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# Center or Centre of a circle

An internal point that maintains equal distance from every point on a circle is called the center (or centre) of a circle.

## Introduction

There are numerous points inside every circle and each point has different distances from every point on the circle.

However, there is only one point, which maintains same distance from every point on the circle and that internal point is called the center of a circle.

Now, it is your time learn the fundamental concept of centre (or center) of a circle from an understandable graphical example.

### Example

Let $A, B, C$ and $D$ be four points inside a circle and they are located at four different positions in the plane.

1. The Point $A$ is very close to the circle at left-top side. So, it cannot maintain equal distance from all the points on the circle.
2. The Point $B$ is near to the right side of the circle. So, it does not maintain equal distance from all the points on the circle.
3. The Point $C$ is exactly middle to the circle in all the directions. So, it can maintain equal distance from every point on the circle.
4. The Point $D$ is close to the circle at left-bottom side. So, it can’t maintain equal distance from all the point on the the circle.

The locations of points $A$, $B$ and $D$ do not allow them to maintain equal distance from all the points on the circle but the location of point $C$ is exactly middle to the circle in all the directions. Hence, the point $C$ is called the center (or centre) of the circle.

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Limited access

A wayward puppy leaves home at $t=0.0 \text{ }s$ and moves along a straight path, with a varying velocity. The velocity of the puppy, in $m/s$, is given by the function $v(t) = -0.036{t}^{2}+1.2t$, where $t$ is measured in seconds.

When, if ever, does the puppy return home?

A

$t=10s$

B

$t=17s$

C

$t=34s$

D

$t=50s$

E

The puppy never returns home.

Select an assignment template

format: plain


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## Algebra 1: Common Core (15th Edition)

Let x = the number of correct answers y = the number of incorrect answers A contestant on a quiz show gets 150 points for every correct answer and loses 250 points for each incorrect answer. The contestant completed 20 questions. Hence, we get the system of equations: $x+y=20$ $150x-250y=200$ The first equation will be: $y=20-x$ Write the second equation. Substitute 10x + 50 for y. $150x-250(20-x)=200$ $150x-5000+250x=200$ $400x=5200$ $x=13$ Solve for $y$: $13+y=20$ $y=7$ The contestant answered 13 questions correctly, and 7 were incorrectly answered.


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# Simple limit question

1. Nov 14, 2007

### uman

Hi all. I encountered the following problem:

lim x->0 x * cot(x). My calculator gives the value of 1 for this limit, which can be corroborated by looking at the graph of y = x * cot(x).

However, I reason that it should be zero, because:

lim x->0 x * cot(x) = (lim x->0 x) * (lim x->0 cot(x))
= 0 * lim x->0 cot(x)
= 0 (because anything multiplied by zero is zero.)

Clearly, my reasoning is flawed. What is wrong with it? And what is the correct way to do this problem?

Thank you.

2. Nov 14, 2007

### Gib Z

In general, $$\lim_{x\to a} f(a)g(a) \neq (\lim_{x\to a} f(a)) (\lim_{x\to a} g(a))$$

For example, as x approaches zero, (x/x) approaches 1. If you split it like you did, we get zero times infinity. Clearly "anything multiplied by zero is zero." does not hold.

3. Nov 15, 2007

### ircdan

it's as Gib Z says, but just so you are clear, lim fg = limf limg IF limf and limg both exist, and the same goes for sums, lim(f + g) = limf + limg if both limf and limg exist, this is a common mistake. so think about this so you don't make it again

Last edited: Nov 15, 2007
4. Nov 15, 2007

### HallsofIvy

Staff Emeritus
That statement is wrong: 0 multiplied by $\infty$ is not 0! It is correct that 0 multiplied by any real number is 0 but that limit is not a real number. The others said that "lim x->0 cot(x)" does not exist which is better wording for the same thing.

cot(x)= cos(x)/sin(x). x cot(x)= x cos(x)/sin(x)= (x/sin(x))(cos(x)). It is proved in Calculus that $\lim_{x\rightarrow 0} sin(x)/x= lim_{x\rightarrow 0}x/sin(x)= 1$. Since cos(x) is continuous and cos(0)= 1, the limit of x cot(x)= 1.

5. Nov 15, 2007

### robert Ihnot

You can use L'Hospital's rule for x/tan(x) goes to 1/(secx)^2 upon derivation top and bottom, which goes to 1 as x goes to 0.

Last edited: Nov 15, 2007
6. Nov 15, 2007

### uman

Thanks a million, both of you!


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1. Feb 17, 2015

bznm

1. The problem statement, all variables and given/known data
I have solved the following exercise, but I have obtained the half of the correct result! I can't understand where is the problem...

$f(x)=\begin {cases} 0, x \in[-\pi, 0]\\cos x, x \in[0, \pi]\end{cases}$

1) Find the Fourier Series (base: ${\frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{\pi}} \cos nx, \frac{1}{\sqrt \pi}\sin nx}$
2) Evaluate the Fourier Series in $x=\pi/4$ to prove $\sum_{k=1}^{\infty} \frac {k sin (k \pi/2)}{4k^2-1}=\frac{\pi}{8}cos \frac{\pi}{4}$

I know that the result that I have to obtain has to be the result that I have just written (so I have to exclude a typo), because the second part of the exercise says that I have to consider only the part of the function in $[0, \pi]$ and do an odd extension. Then, find the new Fourier Series and use it to re-find the relation in point 2). And, for this part of the exercise, I obtain the correct result!

2. Relevant equations
$\displaystyle P(x)=\frac{a_0}{\sqrt{2 \pi}}+ \sum_{n=1}^{\infty} \frac{a_n}{\sqrt{\pi}} cos nx+\sum_{n=1}^{\infty}\frac{b_n}{\sqrt \pi} sin nx$

$\displaystyle a_0=\frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} f(x) dx$
$\displaystyle a_n=\frac{1}{\sqrt\pi} \int_{-\pi}{\pi} f(x) cos nx dx$
$\displaystyle b_n=\frac{1}{\sqrt\pi} \int_{-\pi}{\pi} f(x) sin nx dx$

3. The attempt at a solution
Considering this particular function, I have:

$\displaystyle a_0= \frac{1}{\sqrt{2\pi}} \int_{0}^{\pi}cosx dx=\frac{1}{\sqrt{2\pi}} [sinx]_{0}^{\pi}=0$

$a_n=\frac{1}{\sqrt\pi} \int_{0}^{\pi} \cos nx \cdot \cos x dx=0$ (I don't post the proceeding, it's to long to write in latex code) (http://www.wolframalpha.com/input/?i=cos+nx*cos+x+dx+from+0+to+pi)

$\displaystyle b_n=\frac {1}{\sqrt \pi} \int_{0}{\pi} \cos n \cdot \sin nx dx=\frac {1}{\sqrt \pi} \frac {((-1)^n+1)n}{n^2-1}=$ (http://www.wolframalpha.com/input/?i=sin+nx*cos+x+dx+from+0+to+pi)

Considering only even $n$, I obtain:

$P(x)=\displaystyle \sum_{k=1}^{\infty} \frac{1}{\pi}\frac{4k}{4k^2-1} sin (2kx)$

If $x=\pi/4$: $\displaystyle \sum_{k=1}^{\infty}\frac{k}{4k^2-1} sin (k\pi/2)= \frac{\pi}{4}cos{\pi/4}$ :(

Many thanks for the help!

2. Feb 17, 2015

vela

Staff Emeritus
One problem I see is with your calculation of $a_n$. What happens when $n=1$? The integral clearly doesn't vanish.

You have a similar problem with $b_n$. Your expression isn't defined for $n=1$.

3. Feb 17, 2015

bznm

oh, wonderful! you're right! Thanks a lot!!! :)

($b_n$ for n=1 is 0)


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# Aerosol black carbon radiative forcing at an industrial city in northern India

Tripathi, SN and Dey, Sagnik and Tare, V and Satheesh, SK (2005) Aerosol black carbon radiative forcing at an industrial city in northern India. In: Geophysical Research Letters, 32 (8).

 PDF Aerosol.pdf - Published Version Restricted to Registered users only Download (140Kb) | Request a copy

## Abstract

During a comprehensive aerosol field campaign as part of ISRO-GBP,extensive measurements of aerosol black carbon were made during December 2004, for the first time, at Kanpur, an urban continent allocation in northern India. BC diurnal variation is associated with changes in boundary layer mixing and anthropogenic activities. BC concentration in Kanpur is comparable to those measured in other megacities of India but much higher than in similar locations of Europe,USA and Asia. High BC concentration is found both in absolute terms $(6-20 \mu g m^{-3} )$ and mass fraction (similar to 10%) yielding very low single scattering albedo (0.76). The estimated surface forcing is as high as $-62 +/- 23 W m^{-2}$ and top of the atmosphere (TOA) forcing is $+9 +/- 3 W m^{-2}$, which means the atmospheric absorption is $+71 Wm^{-2}$. The short wave atmospheric absorption translates to a lower atmospheric heating of similar to 1.8 degrees K/day. Large surface cooling and lower atmospheric heating may have impacts to regional climate.

Item Type: Journal Article Copyright for this article belongs to American Geophysical Union. Division of Mechanical Sciences > Centre for Atmospheric & Oceanic Sciences 17 May 2005 20 Jan 2012 09:51 http://eprints.iisc.ernet.in/id/eprint/3183


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# Knowledge base: Adam Mickiewicz University

Back

## Functions of calls in a tropical duetting bird species, the Yellow-breasted boubou (Laniarius atroflavus)

### Amie Wheeldon

#### Abstract

I have examined various aspects of the sex specific acoustic repertoire of the duetting Yellow-breasted boubou, an endemic bird species of the Afro-montane region of Western Africa. The first area of examination was the natural vocalisation activity of the study species which highlighted the male dominant vocal behaviour and the use of sex specific repertoires. I then experimentally examined the intrusion into focal territories of either duets or the sex specific solo types. This suggested that female vocalisations are likely used in intra-pair communications whereas males are using vocalisations for defending a territory and their mate. A further series of experiments were used to assess the aggressive motivation of male vocalisation types. This suggested that males are able to display aggressive motivation through the use of matching behaviours as well as vocalisation type used. Finally, the propagation properties of all vocalisation types for different habitat types were examined to further assess the functions of song and call types. Male vocalisation types transmit further, adding more evidence to the intended receivers being neighbours and strangers, whereas females intended receivers are most likely their own mates. The abiotic source of noise caused by a network of streams, also has a negative effect on the propagation of signals, which is important due to the human induced habitat change occurring in this specific region.
Record ID
UAM1ea23723e5a24b828f08ccfc222ecb5e
Diploma type
Doctor of Philosophy
Author
Title in Polish
Funkcje głosów u duetującego ptaka tropikalnego, dzierzyka żółtobrzuchego (Laniarius atroflavus)
Title in English
Functions of calls in a tropical duetting bird species, the Yellow-breasted boubou (Laniarius atroflavus)
Language
eng (en) English
Certifying Unit
Wydział Biologii [nowa struktura organizacyjna] (SNP/WB)
Scientific discipline (2.0)
6.4 biological sciences
Year of creation
2021
Start date
22-02-2019
Defense Date
15-10-2021
Title date
22-10-2021
Supervisor
Pages
110
URL
https://hdl.handle.net/10593/26431 Opening in a new tab
Keywords in English
acoustic communication; bird song; duet functions; territory defence; tropical forest
Abstract in English
I have examined various aspects of the sex specific acoustic repertoire of the duetting Yellow-breasted boubou, an endemic bird species of the Afro-montane region of Western Africa. The first area of examination was the natural vocalisation activity of the study species which highlighted the male dominant vocal behaviour and the use of sex specific repertoires. I then experimentally examined the intrusion into focal territories of either duets or the sex specific solo types. This suggested that female vocalisations are likely used in intra-pair communications whereas males are using vocalisations for defending a territory and their mate. A further series of experiments were used to assess the aggressive motivation of male vocalisation types. This suggested that males are able to display aggressive motivation through the use of matching behaviours as well as vocalisation type used. Finally, the propagation properties of all vocalisation types for different habitat types were examined to further assess the functions of song and call types. Male vocalisation types transmit further, adding more evidence to the intended receivers being neighbours and strangers, whereas females intended receivers are most likely their own mates. The abiotic source of noise caused by a network of streams, also has a negative effect on the propagation of signals, which is important due to the human induced habitat change occurring in this specific region.

Uniform Resource Identifier
https://researchportal.amu.edu.pl/info/phd/UAM1ea23723e5a24b828f08ccfc222ecb5e/
URN
urn:amu-prod:UAM1ea23723e5a24b828f08ccfc222ecb5e

Confirmation
Are you sure?


text:
## Marble Dropping

A forum for good logic/math puzzles.

Moderators: jestingrabbit, Moderators General, Prelates

Andy
Posts: 45
Joined: Mon Aug 28, 2006 10:23 am UTC

### Marble Dropping

Hey, this is my first non-how-are-you-going style post. I hope it is appropriate:

You have two identical glass marbles and a 100 storey building. You want to find out the highest floor in the building that you can drop marbles off without them breaking using the fewest number of drops.

After dropping a marble, if it breaks you can't use it any more. If it survives, then it isn't weakened or anything like that, so you will always get the same result if you drop a marble from the same floor.

EDIT:
Solution/spoiler discussion: http://forums.xkcd.com/viewtopic.php?t=201
Last edited by Andy on Tue Aug 29, 2006 1:15 pm UTC, edited 1 time in total.

mister k
Posts: 643
Joined: Sun Aug 27, 2006 11:28 pm UTC
Contact:
Hmm, I don't really understand how you can be very efficient here. You could go up in twos, and if your marble breaks you simply go down a level and test the other marble- although that might leave you with two broken marbles, you will know where they break, but I don't see how you can do any better, unless theres something you're missing.

Torn Apart By Dingos
Posts: 817
Joined: Thu Aug 03, 2006 2:27 am UTC
Are we looking for the best average case or the best worst case?

I guess it's okay to break both if that'll tell you what storey you're looking for? If not, you could only use the first marble and try floors 1,2,3,4,5... in order.

If there were infinitely many floors, I'd probably do this: try the first marble on floors 2^n, until it breaks, and then use the second marble to try consecutive floors from the last floor that was okay. If I do this with 100 floors, and try the hypothetical floor 128 if I have to, where I assume it will break, it'd take an average of 17.97 tries, and a worst case of 42. That's not so good.

But there's certainly better than that for 100 floors... one being trying floors 10,20,30,...,100 with the first marble and then trying floor by floor with the second marble. Worst case 18, average 10.

I'll have to think about it some more.

Torn Apart By Dingos
Posts: 817
Joined: Thu Aug 03, 2006 2:27 am UTC
I've got another solution, but unfortunately it wasn't better than my last one. It was easier to analyze, though (I used Python to calculate averages with my two first methods).

It seems the only scheme you can use is to try to exclude as many floors as possible with the first marble, and use the second marble floor-by-floor. If we first try floor above floor n_1 (*), then the floor above n_1+n_2, then the floor above n_1+n_2+n_3, etc, the number of tries we need are in this list:
[2,3,...,n_1,1; 3,4,...,n_2,2; 4,5,...,n_3,3; ...]
(the nth element in the list is the number of tries we need if floor n is the first floor from which the marble will break).

I thought it was natural to give each segment the same average, so I tried with n_1 = 19, n_2 = 18, n_3 = 17, ..., n_10 = 10. Then we have 10 segments, each with average 10, so the average number of tries is 10, and the worst case is 19.

Apparently there are better choices for n_k (my earlier post had a better worst-case), but now it's clearer how to find the solution.

EDIT:
(*) Actually, my convoluted explanation of n_k isn't quite right. If you can make out from the list what it represents, you've got it. I'll maybe fix this later.
Last edited by Torn Apart By Dingos on Mon Aug 28, 2006 5:07 pm UTC, edited 1 time in total.

RealGrouchy
Nobody Misses Me As Much As Meaux.
Posts: 6704
Joined: Thu May 18, 2006 7:17 am UTC
Contact:
Is it implied that the marble won't break when dropped from the first storey, and will break from the 100th?

Not sure if it matters, but it keeps my simple brain occupied.

- RG>
Mighty Jalapeno wrote:At least he has the decency to REMOVE THE GAP BETWEEN HIS QUOTES....
Sungura wrote:I don't really miss him. At all. He was pretty grouchy.

GreedyAlgorithm
Posts: 286
Joined: Tue Aug 22, 2006 10:35 pm UTC
Contact:
RealGrouchy wrote:Is it implied that the marble won't break when dropped from the first storey, and will break from the 100th?

Not sure if it matters, but it keeps my simple brain occupied.

- RG>

It's not implied that it will break from the 100th, but it seems to be implied that it will not break from the 1st (since it seems to imply such a floor exists). I'm going to assume that it can break from the first, though.
GENERATION 1-i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.

ulnevets
Posts: 186
Joined: Wed Aug 09, 2006 1:45 am UTC
Contact:
what if you give it a very slight upward velocity?

then it wouldn't be counted as a drop.

i win.

Shoofle
Posts: 409
Joined: Sun Apr 09, 2006 9:28 pm UTC
Location: Location, Location.
Contact:
ulnevets wrote:what if you give it a very slight upward velocity?

then it wouldn't be counted as a drop.

i win.

That doesn't actually change anything; the amount gravity accelerates it downwards while it gets to the peak is precisely countered by the acceleration it takes to get back down, so tossing it up is exactly the same as tossing it down with the same force.

Andy
Posts: 45
Joined: Mon Aug 28, 2006 10:23 am UTC
Are we looking for the best average case or the best worst case?

Sorry, we were optimising the worst case.

ulnevets
Posts: 186
Joined: Wed Aug 09, 2006 1:45 am UTC
Contact:
Shoofle wrote:
ulnevets wrote:what if you give it a very slight upward velocity?

then it wouldn't be counted as a drop.

i win.

That doesn't actually change anything; the amount gravity accelerates it downwards while it gets to the peak is precisely countered by the acceleration it takes to get back down, so tossing it up is exactly the same as tossing it down with the same force.

nonono

i mean, a drop is defined as starting with zero velocity

RealGrouchy
Nobody Misses Me As Much As Meaux.
Posts: 6704
Joined: Thu May 18, 2006 7:17 am UTC
Contact:
That's rediculous. Presumably, you have to lift the marble to some degree, and when you drop it, it will be from zero vertical velocity.

If you throw it up, it will reach zero vertical velocity and continue to accelerate downwards...

Oh, I give up. I'm really too tired to be discussing marble droppings.

- RG>
Mighty Jalapeno wrote:At least he has the decency to REMOVE THE GAP BETWEEN HIS QUOTES....
Sungura wrote:I don't really miss him. At all. He was pretty grouchy.

ulnevets
Posts: 186
Joined: Wed Aug 09, 2006 1:45 am UTC
Contact:
RealGrouchy wrote:That's rediculous. Presumably, you have to lift the marble to some degree, and when you drop it, it will be from zero vertical velocity.

If you throw it up, it will reach zero vertical velocity and continue to accelerate downwards...

Oh, I give up. I'm really too tired to be discussing marble droppings.

- RG>

you guys don't understand

when you drop something, it leaves your hand at zero velocity. throwing it upwards slightly would be considered a throw, and any amount of these are allowed.

if you're concerned with the slight upward velocity, you could always move your hand a little lower to account for it.

Jesse
Vocal Terrorist
Posts: 8635
Joined: Mon Jul 03, 2006 6:33 pm UTC
Location: Basingstoke, England.
Contact:
Even if it wouldn't be counted as a drop, you would still have two broken marbles.

ulnevets
Posts: 186
Joined: Wed Aug 09, 2006 1:45 am UTC
Contact:
Jesster wrote:Even if it wouldn't be counted as a drop, you would still have two broken marbles.

strategy: start at bottom floor and move up one at a time until it breaks. best case: 0 drops. worst case: 0 drops.

xkcd
Site Ninja
Posts: 365
Joined: Sat Apr 08, 2006 8:03 am UTC
Contact:
ulnevets wrote:
Jesster wrote:Even if it wouldn't be counted as a drop, you would still have two broken marbles.

strategy: start at bottom floor and move up one at a time until it breaks. best case: 0 drops. worst case: 0 drops.

Why do people always forget the worst case of "raptor attack"?

Jesse
Vocal Terrorist
Posts: 8635
Joined: Mon Jul 03, 2006 6:33 pm UTC
Location: Basingstoke, England.
Contact:
Because they are uninformed. Also, at what speed can raptors climb stairs?

Ooh, new problem. At what height does a dropped marble gain maximum raptor killing velocity?

Penguin
Posts: 98
Joined: Wed Jul 05, 2006 2:54 pm UTC
Location: Cambridge, MA
Contact:
I like this game better! What is the minimum floor you have to drop a marble from to kill a raptor?

If the marble doesn't break, you get a bonus!
<3!

Jesse
Vocal Terrorist
Posts: 8635
Joined: Mon Jul 03, 2006 6:33 pm UTC
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Contact:
Although, the marble wouldn't be retrievable.

Don't forget, raptors are intelligent and one would be hiding around the corner until you came to retrieve them.

ulnevets
Posts: 186
Joined: Wed Aug 09, 2006 1:45 am UTC
Contact:
xkcd wrote:
ulnevets wrote:
Jesster wrote:Even if it wouldn't be counted as a drop, you would still have two broken marbles.

strategy: start at bottom floor and move up one at a time until it breaks. best case: 0 drops. worst case: 0 drops.

Why do people always forget the worst case of "raptor attack"?

would it be worse if the raptors had magic powers?

Jesse
Vocal Terrorist
Posts: 8635
Joined: Mon Jul 03, 2006 6:33 pm UTC
Location: Basingstoke, England.
Contact:

hassellhoff
Posts: 30
Joined: Mon Jul 28, 2008 8:40 am UTC
Contact:

### Re: Marble Dropping

gah..

i was hoping for a little riddle i could do at 4 in the morning.. but then i see 1 guy had to use python.

i give up

consolation prize?
Exenon wrote: I play Call of Duty. That's for real men !

hyperion
"I'll show ye...."
Posts: 1569
Joined: Wed Nov 29, 2006 2:16 pm UTC
Location: Perth

### Re: Marble Dropping

therapist wrote:gah..

i was hoping for a little riddle i could do at 4 in the morning.. but then i see 1 guy had to use python.

i give up

consolation prize?

The prize is a smack in the head for dragging up a two year old thread and adding nothing.
Peshmerga wrote:A blow job would probably get you a LOT of cheeseburgers.
But I digress.

Crosby
Posts: 21
Joined: Tue Aug 05, 2008 6:46 pm UTC

### Re: Marble Dropping

first marble, try: 10, 20, 30, ... 100.
If it is still unbroken, then the answer is 100.
If it breaks at any point, then go back down 9 floors, and go up one-at-a time until it breaks.
If it does, then the floor under that is the solution.
If it does not break, then the last floor tried is the solution.

Coupla scenarios:
10,20,30,40,(break), 31,32,33,34,35 (break) => answer: 24 (9 tries)(also, the avg. number of tries need to solve with this solution.)
10,20,30,40,50,60,70,80,90,100 (no break) => answer: 100 (10 tries)
1 (break) => answer: 0 (1 try)(min solution)
10,20,30,40,50,60,70,80,90,100 (break),91,92,93,94,95,96,97,98,99 => answer: 99 (18 tries)(also, the max tries need to find a solution.)

We could try different ways to break up the floors, but I think this is optimal because we have equalized the max number of tries for the first marble and second marble (9 tries). Any solution using more floors for the first marble (and less tries), would result in more tries for the second marble. I'm not smart enough to "proof" this. Or maybe I'm just lazy.

crzftx
Posts: 371
Joined: Tue Jul 29, 2008 4:49 am UTC
Location: Rockford, IL

### Re: Marble Dropping

You can do it in 15. There's an answers thread with the answer posted, along with much more difficult math. I was happy about solving an unsolved puzzle until I realized this. At least I got the answer right

BoomFrog
Posts: 1069
Joined: Mon Jan 15, 2007 5:59 am UTC
Location: Seattle

### Re: Marble Dropping

crzftx wrote:You can do it in 15. There's an answers thread with the answer posted, along with much more difficult math. I was happy about solving an unsolved puzzle until I realized this. At least I got the answer right

Actually you can do it in 14
"Everything I need to know about parenting I learned from cooking. Don't be afraid to experiment, and eat your mistakes." - Cronos

crzftx
Posts: 371
Joined: Tue Jul 29, 2008 4:49 am UTC
Location: Rockford, IL

### Re: Marble Dropping

14?
How was this possible. I guess I may not have understood the answer then. I thought you could take the first marble first by 14 floors, then 13, etc. The other marble would then go 1 by 1 as needed to figure out which floor it is. This method can take up to 15 tries. It'd be 15 if the critical floor is 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, or 99.

I guess it only takes 14, since I don't need to break both marbles, necessarily, to find the floor. Oops.

afarnen
Posts: 157
Joined: Mon May 05, 2008 12:12 pm UTC

### Re: Marble Dropping

Spoiler:

Code: Select all

using the increment algorithm where you go up by n stories, starting with n, and when the first marble breaks, you go up starting at one plus the last story at which the first marble survived the fall, here's a graph i made:example: 10, 20, 30 (break), 21, 22, 23 (break).. that's 6 drops         increment       +----------#      | 1 2 3 4 5   +---+----------d  | 1 | 1 2 2 2 2r  | 2 | 2 2 3 3 3o  | 3 | 3 3 3 4 4p  | 4 | 4 3 3 4 5s  | 5 | 5 4 4 3 5inc.  avg drops  pattern---------------------------------------------1     50.5       1,2..1002     26.5       2,2,3,3..51,513     18.83      2,3,3,3,4,4..34,35,35,354     15.25      2,3,4,4,3,4,5,5..26,27,28,285     13.3       2,3,4,5,5,3,4,5,6,6..21,22,23,24,246     12.14      2,3,4,5,6,6,3,4,5,6,7,7..17,18,19,20,21,21,18,19,20,217     11.46      2,3,4,5,6,7,7,3,4,5,6,7,8,8..15,16,17,18,19,20,20,16,178     11.06      2,3,4,5,6,7,8,8,3,4,5,6,7,8,9,9..13,14,15,16,17,18,19,19,14,15,16,179     10.91      2,3,4,5,6,7,8,9,9,3,4,5,6,7,8,9,10,10..12,13,14,15,16,17,18,19,19,1310    10.9       2,3,4,5,6,7,8,9,10,10..11,12,13,14,15,16,17,18,19,1911    10.91      2,3,4,5,6,7,8,9,10,11,11..10,11,12,13,14,15,16,17,18,19,19,1112    10.94      2,3,4,5,6,7,8,9,10,11,12,12..9,10,11,12,13,14,15,16,17,18,19,19,10,11,12,13the most efficient increment algorithm is 10.

ctxcm2002
Posts: 9
Joined: Sun Aug 17, 2008 6:23 pm UTC

### Re:

ulnevets wrote:
RealGrouchy wrote:That's rediculous. Presumably, you have to lift the marble to some degree, and when you drop it, it will be from zero vertical velocity.

If you throw it up, it will reach zero vertical velocity and continue to accelerate downwards...

Oh, I give up. I'm really too tired to be discussing marble droppings.

- RG>

you guys don't understand

when you drop something, it leaves your hand at zero velocity. throwing it upwards slightly would be considered a throw, and any amount of these are allowed.

if you're concerned with the slight upward velocity, you could always move your hand a little lower to account for it.

He's right you know...

Last I checked I don't drop a basket ball, or baseball. They all reach zero vertical velocity. Granted their parabolic paths are much larger than in the marble dropping experiment. As far as this experiment, the only way I can come up with this being feasible is making a contraption out of cardboard toilet paper rolls, some scotch tape, 3 feet of floss, 3 cotton balls and 5 straws. 'Taint any rules abou' that!

Cauchy
Posts: 602
Joined: Wed Mar 28, 2007 1:43 pm UTC

### Re: Marble Dropping

Due to very fierce winds, your hand gets cut off if you try to make a measurement without dropping the marble.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

Penitent87
Posts: 28
Joined: Wed Apr 15, 2009 2:30 am UTC
Location: London

### Re: Marble Dropping

afarnen wrote:
Spoiler:

Code: Select all

using the increment algorithm where you go up by n stories, starting with n, and when the first marble breaks, you go up starting at one plus the last story at which the first marble survived the fall, here's a graph i made:example: 10, 20, 30 (break), 21, 22, 23 (break).. that's 6 drops         increment       +----------#      | 1 2 3 4 5   +---+----------d  | 1 | 1 2 2 2 2r  | 2 | 2 2 3 3 3o  | 3 | 3 3 3 4 4p  | 4 | 4 3 3 4 5s  | 5 | 5 4 4 3 5inc.  avg drops  pattern---------------------------------------------1     50.5       1,2..1002     26.5       2,2,3,3..51,513     18.83      2,3,3,3,4,4..34,35,35,354     15.25      2,3,4,4,3,4,5,5..26,27,28,285     13.3       2,3,4,5,5,3,4,5,6,6..21,22,23,24,246     12.14      2,3,4,5,6,6,3,4,5,6,7,7..17,18,19,20,21,21,18,19,20,217     11.46      2,3,4,5,6,7,7,3,4,5,6,7,8,8..15,16,17,18,19,20,20,16,178     11.06      2,3,4,5,6,7,8,8,3,4,5,6,7,8,9,9..13,14,15,16,17,18,19,19,14,15,16,179     10.91      2,3,4,5,6,7,8,9,9,3,4,5,6,7,8,9,10,10..12,13,14,15,16,17,18,19,19,1310    10.9       2,3,4,5,6,7,8,9,10,10..11,12,13,14,15,16,17,18,19,1911    10.91      2,3,4,5,6,7,8,9,10,11,11..10,11,12,13,14,15,16,17,18,19,19,1112    10.94      2,3,4,5,6,7,8,9,10,11,12,12..9,10,11,12,13,14,15,16,17,18,19,19,10,11,12,13the most efficient increment algorithm is 10.

I think you forgot to look carefully enough at the final few marbles in some lists. You just kept the pattern going instead of actually thinking about it. (e.g. the number of marble drops required to identify the 100th floor, given an increment of 9, is 12, not 13. 9,18,27,36,45,54,63,72,81,90,99,100.) These alterations mean that 9, 10, and 11 all have the same average case of 10.9 drops.

GeneralFailure
Posts: 1
Joined: Mon Jun 29, 2009 10:40 pm UTC

### Re: Marble Dropping

May we assume that the building is of such a height that the marble, having been dropped from a height below the top floor, will not reach terminal velocity? Otherwise, we could optimize our algorithm given the height of each floor, the force of air resistance on the marble, and the force of gravity by determining the highest floor which it even makes sense to try (since all higher floors would presumably yield the same result).

Soljer
Posts: 29
Joined: Fri Feb 27, 2009 6:31 pm UTC

### Re: Marble Dropping

This is a pretty easy problem, and you can even generalize it to an arbitrary number of marbles.

Spoiler:
For 1 marble, you can't do better than a linear search.

For 2 marbles, and n floors, drop the first marble at increments of root(n) - you'll make at most root(n) drops. Once the first one breaks (let's call it floor p*root(n)), go to floor (p-1)*root(n), where you know the first marble was unbroken, and linearly search upwards till the second one breaks, you know the highest floor is one less than where it broke. This step will also take at most root(n), for a final run time of 2*root(n) or O(root(n)).

For three marbles, drop the first marble in increments of n^(2/3). This will take be at most n^(1/3) drops. Once that breaks (at floor p*n^(2/3)), go to floor (p-1)*n^(2/3) and use the two-marble algorithm as described above. That is, walk from (p-1)*n^(2/3) to p*n^(2/3) in increments of root(n^2/3) or, n^(1/3). This will take at most n^(1/3) steps. Once the second marble breaks, do a linear search over at most n^(1/3) floors. So our run time is 3*n^(1/3) or O(n^1/3).

In general, for k marbles, you can get a run time of k*n^(1/k).

If you have log(n) marbles, you can do a binary search for log(n) run-time.

I don't know if that will always be the best policy for all cases, but it is a sub-linear policy for an arbitrary number of floors and marbles (assuming more than one marble, anyways).

WarDaft
Posts: 1583
Joined: Thu Jul 30, 2009 3:16 pm UTC

### Re: Marble Dropping

Soljer wrote:This is a pretty easy problem, and you can even generalize it to an arbitrary number of marbles.

Spoiler:
For 1 marble, you can't do better than a linear search.

For 2 marbles, and n floors, drop the first marble at increments of root(n) - you'll make at most root(n) drops. Once the first one breaks (let's call it floor p*root(n)), go to floor (p-1)*root(n), where you know the first marble was unbroken, and linearly search upwards till the second one breaks, you know the highest floor is one less than where it broke. This step will also take at most root(n), for a final run time of 2*root(n) or O(root(n)).

For three marbles, drop the first marble in increments of n^(2/3). This will take be at most n^(1/3) drops. Once that breaks (at floor p*n^(2/3)), go to floor (p-1)*n^(2/3) and use the two-marble algorithm as described above. That is, walk from (p-1)*n^(2/3) to p*n^(2/3) in increments of root(n^2/3) or, n^(1/3). This will take at most n^(1/3) steps. Once the second marble breaks, do a linear search over at most n^(1/3) floors. So our run time is 3*n^(1/3) or O(n^1/3).

In general, for k marbles, you can get a run time of k*n^(1/k).

If you have log(n) marbles, you can do a binary search for log(n) run-time.

I don't know if that will always be the best policy for all cases, but it is a sub-linear policy for an arbitrary number of floors and marbles (assuming more than one marble, anyways).
That isn't the solution for worst case.
Spoiler:
You can improve your worst worst case scenario by decreasing the effectiveness of your slightly better worst case scenario. Drop it from the 14th floor, if it breaks, you have up to 13 more drops for worst case 14 total to find the floor. If it doesn't break at 14 drop it from the 27th floor, if it breaks, you have up 12 more drops for... worst case 14. The floors (or at least solution) for your first marble are 14, 27, 39, 50, 60, 69, 77, 84, 89, 94, 97,98,99,100.

Furthermore, k*n^(1/k) is not even the run time for a radix style searching. If you have 10 marbles and 100 floors, then you don't need 15 drops, that's worse than worst case for 2 marbles.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.

Qaanol
The Cheshirest Catamount
Posts: 3062
Joined: Sat May 09, 2009 11:55 pm UTC

### Re: Marble Dropping

If you have m marbles and an s-story building,
Spoiler:
Find the least n such that
${n \choose m} \geq s$
The maximum number of drops can be bounded by n-1 if m>1, or by n if m=1.

Drop the first marble from floor [imath]\large{{n-1}\choose{m-1}}[/imath]. If it doesn't break, increment the floor by [imath]\large{{n-2}\choose{m-1}}[/imath]. Continue this until the first marble breaks, say after incrementing by [imath]\large{{n-k_1}\choose{m-1}}[/imath]. Starting from the last floor where the first marble didn't break, increment by [imath]\large{{n-k_1-1}\choose{m-2}}[/imath] for the second marble. Keep decreasing the value in the top of the choose function by one each time, and adding that to the floor from which you drop the second marble. Keep doing this until the second marble breaks, say after incrementing by [imath]\large{{n-k_2}\choose{m-2}}[/imath]. Then start the increments for the third marble at [imath]\large{{n-k_2-1}\choose{m-3}}[/imath]. What we're doing is using the columns of Pascal's triangles as increment values, moving upwards. The column of Pascal's triangle (the number in the bottom of the choose function) is the number of marbles we still have left not counting the one currently being dropped. Note that the columns are indexed from 0, as are the rows.

Here's the beginning of Pascal's triangle (left-aligned).
$\begin{eqnarray}1\\ 1 & 1\\ 1 & 2 & 1\\ 1 & 3 & 3 & 1\\ 1 & 4 & 6 & 4 & 1\\ 1 & 5 & 10 & 10 & 5 & 1\\ 1 & 6 & 15 & 20 & 15 & 6 & 1\\ 1 & 7 & 21 & 35 & 35 & 21 & 7 & 1\\ 1 & 8 & 28 & 56 & 70 & 56 & 28 & 8 & 1\\ 1 & 9 & 36 & 84 & 126 & 126 & 84 & 36 & 9 & 1\\ 1 & 10 & 45 & 120 & 210 & 252 & 210 & 120 & 45 & 10 & 1\\ 1 & 11 & 55 & 165 & 330 & 462 & 462 & 330 & 165 & 55 & 11 & 1\\ 1 & 12 & 66 & 220 & 495 & 792 & 924 & 792 & 495 & 220 & 66 & 12 & 1\\ 1 & 13 & 78 & 286 & 715 & 1287 & 1716 & 1716 & 1287 & 715 & 286 & 78 & 13 & 1\\ 1 & 14 & 91 & 364 & 1001 & 2002 & 3003 & 3432 & 3003 & 2002 & 1001 & 364 & 91 & 14 & 1\\ 1 & 15 & 105 & 455 & 1365 & 3003 & 5005 & 6435 & 6435 & 5005 & 3003 & 1365 & 455 & 105 & 15 & 1 \end{eqnarray}$
Say we have 6 marbles and a 5000-floor building. Just for kicks let's say the marbles will first break when dropped from the 3600th floor. Since we have 6 marbles, we look in column 6 for the first number over 5000. That's [imath]\large{15\choose 6} = 5005[/imath]. We locate that in Pascal's triangle (on the left half). Then we move one place to the left and one place up. That column will be our increments for the first marble.

First marble:
floor 2002 does not break
floor 2002 + 1287 = 3289 does not break
floor 2002 + 1287 + 792 = 4081 breaks

Second marble:
floor 3289 + 330 = 3619 breaks

Third marble:
floor 3289 + 120 = 3409 does not break
floor 3289 + 120 + 84 = 3493 does not break
floor 3289 + 120 + 84 + 56 = 3549 does not break
floor 3289 + 120 + 84 + 56 + 35 = 3584 does not break
floor 3289 + 120 + 84 + 56 + 35 + 20 = 3604 breaks

Fourth marble:
floor 3584 + 10 = 3594 does not break
floor 3584 + 10 + 6 = 3600 breaks

Fifth marble:
floor 3594 + 3 = 3597 does not break
floor 3594 + 3 + 2 = 3599 does not break

And we are done. In this particular case we didn't need to use all 6 marbles. The point is, since at every drop we rise up one row in Pascal's triangle for our increment value, the maximum number of drops equals the row in which we began. In this case we began on row 14, so we had an upper bound of 14 drops.

For the classic case of 2 marbles and 100 floors, clearly 105 is the first value in the 2-marble column that's at least 100, so we begin there and move diagonally up and to the left to find 14 as the place to start. Since that's on row 14, that's also the upper bound on number of drops.
wee free kings


text:
Letter | Published:

# A strongly interacting polaritonic quantum dot

## Abstract

Polaritons are promising constituents of both synthetic quantum matter1 and quantum information processors2, whose properties emerge from their components: from light, polaritons draw fast dynamics and ease of transport; from matter, they inherit the ability to collide with one another. Cavity polaritons are particularly promising as they may be confined and subjected to synthetic magnetic fields controlled by cavity geometry3, and furthermore they benefit from increased robustness due to the cavity enhancement in light–matter coupling. Nonetheless, until now, cavity polaritons have operated only in a weakly interacting mean-field regime4,5. Here we demonstrate strong interactions between individual cavity polaritons enabled by employing highly excited Rydberg atoms as the matter component of the polaritons. We assemble a quantum dot composed of approximately 150 strongly interacting Rydberg-dressed 87Rb atoms in a cavity, and observe blockaded transport of photons through it. We further observe coherent photon tunnelling oscillations, demonstrating that the dot is zero-dimensional. This work establishes the cavity Rydberg polariton as a candidate qubit in a photonic information processor and, by employing multiple resonator modes as the spatial degrees of freedom of a photonic particle, the primary ingredient to form photonic quantum matter6.

## Main

Recent years have seen a growing effort to create matter from light1,7. The speed of the photon may be employed to realize reduced-dimensional models with near-arbitrary single-particle dynamics, and recent breakthroughs have even enabled the creation of synthetic gauge fields that impose a Lorentz-like force on the photon8,9,10. While a number of groundbreaking experiments have probed the physics of photon condensation4,11, emergent crystallinity from long-range interactions12,13,14, and opto-mechanics15, the challenge of achieving strong interactions between individual photons for exploration of entangled photonic quantum matter persists. The demonstration of a single-photon nonlinearity mediated by a single atom coupled to a Fabry–Pérot cavity16 spurred an outpouring of theoretical work exploring Hubbard physics in coupled nonlinear resonators17,18, but the severe technical challenges associated with the deterministic preparation of individual atoms in resonator arrays remain.

Two major efforts have begun to address the technical challenges of mediating strong photonic interactions using atoms. In the first, researchers successfully demonstrated the loading of an individual atom into the near-field of a photonic crystal cavity19, and this capability was subsequently extended to the coupling of atomic ensembles to waveguides20. In the second, the challenges associated with trapping and cooling single atoms within optical resonators are eliminated by using Rydberg electromagnetically induced transparency (EIT)21. With Rydberg EIT, an entire thermal gas of atoms may act as a single ‘super atom’ that has the cross-section of the ensemble and the nonlinearity of a single atom22,23,24,25,26. Nonetheless, the propagation of these free-space excitations complicates their application to quantum information processors or materials compared to their cavity-confined counterparts.

Here, we present an approach combining Rydberg EIT and optical resonators27,28, which benefits from the best aspects of each previous approach: Rydberg EIT provides a robust atomic-ensemble-mediated single-photon nonlinearity, while the resonator enhances the light–matter coupling and allows us to work at substantially lower atomic densities while maintaining a high optical depth per blockade volume (ODB). This quantity is crucial, because ODB determines the number of collisions per polariton lifetime in a synthetic quantum material and the gate fidelity in a photonic quantum information processor29 (see Supplementary Information B). Perhaps most significantly, in the future, a quantum nonlinear resonator may be tuned near a mode-degeneracy point where interactions couple the transverse modes and the system behaves as a continuum-equivalent of an array of tunnel-coupled resonators. This approach would enable the creation of a strongly interacting quantum fluid within an individual, multi-mode cavity. Furthermore, the fluid can be trapped and subjected to strong magnetic fields3, providing access to the physics of photon crystals and Laughlin puddles6.

Here, we use this platform to demonstrate the first strongly interacting polaritonic quantum dot. We create a discrete mode spectrum by confinement of the photon in a cavity and strong interactions by tight confinement of the assembled atom cloud. In what follows, we describe the quantum dot in more detail and show that it exhibits the characteristic features of strong nonlinearity at the single-photon level, in contrast to related works exhibiting only weak nonlinearities5,30. First, we demonstrate strong light–matter coupling by measuring a well-resolved dark-polariton resonance in the transmission spectrum. We next explore transport via the statistics of photons tunnelling through the cavity, observing deep anti-bunching indicative of blockade due to strong interactions between individual polaritons. We further demonstrate that our polaritonic dot is zero-dimensional by detecting coherent Rabi oscillations between an empty dot and a dot containing a single polariton, as well as saturation of the intracavity dark polariton number at 0.47(3). We conclude with a discussion of applications in quantum information and strongly correlated, topological phases of matter.

Our quantum dot is formed from a sample of 2,600(500) atoms loaded into the 12 μm × 14 μm waist of a high-finesse single-mode optical resonator (Fig. 1a,c, and Supplementary Information D for details), at a peak density of $$1\times 1{0}^{11}$$ cm−3 (see Methods in Supplementary Information A). These atoms are distributed over a 35 μm r.m.s. axial length that may be reduced to 10 μm r.m.s. by spatially selective optical depumping, or ‘slicing’ (see Supplementary Information C). Due to strong light–matter coupling ($${{\rm{OD}}}_{{\rm{B}}}\approx 13$$), the modes of the system hybridize, forming polaritons (Fig. 1b): composite states of a resonator photon with an atomic excitation. Two of the hybridized modes are ‘bright’ polaritons, composed primarily of an admixture of a resonator photon and a short-lived P-state excitation; they are weakly interacting due to their small Rydberg component and short lifetime. The third polariton, called the ‘dark’ polariton, is composed of a coherent superposition of a Rydberg excitation and a resonator photon, with the mixing ratio set by θ, the dark-state rotation angle31. For our typical conditions, $${\rm{\tan }}\theta =\sqrt{N}g/\Omega =2.9(5)$$, where N, g and Ω are the atom number, single-atom Rabi frequency and control-field Rabi frquency, respectively. The Rydberg component of the dark polariton enables it to strongly repel other dark polaritons within a surrounding ‘blockade volume’, which scales with $${n}^{11/2}$$ (for van der Waals interactions), where n is the Rydberg level’s principal quantum number. It is for this reason that we typically employ the $$n=100$$ Rydberg state, as the atoms' extreme properties make this blockade volume comparable to the sample size. Confining the region of overlap between the resonator mode and the atomic gas to be smaller than the blockade volume gives rise to a zero-dimensional, strongly interacting polaritonic quantum dot; like its solid-state counterpart, the electronic quantum dot32, it is expected to exhibit blockaded transport (Fig. 1d).

To investigate the polaritonic excitations of our quantum dot, we first probe the transmission spectrum of the laser-dressed cavity–atom system. In Fig. 1b, we plot the quantum dot’s transmission spectrum on the $$\{5{S}_{1/2},\,F=2\}\to \{5{P}_{3/2},\,F=3\}\to \{100{S}_{1/2}\}$$ transition, without slicing the cloud. We observe two broad resonances that we identify as bright polaritons, whose spectral separation is an indication of strong light–matter coupling. The central feature is the spectrally narrow dark polariton. As we increase the probe power, the fractional transmission on the dark polariton resonance drops, as shown in Fig. 2, indicating that the repulsion between these polaritons suppresses their simultaneous injection into the dot.

To demonstrate that individual dark polaritons interact strongly with one another within the quantum dot, we explore transport dynamics. We test transport by injecting photons into the resonator at the energy of the dark polariton, such that most photons become dark polaritons on entering. The strong repulsion between polaritons shifts the energy and reduces the lifetime of a second polariton in the dot, thereby precluding its injection. In an electronic quantum dot, the analogous Coulomb-blockade physics is typically ascertained from the dependence of transport on bias voltage32; by contrast, we directly observe suppression of simultaneous polariton transit by detecting when photons tunnel through the dot. We characterize the tunnelling times via the temporal intensity autocorrelation function, $${g}_{2}(\tau )$$, where τ is the delay time, of the cavity transmission. This function compares the rate of photon pairs leaking out of the resonator separated by a time τ to a Poisson process with the same average rate. A reduction near zero delay ($$\tau =0$$) indicates interaction-driven suppression of double-occupancy of the dot, while the rise time back to steady state is dominated by the injection Rabi frequency.

After slicing the cloud so that it can hold only a single polariton at a time, we observe almost complete suppression ($${g}_{2}(0)=0.27(8)$$) of simultaneous tunnelling through the dot, indicative of transport blockade due to strong interactions (Fig. 3a). Indeed, strong anti-bunching at zero-time separation validates a model where a single intracavity polariton shifts and broadens the energy for the injection of the next polariton by more than the polariton linewidth, strongly suppressing its tunnelling into the resonator until the first polariton tunnels out. The observed $${g}_{2}(\tau )$$ is consistent with a master-equation effective theory bootstrapped by a full model of interacting three-level atoms coupled to a single resonator mode (see Supplementary Information J). Note that the linear growth of $${g}_{2}(\tau )$$ near zero delay, as opposed to the quadratic growth anticipated from a single-mode theory, arises from virtual excitation of bright polaritons.

We verify that the observed transport blockade results from strongly interacting polaritons by comparing the observations to transport with weaker interactions. First, we test the unsliced cloud (Fig. 3a), which exhibits only weak suppression of g2 near $$\tau =0$$ because the cloud is large enough to hold multiple polaritons along the resonator axis simultaneously. Similarly, a sliced sample with Rydberg atoms in the smaller 85S state can fit multiple excitations transversely and also exhibits weak suppression of g2 (Fig. 3b). Furthermore, we find that transport of bright polaritons is not blockaded at all (Fig. 3b), as these polaritons are essentially non-interacting33. The blockade of the dark polaritons can even be `reversed' by detuning from the atomic 5P state to make the polariton–polariton scattering more elastic and probing transport on the two-polariton resonance (Fig. 3c). In this case, we observe photon bunching, $${g}_{2} > 1$$, reflecting the preferential transit of photon pairs34.

To investigate the dimensionality of the polaritonic dot, we probe the cavity with pulses of substantially higher intensity and observe the transmitted light as the intracavity field rings up and then down. A zero-dimensional system may be coherently excited and then de-excited by the drive field, exhibiting Rabi oscillations, while in a one-dimensional system the excitation would propagate away before it can be removed. Figure 4a shows the detuning dependence of the ring-up process, exhibiting rapid, detuned Rabi oscillations away from the dark-polariton resonance, and slower oscillations on-resonance. Figure 4b shows the ring-up dynamics on the dark-polariton resonance for various probe intensities. We observe Rabi oscillations between zero and one dark polaritons at the highest intensities, indicative of a strongly blockaded, zero-dimensional dot interacting with many photons within the polariton lifetime. The dark-polariton oscillations exhibit a Rabi frequency of $${\Omega }_{{\rm{polariton}}}=2{\rm{\pi }}\times 245$$ kHz, in agreement with a first-principles calculation based on the probe power (see Supplementary Information G). Note that the fast ~100 ns oscillations arise from off-resonant excitation of non-interacting bright polaritons.

Strong interactions should cause the number of dark polaritons in the dot to saturate at large probe powers. However, the off-resonant excitation of bright polaritons does not saturate, resulting in a high steady-state leakage rate of bright polaritons that overwhelms the dark-polariton signal at high probe powers. To disentangle the bright and dark contributions to the cavity transmission, Fig. 4c shows the ring-down of the dot once the probe beam is turned off. At each probe intensity, the ring-down consists of slow and fast exponential decays; the slow feature (with a time constant of 730(40) ns) reflects the dynamics of the dark polariton, while the fast feature reflects the decay of the interfering bright polaritons. We isolate the steady-state dark-polariton number, shown in Fig. 4d, by extrapolating the slowly decaying tail of the ring-down to its zero-time population. The strong saturation of the dark-polariton number with drive power again indicates blockade, while the saturation value of $${n}_{{\rm{D}}}=0.47(3)$$ reflects the behaviour of a single zero-dimensional two-level emitter, as described by the optical Bloch equations. This stationary, strongly coupled, two-level emitter is an excellent candidate qubit for a photonic quantum information processor35. To verify saturation, we measure the g2 of the slowly decaying tail of the ring-down, after the bright polaritons have all decayed away. We find $${g}_{2}=0.31(7)$$, confirming that the dot rarely hosts multiple dark polaritons. Moving forward, blockade-enhanced detection of dark polaritons will directly boost the dark-polariton detection signal-to-noise above the bright-polariton background36, while enhanced control-field and light–matter coupling will suppress the background (see Supplementary Information H), enabling the dot to act as an on-demand single-photon source.

Cavity Rydberg polaritons are now ripe for applications in synthetic quantum materials; recent experiments have demonstrated synthetic magnetic fields for resonator photons3 and in conjunction with the present work, there is now a clear path to topological many-photon states6. More broadly, the numerous proposals to study synthetic quantum matter in cavity arrays and continua1 can now be explored in the modes of an individual, multi-mode optical cavity, employing Rydberg atoms to mediate strong photon–photon interactions. Achieving stronger light–matter coupling will require only additional atomic density, rather than increasingly challenging advances in optical super-mirror technology necessary for single-emitter approaches. Indeed, densities routinely achieved in free-space experiments ($$\rho \approx 1{0}^{12}$$ cm3)23 and finesses of single-atom experiments ($$F\approx 1{0}^{5}$$) would provide $${{\rm{OD}}}_{{\rm{B}}}\approx 88,000$$, more than sufficient for few-particle synthetic materials6. Therefore, upcoming challenges centre on harnessing these developments to answer questions about dissipative preparation of many-body quantum states37,38 and about the exploration of the resulting phase diagrams39 using the quasi-particle manipulation and detection tools unique to quantum optics40,41.

### Data availability

The data that support the plots within this paper and other findings of this study are available from the corresponding author upon reasonable request.

Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

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Carusotto, I. & Ciuti, C. Quantum fluids of light. Rev. Mod. Phys. 85, 299–366 (2013).

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Saffman, M., Walker, T. G. & Mølmer, K. Quantum information with Rydberg atoms. Rev. Mod. Phys. 82, 2313–2363 (2010).

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Schine, N., Ryou, A., Gromov, A., Sommer, A. & Simon, J. Synthetic Landau levels for photons. Nature 534, 671–675 (2016).

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Deng, H., Haug, H. & Yamamoto, Y. Exciton–polariton Bose–Einstein condensation. Rev. Mod. Phys. 82, 1489–1537 (2010).

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Parigi, V. et al. Observation and measurement of interaction-induced dispersive optical nonlinearities in an ensemble of cold Rydberg atoms. Phys. Rev. Lett. 109, 233602 (2012).

6. 6.

A. Sommer, Büchler, H. P. & Simon, J. Quantum crystals and Laughlin droplets of cavity Rydberg polaritons. Preprint at https://arxiv.org/abs/1506.00341 (2015).

7. 7.

Douglas, J. S. et al. Quantum many-body models with cold atoms coupled to photonic crystals. Nat. Photon. 9, 326–331 (2015).

8. 8.

Haldane, F. D. M. & Raghu, S. Possible realization of directional optical waveguides in photonic crystals with broken time-reversal symmetry. Phys. Rev. Lett. 100, 013904 (2008).

9. 9.

Hafezi, M., Mittal, S., Fan, J., Migdall, A. & Taylor, J. M. Imaging topological edge states in silicon photonics. Nat. Photon. 7, 1001–1005 (2013).

10. 10.

Rechtsman, M. C. et al. Photonic Floquet topological insulators. Nature 496, 196–200 (2013).

11. 11.

Klaers, J., Schmitt, J., Vewinger, F. & Weitz, M. Bose–Einstein condensation of photons in an optical microcavity. Nature 468, 545–548 (2010).

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Gopalakrishnan, S., Lev, B. L. & Goldbart, P. M. Emergent crystallinity and frustration with Bose–Einstein condensates in multimode cavities. Nat. Phys. 5, 845–850 (2009).

13. 13.

Baumann, K., Guerlin, C., Brennecke, F. & Esslinger, T. Dicke quantum phase transition with a superfluid gas in an optical cavity. Nature 464, 1301–1306 (2010).

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Kollár, A. J. et al. Supermode-density-wave-polariton condensation with a Bose–Einstein condensate in a multimode cavity. Nat. Commun. 8, 14386 (2017).

15. 15.

Purdy, T. P. et al. Tunable cavity optomechanics with ultracold atoms. Phys. Rev. Lett. 105, 133602 (2010).

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Birnbaum, K. M. et al. Photon blockade in an optical cavity with one trapped atom. Nature 436, 87–90 (2005).

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Hartmann, M. J., Brandao, F. G. S. L. & Plenio, M. B. Strongly interacting polaritons in coupled arrays of cavities. Nat. Phys. 2, 849–855 (2006).

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Angelakis, D. G., Santos, M. F. & Bose, S. Photon-blockade-induced Mott transitions and x y spin models in coupled cavity arrays. Phys. Rev. A 76, 031805 (2007).

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Thompson, J. D. et al. Coupling a single trapped atom to a nanoscale optical cavity. Science 340, 1202–1205 (2013).

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Goban, A. et al. Superradiance for atoms trapped along a photonic crystal waveguide. Phys. Rev. Lett. 115, 063601 (2015).

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Mohapatra, A. K., Jackson, T. R. & Adams, C. S. Coherent optical detection of highly excited Rydberg states using electromagnetically induced transparency. Phys. Rev. Lett. 98, 113003 (2007).

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Pritchard, J. D. et al. Cooperative atom–light interaction in a blockaded Rydberg ensemble. Phys. Rev. Lett. 105, 193603 (2010).

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Peyronel, T. et al. Quantum nonlinear optics with single photons enabled by strongly interacting atoms. Nature 488, 57–60 (2012).

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Dudin, Y. O. & Kuzmich, A. Strongly interacting Rydberg excitations of a cold atomic gas. Science 336, 887–889 (2012).

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Tiarks, D., Baur, S., Schneider, K., Dürr, S. & Rempe, G. Single-photon transistor using a Förster resonance. Phys. Rev. Lett. 113, 053602 (2014).

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Gorniaczyk, H., Tresp, C., Schmidt, J., Fedder, H. & Hofferberth, S. Single-photon transistor mediated by interstate Rydberg interactions. Phys. Rev. Lett. 113, 053601 (2014).

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Guerlin, C., Brion, E., Esslinger, T. & Mølmer, K. Cavity quantum electrodynamics with a Rrydberg-blocked atomic ensemble. Phys. Rev. A 82, 053832 (2010).

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Boddeda, R. et al. Rydberg-induced optical nonlinearities from a cold atomic ensemble trapped inside a cavity. J. Phys. B 49, 084005 (2016).

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Gorshkov, A. V., Otterbach, J., Fleischhauer, M., Pohl, T. & Lukin, M. D. Photon–photon interactions via Rydberg blockade. Phys. Rev. Lett. 107, 133602 (2011).

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Ningyuan, J. et al. Observation and characterization of cavity Rydberg polaritons. Phys. Rev. A 93, 041802 (2016).

31. 31.

Fleischhauer, M., Imamoglu, A. & Marangos, J. P. Electromagnetically induced transparency: Optics in coherent media. Rev. Mod. Phys. 77, 633–673 (2005).

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Kouwenhoven, L. P. et al. in Mesoscopic Electron Transport (eds Sohn, L. L., Kouwenhoven, L. P. & Schön, G.) 105–214 (Springer, Dordrecht, 1997).

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Shore, B. W. & Knight, P. L. The Jaynes–Cummings model. J. Mod. Opt. 40, 1195–1238 (1993).

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Verger, A., Ciuti, C. & Carusotto, I. Polariton quantum blockade in a photonic dot. Phys. Rev. B 73, 193306 (2006).

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Brion, E., Carlier, F., Akulin, V. M. & Mølmer, K. Quantum repeater with Rydberg-blocked atomic ensembles in fiber-coupled cavities. Phys. Rev. A 85, 042324 (2012).

36. 36.

Günter, G. et al. Interaction enhanced imaging of individual Rydberg atoms in dense gases. Phys. Rev. Lett. 108, 013002 (2012).

37. 37.

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Lebreuilly, J. et al. Stabilizing strongly correlated photon fluids with non-Markovian reservoirs. Phys. Rev. A 96, 033828 (2017).

39. 39.

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41. 41.

Umucallar, R. O. & Carusotto, I. Many-body braiding phases in a rotating strongly correlated photon gas. Phys. Lett. A 377, 2074–2078 (2013).

## Acknowledgements

We would like to thank M. Fleischhauer and H. P. Buechler for fruitful conversations. This work was supported by DOE grant DE-SC0010267 for apparatus construction, DARPA grant W911NF-15-1-0620 for modelling, and MURI grant FA9550-16-1-0323 for data collection and analysis. A.G. acknowledges support from the UChicago MRSEC grant NSF-DMR-MRSEC 1420709. A.R. acknowledges support from the NDSEG Fellowship.

## Author information

### Author notes

• Albert Ryou

Present address: Department of Electrical Engineering, University of Washington, Seattle, WA, USA

• Ariel Sommer

Present address: Department of Physics, Lehigh University, Bethlehem, PA, USA

### Affiliations

1. #### Department of Physics and James Franck Institute, University of Chicago, Chicago, IL, USA

• Ningyuan Jia
• , Nathan Schine
• , Alexandros Georgakopoulos
• , Albert Ryou
• , Logan W. Clark
• , Ariel Sommer
•  & Jonathan Simon

### Contributions

The experiment was designed and built by all authors. J.N., N.S., L.W.C. and J.S. collected and analysed the data. All authors contributed to the manuscript.

### Competing interests

The authors declare no competing interests.

### Corresponding author

Correspondence to Jonathan Simon.

## Supplementary information

1. ### Supplementary Information

Supplementary notes, supplementary figures 1–9, supplementary references


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# Uncertainty quantification¶

The Uncertainty quantification block performs sensitivity and reliability analysis studies. It is based on the OpenTURNS methodology. Supported methods are:

• Simulation: the block generates randomized inputs for the studied model and analyzes its responses to estimate their statistical characteristics. Optionally it can also fit each output to known distributions, finding the distribution parameters and the best matching distribution. Input values can be either taken from parametric distributions which you specify in the block’s configuration, or generated using an input sample which the block automatically fits to some distribution, and then takes values from this internally defined distribution.

• Reliability analysis: the block generates inputs for the model and analyzes responses, which have bounds specified in the block’s configuration. The main result of this study is a reliability estimate which is the probability that the value of a response is within its specified bounds (in case of multiple responses, estimates probability that all responses satisfy bounds simultaneously). That is, the estimate shows how robust or reliable the model is when it receives inputs from a certain range. The range is defined, implicitly or explicitly, by the distribution you specify for each input, or by the given input sample.

Reliability analysis may use the following methods:

• Sampling: the direct sampling method which generally works as described above. Input generation can be random (Monte-Carlo sampling) or use values taken from isoprobabilistic intervals (LHS sampling).
• FORM: the first order reliability method. This method performs blackbox sampling but uses the collected data to create an internal probability model. The internal model is then used to estimate reliability.

To perform analysis, the Uncertainty quantification block has to be connected to a blackbox block — the block which evaluates responses of some computational model. Basic configuration is:

• Add variables and responses to the Uncertainty quantification block configuration.
• For each variable, the block adds an output port named Blackbox.<variable_name>. Connect these ports to your blackbox inputs.
• For each response, the block adds an input port named Blackbox.<response_name>. Connect output ports of your blackbox to these inputs.
• Select the analysis method. Note that the Uncertainty quantification block supports SmartSelection, so by default it will select an appropriate method automatically based on properties of variables and responses.
• Set Number of designs. This is the maximum number of design points which the block is allowed to generate during analysis, or the maximum number of model evaluations (blackbox calls).

The block also provides additional input ports which you can use to set properties of variables and responses, instead of specifying these properties in the configuration dialog.

For variables, the <variable_name>.Sample input ports accept samples with variable values. Note that when you send a sample for some variable, the block ignores the distribution which you set for the variable in its configuration, and always derives the distribution from the sample.

Main ports which output analysis results are the following:

• Report — outputs an analysis report in the HTML format.
• All designs — outputs all data collected during the analysis as a single matrix. Each matrix row represents a single design point. Matrix columns contain values of variables and responses sorted in the same order as they are listed on the Variables and Responses panes in block’s configuration (variables first). Variables and responses with Size 2 or more occupy multiple columns.
• Variable designs and Response designs — similar to All designs, but output the variables and responses data as two separate matrices.

Monitoring is enabled by default for the main result ports. To view the block’s report, which is output to the Report port, you can use the Page viewer in Analyze:

1. Create a new report and open the Report database pane.
2. Drag the record with the report from the Project database pane to the Data series pane in the report database. This creates a string data series, which usually contains 1 element — a long string with the HTML code.
3. Select this new data series and click the Page viewer button on the report toolbar to load the report into a new viewer.

To export the report as HTML, you can either use the Export page… command in a Page viewer, or configure the block to save the report to a file on disk by specifying the export path to the Report path port. The path can be relative to the project directory — for example, ./results/My report.html saves the file to the results subdirectory in the project. Absolute paths can be used to save the report outside the project. In both cases, if the specified file already exists, it will be replaced.

In addition to the main ports, there are the following outputs:

• Reliability estimate — outputs the reliability estimate value, if the block runs in the reliability analysis mode.
• Confidence interval — outputs the confidence interval for the above estimate as a RealVector with 2 values. The confidence level used when calculating the interval is set by the Confidence level option.
• Result — outputs a Dict containing all result data so it can be post-processed in various blocks, for example PythonScript.

The additional ports listed above are always present on the block, but monitoring is disabled for them by default. To enable it, use the Ports and parameters dialog.

## Known Issues¶

This section describes known issues in the Uncertainty quantification block.

### Reliability Analysis (FORM)¶

The Reliability analysis (FORM) method sometimes does not respect the Number of designs setting and performs a few more evaluations when it should. This is a bug in the OpenTURNS library used by the block, and will be fixed in future versions.

### Column Order in Results¶

In pSeven 6.16 and older versions, a bug in the block configuration dialog could invalidate the configuration in some cases — for example, when you reordered variables, then deleted one of them and created a new one with the same name. Due to the bug, pSeven did not correctly save the order of variables and responses, making their order shown on the Variables and Responses panes inconsistent with the column order in result samples. This led to various obscure issues in results, such as “mixed” columns in the matrix output to All designs, and also caused incorrect behavior of the variable and response reordering commands (the and buttons).

This bug is fixed since pSeven 6.16.1, but workflows created in pSeven 6.16 and older can still contain a block saved with invalid configuration. When you open such a workflow in pSeven 6.16.1 or newer, you will get validation errors in the Issues pane, like the following:

The order of variables may be inconsistent due to a configuration bug. Reopen the block to resolve the issue.

The order of responses may be inconsistent due to a configuration bug. Reopen the block to resolve the issue.

The workflow cannot run unless these errors are fixed. To remove the errors:

1. Open the block which reports the error — see its name in the Source column in the Issues pane.

2. When you open the block, an error dialog will pop up:

The error resolution dialog shown when you open a block with errors in variable or response configuration. pSeven can automatically fix these errors, but you will have to verify configuration afterwards, as the order of variables or responses may change.

3. Click in the error dialog to fix the block. Note that the order of variables or responses may change after this: unfortunately, it is not possible to restore the correct order from the invalid configuration.

• If you want to check the configuration without any changes, click in the error resolution dialog.
4. After pSeven resolves errors, verify the order of variables and responses in the block configuration dialog, and make your adjustments. Usually there is a problem with one variable or response, and it is automatically moved to the bottom position on the pane.

5. Click or to save the fixed configuration, then save the workflow.

Note that clicking in the error resolution dialog or the block configuration dialog keeps the current (invalid) configuration. It does not remove the validation error, so pSeven will continue to show the error in the Issues pane and the block’s validation pane. No changes are done unless you select and save the block afterwards, so you can review the issue and make backup copies of the block or workflow before allowing pSeven to fix errors.


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Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Three resonant frequencies of a string are

Question:

Three resonant frequencies of a string are $90,150,210 \mathrm{~Hz}$.

a. Find the highest possible fundamental frequency of vibration of this string.

b. Which harmonics of the fundamental are the given frequencies?

c. Which overtones are these frequencies

d. If the length of the string is $80 \mathrm{~cm}$, what would be the speed of the transverse wave on this string?

Solution:


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Introduction

A quadratic equation is one which can be written in the form $a{x}^{2}+bx+c=0$ where $a$ , $b$ and $c$ are numbers, $a\ne 0$ , and $x$ is the unknown whose value(s) we wish to find. In this Section we describe several ways in which quadratic equations can be solved.

Prerequisites

• be able to solve linear equations

Learning Outcomes

• solve a quadratic equation by factorisation
• solve a quadratic equation using the standard formula
• solve a quadratic equation by completing the square
• interpret the solution of a quadratic equation graphically


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# Polyconic projection

American polyconic projection of the world

Polyconic can refer either to a class of map projections or to a specific projection known less ambiguously as the American Polyconic. Polyconic as a class refers to those projections whose parallels are all non-concentric circular arcs, except for a straight equator, and the centers of these circles lie along a central axis. This description applies to projections in equatorial aspect.[1]

As a specific projection, the American Polyconic is conceptualized as "rolling" a cone tangent to the Earth at all parallels of latitude, instead of a single cone as in a normal conic projection. Each parallel is a circular arc of true scale. The scale is also true on the central meridian of the projection. The projection was in common use by many map-making agencies of the United States from the time of its proposal by Ferdinand Rudolph Hassler in 1825 until the middle of the 20th century.[2]

The projection is defined by:

$x = \cot(\varphi) \sin((\lambda - \lambda_0)\sin(\varphi))\,$
$y = \varphi-\varphi_0 + \cot(\varphi) (1 - \cos((\lambda - \lambda_0)\sin(\varphi)))\,$

where $\lambda$ is the longitude of the point to be projected; $\varphi$ is the latitude of the point to be projected; $\lambda_0$ is the longitude of the central meridian, and $\varphi_0$ is the latitude chosen to be the origin at $\lambda_0$. To avoid division by zero, the formulas above are extended so that if $\varphi = 0$ then $x = \lambda$ and $y = 0$.


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Z Transform Solved Problems Pdf

If the ROC contains the unit circle (i.e., |z| = 1) then the system is stable.

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It was later dubbed "the z-transform" by Ragazzini and Zadeh in the sampled-data control group at Columbia University in 1952.

The idea contained within the Z-transform is also known in mathematical literature as the method of generating functions which can be traced back as early as 1730 when it was introduced by de Moivre in conjunction with probability theory.

The following substitution is used: from the Z-domain to the Laplace domain.

Through the bilinear transformation, the complex s-plane (of the Laplace transform) is mapped to the complex z-plane (of the z-transform).

This extends to cases with multiple poles: the ROC will never contain poles.

In example 2, the causal system yields an ROC that includes |z| = ∞ while the anticausal system in example 3 yields an ROC that includes |z| = 0. The stability of a system can also be determined by knowing the ROC alone.In the case where the ROC is causal (see Example 2), this means the path C must encircle all of the poles of is stable, that is, when all the poles are inside the unit circle.With this contour, the inverse Z-transform simplifies to the inverse discrete-time Fourier transform, or Fourier series, of the periodic values of the Z-transform around the unit circle: The Z-transform with a finite range of n and a finite number of uniformly spaced z values can be computed efficiently via Bluestein's FFT algorithm.To learn more or modify/prevent the use of cookies, see our Cookie Policy and Privacy Policy.Find the response of the system $s(n 2)-3s(n 1) 2s(n) = \delta (n)$, when all the initial conditions are zero.For example, the location of zeros and poles move from inside the unit circle using one definition, to outside the unit circle using the other definition.where C is a counterclockwise closed path encircling the origin and entirely in the region of convergence (ROC).The discrete-time Fourier transform (DTFT)—not to be confused with the discrete Fourier transform (DFT)—is a special case of such a Z-transform obtained by restricting z to lie on the unit circle.The region of convergence (ROC) is the set of points in the complex plane for which the Z-transform summation converges.Due to periodicity, there are only a finite number of unique amplitudes, which are readily computed by the much simpler discrete Fourier transform (DFT).(See DTFT; periodic data.) The bilinear transform can be used to convert continuous-time filters (represented in the Laplace domain) into discrete-time filters (represented in the Z-domain), and vice versa.

Comments Z Transform Solved Problems Pdf

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• The z-transform and Analysis of LTI Systems

The z-transform of a signal is an innite series for each possible value of z in the complex plane. Typically only some of those innite series will converge. We need terminology to distinguish the ﬁgoodﬂ subset of values of z that correspond to convergent innite series from the ﬁbadﬂ values that do not.…

• Chapter 6 - The Z-Transform

Gnz n. It is Used in Digital Signal Processing Used to De ne Frequency Response of Discrete-Time System. Used to Solve Di erence Equations { use algebraic methods as we did for di erential equations with Laplace Transforms; it is easier to solve the transformed equations since they are algebraic.…

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• Z-transform - Department of Electrical and Electronic.

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# Riesz measure associated with a subharmonic function

In page 101, corollary 4.3.3., from Armitage and Gardiner's book on potential theory, the authors prove that any subharmonic function, can be identified with a positive measure (Riesz measure).

In doing so, they consider the functional $L_s(\phi)=\int_\Omega s(x)\Delta \phi(x)$, $\Omega$ a open set, $\phi\in C_0^\infty(\Omega)$ and $s$ the subharmonic function.

For any $\phi\in C_0(\Omega)$, he consider the mollified sequence $\phi_n$, which converges to $\phi$ uniformly and claim that $L_s(\phi_n)$ is Cauchy. Maybe, as he says, this is evident, however I fail to see it.

If, for instance the regularized sequence $s_n$, is such that, $\Delta s_n$ is bounded uniformly in $L^1_{loc}(\Omega)$ then, the above is true, however, I also fail to see it. Any help is appreciated.

Remark: I know how to extend $L_s$ to $C_0(\Omega)$, but with another approach, although the extension is the same, because of unicity.

Update: For $x\in \Omega$, fix some $r>0$ such that $\overline{B(x,r)}\subset \Omega$.The regularization $s_n$ of $s$, is a sequence of subharmonic functions in $B(x,r)$ which converges decreasingly to $u$ in $B(x,r)$. Note that (Green's theorem)

\begin{eqnarray} \int_{B(x,r)}|\Delta s_n| &=& \int_{B(x,r)} \Delta s_n \nonumber \\ &=& \int_{\partial B(x,r)} \frac{\partial s_n}{\partial\eta} \nonumber \\ &=& r^{N-1}\frac{\partial}{\partial r}\left(\frac{1}{r^{N-1}}\int_{\partial B(x,r)}s_n\right) \end{eqnarray}

Does anyone knows how to bound the right hand side uniformly in $n$? Also, note that the right hand side, can be written in terms of the average integral of $s_n$.

This is the "continuity from positivity" trick. First, recall that the authors use $C_0$ to denote the space of compactly supported functions, not those vanishing on the boundary. Since $\phi$ is compactly supported, so are $\phi_n$ for large $n$, and we can take a compact set $K\subset \Omega$ such that all large $n$ the functions $\phi_n$ and $\phi$ vanish outside of $K$.
Let $\chi\in C_0^\infty(\Omega)$ be a nonnegative function such that $\chi\equiv 1$ on $K$. Let $\delta_{nm}=\sup|\phi_n-\phi_m|$. Then $$- \delta_{nm} \chi \le \phi_n-\phi_m\le \delta_{nm} \chi$$ everywhere in $\Omega$. Since $L_s$ is a positive functional, it respects pointwise inequalities; thus $$-\delta_{nm} L_s(\chi) \le L_s(\phi_n)-L_s(\phi_m)\le \delta_{nm} L_s(\chi)$$ Here $L_s(\chi)$ is a fixed number, independent of $m,n$. Since $\delta_{nm}\to 0$ as $m,n\to\infty$, the conclusion follows.
• Tomás. I can answer you old question: how to bound the integral. Since the sequence $(s_{n})$ is decreasing, $s_{n}\leq s_{0}$. But how did you go from the 2end equality to the third one? – M. Rahmat Jul 1 at 23:55


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# Problem: A sample of gas has an initial volume of 13.8 L at a pressure of 1.26 atm .If the sample is compressed to a volume of 10.8 L , what is its pressure?

###### FREE Expert Solution

The initial and final pressure and volume of the gas are related by Boyle’s Law:

$\overline{){{\mathbf{P}}}_{{\mathbf{1}}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{P}}}_{{\mathbf{2}}}{{\mathbf{V}}}_{{\mathbf{2}}}}$

We can use Boyle’s law as long as the units for the initial and final values are the same.

We can rearrange the combined gas law to solve for P2:

$\overline{){{\mathbf{P}}}_{{\mathbf{2}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{V}}_{\mathbf{2}}}}$

94% (245 ratings)
###### Problem Details
A sample of gas has an initial volume of 13.8 L at a pressure of 1.26 atm .

If the sample is compressed to a volume of 10.8 L , what is its pressure?


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• Sep 15th 2009, 07:35 AM
reiward

http://img183.imageshack.us/img183/2280/math1.jpg

http://img197.imageshack.us/img197/2851/math2g.jpg
• Sep 15th 2009, 12:22 PM
masters
Quote:

Originally Posted by reiward

http://img183.imageshack.us/img183/2280/math1.jpg

http://img197.imageshack.us/img197/2851/math2g.jpg

Hi reiward,

This is truly some industrial strength algebra. I got the first one down fairly simple, but couldn't manipulate the last term to very much. Maybe it's stated incorrectly or something.

$\sqrt[6]{32 \sqrt{4n^{12}}} - \frac{n\sqrt{4n^2+16n+16}}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$\sqrt[6]{32 \cdot 2n^6} - \frac{n\sqrt{(2n+4)^2}}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$\sqrt[6]{2^6 \cdot n^6} - \frac{n(2n+4)}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$2n - \frac{2n(n+2}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$2n - 2n +\frac{n-1}{\sqrt[3]{n}-1}$

$\frac{n-1}{\sqrt[3]{n}-1}$

I messed around with this last term a while, but could never get it to anything simpler than it already is.

Now, for the second one.

$\frac{\sqrt{z^6}\cdot \sqrt[5]{2x^3y^2}}{\sqrt[4]{4x^2y^4z^8}} \div \frac{\sqrt[5]{64x^8y^{-3}}}{\sqrt[6]{8x^3}}$

$\frac{z^3 \cdot 2^{\frac{1}{5}}x^{\frac{3}{5}}y^{\frac{2}{5}}}{2^{ \frac{1}{2}}x^{\frac{1}{2}}yz^2}\cdot \frac{2^{\frac{1}{2}}x^{\frac{1}{2}}y^{\frac{3}{5} }}{2\cdot 2^{\frac{1}{5}}x\cdot x^{\frac{3}{5}}}=\frac{zy}{2xy}=\frac{z}{2x}$
• Sep 15th 2009, 02:20 PM
reiward
thanks!
thank you! maybe thats the final answer for number 1. oh well, thank you very much :)
• Sep 17th 2009, 08:42 AM
reiward
hi, we discussed the answers already, just wanna share. for the last term this is what we did.

$\frac{n-1}{\sqrt[3]{n}-1} * \frac{\sqrt[3]{n}^2 + \sqrt[3]{n} + 1}{\sqrt[3]{n}^2 + \sqrt[3]{n} + 1}$

then

$\frac{n-1} {n-1} \sqrt[3]{n}^2 + \sqrt[3]{n} + 1
$

$\sqrt[3]{n}^2 + \sqrt[3]{n} + 1


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What is the status of the Gauss Circle Problem?

For $r > 0$, let $L(r) = \# \{ (x,y) \in \mathbb{Z}^2 \ | \ x^2 + y^2 \leq r^2\}$ be the number of lattice points lying on or inside the standard circle of radius $r$. It is easy to see that $L(r) \sim \pi r^2$ as $r \rightarrow \infty$. The Gauss circle problem is to give the best possible error bounds: put

$E(r) = |L(r) - \pi r^2|$.

Gauss himself gave the elementary bound $E(r) = O(r)$. In 1916 Hardy and Landau showed that it is not the case that $E(r) = O(r^{\frac{1}{2}})$. It is now believed that this is "almost" true: i.e.:

Gauss Circle Conjecture: For every $\epsilon > 0$, $E(r) = O_{\epsilon}(r^{\frac{1}{2}+\epsilon})$.

So far as I know the best published result is a 1993 theorem of Huxley, who shows one may take $\epsilon > \frac{19}{146}$.

In early 2007 I was teaching an elementary number theory class when I noticed that Cappell and Shaneson had uploaded a preprint to the arxiv claiming to prove the Gauss Circle Conjecture:

http://arxiv.org/abs/math/0702613

Two more versions were uploaded, the last in July of 2007.

It is now a little more than three years later, and so far as I know the paper has neither been published nor retracted. This seems like a strange state of affairs for an important classical problem. Can someone say what the status of the Gauss Circle Problem is today? Is the argument of Cappell and Shaneson correct? Or is there a known flaw?

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Cappell's Wikipedia page says that the paper "is still being vetted by experts." This was originally mentioned on 29 April 2008, but it has not been changed since. –  Steve Huntsman Mar 23 '10 at 2:31
@SH: Right, that was almost two years ago. I'm asking for an update from an expert. "Still vetting" is a possible answer, although in that case I'd be interested to know which part is taking so long to check. –  Pete L. Clark Mar 23 '10 at 2:40
I've sent an email to Cappell. I took a few courses from him in the nineties and I think he'll remember me. –  Steve Huntsman Mar 23 '10 at 2:58
I'll talk to Shaneson about it and forward the link. –  Justin Curry Mar 23 '10 at 15:52
Bruce Berndt gave a talk last week at Gainesville that gave history and current status, perhaps not from exactly the same viewpoint as yours of course. See math.ufl.edu/~fgarvan/antc-program/2009-10/mar-focused-week/… –  Will Jagy Apr 1 '10 at 4:50

When Cappell visited UWM a few months ago, one of my colleagues asked him about the status of the paper. The answer was that it is "still in works", which in plain English, probably, means "having severe problems with some remote hope to fix them". The point is that it contains no idea that hadn't been well-known to experts before it appeared, just an enormous amount of "brute forcing" (which, by the way, makes it very hard to read). Sometimes you can succeed by being just more persistent than others but it doesn't seem to be the case here. The concensus is that the existing methods have already been brought to their extreme and to proceed some fresh idea is required.

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Thanks for this answer. What is UWM? –  Pete L. Clark Apr 10 '10 at 16:49
University of Wisconsin-Madison –  fedja Apr 10 '10 at 18:04
Thanks. Google thought "UWM" stood for University of Wisconsin-Milwaukee. –  Pete L. Clark Apr 10 '10 at 21:41
I think that is a normal convention. Unfortunately, UW is acknowledged by all reasonable people to denote the University of Washington. –  Ben Webster Jan 27 '11 at 6:17
That really depends on whether you are a West coaster or a Midwesterner but I always love to be classified into the same category as the Cheshire Cat. :) –  fedja Jan 27 '11 at 11:25


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# Atmospheric physics - change of temperature of parcel

1. May 3, 2007

"A parcel of air is lifted slowly from the ground, where the temperature is 295K, to an elevation of 5km, and then returned rapidly to the ground. Estimate the air parcel temperature at 5km and after it returns to the groundm explanation any assumptions."

I assumed an adiabatic process both ways. Upwards the process is "slow" so we can assume reversibility, so the parcel should follow the adiabatic lapse rate. Therefore dT = -g/cp * 5km

assume g is constant, assume dry air and use cp = 1.01KJ per K per mole

This gives T = 246K at the top.

On the way down we can't assume reversibility. I tried setting dU = dW + dQ, where dQ is equal to zero. dW = mg*dz, where we assume that g is constant again, and this time dz = -5km. Therefore assuming the ideal gas law,

n*Cv*dT = mg*dz

m equals the molar mass* no of moles. I assume the gas is diatomic Nitrogen, so the molar mass is 34*10^-3 kg per mole. This gives a change in temperature of -133K which seems far too much (in fact this is close to the temperature at which nitrogen condenses). I'm not sure where I have gone wrong. Any hints appreciated.


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### Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

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# Root Hunter

##### Stage: 5 Challenge Level:

.

This graph is positive for $x = 5$ and negative for $x = 3$. This means that the graph must cut the $x$ axis somewhere between $3$ and $5$.
Although in this case the result is obvious (because we have the whole graph to look at!), we can also use this idea to show that more tricky functions also have roots.

Use this idea to show that these functions possess at least one solution $f(x) = 0$:

$$f(x)=\frac{1}{x-2}+\frac{1}{x-3}$$ $$f(x)= x^x - 1.5 x$$ $$f(x)= x^{1000000}+{1000000}^x - 17$$ $$f(x)=\cos(\sin(\cos x)) - \sin(\cos(\sin x))$$

Optional extension activity: Can you make a spreadsheet that helps you find the numerical values of the roots to, say, four decimal places?


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• Definition for "pocketful"
• the quantity that a pocket can take.
• Sentence for "pocketful"
• By the time he ranked the…
• Phrases for "pocketful"
• a pocketful
• Hypernym for "pocketful"
• containerful
• Same Context for "pocketful"
• hatful

# pocketful definition

• noun:
• the quantity that a pocket can take.
• the quantity that a pocket would typically hold
• up to a pocket will hold; adequate to fill a pocket.
• adequate to fill a pocket; just as much as a pocket will hold.
• the number a pocket will hold


text:
###### Question:

Given right triangle ABCABC with altitude \overline{BD} BD
drawn to hypotenuse ACAC. If AD=5AD=5 and AC=16,AC=16, what is the length of \overline{AB}
AB
in simplest radical form? (Note: the figure is not drawn to scale.)

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