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# Let's make one big, happy "Congratulations!" thread.

Update: The proposal in question has been implemented.

We recently had a discussion about the tag that raised some important issues, but did not result in a consensus. I'd like to pitch the following solution to the community.

Let's consolidate all celebration posts into one bumpable, meta question, similar to the thread for reopen requests.

These are several reasons that I want to propose this.

• These posts are not questions, and they are all very similar. I know we allow other unconventional threads to be posted as questions, but this is very clearly a family of such posts. It would make sense to aggregate them into one place to be viewed all at once, like a .

• Celebrations make more sense as answers than questions. Congratulations posts don't need to be answered. Sometimes those being recognized post an answer to say thanks, but it makes just as much sense to express this sentiment as a comment. Since comments can be upvoted, this presents no loss of visibility.

• Bumping an answer list allows for continuous recognition of past achievements. Many users may not think to use the search to read through past celebrations, and even so, the interface is not as good as a list of answers. This way, every time a new celebration occurs, we not only get to celebrate the occasion at hand, but remember the reasons for past celebrations, too.

• Shaking up tradition broadens the format to allow celebration of other cool achievements. Two months ago, Martin Sleziak opened this chat room as an experiment, and I think it's really great. The informality of the chat format has allowed a variety of "quirkier" achievements to be recognized, such as high "meta-rep" and crazy good editing. It's really cool to see these celebrations of individuality alongside those of pure numbers. However, the chat format has its problems. Discussion is intermingled with the list of results, which makes them more difficult to read. The visibility is inarguably lower, there's no voting to show appreciation, it's less accessible to newcomers, and, as demonstrated by the comments in this thread, some users just plain don't like to use chat. A consolidated celebrations thread would be allow us to adopt some of this flexibility, in addition being more unobtrusive to those who consider these posts off-topic. I do think we should still use this room- it's a great place to talk about data explorer, and a great platform to celebrate smaller achievements that are not big enough to warrant a meta post. But we need something else, too.

• The main thing we'd lose would be having the name of the person being celebrated in the title of the question, which would then get less attention on the meta front page. One way we might compensate for this could be to change the accepted answer whenever a new celebration is posted, thus identifying the "incumbent celebration."

What do you think, community?

• Re the last point: Or edit the title, "Congratulations thread, [whoever caused the last bump] edition". Nov 25 '14 at 11:47
• Re chat and "there's no voting to show appreciation" I do not get that at all. One can "star" a message, which seems as good as an upvote for the purpose at hand. The issue of downvotes (that seem somewhat unpleasant in that context) is gone. And starred messages can be found among the rest. To be clear, I do not argue against that thread, just this particular reason seems like a nonreason to me, since in fact for this purpose in my opinion there is a better suited form of voting in chat then here.
– quid Mod
Nov 25 '14 at 13:01
• Let's just not have user specific congratulatory threads instead.
– Asaf Karagila Mod
Nov 25 '14 at 13:49
• Agreeing with @AsafKaragila: let's just not. Simply downvote, close, and delete off-topic threads as they appear -- business as usual.
– user147263
Nov 25 '14 at 15:13
• @Asaf The issue is that not everyone thinks they are off topic. I do, but others' opinions matter. To me, this seems like a good compromise.
– Alexander Gruber Mod
Nov 25 '14 at 18:49
• Alexander, of course that I don't want to have another divisive issue on meta, there are plenty of serious issues to go around. And I do understand this sort of activity and why people do it. But just like we don't congratulate people on earning gold badges (or a particular number thereof), we shouldn't congratulate them on some reputation barrier. This community is very large as it is, and growing larger with each day. This means that eventually more and more people get to 100k in faster rates, which means that at some point the congratulation thread will be bumped all too often too.
– Asaf Karagila Mod
Nov 25 '14 at 19:01
• Just rip this band-aid (or adhesive strip, if you prefer) now, and get it over with.
– Asaf Karagila Mod
Nov 25 '14 at 19:01
• We might see how good a compromise this idea is once we will have had the first discussion whether it was not a bad idea to replace that really noteworthy achievement in the title by a much less worthy one after so little time (and perhaps if this was not done for sinister motives).
– quid Mod
Nov 25 '14 at 19:46
• @quid Well, we could just agree that the accepted answer denotes the most recent post, not the most important. Or we could just forget about the accepted answer / title change thing. That would be fine by me.
– Alexander Gruber Mod
Nov 26 '14 at 0:18
• My comment is based on "title/accept is most recent." My point is: some things might be "headline" for a long time and some others only for 20 minutes. This could cause problems. Especially since everybody can generate "news" essentially as they see fit. So somebody could come along and "push away" somebody they do not like from the "headline" via posting something. Or somebody could suspect it happened. If at all, yes, I would strongly prefer no title changes or accept (except perhaps at a later point some overview answer could be generated and accept, but this could also be in the post).
– quid Mod
Nov 26 '14 at 7:18
• Do these posts come up so often as to be actively cluttering up meta? It seems as though they're on the order of, say, one per month if that - as opposed to something like the reopen thread, which gets a new note every few days. I just don't see what problem this is trying to solve. Nov 29 '14 at 8:38
• I think it is a nice tradition for the first $n$ people to hit some barrier. Then around $n$ threads need to be kept open and the barrier number updated as appropriate (assuming that, for the most part, the same $n$ people will be the ones hitting the thresholds). I don't find the threads bothersome in any way. Dec 1 '14 at 7:02

Personally, I find these posts not bad. However, if a somewhat complicated procedure should be needed to fit them in (edit the title, change the accpeted answer, and what not), then I would say the thing has run its course and it is better to stop that tradition, as oppose to continuing it in this rigid form.

Perhaps this is a pity for some (including me), but the alternative seems a lot worse to me, be it either the constant quibble about the posts or this rigid form that gets proposed here.

Update: The proposal in question has been implemented.

This is a good compromise and I think its score (presently 31) indicates a clear positive community response.

But I think the thread should be simple rather than complicated. In particular,

• Don't try to have an accepted answer mean anything. This is not what accepting an answer means and this would be confusing.

• Don't try to merge old congratulations threads with this new thread. Just use it for future congratulations.

• Don't edit the title of the question. Leave it as something generic.

• Don't argue about the details. Just make the thread and allow people to make answers congratulating specific users.

The point is that we don't want a rigid and complicated procedure; this is what quid's answer complains about:

if a somewhat complicated procedure should be needed to fit them in (edit the title, change the accpeted answer, and what not), then I would say the thing has run its course and it is better to stop that tradition, as oppose to continuing it in this rigid form.

The main problem with the current congratulations threads is that they are recurring, uninteresting, and a much bigger deal than they really should be. I would find it a bit sad to ban the sentiment of such posts altogether, but on the other hand, I think it is silly to keep congratulating people on this level (and the Hot Meta Posts list should be reserved for more important things). This proposal solves all these issues effectively.

• Votes have many rationals, it is not easy to tell what they represent unless there is a clear statement in the post about how votes will be interpreted. I upvoted the question because I think it worth discussing, but I am leaning more towards what quid wrote. Jan 14 '15 at 0:40
• @Kaveh Upvotes on meta mean "I agree". Downvotes mean "I disagree".
– 6005
Jan 14 '15 at 1:33
• No, they don't. People use votes for various reasons, e.g. I here upvoted because I think it is a useful discussion, not that I agree with the proposed solution. Jan 14 '15 at 1:46
• @Kaveh They generally do. I'm not sure how to convince you of this, maybe you simply haven't spent much time on meta. Here is another post; notice nobody argues with the premise of the question.
– 6005
Jan 14 '15 at 2:06
• I have seen countless times new users on meta get upset by downvotes, before it is explained by more experienced users that downvotes mean disagreement and upvotes mean agree. I didn't really know this was subject to debate. Of course, some users (e.g. you) are not voting this way, and I will contend this might be a significant portion of the meta population, but if you ask anyone here how voting works on meta, they will explain what I have said.
– 6005
Jan 14 '15 at 2:08
• @Goos: Checking the highest voted answer to the first link you gave, you will see a couple of reasons to downvote that don't mean "I disagree." I notice that I long ago downvoted the answer you linked directly to, because I disagree with it; it is sometimes right, but overgeneralized. Jan 14 '15 at 2:52
• I have been visiting this meta (and other metas) for years now and I can tell you that it is not the case, people may vote for usefulness/suitability of a topic, for agreement with a proposal, or even because of general behavior of the user among other things. If someone wants to have a poll on a specific issue it should be stated as so in the post like Asaf does in his answer here. Jan 14 '15 at 4:42
• @Kaveh I wasn't really arguing about how people actually vote, only about how votes are to be interpreted. I should concede that "I agree" is a vague way of saying what I mean; for example, "I agree" can mean that you agree with the usefulness/suitability of a topic.
– 6005
Jan 14 '15 at 5:00
• @Kaveh I don't agree that a separate poll is needed unless the voting looks close. I stand by that 41 upvotes, 5 downvotes (or whatever it was at the time) indicates a "clear positive community response". If you think my interpreting this as clear positive support for the proposal and you think it important enough I suggest you take it up on another thread. I would not see, for instance, 10 upvotes and 7 votes, as a clear positive response, or even as a positive response. But 41 upvotes is clear approval in my mind.
– 6005
Jan 14 '15 at 5:03
• More concretely: If only 5 people downvote a proposal, then only 5 people expressly disagreed with any aspect of it. In particular at most 5 people disagreed with the proposal. And 41 upvotes I would assume includes at least 20 (being safe) votes in favor of the proposal itself.
– 6005
Jan 14 '15 at 5:04

This is a poll answer.

Please UPVOTE this if you think there shouldn't be any more personal congratulatory meta posts.

Please DOWNVOTE this if you think there should be some solution (regardless to a single thread, or continuing in this fashion) for personal congratulatory posts on meta.

If you don't care, please don't vote.

(I will occasionally update the current score for the benefit of those that did not reach 1,000 points yet.)

Update: Current score, +6, -8 (12 Jan 2015 09:51:10 PM UTC)

• Whatever the score is, please add +1 from me on the upvote side.
– Asaf Karagila Mod
Dec 8 '14 at 8:47
• I've voted on your behalf! Dec 8 '14 at 10:11
• @Integrator: Does that mean that you're abstaining?
– Asaf Karagila Mod
Dec 8 '14 at 10:12
• I'm unable to choose sides. Dec 8 '14 at 10:17
• @Integrator: Is this because you don't want to assume the axiom of choice?
– Asaf Karagila Mod
Dec 8 '14 at 10:17
• @quid: It might take another couple of days. I don't expect this to be similar to the "Let's kill the [homework] tag" or "Let's require registration for asking questions" or "Let's have silly hats", but I'm not in a hurry.
– Asaf Karagila Mod
Dec 8 '14 at 18:02
• Let us see. What I am worried about is that it does not get enough visibility anymore. After all 40+ voted OP. Maybe let us make it an "issue" in the election? This is only semi-serious. [Deleted the first comment to avoid excessive clutter.] Current count +3/-3
– quid Mod
Dec 9 '14 at 7:03
• I think the number of votes matches the importance of the issue. Those who want congratulatory posts can always create them. Those who don't want them can vote accordingly. Whichever side has more support will determine the outcome. The system works.
– user147263
Dec 9 '14 at 7:07
• @quid, it is a possibility, yes.
– Asaf Karagila Mod
Dec 9 '14 at 7:20
• @Beh, no this is when it doesn't work. Since meta is not active enough to drown these things in the noise, it becomes annoying when this gets done, closed and reopened and deleted and undeleted. This is where the system needs fixing, and we need to decide on some result and empower the moderators to act, if need be, without hindrance.
– Asaf Karagila Mod
Dec 9 '14 at 7:23
• @AsafKaragila You know there is the (celebration) tag for hiding those posts, if they are annoying.
– user147263
Dec 9 '14 at 7:33


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# in triangle ABC,if AD is perpendicular to BCand AD*2=BD.DC,prove that angle BAC=90 degree

Given : In triangle ABC , AD is perpendicular to BC & AD2 = BD.DC

To proove : BAC = 90 degree

Proof : in right triangles ADB & ADC we have AB2 = AD2 +BD2 . . . . . . . . .1

& AC2 = AD2+ DC2 . . . . . . . .2

from 1 & 2 , AB2 + AC2 = 2AD2 . BD2 + DC2

= 2BD . CD + BD2 + CD2         { AD*2 = BD . CD (GIVEN ) }

= (BD + CD )2 = BC2

Thus in triangle ABC we have , AB2 + AC2 = BC2

hence triangle ABC is a right triangle right angled at  A

so BAC = 90 degree      ( proved )

• 246
What are you looking for?

format: plain


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# Decimal representation

A decimal representation of a non-negative real number r is an expression in the form of a sequence of decimal digits traditionally written with a single separator

${\displaystyle r=b_{k}b_{k-1}\ldots b_{0}.a_{1}a_{2}\ldots \,,}$

where k is a nonnegative integer and ${\displaystyle b_{0},\ldots ,b_{k},a_{1},a_{2},\ldots }$ are integers in the range 0, ..., 9, which are called the digits of the representation.

This expression represents the infinite sum

${\displaystyle r=\sum _{i=0}^{k}b_{i}10^{i}+\sum _{i=1}^{\infty }{\frac {a_{i}}{10^{i}}}.}$

The sequence of the ${\displaystyle a_{i}}$—the digits after the dot—may be finite, in which case the lacking digits are assumed to be 0.

Every nonnegative real number has at least one such representation; it has two such representations if and only one has a trailing infinite sequence of zeros, and the other has a trailing infinite sequence of nines. Some authors forbid decimal representations with a trailing infinite sequence of nines because this allows a one-to-one correspondence between nonnegative real numbers and decimal representations.[1]

The integer ${\displaystyle \sum _{i=0}^{k}b_{i}10^{i}}$, denoted by a0 in the remainder of this article, is called the integer part of r, and the sequence of the ${\displaystyle a_{i}}$ represents the number

${\displaystyle 0.a_{1}a_{2}\ldots =\sum _{i=1}^{\infty }{\frac {a_{i}}{10^{i}}},}$

which is called the fractional part of r.


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# Definition:Pi/Linguistic Note

## Linguistic Note on $\pi$ (Pi)

While the conventional contemporary prounciation of $\pi$ in Western English is pie, it is worth noting that the "correct" Greek pronunciation of the name of the letter $\pi$ is in fact the same as the letter p is pronounced in English.

It is just as well that $\pi$ is pronounced pie, otherwise the opportunity for confusion between $\pi$ and $p$ in spoken language would be too great.


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Math Help - Maximal ideals in Banach algebras

1. Maximal ideals in Banach algebras

Hi!
I've a little problem with understanding proof of the following theorem:

In the red frames are this steps with which I've problems. In the first one I don't know from where we know that X is a zero set. In the second frame - how can I explain this in the simplest way.
Maybe there are simple questions but I'm rather algebraist and I always have problems with functionala analysis
I'll be grateful for any help.
Best regards

2. For the first question, if X has codimension 1, it means that you can fix an element $z\notin X$ and then any element y in A can be uniquely expressed in the form $y = x + \lambda z$, where $x\in X$ and $\lambda$ is a scalar. Define $f(y) = \lambda$. Then f is a linear functional whose zero set is X.

For the second question, if $y=x-f(x)e$ then $f(y) = f(x) -f(x)=0$ (since f(e)=1). Thus y is in the zero set of f, which by hypothesis consists of non-invertible elements. But if an element is non-invertible then 0 is in its spectrum.


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# Difference between revisions of "2003 AIME I Problems/Problem 4"

## Problem

Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$

## Solution

Using the properties of logarithms, we can simplify the first equation to $\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1$. Therefore, $$\sin x \cos x = \frac{1}{10}.\qquad (*)$$

Now, manipulate the second equation. \begin{align*} \log_{10} (\sin x + \cos x) &= \frac{1}{2}(\log_{10} n - \log_{10} 10) \\ \log_{10} (\sin x + \cos x) &= \left(\log_{10} \sqrt{\frac{n}{10}}\right) \\ \sin x + \cos x &= \sqrt{\frac{n}{10}} \\ (\sin x + \cos x)^{2} &= \left(\sqrt{\frac{n}{10}}\right)^2 \\ \sin^2 x + \cos^2 x +2 \sin x \cos x &= \frac{n}{10} \\ \end{align*}

By the Pythagorean identities, $\sin ^2 x + \cos ^2 x = 1$, and we can substitute the value for $\sin x \cos x$ from $(*)$. $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$.


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Department of

# Mathematics

Seminar Calendar
for events the day of Friday, February 25, 2022.

.
events for the
events containing

Questions regarding events or the calendar should be directed to Tori Corkery.
     January 2022          February 2022            March 2022
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
1          1  2  3  4  5          1  2  3  4  5
2  3  4  5  6  7  8    6  7  8  9 10 11 12    6  7  8  9 10 11 12
9 10 11 12 13 14 15   13 14 15 16 17 18 19   13 14 15 16 17 18 19
16 17 18 19 20 21 22   20 21 22 23 24 25 26   20 21 22 23 24 25 26
23 24 25 26 27 28 29   27 28                  27 28 29 30 31
30 31

Friday, February 25, 2022

1:00 pm in Zoom,Friday, February 25, 2022

#### Dimension distortion and applications

###### Efstathios Konstantinos Chrontsios Garitsis (UIUC Math)

Abstract: Since Hausdorff dimension was first introduced in 1918, many different notions of dimension have been defined and used throughout many areas of Mathematics. An interesting topic has always been the distortion of said dimensions of a given set under a specific class of mappings. More specifically, Gehring and V\"ais\"al\"a proved in 1973 a theorem concerning the distortion of Hausdorff dimension under quasiconformal maps, while Kaufman in 2000 proved the analogous result for Box-counting dimension. In this talk, an introduction to the different types of dimensions will be presented, along with the results of of Gehring, V\"ais\"al\"a and Kaufman. We will then proceed to discuss analogous theorems we proved for the Assouad dimension and spectrum, which describe how K-quasiconformal maps change these notions of a given subset of $\mathbb{R}^n$. We will conclude the talk by demonstrating how said theorems can be applied to fully classify polynomial spirals up to quasiconformal equivalence. Contact: ryanm12@illinois.edu for Zoom coordinates.

1:00 pm in Zoom,Friday, February 25, 2022

#### Dimension distortion and applications

###### Efstathios Konstantinos Chrontsios Garitsis (UIUC Math)

Abstract: Since Hausdorff dimension was first introduced in 1918, many different notions of dimension have been defined and used throughout many areas of Mathematics. An interesting topic has always been the distortion of said dimensions of a given set under a specific class of mappings. More specifically, Gehring and V\"ais\"al\"a proved in 1973 a theorem concerning the distortion of Hausdorff dimension under quasiconformal maps, while Kaufman in 2000 proved the analogous result for Box-counting dimension. In this talk, an introduction to the different types of dimensions will be presented, along with the results of of Gehring, V\"ais\"al\"a and Kaufman. We will then proceed to discuss analogous theorems we proved for the Assouad dimension and spectrum, which describe how K-quasiconformal maps change these notions of a given subset of $\mathbb{R}^n$. We will conclude the talk by demonstrating how said theorems can be applied to fully classify polynomial spirals up to quasiconformal equivalence. Contact: ryanm12@illinois.edu for Zoom coordinates.

3:00 pm in 341 Altgeld Hall,Friday, February 25, 2022


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# First-order reaction

First-order reaction may refer to:

A--> product

$\frac{dC}{dt} =-kC$

Solving the above differential equation, we get

$C=C_ie^{-kt}$

From Arrhenius equation,

$k=k_0e^{\frac{-E_a}{RT}}$

     $t$   = time in seconds (s)
$C$  = the concentration of the reactant (mol/m^3, other units are also used, e.g. kg/m^3)
$C_i$  = the initial concentration of the reactant ($mol m^{-3}$)
$k_0$ = pre-exponential factor (1/s)
$E_a$ = the activation energy (J/mol)
$k$  = reaction rate constant (1/s) which is usually described by the Arrhenius equation: .
$R$  = the universal gas constant (J/mol.K, K is temperature in kelvin not to be confused with little k, the reaction constant)


text:
Blog Entries: 27
Recognitions:
Gold Member
Homework Help

## Geometry and trig proofs, with diagrams

 PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract

 Quote by http://www.mathsisfun.com/geometry/degrees.html Why 360 degrees? Probably because old calendars (such as the Persian Calendar) used 360 days for a year - when they watched the stars they saw them revolve around the North Star one degree per day.
I thought that was a fairly interesting statement. If I was a student, I would want my professor to state things like this. Liekwise, if I was a professor, I would introduce things like this to my class.
 Admin Theorems and Postulates for Geometry http://regentsprep.org/regents/math/...B/theorems.htm http://regentsprep.org/regents/math/...h-GEOMETRY.htm

## Geometry and trig proofs, with diagrams

I was messing with the triangle on the interactive thing, and made the triangle just a line. It said the line is an obtuse isosceles triangle. Really? Is "obtuse isosceles triangle" really another way to say "a line"?

Edit: I was thinking about this and I think the definition of triangle should be (if it's not already) a shape with 3 angles, each greater than 0.

 Quote by leroyjenkens I was messing with the triangle on the interactive thing, and made the triangle just a line. It said the line is an obtuse isosceles triangle. Really? Is "obtuse isosceles triangle" really another way to say "a line"? Edit: I was thinking about this and I think the definition of triangle should be (if it's not already) a shape with 3 angles, each greater than 0.
I haven't used this tool, so I can't check what you were looking at. However, in mathematics you will find many examples of what are called "degenerate" forms of familiar objects. Yes, you can flatten a triangle into a straight line and it is still technically a triangle. You can also squash a cube into a flat plane and you can do many other strange things. When first learning the subject, these special cases are just confusing so teachers avoid them. However, they turn out to be important further on in mathematics so it is useful to get comfortable with degenerate cases of geometric and algebraic objects.

 Tags geometry proofs

 Similar discussions for: Geometry and trig proofs, with diagrams Thread Forum Replies Precalculus Mathematics Homework 5 General Math 2 Differential Geometry 5 Precalculus Mathematics Homework 2 General Math 6


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# In terms of covariance matrices, are partial measurement and partial trace equivalent?

Partial measurement and partial trace

There is a connection between a measurement of a part of a system and tracing this subsystem out. Say, we have a system composed of subsystems $A$ and $B$ in a pure state $|\psi\rangle = \sum_{i,j}a_{ij}|i\rangle_A|j\rangle_B$ and do a projective measurement on part $B$, projecting it on a state $|k\rangle_B$. The resulting state of subsystem $A$, conditioned on the measurement result, will be (up to normalization) $$|\psi^{(k)}\rangle_A = \sum_i a_{ik}|i\rangle_A.$$ Taking partial trace over the subsystem $B$ then corresponds to taking a statistical mixture of all possible measurement outcomes (this can be shown by a simple explicit calculation), $$\mathrm{Tr}_B(|\psi\rangle\langle\psi|) = \sum_{k}|\psi^{(k)}\rangle_A\langle\psi^{(k)}|.$$

Gaussian states

When dealing with Gaussian states in quantum optics [i.e., states with Gaussian phase space representation, where the phase space is spanned by quadrature operators $x = a+a^\dagger$, $p = i(a^\dagger-a)$], one often uses the description by first and second statistical moments. The first moment is, as far as entanglement is concerned, irrelevant and one then works just with the covariance matrix, i.e., the second moment. It can then be shown that a partial measurement affects the covariance matrix of the remaining subsystem in such a way that the resulting covariance matrix does not depend on the particular measurement result.

The question

Does this mean that, when dealing with covariance matrices of Gaussian states, partial measurement and partial trace are effectively the same? This is something one could intuitively (and perhaps naively) expect from the observations above. However, there is a difference in the purity of the state after partial measurement (pure for pure initial state) and partial trace (generally mixed). For Gaussian states, state purity is directly connected to determinant of the covariance matrix (the purity scales as $P\propto 1/\sqrt{\det\gamma}$, where $\gamma$ is the covariance matrix). Therefore, there must be a difference between covariance matrix after a partial measurement and partial trace. What am I missing?

-

In fact, the covariance matrix after partial trace is given by the respective block in the original covariance matrix. If we have a covariance matrix of the composite system in the block form given by $$\gamma_\mathrm{AB} = \left(\begin{array}{cc} A&C\\C^T&B\end{array}\right),$$ the covariance matrix of subsystem A after tracing out subsystem B is given by the block $A$.
Is there a place where this calculation is done explicitly? In particular, I'd like a rigid framework for translating back and forth between Gaussian POVMs and operations on the $(d,\gamma)$ representation, where $d$ is the first moment and $\gamma$ is the second. –  Jess Riedel May 27 '14 at 20:52


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# Disdyakis dodecahedron

Disdyakis dodecahedron

Click on picture for large version

Spinning version

Type Catalan solid
Conway notation mC
Coxeter diagram
Face polygon
scalene triangle
Faces 48
Edges 72
Vertices 26 = 6 + 8 + 12
Face configuration V4.6.8
Symmetry group Oh, B3, [4,3], *432
Dihedral angle 155° 4' 56"
${\displaystyle \arccos(-{\frac {71+12{\sqrt {2}}}{97}})}$
Dual polyhedron truncated cuboctahedron
Properties convex, face-transitive

Net

In geometry, a disdyakis dodecahedron, (also hexoctahedron[1], hexakis octahedron, octakis cube, octakis hexahedron, kisrhombic dodecahedron[2]), is a Catalan solid with 48 faces and the dual to the Archimedean truncated cuboctahedron. As such it is face-transitive but with irregular face polygons. It superficially resembles an inflated rhombic dodecahedron—if one replaces each face of the rhombic dodecahedron with a single vertex and four triangles in a regular fashion one ends up with a disdyakis dodecahedron. More formally, the disdyakis dodecahedron is the Kleetope of the rhombic dodecahedron. It is the net of a rhombic dodecahedral pyramid.

## Symmetry

It has Oh octahedral symmetry. Its collective edges represent the reflection planes of the symmetry. It can also be seen in the corner and mid-edge triangulation of the regular cube and octahedron, and rhombic dodecahedron.

 Disdyakis dodecahedron Spherical Cubic Octahedral Rhombic dodecahedral

Seen in stereographic projection the edges of the disdyakis dodecahedron form 9 circles (or centrally radial lines) in the plane. The 9 circles can be divided into two groups of 3 and 6 (drawn in purple and red), representing in two orthogonal subgroups: [2,2], and [3,3]:

Orthogonal Stereographic
[4] [3] [2]

## Dimensions

If its smallest edges have length a, its surface area and volume are

{\displaystyle {\begin{aligned}A&={\tfrac {6}{7}}{\sqrt {783+436{\sqrt {2}}}}\,a^{2}\\V&={\tfrac {1}{7}}{\sqrt {3\left(2194+1513{\sqrt {2}}\right)}}a^{3}\end{aligned}}}

## Orthogonal projections

The truncated cuboctahedron and its dual, the disdyakis dodecahedron can be drawn in a number of symmetric orthogonal projective orientations. Between a polyhedron and its dual, vertices and faces are swapped in positions, and edges are perpendicular.

 Projective Image Dual symmetry image [4] [3] [2] [2] [2] [2] [2]+

## Related polyhedra and tilings

 Polyhedra similar to the disdyakis dodecahedron are duals to the Bowtie octahedron and cube, containing extra pairs triangular faces .[3]

The disdyakis dodecahedron is one of a family of duals to the uniform polyhedra related to the cube and regular octahedron.

It is a polyhedra in a sequence defined by the face configuration V4.6.2n. This group is special for having all even number of edges per vertex and form bisecting planes through the polyhedra and infinite lines in the plane, and continuing into the hyperbolic plane for any n ≥ 7.

With an even number of faces at every vertex, these polyhedra and tilings can be shown by alternating two colors so all adjacent faces have different colors.

Each face on these domains also corresponds to the fundamental domain of a symmetry group with order 2,3,n mirrors at each triangle face vertex.


text:
# yt.testing module¶

Utilities to aid testing.

yt.testing.amrspace(extent, levels=7, cells=8)[source]

Creates two numpy arrays representing the left and right bounds of an AMR grid as well as an array for the AMR level of each cell.

Parameters: extent (array-like) – This a sequence of length 2*ndims that is the bounds of each dimension. For example, the 2D unit square would be given by [0.0, 1.0, 0.0, 1.0]. A 3D cylindrical grid may look like [0.0, 2.0, -1.0, 1.0, 0.0, 2*np.pi]. levels (int or sequence of ints, optional) – This is the number of AMR refinement levels. If given as a sequence (of length ndims), then each dimension will be refined down to this level. All values in this array must be the same or zero. A zero valued dimension indicates that this dim should not be refined. Taking the 3D cylindrical example above if we don’t want refine theta but want r and z at 5 we would set levels=(5, 5, 0). cells (int, optional) – This is the number of cells per refinement level. left (float ndarray, shape=(npoints, ndims)) – The left AMR grid points. right (float ndarray, shape=(npoints, ndims)) – The right AMR grid points. level (int ndarray, shape=(npoints,)) – The AMR level for each point.

Examples

>>> l, r, lvl = amrspace([0.0, 2.0, 1.0, 2.0, 0.0, 3.14], levels=(3,3,0), cells=2)
>>> print l
[[ 0.     1.     0.   ]
[ 0.25   1.     0.   ]
[ 0.     1.125  0.   ]
[ 0.25   1.125  0.   ]
[ 0.5    1.     0.   ]
[ 0.     1.25   0.   ]
[ 0.5    1.25   0.   ]
[ 1.     1.     0.   ]
[ 0.     1.5    0.   ]
[ 1.     1.5    0.   ]]

yt.testing.assert_allclose_units(actual, desired, rtol=1e-07, atol=0, **kwargs)[source]

Raise an error if two objects are not equal up to desired tolerance

This is a wrapper for numpy.testing.assert_allclose() that also verifies unit consistency

Parameters: actual (array-like) – Array obtained (possibly with attached units) desired (array-like) – Array to compare with (possibly with attached units) rtol (float, optional) – Relative tolerance, defaults to 1e-7 atol (float or quantity, optional) – Absolute tolerance. If units are attached, they must be consistent with the units of actual and desired. If no units are attached, assumes the same units as desired. Defaults to zero.

Notes

Also accepts additional keyword arguments accepted by numpy.testing.assert_allclose(), see the documentation of that function for details.

yt.testing.assert_fname(fname)[source]

Function that checks file type using libmagic

yt.testing.assert_rel_equal(a1, a2, decimals, err_msg='', verbose=True)[source]
yt.testing.check_results(func)[source]

This is a decorator for a function to verify that the (numpy ndarray) result of a function is what it should be.

This function is designed to be used for very light answer testing. Essentially, it wraps around a larger function that returns a numpy array, and that has results that should not change. It is not necessarily used inside the testing scripts themselves, but inside testing scripts written by developers during the testing of pull requests and new functionality. If a hash is specified, it “wins” and the others are ignored. Otherwise, tolerance is 1e-8 (just above single precision.)

The correct results will be stored if the command line contains –answer-reference , and otherwise it will compare against the results on disk. The filename will be func_results_ref_FUNCNAME.cpkl where FUNCNAME is the name of the function being tested.

If you would like more control over the name of the pickle file the results are stored in, you can pass the result_basename keyword argument to the function you are testing. The check_results decorator will use the value of the keyword to construct the filename of the results data file. If result_basename is not specified, the name of the testing function is used.

This will raise an exception if the results are not correct.

Examples

>>> @check_results
... def my_func(ds):
...     return ds.domain_width

>>> my_func(ds)

>>> @check_results
... def field_checker(dd, field_name):
...     return dd[field_name]

>>> field_checker(ds.all_data(), 'density', result_basename='density')

yt.testing.disable_dataset_cache(func)[source]
yt.testing.expand_keywords(keywords, full=False)[source]

expand_keywords is a means for testing all possible keyword arguments in the nosetests. Simply pass it a dictionary of all the keyword arguments and all of the values for these arguments that you want to test.

It will return a list of kwargs dicts containing combinations of the various kwarg values you passed it. These can then be passed to the appropriate function in nosetests.

If full=True, then every possible combination of keywords is produced, otherwise, every keyword option is included at least once in the output list. Be careful, by using full=True, you may be in for an exponentially larger number of tests!

Parameters: keywords (dict) – a dictionary where the keys are the keywords for the function, and the values of each key are the possible values that this key can take in the function full (bool) – if set to True, every possible combination of given keywords is returned An array of dictionaries to be individually passed to the appropriate function matching these kwargs. array of dicts

Examples

>>> keywords = {}
>>> keywords['dpi'] = (50, 100, 200)
>>> keywords['cmap'] = ('arbre', 'kelp')
>>> list_of_kwargs = expand_keywords(keywords)
>>> print list_of_kwargs

array([{‘cmap’: ‘arbre’, ‘dpi’: 50},
{‘cmap’: ‘kelp’, ‘dpi’: 100}, {‘cmap’: ‘arbre’, ‘dpi’: 200}], dtype=object)
>>> list_of_kwargs = expand_keywords(keywords, full=True)
>>> print list_of_kwargs

array([{‘cmap’: ‘arbre’, ‘dpi’: 50},
{‘cmap’: ‘arbre’, ‘dpi’: 100}, {‘cmap’: ‘arbre’, ‘dpi’: 200}, {‘cmap’: ‘kelp’, ‘dpi’: 50}, {‘cmap’: ‘kelp’, ‘dpi’: 100}, {‘cmap’: ‘kelp’, ‘dpi’: 200}], dtype=object)
>>> for kwargs in list_of_kwargs:
...     write_projection(*args, **kwargs)

yt.testing.fake_amr_ds(fields=('Density', ), geometry='cartesian', particles=0)[source]
yt.testing.fake_hexahedral_ds()[source]
yt.testing.fake_particle_ds(fields=('particle_position_x', 'particle_position_y', 'particle_position_z', 'particle_mass', 'particle_velocity_x', 'particle_velocity_y', 'particle_velocity_z'), units=('cm', 'cm', 'cm', 'g', 'cm/s', 'cm/s', 'cm/s'), negative=(False, False, False, False, True, True, True), npart=4096, length_unit=1.0)[source]
yt.testing.fake_random_ds(ndims, peak_value=1.0, fields=('density', 'velocity_x', 'velocity_y', 'velocity_z'), units=('g/cm**3', 'cm/s', 'cm/s', 'cm/s'), particle_fields=None, particle_field_units=None, negative=False, nprocs=1, particles=0, length_unit=1.0, unit_system='cgs', bbox=None)[source]
yt.testing.fake_tetrahedral_ds()[source]
yt.testing.fake_vr_orientation_test_ds(N=96, scale=1)[source]

create a toy dataset that puts a sphere at (0,0,0), a single cube on +x, two cubes on +y, and three cubes on +z in a domain from [-1*scale,1*scale]**3. The lower planes (x = -1*scale, y = -1*scale, z = -1*scale) are also given non-zero values.

This dataset allows you to easily explore orientations and handiness in VR and other renderings

Parameters: N (integer) – The number of cells along each direction scale (float) – A spatial scale, the domain boundaries will be multiplied by scale to test datasets that have spatial different scales (e.g. data in CGS units)
yt.testing.periodicity_cases(ds)[source]
yt.testing.requires_file(req_file)[source]
yt.testing.requires_module(module)[source]

Decorator that takes a module name as an argument and tries to import it. If the module imports without issue, the function is returned, but if not, a null function is returned. This is so tests that depend on certain modules being imported will not fail if the module is not installed on the testing platform.

yt.testing.run_nose(verbose=False, run_answer_tests=False, answer_big_data=False, call_pdb=False, module=None)[source]
yt.testing.small_fake_hexahedral_ds()[source]
yt.testing.units_override_check(fn)[source]


text:
# SLATEC rouitne dslucs() and MKL correspondence

I am looking for a routine (or set of routines) in the Intel MKL that that can replace dslucs (Incomplete LU BiConjugate Gradient Squared Ax=b Solver) in SLATEC (http://www.netlib.org/slatec/lin/).

Looking at the MKL documentation, I see that MKL has only CG and GMRES solvers, along with ILU preconditioners. It may be that the GMRES solver will be adequate for your purposes.


text:
# I'm not able to leave a certain selection tool

I did, somehow (think I pressed a hotkey), end up with this circle around the mouse cursor, which could select multiple vertices. I do not know how to leave this mode/tool, and I'm not able to do anything but to use this certain tool. I can't use shortcuts, for example cmd+S to save. I cant open any menu, I cant rotate the view, I cant do anything but to change the size of the circle (using the scroll wheel on the mouse) and to select vertices using it. No view hotkeys works (1-9)

This is what the tool looks like:

Would love to at least get to know the name of the tool I'm stuck with, so that I could ask a more accurate question. Since I can't save the file, I don't want to close it. Therefore, if I could get some help soon, that would be very appreciated.

• That looks like circle select. Just press escape or right click Apr 8 '17 at 14:00
• right mouse button to escape circle select mode Feb 6 at 22:00

Try pressing "W", it will change the pointer type from select circle to select box, hope that helps.

i accidentally pressed it and had the same issue, pressing it again fixed it for me.

• this was the answer for my problem. It had gotten stuck in circle select mode, which seems to be different than the mode that you get when pressing "C" Aug 10 '19 at 20:09

That is the likely the circle select tool. Quoting from the page:

To cancel the selection use RMB or key Esc or Enter.

• Nah, it's a different circle select mode. It's not the same as the one you get when you press C (Escape and Enter were the first things I tried). -- Instead to get out of it you have to press W as the other answer suggests. Nov 10 '20 at 8:52

in 2.8 press and hold on the selector in the top left until the options appear

For anyone who still has this issue you need to go to the top where it says "Drag" next to "Drag" should be a drop down box, simply pick "box tool" or "Select" or whatever you need to use to go back to the normal tool selection. Hope this helps! :))

Press Esc key to exit, the C select mode


text:
# Thread: Second Order Differential Equation

1. ## Second Order Differential Equation

how to solve the second order differential equation in the form:
ay" + by' + cy = R(x)?

the example is this:
(d2i/dt2) + 6(di/dt) +25i = -292sin(4t)

First you find the general solution of the corresponding homogeneous equations ie. $ay'' + by' + cy = 0$.
Then you look for a particular solution, since in this case $g(t) = -292\sin{4t}$ we can guess that the particular solution will be of the form $Y(t) = A\cos{4t} + B\sin{4t}$. Now you solve for $A$ and $B$ obviously, by substituting in the appropriate terms ie. $y''$ and $y'$ and $y$ into your original differential equation. Then you know the general form of the solution of nonhomogeneous equation is simply $y = \phi(t) = c_1y_1(t) + c_2y_2(t) + Y(t)$, where $y_1(t)$ and $y_2(t)$ are solutions of the homogeneous equation and $Y(t)$ is the particular solution we guessed and solved the constants $A$ and $B$ for. And $c_1$ and $c_2$ are arbitrary constants.


text:
# .h5 file size is same before and after training?

learner = ConvLearner.pretrained(arch, md, ps=0.5) #dropout 50%
learner.fit(lr,1)
learner.save('ResNet34_256_1')

h5 file in load and save is having same size. Should it increase after training? How do I know that saved model is better than loaded on?

This is not surpising. h5 is the save file of the model's weights. The number of weights does not change before and after training (they are modified, though), therefore, your file should have the same size.

To know if your model after training is better, you have to measure it : you have to make predictions on a test set, and measure it, with usual metrics like accuracy or F score and so on. look at this to see usual metrics used in machine learning.

You're basically just modifying the weights you've loaded from the initial .h5 file during training, thus, you will end up with a file of same size but with different weights. The size of the file will change if you modify the construction of the network model in your code. Imagine a model having the weights

[0.1234567, 0.2345678, 0.3456789]

saved as an .h5 file. When you load this file together with the neural network model file (.json) and retrain it, it modifies the weights based on your training data. In our example, let's say the weights were modified into

[0.1212121, 0.2323321, 0.1221231]

When you save the modified weights into a new .h5 file, it will have the same size, just different values.


text:
contributes 3 of its valence electrons to form three bond pairs with chlorine Textbook content produced by OpenStax is licensed under a = 2.1) and oxygen (e.n. Coordinate covalent bond: Some times, In Lewis terms a covalent bond is a shared electron pair. triple bond is formed when the atoms share three bond pairs. covalent bond. * Therefore, during the formation of H2 molecule, each hydrogen atom contributes one electron In the case of CH2O, the O and C atoms share two pairs of electrons, with the following Lewis electron dot diagram as a result: By circling the electrons around each atom, we can now see that the O and C atoms have octets, while each H atom has two electrons: Each valence shell is full, so this is an acceptable Lewis electron dot diagram. 1.7. The two H atoms can share their electrons: We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom's valence shell: Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. electrons to form 3 bond pairs, which in turn are shared by two nitrogen atoms. Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration. 6. • Draw Lewis dot diagrams to represent valence electrons in elements and draw Lewis dot structures to show covalent bonding. covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may There is no single vitamin B; there is a group of substances called the B complex vitamins that are all water soluble and participate in cell metabolism. by only one atom This is called as coordinate covalent * In the formation of water molecule, the oxygen atom contributes two of its However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. 12.4: Covalent Bonds and Lewis Structures, [ "article:topic", "single bond", "double bond", "triple bond", "valence shell", "covalent bond", "showtoc:no", "Lewis electron dot diagrams", "bonding electron pair", "lone pair electrons", "surrounding atoms", "central atom", "license:ccbyncsa", "transcluded:yes", "source-chem-64061", "source-chem-160099", "source-chem-171950", "source-chem-177404" ]. citation tool such as, Authors: Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson, PhD. 2. to form a pair of electrons. We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. A covalent bond is formed between two atoms when their electronegativity Covalent bond –Lewis Langmuir concept :. * Remember that, in the Lewis dot structures, only the * There are no lone pairs on carbon in methane. during the formation of covalent bond, the shared pair is entirely contributed Whereas, the atom with less electronegativity gets * Lewis theory could not explain the geometry of molecules and bond angle in Note: The beryllium atom may lose two of its valence electrons to get the

configuration of hydrogen is 1s1. hydrogen (e.n. Thus it gets octet Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. It forms 4 bonds with four hydrogen atoms. Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. Note that each hydrogen gets two electrons after forming the bond. To explain the formation of covalent bond, a simple qualitative model was developed by * The electronic configuration of nitrogen is [He]2s22p3. Draw a skeleton joining the atoms by single bonds. = 2.1) is not considerable. 5. * This is a qualitative explanation for covalent bond only. Interestingly, most minerals are consumed in ionic form, rather than as elements or from covalent molecules. Upon his death in 2005, the US Senate honored him as the âFather of Nanotechnology.â (credit: United States Department of Energy), https://openstax.org/books/chemistry-2e/pages/1-introduction, https://openstax.org/books/chemistry-2e/pages/7-3-lewis-symbols-and-structures, Creative Commons Attribution 4.0 International License, Write Lewis symbols for neutral atoms and ions, Draw Lewis structures depicting the bonding in simple molecules. contributed by the atom of an element in the formation of covalent compound is
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# IT Blog

* The carbon atom contributes four of its valence electrons, whereas each slightly towards the atom with higher electronegativity by creating partial (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:). Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration. Note: The bond between boron (e.n. There is a simple set of steps for determining the Lewis electron dot diagram of a simple molecule. * However there are two lone pairs on each oxygen atom. This is known as a ‘single covalent bond.’ Notice how the two electrons can be found in a region of space between the two atomic nuclei. electronegativity difference is zero. bond. excess of electron density and hence carry no charge. After the central and surrounding atoms have been identified, follow these steps: Let us try these steps to determine the electron dot diagram for BF4−. * Thus Cl2 molecule is formed with a covalent bond between two The them. * The electronic configuration of hydrogen is 1s1. need of one electron to complete the shell. * There is also one lone pair on nitrogen atom. * Covalency of phosphorus in this molecule is 5. The hydrogen atom gets partial positive charge, whereas the chlorine atom Hence the bond between them has You can find more For example, NH3 reacts with BF3 because the lone pair on nitrogen can be shared with the boron atom: Elements in the second period of the periodic table (n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2s and three 2p orbitals). Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules. The other halogen molecules (F2, Br2, I2, and At2) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. However the Beryllium chloride is a relatively stable molecule. Thus two bonds are formed by oxygen atom to get the configuration of An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C60. * The electronic configuration of hydrogen is 1s1. There are three such lone If we were to use lines to represent the bonds, we would use two lines between the C and O atoms: The bond between the C and O atoms is a double bond and represents two bonding pairs of electrons between the atoms. This is how a water molecule, H2O, is made. The Helium atom with 2 The distance between the nuclei of two atoms when their potential energy polar covalent bond. When electrons are shared between two atoms, they form a covalent bond. these two atoms are maximum at this point. Creative Commons Attribution License 4.0 license. the bond pair is no longer shared equally between the atoms. * Electronic configuration of chlorine is [Ne]3s23p5. The sharing of pair of electrons between two atoms is  referred to as a covalent bond. Lewis dot diagrams are used to represent valence electrons in an element. Note that each hydrogen gets two electrons after forming the bond. Add extra if the species has negative charges and remove some for every positive charge on the species. * The electronic configuration of Cl is [Ne]3s23p5. Essential minerals that the body needs in tiny quantities (so-called trace elements) include manganese, iron, cobalt, nickel, copper, zinc, molybdenum, selenium, and iodine. Put a pair of electrons between the central atom and each surrounding atom. One example is HF. However, the BâF bonds are slightly shorter than what is actually expected for BâF single bonds, indicating that some double bond character is found in the actual molecule. = 2.0) and chlorine (e.n. Calculate the number of valence electrons: XeF 2: 8 + (2 × × 7) = 22 XeF 6: 8 + (6 × × 7) = 50; Step 2. * The Lewis dot structure for H2 molecule is shown below. Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds: As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. Thus nitrogen forms 3 single bonds with three hydrogen atoms and gets the It requires one electron to Hence it There is another mechanism for obtaining a complete valence shell: sharing electrons. Minerals that the body needs in quantity include sodium, potassium, magnesium, calcium, phosphorus, sulfur, and chlorine. For example, in the Lewis structures of beryllium dihydride, BeH2, and boron trifluoride, BF3, the beryllium and boron atoms each have only four and six electrons, respectively. It
electrons in its outer shell is also stable. hydrogen (e.n. octets. * Thus there are 10 electrons in the valence shell of phosphorus in this referred to as polar * In the formation of dioxygen molecule, each oxygen atom contributes 2 bond. Note: The O-H bond is also considerably polar since the electronegativity difference between

carbon (e.n. However, still this molecule is stable. Richard Smalley (1943â2005), a professor of physics, chemistry, and astronomy at Rice University, was one of the leading advocates for fullerene chemistry. The OpenStax name, OpenStax logo, OpenStax book * In Cl2 molecule, each Cl atom gets 8 electrons in its outer Hydrogen needs only two electrons to fill its valence, which it gets through the covalent bond. Each H atom starts with a single electron in its valence shell: $\mathbf{H\, \cdot }\; \; \; \; \; \mathbf{\cdot \: H}\nonumber$. The central atom is usually written first in the formula of the compound (H2O is the notable exception). The word vitamin comes from "vital amine" because it was once thought that all these compounds had an amine group (NH2) in it. Being highly electronegative than hydrogen, the nitrogen atom gets partial This is an acceptable Lewis electron dot diagram for CO2. with six fluorine atoms. Vitamins are nutrients that our bodies need in small amounts but cannot synthesize; therefore, they must be obtained from the diet. Molecules formed from these elements are sometimes called hypervalent molecules. molecule. Note that each F atom has a complete octet around it now: We can also write this using a dash to represent the shared electron pair: There are two different types of electrons in the fluorine diatomic molecule. nearest inert gas: Helium's configuration and form Be2+ ion. Free LibreFest conference on November 4-6! configuration. These molecules fall into three categories: We call molecules that contain an odd number of electrons free radicals. This book is Creative Commons Attribution License The bond Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium: Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur: Figure 7.10 demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH2O (formaldehyde) and between the two carbon atoms in C2H4 (ethylene): A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CNâ): For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms.

contributes 3 of its valence electrons to form three bond pairs with chlorine Textbook content produced by OpenStax is licensed under a = 2.1) and oxygen (e.n. Coordinate covalent bond: Some times, In Lewis terms a covalent bond is a shared electron pair. triple bond is formed when the atoms share three bond pairs. covalent bond. * Therefore, during the formation of H2 molecule, each hydrogen atom contributes one electron In the case of CH2O, the O and C atoms share two pairs of electrons, with the following Lewis electron dot diagram as a result: By circling the electrons around each atom, we can now see that the O and C atoms have octets, while each H atom has two electrons: Each valence shell is full, so this is an acceptable Lewis electron dot diagram. 1.7. The two H atoms can share their electrons: We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom's valence shell: Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. electrons to form 3 bond pairs, which in turn are shared by two nitrogen atoms. Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration. 6. • Draw Lewis dot diagrams to represent valence electrons in elements and draw Lewis dot structures to show covalent bonding. covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may There is no single vitamin B; there is a group of substances called the B complex vitamins that are all water soluble and participate in cell metabolism. by only one atom This is called as coordinate covalent * In the formation of water molecule, the oxygen atom contributes two of its However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. 12.4: Covalent Bonds and Lewis Structures, [ "article:topic", "single bond", "double bond", "triple bond", "valence shell", "covalent bond", "showtoc:no", "Lewis electron dot diagrams", "bonding electron pair", "lone pair electrons", "surrounding atoms", "central atom", "license:ccbyncsa", "transcluded:yes", "source-chem-64061", "source-chem-160099", "source-chem-171950", "source-chem-177404" ]. citation tool such as, Authors: Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson, PhD. 2. to form a pair of electrons. We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. A covalent bond is formed between two atoms when their electronegativity Covalent bond –Lewis Langmuir concept :. * Remember that, in the Lewis dot structures, only the * There are no lone pairs on carbon in methane. during the formation of covalent bond, the shared pair is entirely contributed Whereas, the atom with less electronegativity gets * Lewis theory could not explain the geometry of molecules and bond angle in Note: The beryllium atom may lose two of its valence electrons to get the

configuration of hydrogen is 1s1. hydrogen (e.n. Thus it gets octet Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. It forms 4 bonds with four hydrogen atoms. Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. Note that each hydrogen gets two electrons after forming the bond. To explain the formation of covalent bond, a simple qualitative model was developed by * The electronic configuration of nitrogen is [He]2s22p3. Draw a skeleton joining the atoms by single bonds. = 2.1) is not considerable. 5. * This is a qualitative explanation for covalent bond only. Interestingly, most minerals are consumed in ionic form, rather than as elements or from covalent molecules. Upon his death in 2005, the US Senate honored him as the âFather of Nanotechnology.â (credit: United States Department of Energy), https://openstax.org/books/chemistry-2e/pages/1-introduction, https://openstax.org/books/chemistry-2e/pages/7-3-lewis-symbols-and-structures, Creative Commons Attribution 4.0 International License, Write Lewis symbols for neutral atoms and ions, Draw Lewis structures depicting the bonding in simple molecules. contributed by the atom of an element in the formation of covalent compound is


text:
My watch list
my.chemeurope.com

# Thermionic emission

Thermionic emission is the flow of charge carriers from a surface or over some other kind of electrical potential barrier, caused by thermal vibrational energy overcoming the electrostatic forces restraining the charge carriers. The charge carriers can be electrons or ions, and are sometimes referred to as "thermions". The total charge of the emitted carriers (either positive or negative) will be equal in magnitude and opposite in sign to the charge left in the emitting region. The most classical example of thermionic emission is the emission of electrons from a hot metal cathode into a vacuum (archaically known as the Edison effect), but the term is now used to refer to any thermally excited charge emission process, even when the charge is emitted from one solid-state region into another. This process is crucially important in the operation of a variety of electronic devices and can be used for power generation or cooling. The magnitude of the charge flow increases dramatically with increasing temperature and for vacuum emission from metals tends to only become significant for temperatures over 1000 K. The science dealing with this phenomenon is thermionics.

## History

The phenomenon was initially reported in 1873 by Daniel Lordan in Britain. While doing work on charged objects, Lordan discovered that a red-hot iron sphere with a negative charge would lose its charge (discharging electrons into vacuum). He also found that this did not happen if the sphere had a positive charge. He didn't understand what any of this meant.[citation needed] Other early contributors included Hittorf (1869–1883), Goldstein (1885), and Elster and Geitel (1882–1889).

The effect was rediscovered by Thomas Edison on February 13, 1880, while trying to discover the reason for breakage of lamp filaments and uneven blackening (darkest near one terminal of the filament) of the bulbs in his incandescent lamps.

Edison built several experiment bulbs, some with an extra wire, a metal plate, or foil inside the bulb which was electrically separate from the filament. He connected the extra metal electrode to the lamp filament through a galvanometer. When the foil was given a more negative charge than the filament, no current flowed between the foil and the filament because the cool foil emitted few electrons. However, when the foil was given a more positive charge than the filament, the many electrons emitted from the hot filament were attracted to the foil, causing current to flow. This one-way flow of current was called the Edison effect (although the term is occasionally used to refer to thermionic emission itself). He found that the current emitted by the hot filament increased rapidly with increasing voltage, and filed a patent application for a voltage regulating device using the effect on November 15, 1883 (U.S. patent 307,031, the first US patent for an electronic device). He found that sufficient current would pass through the device to operate a telegraph sounder. This was exhibited at the International Electrical Exposition in Philadelphia in September 1884. William Preece, a British scientist took back with him several of the Edison Effect bulbs, and presented a paper on them in 1885, where he referred to thermionic emission as the "Edison Effect." [1] The British physicist John Ambrose Fleming, working for the British "Wireless Telegraphy" Company, discovered that the Edison Effect could be used to detect radio waves. Fleming went on to develop the two-element vacuum tube known as the diode, which he patented on November 16, 1904.

The thermionic diode can also be configured as a device that converts a heat difference to electric power directly without moving parts (a thermionic converter, a type of heat engine).

Owen Willans Richardson worked with thermionic emission and received a Nobel prize in 1928 "for his work on the thermionic phenomenon and especially for the discovery of the law named after him".

## Richardson's Law

In any metal, there are one or two electrons per atom that are free to move from atom to atom. This is sometimes referred to as a "sea of electrons". Their velocities follow a statistical distribution, rather than being uniform, and occasionally an electron will have enough velocity to exit the metal without being pulled back in. The minimum amount of energy needed for an electron to leave the surface is called the work function. The work function is characteristic of the material and for most metals is on the order of several electronvolts. Thermionic currents can be increased by decreasing the work function. This often-desired goal can be achieved by applying various oxide coatings to the wire.

In 1901 Owen Willans Richardson published the results of his experiments: the current from a heated wire seemed to depend exponentially on the temperature of the wire with a mathematical form similar to the Arrhenius equation. The modern form of this law (demonstrated by Saul Dushman in 1923, and hence sometimes called the Richardson-Dushman equation) states that the emitted current density J (A/m2) is related to temperature T by the equation:

$J = A T^2 e^{-W \over k T}$

where T is the metal temperature in kelvin, W is the work function of the metal, k is the Boltzmann constant. The proportionality constant A, known as Richardson's constant, given by

$A = {4 \pi m k^2 e \over h^3} = 1.20173 \times 10^6 A m^{-2}K^{-2}$

where m and -e are the mass and charge of an electron, and h is Planck's constant.

Because of the exponential function, the current increases rapidly with temperature when kT is less than W. (For essentially every material, melting occurs well before kT=W.)

The thermionic emission equations are of fundamental importance in electronics, significantly affecting both older vacuum tube technology (e.g. CRT applications, like television picture tubes and computer monitors, as well as high end radio and microwave applications requiring the high power intrinsic to tube technology), and more modern semiconductor designs.

While A theoretically has a value of 1.20.106 A m-2 K-2, in practice it strongly depends on material used. See work function for some practical values for A and W for some commonly used materials.

## Enhancement of thermionic emission

In most electronic devices, especially electron guns, the thermionic emitter will be given an electrical potential bias (or voltage), which creates an electric field and increases the emission current if it has the same sign as the thermion charge or decreases it if the signs are opposite. The effect of the potential can usually be modeled by a simple modification of the Richardson-Dushman equation; the current emitted from the metal cathode into the vacuum depends on the metal's thermionic work function, and this function is lowered from its normal value by the presence of image forces and by the electric field at this cathode. This field enhancement is given by the field-enhanced thermionic emission (FEE) equation:

$J (E_s,T,W) = A T^2 e^{ - (W - \Delta W) \over k T}$
$\Delta W = \left[{e^3 E_c \over (4 \pi \epsilon_0)}\right]^{1/2}$,

where Ec is the electric field strength at the cathode spot, ε0 is the vacuum permittivity.

This equation is relatively accurate for electric field strengths lower than about 108 V m−1, in which range the enhancement is known as the Schottky effect. For electric field strengths higher than 108 V m−1, quantum tunneling begins to contribute some significant emission current, which is called field emission. In this regime, the combined effects of field-enhanced thermionic and field emission can be modeled by the Murphy-Good equation for thermo-field (T-F) emission.[2] At even higher fields, field emission can become dominant and thermionic emission will no longer be significant.

Thermionic emission can also be enhanced by interaction with other forms of excitation such as light.[3]

## References

1. ^ "Edison" by Matthew Josephson. McGraw Hill, New York, 1959, ISBN 07-033046-8
2. ^ E.L. Murphy and R.H. Good, "Thermionic Emission, Field Emission, and the Transition Region", Phys. Rev. 102(6), pp. 1464-1473 (1956).
3. ^ A.G. Mal'shukov1 and K.A. Chao, "Opto-Thermionic Refrigeration in Semiconductor Heterostructures," Phys. Rev. Lett. 86, pp. 5570-5573 (2001).


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Article

# On the Cohomology of Galois Groups Determined by Witt Rings

Department of Mathematics, University of Western Ontario, London, Ontario, Canada, N6A 5B7f2E-mail: minac@uwo.caf2
(Impact Factor: 1.29). 08/1998; 148(1):105-160. DOI: 10.1006/aima.1999.1847

ABSTRACT

Let F denote a field of characteristic different from two. In this paper we describe the mod 2 cohomology of a Galois group F (called the W-group of F) which is known to essentially characterize the Witt ring WF of anisotropic quadratic modules over F. We show that H*(F, 2) contains the mod 2 Galois cohomology of F and that its structure will reflect important properties of the field. We construct a space XF endowed with an action of an elementary abelian group E such that the computation of the cohomology of F reduces to calculating the equivariant cohomology H*E(XF, 2). For the case of a field which is not formally real this amounts to computing the cohomology of an explicit Euclidean space form, an object which is interesting in its own right. We provide a number of examples and a substantial combinatorial computation for the cohomology of the universal W-groups.

### Full-text

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##### Article: Additive Structure of Multiplicative Subgroups
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ABSTRACT: One of the fundamental questions in current field theory, related to Grothendieck's conjecture of birational anabelian geometry, is the investigation of the precise relationship between the Galois theory of fields and the structure of the fields themselves. In this paper we initiate the classification of additive properties of multiplicative subgroups of fields containing all squares, using pro-2-Galois groups of nilpotency class at most 2, and of exponent at most 4. This work extends some powerful methods and techniques from formally real fields to general fields of characteristic not 2.
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##### Article: Field Theory and the Cohomology of Some Galois Groups
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ABSTRACT: We prove that two arithmetically significant extensions of a field F coincide if and only if the Witt ring WF is a group ring Z/n[G]. Furthermore, working modulo squares with Galois groups which are 2-groups, we establish a theorem analogous to Hilbert's Theorem 90 and show that an identity linking the cohomological dimension of the Galois group of the quadratic closure of F, the length of a filtration on a certain module over a Galois group, and the dimension over Z/2 of the square class group of the field holds for a number of interesting families of fields. Finally we discuss the cohomology of a particular Galois group in a topological context.
Full-text · Article · Oct 2000 · Journal of Algebra
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##### Article: On the cohomology of central Frattini extensions
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ABSTRACT: In this paper we provide calculations for the cohomology of certain p-groups, using topological methods. More precisely, we look at p-groups G defined as central extensions 1→V→G→W→1 of elementary abelian groups such that and the defining k-invariants span the entire image of the Bockstein. We show that if p>dimV−dimW+1, then the cohomology of G can be explicitly computed as an algebra of the form where is a polynomial ring on two-dimensional generators and A is the cohomology of a compact manifold which in turn can be computed as the homology of a Koszul complex. As an application we provide a complete determination of the cohomology of the universal central extension provided , where n=dimW.
Full-text · Article · May 2001 · Journal of Pure and Applied Algebra


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# Laplace operator help

#### Ed Aboud

1. Homework Statement

Let x = (x,y,z) .
Recall that the vector x is determined by its direction and length
r = |x| = $\sqrt{x^2 + y^2 + z^2}$

and assume we are given a function f which depends only on the length of x

f = f(r)

Show that

$$\Delta f = f'' + \frac{2}{r} f'$$

where $$f' = \frac{\partial f}{\partial r}$$

2. Homework Equations

3. The Attempt at a Solution

$$u = x^2 + y^2 + z^2$$
$$r = \sqrt{u}$$
$$\frac{\partial r}{\partial x} = \frac{\partial \sqrt{u}}{\partial u} \frac{\partial u}{\partial x} = \frac{1}{2}(\frac{1}{\sqrt{r}})(2x) = \frac{x}{\sqrt{r}}$$

$$\frac{\partial ^2 r}{\partial x^2} = \frac{\partial (x)}{\partial x } \frac{1}{\sqrt{u}} + x \frac{\partial \frac{1}{\sqrt{u}}}{\partial u} \frac{\partial u }{\partial x}$$

$$= \frac{1}{\sqrt{u}} - x^2 \frac{1}{\sqrt{u^3}}$$

Since f and u are symmetric in x,y,z

$$\frac{\partial ^2 r}{\partial y^2} = \frac{1}{\sqrt{u}} - y^2 \frac{1}{\sqrt{u^3}}$$

$$\frac{\partial ^2 r}{\partial z^2} = \frac{1}{\sqrt{u}} - z^2 \frac{1}{\sqrt{u^3}}$$

$$x^2 + y^2 + z^2 = u$$

$$\Delta r = (\frac{1}{\sqrt{u}}) - x^2 (\frac{1}{\sqrt{u^3}}) + (\frac{1}{\sqrt{u}}) - y^2 (\frac{1}{\sqrt{u^3}}) + (\frac{1}{\sqrt{u}}) - z^2 (\frac{1}{\sqrt{u^3}})$$

$$= \frac{3}{\sqrt{u}} - (x^2 + y^2 + z^2) \frac{1}{\sqrt{u^3}}$$

$$= \frac{2}{\sqrt{u}}$$

$$= \frac{2}{r}$$

I see that this is a part of the solution but I have no idea what to do to get the rest.

Any help would be greatly appreciated because I'm lost and it has to be in tomorrow morning.

Related Calculus and Beyond Homework Help News on Phys.org

#### yyat

Compute

$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)f(r(x,y,z))$$

using the chain rule.

#### Ed Aboud

I'm not really sure how to apply the chain rule in this case. Is there any general formula that I can use?

#### Ed Aboud

Actually its cool, I got it.

Thanks for the help!

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# Permutations and Combinations Questions

FACTS  AND  FORMULAE  FOR  PERMUTATIONS  AND  COMBINATIONS  QUESTIONS

1.  Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.

Examples : We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

2.  Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:

$P_{r}^{n}=n\left(n-1\right)\left(n-2\right)....\left(n-r+1\right)=\frac{n!}{\left(n-r\right)!}$

Ex : (i) $P_{2}^{6}=\left(6×5\right)=30$   (ii) $P_{3}^{7}=\left(7×6×5\right)=210$

Cor. number of all permutations of n things, taken all at a time = n!.

Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,

such that $\left({p}_{1}+{p}_{2}+...+{p}_{r}\right)=n$

Then, number of permutations of these n objects is :

3.  Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note that AB and BA represent the same selection.

Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.

Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.

Ex.5 : Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combinations of n things, taken r at a time is:

Note : (i)     (ii)$C_{r}^{n}=C_{\left(n-r\right)}^{n}$

Examples : (i) $C_{4}^{11}=\frac{11×10×9×8}{4×3×2×1}=330$      (ii)$C_{13}^{16}=C_{\left(16-13\right)}^{16}=C_{3}^{16}=560$

Q:

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

 A) 4050 B) 3600 C) 1200 D) 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

$10P4$

= 5040.

27 23475
Q:

The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and atleast 4 bowlers?

 A) 1024 B) 1900 C) 2000 D) 1092

Explanation:

We are to choose 11 players including 1 wicket keeper and 4 bowlers  or, 1 wicket keeper and 5 bowlers.

Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in $2C1*5C4*9C6$ = 840

Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in $2C1*5C5*9C5$ =252

Total number of ways of selecting the team = 840 + 252 = 1092

23 22264
Q:

How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

 A) 216 B) 45360 C) 1260 D) 43200

Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = $9!2!×2!×2!$ = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in $6!2!×2!$ = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $4!2!$ = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

21 21541
Q:

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

 A) 135 B) 63 C) 125 D) 64

Explanation:

Required number of ways =

28 19926
Q:

How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane?

 A) 3 B) 6 C) 2 D) 4

Explanation:

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

41 19084
Q:

When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

 A) 620 B) 671 C) 625 D) 567

Explanation:

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

25 18707
Q:

If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?

 A) 32 B) 24 C) 72 D) 36

Explanation:

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

7 16840
Q:

There are 7 non-collinear points. How many triangles can be drawn by joining these points?

 A) 10 B) 30 C) 35 D) 60

Explanation:

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = $7C3$ = 35


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# Simple indefinite integral question

Discussion in 'Math' started by MichealY, May 23, 2009.

1. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
$$\int cost^2dt$$ = ?

Thanks.
MichealY.

2. ### Ratch New Member

Mar 20, 2007
1,068
4
MichealY,

What are you trying to integrate? cos^2 (t)dt, cos(t^2), or what? Let's use some parentheses so we know what you what you are talking about. Can't you look that up in a table of integrals?

Ratch

3. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
Ratch:
It is cos(t^2) not cos^2(t).
No I cannot look up in a table of integrals.
I want to know how to solve it by us,altough it could be solved by MATLAB/Mathematica.

MichealY

4. ### Nukular New Member

May 16, 2009
3
0
Change variables from t^2 to x, then integrate by parts?

Last edited: May 23, 2009
5. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
Nukular:
What do you mean by integrating by parts?

MichealY

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Presumably you realize the solution is an infinite series.

7. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
t n k,
Bingo.I asked someone of mathematic major the other day.He said it is a problem of infinite series,But I do not even know what's infinite series,so I need the help of you,and at the same time I will work hard to learn infinite series

MichealY

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Hi MichealY,

As Nukular rightly points out the solution probably requires an understanding of both substitution of variables and integration by parts. It's the approach I would use anyway. Your earlier post suggests you are not familiar with the latter method. It would be useful to do some background study on that first and then return to this problem. Are you studying integration as part of formal course or is this something simply of personal interest?

9. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
Nukular,
I have understood a bit of what you mean.

t n k,
Yes, I am not familiar with integration by parts,maybe I just don't know this English phrase I will do some background study and then return to this problem.I'm studying integration as part of formal course(I failed the final exam,now trying to make up),and it is something of personal interest too.

Thanks to both of you.

MichealY

10. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
Here is integration by parts in mathematics notation:
$\int u dv = uv - \int v du$

It is the product rule of differentiation in reverse.
The trick to using this technique is to identify what part of the original integrand is u, which the rest being called dv. From this you antidifferentiate dv to get v, and take the differential of u to get du. If you have chosen u and dv correctly, the new integral will be easier to find than the one you started with.

11. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
Mark44,
Thanks.I have had learnt integration by parts before.I thought it means,first,reduce cost^2 to many parts,then integrate them by parts.

I have also tried this method,but it seems that it won't work.

MichealY

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
It does work.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Recognizing that the solution is an infinite series, the other approach would be to expand cos(t^2) as an infinite series and integrate that wrt t. The trouble with any solution one obtains using any approach is getting the answer into an agreed "standard form" - i.e. as agreed by mathematicians. I haven't looked up a standard handbook of integral tables for this function, but I would think the agreed solution has long been cast in stone. Finding a derivation might be more difficult - I guess it's out there somewhere!

One could probably better spend time doing practice on any number of functions with finite solutions of increasing degree of difficulty.

Then return to this at leisure........if one has such luxury.

Mar 6, 2009
5,448
784

Nov 9, 2007
5,005
515
16. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Good find Studiot! - hopefully MichealY will find this helpful.

17. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
Thanks,you are really kind.

I mean I have tried method of integrating by parts.

And by using MATLAB,the result returned contain Fersnel C.

MichealY

Last edited: May 27, 2009
18. ### rin-elt New Member

Jun 18, 2009
9
0
Hi I also do think this is related to infinite series
try and use sandwicth theorem to find whether it is convergent and find it's limit,if i'm not mistaken this limit is the value of the integral

19. ### MichealY Thread Starter Active Member

Apr 9, 2009
49
0
Starting to study numerical analysis,I have learnt a bit
It seems that langarage polynomail is better than taylor polynomial.


text:
# 12.10: Ecosystems

### What nonliving things are essential for life?

Living organisms cannot exist without the nonliving aspects of the environment. For example: air, water, and sunlight, which are all nonliving, are all essential to living organisms. Both nonliving and living things make up an ecosystem.

## What is an Ecosystem?

Ecology is the study of ecosystems. That is, ecology is the study of how living organisms interact with each other and with the nonliving part of their environment. An ecosystem consists of all the nonliving factors and living organisms interacting in the same habitat. Recall that living organisms are biotic factors. The biotic factors of an ecosystem include all the populations in a habitat, such as all the species of plants, animals, and fungi, as well as all the micro-organisms. Also recall that the nonliving factors are called abiotic factors. Abiotic factors include temperaturewater, soil, and air.

You can find an ecosystem in a large body of fresh water or in a small aquarium. You can find an ecosystem in a large thriving forest or in a small piece of dead wood. Examples of ecosystems are as diverse as the rainforest, the savanna, the tundra, or the desert (Figure below). The differences in the abiotic factors, such as differences in temperature, rainfall, and soil quality, found in these areas greatly contribute to the differences seen in these ecosystems. Ecosystems can include well known sites, such as the Great Barrier Reef off the coast of Australia and the Greater Yellowstone Ecosystem of Yellowstone National Park, which actually includes a few different ecosystems, some with geothermal features, such as Old Faithful geyser.

Ecosystems need energy. Many ecosystems get their energy in the form of sunlight, which enters the ecosystem through photosynthesis. This energy then flows through the ecosystem, passed from producers to consumers. Plants are producers in many ecosystems. Energy flows from plants to the herbivores that eat the plants, and then to carnivores that eat the herbivores. The flow of energy depicts interactions of organisms within an ecosystem.

Matter is also recycled in ecosystems. Biogeochemical cycles recycle nutrients, like carbon and nitrogen, so they are always available. These nutrients are used over and over again by organisms. Water is also continuously recycled. The flow of energy and the recycling of nutrients and water are examples of the interactions between organisms and the interactions between the biotic and abiotic factors of an ecosystem.

## Summary

• An ecosystem consists of all the living things and nonliving things interacting in the same area.
• Matter is also recycled in ecosystems; recycling of nutrients is important so they can always be available

## Explore More

Use the resource below to answer the questions that follow.

1. How do land plants generate the energy they need for their metabolic energy? What do they do with excess energy?
2. Where do scavengers in an ecosystem obtain their energy from? How can scavenging be a beneficial strategy for an organism?
3. What is the role of decomposers?
4. What kind of problems can you foresee if every speck of carbon were turned into biomass? Why?
5. Complete this statement: Energy ___________ through an ecosystem, whereas nutrients are ____________.

Review

1. Define an ecosystem.
2. Distinguish between abiotic and biotic factors. Give examples of each.
3. Where does the energy come from for many ecosystems?
4. Name two nutrients that are recycled through an ecosystem.

12.10: Ecosystems is shared under a CC BY-NC license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


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### A constructive and formal proof of Lebesgue’s Dominated Convergence Theorem in the interactive theorem prover Matita

Claudio Sacerdoti Coen & Enrico Tassi
We present a formalisation of a constructive proof of Lebesgue’s Dominated Convergence Theorem given by the Sacerdoti Coen and Zoli in [CSCZ]. The proof is done in the abstract setting of ordered uniformities, also introduced by the two authors as a simplification of Weber’s lattice uniformities given in [Web91, Web93]. The proof is fully constructive, in the sense that it is done in Bishop’s style and, under certain assumptions, it is also fully predicative. The...


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### Archive

Archive for January 13th, 2014

## More On Collections of Powers (via Starring) of Functions

These proofs really belong to the Compendium, but I have no access to the tex file at the moment, so I am using this post to jot down ideas.

If we accept powering of functions via the defintions of the star product, then the following will make sense. Â On the other hand if we view this from the standpoint of simple matrices (and replace the star product by matrix multiplication), the following claims may also make sense.

Definition 1.Â  For a countably infinite collection $\mathbb{F} = \{F^1, F^2, \ldots, F^j, \ldots \}_{j \in \mathbb{Z}^+}$, the "generator" isÂ $F^1$ since all other elements in the collection are powers of such. Â (As an aside, we may contemplate finite collections in the event of powering periodicity or nilpotency, but not presently).

Claim 1. Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^1 = G^1$, then $\mathbb{F} = \mathbb{G}$.

Proof. Â By hypothesis we have thatÂ $F^1 = G^1$, and let us assume thatÂ $F^k = G^k$. Then:

Â Thus we have shown that Â $\mathbb{F} \subset \mathbb{G}$ and $\mathbb{G} \subset \mathbb{F}$ and so naturally we concludeÂ $\mathbb{F} = \mathbb{G}$.

Claim 2.Â Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^1 = G^n$ for some $n \in \mathbb{Z}^+$, then $\mathbb{F} \subset \mathbb{G}$.

Proof.Â By hypothesisÂ $F^1 = G^n$ for some positive integer $n$. Â Also,

using Claim 1.10 of the Compendium. Â Next suppose thatÂ $F^k = G^{kn}$. Â Then:Â

Â Thus $\mathbb{F} \subset \mathbb{G}$ since all elements of the collection $\mathbb{F}$ are contained in $\mathbb{G}$.

Definition 2.Â Â For a countably infinite collectionÂ $\mathbb{F} = \{F^1, F^2, \ldots, F^j, \ldots \}_{j \in \mathbb{Z}^+}$, we call the subset $\mathbb{F}_{n} \subset \mathbb{F}$ the "subsequent power subset at $n$", and it is composed of all those powers greater than or equal to $n \in \mathbb{Z}^+$.

Claim 3. Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^m = G^n$ for some $m, n \in \mathbb{Z}^+$ and $m (since we haven't defined "roots" it must also be true that $\frac{n}{m} \in \mathbb{Z}^+$), then $\mathbb{F}_m \subset \mathbb{G}_n \subset \mathbb{G}$. (At this point we don't really care about containment of powers less than $n$, since we haven't established a mechanism to obtain them by "unpowering" or "taking roots").

Proof. Â By hypothesisÂ $F^m = G^n$. Â In particular, we have the statement $F^{m+1} = G^{n + p}$ for some integer $p \in \mathbb{Z}^+$. Â (Advancing one element in the collection $\mathbb{F}$ starting from $m$, advances us $p$ units in the collection $\mathbb{G}$ starting from $n$.) Â Taking this statement to the logical consequence, $F^m \star F^1 = G^n \star G^p \Rightarrow F^1 = G^p$. Â (Notice that since we have not defined negative powers of functions, $p$ is necessarily a positive integer).

Next let us assume thatÂ $F^{m+k} = G^{n + kp}$. Â Then:Â

It can be seen now thatÂ Â $\mathbb{F}_m \subset \mathbb{G}_n \subset \mathbb{G}$.

Corollary 1. Â Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$.Â Â $\mathbb{F}_m = \mathbb{G}_n \subset \mathbb{G}$ if and only if $m = n$ and $p = 1$.

Proof.

$\Rightarrow$ SinceÂ $\mathbb{F}_m = \mathbb{G}_n$ there is no element of either collection that does not match one in the other collection. Â Thus the step between elements must be 1, and $p = 1$. Â Furthermore the elements in a collection can be put in bijective correspondence (by definition of equality of sets) with the elements in the other collection, and the indices must exactly coincide by the ordering being the usual order on the positive integers (we order by power). Â Thus $m=n$.

$\Leftarrow$ Since $m = n$ we have that $F^m = G^n \Rightarrow F^m = G^m$ for all $m \in \mathbb{Z}^+$, and also $F^{m+1} = G^{n + p} \Rightarrow F^{m+1} = G^{m + 1}$Â since $p = 1$. Â So now assume that this works for the $k$th element $k>m+1$, so that $F^{m+k} = G^{m+k} \Rightarrow F^k = G^k$. Â Then we have that:

and we are done.

The following claims and proofs are applicable to Pasquali patches as they are to Â Markov matrices, where we shift from a function point-of-view to a matricial point-of-view using in one the star product and in the other normal matrix multiplication.

Claim 4. Take the countably infinite collections of Pasquali patch self-powers $\mathbb{P} = \{ P^1, P^2, \ldots, P^j, \ldots \}_{j \in \mathbb{Z}^+}$ andÂ $\mathbb{Q} = \{ Q^1, Q^2, \ldots, Q^k, \ldots \}_{k \in \mathbb{Z}^+}$ with stationary limiting $P^\infty$ and $Q^\infty$ respectively. If $\mathbb{P}$ and $\mathbb{Q}$ coincide at ($m, n$) so that $P^m = Q^n$ ($\frac{n}{m} \in \mathbb{Z}^+$) then $P^\infty = Q^\infty$.

Proof.

The infinite collections coincide at $(m,n)$ for $m, n, \frac{n}{m} \in\mathbb{Z}^+$, thus we have that $P^m = Q^n$. Â By Claim 3 we have thatÂ $\mathbb{P}_m \subset \mathbb{Q}_n$. Â Now the limit $\lim_{n \to \infty} Q_n = Q^\infty$ by hypothesis of the existence of the stationary patch, and sinceÂ $\mathbb{P}_m$ is a subset of $\mathbb{Q}_n$, it follows that$\lim_{m \to \infty} P_m = Q^\infty$. Â We also know thatÂ $\lim_{m \to \infty} P_m = P^\infty$ however (again by hypothesis of the existence of the stationary patch), so it must therefore be true thatÂ Â $P^\infty = Q^\infty$.

The reverse claim is not (always) true:

False Claim 1. Â IfÂ $P^\infty = Q^\infty$, thenÂ $\mathbb{P}$ and $\mathbb{Q}$ coincide at ($m, n$) so that $P^m = Q^n$ ($\frac{n}{m} \in \mathbb{Z}^+$).

Disproof. Â Take the collection of Pasquali patches $\mathbb{P} = \{P^1 = 1, P^2 = 1, \ldots, P^j = 1, \ldots\}_{j \in \mathbb{Z}^+}$ with $P^\infty = 1$, and any collection of Pasquali patches $\mathbb{Q} = \{Q^1, Q^2, \ldots, Q^k, \ldots\}_{k \in \mathbb{Z}^+}$ so that the generator $Q^1 = q(x,y)$ is an explicit function of $y$ and $Q^\infty = 1$, as in an example in the Compendium. Â By a Claim in the Compendium, all elements in the collectionÂ $\mathbb{Q}$ are explicit functions of $y$ (even as $Q^\infty$ is not). Â Clearly, no $P \in \mathbb{P}$ is equal to an element $Q \in \mathbb{Q}$ (all $P$ are constant where all $Q$ vary with $y$), yet they have the same stationary patch.


text:
# Why quantum map must be hermitian?

Quantum maps transform a density matrix into another one, Assume we are in the Hilbert space :$$H_A$$ the quantum map on the density matrix $$\rho_A$$ living in $$H_A$$ is : $$\mathcal{L}_A$$

Why $$\mathcal{L}_A$$ must conserve the hermiticity : $$\mathcal{L}_A(\rho_A)^{\dagger}=\mathcal{L}_A(\rho_A)$$ = $$\mathcal{L}_A(\rho_A)^{\dagger}=\mathcal{L}_A(\rho_A)$$ ?

• Your last equality seems duplicated. I'm not sure if the answer "Density matrices are Hermitian and therefore the map must conserve hermiticity to be a valid quantum map" is what you're looking for? – user1936752 Jan 23 at 14:03

A sensible density matrix needs to be hermitian, i.e. $$\rho^{\dagger} = \rho$$. This is because the diagonal elements of the density matrix can be interpreted as probabilities in any basis and therefore they should be real. This is guaranteed for hermitian matrices. Furthermore, they should sum to $$1$$, which is why we also require $$\textrm{tr}\ \rho = 1$$.
A map $$\mathcal{L}$$ from one valid density matrix to another one must preserve these properties, i.e. it must satisfy (among other things) $$\mathcal{L}(\rho)^{\dagger} = \mathcal{L}(\rho)$$ if $$\rho^{\dagger} = \rho$$ and $$\textrm{tr}\ \mathcal{L}(\rho) = 1$$ if $$\textrm{tr}\ \rho = 1$$. Otherwise $$\mathcal{L}(\rho)$$ would not be a proper density matrix.