text:
# American Institute of Mathematical Sciences

ISSN:
1531-3492

eISSN:
1553-524X

All Issues

## Discrete & Continuous Dynamical Systems - B

2010 , Volume 13 , Issue 4

Special Issue on Oceanography and Mathematics

Select all articles

Export/Reference:

2010, 13(4): i-ii doi: 10.3934/dcdsb.2010.13.4i +[Abstract](285) +[PDF](30.2KB)
Abstract:
The motivation for this special edition arose during the conference "Oceanography and Mathematics" organized at the Ecole Normale Supérieure of Paris, in January 2009. The goal of this conference was to bring together mathematicians and physicists working on research fields related to oceanography, and to offer the opportunity for cooperations between researchers of different cultures.
If one thing was certainly felt by all the participants to this conference, it is that the range of open problems that stem from the study of oceanographic phenomena is huge. The difficulties that remain to be understood offer very challenging problems in physical modelling, numerical simulation, and analysis of partial differential equations.

2010, 13(4): 729-738 doi: 10.3934/dcdsb.2010.13.729 +[Abstract](432) +[PDF](604.2KB)
Abstract:
In this paper, we investigate the mechanisms which control the generation of wave-induced mean current vorticity in the surf zone. From the vertically-integrated and time-averaged momentum equations given recently by Smith [21], we obtain a vorticity forcing term related to differential broken-wave energy dissipation. Then, we derive a new equation for the mean current vorticity, from the nonlinear shallow water shock-wave theory. Both approaches are consistent, under the shallow water assumption, but the later gives explicitly the generation term of vorticity, without any ad-hoc parametrization of the broken-wave energy dissipation.
2010, 13(4): 739-758 doi: 10.3934/dcdsb.2010.13.739 +[Abstract](424) +[PDF](294.7KB)
Abstract:
The numerical resolution of the multi-layer shallow water system encounters two additional difficulties with respect to the one-layer system. The first is that the system involves nonconservative terms, and the second is that it is not always hyperbolic. A splitting scheme has been proposed by Bouchut and Morales, that enables to ensure a discrete entropy inequality and the well-balanced property, without any theoretical difficulty related to the loss of hyperbolicity. However, this scheme has been shown to often give wrong solutions. We introduce here a variant of the splitting scheme, that has the overall property of being conservative in the total momentum. It is based on a source-centered hydrostatic scheme for the one-layer shallow water system, a variant of the hydrostatic scheme. The final method enables to treat an arbitrary number $m$ of layers, with arbitrary densities $\rho_1$,...,$\rho_m$, and arbitrary topography. It has no restriction concerning complex eigenvalues, it is well-balanced and it is able to treat vacuum, it satisfies a semi-discrete entropy inequality. The scheme is fast to execute, as is the one-layer hydrostatic method.
2010, 13(4): 759-772 doi: 10.3934/dcdsb.2010.13.759 +[Abstract](399) +[PDF](2328.2KB)
Abstract:
In this paper, shape optimization tools are applied to coastal engineering. A barred beach protection study is described, using a new type of water wave attenuator device, namely geotextile tube, to reduce the mobilization of sediments. A global optimization algorithm, able to pursue beyond local minima, is used to search for the optimum properties of the geotextile tube.The optimal configuration is then post-validated for generated currents.
2010, 13(4): 773-782 doi: 10.3934/dcdsb.2010.13.773 +[Abstract](317) +[PDF](1302.1KB)
Abstract:
The influence of wind on extreme wave events in shallow water is investigated numerically. A series of numerical simulations using a pressure distribution over the steep crests given by the modified Jeffreys' sheltering theory shows that wind blowing over a strongly modulated wave group due to the dispersive focusing of a chirped long wave packet increases the time duration and maximal amplitude of the extreme wave event. These results are coherent with those obtained within the framework of deep water. However, steep wave events are less unstable to wind perturbation in shallow water than in deep water.
2010, 13(4): 783-798 doi: 10.3934/dcdsb.2010.13.783 +[Abstract](410) +[PDF](236.0KB)
Abstract:
We study the low Mach number limit for the compressible Navier-Stokes system supplemented with Navier's boundary condition on an unbounded domain with compact boundary. Our main result asserts that the velocities converge pointwise to a solenoidal vector field - a weak solution of the incompressible Navier-Stokes system - while the fluid density becomes constant. The proof is based on a variant of local energy decay property for the underlying acoustic equation established by Kato.
2010, 13(4): 799-818 doi: 10.3934/dcdsb.2010.13.799 +[Abstract](459) +[PDF](619.1KB)
Abstract:
The classical dam break problem has become the de factostandard in validating the nonlinear shallow water equations solvers. Moreover, the NSWE are widely used for flooding simulations. While applied mathematics community is essentially focused on developing new numerical schemes, we tried to examine the validity of the mathematical model under consideration. The main purpose of this study is to check the pertinence of the NSWE for flooding processes. From the mathematical point of view, the answer is not obvious since all derivation procedures assumes the total water depth positivity. We performed a comparison between the two-fluid Navier-Stokes simulations and the NSWE solved analytically and numerically. Several conclusions are drawn out and perspectives for future research are outlined.
2010, 13(4): 819-840 doi: 10.3934/dcdsb.2010.13.819 +[Abstract](475) +[PDF](4033.2KB)
Abstract:
This review paper is devoted to a presentation of recent progress in wave turbulence. I first present the context and state of the art of this field of research both experimentally and theoretically. I then focus on the case of wave turbulence on the surface of a fluid, and I discuss the main results obtained by our group: caracterization of the gravity and capillary wave turbulence regimes, the first observation of intermittency in wave turbulence, the occurrence of strong fluctuations of injected power in the fluid, the observation of a pure capillary wave turbulence in low gravity environment and the observation of magnetic wave turbulence on the surface of a ferrofluid. Finally, open questions in wave turbulence are discussed.
2010, 13(4): 841-857 doi: 10.3934/dcdsb.2010.13.841 +[Abstract](294) +[PDF](399.9KB)
Abstract:
This works aims at studying the impact of the cosine terms of the Coriolis force, that are usually neglected in geophysical fluid dynamics, leading to the so-called traditional approximation. Mathematical well-posedness arguments for simplified models, as well as numerical simulations, are presented in order to suggest the use of these terms in large scale ocean modelling.
2010, 13(4): 859-872 doi: 10.3934/dcdsb.2010.13.859 +[Abstract](299) +[PDF](205.1KB)
Abstract:
This text represents the content of a talk given by the second author at ENS Paris on January 28, 2009 at the conference "Mathematics and Oceanography". We are grateful to David Gérard-Varet, David Lannes and Laure Saint-Raymond for the kind invitation to present our recent work on the water waves. The proofs of the results announced here will appear in [18].

2017  Impact Factor: 0.972


text:
# Math Help - Sum Game

1. ## Sum Game

James is playing a card game called sum game. He holds 8 cards, numbered 1 to 8. Each time he places a card on the table he finds the sum of all the cards he has placed on the the table so far. To win the game , alternate sums must be divisible by two different numbers. For example, he plays the cards in order 5, 1, 4, 2, 8, 7, 3, 6, the sums are 5, 6,10, 12, 20, 27, 30, 36 which are alternately divisible by 5 and 3, starting with 5.

1. James now plays with 17 cards, numbered 1 to 17. SHow that he cannont place the cards so that the alternate sums are divisible by 4 and 9, starting with 4

2. James has cards numbered 1 to n. He places them so that the alternate sums are divisible by positive integers a and b, starting with a. Show that, if the card numbered 1 is placed in an even position, the maximum value for a + b is 2n + 1

2. Originally Posted by souksfxc
James is playing a card game called sum game. He holds 8 cards, numbered 1 to 8. Each time he places a card on the table he finds the sum of all the cards he has placed on the the table so far. To win the game , alternate sums must be divisible by two different numbers. For example, he plays the cards in order 5, 1, 4, 2, 8, 7, 3, 6, the sums are 5, 6,10, 12, 20, 27, 30, 36 which are alternately divisible by 5 and 3, starting with 5.

1. James now plays with 17 cards, numbered 1 to 17. SHow that he cannont place the cards so that the alternate sums are divisible by 4 and 9, starting with 4

2. James has cards numbered 1 to n. He places them so that the alternate sums are divisible by positive integers a and b, starting with a. Show that, if the card numbered 1 is placed in an even position, the maximum value for a + b is 2n + 1
1. Even if we drop the restriction that the first, third, fifth, etc. sums are divisible by 4, no such arrangement is possible. Here is why.

Let's say the cards played are numbered $x_1, x_2, x_3, ... , x_{17}$ and satisfy the condition that the sum of the first two, four, six, .., sixteen cards are all multiples of 9.

Working modulo 9, we must have
$x_1 + x_2 \equiv 0$
$x_1 + x_2 + x_3 + x_4 \equiv 0$
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \equiv 0$
...
$x_1 + x_2 + x_3 + \dots x_{16} \equiv 0$

So
$x_1 + x_2 \equiv 0$
$x_3 + x_4 \equiv 0$
$x_5 + x_6 \equiv 0$
...
$x_{15} + x_{16} \equiv 0$

Where is 9 in this list? There is no solution to $x_i + 9 \equiv 0$ with $1 \leq x_i \leq 17$ and $x_i \neq 9$, so we must have $x_{17} = 9$.

So all the integers from 1 to 17, except 9, must be among the first 16 numbers in the list: $x_1, x_2, x_3, \dots , x_{16}$. Let's say $x_1$ and $x_2$ are "friends", as are $x_3$ and $x_4$,
$x_5$ and $x_6$ , ... $x_{15}$ and $x_{16}$. Every number from 1 to 17, except for poor lonely 9, must have a friend. A pair of friends must sum to a multiple of 9.

Who are the possible friends of 1? Only 8 and 17. Let's assume 1's friend is 8. Then 17's friend must be 10 because 1 is already taken. Who can be 10's friend? Nobody is left; 8 and 17 are already taken. So 8 can't be 1's friend. It's left to the reader to show a similar contradiction arises if 1's friend is 17.

So no such arrangement is possible.

format: plain


text:
# The Identity Theorem for real analytic functions

What is the condition for two real analytic functions to be identically equal? We know that there is a nice condition (Identity Theorem) for holomorphic function to check if they are the same. What is its version for real analytic functions?

• If $U$ is a domain, and $f,g$ are two real-analytic functions defined on $U$, and if $V\subset U$ is a nonempty open set with $f\lvert_V \equiv g\lvert_V$, then $f \equiv g$. If the domain is one-dimensional (an interval in $\mathbb{R}$), then it suffices that $f\lvert_M \equiv g\lvert_M$ for some $M\subset U$ that has an accumulation point in $U$. Apr 4, 2014 at 12:09
• @DanielFischer Thanks, I also think it should be like this way, but would you give me some references about it?
– Yui
Apr 4, 2014 at 12:12
• Can't list a reference, but it's proved exactly like the holomorphic case. For the nonempty open $V$ case (without loss of generality $g\equiv 0$), you consider the set $W = \{ x : f\lvert_N \equiv 0 \text{ for some neighbourhood } N \text{ of } x\}$, and show $W$ is open (by definition) and closed. For the $M$ has an accumulation point case, consider the power series expansion in the accumulation point, derive all coefficients are $0$, and you're in the first case. Apr 4, 2014 at 12:19
• @DanielFischer Right,thank you.
– Yui
Apr 4, 2014 at 12:25

As a reference, I suggest A Primer of Real Analytic Functions by Krantz and Parks.

The two versions of the Identity Theorem stated by Daniel Fisher can be unified at the expense of more complicated statement.

If $U$ is a domain, and $f,g$ are two real-analytic functions defined on $U$, and if $V\subset U$ is a nonempty open set with $f\lvert_V \equiv g\lvert_V$, then $f \equiv g$. If the domain is one-dimensional (an interval in $\mathbb{R}$), then it suffices that $f\lvert_M \equiv g\lvert_M$ for some $M\subset U$ that has an accumulation point in $U$.

Claim. If $f,g$ are real analytic and there is a point $p\in \mathbb R^n$ such that the set of all limits $$\left\{\lim_{n\to \infty} \frac{x_n-p}{|x_n-p|} : \qquad f(x_n)=g(x_n),\ x_n\to p, \ x_n\ne p\right\} \tag{1}$$ has an interior point in the topology of the sphere $S^{n-1}$, then $f\equiv g$.

Indeed, suppose $f-g$ is not identically zero. Express its Taylor series at $p$ as the sum of homogeneous polynomials $P_d$. Let $d$ be the smallest degree for which $P_d$ is not identically zero. Then the set defined by (1) can be shown to be precisely $$S^{n-1} \cap \{ P_d =0\} \tag{2}$$ Since the zero set of a polynomial has empty interior in $\mathbb R^n$, it follows that (2) has empty interior in $S^{n-1}$. $\quad \Box$

When $n=1$, the sphere $S^0$ is a two-point set, so any nonempty subset of it has nonempty interior. We thus recover the one-dimensional result.

• How to prove the claim?
– Yui
Apr 8, 2014 at 4:59
• @Yui The part beginning with "Indeed" is meant to be a proof, at least in an outline. Apr 8, 2014 at 5:05
• I know this is an old post. But does this conclusion hold in $\mathbb{R}^n$?
– Boby
Apr 28, 2017 at 20:19
• I think there is something wrong with the claim. In $\mathbb{R}^2$ the analytic function $\sin(2\pi (x^2+y^2))$ vanishes on the unit circle. What am I missing?
– Kosh
Aug 11, 2020 at 18:07
• @Kosh, the set points where that function vanishes in R^2 has an empty interior. You need this to hold over an entire area in R^2.
– Joel
Apr 10, 2021 at 19:33


text:
## Computational Linguistics

March 2009, Vol. 35, No. 1, Pages 29-46
(doi: 10.1162/coli.07-036-R2-06-22)
© 2008 Association for Computational Linguistics
Evaluating Centering for Information Ordering Using Corpora
Article PDF (118.74 KB)
Abstract

In this article we discuss several metrics of coherence defined using centering theory and investigate the usefulness of such metrics for information ordering in automatic text generation. We estimate empirically which is the most promising metric and how useful this metric is using a general methodology applied on several corpora. Our main result is that the simplest metric (which relies exclusively on NOCB transitions) sets a robust baseline that cannot be outperformed by other metrics which make use of additional centering-based features. This baseline can be used for the development of both text-to-text and concept-to-text generation systems.


text:
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# Better

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Feeling a bit better today. Reckon by the time the weekend is over I'll be fine again. Quite a scary week at times though.

Anyway, I've started working on Udo again but I have some bad news for fans of my leaf monsters and rustling bushes - both features have been ceremoniously dumped from the game.

I know people seemed to really like them, but somehow they just weren't fitting in my mind into the overall game. It's a bit abstract, but they just didn't feel right.

I've made some minor improvements to my map editor to make it slightly easier to work with these multi-layer maps.

Haven't posted any map editor screenshots for ages actually...

Lot easier posting pictures here than on GGE. Have to say I really didn't get on well with that site. I haven't been there once since GDNet arose from the flames.

EDIT

Just finished implementing the autosorting of items order thingy, so it now doesn't matter what order you place items in the level editor.

The basic Item interface had a virtual GetZ() method added, which is then implemented in each class type, just returning constants at the moment, although of course individual items could store a personal Z value if needed later.

Then an InsertSorted() method was added to the ItemList class that replaced the Add() method that was being used by the level loader.

Simple as that really. Now you can declare items in any order in the editor and they will be drawn in the correct order during the game.

The only actual change this forced was to convert the door frames from being instances of a CStaticBlock type into their own CDoorFrame class. Even simple static blocks will now need specific classes to describe them so that there can be an implicit Z value for the class. I could have added a Z field to the StaticBlock class and set it in the editor, but it is precisely this kind of low-level fiddling about in the editor that I am currently trying to improve upon.

Next issue to tidy up is the management of all this messy additional data that each level file needs.

The level editor looks great! And I don't see how it would be hard to work with, at least not from that screen shot.

Well, the actual placing of blocks and stuff is quite easy, especially now that the layer placement is handled automatically. A slightly trickier bit is having to make sure you keep the item list in the correct draw order, but that is not too bad.

Biggest problem would be if someone else wanted to create a complete new level containing different graphics or enemies, since that would involve creating some quite awkward files using my slightly esoteric virtual assembler, which I wouldn't wish on anyone else, so I guess I'll need to create the templates for all the different types of world files before I ship the editor to anyone.

Also a bit inconvienient is the fact that each level's imp script (my homemade scripting engine) needs to be written and compiled seperately from the level, which for me involves jumping between the map editor and notepad, and having the imp compiler set up in the right-click context menu for .ic files in my windows explorer.

Really I guess I need to think about bringing the script creation into the map editor itself and embedding the compiled script into the level file rather than have it as a seperate file as it is at the moment.

I must admit I may have got too complex with the structure of the level files for this project though. Too late now to worry about it. Always trying to balance between simplicity and flexibility and my nature is to always sacrifice simplicity for flexibility, and then never take advantage of the flexibility. Daft really.

This is the current simple imp script for the demo level

include "udo.ic";

proc 0
{
}

proc 1
{
int x,y; getpcxy(x,y);

if(x>384)
{
activate(1005);
activate(1006);
}

if(x>1280) activate(1018);

if(x>3264)
{
activate(1033);
activate(1035);
}
}

The included udo.ic file just defines the function getpcxy(int&,int&) in terms of a service call that calls back the main application.

I'm thinking about extending out my scripting language into something a bit more generally useful and releasing it for other people to use if they wanted as it strikes a nice balance between ease of integration into a project and powerful language features.

Couldn't you sort the item list in some sort of smart way? I guess not because then you would already be doing that, what do you mean with keeping them in order, I guess some objects in the level obscure another one if you change the later and etc.?

You mean the making of bitmaps you talked about in my journal? That didn't sound too complicated, but I see the template making can be a hassle, and time consuming witch is the worst :)

Yes I've been thinking a bit on scripting for my level editor, I will be including the script in the map, witch is more secure and more convenient IMHO. But I haven't even started on the scripting part yet, so there may be a lot of changes in my plans.

Well, because this is a generic level creator and not specific to Udo, each level has to have the scheme file I was mentioning before specified, which defines the different blocks and types of item and so on.

However, Udo levels require a load of other binary information prefixed to the start of the file so the map editor has the facility to allow the scheme file to specify a header file that is automatically prepended to the start of the exported level file.

Currently this header file is created by using my virtual assembler script. This is the script for the current demo level:

"Resource/gamefont.spf" 0b
"Resource/sky.spr" 0b

"back.ubm" 0b

"Resource/level.sfx" 0b

11
"Resource/system.spr" 0b
"Resource/blocks.spr" 0b
"Resource/udo.spr" 0b
"Resource/star.spr" 0b
"Resource/platform.spr" 0b
"Resource/wave.spr" 0b
"Resource/particle.spr" 0b
"Resource/sheep.spr" 0b
"Resource/fstar.spr" 0b
"Resource/fish.spr" 0b
"Resource/pickup.spr" 0b

17
"[0,1]<" 0b
"[1,4][2,4][3,4][2,4][1,4][4,4][5,4][4,4]<" 0b
"[3,1]<" 0b
"[6,1]<" 0b

"[7,1]<" 0b
"[8,4][9,4][10,4][9,4][8,4][11,4][12,4][11,4]<" 0b
"[10,1]<" 0b
"[13,1]<" 0b

"[0,4][1,4][2,4][3,4][4,4][5,4][6,4][7,4]<" 0b

"[0,4][1,4][2,4][3,4]" 0b

"[14,4][15,4][16,4][15,4]<" 0b

"[0,6][1,6][0,6][2,6]<" 0b
"[3,6][4,6][3,6][5,6]<" 0b

"[0,6][1,6][0,6][2,6]<" 0b
"[3,6][4,6][3,6][5,6]<" 0b

"[17,6][18,6]<" 0b
"[19,6][20,6]<" 0b

5
3 8 15 15
6 0 15 15
6 1 30 30
6 2 15 15
6 3 21 21

0 // star particle
1 // fire particle
2 // fireball particle
3 // greystar particle
4 // shard particle

12
00 01 00 0 00 0 0   0
01 01 00 0 00 0 0   0
02 01 01 1 00 0 0   0
03 01 02 2 00 0 0   0
05 05 09 0 01 0 0   0
06 01 01 2 00 0 0   0
07 01 03 2 02 0 0   0
08 01 04 0 00 0 0   0
09 01 08 2 08 0 0   0
10 01 09 2 00 0 0   0
11 01 10 0 00 0 0   0
12 01 00 2 00 0 0   0

It basically defines all the graphics, animations, block types, particle types, background, music - everything that can vary from level to level.

The actual scheme file that the map editor loads looks like this:

header "header.dat"

layers 3

defaultlayer 1

width 64
height 64

filter "Udo Files" "*.udo"

autoid 1000

block 1 "Rock" img rock.bmp 0
block 2 "Rock Top" img rocktop.bmp 1
block 3 "Grey Box" img grey.bmp 1
block 5 "Wave" img wave.bmp 2
block 6 "Solid Rock Top" img rocktop.bmp 1
block 12 "Solid Rock" img rock.bmp 1
block 7 "Spikes" img spike.bmp 1
block 8 "Spring base" img springbase.bmp 0
block 9 "Glass" img glass.bmp 1
block 10 "Brick" img brick.bmp 1
block 11 "Bush" img bush.bmp 0

item 01 "Udo" 16 32
Class 01 ID 0 Bank 2 Width 32 Height 32 X 0 Y 0 Dir 1
AnStandL 0 AnRunL 1 AnFallL 2 AnDuckL 3
AnStandR 4 AnRunR 5 AnFallR 6 AnDuckR 7
AnDie 10 AnOut 15 AnIn 16
end
img "udo.bmp"

item 02 "Present" 0 0
Class 3 ID 0 Bank 1 Frame 5 Width 64 Height 64 X 0 Y 0
end
img "present.bmp"

item 03 "Horizontal Platform" 0 0
Class 04 ID 0 Bank 4 Frame 0 Width 64 Height 32 X 0 Y 0
Dir 1 Min -1 Max -1
end
img "plat.bmp"

item 04 "Springboard" 0 32
Class 05 ID 0 Bank 4 Frame 1 Width 64 Height 20 X 0 Y 0
end
img "spring.bmp"

item 05 "Star" 16 16
Class 06 ID 0 Bank 3 Anim 8 Width 32 Height 32 X 0 Y 0
end
img "star.bmp"

item 06 "Sheep" 8 24
Class 07 ID 0 Bank 7 Width 48 Height 40 X 0 Y 0
Dir 0 Speed 2 Turns 1
AnLeft 11 AnRight 12
Active 0
end
img "sheep.bmp"

item 07 "FireStar" 16 16
Class 08 ID 0 Bank 8 Anim 8 Width 32 Height 32 X 0 Y 0
end
img "fstar.bmp"

item 08 "Fish" 12 13
Class 09 ID 0 Bank 9 Width 40 Height 38 X 0 Y 0
TimeTarget 20 Jump 8 Speed 0 Dir 0
AnLeft 13 AnRight 14
Active 1
end
img "fish.bmp"

item 09 "Door" 0 0
Class 10 ID 0 Bank 1 Frame 6 Width 64 Height 64 X 0 Y 0
Open 0
end
img "door.bmp"

item 10 "Door Frame" -9 -10
Class 2 ID 0 Bank 1 Frame 7 Layer 0 Width 82 Height 74 X 0 Y 0
end
img "doorframe.bmp"

item 11 "CheckPoint" 8 8
Class 11 ID 0 Bank 10 Frame 0 Width 48 Height 48 X 0 Y 0
end
img "checkpoint.bmp"

Creating a new kind of level with different graphics or blocks or items would involve creating a new script based on the one above and a new scheme file for the map editor.

As you can see, the script is not very self-descriptive and so I don't really think that it would be fair to expect a modder to be able to modify or create these files, or link them into map schemes manually.

Creating new headers and schemes will largely be based upon modifying the existing ones, but it is not very user friendly. Perhaps I should be writing some companion tools to help this process. I don't know.

The map editor has evolved into a fairly complex piece of software now and I hope it isn't running away with me. [smile]

Regarding "smart" sorting the item list, I have thought about it. It should actually be possible really. The order I guess needs to be stars, then enemies, then Udo. The doors and doorways are special cases. The problem is that the doorways are just general-purpose static blocks, which could vary in terms of where in the layer order they should appear.

I suppose if I made the doorways objects in their own right, the order would just be doorways, doors, pickups, enemies then Udo, and I could implement some kind of ordering based on that. Rather than having general purpose static objects, I could enforce that any type of object had to have its own class, then an ordering strategy could be applied by the game when loading so that it didn't actually matter what order objects are declared in the editor.

That's probably the best solution.

Cheers for the input, walle. Really helps to discuss this stuff with someone.

Yea I can see how it can be fairly complex to use it with additional data :)
And no it isn't very self explaining. But a simple paper that describes what goes were would suffice for most coders I guess.

Regarding the sorting, maybe you could initialize the items with a z-index, used in the layer, like start = 1, enemies = 2 and udo = 3 and then make a function in the item list to change the z-index in case the default ordering isn't working.
Don't know if this is applicative but may give you some idea.

I like discussing this too :)
And I believe discussing this will help me a lot in my own development.

Walle is right: introducing a 'Z preference' for each sprite type would take care of the ordering problem for sure, and is probably the most elegant solution. To make life even easier, you can just use the built-in hardware-accelerated depth buffer. [smile]

As for making your data more easily extensible, I think that somesort of data editor would work just fine. How much of a pain in the arse it'll be to write is arguable, but if you want people to wind up modifying Udo in the future, then opting for user friendliness is definitely the way to go. Plus, keeping the low-level access (ie. text editing the script files) still gives flexibility beyond the simplicity of the editor. Best of both worlds!

Generally, it sounds like you have things in a better position than you may think you do. Zip up those two issues and things should feel a lot cleaner. And I know for a fact that developer motivation is directly proportional to how clean one's game feels. [grin]

Yeah, maybe that's why I've felt a bit flat on Udo recently. Perhaps I need to do a proper cleaning phase instead of developing new bits.

I'm not going to use a z buffer and let the hardware to my depth sorting because I found with some tests last year that that can really cripple the framerate and also increases the minimum specs for the game.

I think I am going to come up with a system whereby the game does an insertion sort of the items as it loads them from the level file though, so that you can declare them in any order in the map editor.

Actually, that should be quite easy really, thinking about it.

Quote:
 Original post by EasilyConfused... I think I am going to come up with a system whereby the game does an insertion sort of the items as it loads them from the level file though, so that you can declare them in any order in the map editor. Actually, that should be quite easy really, thinking about it.

Yea, that was what I meant with using a z-index in the layer level editor. Probably should have used "z-index" because I didn't mean a real one. :)

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text:
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Calderón reproducing formulas and new Besov spaces associated with operators. (English) Zbl 1241.46020
Let $$\mathcal{X}$$ be a space of polynomial upper bound on volume growth, which might not be doubling, and $$L$$ be the generator of an analytical semigroup $$\{e^{-tL}\}_{t>0}$$ acting on $$L^2(\mathcal{X})$$ whose heat kernel satisfies an upper bound of Poisson type. In this paper, for $$-1<\alpha<1$$ and $$1\leq p,\,q\leq \infty$$, the authors introduce a new class of Besov spaces $$B_{p,q}^{\alpha,L}(\mathcal{X})$$ associated with the operator $$L$$. This new Besov space $$B_{p,q}^{\alpha,L}(\mathcal{X})$$ is defined to be the space of all elements belonging to some “distribution space” with the norm $\|f\|_{B_{p,q}^{\alpha,L}(\mathcal{X})}:=\left\{\int_{0}^\infty \left( t^\alpha\left\| tLe^{-tL} (f)\right\|_{L^p(\mathcal{X})} \right)^q\frac{dt}{t}\right\}^{1/q}<\infty,$ which generalizes the notion of the classical Besov space and has wide applications.
One of the main aims of this paper is to find the relationship between the classical and the new Besov spaces. By assuming that $$L$$ has some conservation property, the authors show that the classical Besov space $$B_{p,q}^{\alpha}(\mathbb R^n)$$ is continuously embedded in $$B_{p,q}^{\alpha,L}(\mathbb R^n)$$. Moreover, if the heat kernels $$\{p_t(x,\,y)\}_{t>0}$$ of $$\{e^{-tL}\}_{t>0}$$ further satisfy some Hölder continuity, the authors prove the equivalence $$B_{p,q}^{\alpha}(\mathbb R^n)=B_{p,q}^{\alpha,L}(\mathbb R^n)$$. Also, by establishing four different Calderón reproducing formulas, which are of independent interest, the authors obtain many basic properties of $$B_{p,q}^{\alpha,L}(\mathcal{X})$$, such as the embedding theorem, some norm equivalences and the boundedness of the fractional integral $$L^{-\gamma}$$ for $$\gamma\in(0,\,1)$$.
As an application, when the operator $$L:=-\Delta+V$$ is a Schrödinger operator on the Euclidean space $$\mathbb{R}^n$$ with the nonnegative potential $$V$$ satisfying some reverse Hölder estimates, the authors study the Besov space $$B_{1,1}^{0,L}(\mathbb{R}^n)$$ and prove that the classical Besov space $$B_{1,1}^{0}(\mathbb{R}^n)$$ is strictly contained in $$B_{1,1}^{0,L}(\mathbb{R}^n)$$. Moreover, by using one of the Calderón reproducing formulas, a new molecular characterization of $$B_{1,1}^{0,L}(\mathbb{R}^n)$$ is also obtained.

MSC:
 46E35 Sobolev spaces and other spaces of “smooth” functions, embedding theorems, trace theorems 42B25 Maximal functions, Littlewood-Paley theory 42B35 Function spaces arising in harmonic analysis 47F05 General theory of partial differential operators (should also be assigned at least one other classification number in Section 47-XX)
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# 2.2: The Probability of An Energy Level at Constant N, V, and T

If only pressure–volume work is possible, the state of a closed, reversible system can be specified by specifying its volume and temperature. Since the system is closed, the number, $$N$$, of molecules is constant. Let us consider a closed, equilibrated, constant-volume, constant-temperature system in which the total number of molecules is very large. Let us imagine that we can monitor the quantum state of one particular molecule over a very long time. Eventually, we are able to calculate the fraction of the elapsed time that the molecule spends in each of the quantum states. We label the available quantum states with the wavefunction symbols, $${\psi }_{i,j}$$.

We assume that the fraction of the time that a molecule spends in the quantum state $${\psi }_{i,j}$$ is the same thing as the probability of finding the molecule in quantum state $${\psi }_{i,j}$$. We denote this probability as $$\rho \left({\psi }_{i,j}\right)$$. To develop the theory of statistical thermodynamics, we assume that this probability depends on the energy, and only on the energy, of the quantum state $${\psi }_{i,j}$$. Consequently, any two quantum states whose energies are the same have the same probability, and the $$g_i$$-fold degenerate quantum states, $${\psi }_{i,j}$$, whose energies are $${\epsilon }_i$$, all have the same probability. In our imaginary monitoring of the state of a particular molecule, we observe that the probabilities of two quantum states are the same if and only if their energies are the same; that is, we observe $$\rho \left({\psi }_{i,j}\right)=\mathrm{\ }\rho \left({\psi }_{k,m}\right)$$ if and only if $$i=k$$.

The justification for this assumption is that the resulting theory successfully models experimental observations. We can ask, however, why we might be led to make this assumption in the first place. We can reason as follows: The fact that we observe a definite value for the energy of the macroscopic system implies that quantum states whose energies are much greater than the average molecular energy must be less probable than quantum states whose energies are smaller. Otherwise, the sum of the energies of high-energy molecules would exceed the energy of the system. Therefore, we can reasonably infer that the probability of a quantum state depends on its energy. On the other hand, we can think of no plausible reason for a given molecule to prefer one quantum state to another quantum state that has the same energy.

This assumption means that a single function suffices to specify the probability of finding a given molecule in any quantum state, $${\psi }_{i,j}$$, and the only independent variable is the quantum-state energy, $${\epsilon }_i$$. We denote the probability of a single quantum state, $${\psi }_{i,j}$$, whose energy is $${\epsilon }_i$$, as $$\rho \left({\epsilon }_i\right)$$. Since this is the probability of each of the $$g_i$$-fold degenerate quantum states, $${\psi }_{i,j}$$, that have energy $${\epsilon }_i$$, the probability of finding a given molecule in any energy level, $${\epsilon }_i$$, is $$P\left({\epsilon }_i\right)=g_i\rho \left({\epsilon }_i\right)$$. We find it convenient to introduce “$$P_i$$” to abbreviate this probability; that is, we let

$P_i=\sum^{g_i}_{j=1}{\rho \left({\psi }_{i,j}\right)}=P\left({\epsilon }_i\right)=g_i\rho \left({\epsilon }_i\right)$

(the probability of energy level $$\mathrm{\epsilon}_{\mathrm{i}}$$)

There is a $$P_i$$ for every energy level $${\epsilon }_i$$. $$P_i$$ must be the same for any molecule, since every molecule has the same properties. If the population set $$\{N^{\textrm{⦁}}_1,\ N^{\textrm{⦁}}_2,\dots ,N^{\textrm{⦁}}_i,\dots \}$$ characterizes the equilibrium system, the fraction of the molecules that have energy $${\epsilon }_i$$ is $${N^{\textrm{⦁}}_i}/{N}$$. (Elsewhere, an energy-level population set is often called a “distribution.” Since we define a distribution somewhat differently, we avoid this usage.) Since the fraction of the molecules in an energy level at any instant of time is the same as the fraction of the time that one molecule spends in that energy level, we have

$P_i=P\left({\epsilon }_i\right)=g_i\rho \left({\epsilon }_i\right)=\frac{N^{\textrm{⦁}}_i}{N}$

As long as the system is at equilibrium, this fraction is constant. In Chapter 21, we find an explicit equation for the probability function, $$\rho \left({\epsilon }_i\right)$$.

The energy levels, $${\epsilon }_i$$, depend on the properties of the molecules. In developing Boltzmann statistics for non-interacting molecules, we assume that the probability of finding a molecule in a particular energy level is independent of the number of molecules present in the system. While $$P_i$$ and $$\rho \left({\epsilon }_i\right)$$ depend on the energy level, $${\epsilon }_i$$, neither depends on the number of molecules, $$N$$. If we imagine inserting a barrier that converts an equilibrated collection of molecules into two half-size collections, each of the new collections is still at equilibrium. Each contains half as many molecules and has half the total energy of the original. In our model, the fraction of the molecules in any given energy level remains constant. Consequently, the probabilities associated with each energy level remain constant. (In Chapter 25, we introduce Fermi-Dirac and Bose-Einstein statistics. When we must use either of these models to describe the system, $$P_i$$ is affected by rules for the number of molecules that can occupy an energy level.)

The number of molecules and the total energy are extensive properties and vary in direct proportion to the size of the system. The probability, $$P_i$$, is an intensive variable that is a characteristic property of the macroscopic system. $$P_i$$ is a state function. $$P_i$$ depends on $${\epsilon }_i$$. So long as the thermodynamic variables that determine the state of the system remain constant, the $${\epsilon }_i$$ are constant. For a given macroscopic system in which only pressure–volume work is possible, the quantum mechanical energy levels, $${\epsilon }_i$$, are constant so long as the system volume and temperature are constant. However, the $${\epsilon }_i$$ are quantum-mechanical quantities that depend on our specification of the molecule and on the boundary values in our specification of the system. If we change any molecular properties or the dimensions of the system, the probabilities, $$P_i$$, change.


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MSBuild hangs while copying files

Consider this command line

msbuild R:\Build\trunk\19973\Deploy-Admin\Deploy-Admin.wdproj /p:Configuration="Release" /property:OutputPath=R:\Build\trunk\19973\!Output\Admin

What we see here is the msbuild compiler building a Web Deployment Project with some parameters manipulating the configuration and the OutputPath.

Everything worked fine. The time the MSBuild took to compile was always around 8 minutes.

Something happened then. The MSBuild now takes 30 minutes to compile the project and it seems it's doing "something" just for 22 minutes.

It was then when I noticed that the MSBuild is working 2 minutes well until it gets to the part of copying files. The MSBuild output stops outputing anything and I see the csc.exe compiler taking some percentage of the processor.

This takes approx these 22 minutes, the msbuild output unfreezes after that and the compiler finishes.

This is how the last lines from the output look like.

ResolveAssemblyReferences:

Add /v:diag switch to generate diagnostic log. Also consider /m:n where n=number of processors to increase build speed. –  KMoraz Feb 27 '12 at 15:52
Is R: a mapped network share? –  Christian.K Feb 28 '12 at 10:02


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# How do you multiply 4 * 4sqrt4?

$\sqrt{4} = 2$- Knowing that, the question turns into a simple multiplication problem:
$4 \cdot 4 \cdot \sqrt{4} = 4 \cdot 4 \cdot 2 = 32$


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This is an archived post. You won't be able to vote or comment.

[–] 2 points3 points  (0 children)

In my opinion, $50k is exceedingly unrealistic until you have several years under your belt and you've built a name for yourself. I don't think it's impossible, but be prepared for some lean times until it picks up. Independent tutoring is a very word-of-mouth driven business. Meeting with students at their homes means that you have more travel time and less time for tutoring. If you can arrange for students to meet you at a central location like a library, you can go from one appointment to the next with minimal down time. Being a full-time tutor is difficult. If you tutor high school students, they'll be in school most of the time, so you realistically have 4-5 hours on weekdays to work. Are you prepared to work nights during the week, and then during the weekend as well? Tutoring drops to almost zero during school vacations. You'll get a few random appointments, but there will be long stretches like winter vacation and the summer where tutoring will drop dramatically. Don't e-mail anybody, make a phone call at a minimum. I'd stay away from teachers altogether, but try to make appointments with counselors to let them know about your services. Even if they can't recommend you specifically, they may just keep lists of available tutors for parents to go through. Private schools are probably better targets than public. For high-school students, keep in mind that the student isn't actually your customer. Their parent is. It's not enough to make sure the student understands the material, you'll have to keep the parent informed as to how their kid is doing. You should also consider doing something like SAT-math prep. This will require some additional research on your part to make sure you can teach the test in addition to the subject matter, but parents will pay more for standardized test-prep versus normal tutoring. I was in charge of operations for a tutoring program, so let me know if you have any questions and I'll see if I can add anything. [–] 0 points1 point (0 children) Look what other tutors in the local area charge and see how well qualified they are is a good start on how much to charge. It may be worth contacting a few schools and offering them a discounted rate etc if they want to offer your services to their pupils etc. It may be worth considering taking on some children on a reasonably reduced rate to start off with so you gain a bit of experience and a few clients who will generate more clients etc. And most importantly do a good job, things like tutoring bring in business heavily based on word of mouth... a few parents saying how good you are/how much their child has improved with your help is probably the best form of advertising. Make sure you get testimonies off happy parents and keep a note of how well students are improving so you can use those stats when talking to new prospective parents. [–] 0 points1 point (0 children) I recommend universitytutor.com. I work a full time job separately so tutoring is only on the side for me but I get more business from this website than I can handle. During the summer was very busy for me. The only down time has been the first couple weeks of each semester when no one feels like they are struggling with class. [–] 0 points1 point (0 children) I'd start off at a lower price, maybe$30/hr in the first year. Then every year raise your price by $5 ($10 if you see yourself getting too busy).

As for advertising, charge a lower price at first as per above and let the word of mouth carry it from there. If you're good you will get busy pretty fast.

Also, give out your name to your local schools and possibly put out flyers where people are found.

[–] 0 points1 point  (0 children)

At something like 25 hours a week (3-6 and 7-9 every weekday) and 40 weeks a year you get a flat 1000 hours. \$50 an hour is probably reasonable for someone with your experience but may be a hard sell given that you don't need to be that superqualified to help highschool students - you're basically aiming for the most elite students and/or need to convince people that you are worth paying twice as much as e.g. an undergrad maths student might charge for essentially the same thing. I think you are more likely to have success tutoring university students, particularly if you have a large university nearby - lots of people take big first year calculus or stats subjects and have comparatively less access to 1-on-1 help than they do in school, and more importantly you are likely to be able to schedule said tutoring during the day (particularly if you can e.g. meet your students at uni libraries or such).

[–]Mathematical Physics -2 points-1 points  (0 children)

Just for reference, remember that you are starting a business and will need to register with your locality/state as such. That's how it works in the US, anyway. Many people just ignore this step, but it can really come back to bite you around tax time.


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# Does logarithmic decrement depend on initial amplitude?

by guest1234
Tags: amplitude, decrement, depend, initial, logarithmic
 P: 30 Lets say we have a weight attached to a spring. When releasing it under water (whole motion occours under water) at different initial amplitudes, will logarithmic decrement be the same? And will the period change? I think it should be the same for different initial amplitudes. But $\Theta=\frac{T_1}{t}\ln\frac{A_1}{A(t)}$, where $T_1$ is period of first cycle, $t$ is time until amplitude is $A(t)$ and $A_1$ is initial amplitude, suggests that when initial amplitude increases, logarithmic decrement must increase as well, or the ratio $\frac{T_1}{t}$ must decrease. So... which one is correct?


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# distillation apparatus

## Recommended Posts

I've been wanting for a while to try out extracting different scents from different plants but I'm lacking one crucial piece of apparatus, namely distillation set (or more precisely, the condenser part). They are available for purchase at very expensive prices so I was wondering if anyone has any ideas on how to make one that would work for my purpose...

Considering other equipment, I have beakers, flasks, round bottom flasks, funnels, tubes, I have water and heat source etc... I was thinking of using polystyrene test tubes and then cutting and melting them such to make something that resembles a condenser, but before I go on with that I just wanted to see if anyone has any other ideas.

I've looked around and found some interesting setups, this being one of them: http://orgchem.colorado.edu/hndbksupport/dist/distsetup/distsetmain.html

and these being many of them; http://images.google.com.au/images?q=distillation+apparatus&hl=en&um=1&sa=X&oi=images&ct=title

Wiki also has a nice diagram of how everything should look like. Basically I need to extract some of these; http://en.wikipedia.org/wiki/Essential_oils using http://en.wikipedia.org/wiki/Distill .

And one other question, rather noobish, am I heading in the right direction here? Is this how scents are usually extracted in perfumery? I have a notion that a lot of scents are lab synthesized these days...

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although crude serves me quite well, from alcohols to bromine or acids.

the vapor gets chance to cool a little in the glass pipe but since the end of the tube is in a receiver vessel that IS kept cool, all the real condensing takes place in there

thats certainly One way you could do it, and the glass tubing I bought in 500mm lengths and are quite easy to shape when heated over a bunsen flame.

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Nice setup. Simple too. The thing is I don't have any glass tubing and it might prove tricky to get some.

The place where you say "real condensing" apparently takes place you seem to have a large glass .. is that a glass stand or something... and water in it, with another flask in it used to collect the liquid? That's pretty cool anyhow, can I use a bucket of water instead?

One other thing I found tricky in distillation is that there are no holes anywhere... Doesn't the internal pressure increase due to heat? Does the condensation counter-act it by decreasing the pressure?

If that what you got is all that's needed, distillation doesn't look too hard then.

Would it be feasible to use rubber or some other tubing instead of the glass tubing?

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yeah, the receiver is a 250ml graduated volumetric cylinder floating in a gas jar full of ice water.

I wouldnt use rubber or anything that could A) React and B) go soft at temperatures

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Ah I see, I'll try and get some glass tubing, something cheap pops up on ebay every now and then.

Ah cool. I was wondering if you also put ice in there

Just an idea, would it help if you added ammonium nitrate to that icey water there since that would result in a nice endothermic reaction?

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sure you could, it makes a great freezing mixture, and its easily recoverable too, esp since the summers getting closer, a few days out in the sun and it should recrystalize again quite nicely

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Nice. I didn't realize it's reusable! In my case winter's getting closer so would it be feasible to dry it in the oven or on a hot plate ?

I'll report back when (and if) I find cheap glass tubing

Thanks.

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No, dont use direct heat to dry AN, the oven on low with the door open a little would be the safest of the 2 you mentioned, it doesnt need to be Bone dry either, just nice and crystalised.

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Ah I see. No probs, will do, thanks.

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I received my glass tubing today, 4 tubes around 50cm each, but I can't seem to be able to bend a piece as required using the best heat source I have available at home - gas stove. I'm assuming it can't provide the necessary temperature? As I haven't had any experience with glassworks, is there a method to this? Is the borosilicate glass supposed to turn 'orange hot' for me to be able to bend it?

I don't think I'm in the mood to buy propane torch right now, so just a question, will I be able to do what I want using Bunsen burner at uni lab?

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Borosilicate is Very hard to work with using ordinary Methane and or Propane, yes it need to glow orange/yellow and also only Very gentle pressure, Soda Glass tubing is what I use.

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Ah ok. I'm not entirely sure if it's borosilicate. Actually this is the one can you tell me what it is? http://cgi.ebay.com/SIMIX-12-MIL-PYREX-TYPE-GLASS-TUBING-24-24-feet_W0QQitemZ250101770739QQihZ015QQcategoryZ11812QQrdZ1QQcmdZViewItem

The seller claims its easily workable with propane torch...

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it is yeah, not too bad a deal either.

all you need is Blowtorch really, you dont need anything fancy or special like he has on his site, even a Cooks blowtorch works well (Butane gas).

it is workable, but I dont like working boro-glass personally.

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Awesome. Well it's the only thing I've got available so I'll try to work it out. I might buy one of those cooking torches since they're not too expensive but I'll first check if they'll let me use uni's lab - which I doubt, but remain hopeful. Thanks YT, I'll report again when I make progress

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Aight. As an update, lab staff said they don't have anything to work with boro glass (equipment on other campus ), Bunsen flame is too weak they say, so I have to come up with a blowtorch... Where do you think I can find one suitable... cooking supplies? I'll hunt around..

I also recently realised that I can't use flat bottom flask for distillation (lol), so.. as I haven't got any round bottom ones, I thought this seller has some good stuff; http://stores.ebay.com.au/UNITED-GLASS-TECHNOLOGIES_FLASKS_Round-Bottom-Flasks_W0QQcolZ4QQdirZ1QQfsubZ9395357QQftidZ2QQtZkm however these are fairly expensive.. what's their value in retail shops?

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Just to add a few notes to this thread; I've managed to bend the glass with propane torch, make the cooler/condenser from a 2L coke bottle and some tubing attached to it (letting the water through it very slowly) and that's as much as I'll say about that (that's the part that worked out ok ) because I soon realized that the setup was not very efficient at all, though it did work (this was a couple of months back, forgot to update ).

I then decided to get proper condenser and few other things and I was delighted at the workings of the new setup which was test-run to produce some lavander oil. As I said, worked very well and got me some very nice oil with greenish tint. Will do many distillations in the coming weeks

I recycled water for cooling and used good old alcohol burner as heat source

Here's photo

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Ammonium nitrates getting pricey and difficult to find so I tend to use all my AN in experiments but a good cheap substitute for an ice bath is UREA available as fertilizer (45-0-0) at ace hardware and others also available in handy 4lb sacks IIRC at around $5-$8.If they dont have it in stock they can order it if your local stores friendly o google 'Mastergardener'.Good thread ive put together some pretty crude distillation apparatus that neeed rebuilt nearly every run.Some handy ideas for an all glass/PVC/teflon imp. distillation apparatus,

Given the cost of maintaining an imp set-up ive seen e-bay of all places has all glass dstillation kits with a teflon thermometer adapter and thermometer complete 2000ml distillation kit (all glass/teflon) complete kit only no stand etc. for less than or a little more than $100.If nescessary the stand is far easier to imp than the distillation set-up and far safer depending on distillate. Besides having a limited$ supply for this one of many  hobbys I do enjoy improvsing what I can safely.At the momet im rigging an 'agitator' hotplate using a $12 hotplate and the 2 vibrator motor assemblys from a massaging recliner attached to the hotplate base.Not a magnetic stirrer but it does heck of a job mixing w/the agitation saving much elbow grease although not in the league of a$300

magnetic stirrer but recall its total cost was $12.00 plus a piece of scrap plywood and 79cent L brackets to secure the hotplate Edited by grndpndr #### Share this post ##### Link to post ##### Share on other sites I found that by using large volume of water the heat generated in such small setup would just not lift the temperature of the water by much at all. Just let the water come through the condenser and more than half of it would feel cold to the touch. On the other hand if you had a large condenser in a barrel or something then you could use the AN if you wanted. Also, I've made some improvements to the setup in the picture. For distillation of raw plant material I used steam distillation by boiling water in another flask and letting it through into the main flask (one with plant material) via rubber tubing. Main flask is enclosed by some aluminium foil to keep the heat inside and prevent condensation. Additional heat source can also be used for the main flask. Steam+oil molecules then move into the condenser as usual. With that$12 hotplate of yours, maybe you can implement a little motor below the plate and maybe make a magnetic stirrer similar to the one I described in the other thread.

I need to get some larger round bottom flasks too, at least 2L. I hear they're going to get illegal pretty soon, which is a damn shame.

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text:
# Some isomorphism conditions

Here I want to ask some true or false problems regarding isomorphisms (and if false, is there some extra condition to make it true). I do not what is the correct title for these problem. And I also want some brief proofs if possible.

1. Let $R$ and $R'$ be two rings with $|R| =|R'|< \infty$. And each proper ideal in $R$ is isomorphic to some ideal in $R'$. Then is it true that $R\cong R'$? (and if false, is there some extra condition to make it true).

2. Let $G$ and $G'$ be two groups with the same order ($< \infty$). If their abelianizations are isomorphic, is it true that $G\cong G'$.

3. Let $\mathcal{D}$ be a subcategory of a category $\mathcal{C}$. (1) If $u$ is an isomorphism of $\mathcal{D}$, is $u$ an isomorphism of $\mathcal{C}$? (2) If $v$ is an isomorphism of $\mathcal{C}$, is $v$ an isomorphism of $\mathcal{D}$? (and if false, is there some extra condition to make it true).

-
What does it mean for two ideals to be isomorphic? –  Qiaochu Yuan Apr 29 '11 at 4:53
3. (1) is obviously true; (2) make $\mathcal{D}$ a full subcategory. 2. there are simple groups of the same order. It remains to answer 1, which I leave to the algebraists after you clarified Qiaochu's question. –  t.b. Apr 29 '11 at 4:57
@Qiaochu Yuan: I mean there exists ring homomorphisms $\psi: I\to I'$ and $\phi: I'\to I$ such that $\phi\circ\psi = id$ and $\psi\circ\phi = id$. –  Junyu Apr 29 '11 at 4:59
(1) is not true. take $R=\mathbb{Z}/4\mathbb{Z}$ and $R'=\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. The only proper ideals of R are the zero ideal and $\mathbb{Z}/2\mathbb{Z}$ and they both appear in R'. Also, you might want to ask about proper subrings and not ideals (though it is still not true, but might be with extra conditions) –  Prometheus Apr 29 '11 at 5:00
@user5980: what do you mean by a ring homomorphism between two ideals? (Is this a homomorphism of not necessarily unital rings, i.e. a rng homomorphism?) –  Qiaochu Yuan Apr 29 '11 at 5:39

The answer to #2 is no. For example, the dihedral group $D_4$ and the quaternion group $Q_8$ both have order $8$ and abelianization $(\mathbb{Z}/2\mathbb{Z})^2$. (This implies, among other things, that they have the same character table.) Group theory would be very boring if anything like this was true.


text:
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# Prove the Following. Tan θ S E C θ + 1 = S E C θ − 1 Tan θ - Geometry

ConceptApplication of Trigonometry

#### Question

Prove the following.
$\frac{\tan\theta}{sec\theta + 1} = \frac{sec\theta - 1}{\tan\theta}$

#### Solution

$\frac{\tan\theta}{\sec\theta + 1}$
$= \frac{\tan\theta}{\sec\theta + 1} \times \frac{\sec\theta - 1}{\sec\theta - 1}$
$= \frac{\tan\theta\left( \sec\theta - 1 \right)}{\sec^2 \theta - 1}$
$= \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \left( 1 + \tan^2 \theta = \sec^2 \theta \right)$
$= \frac{\sec\theta - 1}{\tan\theta}$

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 6: Trigonometry
Problem set 6 | Q: 5.08 | Page no. 138
Solution Prove the Following. Tan θ S E C θ + 1 = S E C θ − 1 Tan θ Concept: Application of Trigonometry.
S


text:
These are chat archives for brunch/brunch

8th
Dec 2015
David
@dlepaux
Dec 08 2015 08:07
Hi you can try to use this /^(bower_components[\/\\]MathJax)/
@pprzetacznik
Piotr Przetacznik
@pprzetacznik
Dec 08 2015 12:31
that provides me only MathJax.js file without configurations
@dlepaux: and configurations are important to MathJax, I think
David
@dlepaux
Dec 08 2015 12:55
Do you want to keep this config ? ?config=TeX-AMS-MML_HTMLorMML
I think you can't but if you set this config in option probably
@pprzetacznik
David
@dlepaux
Dec 08 2015 13:04
You need to compile this file too config/TeX-AMS-MML_HTMLorMML.js
I'm not sure, but you can try to override your bower.json with something like this :
"overrides": {
"mathjax": {
"main": [ "./MathJax.js", "./config/TeX-AMS-MML_HTMLorMML.js"]
}
}
Piotr Przetacznik
@pprzetacznik
Dec 08 2015 13:34
@dlepaux the problem is TeX-AMS-MML_HTMLorMML.js config loads a lot more js files
so I have to provide all of them, there are plenty of them
David
@dlepaux
Dec 08 2015 13:35
yes i see it
Piotr Przetacznik
@pprzetacznik
Dec 08 2015 13:37
is there any reasonable solution for this kind of problem?
David
@dlepaux
Dec 08 2015 13:38
i don't know how you can add all this files...
Maybe to adding all of them in the override of bower.json... And next to order files in your brunch-config
Piotr Przetacznik
@pprzetacznik
Dec 08 2015 13:40
so add all of them manually?
David
@dlepaux
Dec 08 2015 13:41
Yes like this :
mhh
there is some script at the end of ./config/TeX-AMS-MML_HTMLorMML.js
If i were you, i will pepare properly all this as a vendor, this lib don't seem to be optimized for brunch..
you can't build your lib without using the GET parameter...


text:
# Update DQN at each steps or once every K steps

I have a question about the correct way of implementing replay memory for a DQN. In the pseudo code provided in the original DQN network, it seems they update the network at each step (after adding a new experience to the replay memory and taking a batch of samples from it). But, as claimed by this paper here, they update the network every 4 steps:

Another parameter worth noting is the network update frequency. The original DQN implementation only chose to take a gradient descent step every 4 environment steps of the algorithm as opposed to every step, as Algorithm 1 might suggest. Not only does this greatly increase the training speed (since learning steps on the network are far more expensive than forward passes), it also causes the experience memory to more closely resemble the state distribution of the current policy (since 4 new frames are added to the memory between training steps as opposed to 1) and may prevent the network from over-fitting.

Is this really correct? If yes, why 4? Assume the environment we are going to learn is such that at least every $K$ steps, we might observe a noticeable positive reward. Can we change 4 to $K$?

• 4 is just another hyperparameter to tune -- it's not necessarily the number of steps to receive a reward or anything like that. – shimao Jun 21 '18 at 6:26


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# How to use HKDF to combine two keys

I have heard that to combine two keys HKDF is better than using XOR function. But if you look at the HKDF RFC I could not figure out how to use it for concatenating two keys. What I can see is HKDF can be used to derive a key from another key. So I want to know how to use HKDF to concatenate two keys.

$$combined\_key = \text{SHA-2}(k_1 || k_2)$$


text:
# Which family in glm() in R is most appropriate for a response variable of proportions?

I currently have a data set where my dependent variable is a proportion (ex: the percent of a success). I currently have it set up where my y-values are between 0 and 1 (ex: 54.6% is 0.546). I decided to use logistic regression because my y-variable is between 0 and 1. However, since my y-variable is not binary, I assume I cannot use the family "binomial" in glm() R.

It is also worthy to note that I am doing a class project so dispersion value is important as well as AIC. When I use the binomial family, R automatically makes the dispersion value as 1, which does not help me figure out if my data is overdispersed.

Questions:

1. Is logistic regression appropriate to use in this scenario?
2. If the binomial family (link = logit) is appropriate as well for logistic reg, how should I scale/change the weight so that I can get an accurate dispersion value, as opposed to the assumed "1"?

UPDATE: I was able to do some more research, and one of my independent variable "total population" is the total for the percentage proportion (had to calculate it myself to make sure it aligns). My y-variable is still the percentage/proportion of success that the event happened within the population.

• Where do the proportion values come from? Are any exactly 0 or 1? What are their distributions between 0 and 1? Error estimates in binomial logistic regression depend on the numbers of cases in each group, not just the proportion. Depending on the nature of your data, other types of analysis might be possible. Please provide that information by editing your question, as comments are easily overlooked and can be deleted.
– EdM
Jul 30 at 15:40
• If you really don't have the counts/denominators, look into Fractional outcome regression. The tutorial I linked to shows examples with R and contrasts them with outputs from Stata. In short: glm with family = binomial is probably fine but the use robust standard errors is encouraged. Jul 30 at 15:45
• Jul 30 at 23:39

As far as I know, you can use the binomial family if you make use of the weights argument. However, this only makes sense, if your dependent variable comes from a binomial distribution and is already weighted by the number of trials. A similar discussion can be found here: How to apply binomial GLMM (glmer) to percentages rather than yes-no counts?.
Alternatively, you could use a fractional logit model, where the dependent variable is a fraction. A reference on that can be found here: ECONOMETRIC METHODS FOR FRACTIONAL RESPONSE VARIABLES WITH AN APPLICATION TO 401 (K) PLAN PARTICIPATION RATES. For the glm model type this can be implemented by changing the family to quasibinomial.


text:
Problems in homotopy theory Applications

• (J. McClure) Create elliptic cohomology for ${C}^{*}$ algebras. K-theory is one of the basic tools for ${C}^{*}$ algebras, but so far no one seems to have considered elliptic cohomology. There is already quite a bit of work on model categories for ${C}^{*}$ algebras, e.g. Østvaer Homotopy theory of ${C}^{*}$-algebras, MR2723902.

• (J. McClure) Create an intersection form of K-theory (and other generalized cohomology theories). This is a very old problem, e.g. p. 222 of the book Intersection Cohomology edited by Borel. Problems 11 and 12 on page 229 are also interesting.

• (C. Westerland) Problems on moduli spaces.

The cohomology of the moduli space ${M}_{g,n}$ of genus $g$ curves with $n$ marked points is known to stabilize as $g$ tends to infinity (Harer). Madsen-Weiss’s proof of the Mumford conjecture computes this stable cohomology rationally (for the answer over ${𝔽}_{p}$, see Søren Galatius’ thesis). Very little is known about the unstable cohomology of ${M}_{g,n}$, e.g., the cohomology in degrees greater than approximately $\frac{2g}{3}$ (using Søren Boldsen’s improvement of Harer’s argument).

Problem: compute the unstable cohomology of ${M}_{g,n}$.

A possible route to this is through field theories. If we believe that it is possible to understand a group through its representations, then the same should be true of a category. The homology of all of the ${M}_{g,n}$ fit together into a categorical construct whose representations are called homological conformal field theories (HCFT); there is a closely related notion, cohomological field theories (CohFT) which uses the compactified moduli. Gromov-Witten theory is an example of a CohFT, string topology and Hochschild cohomology of ${A}_{\infty }$ algebras are examples of HCFT’s.

Constantin Teleman’s classification of semisimple CohFT’s tells us that they only “see” the stable cohomology of moduli space, and so will be unhelpful in this problem; one must examine non-semisimple algebras. Conversely (or perhaps dually?) Tamanoi has proven that the stable classes in the homology of ${M}_{g,n}$ act trivially in string topology. Perhaps then string topology is exactly the “right” place to find information about the unstable homology of moduli space. Lastly, it’s a result of Wahl, Costello, and others that in some sense all of the homology of ${M}_{g,n}$ should be detected in its action on ${\mathrm{HH}}_{*}\left(A,A\right)$ for some ${A}_{\infty }$ Frobenius algebra $A$. So one could hope to cook up good examples of such $A$ to probe the structure of ${H}_{*}\left({M}_{g,n}\right)$.

• Problems from Ismar Volić on applications of homotopy theory to knot theory.


text:
Holomorphic cusp forms and cohomology of GL(2,Z)

Let $V_{k}$ denote the complex representation of $\mathrm{GL}(2)$ given by $\mathrm{Sym}^k(V)$, where $V$ is the defining 2-dimensional representation. Assume that $k$ is even. I would like to compute the cohomology group $$H^1(\mathrm{GL}(2,\mathbf Z),V_{k}).$$ By the Lyndon-Hochschild-Serre spectral sequence, this is the same as the $\mathbf Z/2$-invariants in $H^1(\mathrm{SL}(2,\mathbf Z),V_k)$. The latter cohomology group is a direct sum of three terms, each of which corresponds to a space of modular forms of weight $k+2$: one to holomorphic cusp forms, one to antiholomorphic cusp forms, and one to Eisenstein series. I think that the $\mathbf Z/2$-invariants consist exactly of the holomorphic cusp forms, but I can't seem to figure out why. Can someone give an argument and/or a reference that explains this, preferably in geometric terms?

Thanks to Joël's great answer I think I understand what's going on. Let me try to make really explicit where my confusion came from. If you write the Eichler-Shimura isomorphism $$H^1_{\mathrm{cusp}}(\mathrm{SL}(2,\mathbf Z),V_k) \cong S_{k+2} \oplus \overline S_{k+2}$$ then it's not true as I thought (and which led to nonsensical answers) that the subspace $S_{k+2}$ is $\mathbf Z/2$-invariant and $\overline S_{k+2}$ is anti-invariant. Instead the $\mathbf Z/2$-action interchanges these two subspaces according to the involution $$f(z) \mapsto -f(-\overline z),$$ which maps holomorphic modular forms to antiholomorphic ones and vice versa. (I guess depending on how you've defined the Eichler-Shimura isomorphism there might be a sign difference here.) Still, there is of course a natural way to identify the invariant subspace with the space of holomorphic cusp forms.

Let me try an answer. Instead of working with $H^1(SL(2,{\bf Z}),V_k)$, I'll work with a space which is naturally isomorphic to it, but more concrete, the space of modular symbols $Symb(V_k)$, defined as $Hom(\Delta^0,V_k)^\Gamma$. Here, $\Delta^0$ is the abelian group of divisors of degree $0$ on the projective rational line $\mathbb P^1(\bf Q)$, and $\Gamma = SL_2(\bf Z)$ for shortness. The isomorphism between those two spaces is standard, follows from the long exact sequence of cohomology attached to the pair (Poincaré half-plane with cusps $P^1(\bf Q)$ attached, $P^1(\bf Q)$), and commute with the action of Hecke operator on both spaces, in particular, what matters to us, with the action of the "Hecke operator at $\infty$" $\iota$ defined by the matrix $\left( \begin{matrix} 1 & 0 \\ 0 & -1\end{matrix}\right)$ (a generator of $GL_2({\bf Z})/SL_2({\bf Z})$). Also I see $V_k$ as the dual of the space of polynomial in one variable z of degree $\leq k$, with the natural homographic action of $\Gamma$. That's equivalent to your definition as a $Sym^k$.
Now we need to describe explicitly the relation you mention between modular forms and cohomology/modular symbols. Let us begin by cusp forms. In your post, the $\Gamma$-module $V_k$ is a complex vector space, but of course it is naturally defined over $\bf Z$, and I will call $V_k(\bf R)$ the version over $\bf R$. Let $S_{k+2}=S_{k+2}(\Gamma,\bf C)$ be the space of cusp forms. Then there is an injective $\bf R$-linear map $$S_{k+2} \rightarrow Symb(V_k({\bf R})),\ \ f \mapsto \phi_f,$$ where $\phi_f$ is the modular symbol defined by $$\phi_f(D)=(P \mapsto \Re( \int_D f(z) P(z) dz)).$$ Here, for $D=[a] - [b]$, $\int_D$ means the integral along any geodesic in the Poincaré half plane relying the cusp $a$ to the cusp $b$, for more general $D \in \Delta^0$, $\int_D$ is defined by linearity, and we see the right hand side is a linear form on the space of polynomials $P$ of degree $\leq k$ with real coefficients, hence an element of $V_k(\bf R)$ as it should be.
Now if $f = \sum a_n q^n \in S_{k+2}$, let $f'=\sum \bar a_n q^n \in S_{k+2}$. The $\bf R$-linear involution $f \mapsto f'$ is of course the complex conjugation on $S_{k+2}$ associated with its canonical real structure given by the forms defined over $\bf R$ (= with real coefficients). Note that $\overline{f(z)} = \sum \bar a_n \overline{e^{2 i \pi n z}} = \sum \bar a_n e^{-2 i \pi n \bar z}=f'(-\bar z)$. If $D=[\infty]-[a] \in \Delta^0$, $a \in \bf Q$, one has $\iota(D)=[\infty]-[-a]$ one has: $$\Re \left( \int_D f(z) P(z) dz \right) = \Re \left( \int_0^\infty f(a+iy) P(a+iy) i dy \right) \\ = \Re \left( \overline{ \int_0^\infty f(a+iy) P(a+iy) i dy } \right) \\ = - \Re \left( \int_0^\infty f'(- a+iy) P(a-iy) i dy \right) \\ = -\Re \int_{\iota D} f'(z) P(-z) dz,$$ hence $$\phi_{f'} = - \iota (\phi(f)),$$ that is the $\iota$ involution on modular symbols corresponds to minus the complex conjugation on modular cusp forms.
Now we complexify this real linear map, getting an injective map $$S_{k+2}(\Gamma,{\bf C}) \otimes_{\bf R} {\bf C} \hookrightarrow Symb(V_k({\bf C}),$$ where the left hand side can be identified with two copies of $S_{k+2}(\Gamma,{\bf C})$ and clearly from the above and easy linear algebra, one copy is identified with the $+1$-eigenspace, an other with $-1$ eigenspace for $\iota$ on the RHS. So this justifies your claim that the subspace invariant by $\iota$ in $Symb(V_k)$ contains one copy of the holomorphic modular cusp forms.
There is also the question of Eisenstein series. Working in level 1 as you do, there is only one of them for each k, and the question if whether the modular symbol corresponding to it is of eigenvalue $+1$ or $-1$ for $\iota$. This implies unfortunately that to have it right we need to be very careful in choosing a consistent system of conventions, which is a little bit too hard for me. If I interpret correctly my own computations with Samit Dasgupta however in our paper THE p-ADIC L-FUNCTIONS OF EVIL EISENSTEIN SERIES (Prop. 2.6), to appear in compositio, I think this sign is +1, which would mean that the Eisenstein series is also in the invariant subspace of $Symb(V_k)$, in other words also appears in $H^1(GL_2({\bf Z}),V_k)$. but maybe I'm wrong. What makes you think the latter is constituted of only cuspidal nodular forms?


text:
# Distributed load balancing: a new framework and improved guarantees

Innovations in Theoretical Computer Science (2021)

## Abstract

Inspired by applications on search engines and web servers, we consider a load balancing problem with a general \textit{convex} objective function. In this problem, we are given a bipartite graph on a set of sources $S$ and a set of workers $W$ and the goal is to distribute the load from each source among its neighboring workers such that the total load of workers are as balanced as possible. We present a new distributed algorithm that works with \textit{any} symmetric non-decreasing convex function for evaluating the balancedness of the workers' load. Our algorithm computes a nearly optimal allocation of loads in $O(\log n \log^2 d/\eps^3)$ rounds where $n$ is the number of nodes, $d$ is the maximum degree, and $\eps$ is the desired precision. If the objective is to minimize the maximum load, we modify the algorithm to obtain a nearly optimal solution in $O(\log n \log d/\eps^2)$ rounds. This improves a line of algorithms that require a polynomial number of rounds in $n$ and $d$ and appear to encounter a fundamental barrier that prevents them from obtaining poly-logarithmic runtime \cite{berenbrink2005dynamic, berenbrink2009new, subramanian1994analysis, rabani1998local}. In our paper, we introduce a novel primal-dual approach with multiplicative weight updates that allows us to circumvent this barrier. Our algorithm is inspired by \cite{agrawal2018proportional} and other distributed algorithms for optimizing linear objectives but introduces several new twists to deal with general convex objectives.


text:
# Differentiate between molarity and molality of a solution. How can we change the molality value of a solution into molarity value?

$m= frac{1000times M}{1000times (d-M)times M^{‘}}$


text:
# Intersection and union of ArrayLists in Java

Are there any methods to do so? I was looking but couldn't find any.

Another question: I need these methods so I can filter files. Some are AND filters and some are OR filters (like in set theory), so I need to filter according to all files and the unite/intersects ArrayLists that holds those files.

Should I use a different data structure to hold the files? Is there anything else that would offer a better runtime?

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If you didn't want to creayte a new list, Vector.retainAll(Vector) trims your orignal vector to only the intersection with second vector. –  user2808054 Jan 16 at 17:50

Here's a plain implementation without using any third-party library. Main advantage over retainAll, removeAll and addAll is that these methods don't modify the original lists input to the methods.

public class Test {

public static void main(String... args) throws Exception {

List<String> list1 = new ArrayList<String>(Arrays.asList("A", "B", "C"));
List<String> list2 = new ArrayList<String>(Arrays.asList("B", "C", "D", "E", "F"));

System.out.println(new Test().intersection(list1, list2));
System.out.println(new Test().union(list1, list2));
}

public <T> List<T> union(List<T> list1, List<T> list2) {
Set<T> set = new HashSet<T>();

return new ArrayList<T>(set);
}

public <T> List<T> intersection(List<T> list1, List<T> list2) {
List<T> list = new ArrayList<T>();

for (T t : list1) {
if(list2.contains(t)) {
}
}

return list;
}
}

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you can create new list with list1 elements and then call retainAll, addAll methods –  smas Mar 12 '11 at 14:44
why you using strictfp in this solution? –  smas Mar 12 '11 at 14:47
@smas - it was straight from my laboratory, edited now :) –  adarshr Mar 12 '11 at 14:50
Should use a HashSet for intersection so that the average case performance is O(n) instead of O(n^2). –  Zong Zheng Li Mar 16 at 23:30

Collection (so ArrayList also) have:

col.retainAll(otherCol) // for intersection

Use a List implementation if you accept repetitions, a Set implementation if you don't:

Collection<String> col1 = new ArrayList<String>(); // {a, b, c}
// Collection<String> col1 = new TreeSet<String>();

Collection<String> col2 = new ArrayList<String>(); // {b, c, d, e}
// Collection<String> col2 = new TreeSet<String>();

System.out.println(col1);
//output for ArrayList: [a, b, c, b, c, d, e]
//output for TreeSet: [a, b, c, d, e]

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There's been a suggested edit that this union "is incorrect since it will contain common elements twice". The edit recommended to use a HashSet instead. –  Kos Nov 30 '12 at 17:26
Actually it was edited, see: "Use a List implementation if you accept repetitions, a Set implementation if you don't:" –  smas Sep 12 '13 at 10:53
List1.retainAll(List2) - is intersection

union will be removeAll and then addAll.

-

Unions and intersections defined only for sets, not lists. As you mentioned.

Check guava library for filters. Also guava provides real intersections and unions

 static <E> Sets.SetView<E >union(Set<? extends E> set1, Set<? extends E> set2)
static <E> Sets.SetView<E> intersection(Set<E> set1, Set<?> set2)

-

You can use CollectionUtils from apache commons.

-

The solution marked is not efficient. It has a O(n^2). What we can do is to sort both lists, and the execute an intersection algorithm as the one below.

private  static ArrayList<Integer> interesect(ArrayList<Integer> f, ArrayList<Integer> s) {
ArrayList<Integer> res = new ArrayList<Integer>();

int i = 0, j = 0;
while (i != f.size() && j != s.size()) {

if (f.get(i) < s.get(j)) {
i ++;
} else if (f.get(i) > s.get(j)) {
j ++;
} else {
i ++;  j ++;
}
}

return res;
}

This one has a complexity of O(n log n + n) \in O(n \log n). The union is done in a similar manner. Just make sure you make the suitable modifications on the if-elseif-else statements.

You can also use iterators if you want (I know they are more efficient in C++, I dont know if this is true in Java as well).

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I think you should use a Set to hold the files if you want to do intersection and union on them. Then you can use Guava's Sets class to do union, intersection and filtering by a Predicate as well. The difference between these methods and the other suggestions is that all of these methods create lazy views of the union, intersection, etc. of the two sets. Apache Commons creates a new collection and copies data to it. retainAll changes one of your collections by removing elements from it.

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Here is a way how you can do an intersection with streams (remember that you have to use java 8 for streams):

List<foo> fooList1 = new ArrayList<>(Arrays.asList(new foo(), new foo()));
List<foo> fooList2 = new ArrayList<>(Arrays.asList(new foo(), new foo()));
fooList1.stream().filter(f -> fooList2.contains(f)).collect(Collectors.toList());

An example for lists with different types. If you have a realtion between foo and bar and you can get a bar-object from foo than you can modify your stream:

List<foo> fooList = new ArrayList<>(Arrays.asList(new foo(), new foo()));
List<bar> barList = new ArrayList<>(Arrays.asList(new bar(), new bar()));

fooList.stream().filter(f -> barList.contains(f.getBar()).collect(Collectors.toList());

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There is no google collections anymore. Enjoy guava) –  Stas Kurilin Mar 12 '11 at 14:37
• retainAll will modify your list
• Guava doesn't have APIs for List (only for set)

I found ListUtils very useful for this use case.

Use ListUtils from org.apache.commons.collections if you do not want to modify existing list.

ListUtils.intersection(list1, list2)

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This post is fairly old, but nevertheless it was the first one popping up on google when looking for that topic.

I want to give an update using Java 8 streams doing (basically) the same thing in a single line:

List<T> intersect = list1.stream().filter(list2::contains).collect(Collectors.toList());

List<T> union = Stream.concat(list1.stream(), list2.stream()).distinct().collect(Collectors.toList());

If anyone has a better/faster solution let me know, but this solution is a nice one liner that can be easily included in a method without adding a unnecessary helper class/method and still keep the readability.

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text:
# pushforward of locally free sheaf is locally free?

Hi,

Is there an example of a proper smooth map of schemes $f:X\to Y$ and a vector bundle $E$ on $X$ such that $f_*E$ is not locally free on $Y$?

Thanks

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Take $Y$ to be an elliptic curve and let $X=Y\times Y$, with $f$ a projection. Let $Z\in X$ denote the zero section, $\Delta$ the diagonal and let $E=O(\Delta-Z)$. I'll let you compute what happens. –  Donu Arapura Sep 27 '11 at 21:59
Actually, I realize the above ex. doesn't work, the derived dir, image $Rf_*E$ isn't loc. free, but $0$th is. I'll let someone else take over -- too much on my plate these days... –  Donu Arapura Sep 27 '11 at 23:00
According to Corollary II.13.1 of Mumford's Abelian varieties, $E$ has no global sections (or any cohomology at all outside of degree 1), which wouldn't be the case if $pr_{2*}E$ were a skyscraper sheaf. –  Keerthi Madapusi Pera Sep 28 '11 at 2:47
$f_*E$ is torsion free so there cannot be a counterexample with $Y$ a smooth curve. –  ulrich Sep 28 '11 at 4:33
It's local on the target, not the source. –  Keerthi Madapusi Pera Sep 28 '11 at 12:08

Here is a an example, albeit with $Y$ non reduced:

Let $C$ be a smooth projective curve of genus $g > 0$ over a field $k$ and let $C_{\epsilon} = C \times_{k} Spec(k[\epsilon])$ where $k[\epsilon] = k[x]/(x^2)$ is the ring of dual numbers. Let $\mathcal{L}$ be a non-trivial line bundle on $C_{\epsilon}$ such that its restriction to $C$ is trivial. Such line bundles exist since $g > 0$.

Multiplication by $\epsilon$ gives us an exact sequence of sheaves on $C$ $$0 \to \epsilon \mathcal{L} \to \mathcal{L} \to \epsilon\mathcal{L} \to 0 \ .$$

Since $\epsilon \mathcal{L} = \mathcal{L}/\epsilon \mathcal{L}$, it follows from the construction that this sheaf is just $\mathcal{O}_C$. Since $\mathcal{L}$ is non-trivial the boundary map in the long exact sequence of cohomology from $H^0(C, \mathcal{O}_C)$ to $H^1(C, \mathcal{O}_C)$ is non-trivial. Thus $H^0(C, \mathcal{L}) = H^0(C, \mathcal{O}_C)= k$.

If we let $X = C_{\epsilon}$, $Y = Spec(k[\epsilon])$, $f$ the natural map $C_{\epsilon} \to Spec(k[\epsilon])$, and $E = \mathcal{L}$, it follows from the above that $f_*E$ is not locally free.

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Thanks for the answer. Why is the boundary map nonzero? –  Nicolás Sep 28 '11 at 19:37
Line bundles on $C_{\epsilon}$ with a trivialisation over $C$ correspond bijectively to $Ext^1_C(\mathcal{O}_C,\mathcal{O}_C) = H^1(C, \mathcal{O}_C)$. This identification is given precisely by the boundary map, so a line bundle is trivial iff the corresponding boundary map is zero. –  ulrich Sep 29 '11 at 5:16

Here is an example with $Y$ smooth!

Let $C$ be a smooth projective curve of genus $g \geq 3$. Denote by $Y$ the 'translated Jacobian' $J = {\rm Pic}^{2g-2}C$. Set $X = Y \times C$. Let $f \colon X \to Y$ and $g \colon X \to C$ be the two projections. Let $L$ be a Poincare bundle on $X$, namely a universal line bundle of degree $2g-2$. Denote by $O \in Y$ the point corresponding to the canonical bundle $K_C$. I claim that $f_*L$ is not locally free at $O$.

Let $y$ denote a point in $y$ and $L|_y$ the restriction of $L$ to $y \times C$, the fiber of $f$ over $y$. Note that $$h^0(L|_y) = \begin{cases} g-1 & \text{ if y \neq O} \\\\ g & \text{ if y = O} \end{cases}.$$ Similarly, $$h^1(L|_y) = \begin{cases} 0 & \text{ if y \neq O} \\\\ 1 & \text{ if y = O} \end{cases}.$$

To see that $f_*L$ is in fact not locally free at $O$, choose a point $p \in C$ and let $P = Y \times p$. We have an exact sequence

$$0 \to L \to L \otimes O_Y(P) \to L \otimes O_Y(P)|_P \to 0.$$

Applying $f_\*$, we get

$$0 \to f_\* L \to f_\*(L \otimes O_Y(P)) \to f_\*(L \otimes O_Y(P)|_P) \to R^1f(L) \to 0$$

The zero on the far right is because $R^1f_\*(L \otimes O_Y(P)) = 0$, by Grauert's theorem. Indeed, $H^1(L|_y \otimes O_C(P)) = 0$ for any $y \in Y$, since $L|_y \times O_C(p)$ is a line bundle on $C$ of degree $2g-1$. Also, note that the middle two terms $f_\*(L \otimes O_Y(P))$ and $f_\*(L \otimes O_Y(P)|_P)$ are locally free on $Y$. For the first, we again apply Grauert's theorem and for the second we see that $f \colon P \to Y$ is an isomorphism. Finally, the jumping of $h^0$ and $h^1$ mentioned above and a theorem on cohomology and base change (i.e. Hartshorne Ch 3 Thm 12.11) implies that $R^1f(L)$ is supported (set-theoretically) at $O$.

I claim that if $f_*L$ were locally free, we would get a contradiction. Namely, we will have a resolution of a sheaf supported at a point that is too small in length. To make this precise, we use the Auslander-Buchsbaum formula. If $f_*L$ is locally free, then the projective dimension of $R^1f(L)$ is at most two. Hence its depth is at least $g-2 \geq 1$. However, the dimension of its support is zero! This is a contradiction. We conclude that $f_*L$ is not locally free.

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I read rita's comment after writing all this! It's basically the same idea, but I need a curve of genus $g \geq 3$. –  anand Sep 29 '11 at 2:32
Very nice example! –  Rex Sep 29 '11 at 6:40
In fact in the meantime I had also realized that one needs $g>2$. I had made a very down to earth computation that amounted to your argument. It's nice to have it all sorted out. –  rita Sep 29 '11 at 7:40
I concur, it's very nice. –  Donu Arapura Sep 29 '11 at 12:19

There is even a more simple example (although essentially the same) than in the Anand's answer. Take $C = P^1 = P(V)$, where $\dim V = 2$, let $Y = S^2V$ and $X = C\times Y$ with the map $f:X \to Y$ being just the projection. Take $E$ to be the universal extension of $O(2)$ by $O(-2)$. This means that $E$ fits into exact sequence $$0 \to p^*O(-2) \to E \to p^*O(2) \to 0,$$ where $p$ is the projection $X \to C$, and the extension is given by the canonical element in $S^2V\otimes S^2V^* \subset S^2V\otimes k[S^2V] = Ext^1(p^*O(2),p^*O(-2))$. Applying the pushforward via $f$ to the above sequence one gets $$0 \to f_*E \to S^2V^*\otimes O_Y \to O_Y \to R^1f_*E \to 0,$$ and it is clear that the middle map is given by the canonical embedding $S^2V^* \to k[S^2V]$. Thus $R^1f_*E$ is the structure sheaf of the point $0 \in Y = S^2V$ and $f_*E$ is the second syzygy sheaf of a point on a 3-dimensional variety, which is the simplest example of a reflexive non-locally-free sheaf.

By the way, the pushforward of a vector bundle under a smooth morphism is always reflexive. This is why the above example is the simplest possible.

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This is also very nice. –  Donu Arapura Sep 29 '11 at 18:32

EDIT: the example below does not answer the question because the map is not smooth. (I had not read the question carefully, sorry!). I don't remove the answer since it might still be useful to somebody.

Let $Y\subset \mathbb A^3$ be the quadric cone, defined by $xy-z^2=0$ and take the map $X={\mathbb A}^2\to Y$ given by $(u,v)\mapsto (u^2,v^2, uv)$. This is a (non flat) double cover and the direct image of ${\mathcal O}_X$ is of the form ${\mathcal O}_Y\oplus F$, where $F$ is a rank 1 reflexive sheaf that is not locally free.

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In the question $f$ is supposed to be a smooth proper morphism. –  ulrich Sep 28 '11 at 10:38
I had missed the smooth''. Thank you! –  rita Sep 28 '11 at 11:01

When $E$ is trivial and $X$ is smooth, the following result of De Bois gives a positive answer:

Let $f:X\to Y$ be a proper, flat morphism of algebraic varieties over a field of characteristic zero, such that the fibers $X_y$ have du Bois singularities for all $y\in Y$ (this holds in particular if $f$ is smooth). Then $R^if_*\mathcal{O}_X$ is locally free for all $i\ge 0$.

See Theorem 4.6 of Du Bois, P. Complexe de de Rham filtré d'une variété singulière. Bulletin de la Société Mathématique de France, 109 (1981), p. 41-81

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I haven't looked at that paper in a while, but don't need to assume that $f$ is semistable? Should be OK in general when $\dim Y=1$, by semistable reduction. –  Donu Arapura Sep 28 '11 at 17:41
For $i=0$ only, a similar result is given in EGA III, 7.8.7. –  user2035 Sep 28 '11 at 17:45
–  user2035 Sep 28 '11 at 17:54
JC: I'm sorry if this sounds critical. In fact, I would be happier if it was true unconditionally since I use these sorts of results, but note the sentence: "Si pour tout point geometrique s de la fleche $O_{X_s}\to \Omega_{X_s}^0$ est isomorphisme" Nowadays, we would say the fibres are Du Bois. It is quite a strong condition. –  Donu Arapura Sep 28 '11 at 18:30
Sure, with smooth maps there is no problem. It is also OK for semistable maps, which is why I remembered it that way. –  Donu Arapura Sep 28 '11 at 18:47