text:
Proposition 51.14.2. Let $A$ be a Noetherian ring which has a dualizing complex. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. Let $s \in \mathbf{Z}$. Let $K \in D_{\textit{Coh}}^+(A)$. The following are equivalent

1. $H^ i_ T(K)$ is a finite $A$-module for $i \leq s$, and

2. for all $\mathfrak p \not\in T$, $\mathfrak q \in T$ with $\mathfrak p \subset \mathfrak q$ we have

$\text{depth}_{A_\mathfrak p}(K_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > s$

Proof. Formal consequence of Proposition 51.13.2 and Lemma 51.14.1. $\square$

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text:
## History is too important to be left to the history professors

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## Marie Curie says stop hating on quilt plots already.

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## Video Tutorial: Breaking Down the Definition of the Hazard Function

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## Ulam spirals: Visualizing properties of prime numbers with SAS

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## Data Stories Podcast: 2013 in Review, Outlook to 2014

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## Converting plots to data II

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## Introduction to the particle Gibbs sampler

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# Why is the speed of light?

1. ### rogerp

5
Hi

I'm a confused layman, and was hoping to get some answers to the question: why is the speed of light? That's not a typo - I mean, why is it what it is? The questions below are specific ways to ask essentially the same question.

1. The speed of light is 299,792,458 metres per second. What stops it from reaching 299,792,459 m/s?

2. If light has no mass, what stops it from having infinite speed? What’s slowing it down?

3. Light from the sun takes 8 minutes 24 seconds to reach the earth – why can’t it get here faster?

.

Last edited: Apr 17, 2007
2. ### HallsofIvy

41,050
Staff Emeritus
I'm not sure what your question is. Is "why is the speed of light that particular number and not some other number?"

The speed of light happens to be that number because of a particular choice of the length of a meter- which is completely arbitrary. If you measure distance in km and time in hours, or distance in feet and time in minutes, you would get some other number. Nothing very deep about that!

### Staff: Mentor

The Universe has a fundamental symmetry, called Lorentz symmetry, which causes to exist an asymptotic upper limit on all relative speeds. Objects with mass can never reach that upper limit. Massless objects (such as photons) must travel at that limiting speed, and no other. These are consequences of the mathematics of Lorentz symmetry.

This obviously begs the question, "why does the universe have Lorentz symmetry?" Nobody knows. (Or at least there is no generally accepted answer, as far as I know.) This is the ultimate answer to all "why?" questions in physics, by the way. The answer to any "why?" question leads to another "why?" question, and ultimately we always come to one that we cannot answer in the context of physics, at least not yet.

4. ### Gib Z

3,347
1. We defined light to go that distance in 1 second. If somehow miraculously, and it cant, light went faster, that new distance is still 299,792,458m.

2. Nothings "slowing" it down, thats the fastest it can go. It can't have an infinite speed because the fabric of spacetime, and infact everything, has 2 quantities. Permittivity and Permeability. The former is sort of a measure of how well Electric fields transmit in the material, in this case spacetime. The latter is a similar measure, except of magnetic fields. Both these quantities must be finite, by definition. Something can't transmit something infinity well.

Since Light is just magnetic and electric fields travelling through space, the speed must be finite as the ability for the spacetime to transmit it is finite.

It may interest you that this equation holds:

$$c= \frac {1} {\sqrt{\varepsilon_0\mu_0}}$$

Where c is the speed of light, episilon_0 is the permittivity of free space, and mu_0 is the permeability.

3. As I said, light can't go faster than spacetime allows it to. To find out why spacetime has the values of permeability and permittivity that it does, you would need to do some serious study into both physics and mathematics so you can Understand Maxwells Famous Equation and Einsteins General Relativity.

5. ### Gib Z

3,347
Didn't you read my post >.<
EDIT: or jtbells...he's got it in the first line lol.

6. ### rogerp

5
Yes, thanks Gib , I did! That's why I deleted my reply.

May I ask one more question?

I'm really interested by this idea of the Permittivity and Permeability of the fabric of spacetime. But am I right that this is a circular theory - i.e. that the speed of light is defined by the Permittivity and Permeability of spacetime, and the Permittivity and Permeability of spacetime are defined by the speed of light?

I'm not sure that physics is explaining here so much as describing. I suspect this question may have been answered by the second paragraph in jtbell's post.

.

7. ### Gib Z

3,347
It is somewhat circular, good spot :D

However, only in its definitions. I only when I say "only" you're thinking I'm quite an idiot :) You see, Permittivity and Permeability can be experimentally measured!

The Permittivity and Permeability of Space are defined exactly, as a consequence of the mathematical relation I posted previously. However, one can experimentally measure Permittivity and Permeability without measuring the speed of light, and use the equation to find the speed of light. Measurements are not definitions, but they do show us approximations and that the speed of light, what ever it is, in finite.

8. ### rogerp

5
Gib, thanks again - believe me, I think you're very, very far from an idiot; after all, you just explained some very difficult concepts to a layman in a way which he understood.

I've got to think about this some more, and come back with further impertinent questions later. One is likely to be about how there can ever be 'inertial observers' (which, as I understand it, is a postulate of special relativity) when we're all falling towards one massive body or another by the operation of gravity...

I thought of this while reading this paper:

http://www.cfa.harvard.edu/Walsworth/pdf/PhysicsWorld_2004.pdf

Anyway, thanks again

.

9. ### reilly

1,071
The plain fact is that nobody knows why c is what it is. It is built into physics from the beginning -- Maxwell's equations require knowledge of the speed of light. All the stuff about Permittivity and Permeability is after the fact, and provides no clue about c.

In short, like the charge of the electron, the gravitational constant, ... , nobody knows from c -- Nature gives us things for which we have no explanation.

Regards,
Reilly Atkinson

10. ### exponent137

374
good question

...Which we have no explanation, but one explanation should be.

In next reference there are clues.

Last edited: Apr 17, 2007
11. ### rbj

the gravitational constant, the speed of propagation (EM, gravitation, information, whatever is your "instantaneous" action), Planck's constant, permitivity of free space, these are all numbers that are purely human constructs resulting on how humans chose to define units of time, length, mass, and charge. note the 4 constraints and the 4 unknowns (that are eventually measured in terms of these anthropometric units). i know that the meter is now defined to be the distance traversed by light in 1/299792458 second, but that wasn't the original definition, for the sake of illustration, let's revert the definition of the meter back before 1960.

now, if instead, you measure everything in terms of Planck units, the values for all these constants become 1, 1, 1, and $1/(4 \pi)$. the only numbers given to us by Nature are dimensionless numbers. so, to ask "why is the speed of light equal to 299792458 meters per second?" causes us to ask the more basic questions that are "why are there about 6.1821 x 1034 Planck lengths in a meter?" and "why are there about 1.8549 x 1043 Planck times in a second?" those are the meaningful questions.

you see, if we measure and describe everything in Planck units, there simply is no c, or G, or $\hbar$, or $4 \pi \epsilon_0$. those numbers just go away from all of our algebraic equations of physical law.

we know that the meter and second are determined to be related to our experience of reality. a meter is approximately how big we are. and a second is, well, not the absolute shortest period of time in our bioological perception, but close to it. somewhere i read that, at our prime, we can do about 20 basic logical operations or computations (crude compare operations) per second in our conscious mind, dunno if that is true or not. when we get older, our CPU slows down but we got a better database.

so then we might start asking, why are there about 1025 Planck lengths in the Bohr radius (about the size of atoms)? and why are there about 105 atoms in the length of a biological cell? and why are there about the same number of cells in the length of a sentient organism like us?

you could construct similar questions about physiological processes regarding why it takes about 1040 Planck times for us to do anything with our bodies (without tools). there is a relationship of the speed of our consciousness and the time around a second. if we were tiny insects, a second might seem like a long period of time. but then we wouldn't be thinking about why the speed of light is what it is. suppose we lived for 1000 years and it took us what we now consider a minute to think every new thought. we wouldn't be manuvering cars at 100 km/hr and i don't think a second would be our unit time and the speed of light would seem even faster to us.

you answer those questions, then you'll get close to why the meter and the second are as big as they are (relative to some Natural units), and, from that, you'll have an idea why the speed of light (which as far as Planck units are concerned is just 1, not some dumb and arbitrary number like 299793458) is what it is, from our perspective.

the speed of light (and of all things instantaneous) is just the natural speed of things in the universe of which to reference all other speeds against.

now the Elementary Charge, that's something else, since the natural unit of charge is already defined. it's interesting (to me at least) that the electron charge, relative to the Planck charge, is just $$\sqrt{\alpha}$$. one can say that e is what it is because of the value of the fine-structure constant (this important dimensionless number that the universe does give us), or (what i prefer) the fine-structure constant is what it is because of the amount of charge that Nature has bestowed upon electrons and other charged particles. because i think that they should have normalized $\epsilon_0$ and $4 \pi G$ instead of $4 \pi \epsilon_0$ and $G$, i think the most natural units would come out slightly different than the Planck units (but be the same order of magnitude) and then, measured in these natural units, the electron charge would be $\sqrt{4 \pi \alpha}$ which is about 0.30282212 . VERY close (as far as orders of magnitude go) to unity. i think 0.30282212 is the number theoretical physicists should put on their walls instead of 137.035999679 . i think the latter flows from the former.

those are the sorta numerical "why?" questions i would be wondering about.

Last edited: Apr 17, 2007
12. ### rbj

i took a look at it. G is not dimensionless and i agree with Michael Duff (and a bunch of other physicists who dispute VSL and varying-G) about the meaninglessness of the varying G or c or any other dimensionful parameter in and of themselves. it's only the dimensionless parameters (which often come out as ratios of like dimensioned quantities) that matter. variable $\alpha$ has meaning. if the fine-structure changed, we would actually notice, but not if G did or c.

as stated previously, you measure everything in Planck units and there simply is no G or c or $\hbar$ to vary. they're just not there in physical law.

here's a nice quote from John Barrow:

Last edited: Apr 17, 2007
13. ### windscar

29
I think if you were to answer the question; Why does all matter move exactly the speed of light slower than light?; would give more insight into this matter. What property of matter makes everything with mass travel exactly the speed of light slower than light, that travels at one truely constant speed.

14. ### HallsofIvy

41,050
Staff Emeritus
What?? That doesn't even make sense. It is not true that all matter moves "exactly the speed of light slower than light". Literally interpreted that would mean that everything has speed 0! I presume you mean "everything travels slower than the speed of light", not "exactly the speed of light slower".

I'm not sure you will accept it as a property of mass, by the Lorentz formula an object moving at speed v has mass
$$\frac{m0}{\sqrt{1- v^2/c^2}}$$
where m0 is the "rest mass" or "invariant mass". As long as m0 is not 0, an object cannot move at the speed of light because that would mean infinite mass. Or, since acceleration= force/mass, the increasing mass requires constantly increasing force to accelerat further. Close to the speed of light requires almost infinite force to accelerate any more.

15. ### windscar

29
Everything with mass measure's the speed of light being exactly the speed of light. It is one of the basis that started relativity. So if every object see's light traveling exactly the speed of light, then the difference in velocity of every object relative to light is exactly the speed of light. I did not say that matter can travel the speed of light, but all matter travel's the speed of light slower than light. Because they all measure the speed of light to be C.

16. ### cristo

8,384
Staff Emeritus
Correct.

This is an invalid comment. I can measure the speed of light in my rest frame, and I will find it to be c. But, you are saying that light can measure my velocity relative to itself. However, this is not true, since light has no inertial frame of reference, and so the notion of travelling on a photon and measuring a body's speed relative to the photon does not make sense.

Last edited: Apr 19, 2007
17. ### windscar

29
You don't have to sadle up on any photons to see this. All you must do is realize that light is always the speed of light faster than you. The difference in velocity of light and matter is always C. And I was just putting a brain teaser out there that matter always travels a speed of zero relative to light when traveling at a constant speed.

18. ### masudr

932
In which reference frame?

3,347
In its own.

20. ### masudr

932
Good, good.

Also, what does it mean when a username has a line through it in these forums (as windscar's appears to now do in all of his/her posts)?


text:
When I type my papers, I need to include a table listing all the figures. When I change the order of the pages, the table is no longer correct. Is there any way to make the page numbers in the table of figures update automatically?

Yi Chen

To do this, you must ensure that each figure is identified by a caption field, not just by a caption typed below it. Go through your text from start to finish, click on the line just after each figure, and choose Caption... from the Insert menu. Enter the desired caption. If you already had a plain-text caption below the figure, delete it.

Now navigate to the page where you want the figure table. Select Index and Tables from the Insert | Reference menu, then click the Table of Figures tab. Set the options for the table's format and click OK. Any time the page numbers change, just click on the table and press F9 to update the table with current information.


text:
# naginterfaces.library.specfun.opt_lookback_fls_greeks¶

naginterfaces.library.specfun.opt_lookback_fls_greeks(calput, sm, s, t, sigma, r, q)[source]

opt_lookback_fls_greeks computes the price of a floating-strike lookback option together with its sensitivities (Greeks).

For full information please refer to the NAG Library document for s30bb

https://www.nag.com/numeric/nl/nagdoc_28.5/flhtml/s/s30bbf.html

Parameters
calputstr, length 1

Determines whether the option is a call or a put.

A call; the holder has a right to buy.

A put; the holder has a right to sell.

smfloat, array-like, shape

must contain , the th minimum observed price of the underlying asset when , or , the maximum observed price when , for .

sfloat

, the price of the underlying asset.

tfloat, array-like, shape

must contain , the th time, in years, to expiry, for .

sigmafloat

, the volatility of the underlying asset. Note that a rate of 15% should be entered as .

rfloat

The annual risk-free interest rate, , continuously compounded. Note that a rate of 5% should be entered as .

qfloat

The annual continuous yield rate. Note that a rate of 8% should be entered as .

Returns
pfloat, ndarray, shape

contains , the option price evaluated for the minimum or maximum observed price or at expiry for and .

deltafloat, ndarray, shape

The leading part of the array contains the sensitivity, , of the option price to change in the price of the underlying asset.

gammafloat, ndarray, shape

The leading part of the array contains the sensitivity, , of to change in the price of the underlying asset.

vegafloat, ndarray, shape

, contains the first-order Greek measuring the sensitivity of the option price to change in the volatility of the underlying asset, i.e., , for and .

thetafloat, ndarray, shape

, contains the first-order Greek measuring the sensitivity of the option price to change in time, i.e., , for and , where .

rhofloat, ndarray, shape

, contains the first-order Greek measuring the sensitivity of the option price to change in the annual risk-free interest rate, i.e., , for and .

crhofloat, ndarray, shape

, contains the first-order Greek measuring the sensitivity of the option price to change in the annual cost of carry rate, i.e., , for and , where .

vannafloat, ndarray, shape

, contains the second-order Greek measuring the sensitivity of the first-order Greek to change in the volatility of the asset price, i.e., , for and .

charmfloat, ndarray, shape

, contains the second-order Greek measuring the sensitivity of the first-order Greek to change in the time, i.e., , for and .

speedfloat, ndarray, shape

, contains the third-order Greek measuring the sensitivity of the second-order Greek to change in the price of the underlying asset, i.e., , for and .

colourfloat, ndarray, shape

, contains the third-order Greek measuring the sensitivity of the second-order Greek to change in the time, i.e., , for and .

zommafloat, ndarray, shape

, contains the third-order Greek measuring the sensitivity of the second-order Greek to change in the volatility of the underlying asset, i.e., , for and .

vommafloat, ndarray, shape

, contains the second-order Greek measuring the sensitivity of the first-order Greek to change in the volatility of the underlying asset, i.e., , for and .

Raises
NagValueError
(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry with a put option, .

Constraint: for put options, for all .

(errno )

On entry with a call option, .

Constraint: for call options, for all .

(errno )

On entry, .

Constraint: for all .

(errno )

On entry, .

Constraint: and .

(errno )

On entry, .

Constraint: for all .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, and .

Constraint: , where is the machine precision.

Notes

opt_lookback_fls_greeks computes the price of a floating-strike lookback call or put option, together with the Greeks or sensitivities, which are the partial derivatives of the option price with respect to certain of the other input parameters. A call option of this type confers the right to buy the underlying asset at the lowest price, , observed during the lifetime of the contract. A put option gives the holder the right to sell the underlying asset at the maximum price, , observed during the lifetime of the contract. Thus, at expiry, the payoff for a call option is , and for a put, .

For a given minimum value the price of a floating-strike lookback call with underlying asset price, , and time to expiry, , is

where . The volatility, , risk-free interest rate, , and annualised dividend yield, , are constants.

The corresponding put price is

In the above, denotes the cumulative Normal distribution function,

where denotes the standard Normal probability density function

and

where is taken to be the minimum price attained by the underlying asset, , for a call and the maximum price, , for a put.

The option price is computed for each minimum or maximum observed price in a set or , , and for each expiry time in a set , .

References

Goldman, B M, Sosin, H B and Gatto, M A, 1979, Path dependent options: buy at the low, sell at the high, Journal of Finance (34), 1111–1127


text:
# How can I enable use of math in my html web page

I have a blog where people can post their comments. My problem is that I need to enable people to be able to use math formulas using latex syntax. I cannot make any changes to the server. I can only embed a script in my HTML page. Kindly help.

-

You can use jsMath to solve this problem. It is all client side javascript, but you must reference the jsMath library to use its functions.

-
There's also MathJax, which seems to be shaping up as the successor to jsMath. –  Jim Lewis Jul 28 '10 at 18:07
I can't install in the server. –  Anonymous Jul 28 '10 at 18:27
You can link to it from any accessible location. The library doesn't necessarily need to be on the same server. –  recursive Jul 28 '10 at 23:45

You can use MathJax. Sample Blogger post: MathJax testing

<script src="http://www.mathjax.org/MathJax/MathJax.js">
//
//  This script call is what gets MathJax loaded and running
//
MathJax.Hub.Config({
// Only needed if to do more configuration within the page
// input is TeX and output is HTML-CSS
jax: ["input/MathML", "input/TeX", "output/HTML-CSS"],    format
// use the tex2jax preprocessor
extensions: ["tex2jax.js"],
tex2jax: {
// uncomment to use $...$ for inline math
inlineMath: [['$','$'], ['\$$','\$$']],
// set to 1 to allow \\$ to produce a dollar sign
processEscapes: 1
}
});
</script>

-

You can also try the LaTeX 4 Web library, which converts LaTeX code into HTML.

-
I tried it and it's very limited. –  Artefacto Jul 28 '10 at 18:06

I searched for "javascript latex" and found:

I'd suggest trying one of these, or finding another.

-


text:
# Relationship Between Force of Limiting Friction and Normal Reaction

The force which is opposing the object will arise as a reaction to the applied force and acts in the opposite direction when a body makes an attempt to slide over another body.

Static friction exists when two objects touch each other at rest. This friction has a limiting value which is called limiting friction, and which is equal to the least force required to move the body from rest. When the external force F is increased, it makes the body just on the verge of moving. The force of friction is maximum at this stage, and that is called limiting friction.

The limiting frictional force is called the coefficient of static friction, which is found experimentally. The coefficient of friction is a dimensionless constant, but it depends on the nature of surfaces in contact.

### Experiment For Finding The Coefficient Of Friction

Objective

The aim of the study is to identify the relationship between the force of limiting friction and the normal reaction and to find the coefficient of friction between a block and a horizontal surface.

### Apparatus Required:

1. A Wooden block

2. A 50g or 20g weights

3. A horizontal plane fitted with a frictionless pulley at one end

4. A pan

5. A spring balance

7. A spirit level

### Theory:

The friction between two surfaces when the bodies are sliding over them is known as sliding friction. It is also called kinetic friction.

The least force required to make a body slide over the surface is known as the force of the sliding friction.

Force of friction,

F ∝ RF

F= μR

Where,

μ is the coefficient of friction

R is the normal reaction

F is the force of sliding

At the equilibrium, the net force should be zero

F = P + p

R =W + w

Diagram:

Study for the force of sliding friction

### Procedure:

1. The pulley should be free of friction, and the tabletop should be clean.

2. The block should be passed on to the pulley by tieing one end of the thread on the hook of the wooden block.

3. The weight of the pan should be found.

4. The pan should be tied to the free end of the thread and left hanging vertically.

5. To pull the block easily, some weights should be put on the pan.

6. The table top should be tapped, to make the block slide, by adding the weights little by little, continue the tapping.

7. The total weights of the pan in the observation table should be recorded.

8. Steps 8, and 9, should be repeated, by adding one 50g or 20g weight on the wooden block, and repeat steps 8, 9, and 10 six times.

### Observations:

1. The weight of the wooden block (w) should be recorded in  g wt

2. The weight of the pan (P) should be recorded in g wt

3. The weight on the wooden block (w) should be recorded in g wt

4. The weight being pulled (W+w) should be recorded in g wt

5. Weight on the pan (p) should be recorded in g wt

6. The total weight being pulled (block and weights) (P+p) should be recorded in g wt, which is the limiting friction (F) in  g wt

Calculations:

1. The value of the force of sliding friction is being given by the total weight.

2. The normal reaction R should be given by the total weights being pulled on the horizontal surface.

3. A graph between normal reaction R and limiting friction F should be plotted, taking R along x-axis and F along the y-axis.

As shown below the graph comes to be a straight line:

The graph between R and F

### Result:

1. From the graph, it's clear that the limiting friction F is directly proportional to the normal reaction R. It's an agreement with laws of limiting friction.

2. The constant ratio is also called the coefficient of friction (). It can be calculated by finding the slope of the graph.

In ABC,

Slope of the straight line AB = tan$\theta$ = $\frac{BC}{AC}$ = $\frac{F}{R}$     --(1)

But, F= μR

$\frac{F}{R}$ = μ                               --(2)

μ can be calculated from equation (1) and (2)

### Fun Facts

1. Friction is a force that opposes when two objects slide against each other and is a contact force like the normal force. Friction acts in the flat surface direction of an object, while the normal force acts perpendicular to the flat surface.

2. It is found that the total weight pulled, and the force of limiting friction both increase simultaneously.

3. The thread part between block and pulley is kept horizontal for the equal distribution of the total weight of pan and weights.

4. The roads become slippery after rains, due to the reduction of friction between feet and the road, for the thin layer of rainwater between them.

5. The surface of the brake is flat because, when the brake is pressed, it gets pushed from hydraulic pistons and the surface of the brake rotor. So the surface should be flat since it is getting pushed from both sides.


text:
# Math Help - Integrals

1. ## Integrals

Hi everyone, a newbie here

I'm having trouble with the following problems:

Thanks

SL

2. Originally Posted by soundlift
Hi everyone, a newbie here

I'm having trouble with the following problems:

3. unless a is -1, you should be using $\int x^adx={x^{a+1}\over a+1}+C$


text:
# Combining 1D normal distributions into a 2D distribution

First of all, apologies for my poor terminology - I have a particular problem which I understand in own terms, but I am having difficulty in applying the mathematics in the correct manner.

My problem is that I have a point in 2D space with an uncertainty in two directions. These directions happen to be perpendicular to one another, however they are not necessarily perpendicular to the axes. (If it helps, the point represents a top-down view of an object from an image - there is uncertainty horizontally in the image as well as a larger uncertainty in the depth of the object). I can find the standard deviation of the two components and represent them as univariate distributions, but I'd instead like to represent them as a 2D distribution as they should be.

I can completely see that this is possible and probably very simple but I just can't quite make the leap from my visualisation of the problem to the mathematical formalism - I only need to calculate the parameters of the 2D distribution (specifically, I guess, the covariance - I know the mean). If someone could just give me a prod in the right direction I would be very grateful.

-
This looks as if it may be a bivariate normal distribution –  Henry Feb 16 '12 at 23:29

## 1 Answer

If the components in the two directions you refer to are independent and both have the same standard deviation, then it's easy. If they're independent and the standard deviations in those directions differ, then it's more involved. In the former case you have circular symmetry, so the standard deviations in all directions are the same and components in directions at right angles to each other will always be independent. In the latter case, the $x$- and $y$-coordinates would be correlated.

Suppose one of the directions you refer to is at an angle $\theta$ to the $x$-axis. The unit vector in that direction is $\begin{bmatrix}\cos\theta \\ \sin\theta\end{bmatrix}$, but the angle is not what we're concerned with, I'll call it $\begin{bmatrix} a \\ b \end{bmatrix}$, where $a^2+b^2=1$. The one perpendicular to it will be $\begin{bmatrix} -b \\ a \end{bmatrix}$. Suppose the standard deviation in the first direction is $\sigma$ and in the second is $\tau$. So let $U \sim N(0,\sigma^2)$ and $V \sim N(0,\tau^2)$ and suppose $U$ and $V$ are independent.

Then $$\begin{bmatrix} X \\ Y \end{bmatrix} = U \begin{bmatrix} a \\ b \end{bmatrix} + V \begin{bmatrix} -b \\ a \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} U \\ V \end{bmatrix}.$$ The variance of this random vector is $$\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \left( \operatorname{var} \begin{bmatrix} U \\ V \end{bmatrix} \right) \begin{bmatrix} a & -b \\ b & a \end{bmatrix}^T = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} \sigma^2 & 0 \\ 0 & \tau^2 \end{bmatrix} \begin{bmatrix} a & b \\ -b & a \end{bmatrix}$$ $$= \begin{bmatrix} \sigma^2 a^2 + \tau^2 b^2, & (\sigma^2-\tau^2)ab \\ (\sigma^2-\tau^2)ab, & \sigma^2 b^2 + \tau^2 a^2 \end{bmatrix}.$$ The entries in this matrix are the variances of the $x$- and $y$-components and the covariance.

The correlation is therefore $$\frac{(\sigma^2-\tau^2)ab}{\sqrt{(\sigma^2 a^2 + \tau^2 b^2)(\sigma^2 b^2 + \tau^2 a^2)}}.$$

You might consider the advantages of working in the rotated coordinate system in which the components in perpendicular directions are independent, and then, if necessary, converting back to the $xy$-coordinate system only at the end.

-
This is an excellent answer, thanks very much! –  jazzbassrob Feb 17 '12 at 0:42


text:
# Math Help - polar coordinate conversion.

1. ## polar coordinate conversion.

find polar coordinates of the points whose cartesian coordinates are given.

heres the point: (3sqrt(3), 3)

well i know that r^2 = (sqrt(a^2 + b^2))
so the answer here is : 6

and if we use tan(theta) = o/a = 3/(3sqrt(3)
so the answer here is: 1/sqrt(3)

so theta is pi/6

so how do i know its pi/6?
how do i convert to get this answer with pi? radians?

2. Hello, rcmango!

Find polar coordinates of the point . . _
whose cartesian coordinates are: .(3√3, 3)

Well, I know that: r² = sqrt(a² + b²)
. . so: r = 6
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
and if we use: .tanθ .= .opp/adj /= .3/(3√3)
. . so: .tanθ = 1/√3

Then: .θ = π/6

So how do i know its π/6? . The point is in Quadrant I

How do i convert to get this answer with pi? radians?
I don't understand your question . . . the answer is in π.

Polar coordinates are: .(r, θ) .= .(6, π/6)


text:
Difference between absolutely continuous measures

Suppose we are given two probability measures $$\mathbb{G}^{H}$$ and $$\mathbb{G}^{L}$$ with same support $$\text{supp}(\mathbb{G})$$.

Suppose as well that there exists an integrable function $$\gamma$$ such that $$\mathbb{G}^{H}(B)=\int_{B}\gamma(r)\mathbb{G}^{L}(dr)$$ for every subset $$B\in \mathcal{B}$$, the Borel $$\sigma$$-algebra of $$[0,1]$$. This function is a.s. positive, finite and different than 1. It is also increasing.

Question: Can we assert that $$|\mathbb{G}^{H}(B)-\mathbb{G}^{L}(B)|>0$$ for any half-closed interval $$B=(a,b]$$ such that $$0<\mathbb{G}^{L}(B)<1$$?

1 Answer

Let $$G^{L}$$ b e Lebesgue measure on $$[0,1]$$ and $$G^{B}(E)=\int_E 3x^{2} dx$$. Then the condition $$G^{B}((a,b])=G^{L}((a,b])$$ reduces to $$b^{3}-a^{3}=b-a$$. There are plenty of intervals with this property; in fact for any $$a\in (0,\frac 1 {\sqrt 3})$$ we can find $$b \in (a,1)$$ such that $$a^{2}+b^{2}+ab=1$$ which implies $$G^{B}((a,b])=G^{L}((a,b])$$.

• Thank you! I edited the question, sorry. What I should have writtten actually was whether we can assert for every half-closed intervals. – Caio Lorecchio Apr 2 at 0:06
• @CaioLorecchio I have edited the answer accordingly. – Kavi Rama Murthy Apr 2 at 0:18
• $G^{H}(B)=1/8$ and $G^{L}(B)=1/2$, so the measures do not coincide. Am I missing something here? – Caio Lorecchio Apr 2 at 0:52
• @CaioLorecchio Sorry, that example was wrong. – Kavi Rama Murthy Apr 2 at 5:34
• @CaioLorecchio I have revised my answer. – Kavi Rama Murthy Apr 2 at 7:32


text:
# Surface current on a spherical superconductor

Homework Statement:
Find the surface current on a spherical superconductor.
Relevant Equations:
K=sigma * v
I need help with part b.

My solution:

Have I done it right?

vanhees71


text:
# Dual object

In category theory, a branch of mathematics, it is possible to define a concept of dual object generalizing the concept of dual space in linear algebra.

A category in which each object has a dual is called autonomous or rigid.

## Definition

Consider an object $X$ in a monoidal category $(\mathbf{C},\otimes, I, \alpha, \lambda, \rho)$. The object $X^*$ is called a left dual of $X$ if there exist two morphsims

$\eta:I\to X\otimes X^*$, called the coevaluation, and $\varepsilon:X^*\otimes X\to I$, called the evaluation,

such that the following two diagrams commute

 and

The object $X$ is called the right dual of $X^*$. Left duals are canonically isomorphic when they exist, as are right duals. When C is braided (or symmetric), every left dual is also a right dual, and vice versa.

If we consider a monoidal category as a bicategory with one object, a dual pair is exactly an adjoint pair.

## Categories with duals

A monoidal category where every object has a left (resp. right) dual is sometimes called a left (resp. right) autonomous category. Algebraic geometers call it a left (resp. right) rigid category. A monoidal category where every object has both a left and a right dual is called an autonomous category. An autonomous category that is also symmetric is called a compact closed category.


text:
# Is $a_n = f(a + b c^n)$ when $a_0 = \ln(2)$ and $a_{n+1} = (e+1) a_n (a_n - 1)$?

Definition

Let $a_n$ be a real sequence.

Assume there exists a continuous real-periodic function $f(x)$ such that

$f(n) = a_n$

And $f(x)$ has the period $t$ , where $t$ is An irrational real number.

Then we say $a_n$ is periodic and has An irrational period. Or we say $a_n$ Has An irrational periodic orbit.

Let $a_0 = \ln(2)$ and $a_{n+1} = (e+1) a_n (1 - a_n)$.

Does there exist an $f$ such that :

$$a_n = f( a + b c^n)$$

Where $a,b,c$ are (fixed) real values and $f$ is a real-continuous periodic function with real irrational period $t$ ?

• I changed " is " $a_n = f$ ... To : Does there exist An $f$ such that $a_n = f$ ... Although imho the " is " was sufficient to be clear and consistent. Not sure if that improves , but it can not be worse I assume. Thanks to hippocampus from chat for the suggestion.
– mick
Dec 13, 2015 at 21:49
• That is hippalectryon. Not hippocampus. en.m.wikipedia.org/wiki/Seahorse
– mick
Dec 13, 2015 at 22:05


text:
# Adding Fill to a ContourPlot

I am trying to implement the kind of solution Bob Hanlon provided here, but on another equation. There are two problems. First, this time, the gap at the top-right is not fully closed by the method, and second, I would like to add a standard blue fill to the regions below zero in the contour plot, while the above zero region should remain white.

Can these things be achieved?

Here is my code:

Clear["Global*"]

expr2 = 1/(32 (p1 - p2)^2 (-(-1 + p2) p2)^(3/2) (-1 + p1 + p2)^2) (-1 + p1)^2 p1^2 (-64 Sqrt[-(-1 + p1) p1] (-1 + p2)^2 p2^2 - 32 (-1 + p1) p1 (-(-1 + p2) p2)^(3/2) - 32 (-(-1 + p2) p2)^(5/2) + (32 (-(-1 + p2) p2)^(7/2))/((-1 + p1) p1) - (192 (p1 - p2)^2 (-1 + p1 + p2)^2)/Sqrt[1/(p2 - p2^2)] + (64 (-1 + p2)^3 p2^3 (-3 + 2 Sqrt[((-1 + p2) p2)/((-1 + p1) p1)]))/Sqrt[-(-1 + p1) p1] + 32 (1/(p1 - p1^2))^(3/2) (6 (-1 + p1) p1 (-1 + p2)^3 p2^3 - 2 (-1 + p2)^4 p2^4 + Sqrt[1/(p1 - p1^2)] (-(-1 + p2) p2)^(9/2) + (2 (-1 + p2)^4 p2^4)/(((-1 + p2) p2)/((-1 + p1) p1))^(3/2) + 6 (-1 + p1)^4 p1^4 Sqrt[((-1 + p2) p2)/((-1 + p1) p1)] - 12 (-1 + p1)^4 p1^4 (((-1 + p2) p2)/((-1 + p1) p1))^(3/2))) // Simplify;

fd = FunctionDomain[expr2, {p1, p2}]; expr3 = expr2 // FullSimplify[#, fd] &; ContourPlot[Numerator@expr3 == 0, {p1, 0.5, 1}, {p2, 0.5, 1}, MaxRecursion -> 5, FrameLabel -> {p1, p2}, WorkingPrecision -> 50] // Quiet

• Try the option PlotPoints ->50?
– kglr
Jan 4 '19 at 20:34

You may do the contour plot for expr3 (instead of expr3==0), then use a region function to pick out the zero line, using 1 contour to just draw the boundary. The fill color is done via ContourShading, like this:

b = Function[{p1, p2}, Numerator@expr3 // Evaluate];
ContourPlot[Numerator@expr3, {p1, 0.5, 1}, {p2, 0.5, 1},
MaxRecursion -> 5, FrameLabel -> {p1, p2}, Contours -> 1,
RegionFunction -> Function[{x, y, z}, b[x, y] < 0]] // Quiet

To close off at the {1,1} point, you may try changing MaxRecursion or PlotPoints. To get a quick result, I've set MaxRecursion to 1:

ContourPlot[Numerator@expr3, {p1, 0.5, 1}, {p2, 0.5, 1},
MaxRecursion -> 1, PlotPoints -> 60, FrameLabel -> {p1, p2},
Contours -> 1, RegionFunction -> Function[{x, y, z}, b[x, y] < 0],

To get a smoother boundary, try larger MaxRecursion or PlotPoints.

Edit

To change the contour fill/line style for this plot, one way to get around the contour lines being not smooth/easy to see, plot one without the outline, just the fill color; plot another that shows just the outline, then Show them together, like this:

p0 = ContourPlot[Numerator@expr3, {p1, 0.5, 1}, {p2, 0.5, 1},
PlotPoints -> 50, FrameLabel -> {Subscript[p, 1], Subscript[p, 2]},
Contours -> 1, RegionFunction -> Function[{x, y, z}, b[x, y] < 0],
`


text:
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Yale SOM 2013 - Calling All Applicants

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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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21 Oct 2012, 01:35
Guys, a quick question. I am an international applicant from Europe and would like to fly over for the campus interview since I want to visit Harvard and Columbia anyways (and took some holidays for this time before I knew anything about my Yale application)

What is better, to fly directly to New Heaven airport or JFK? JFK is cheaper and only 80 miles away. Nevertheless, how could I get from JFK to New Heaven? Is there any shuttle service I could book?

What is the dresscode for the campus visit and the interview? I assume that everything happens during one day so there is no chance to wear something casual and then a suit....
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21 Oct 2012, 07:00
BankerRUS wrote:
Guys, a quick question. I am an international applicant from Europe and would like to fly over for the campus interview since I want to visit Harvard and Columbia anyways (and took some holidays for this time before I knew anything about my Yale application)

What is better, to fly directly to New Heaven airport or JFK? JFK is cheaper and only 80 miles away. Nevertheless, how could I get from JFK to New Heaven? Is there any shuttle service I could book?

What is the dresscode for the campus visit and the interview? I assume that everything happens during one day so there is no chance to wear something casual and then a suit....

Well, JFK is about a 1 hour 45 min drive from Yale. There apparently are these shuttle services:

http://www.yale.edu/travel/ground/GoAir ... leInfo.htm
http://www.yale.edu/travel/ground/QConnShuttleInfo.htm
http://ridethedot.com/

I'm not sure if there are others. I just did a quick google search.

if the flight is any cheaper, you could fly into Boston, which is about a 2 1/2 hour drive from New Haven. I didn't look for Boston shuttles, but I would guess there is at least one service.

Obviously, flying directly into New Haven would be more convenient, but I can imagine that the flight would be more expensive. You'll have to weigh that against the time/cost for a shuttle.

In terms of campus visit and the interview, you are correct, everything happens in one day. Dress code for campus visit is casual, although some people always wear suits. Dress code for interview is business formal
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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21 Oct 2012, 07:05
Hi there -

If your plan is to visit Columbia, HBS, and SOM I'd fly to either NYC or Boston visit Columbia/HBS respectively then take the train (Metro North from NYC or Amtrak from Boston) to New Haven and then take the train to whichever school is left to visit. The train from NYC is ~2hrs and then you can walk (its a fairly long walk) or cab to SOM campus. You can definitely just have a day in New Haven. When I visited some folks left their baggage (small roll-aboard/duffle) with the kind admissions folk in the office for the day.

Have fun!
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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21 Oct 2012, 07:53
mba2527 wrote:
Hi there -

If your plan is to visit Columbia, HBS, and SOM I'd fly to either NYC or Boston visit Columbia/HBS respectively then take the train (Metro North from NYC or Amtrak from Boston) to New Haven and then take the train to whichever school is left to visit. The train from NYC is ~2hrs and then you can walk (its a fairly long walk) or cab to SOM campus. You can definitely just have a day in New Haven. When I visited some folks left their baggage (small roll-aboard/duffle) with the kind admissions folk in the office for the day.

Have fun!

I agree. There are only a few commercial flights that fly into and out of New Haven and it does drive up the cost. You'll have far more options with Boston and NYC - traveling between all three cities on the train is easy.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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22 Oct 2012, 11:46
Taxi is also an option:

The New York area airports Kennedy (JFK) and LaGuardia (LGA), as well as Newark Airport (EWR) in New Jersey, are two to three hours from campus. Connecticut Limousine (800 472-5466) is available from all three airports into New Haven (approximately $68–$92 one way). The company offers a discounted rate of $95 round trip for pickup and drop-off service from Phelps Gate on the Old Campus. GO Airport Shuttle Connecticut (866 284-3247) also offers service to LaGuardia and Kennedy (approximately$54 one way), but not to Newark. Amtrak trains are available from Newark Airport for approximately $67–$75 one way (800 872-7245).

Check it out: http://yalecollege.yale.edu/content/tra ... and-travel
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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22 Oct 2012, 12:56
Fixed my off-campus (skype) interview for 1st Nov. Any ideas on how to prepare? Are there any resources around?

Anyone up for mock practice?

Cheers
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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23 Oct 2012, 13:59
OjilEye wrote:
I JUST GOT AN E-MAIL...from my Dad wishing me luck in this process.

You are such a dork. Hahahaha. I haven't even gotten THAT email!
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http://www.mybreakaway.com/ Recent post: September 20, "Transitions"

Yale School of Management MBA Candidate, 2015.

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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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24 Oct 2012, 10:32
Sorry if this sounds like a dumb question, but what should one wear at a Skype interview?

Coat and tie?
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24 Oct 2012, 11:03
jason12e wrote:
Sorry if this sounds like a dumb question, but what should one wear at a Skype interview?

Coat and tie?

I think the rule of thumb is to prepare for a Skype interview the way you would for an in-person interview. I'd wear a suit. I've interviewed people on Skype before where something happened and they had to stand up or move out of the frame - if you were only wearing a coat and tie plus jeans, it might not look that great.

It also helps put you in more of an interview mindset, as opposed to a chatting-with-your-friends-on-Skype mindset, if you dress the part.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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25 Oct 2012, 07:45
Does anyone know roughly how long the admissions committee spends with an application before deciding whether to interview a candidate? Just curious.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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25 Oct 2012, 08:14
Crmson10 wrote:
Does anyone know roughly how long the admissions committee spends with an application before deciding whether to interview a candidate? Just curious.

That is something hard to guess.

What we know is (from the adcom of course): 2 real people (humans) review your application separately. If you are still rejected another review is done by someone else. So, I think it is fair to say that 'there are no accidents' (kung fu panda?) as such. No fluke interview calls/acceptances (I would like to believe it anyway)

My take - should be around 15-20 mins at MAX.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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25 Oct 2012, 11:41
Dbalks wrote:
I will not be fooled and excited by todays 4:15 letter from the admissions office telling me all of the exciting things going on at Yale. I will not be fooled, I will not be fooled. It is not an interview invite, just an admissions newsletter.

Totally forgot about that. Thanks for the warning 'mate!
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25 Oct 2012, 11:46
Seems like every interview invite this year has been sent out on a Friday...crossing my fingers tomorrow.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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25 Oct 2012, 11:59
Dbalks wrote:
I will not be fooled and excited by todays 4:15 letter from the admissions office telling me all of the exciting things going on at Yale. I will not be fooled, I will not be fooled. It is not an interview invite, just an admissions newsletter.

Thanks for the reminder!
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25 Oct 2012, 12:26
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Last edited by breakthelaw on 05 Mar 2013, 10:17, edited 1 time in total.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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25 Oct 2012, 13:20
Chupacabra wrote:
Seems like every interview invite this year has been sent out on a Friday...crossing my fingers tomorrow.

Any idea what time day they might be sending these out? Thanks.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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25 Oct 2012, 13:54
Hey guys,
Could anyone throw some light on how Yale hands out dings? If not selected to interview, can I expect a rejection mail in my inbox anytime now, or will that decision also be communicated only on the last day?

The wait only gets more and more excruciating.
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25 Oct 2012, 13:57
Again, as a 1st year, if anyone is coming in for an interview and want to meet some of us, relax before the interview, or hang out after, please message me. More than happy to help.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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25 Oct 2012, 17:38
Congrats on the invite! Still feverishly checking my e-mail and application status for that elusive interview.
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Re: Yale SOM 2013 - Calling All Applicants [#permalink]

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26 Oct 2012, 12:00
Anyone get invites yet? Man... I'm getting antsy.
Re: Yale SOM 2013 - Calling All Applicants   [#permalink] 26 Oct 2012, 12:00

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# If-Then Statements

## Hypothesis followed by a conclusion.

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If-Then Statements

If-then statements are very common outside formal mathematics. In many cases, your familiarity with these types of statements will help you to interpret their meaning and truth value. In other cases, the truth value of some statements may be unclear without a formal understanding of logic. Consider the following two statements:

• If it rains today then you will stay at home and read a book.
• You stayed at home and read a book.

Did it or did it not rain?

### If-Then Statements

A mathematical set is just a group of things. The group can include anything: letters, numbers, objects or monkeys. Set theory focuses on the relationships between sets as they overlap or are completely within each other. For the purposes of if-then statements, set theory provides a perfect framework in which to reason.

Consider set \begin{align*}P\end{align*} and set \begin{align*}Q\end{align*} that are just collections of things represented by circles. If something is not in the set, then it is not in the circle.

In this case set \begin{align*}P\end{align*} is a subset of set \begin{align*}Q\end{align*} since it is entirely included within set \begin{align*}Q\end{align*}. Mathematically we write the statement “\begin{align*}P\end{align*} is a subset of \begin{align*}Q\end{align*}” as:

\begin{align*}P \subseteq Q\end{align*}

This can be translated to an if-then statement, and simplified using symbols:

If it is an element in \begin{align*}P\end{align*}, then it is an element in \begin{align*}Q\end{align*}.

If \begin{align*}P\end{align*}, then \begin{align*}Q\end{align*}.

\begin{align*}P\rightarrow Q\end{align*}

If-then statements are examples of conditional statements. Sometimes conditional statements are written without an “if” or a “then”, but can be rewritten. The “if” part of the statement (represented by \begin{align*}P\end{align*} above) is called the hypothesis, antecedent or protasis. The “then” part of the statement (represented by \begin{align*}Q\end{align*} above) is called the conclusion, consequent or apodosis.

In order to precisely define the truth value of a conditional statement, we need to consider the four different combinations of the truth value for \begin{align*}P\end{align*} and \begin{align*}Q\end{align*} in relation to the diagram

1. If \begin{align*}P\end{align*} is true, then \begin{align*}Q\end{align*} is true. This statement is true because if an object is inside circle \begin{align*}P\end{align*}, then it is definitely inside circle \begin{align*}Q\end{align*}.
2. If \begin{align*}P\end{align*} is true, then \begin{align*}Q\end{align*} is false. This statement is false because there is no possible way an object could be inside circle \begin{align*}P\end{align*} and yet outside circle \begin{align*}Q\end{align*}.
3. If \begin{align*}P\end{align*} is false, then \begin{align*}Q\end{align*} is true. This statement is considered true because if an object is outside circle \begin{align*}P\end{align*} then it may or may not be in circle \begin{align*}Q\end{align*}. There is no contradiction.
4. If \begin{align*}P\end{align*} is false, then \begin{align*}Q\end{align*} is false. This statement is also considered true because if an object is outside circle \begin{align*}P\end{align*}, then it can be outside circle \begin{align*}Q\end{align*}. Like the previous statement, there is no contradiction.

The truth values of the four combinations can be summarized in a truth table. Recall that truth tables are extremely useful for summarizing complicated logical sentences and identifying whether the statements are true or false.

 \begin{align*}P\end{align*} \begin{align*}Q\end{align*} \begin{align*}P \rightarrow Q\end{align*} T T T T F F F T T F F T

Note that a conditional statement is only false when the hypothesis is true and the conclusion is false. Also note that any conditional statement with a false hypothesis is trivially true. The following statement is trivially true because the hypothesis is false.

If pigs can fly then butterflies eat elephants.

The truth of this statement confuses many people the first time they look at it. One way to frame it in your mind is to realize that a statement is false only when it results in a logical contradiction. In a world where pigs could fly perhaps butterflies could eat elephants, who knows? It would be ridiculous for a person to argue that in the hypothetical world where pigs could fly that there is no way that butterflies could eat elephants.

### Examples

#### Example 1

Earlier, you were asked if it rained given the following two statements:

• If it rains today then you will stay at home and read a book.
• You stayed at home and read a book.

Use a diagram to represent the conditional statement.

If it rains today then you will stay at home and read a book.

• \begin{align*}P=\ it \ rains \ today\end{align*}
• \begin{align*}Q=\ you \ will \ stay \ at \ home \ and \ read\ a \ book\end{align*}.

You stayed at home and read a book. This implies that \begin{align*}Q\end{align*} is true. According to the diagram, if an object is inside \begin{align*}Q\end{align*} it may or may not be inside \begin{align*}P\end{align*}. Thus, you can conclude nothing about the rain. Many people will want to incorrectly conclude that it must have rained, but conditional statements only flow in one direction.

#### Example 2

Rewrite the following conditional statements in if-then form.

1. If you go to the show, you will be amazed.
2. Unless you buy firewood you will be cold.
3. Come here and you will get a present.
4. Kicking a soccer ball makes it bounce.
5. Give me your lunch money or I’ll put you in a locker.
6. Anyone who wears orange likes Halloween.
7. Without my sunglasses on I can’t drive.
8. Buy this product and you’ll be beautiful and popular.

Even though these statements have words like “and”, “or” and “not” they are still just conditional statements. In each case, consider which action or event leads to another action or event.

1. If you go to the show, then you will be amazed.
2. If you do not buy firewood, then you will be cold.
3. If you come here, then you will get a present.
4. If you kick a soccer ball, then it will bounce.
5. If you don’t give me your lunch money, then I’ll put you in a locker.
6. If a person wears orange, then that person likes Halloween.
7. If I do not wear my sunglasses, then I can’t drive.
8. If you buy this product, then you will be beautiful and popular.

#### Example 3

Evaluate the truth value of the following conditional statement using a truth table.

If when I go somewhere I always run, then when I run I always go somewhere.

This statement actually has two layers of conditional statements because both the hypothesis and the conclusion are conditional statements themselves.

• Let \begin{align*}R\end{align*} be the statement “I run”.
• Let \begin{align*}S\end{align*} be the statement “I go somewhere”.

The original sentence can be rewritten in symbols as: \begin{align*}(S\rightarrow R) \rightarrow (R \rightarrow S)\end{align*}. To make the truth table for the sentence, start with all possible truth combinations of \begin{align*}S\end{align*} and \begin{align*}R\end{align*} (both true, \begin{align*}S\end{align*} true/\begin{align*}R\end{align*} false, \begin{align*}S\end{align*} false/\begin{align*}R\end{align*} true, both false). Then, find the truth values for each piece working up to the full sentence.

 \begin{align*}S\end{align*} \begin{align*}R\end{align*} \begin{align*}S\rightarrow R\end{align*} \begin{align*}R \rightarrow S\end{align*} \begin{align*}(S\rightarrow R) \rightarrow (R \rightarrow S)\end{align*} T T T T T T F F T T F T T F F F F T T T

The statement is always true unless you run but do not go anywhere.

#### Example 4

If all of the following statements are true, what can you conclude?

• All babies are cute.
• Laura likes cute people.
• Laura is a baby.

First translate each of the statements into conditional statements (even if they sound awkward!). This is helpful for determining an if-then chain of events.

• \begin{align*}A\end{align*}: If a person is a baby then the person is cute.
• \begin{align*}B\end{align*}: If a person is cute then Laura likes that person.
• \begin{align*}C\end{align*}: If a person is named Laura then that person is a baby.

You should notice the circular structure of these three statements.

\begin{align*}A \rightarrow B, B \rightarrow C, C \rightarrow A\end{align*}

\begin{align*}A \rightarrow B \rightarrow C \rightarrow A\end{align*}

While many conclusions could be made, one conclusion about Laura is that she likes herself.

#### Example 5

There are 53 people in the marching band and 49 people on the swim team. If 84 people belong to either or both teams, how many people are on both teams?

While this problem is not specific to if-then statements, it can be solved using a set theory representation and if-then logic.

Students on both teams would be double counted if you simply added up the number of students in band and the number of students on the swim team.

\begin{align*}53+49=102\end{align*}

Since there are only 84 people total, then \begin{align*}102-84=18\end{align*} students must have been counted twice. Therefore, there are 18 students on both squads.

### Review

1. What are the three names for the “if” part of an if-then statement?

2. What are the three names for the “then” part of an if-then statement?

Rewrite each of the following statements in if-then form.

3. If you like Pepsi, you will like Coke.

4. Do your homework and you will get candy.

5. Anyone who goes to the mall likes to shop.

6. Unless you cook dinner, you will be hungry.

7. Join this program and you will lose weight.

8. Shoveling snow makes your back sore.

9. Without knowing how to drive, you will not get your license.

10. Be nice to your sister or you will be punished.

11. When is the following statement false? If you are a kid, then you like pizza.

12. If all of the following statements are true, what can you conclude?

• All kids like pizza.
• Sam is 8 years old.
• 8 year olds are kids.

13. There are 15 people in the math club and 27 people in the debate club. If 33 people belong to either or both clubs, how many people are just in the debate club?

14. There are 32 people on the bus in the morning and 40 people on the bus in the afternoon. If 20 people only ride the bus in the afternoon, how many people only ride the bus in the morning?

15. There are 24 people in your math class and 27 people in your English class. If 7 people are in your English class but not your math class, how many people are in your math class but not your English class?

To see the Review answers, open this PDF file and look for section 16.2.

### My Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

Angle

A geometric figure formed by two rays that connect at a single point or vertex.

Subset

A subset is a collection of numbers or objects within a larger set.


text:
• Corpus ID: 203594080

# Explicit formulas for Grassmannian polylogarithms

@article{Charlton2019ExplicitFF,
title={Explicit formulas for Grassmannian polylogarithms},
author={Steven Charlton and Herbert Gangl and Danylo V. Radchenko},
journal={arXiv: Number Theory},
year={2019}
}
• Published 30 September 2019
• Mathematics
• arXiv: Number Theory
We give a new explicit formula for Grassmannian polylogarithms in terms of iterated integrals. We also explicitly reduce the Grassmannian polylogarithm in weight 4 and in weight 5 each to depth 2. Furthermore, using this reduction in weight 4 we obtain an explicit, albeit complicated, form of the so-called 4-ratio, which gives an expression for the Borel class in continuous cohomology of $\mathrm{GL}_4$ in terms of $\mathrm{Li}_4$.
2 Citations

### Clean Single-Valued Polylogarithms

• Mathematics, Computer Science
Symmetry, Integrability and Geometry: Methods and Applications
• 2021
We define a variant of real-analytic polylogarithms that are single-valued and that satisfy ''clean'' functional relations that do not involve any products of lower weight functions. We discuss the

### On the Goncharov depth conjecture and a formula for volumes of orthoschemes

We prove a conjecture of Goncharov, which says that any multiple polylogarithm can be expressed via polylogarithms of depth at most half of the weight. We give an explicit formula for this

## References

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### Identities arising from coproducts on multiple zeta values and multiple polylogarithms

In this thesis we explore identities which can be proven on multiple zeta values using the derivation operators $D_r$ from Brown's motivic MZV framework. We then explore identities which occur on

### On the Goncharov depth conjecture and a formula for volumes of orthoschemes

We prove a conjecture of Goncharov, which says that any multiple polylogarithm can be expressed via polylogarithms of depth at most half of the weight. We give an explicit formula for this

### On functional equations for Nielsen polylogarithms

• Mathematics
Communications in Number Theory and Physics
• 2021
We derive new functional equations for Nielsen polylogarithms. We show that, when viewed modulo $\mathrm{Li}_5$ and products of lower weight functions, the weight $5$ Nielsen polylogarithm $S_{3,2}$