orbit. The density of this matter concentration exceeds that of the densest known star clusters by a factor of nearly a million. The only known object with such a high density and total mass is a black hole.

### 25.5Stellar Populations in the Galaxy

We can roughly divide the stars in the Galaxy into two categories. Old stars with few heavy elements are referred to as population II stars and are found in the halo and in globular clusters. Population I stars contain more heavy elements than globular cluster and halo stars, are typically younger and found in the disk, and are especially concentrated in the spiral arms. The Sun is a member of population I. Population I stars formed after previous generations of stars had produced heavy elements and ejected them into the interstellar medium. The bulge stars, most of which are more than 10 billion years old, have unusually high amounts of heavy elements, presumably because there were many massive first-generation stars in this dense region, and these quickly seeded the next generations of stars with heavier elements.

### 25.6The Formation of the Galaxy

The Galaxy began forming a little more than 13 billion years ago. Models suggest that the stars in the halo and globular clusters formed first, while the Galaxy was spherical. The gas, somewhat enriched in heavy elements by the first generation of stars, then collapsed from a spherical distribution to a rotating disk-shaped distribution. Stars are still forming today from the gas and dust that remain in the disk. Star formation occurs most rapidly in the spiral arms, where the density of interstellar matter is highest. The Galaxy captured (and still is capturing) additional stars and globular clusters from small galaxies that ventured too close to the Milky Way. In 3 to 4 billion years, the Galaxy will begin to collide with the Andromeda galaxy, and after about 7 billion years, the two galaxies will merge to form a giant elliptical galaxy.

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## Finding Price Paths from Difference Equations

$Qt^{s}$ = Quantity supplied at time t

$Qt^{d}$ = Quantity demanded at time t

$Pt$ = Price of good at time t

$Qt^{s}$ = -Pt +2Pt-1

$Qt^{d}$ = 50 – 0.5Pt

The market is in equilibrium $Qt^{d} = Qt^{s}$

I am asked to find the price path for Pt if P0 = 30

Im a little unsure as to whether I’ve plugged the right info into the formula

$50-0.5Pt = -Pt + 2Pt-1$

Rearranging that I get

$Pt=A(\frac {-2}{0.5})$ $^{t} + \frac{1+50}{0.5 + 2}$

$= A(\frac{-2}{0.5})^{t}+ 20.4$

$A = [30 - (\frac{1+50}{0.5+2})$

$=[30-20.4] = 9.6$

Therefore I get

$Pt=9.6(\frac{-2}{0.5})^{t} + 20.4$

I’m unsure as to whether I’ve actually done this correctly, as my understanding on this isn’t that great, any help would be very much appreciated.


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# Oracle 12cR2 – How To Create DDL Script Of A View With All Dependencies and Permissions

ddloracleoracle-12cview

I have a view in my production database and I want to create same view in my test database.

I tried this commmand:

select dbms_metadata.get_ddl('VIEW','VIEW_NAME','VIEW_OWNER') from dual;

It generated a create script and I run it. But it didn't copy all of the dependencies, data types, etc.

How can I do this?

Regards,

That would require selecting from dba_dependencies. Since your dependencies may have their own dependencies that query would probably need to be recursive.

Also, if any of the dependencies are in a different schema then you'll also need the grants to go with them which you can get by selecting from dba_tab_privs and dba_col_privs.


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# Find the LCM 21 , 35 , 15

, ,
The LCM is the smallest number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
has factors of and .
has factors of and .
has factors of and .
The LCM of is the result of multiplying all prime factors the greatest number of times they occur in either number.
The LCM of is .
Multiply by .
Multiply by .
Find the LCM 21 , 35 , 15


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# Digital Logic

Design a half-subtractor circuit with inputs x and y and outputs Diff and B out . The circuit

subtracts the bits x – y and places the difference in D and the borrow in B out .

(a) Design a full-subtractor circuit with three inputs x , y , B in and two outputs Diff and

B out . The circuit subtracts x – y – B in , where B in is the input borrow, B out is the output

borrow, and Diff is the difference

• haqrozone
20 orders completed
$5.00 ANSWER Tutor has posted answer for$5.00. See answer's preview

**** is **** please ***** ********** below

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# 6.4 Special products  (Page 2/5)

 Page 2 / 5

Multiply: ${\left(4{u}^{3}+1\right)}^{2}.$

## Solution

 Use the pattern. Simplify.

Multiply: ${\left(2{x}^{2}+1\right)}^{2}.$

$4{x}^{4}+4{x}^{2}+1$

Multiply: ${\left(3{y}^{3}+2\right)}^{2}.$

$9{y}^{6}+12{y}^{3}+4$

## Multiply conjugates using the product of conjugates pattern

We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

What do you notice about these pairs of binomials?

$\begin{array}{ccccccc}\hfill \left(x-9\right)\left(x+9\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(y-8\right)\left(y+8\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(2x-5\right)\left(2x+5\right)\hfill \end{array}$

Look at the first term of each binomial    in each pair.

Notice the first terms are the same in each pair.

Look at the last terms of each binomial in each pair.

Notice the last terms are the same in each pair.

Notice how each pair has one sum and one difference.

A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference has a special name. It is called a conjugate pair and is of the form $\left(a-b\right),\left(a+b\right)$ .

## Conjugate pair

A conjugate pair    is two binomials of the form

$\left(a-b\right),\left(a+b\right).$

The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.

There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier.

Let’s look for the pattern by using FOIL to multiply some conjugate pairs.

$\begin{array}{ccccccc}\hfill \left(x-9\right)\left(x+9\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(y-8\right)\left(y+8\right)\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(2x-5\right)\left(2x+5\right)\hfill \\ \hfill {x}^{2}+9x-9x-81\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{y}^{2}+8y-8y-64\hfill & & & \hfill \phantom{\rule{4em}{0ex}}4{x}^{2}+10x-10x-25\hfill \\ \hfill {x}^{2}-81\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{y}^{2}-64\hfill & & & \hfill \phantom{\rule{4em}{0ex}}4{x}^{2}-25\hfill \end{array}$

Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term.

$\begin{array}{}\\ \hfill \left(a+b\right)\left(a-b\right)={a}^{2}-\text{____}\hfill \\ \hfill \text{To get the}\phantom{\rule{0.2em}{0ex}}\mathbf{\text{first term, square the first term}}.\hfill \end{array}$

The last term came from multiplying the last terms, the square of the last term.

$\begin{array}{}\\ \hfill \left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}\hfill \\ \hfill \text{To get the}\phantom{\rule{0.2em}{0ex}}\mathbf{\text{last term, square the last term}}.\hfill \end{array}$

What do you observe about the products?

The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.

Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.

The product of conjugates is always of the form ${a}^{2}-{b}^{2}$ . This is called a difference of squares.

## Product of conjugates pattern

If $a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b$ are real numbers,

The product is called a difference of squares.

To multiply conjugates, square the first term, square the last term, and write the product as a difference of squares.

Let’s test this pattern with a numerical example.

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\left(10-2\right)\left(10+2\right)\hfill \\ \text{It is the product of conjugates, so the result will be the}\hfill & & & \\ \text{difference of two squares.}\hfill & & & \phantom{\rule{4em}{0ex}}\text{____}-\text{____}\hfill \\ \text{Square the first term.}\hfill & & & \phantom{\rule{4em}{0ex}}{10}^{2}-\text{____}\hfill \\ \text{Square the last term.}\hfill & & & \phantom{\rule{4em}{0ex}}{10}^{2}-{2}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}100-4\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}96\hfill \\ \text{What do you get using the Order of Operations?}\hfill & & & \\ & & & \phantom{\rule{4em}{0ex}}\left(10-2\right)\left(10+2\right)\hfill \\ & & & \phantom{\rule{4em}{0ex}}\left(8\right)\left(12\right)\hfill \\ & & & \phantom{\rule{4em}{0ex}}96\hfill \end{array}$

Notice, the result is the same!

Multiply: $\left(x-8\right)\left(x+8\right).$

## Solution

First, recognize this as a product of conjugates. The binomials have the same first terms, and the same last terms, and one binomial is a sum and the other is a difference.

 It fits the pattern. Square the first term, x . Square the last term, 8. The product is a difference of squares.

p-2/3=5/6 how do I solve it with explanation pls
P=3/2
Vanarith
1/2p2-2/3p=5p/6
James
is y=7/5 a solution of 5y+3=10y-4
yes
James
Lucinda has a pocketful of dimes and quarters with a value of $6.20. The number of dimes is 18 more than 3 times the number of quarters. How many dimes and how many quarters does Lucinda have? Rhonda Reply Find an equation for the line that passes through the point P ( 0 , − 4 ) and has a slope 8/9 . Gabriel Reply is that a negative 4 or positive 4? Felix y = mx + b Felix if negative -4, then -4=8/9(0) + b Felix -4=b Felix if positive 4, then 4=b Felix then plug in y=8/9x - 4 or y=8/9x+4 Felix Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. macadamia nuts cost$9 per pound and almonds cost $5.25 per pound. how many pounds of macadamia nuts and how many pounds of almonds should macario use for the mixture to cost$6.50 per pound to make?
Nga and Lauren bought a chest at a flea market for $50. They re-finished it and then added a 350 % mark - up Makaila Reply the sum of two Numbers is 19 and their difference is 15 Abdulai Reply 2, 17 Jose interesting saw Felecia left her home to visit her daughter, driving 45mph. Her husband waited for the dog sitter to arrive and left home 20 minutes, or 13 hour later. He drove 55mph to catch up to Felecia. How long before he reaches her? Rafi Reply integer greater than 2 and less than 12 Emily Reply 2 < x < 12 Felix I'm guessing you are doing inequalities... Felix Actually, translating words into algebraic expressions / equations... Felix hi Darianna hello Mister Eric here Eric He charges$125 per job. His monthly expenses are $1,600. How many jobs must he work in order to make a profit of at least$2,400?
at least 20
Ayla
what are the steps?
Alicia
6.4 jobs
Grahame
32
Grahame
1600+2400= total amount with expenses. 4000/125= number of jobs needed to make that min profit of 2400. answer is 32
Orlando
He must work 32 jobs to make a profit
POP
what is algebra
repeated addition and subtraction of the order of operations. i love algebra I'm obsessed.
Shemiah
hi
Krekar
Eric here. I'm a parent. 53 years old. I have never taken algebra. I want to learn.
Eric
One-fourth of the candies in a bag of M&M’s are red. If there are 23 red candies, how many candies are in the bag?
they are 92 candies in the bag
POP
rectangular field solutions
What is this?
Donna
t
muqtaar
the proudact of 3x^3-5×^2+3 and 2x^2+5x-4 in z7[x]/ is
?
Choli
a rock is thrown directly upward with an initial velocity of 96feet per second from a cliff 190 feet above a beach. The hight of tha rock above the beach after t second is given by the equation h=_16t^2+96t+190
Usman
Stella bought a dinette set on sale for $725. The original price was$1,299. To the nearest tenth of a percent, what was the rate of discount?
44.19%
Scott
40.22%
Terence
44.2%
Orlando
I don't know
Donna
if you want the discounted price subtract $725 from$1299. then divide the answer by $1299. you get 0.4419... but as percent you get 44.19... but to the nearest tenth... round .19 to .2 and you get 44.2% Orlando you could also just divide$725/\$1299 and then subtract it from 1. then you get the same answer.
Orlando


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## Cryptology ePrint Archive: Report 2020/1590

RandPiper -- Reconfiguration-Friendly Random Beacons with Quadratic Communication

Adithya Bhat and Nibesh Shrestha and Aniket Kate and Kartik Nayak

Abstract: Random beacon protocols provide a continuous public source of randomness and their applications range from public lotteries to zero-knowledge proofs. Existing random beacon protocols in the bounded synchronous model sacrifice either the fault tolerance or the communication complexity for security, or ease of reconfigurability. This work overcomes the challenges with the existing works through a novel communication efficient combination of state machine replication and (publicly) verifiable secret sharing (PVSS/VSS) protocols.

We first design a new Byzantine fault-tolerant state machine replication protocol with $O(\kappa n^2)$ bits communication per consensus decision without using threshold signatures. Next, we design GRandPiper (Good Pipelined Random beacon), a random beacon protocol with bias-resistance and unpredictability, that uses PVSS and has a communication complexity of $O(\kappa n^2)$ always (best and worst cases), for a static adversary. However, GRandPiper allows an adaptive adversary to predict beacon values up to $t+1$ epochs into the future. Therefore, we design BRandPiper (Better RandPiper), that uses VSS and has a communication complexity of $O(\kappa fn^2)$, where $f$ is the actual number of faults, while offering a strong unpredictability with an advantage of only a single round even for an adaptive adversary.

Category / Keywords: cryptographic protocols / Random Beacons

Date: received 19 Dec 2020, last revised 27 Apr 2021

Contact author: bhat24 at purdue edu, nxs4564 at rit edu, aniket at purdue edu, kartik at cs duke edu

Available format(s): PDF | BibTeX Citation

Short URL: ia.cr/2020/1590

[ Cryptology ePrint archive ]


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# Do black holes have Arago spots?

I've been reading about Arago spots and the Aragoscope. Some fascinating concepts! Basically, due to wave diffraction, there is a bright spot in the centre of the shadow behind any circular object.

It strikes me that there are some similarities between a black hole and the disc used for the Arago spot experiments. Black holes (or their event horizons) are spherical, so should cast a circular shadow. Have there been any observations of bright spots in the centre of a black hole? If so, would these be useful for astronomy, like a supersized Aragoscope?

I have read that the shadowing object has to be very precisely circular, so I am not sure if a black hole is circular enough. Oblateness caused by rotation might distort the event horizon? There might also be issues with relativity around the fringes of the black hole. I don't have a clear concept of how gravitational lensing and Fresnel diffraction would work together.

A proper analysis of this should treat light as an EM wave propagating in the curved spacetime background of the black hole. The papers

(1) "Wave optics and image formation in gravitational lensing," https://arxiv.org/abs/1207.6846

(2) "Viewing Black Holes by Waves," https://arxiv.org/abs/1303.5520

(3) "Wave Optics in Black Hole Spacetimes: Schwarzschild Case," https://arxiv.org/abs/1502.05468

present analyses for the case of a non-rotating (Schwarzschild) black hole. The papers (1) and (2) use numerical techniques (solving the wave equation for the massless scalar field using a finite difference method) and the paper (3) uses analytic techniques. The abstract of (2) says

We study scattering of waves by black holes. Solving a massless scalar field with a point source in the Schwarzschild spacetime, waves scattered by the black hole is obtained numerically. We then reconstruct images of the black hole from scattered wave data for specified scattering angles. For the forward and the backward directions, obtained wave optical images of black holes show rings that correspond to the black hole glories associated with existence of the unstable circular photon orbit in the Schwarzschild spacetime.

Figure 7 in (2) shows "Images of black holes reconstructed from scattering waves...", and here is an excerpt from that figure:

Notice the relatively bright spot in the center. To help interpret the picture, page 12 says:

In the geometric optics limit, images of black holes can be obtained by solving null geodesics. For the observer at $$\theta_0=0$$, the primary null rays, which are deflected by the black hole but do not go around it, result in the Einstein ring. The secondary and the higher degrees of null rays that go around the black hole many times also form ring images with smaller angular radius compared to the Einstein ring.

This is called the glory effect. The words "Arago spot" are not used in the paper, but the words "Airy disk" are used. The papers (2) and (3) both say that an extension of this analysis to rotating (Kerr) black holes is underway, but I don't know if it has been published yet.

• Great answer, thanks. That image definitely looks similar to an Arago spot, but perhaps a little dimmer. The authors seem to only be commenting on the rings - which are definitely interesting in themselves! This answer triggered me to look up Einstein Rings, and most of the examples had quite a large bright region in the middle. I guess the bright region is the accretion disc, which would hide the Arago spot in most cases? – craq Jan 9 at 3:57
• @craq I would guess that you're right, that something like a bright accretion disk would hide any diffraction-related spot that might otherwise be visible in the center. I didn't find any analyses that accounted for this. – Chiral Anomaly Jan 9 at 4:03
• I think, given today's news release with the first ever photo of a black hole, I have to retract my comment, at least partially. The Einstein Ring photos I was looking at before were lensed by galaxies, and the bright spot in the middle was the galaxy, not the accretion disc. In the new photo, the centre of the black hole actual looks very black, so perhaps the accretion disc wouldn't be a problem for using a black hole as an aragoscope. – craq Apr 10 at 18:59

I also recently learned about Arago spots and had the very same thought regarding black holes. Concerning your retracting part of your comments, it's true that there isn't evidence black boles generally produce Arago spots. That said, while the variables make it unlikely most black holes would produce an Arago spot they also make it likely that a few black holes would produce Arago spots. After all the conditions would have to be perfect (the perfect black holes and the exact pinning up of the light source, the black holes and us, etc.). In the future, after the photographing of thousands of black holes, I wouldn't be surprised if an Arago spot would be among the photographs.


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# Induction

### Induction

Use the fact that $\frac{1}{k}$ - \frac{1}{k+1} = \frac{1}{k(k+1)} to show that

n
sigma (\frac{1}{k(k+1)} = 1- \frac{1}{n+1}
r=1

How do I solve this?
Guest

### Re: Induction

$\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)} =\frac{k+1-k}{k(k+1} =\frac{1}{k(k+1)}$ (1)
$\Rightarrow A= \sum_{k=1}^{n }\frac{1}{k(k+1)}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}$
From (1) $\Rightarrow A= \frac{1}{1}-\frac{1}{2}+\frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}$
$\Rightarrow A=\sum_{k=1}^{n }\frac{1}{k(k+1)}= 1 - \frac{1}{n+1} = \frac{n}{n+1}$.

nathi123

Posts: 13
Joined: Sun Sep 17, 2017 1:56 pm
Reputation: 12

### Re: Induction

nathi123 wrote:$\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)} =\frac{k+1-k}{k(k+1} =\frac{1}{k(k+1)}$ (1)
$\Rightarrow A= \sum_{k=1}^{n }\frac{1}{k(k+1)}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}$
From (1) $\Rightarrow A= \frac{1}{1}-\frac{1}{2}+\frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}$
$\Rightarrow A=\sum_{k=1}^{n }\frac{1}{k(k+1)}= 1 - \frac{1}{n+1} = \frac{n}{n+1}$.

The induction step is hidden in the ...s. The key idea (using (1) to telescope the sum) that makes the induction work is of course correct, but if your instruction was explicitly to use induction, you'll want to replace the ... handwaving with a formal induction proof like

$\frac{1}{1*2} = 1-\frac{1}{2}$ so it's valid when n = 1.

Suppose that for some $l\ge2$,

$\sum_{k=1}^{l-1 }\frac{1}{k(k+1)}= 1 - \frac{1}{l}$.

Then $\sum_{k=1}^{l}\frac{1}{k(k+1)}= \left(1 - \frac{1}{l}\right) + \frac{1}{l(l+1)}$ (add the next term, corresponding to $k=l$)

$= 1 - \frac{1}{l} + \left(\frac{1}{l} - \frac{1}{l+1}\right)$ (Simplify the sum using the induction hypothesis and apply (1) to the last term)

$= 1-\frac{1}{l+1}$. (Regroup and cancel the $1/l$ terms) Showing that the same identity holds when one additional term is included in the sum.
Guest

### Re: Induction

The way I have demonstrated proves the affirmation without the use of mathematical induction.
I do not think it is appropriate to demonstrate this method using (1).
I suggest the following formulation. To prove that
$\sum_{k=1}^{n }\frac{1}{k(k+1)}=\frac{n}{n+1}$ using mathematical induction.

If n=1$\Rightarrow\frac{1}{1.2}=\frac{1}{2} -$true
Let's assume that $=\frac{n}{n+1}$ (2)and we will prove that
$\sum_{k=1}^{n +1}\frac{1}{k(k+1)}=\frac{n+1}{n+2}$
$\sum_{k=1}^{n +1}\frac{1}{k(k+1)}=\sum_{k=1}^{n }\frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n^{2}+2n+1}{(n+1)(n+2)}$
$\Rightarrow \sum_{k=1}^{n +1}\frac{1}{k(k+1)}=\frac{(n+1)^{2}}{(n+1)(n+2)}=\frac{n+1}{n+2}$.

nathi123

Posts: 13
Joined: Sun Sep 17, 2017 1:56 pm
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# DavMail Office365 Bridge

Recently I started at an institution which uses a groupware which is deliberately incompatible with open groupware protocols such as IMAP,SMTP,CalDAV. This leaves the option to use the provided webinterface which I think is inconvenient.

Fortunately there exists the DavMail project, which is a bridge between the ExchangeWebServices and open protocols. It works by being a client to EWS and then exposing a server interface on the usual ports shifted up by 1000 (e.g. IMAP being at 1143 instead of 143). There are two possible operation modes, one as a local deamon to run on desktop systems and one server/headless version. For a single desktop client, the first is fine, but I wanted to have access from my phone too. Unfortunately the headless version does not yet have an oauth workflow to log in with the two-factor authentication. The workaround is to register through the desktop version and then copy the authentication token to the headless deployment.

## Acquiring the Authentication Token

Install DavMail on a desktop system and keep the default values, except change the authentication method to O365Interactive. The easiest way to trigger the MFA authentication is to open the CalDAV interface at http://localhost:1080/ and log in with your email and password.

This opens a dialog of DavMail with a link which you can use in any browser, authenticate to Microsoft and provide the SMS/auth token and authorize the DavMail application. Copy the final link (with the token in the parameters) showing an empty page, back to the DavMail dialog. After successful authentication, close davmail and locate the davmail.properties file to extract the line which starts with

davmail.oauth.<user@email.tld>.refreshToken={AES}...

This is your oauth token which keeps you logged in. It is encrypted with your password such that on successful authentication to the DavMail server it can decrypt the token and use it to log into the EWS service.

## Preparing a TLS Certificat

When running DavMail as a server, the connection no longer stays on the localhost and must be secured by TLS. Acquire a TLS certificate for the domain you want to run the service on with e.g. Let's Encrypt or ZeroSSL.

This usually returns a key.pem (unencrypted RSA key) and cert.pem (with Certificate-Chain). To prepare it for DavMail, it has to be stored in PKCS12 format. First encrypt the private key with a password of your choice:

openssl rsa -aes256 -in key.pem -out key.encrypted

and then create the PKCS12 container with a fresh container password:

openssl pkcs12 -export -in cert.pem -inkey key.encrypted -certfile cert.pem -out davmail.p12

As ACME certificates usually have a 90 day validity, you need to automate this process in a cron script.

## Deploying a DavMail server

The easiest option to run the DavMail headless version is as docker image with docker-compose and the following configuration:

version: "3"

services:
davmail:
image: connectical/davmail
ports:
- "1080:1080" # caldav
- "1025:1025" # smtp
- "1143:1143" # imap
volumes:
- logs:/var/log/davmail
- ./davmail.properties:/etc/davmail/davmail.properties
- ./davmail.p12:/etc/davmail/davmail.p12
volumes:
logs

The ./davmail.p12 path should point to the previously prepared PKCS12 container.

To prepare the davmail.properties file, download a default version from https://github.com/ogarcia/docker-davmail/blob/master/.circleci/docker/davmail.properties and set up the certificate with the matching passwords:

davmail.ssl.keystoreType=PKCS12
davmail.ssl.keystoreFile=/etc/davmail/davmail.p12

More importantly, enable the headless version to authenticate to EWS. therefore replace davmail.enableEws=auto with

davmail.enableEws=EWS
davmail.authenticator=davmail.exchange.auth.O365Authenticator

davmail.oauth.<user@email.tld>.refreshToken={AES}...

If you want to proxy multiple users, add a line for each.

Run the server with

docker-compose up -d

and make sure that the ports are open in your firewall.

## Clients

To connect clients to the new server, you need the correct settings as there is no autoconfiguration available. As server address it is always the one you used for the certificate creation. As username, use your full email, and the correct password.

Protocol Port Config
IMAP 1143 enable TLS (not STARTTLS) and cleartext auth
SMTP 1025 enable TLS, plain auth
CaDav https://server.domain:1080/ make sure to use https

### Troubleshooting: K9 Mail on Android

When setting up the mail account on K-9 mail or its derivatives, it fails because of an unsupported IMAP command. This is resolved by creating a new empty folder named SPECIAL-USE through the Outlook web interface.


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• # question_answer A gas reacts with $CaO$, but not with $NaHC{{O}_{3}}$. The gas is [AFMC 1987] A) $C{{O}_{2}}$ B) $C{{l}_{2}}$ C) ${{N}_{2}}$ D) ${{O}_{2}}$

$CaO+C{{l}_{2}}\,\to \,CaOC{{l}_{2}}$ $NaHC{{O}_{3}}+C{{l}_{2}}\,\to$ No reaction


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# Electric potential of insulating rod

1. Feb 29, 2004

### pattiecake

This hmwk. problem has got me totally buggin...

A uniformly charged insulating rod of length 20cm is bent into the shape of a semicircle (so it looks like the letter "C"). The rod has a total charge of -8.50e-6C. Find the electric potential at a point P, in the center of the semicircle.

I know crazy things happen to electric potential when dealing with insulators...also, will this problem involve integrals?

If anyone has a clue I'd appreciate the guidance! Thanks in advance!

2. Mar 2, 2004

### pattiecake

Why will noone respond? =(

3. Mar 3, 2004

### paul11273

Yes, this requires and integral.

$$V=ke \int{dq/r}$$
$$V =ke(Q/r)$$

Where Q is the total charge of the rod and r is the radius this rod makes. Remember that circumference=2*pi*r, so use that to find r.

I think no one has responded because you didn't post any sort of work that you did.
Post up what you have and we can take this further.

4. Mar 7, 2004

### pattiecake

thank u so much. i didn't mean to be an answer leech!

i used this formula before: V=Ke(Q/r): using Circumference=2piR to find the radius. The problem was after i got my answer: -2.39, i forgot we were dealing with a semicircle, and that i had to divide by 2. my final answer (which webassign confirmed) was -1.2MV. thanks so much for your help!


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# Conversion of Radians into Degrees and Vice Versa

We know that the circumference of a circle of radius $r$ is $2\pi r = \left( l \right)$ and the angle formed by one complete revolution is $\theta$ radian; therefore,

$\begin{gathered} \theta = \frac{l}{r} \\ \Rightarrow \theta = \frac{{2\pi r}}{r} \\ \Rightarrow \theta = 2\pi \,radian \\ \end{gathered}$

Thus we have the relationship

$\begin{gathered} 2\pi \,radian\, = 360^\circ \\ \Rightarrow \pi \,radian = 180^\circ \\ \Rightarrow 1{\text{ }}radian = \frac{{180^\circ }}{\pi }\,\,\, = \frac{{180^\circ }}{{3.1416}}\,\,\,\, = \,\,\,57.296^\circ \\ \end{gathered}$

Further
$\begin{gathered} 1^\circ = \frac{\pi }{{180}}\,radian \\ 1^\circ = \frac{{3.1416}}{{180}}\,\,\,\,\, = 0.0175\,radian \\ \end{gathered}$

Example:

Convert the following angles into degrees:

(i) $\frac{{2\pi }}{3}$ radians
(ii) $3 radians$

Solution:

(i)
$\begin{gathered} \frac{{2\pi }}{3}\,radians = \frac{2}{3}\left( {\pi \,radian} \right) \\ \frac{{2\pi }}{3}\,radians = \frac{2}{3}\left( {180^\circ } \right)\,\,\,\, = 120^\circ \\ \end{gathered}$

(ii)

$3{\text{ }}radians\; = 3{\text{ }}\left( {1{\text{ }}radian} \right) = {\text{ }}3\left( {57.296^\circ } \right)\,\,\,\, = 171.888^\circ$

Example:

Convert $54^\circ \,\,45’$ into radians.

Solution:

$\begin{gathered} 54^\circ \,\,45′ = \left( {54\frac{{45}}{{60}}} \right)^\circ = \left( {54\frac{3}{4}} \right)^\circ \,\,\, = \frac{{219^\circ }}{4} \\ \Rightarrow 54^\circ \,\,45′ = \frac{{219}}{4}\left( {1^\circ } \right) \\ \Rightarrow 54^\circ \,\,45′ = \frac{{219}}{4}\left( {0.0175} \right) = 54^\circ \,\,45′ = 0.958\,radians \\ \end{gathered}$

Most calculators automatically would convert degrees into radians and radians into degrees.

Example:

An arc subtends an angle of $70^\circ$ at the center of a circle and its length is $132mm$. Find the radius of the circle.

Solution:

$70^\circ \,\,\, = 70 \times \frac{\pi }{{180}} = \frac{{70}}{{180}}\left( {\frac{{22}}{7}} \right) = \frac{{11}}{9}\,\,\,\,\,\,\,\,\,\,\left( {\because \pi = \frac{{22}}{7}} \right)$
$\therefore \theta = \frac{{11}}{9}\,radian$ and $l = 132m.m$
$\therefore \theta = \frac{1}{r} \Rightarrow r = \frac{l}{\theta } = 132 \times \frac{9}{{11}} = 108m.m$

Example:

Find the length of the equatorial arc subtending an angle of $1^\circ$ at the center of the Earth, taking the radius of the Earth as 5400km.

Solution:

$1^\circ = \frac{\pi }{{180}} = \frac{{3.1416}}{{180}}\,radians$

$\therefore \theta = \frac{{3.1416}}{{180}}$ and $r = 6400\,km$
Now
$\theta = \frac{l}{r}$
$\Rightarrow l = r\theta = 6400 \times \frac{{31416}}{{1800000}} = 111.7\,km$