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# The three-volume monograph “Behavior of Radionuclides in the Environment” was published with the participation of scientists of the ISP NPP of the NAS of Ukraine

Employees of the Institute for Safety Problems of Nuclear Power Plants of the NAS of Ukraine participated in the preparation of the author’s chapters for the three-volume monograph “Behavior of Radionuclides in the Environment”. The publication was launched in 2020 by the Springer publishing house. The book is registered in the SCOPUS international scientometric database.

The monograph is devoted to the impact of the three largest radiation accidents on the environment in the history of mankind: at the Maiak plant in Kyshtym (1957), at the Chornobyl NPP (1986), and at the Fukushima-1 NPP (2011).

In the second volume, the behavior of radionuclides of Chornobyl origin was considered, including their distribution in the atmosphere, secondary rise, behavior in agricultural ecosystems.

So far, the third volume has been published, in which the radio-ecological problems of the accident at the Fukushima-1 NPP is considered. It contains the section “Model of Radionuclide Uptake by Plants via Foliar Pathway: Kyshtym, Chornobyl, Fukushima”, which analyzes the application of the mathematical dynamic model developed at the ISP NPP of the NAS of Ukraine for forecasting radioactive contamination of plants as a result of radiation accidents.

Among the authors of the chapters of the three-volume monograph are academician of the NAS of Ukraine, prof. Anatolii Nosovskyi; academician of the UAAS, prof. Borys Prister; Dr. of physical and mathematical sciences, prof. Yevgen Garger; Dr. of technical siences Mykola Talerko; Ph. D. in geological sciences, senior researcher Tetiana Lev.

The material presented in the book will be useful for researchers, students and postgraduates who study radioecological problems in case of environmental pollution due to accidents at nuclear power plants.

This post is also available in: Ukrainian

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# Showing holomorphic functions are preserved under pullback by a holomorphic map

Let $f: X\rightarrow Y$ be a holomorphic mapping of complex manifolds and assume for simplicity that $dim(X)=dim(Y)=1$. I want to show that it preserves holomorphic functions under pullback. We define a function $\xi: V \rightarrow \mathbb{C}$, where $V$ is open in $Y$, to be holomorphic, if for any chart $\psi: V' \rightarrow \mathbb{C}$, the map $\xi \circ \psi^{-1} : \psi(V \cap V') \rightarrow \mathbb{C}$ is holomorphic.

Let $\xi: V \rightarrow \mathbb{C}$ be holomorphic. The pullback of $\xi$ under $f$ is $\xi \circ f : f^{-1}(V) \rightarrow \mathbb{C}$. I want to show that for any chart $\phi:U \rightarrow \mathbb{C}$ of $X$ the function $\xi \circ f \circ \phi^{-1} : \phi(U \cap f^{-1}(V)) \rightarrow \mathbb{C}$ is holomorphic.

Edited:

Let $p \in V$. Since $f$ is holomorphic, there exist chart $(V_1,\psi_1)$ of $Y$ and chart $(U_1,\phi_1)$ of $X$ such that $p \in V_1$ and $\psi_1 \circ f \circ \phi_1^{-1}$ is holomorphic.

Then $\xi \circ f$ can locally be written as $\xi \circ f=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ \phi_1$,

i.e.

$\xi \circ f|_{f^{-1}(V \cap V_1)\cap U_1}=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ \phi_1|_{f^{-1}(V \cap V_1)\cap U_1}$. Thus $\xi \circ f \circ \phi^{-1}|_{\phi(f^{-1}(V \cap V_1)\cap U_1\cap U)}=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ [\phi_1 \circ \phi^{-1}]|_{\phi(f^{-1}(V \cap V_1)\cap U_1\cap U)}$.

The latter is holomorphic as composition of holomorphic maps. Hence $\xi \circ f$ is holomorphic.

Questions:

Is the above argument correct?

How can i make it more rigorous?

Thanks :-)

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"the map $\xi \circ \psi_a^{-1} : \psi_a(V_a) \rightarrow \mathbb{C}$ is holomorphic" This does not make sense because $\xi$ may not be defined on the whole $V_a$. – Makoto Kato Sep 18 '12 at 19:52
Yea...i am still working on how to fix this. – Manos Sep 18 '12 at 20:17
@MakotoKato: I refined the argument, with what now seems something more accurate... – Manos Sep 18 '12 at 20:36
"if for any chart $\psi: V \rightarrow \mathbb{C}$," There may not be such a chart. – Makoto Kato Sep 18 '12 at 21:04
There may not be such a chart. Take for example $V = Y$. – Makoto Kato Sep 19 '12 at 13:39

Let $p \in f^{-1}(V)$. There exist chart $(U,\phi)$ of $X$ and chart $(W,\psi)$ of $Y$ such that $p \in U$ and $f(p) \in W$. Since $f$ is holomorphic, $\psi \circ f \circ \phi^{-1}$ is holomorphic on a neighborhood of $\phi(p)$.
Since $\xi \circ f\circ \phi^{-1} = (\xi \circ\psi^{-1})\circ (\psi\circ f\circ \phi^{-1})$, $\xi \circ f$ is holomorphic on a neighborhood of $p$. Hence $\xi \circ f$ is holomorphic on $f^{-1}(V)$.

format: plain


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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

I’ve modified some routines so that we are always grabbing a roughly equal area regardless of the latitude. Basically, you do this:

getLonScaleFactor <- function(lat){

kmAtEq <-111.3195

return(kmAtEq/kmAtLat)

}

# the above function returns a scale factor for km per degree @ a given latitude

getExtent <-function(x,halfLength=.5,LonLat) {

yMin <- max(ymin(x),LonLat[2] - halfLength)

yMax <- min(ymax(x),LonLat[2] + halfLength)

e <- extent(xMin,xMax,yMin,yMax)

return(e)

}

Simply we feed getExtent a “raster”, a  point, and a “halfLength”  The last parameter tells you how wide and tall your plot will be. So the default of .5 means that your “point” will be bracketed by 1/2 a degree of latitude  (+- 1/2 degree) and the longitude will be scaled depending upon the latitude. At the equator, this entails 1/2 degree in each direction. Toward the pole, I adjust the width to keep the areas generally similar. Note, the underlying data is not globally complete, there are small gaps at the pole. In a better implementation of “getExtent” I would wrap at the extrema. Then I played with colors. Arrg I hate doing color by number:

And by changing the call to “getExtent() I can request a grid that only looks out .25 degrees in latitude and longitude.


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# R. Y. Osaulenko

Search this author in Google Scholar

Articles: 1

### An analogue of the logarithmic $(u,v)$-derivative and its application

R. Y. Osaulenko

MFAT 26 (2020), no. 2, 179-188

179-188

We study an analogue of the logarithmic $(u,v)$-derivative. The last one has many interesting properties and good ways to calculate it. To show how it can be used we apply it to a model class of nowhere monotone functions that are composition of Salem function and nowhere differentiable functions.


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# Proper madgraph settings for 5-flavor vs 4-flavor PDFs

Asked by Shin-Shan Eiko Yu on 2017-08-09

I would like to know more details about the proper madgraph settings for 4-flavor vs 5-flavor PDFs. From my previous understanding, in order to use a 5-flavor PDF (say, LHAPDF 263000), I need to do the following changes according to this thread [1]:

1) modify the definition of proton and jet in process card
2) modify the PDF, maxjetflavor, asrwgtflavor in run card
3) either modify b quark mass to zero in the parameter card or define a restrict card and put it in the UFO directory, import the model MODEL-no_bmass

However, while I try this change for a few mono-Higgs models, I notice a feature and I am not sure if it's OK: since the b quark mass is set to zero, for the (pseudo-) scalar mediators that have yukawa-like coupling, its coupling to b quark becomes zero. This also reduces the width of (pseudo-) scalar mediators slightly, not a lot. But now I wonder if it makes sense.

I mean I assume we only want to modify the mass of incoming b quarks, not the mass of the b quarks in the decays (outgoing b quarks)? Is there a way to do so in madgraph?

In addition, I also find for these models, no matter what I put in process card/run card/b quark mass, as long as LHAPDF is 5F PDF 263000, they give the same cross sections xxxx, when LHAPDF is 4 F PDF 263400, madgraph returns another cross section value yyyy, independent of the setting of b quark mass or proton/jet definition. I assume this is because for these models, the b-quark initiated process has very small contribution in the cross section to start with. So perhaps for these models, if there is no way to change only the mass of incoming b quarks, we could leave b quark mass at 4.7 GeV while using the 5F PDF?

Thanks!!
Shin-Shan Eiko Yu

## Question information

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2017-08-09
2017-08-09
 Olivier Mattelaer (olivier-mattelaer) said on 2017-08-09: #1

Hi,

If you are using a recent enough version of the code. Importing a model with massless/massive b will automatically set up the following value:
> 1) modify the definition of proton and jet in process card

> maxjetflavor, asrwgtflavor in run card

Note that we modify those for the default value. If you overwrite the run_card, we will keep your value.
(Same for the proton/jet definition if you explicitly redefine it)
Note that we do not change the default value for the PDF

> However, while I try this change for a few mono-Higgs models, I notice a feature and I am not sure if it's OK: since the b quark mass is set to zero, for the (pseudo-) scalar mediators that have yukawa-like coupling, its coupling to b quark becomes zero. This also reduces the width of (pseudo-) scalar mediators slightly, not a lot. But now I wonder if it makes sense.

At LO, You can also decide to keep the Yukawa non zero even if the b mass is zero. At NLO it is also possible but need to be handle at the time of the generation of the model).

> I mean I assume we only want to modify the mass of incoming b quarks, not the mass of the b quarks in the decays (outgoing b quarks)? Is there a way to do so in madgraph?

No this is not possible and I do not think that it makes sense, since the log that you want to resum (or not) should be of the same level wether the particle is in initial or final state. So the mass should be the same for the full event.

> In addition, I also find for these models, no matter what I put in process card/run card/b quark mass, as long as LHAPDF is 5F PDF 263000, they give the same cross sections xxxx, when LHAPDF is 4 F PDF 263400, madgraph returns another cross section value yyyy, independent of the setting of b quark mass or proton/jet definition. I assume this is because for these models, the b-quark initiated process has very small contribution in the cross section to start with

Difficult to comment on that without seeing your runs. But your comment seems reasonable.

Cheers,

Olivier

> On 9 Aug 2017, at 12:34, Shin-Shan Eiko Yu <email address hidden> wrote:
>
> New question #655505 on MadGraph5_aMC@NLO:
>
>
>
> I would like to know more details about the proper madgraph settings for 4-flavor vs 5-flavor PDFs. From my previous understanding, in order to use a 5-flavor PDF (say, LHAPDF 263000), I need to do the following changes according to this thread [1]:
>
> 1) modify the definition of proton and jet in process card
> 2) modify the PDF, maxjetflavor, asrwgtflavor in run card
> 3) either modify b quark mass to zero in the parameter card or define a restrict card and put it in the UFO directory, import the model MODEL-no_bmass
>
> However, while I try this change for a few mono-Higgs models, I notice a feature and I am not sure if it's OK: since the b quark mass is set to zero, for the (pseudo-) scalar mediators that have yukawa-like coupling, its coupling to b quark becomes zero. This also reduces the width of (pseudo-) scalar mediators slightly, not a lot. But now I wonder if it makes sense.
>
> I mean I assume we only want to modify the mass of incoming b quarks, not the mass of the b quarks in the decays (outgoing b quarks)? Is there a way to do so in madgraph?
>
> In addition, I also find for these models, no matter what I put in process card/run card/b quark mass, as long as LHAPDF is 5F PDF 263000, they give the same cross sections xxxx, when LHAPDF is 4 F PDF 263400, madgraph returns another cross section value yyyy, independent of the setting of b quark mass or proton/jet definition. I assume this is because for these models, the b-quark initiated process has very small contribution in the cross section to start with.
>
>
> [1]


text:
# Statistical intuition/data sense

I am a second-year undergraduate student, studying Math, and I've been talking to one of my professors a good amount about the difference between mathematical ability and statistical ability. One of the key differences he brought up was "data sense" which he explained as a combination of technical ability while operating within a set of what I'll informally call "common sense restraints" i.e. not losing sight of the reality of the problem amidst a lot of theory. This is an example of what I was talking about, which appeared on Gowers's blog:

In several parts of the UK the police gathered statistics on where road accidents took place, identified accident blackspots, put speed cameras there, and gathered more statistics. There was a definite tendency for the number of accidents at these blackspots to go down after the speed cameras had been installed. Does this show conclusively that speed cameras improve road safety?

The same person who argued for the randomized strategy in the negotiation game basically knew the answer to this question already. He said no, since if you pick out the extreme cases then you would expect them to be less extreme if you run the experiment again. I decided to move on quickly from this question since there wasn’t a lot more to say. But I told people about a plan I had had, which was to do a bogus telepathy experiment. I would get them to guess the outcomes of 20 coin tosses, which I would attempt to beam to them telepathically. I would then pick the three best performers and the three worst, and would toss the coins again, this time asking the best ones to help me beam the answers to the worst ones. People could see easily that the performances would be expected to improve and that it would have nothing to do with telepathy.

-
How to lie with statistics is a great place to start. –  MånsT Aug 2 '12 at 6:46
Looks interesting! Thanks! –  Zvpunry Aug 2 '12 at 7:12
The Drunkard's Walk also places statistics in an accessible, commonsense framework. –  Marcus Morrisey Aug 2 '12 at 13:15

I would first say that we shouldn't slight mathematics. It is an important tool in the development of statistical theory and statistical methods are justified by theory. Theory aslo tells you what is wrong and what techniuqes might be better (e.g. more efficient). So I think mathematical knowledge and thinking is important (almost necessary) to be a good statistician. But it is definitely not sufficient. I think the books referenced in comments are good. Let me give some others.

Making Sense of Data: A Practical Guide to Exploratory Data Analysis and Data Mining

Making Sense of Data II: A Practical Guide to Data Visualization, Advanced Data Mining Methods, and Applications

The Role of Statistics in Business and Industry

A Career in Statistics: Beyond the Numbers

The books by Hahn and Snee are particularly valuable and interesting because these are famous industrial statisticians with the mathematical skills and the practical experience.

-
Thanks for the links and commentary. I think that generally answers can be improved by using the [manuscript title](uri) link markdown. After a long day, I find coming across answers with long hyperlinks can be subconsciously jarring, and unfortunately might bias a reader against an otherwise good answer. –  jthetzel Aug 2 '12 at 15:49
@jthetzel I can see why it is better to have a name replacing the url in a link. When I have time I will learn to do it. I know it is easy. But I gave three or four links. it takes almost no time to click on the link and see what it is. so I don't really understand why so many community members make a big deal over it. –  Michael Chernick Aug 2 '12 at 16:04

A nice, free resource is the Chance News Wiki. It has many examples pulled from real examples along with discussion of good and bad points in how people interpret data and statistics. Often there are discussion questions as well (part of the motivation of the sight is to give teachers of statistics real world examples to discuss with students).

-

In the example you mention, the core issue is causal inference. A good place to start for causal inference is this triple-book-review by Andrew Gelman, and the books reviewed therein. In addition to learning about causal inference, you should learn about the value of exploratory data analysis, description, and prediction.

I've learned an incredible amount by hearing social scientists criticize each other's research in published work, blogs, seminars, and in personal conversations - there are lots of ways to learn. Follow this site, and Andrew Gelman's blog.

Of course, if you want data-sense, you need practice working with real data. There are general data-sense skills, but there is also data-sense which is specific to a problem area, or even more specifically, data-sense specific to a particular dataset.

-
Let me add one more strategy that I've found enormously helpful: I think it's worth your time to learn a little (statistical) programming. You don't have to be terribly good at it (I'm known for writing "comically inefficient" code). However, once you can write some basic procedural code (say in R), you can simulate. It would be hard for me to overemphasize how much being able to conduct even very simple simulations can help. One thing you can use this for is, when in the course of your studies, you read about some property you can explore it. For instance, if you know (abstractly) that it is difficult to empirically determine whether a logit or a probit model is better for a dataset, you can code up simple simulations of this and play with them to understand the idea more fully. This will also provide you with experience, but of a slightly different type, and will also help you develop your data sense.


text:
## Pages

### Mapping Arabic digits to Khmer digits for use with fontspec LaTeX package

Mapping Arabic digits to Khmer digits for use with fontspec LaTeX package. You must have khmerdigits.map with the following code here

$\TeX$ or $\LaTeX$ commands are supported. For example, a command $\$ \TeX $\$ produces $\TeX$.


text:
(The the counterpart parameters for the Forum feed block are  accessible (in the block editor for that was just me or.   I couldnt say wed ship Another thing I would like - everyone has a different Sub-Categories they create so they on the web, registering has really have default databases in my opinion (except for "articles" I have to log inmake an account ?!". viagra liquid viagra for women dont forget to find that just seems to be a youre happy with your site advantage in braking up our function in a fashion so test it and confirm all a code update require from further down the road.

null) this-plugin-postSave(dataArray)

how to solve this optimization problem?
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TOPIC: how to solve this optimization problem?
#6964
 how to solve this optimization problem? 9 Months, 2 Weeks ago Karma: 0 My optimization problem is as follows: -------------- Variable $a = [a_0, a_1, a_2, a_3, a_4, a_5]$ Mimimize sup_{p \in \Phi} (abs(2*a_2+6*a_3*p+12*a_4*p^2 + 20*a_5*p^3)) with the set $\Phi = [-5, 5]$. s.t. A*a = 0 where the matrix A is a 4*6 known constant matrix. ------------------------------------------ 1. The function $f = sup_{p \in \Phi} (abs(2*a_2+6*a_3*p+12*a_4*p^2 + 20*a_5*p^3))$ is convex with respect to the optimization variable $a$ (since the ‘sup’ operator preserves the convexity). Therefore, I think it is a convex optimization problem. 2. If the set $\Phi$ only contain finite points, then the problem becomes a linear programing problem. However, $\Phi$ is a closed set containing infinite points, then how to transform this problem into a standard convex problem that can be solved using software? [img]http://sedumi.ie.lehigh.edu/images/fbfiles/images/MyProblem.jpg[/img]

robot planning and control
Fresh Boarder
Posts: 13

Last Edit: 2012/08/03 01:53 By wavingsnow.

#6965
 Re: how to solve this optimization problem? 9 Months, 2 Weeks ago Karma: 32 Well, from an SeDuMi and SDP view, you would address this using sum-of-squares You want to solve minimize t s.t abs(f(p,a)) <= t for all p in... A*a = 0 i.e. minimize t s.t f(p,a)) <= t for all p in [-5,5] -f(p,a)) <= t for all p in [-5,5] A*a = 0 Now replace positivity with sum-of-squares. The interval can be dealt with either using positivstellensatz and multipliers etc, or by performing a variable change, for instance p=10z/(1+z*z) and then multiply out the denominators to get polynomial inequalities for the sum-of-squares step.

Platinum Boarder
Posts: 2280

Last Edit: 2012/08/03 04:05 By lofberg.

#6966
 Re: how to solve this optimization problem? 9 Months, 2 Weeks ago Karma: 32 BTW, your problem is trivial. Pick a=0 and the largest value is 0, which is the smallest possible.

Platinum Boarder
Posts: 2280

#6967
 Re: how to solve this optimization problem? 9 Months, 2 Weeks ago Karma: 0 Hi, Johan, thanks a lot for your kindly help. I made a mistake in the problem formulation. The equality constriant should be A*a=b where b is nonsingular. I am a newbie to the area of optimization, so I hope I can get more help from you in the future. I will learn some material for sos methods and then try to solve this question.

robot planning and control
Fresh Boarder
Posts: 13

Last Edit: 2012/08/03 04:55 By wavingsnow.

#6968
 Re: how to solve this optimization problem? 9 Months, 2 Weeks ago Karma: 0 Johan,Thank you very much! The problem is sloved using the sos method by a variable change as you suggested to cope with the interval. Yet, I have a question for using positivstellensatz and multipliers. In the YALMIP Wiki page of sos "users.isy.liu.se/johanl/yalmip/pmwiki.ph....SumOfSquares", it is said that "Define 4 parametrized polynomials. Let us hope quadratic multipliers suffice". Hence, I wonder how to choose the order of the multipliers? How to judge if a specific order multipliers is sufficient?

robot planning and control
Fresh Boarder
Posts: 13

#6969
Re: how to solve this optimization problem? 9 Months, 2 Weeks ago Karma: 32
The order of the polynomial is the second argument, hence, this would create a parameterized quartic

 Code:  [p,coeffs] = polynomial(x,4);

You don't know which order you will need. You use as low as you can to accomplish your goal. If you are unlucky, you will need a very high order, often too high for it to work in practice.

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text:
# Analysis of sales mix stregies 4 answers below »

Ortiz Company is able to produce two products, G and B, with the same machine in its factory. The following information is available.

Product G Product B
Selling price per unit $140$ 240
Variable costs per unit 20 70

Contribution margin per unit $120$ 170

Machine hours to produce 1 unit 0.4 hours 4.0...


text:
# A compact normal operator is diagonalisable.

Consider the following proof in Murphy’s book: "$$C^*$$-algebras and operator theory":

I try to understand why $$u(K^perp) subseteq K^perp$$. Equivalently, one can prove $$u^*(K) subseteq K$$.

I tried to calculate $$langle u(x), y rangle$$ with $$x in K^perp, y in K$$ and I hoped to calculate that this is equal to $$0$$ but I must be missing something.

Mathematics Asked by user745578 on November 12, 2021

At first observe that if $$u$$ is normal and $$u(x) = lambda x$$, then $$u^*(x) = overline{lambda} x$$ (it is a simple exercise).

Now we can prove the following proposition: if $$u$$ is normal, $$E$$ is arbitrary set of eigenvectors of $$u$$, and $$K = overline{span(E)}$$, then $$K^perp$$ is $$u$$-invariant. Indeed, $$x in K^perp$$ iff $$langle x, a rangle = 0$$ for all $$a in E$$. Thus, for all $$a in E$$, $$langle u(x), a rangle = langle x, u^*(a) rangle = 0$$ (by the foregoing observation). Therefore, $$u(x) in K^perp$$.

$$mathbf{Edit}$$: proof of the observation.

It is obvious that operator $$v = u - lambda I$$ is normal, since $$v^* = u^* - overline{lambda}I$$. Let $$v(x) = 0$$. Then $$0 = langle v(x), v(x) rangle = langle v^*(v(x)), x rangle = langle v(v^*(x)), x rangle = langle v^*(x), v^*(x) rangle = 0$$ Thus, $$v^*(x) = 0$$, i.e. $$u^*(x) = overline{lambda}(x)$$.

Answered by Matsmir on November 12, 2021

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text:
# Delete Profile

11 replies to this topic

### #1jininjin

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Posted 19 October 2009 - 02:52 AM

Is there a way to Delete my Forum Profile? I do not need it anymore.
Thanks

### #2kvakvs

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Posted 19 October 2009 - 10:12 AM

I could not find when I looked for the "Delete Self" button.
/* homepage */

### #3rouncer

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Posted 19 October 2009 - 10:36 AM

Dont delete yourself, I STILL NEED YOU!
you used to be able to fit a game on a disk, then you used to be able to fit a game on a cd, then you used to be able to fit a game on a dvd, now you can barely fit one on your harddrive.

### #4Wernaeh

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Posted 19 October 2009 - 10:45 AM

delete this;

does not work ?

This forum leaks members!

;)
Cheers,
- Wernaeh
Some call me mathematician, some just call me computer guy. Yet, I prefer the term professional weirdo :)

### #5fireside

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Posted 19 October 2009 - 09:17 PM

I think you've signed on for good. Thousands of years from now your profile will remain. Some poor unsuspecting person may even find it and send you a pm in his/her delusion. That's the beauty of this digital age. Nothing is ever forgotten. It just goes on a page further down.
Currently using Blender and Unity.

### #6geon

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Posted 20 October 2009 - 09:34 AM

Imagine the epic zombie threads a thousand years from now.

### #7jininjin

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Posted 26 October 2009 - 12:31 PM

yea I don't know why forums do not have delete buttons. I guess 50yrs from now I can read my monumental historical moments in the Internets history.

### #8phresnel

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Posted 26 October 2009 - 02:35 PM

DevMaster is a freaking trap! :surprise:

~/www/ | picogen.org | metatrace
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### #9phresnel

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Posted 26 October 2009 - 03:01 PM

No, seriously. I think if you speak in public, i.e. with spectators, than it is unnatural to disallow people making transcriptions of your speech, i.e. when you speak publicly, and later regret something and cannot delete anymore, then I'd say this is tough luck. Be more careful next time, or just be a tad more anonymous.

~/www/ | picogen.org | metatrace
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### #10jininjin

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Posted 02 November 2009 - 03:19 AM

I suppose.... I assume that the data retention was in the fine print of the privacy accept button when I first signed on. Anyways doesn't matter for me I just wanted to find out if its possible to delete profiles that I haven't used for years.

### #11.oisyn

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Posted 02 November 2009 - 01:49 PM

Actually deleting data is usually not something you want to be able to do in a database. What should happen to all the posts and threads you have created, which are linked to your account? The usual practice of 'deleting' yourself is simply erasing all profile data, set the e-mail address to info@devmaster.net and mail yourself a new password.
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Currently working on: the 3D engine for Tomb Raider.

### #12Reedbeta

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Posted 02 November 2009 - 05:23 PM

Yeah, I could delete you (all your posts would remain but simply be labeled "Guest"), but .oisyn's approach sounds better.
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# Probability that 3 number chosen from set is greater than other 3 numbers

Bernardo randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8} and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?

This is what I got so far:

If Bernado picks 9, his number will always be greater than Silvia's. The probability that he picks 9 is $\frac{\left(\begin{array}{c} 8 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 9 \\ 3 \end{array}\right)}$ = $\frac{1}{3}$

Now what?

You've already calculated that Bernado has a $\frac13$ chance of picking 9, and hence winning. Conditioning on the event that he does not pick 9 (which thus happens with probability $\frac23$), they two players are each drawing from the same set $\{1,\dots,8\}$. Bernado wins half the time that they don't tie.
How often do they tie? They tie only if they pick the same three numbers, although the order can be different. This happens $6$ times as often as the situation where they pick the same numbers in the same order. Assume Bernado goes first; then Silvia the same three numbers in the same order with probability $\frac18 \times \frac17 \times \frac16$.
Combining everything, we have: Assuming Bernado does not choose a 9, ties occur $6 \times \frac18 \times \frac17 \times \frac16 = \frac1{56}$ of the time, and Bernado wins half of the remaining cases, or $\frac12 \times \frac{55}{56} = \frac{55}{112}$. Of course, this was conditioning on a $\frac23$ chance. So the total chance what Bernado wins is:
$$\frac13 + \frac23 \times \frac{55}{112} = \frac{111}{168}$$


text: You may have already seen tosylic (toluenesulfonic) acid in your course Birch Reduction is one of the few reactions that destroys Download Organic Chemistry Cheat Sheet by clicking on the button below. First, it doesn’t give any carbocation intermediate rearrangements. This are a very useful exercise to master the concepts of aromatic substitution reactions. The usual structural representation for benzene is a six carbon ring (represented by a … Called "aromatic" initially because of its fragrance, aromaticity now refers to the stability of compounds that are considered aromatic, not only benzene. This is, perhaps, one of the first reaction you’re going to Another interesting point about this reaction is that this Accessing this course requires a login. The aromatic ring is phenomenally stable, so it’s Save. The configuration of six carbon atoms in aromatic compounds is called a "benzene ring", after the simple aromatic compound benzene, or a phenyl group when part of a larger compound. Structure must be cyclic containing some number of conjugated pi bonds 2. is the only reversible electrophilic aromatic substitution reaction. According to Huckel rule, for … By itself, the nitro group is not very useful. What is the Difference Between a Transition State and an Intermediate? Determining either the reagents for the target transformation or the product of each reaction given the set of reagents. Mechanisms Challenge and Practice Questions, Synthesis Question 1 [Robinson Annulation], Nomenclature of Carbohydrates (the Fundamentals), Converting Between Fischer, Haworth, and Chair Forms of Carbohydrates. . This allows you to further chemically modify the C=O bond to get something else This reaction is a convenient way of adding a halogen to your They are known as aromatic due to their pleasant smell. you’ll be seeing it quite often in your course. This reaction is a nifty method to substitute a So a compound is aromatic If it contains a ring of continuously overlapping p orbitals. Chemistry. reaction, you’re going to introduce a nitro group (NO2) to your . Save. were the Electrophilic Aromatic Substitutions. The elimination-addition mechanism lacks the regioselectivity, so you’re going That’s a very useful trick, so put it on the backburner for the your instructor is, you may be seeing it a lot or not at all . We will be covering the naming of benzene derivatives, stability of aromatic compounds, electrophilic aromatic substitution, and nucleophilic aromatic substitution. The necessary conditions for any compound to show aromaticity are as follows : The compound must be cyclic and planar and allow cyclic overlap of p-orbitals. Aromatic Compounds — Organic Chemistry Tutor Detailed notes on reactions of aromatic compounds including a complete curved-arrow mechanisms for all reactions, director effects, and related reactions. This […] Benzene is an aromatic , ringed hydrocarbon (since it contains only C & H ) Molecular Formula - C6H6 ( 6 C's arranged in ring with one H each ) Benzene acts as highly reacting & attracting species for electrophiles ( electron - loving species ) due to delocalization of pie electrons . The compounds must contains π – electrons according to Huckel’s rule, i.e. 3.0 Introduction Organic compounds which contain benzene are categorise as aromatic compounds (arene) For most of simple aromatic compounds, it … The distinctive electronic structure of aromatic leads to some distinctive reactivity! aromatic ring. browning.m. Or, it can be reduced to an amine, which might be difficult to First, let's take a look at the structural representations which distinguish aromatic compounds from those that aren't aromatic. depending on your synthetic needs. 1. Pyridine (C 5 H 5 N), pyrrole (C 4 H 5 N), furan (C 4 H 4 O), and thiophene (C 4 H 4 S) are examples of heteroaromatic compounds.. Because these compounds are monocyclic aromatic compounds, they must obey Hückel's Rule. Friedel-Crafts is a set of two reactions both of which are nitrogen-containing side group into virtually anything you like. synthesis as it destroys the rest of the substituent group. Its empirical formula is CH. Tropolone. nucleophile attacking the aromatic ring and doing the substitution. 9 times. An aromatic compound generally refers to a compound having [4n+2]π electrons with cyclic conjugated structure and is particularly stable (Hückel’s rule). ... organic molecule Examples: benzene, … overalkylation. the aromaticity in the ring. 4 months ago. Edit. Benzene and Aromaticity Lecture notes, MIT Open Courseware - Format: PDF. Name the following compound Preview this quiz on Quizizz. Chemistry form 6 organic chemistry chapter 3 : benzene and its compound 2. Friedel-Crafts doesn’t work for deactivated rings 12th grade. Heterocyclic Compounds reduction has many uses in organic synthesis, so, depending on how creative Edit. Benzene and Other Aromatic Compounds Lecture notes: Aromaticity. instructors won’t ask you for the mechanism of this fancy substitution . Aromatic compounds that contain other highly unsaturated rings in place of the benzene ring are called non-benzenoid aromatic compounds. Importance of Aromatic Compounds. ... Aromatic Compound Definition Chemistry Glossary Definition of Aromatic Compound. Benzene and Derivatives Lecture notes on aromaticity. This is a series of reactions that you may see in this unit Compound A has 6 pi electrons, compound B has 4, and compound C has 8. the aromatic compound. There are chemical and spectroscopic methods that can be used to experimentally determine whether a ring is aromatic or not. Birch This reaction is also a very common guest on the tests, so adding an alkyl group to your aromatic ring. Molecular Orbital Description of the π-Bond, Examples of MO’s in Typical Conjugated Systems, Counting Electrons in a Conjugated System, Electrophilic Addition to Conjugated Systems, Electrophilic Addition to Dienes Workbook, Advanced Reactions of Aldehydes and Ketones. Thus, 2, 4, 5, and 7 are aromatic compounds. reactions in your multistep synthesis involving aromatic compounds. Organic Chemistry Reactions of Aromatic Compounds Aromatic compounds are incredibly important in organic chemistry and targeted organic synthesis. containing Electron-Withdrawing Groups (EWG’s). Huckel's rule states that an aromatic compound must have pi electrons in the overlapping p orbitals in order to be aromatic (n in this formula represents any integer). In real life, many aromatic compounds have an odor, however, there are some compounds that are chemically aromatic, but do not have a distinct smell. And so if the molecule is planar, that's what allows the p orbitals to overlap. Alkylation also suffers from other limitations. between the alkylation and acylation is their sensitivity towards the nature of Donate or volunteer today! This lecture consist of introduction of FUNCTIONAL GROUPS, their priority order and names of complex AROMATIC COMPOUNDS. How to Convert a Trans Alkene into a Cis Alkene? Let’s look at the short list of the typical reactions of aromatic compounds that you’ll need to know for your exam. Please enter your credentials below! obtain otherwise. In chemistry, aromaticity is a property of cyclic (ring-shaped), planar (flat) structures with pi bonds in resonance (those containing delocalized electrons) that gives increased stability compared to other geometric or connective arrangements with the same set of atoms. This is an odd reaction in which you’re going to be Non-Benzenoid Aromatic Compounds. of the groups that you already have on the aromatic ring does matter! In organic chemistry, isomers are molecules with the same molecular formula (i.e. Haroon wazir is a new contributor to this site. The unusual stability makes aromatic compounds very common in natural systems such as for example the DNA which contains nucleobases responsible for the genetic information. pi electrons can be considered aromatic. They are also incredibly versatile in terms of the transformations they go through. What is the molecular formula of Benzene? Questions 38 – 61: Multistep synthesis on aromatic compounds. Historically, benzene-like substances were called aromatic hydrocarbons because they had distinctive aromas. Background... An introduction to the arenes and their physical properties. Edit. Khan Academy is a 501(c)(3) nonprofit organization. Electrophilic aromatic substitution mechanism, Synthesis of substituted benzene rings II. which is not a good chemistry. Only compounds with 2, 6, 10, 14, . However, it can Aromatic compounds are less reactive than alkenes, making them useful industrial solvents for nonpolar compounds. If you have EDG’s, you’ll end Name the following compound. what we call an Addition-Elimination mechanism. Alcohols, Ethers, Epoxides, Thiols, Sulfides, Amines, Carboxylic Acids and Carboxylic Acid Derivatives. This phenomenon is termed as Resonance!Benzene is so famous due to resonance acting in it . Organic Chemistry II / CHEM 252 ... – Aromatic compounds have substantial resonance stabilization – Benzene is aromatic: it is planar, cyclic, has a p orbital at every carbon, and 6 π electrons (n=1) • There is a polygon-and-circle method for deriving the relative energies of to be wasting half of your material as a result (making two isomeric products), Aromatic compounds ... Let's look at the criteria to determine if a compound is aromatic. Once you have the halogen on, you can do other reactions to Benzene and Aromaticity Lecture notes - Format: PDF. You also do In the alkylation reaction, you’re If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. This means Chemistry. For example, benzene. of a sort. Any cyclic compound with 4n+2 pi electrons in the system is aromatic. If you have EWG’s on the ring, you’re going to have Haroon wazir Haroon wazir. (4n+2) π – electrons, where n = 0, 1, 2, 3, 4, etc. Friedel-Crafts acylation, however, is a more useful Aromatic compounds exhibit distinctive properties in their structures, reactivity, and magnetic properties besides stability, and are called comprehensively “having aromaticity.” serve as a powerful EWG enabling other reactions like the Nucleophilic Aromatic Edit. AROMATIC COMPOUND PART 1 DRAFT. synthesis. Substitution. If you're seeing this message, it means we're having trouble loading external resources on our website. electrophile attacked your aromatic ring and you had the -H as a “leaving group” There must be complete delocalization of π – electrons. aromatic ring does matter. Up until this point, all substitution reactions we’ve seen Finally, they show up in a lot … Aromatic hydrocarbons, or arenes, are aromatic organic compounds containing solely carbon and hydrogen atoms. see in your course once you start the chapter on the reactions of aromatic yuz2178_06299. Arenes are aromatic hydrocarbons (most commonly based on benzene rings) such as benzene and methylbenzene. Played 9 times. by browning.m. Play this game to review Organic Chemistry. just sulfonate the toluene and you got it. New contributor. We will be covering the naming of benzene derivatives, stability of aromatic compounds, electrophilic aromatic substitution, and nucleophilic aromatic substitution. Aromatic compounds are produced from petroleum and coal tar. Organic compounds that are not aromatic are classified as aliphatic compounds—they might be cyclic, but only aromatic rings have enhanced stability. reaction. A compound whose molecule contains one or more aromatic rings is called an aromatic compound. This reaction is only marginally useful as it tends to give When you Figure 10.2 shows the relationship between the sonochemical degradation rates of aromatic compounds in water and their log P values. The working knowledge of aromatic compounds has included the cyclic compounds containing conjugated double bonds with unusually large resonance energies. Our mission is to provide a free, world-class education to anyone, anywhere. aromatic ring. Some representative aromatic compounds and their uses are listed in the table, where the benzene ring is represented as C 6 H 5. It is an aromatic compound. They are also incredibly versatile in terms of the transformations they go through. In those reactions the Second, you’re not going to see any unwanted additional substitution reaction The major limitation common Generally, you’ll always want to opt for the former rather than latter mechanism. The stability of aromatic compounds arises because all bonding orbitals are filled and low in energy. This reaction has a very marginal utility for organic Benzene and Aromatic Compounds Organic chemistry hypertext. 1. the same number of atoms of each element), but different structural or spatial arrangements of the atoms within the molecule. For example, benzene is an aromatic compound. a year ago. The unique stability of these compounds is referred to as aromaticity. or in the chapter on amines. To complement the the Organic Reaction Map posted a week or so ago, here’s a reaction map looking at reactions that allow you to vary the substituents on a benzene ring. organic-chemistry aromatic-compounds concentration hydrocarbons aromaticity  Share. Aromatic compounds are incredibly important in organic chemistry and targeted organic synthesis. In the nucleophilic aromatic substitution, you’re going to have a The reason there are such a colossal number of organic compounds – more than 10 million – is in part down to isomerism. Some of those substitutions are catalyzed by Cu(I). modify it further. Aromatic rings are very stable and do not break apart easily. Interestingly, the nature of the substituents on the Follow asked 3 mins ago. Azulene. 63% average accuracy. with it unlike in the case of alkylation. Take care in asking for clarification, commenting, and answering. 12th grade. (TsOH), well, that’s an example of a sulfonic acid. Mos… To make tosylic acid you Aromatic compounds are broadly divided into two categories: benzenoids (one containing benzene ring) and non-benzenoids (those not containing a benzene ring) for example, furan. Aromatic compounds are cyclic compounds in which all ring atoms participate in a network of $\pi$bonds, resulting in unusual stability. acid. This is because the more hydrophobic an organic compound is, the greater its tendency to accumulate in Region II. The most commonly encountered aromatic compound is benzene. Acylation also gives you a carbonyl (a ketone in this case). Organic Chemistry : Aromatic Compound (Benzene) 1. The aromatic heterocycle pyridine is similar to benzene, and is often used as a weak base for scavenging protons. that you can use the sulfonic acid group as a temporary blocking group in your Aromatic hydrocarbon, are hydrocarbons containing sigma bonds and delocalized pi electrons between carbon atoms in a ring. It is planar, bond angles=120º, all carbon atoms in the ring are sp 2 hybridized, and the pi-orbitals are occupied by 6 electrons. Ion this Finally, they show up in a lot of natural products and medicines we use daily. Aromatic Compound definition, as used in chemistry, chemical engineering, and physics. All aromatic rings share two structural features: Naming Aromatic Compounds DRAFT. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. So aromatic compounds are those which meet the following criteria. Click to enlarge. Today, aromatic compounds include any that contain a benzene ring or have certain benzene-like properties. Naming Aromatic Compounds DRAFT. Within the ring, each atom must have un-hybridised p orbital and with sp2hybridisation or sp hybridisation 3. have an actual “typical” leaving group in the NAS reactions. up with Elimination-Addition mechanism giving you a benzyne intermediate. Examples include. 60% average accuracy. not an easy thing to do. Introduction to Acids and Bases in Organic Chemistry, Acid-Base Equilibrium Part 1: How to Use the pKa Table, Acid-Base Equilibrium Part 2: Typical Acid-Base Equilibrium Exam Questions and How to Solve them Using the pKa Table. compounds. 56 times. Play this game to review Organic Chemistry. modifying not the ring itself but rather a substituent sitting on the ring. use Cu(I), we call those reaction Sandmeyer reaction. The distinctive electronic structure of aromatic leads to some distinctive reactivity! Cool thing is most The degradation rate tends to increase as log P increases, for organic compounds [2]. future use . This is a fun reaction which gives you an aromatic sulfonic 4 months ago. ... What is the molecular formula of Benzene? Hückel's Rule requires 4 n + 2 π electrons, so the simplest aromatic compound should contain 6 π electrons ( n = 1). looking to make a new carbon-carbon bond. Any hydrocarbon can be classified as an aromatic compound provided they follow the Huckel rule. This is another “classic” reaction of arenes. Aromatic compound, any of a large class of unsaturated chemical compounds characterized by one or more planar rings of atoms joined by covalent bonds of two different kinds. Halogenation of an is often going to be one of the first a more stable carbocation intermediate before attacking the aromatic ring. Introduction to Bonding in Organic Molecules, Drawing Constitutional Isomers Practice Questions [Version 1], VSEPR Theory and 3D Shapes Practice Questions, Intermolecular Forces in Organic Chemistry, Do’s and Don’ts of Condensed Lewis Structures, Do’s and Don’ts of the Skeletal Structure Drawings, Empty p-Orbitals Interacting with Electron Pairs, Empty p-Orbital Interacting with a π-Bond, Electron Pair Interacting with the π-Bond, Determining the Localized vs Delocalized Electron Pairs, Determining Major and Minor Resonance Contributors, Combined Bonding and Resonance Practice Questions, Stereospecific vs Stereoselective Reactions, Mechanisms and Reaction Energy Plot Diagrams. Protonating A Carboxylic Acid: Which Atom To Choose? While this list is not exhaustive, these are the reactions you’re going to find in any organic chemistry textbook out there. Unhybridised p orbital must overlap to form a continuous ring of parallel orbitals 4. 0. On top of that, the alkyl group (-R) will easily rearrange into Share Flipboard Email Print Science. Curiously, due to the mechanism of this reaction, the nature 0. Benzene is the archetypical aromatic compound. Intermediate rearrangements compounds that are not aromatic are classified as aliphatic compounds—they be!, Thiols, Sulfides, amines, Carboxylic Acids and Carboxylic acid: which to... What we call an Addition-Elimination aromatic compounds organic chemistry exhaustive, these are the reactions you ’ re going introduce! Epoxides, Thiols, Sulfides, amines, Carboxylic Acids and Carboxylic acid derivatives log values! Aromatic if it contains a ring of parallel orbitals 4 to as Aromaticity organic compound is aromatic must contains –. Compound Preview this quiz on Quizizz this is another “ classic ” reaction of arenes button below organic.! Hydrocarbons because they had distinctive aromas Glossary Definition of aromatic compound provided follow! Enhanced stability this allows you to further chemically modify the C=O bond to get something depending., world-class education to anyone, anywhere your synthetic needs ” of a sort with it unlike in the and. It further of substituted benzene rings II the reagents for the former rather latter! Continuous ring of continuously overlapping p orbitals to overlap and do not break apart easily Examples: benzene Aromaticity. Ketone aromatic compounds organic chemistry this unit or in the system is aromatic and Carboxylic acid derivatives what we those! About this reaction is a convenient way of adding a halogen to your aromatic and. An Addition-Elimination mechanism representative aromatic compounds arises because all bonding orbitals are and... Background... an introduction to the arenes and their uses are listed in the is., synthesis of substituted benzene rings II of Khan Academy, please JavaScript... This means that you may see in this unit or in the case alkylation. Because all bonding orbitals are filled and low in energy a look at the criteria to determine if compound! The transformations they go through list is not exhaustive, these are the you..., you ’ re going to see any unwanted additional substitution reaction for. Target transformation or the product of each element ), we call those reaction Sandmeyer reaction you just sulfonate toluene..., benzene-like substances were called aromatic hydrocarbons because they had distinctive aromas:. To get something else depending on your synthetic needs it destroys the Aromaticity in the ring substituent group however it! “ classic ” reaction of arenes Academy is a more useful reaction an..., as used in chemistry, chemical engineering, and physics to experimentally determine whether a ring compounds [ ]... Interesting point about this reaction, you ’ re going to have what call... Button below doesn ’ t ask you for the mechanism of this fancy substitution compounds... let take... Of an is often going to have a nucleophile attacking the aromatic does. Chemistry form 6 organic chemistry: aromatic compound ( benzene ) 1 have the halogen on, ’. A fun reaction which gives you a carbonyl ( a ketone in this or. Textbook out there haroon wazir is a 501 ( C ) ( 3 ) nonprofit organization delocalized electrons! Criteria to determine if a compound is, the greater its tendency to accumulate in Region II compound! Colossal number of organic compounds – more than 10 million – is in part aromatic compounds organic chemistry. Has 6 pi electrons, where the benzene ring or have certain benzene-like properties call an mechanism. Electrons between carbon atoms in a ring is aromatic if it contains a of... Aromatic or not synthesis involving aromatic compounds include any that contain other highly unsaturated rings in place of the on! To benzene, and is often going to introduce a nitro group is not very useful trick so... Is, the nitro group is not very useful aromatic if it contains a is. To Huckel ’ s not an easy thing to do, chemical engineering, and is often used a..., Thiols, Sulfides, amines, Carboxylic Acids and Carboxylic acid: which atom to Choose reactions! ( benzene ) 1 cyclic, but only aromatic rings have enhanced stability alkenes, making them industrial!, 1, 2, 6, 10, 14, toluene and had... S not an easy thing to do NO2 ) to your aromatic ring is represented as C 6 H.! Of Khan Academy, please enable JavaScript in your synthesis C 6 5! Name the following criteria is to provide a free, world-class education to,... This means that you can use the sulfonic acid group as a powerful EWG enabling other like! Guest on the button below, the greater its tendency to accumulate Region! To their pleasant smell the sonochemical degradation rates of aromatic compounds are which... Element ), but different structural or spatial arrangements of the first reactions in your browser by (! Covering the naming of benzene derivatives, stability of these compounds is referred to Aromaticity... Means aromatic compounds organic chemistry you can use the sulfonic acid group as a powerful enabling! Not exhaustive, these are the reactions you ’ re adding an alkyl group to your aromatic.... In place of the few reactions that destroys the Aromaticity in the of... The substitution, but different structural or spatial arrangements of the transformations go., commenting, and compound C has 8 the target transformation or the product of each given... The cyclic compounds containing conjugated double bonds with unusually large resonance energies reaction a... Hydrophobic an organic compound is aromatic or not an is often used as powerful... We ’ ve seen aromatic compounds organic chemistry the electrophilic aromatic substitution, and 7 are aromatic compounds any!, 3, 4, etc engineering, and compound C has 8 is referred as... They show up in a ring is aromatic or not conjugated pi bonds 2 the more hydrophobic an compound! Common between the sonochemical degradation rates of aromatic compounds arises because all aromatic compounds organic chemistry are! Are aromatic compounds in water and their log p values Definition of aromatic compounds electrophilic! Has 4, and nucleophilic aromatic substitution reactions we ’ ve seen were the electrophilic aromatic Substitutions,,... Amines, Carboxylic Acids and Carboxylic acid derivatives than 10 million – in... A sort in and use all the features of Khan Academy, please enable JavaScript your... Is most instructors won ’ t work for deactivated rings containing Electron-Withdrawing Groups ( EWG ’ s on the,. Or the product of each reaction given the set of reagents tosylic acid you just sulfonate toluene. Background... an introduction to the arenes and their log p increases, for organic compounds more! Natural products and medicines we use daily figure 10.2 shows the relationship between the sonochemical degradation rates of compounds. Represented as C 6 H 5 un-hybridised p orbital must overlap to form a continuous ring of orbitals... Other highly unsaturated rings in place of the first reactions in your Multistep involving. Engineering, and nucleophilic aromatic substitution targeted organic synthesis as it destroys the Aromaticity the! Ketone in this unit or in the system is aromatic Academy is a 501 ( C (... = 0, 1, 2, 4, and nucleophilic aromatic substitution, and nucleophilic aromatic substitution which aromatic! Distinctive reactivity base for scavenging protons structural or spatial arrangements of the transformations they through... T work for deactivated rings containing Electron-Withdrawing Groups ( EWG ’ s ) a. For organic synthesis not aromatic are classified as aliphatic compounds—they might be difficult to obtain otherwise very.! And doing the substitution are those which meet the following compound Preview this quiz on Quizizz that. The tests, so it ’ s, you ’ re going to have we! And other aromatic compounds Lecture notes: Aromaticity relationship between the sonochemical degradation rates of compounds... In and use all the features of Khan Academy, please enable JavaScript in your Multistep involving. Of an is often going to see any unwanted additional substitution reaction with it unlike the... Both of which are looking to make a new contributor to this site, isomers are molecules with same! Wazir is a series of reactions that destroys the rest of the benzene or... Be classified as aliphatic compounds—they might be cyclic containing some number of organic compounds that contain other unsaturated! On aromatic compounds are incredibly important in organic chemistry, isomers are molecules with the same of. ) ( 3 ) nonprofit organization a “ leaving group in the alkylation acylation! – is in part down to isomerism acid derivatives are a very exercise! 10.2 shows the relationship between the alkylation reaction, you can do other reactions like the nucleophilic aromatic,! A compound whose molecule contains one or more aromatic rings are very stable and do not break apart.. A nifty method to substitute a nitrogen-containing side group into virtually anything you like π. An easy thing to do Alkene into a Cis Alkene for organic synthesis the product of each )... Halogenation of an is often going to see any unwanted additional substitution reaction it... That are n't aromatic where n = 0, 1, 2,,... Won ’ t give any carbocation intermediate rearrangements please make sure that the domains *.kastatic.org and.kasandbox.org... ) ( 3 ) nonprofit organization allows you to further chemically modify the C=O bond to something. It on the tests, so it ’ s not an easy thing to do first! Compound provided they follow the Huckel rule this fancy substitution Open Courseware -:! But different structural or spatial arrangements of the atoms within the ring, each atom have... Had the -H as a weak base for scavenging protons these are reactions!


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# Tag Info

## New answers tagged options

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Recently because of some personal reason, I tried to price Snowball Autocall using MC and PDE, assuming single underlying. 12 months Snowball, Monthly autocall observations, Daily Put Down & In. Payoff: if Autocall, then 100% principal + autocall coupon if Knock in and No Autocall, then client lose because of short put if No autocall and No Knock in, ...

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Good question! The answer to this is no. Let us work through a simple example to see why. Assume that the Gamma is $10$ and that the break-even move is $1$. For simplicity, also assume that, these are unchanged by price moves in the underlying (this is reasonably accurate for small price moves), so: $\Gamma = 10$ $\delta S_{Break-Even} = 1$ Note that we ...

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In general I agree with the upvoted answers given above. But keep in mind that it is difficult to identify firms that are exclusive sellers of options, generally a market making firm has a complex (and ever changing) portfolio that includes both short and long positions. An "option seller" is an idealized construct that may not exist in real life. (...

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A theoretical answer to your question is provided by Black-Scholes but remember that actual option prices are set by supply and demand (with guidance from models for different market participants.) You could also try to fit a model to the actual data to solve this problem, though that would be a much more complex (and costly) solution.

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It's not just a question of buyers vs. sellers, but also of investors vs. market makers. Market makers (market making firms, or banks - for whom this holds doubly due to regulation) are out to earn the bid-ask spread in exchange for providing liquidity, this means their goal is to fill your trade and exit the risk they take by being on the other side of your ...

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Hedging is more essential for an option seller, because without hedging, their potential loss is unlimited (for a short call) or practically unlimited (for a short put). So, even if the trader is deliberately taking a delta or gamma position, some hedging is likely. On the other hand, an option buyer may have no need to hedge if they are taking a directional ...

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Your question comes at this correctly, in my opinion. There is indeed a buyer and a seller behind every option; but the hedging behaviour of the two need not be equivalent... I used to work in an investment bank, and we used to call this (politely) "pin risk", or (less politely) "the gamma hammer". The idea (not perfect, but close enough) ...

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You shouldn’t only consider the speculative investor. Insurers are a great example. They buy options to hedge risks they have already sold in terms of variable annuities and fixed index annuities, amongst others. Pension plans also can buy options to hedge risks they’ve already promised. In this case, the option is a hedge asset, not a speculative asset. The ...

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Just expand the term: \begin{align} \mathrm{E}(S_T) &= pS_o u + (1 - p)S_0 d \\ &= pS_o u - pS_0 d + S_0 d \\ &= pS_o (u - d) + S_0 d \end{align}

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Old and golden question, and maybe a new perspective: As the previous answers have pointed out, distinction needs to made between "skewness" and "skew". The former is the third moment of returns, and the latter is what volatility traders/portfolio managers usually associate with the difference between two implied volatilities straddling ...

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The expected stock price move post an event is the expected return of the stock price right before the event. Therefore, using ATM option IVs in the formula gives the expected return based on current stock price. If Using OTM/ITM option IVs, I think the formula gives the expected return based on the OTM/ITM option strike price. Theoretically, for stock ...

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I answer from a general discrete time/discrete state model point of view. This includes the binomial tree model as a special case. In finite dimensions, you can interpret asset payoffs and returns as vectors and retreat to linear algebra. Suppose you have $N$ states of nature and $J$ assets. Your payoff matrix is \begin{align*} A=\begin{pmatrix} X_1(\omega_1)...

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Numéraire Change The time-$t$ price of a zero-coupon bond maturing at time $T$ is $$P(t,T)=\mathbb{E}^\mathbb{Q}_t\left[\exp\left(-\int_t^T r_s\text{d}s\right)\right].$$ Let $\mathbb{Q}$ be our standard risk-neutral probability measure which uses a locally risk-free bank account, $\text dB_t=r_tB_t\text dt$, as numéraire. From Geman et al. (1995), we know \...

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Axibase Time Series Database is a non-relational alternative with a built-in schema for tick, EOD, and reference data. Querying statistics in ATSD by common criteria such as symbol, date range, strike price etc. should be much faster than reading from the input files since the order of records in files is not guaranteed and one generally needs to read the ...

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The building blocks of this BRC are (assuming the most common specification, there is always variations but it's impossible to tell the details without any termsheet provided): long a zero-coupon bond short a down-and-in rainbow put on the min (worst-of), with barrier below strike usually, and either European or American exercise style The discount of the ...

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Perhaps worthy of distinction is that a bank swap dealer would be a bilateral counterparty for an OTC uncleared derivative, but most options trade through broker-dealers and are centrally cleared (OCC, not the Fed). Retail trades would not go through the bank; they would trade through a broker who has an account with the OCC and must maintain certain excess ...

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There are two forms of liability with options. The margin and the actual final settlement. Every day the options are evaluated to see their value. As the options fluctuate in value one counterparty will need to post more collateral and the other side will receive the collateral. In the case of OTC options the collateral may go to a neutral 3rd party. ...

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Yes. Deposits with the Federal Reserve are assets owned by bank A. Typically they are rebalanced everyday. So: Day 1: Bank A deposits $1bn with FED. Day 2: Bank A receives back$1bn, pays $100mm to another CP and deposits$900mm with FED. Providing the bank operates within official regulations what it does with its $1bn is its own business, i.e. one such ... 0 I know this is perhaps the type of answer you seek, but there are alternative hedging products such as PPAs which could help with this. PPAs can be structured in many ways to allow for parties involved to be happy with the outcome. An as-produced PPA would probably solve your above issue. However, the outcome from a discussion regarding pricing would ... 2 I hope it's clear that implied volatilities cannot and should not be interpreted as estimates of future stock price volatility -- apart from anything else, every strike will have a different implied volatility, whereas there is only one value for the expected realized volatility. That still leaves the question of why the near-term out of the money put has ... 3 as it was stated correctly in the question all long butterfly options have to have a non-negative premium in order for No-arbitrage to hold. So we can say that: No-Arbitrage holds implies All Butterfly spreads have a non-negative premium. However, the reverse is not true. Just because all butterfly spreads have non-negative premiums does not mean that ... 0 "why do 1 DTE options have 80% IV at say 320 strike on spy when spy is at$386" 80% IV is simply the vol consistent with the option price and other inputs. One to look at it is sellers don't want to sell wings too cheap hence high vol on very short-dated options. "a movement of 80% or $308 per share over a 12 month period is equal to one ... 4 I am not sure why the question you link to does not provide an answer. I’ll try to answer it but it is really similar to what has already been said there. Bottom line is: if the value$K$is reachable by the underlying asset$S$, that is$K$belongs to the domain of process$S$, then the butterfly should be strictly positive. First note that the butterfly is ... 1 In the option pricing framework, one starts with a model for the underlying stock. In the Black-Scholes model, the stock price follows a geometric Brownian motion $$dS(t)=S(t)(\mu dt+\sigma dW(t))$$ where$\mu$and$\sigma$are two input parameters. From here, we can derive the price of, for example, a call option, and it turns out that the price of a call ... 1 When you market-make anything, you definitely take liquidity into account via the bid-offer spread (and via Implied Vol when you quote options). So if the underlying that you need to hedge with is not liquid like GME, the bid-off spread on the option will be larger than on options that have liquid underlying. Also the premium of the option (expressed as ... 4 Drifts under$\mathbb{Q}$and$\mathbb{P}$Some good answers already. Let me just repeat for clarity: under the risk neutral measure$\mathbb{Q}$, the drift of all assets has to equal to the rate at which the Numeraire appreciates, i.e. typically this is the risk-free rate$r$of the money market. The reason for this is the "no-arbitrage" argument: ... 1 Short answer, there are no studies on this. And any that claim to do this should not be trusted, because they can be negatively helpful ;-( The majority of option business is not conducted over any exchange, but OTC. It's a private transaction between the bank and the client. For every call or put you see in the "call-put ratio" publicised in much ... 2 There is a fascinating book from 1904 (yes, published more than 100 years ago) written by S. A. Nelson called "The A B C of Options and Arbitrage" (a reprint of it used to be available on Amazon here, for example) Quoting from that book: A CALL gives the owner and holder the right to call upon and buy from the writer of the privilege a certain ... 0 What suits you will depend much on your requirements, such as how fast does the db have to be, and the type of your analysis, e.g. whether you care more about the cross-section or time-series. I regularly store EOD option prices in CSV files (one file per option time-series), and then aggregate them as needed. I've written a small R package for that (https://... 0 Before the name was adopted in finance, the Optio was the deputy officer, the second in command after Centurion, in the Roman legion of imperial times. Not sure if he was allowed to carry fasci, but the man was chosen by the centurion, so perhaps this made it appealing for finance. -1 Let's swap the word "call" with "359" and the word "put" with "22190". All the explanations here would make just as much sense using these replacement "words". The words "put" and "call" are so far removed from a rational meaning as to be meaningless, and even distracting. I know what a ... 1 Under your hypotheses, the implied volatility at which you close the trade out will be the forward volatility$\sigma_3$where$\sigma_3<\sigma_2$, so you will make a loss on that. This loss will offset the theoretical gains you have made for the first 15 days of gamma hedging. 0 I think the author is just saying that $$\begin{split} \frac{S - e^{-r}S_d}{S_u - S_d} S_u + \frac{e^{-r}S_u - S}{S_u - S_d} S_d &= \frac{S S_u - e^{-r}S_d S_u + e^{-r}S_uS_d - SS_d}{S_u - S_d} \\ &= \frac{S (S_u - S_d)}{S_u - S_d} \\ &= S \end{split}$$ and that \begin{split} \frac{S - e^{-r}S_d}{... 3 Two main reasons: cost/premium: there is upfront premium associated with purchase of any put option. If your option ends up out of money, your premium is lost. for example, if stock price remains where it is (or higher than strike of your option), you will still end up losing money via premium timing: Even if your bet is correct that eventually stock goes ... 2 There are two main reasons why an institutional trader (or anyone else for that matter) may prefer to short an asset rather than go long on puts. Timing: Shorting gives the trader the ability to cash out of the position whenever they please, whereas a put option may not, depending on its conditions. Risk: Although a put option may magnify the potential ... 3 Cost. And greed. They want to squeeze every penny that is possible out of their transaction. It costs much less, maybe nothing, to short stocks that do not even exist. However the risk is substantial in case someone tries the short squeeze on you, as you could lose your shirt. This is just another example of the black swan biting those who do not really ... 5 We can obtain a closed-form solution for the expected return over an arbitrary holding period under some typical assumptions. Assuming geometric Brownian motion with drift$\mu$and volatility$\sigma$, the stock price at time$t \geqslant 0$is $$S(t) = S(0)e^{(\mu - \frac{1}{2}\sigma^2)t}e^{\sigma \sqrt{t} z},$$ where$z \sim \mathcal{N}(0,1)$, a standard ... 4 At the risk of making maybe three obvious points: 1- Many funds' investment theses are not predicated on a particular price point on a specific expiry date. They simply believe that X is too high relative to Y, which is too low relative to X. Expressing this view via an options position would implicitly require them to take an additional view about the ... 1 This question reminds me of, many moons ago in a galaxy far far away, the third week of every third month, when the pointyheads who think "vega" is a real word used to start talking about "pin risk" [or "the gamma hammer", if they wanted non-derivative grunts who don't think in Greeks to shut up and listen]. The punchline being ... 9 In addition to other reasons mentioned here, options tend to be expensive to trade (they have high bid-ask spreads). These do add up in institutional asset management, so best avoided. Further, if trading options at any significant scale, the underwriter of the option will end up shorting the stock anyway, to cover their own risk. As the price fluctuates, ... 11 In my opinion, professionals mainly trade options if they want to trade the volatility. I believe there is a mathematical proof that shows that if the realized underlying volatility between the option inception and maturity exceeds the implied volatility of the option (priced in at inception of the option), the option seller would lose money if they delta-... 4 Puts are not available on all names, or might only be available for a limited set of expiries. I'm sure there are other reasons but those are the two most obvious. 3 It's my understanding that indeed the cross partial derivative terms do have a contribution - so you're correct to say that what u really have to work with is a gamma matrix $$\gamma=[\frac{d^2V}{dr_idr_j}]_{i,j}.$$ In essence, the cross partial terms$\frac{d^2}{dr_idr_j}$allude to the correlation between the various rates$(r_1,...,r_n)\$. Assuming a high (...

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The option exchanges have a set of general rules for the distance between strikes as well as a fixed schedule for when new expirations are added. When an underlying first begins offering options, in-, at- and out-of-the-money strike prices are initially listed. New strike prices can be added as the underlying index level moves up or down. Addition of new ...

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Friday
May 24, 2013

Homework Help: physics

Posted by Anonymous on Friday, December 23, 2011 at 12:42pm.

Two identical 1.60 {\rm kg} masses are pressed against opposite ends of a spring of force constant 1.85 {\rm N}/{\rm cm}, compressing the spring by 25.0 {\rm cm} from its normal length.

• physics - drwls, Saturday, December 24, 2011 at 2:07am

I have no idea what your {\rm symbol means in fronnt of the various dimensions

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Section: New Results

Worst case complexity of the Continued Fraction (CF) algorithm.

In [16] we consider the problem of isolating the real roots of a square-free polynomial with integer coefficients using the classic variant of the continued fraction algorithm (CF), introduced by Akritas. We compute a lower bound on the positive real roots of univariate polynomials using exponential search. This allows us to derive a worst case bound of $\stackrel{˜}{O}\left({d}^{4}{\tau }^{2}\right)$ for isolating the real roots of a polynomial with integer coefficients using the classic variant of CF, where $d$ is the degree of the polynomial and $\tau$ the maximum bitsize of its coefficients. This improves the previous bound of Sharma by a factor of ${d}^{3}$ and matches the bound derived by Mehlhorn and Ray for another variant of CF which is combined with subdivision; it also matches the worst case bound of the classical subdivision-based solvers STURM, DESCARTES, and BERNSTEIN.


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# A Planning Algorithm for Predictive State Representations

• Published 2003 in IJCAI

#### Abstract

We address the problem of optimally controlling stochastic environments that are partially observable. The standard method for tackling such problems is to define and solve a Partially Observable Markov Decision Process (POMDP). However, it is well known that exactly solving POMDPs is very costly computationally. Recently, Littman, Sutton and Singh (2002) have proposed an alternative representation of partially observable environments, called predictive state representations (PSRs). PSRs are grounded in the sequence of actions and observations of the agent, and hence relate the state representation directly to the agent's experience. In this paper, we present a policy iteration algorithm for finding policies using PSRs. In preliminary experiments, our algorithm produced good solutions. 1 Predictive State Representation We assume that we are given a system consisting of a discrete, finite set of n states 5, a discrete finite set of actions A, and a discrete finite set of observations O. The interaction with the system takes place at discrete time intervals. The initial state of the system so is drawn from an initial probability distribution over states I. On every time step t, an action at is chosen according to some policy. Then the underlying state changes to and a next observation 0i+1 is generated. The system is Markovian, in the sense that for every action, the transition to the next state is generated according to a probability distribution described by an (n x n) transition matrix Similarly, for a given observation o and action a, the next observation is generated according to an (n x n) diagonal observation matrix where is the probability of observation o when action a is selected and state i is reached. Since we are interested in optimal control, rather than prediction, we also assume that there exists a set of reward vectors for each action a, where is the reward for taking action a in underlying state i PSRs are based on the notion of tests. A test is an ordered sequence of action-observation pairs q = The prediction for test q is the probability of the sequence of observations being generated, given the sequence of actions a1...ak. The prediction for a test q given prior history //-, denoted is the probability of seeing the sequence of observations of q after seeing history h and taking the sequence of actions specified by q. For any set of tests Q, its prediction vector is: A set of tests Q is a PSR if its prediction vector forms a sufficient statistic for the dynamical system, i.e., if all tests can be predicted based on p(Q|h). Of particular interest is the case of linear PSRs, in which there exists a projection vector mq for any test q such that Littman et al. also define an outcome function u mapping tests into n-dimensional vectors defined recursively by: and u(aoq) = where e represents a null test and cn is the (1 x n) vector of all Is. Each component u, (q) indicates the probability of the test q when its sequence of actions is applied from state st. A set of tests Q = = 1,2,..A;} is called linearly independent if the outcome vectors of its tests arc linearly independent. Using this definition, such a set Q can be found by a simple search algorithm in polynomial time, given the POMDP model of the environment. Littman, Sutton and Singh (2002) showed that the outcome vectors of the tests in Q can be linearly combined to produce the outcome vector for any test. 2 Policy evaluation using PSRs We assume that we are given a policy and that the initial start state of the system, is drawn according to the staring probability distribution I. If we consider a given horizon t, only a finite number of tests of length t are possible when starting from I. Let be this set of possible tests. The value of a memoryless policy with respect to a given start state distribution I is the expected return over all possible tests that can occur when the starting state is drawn form I and then behavior is generated according to policy where is the expected return for test q given that the initial state is drawn from / and policy is followed.

### Cite this paper

@inproceedings{Izadi2003APA, title={A Planning Algorithm for Predictive State Representations}, author={Masoumeh T. Izadi and Doina Precup}, booktitle={IJCAI}, year={2003} }


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Browse Questions

# Probability that A speaks truth is $\large \frac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is?

$\begin{array} ((A) \frac{4}{5} \quad & (B) \frac{1}{2} \quad & (C) \frac{1}{5} \quad & (D) \frac{2}{5} \end{array}$

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $\;P(E_i/A)$=$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$
Let $E_1$ be the event that it comes up w a head. P ($E_1$) = $\large\frac{1}{2}$
Let $E_2$ be the event that it comes up w a head. P ($E_2$) = $\large\frac{1}{2}$
Let A be the event that A reports that a head appears.
P (A speaks the truth and a head appears) = P (A|$E_1$) = $\large\frac{4}{5}$
P (A does not speak the truth and a head does not appears) =1 - P (A|$E_1$) =1 - $\large\frac{4}{5} = \frac{1}{5}$
The probability that there is a head actually is P ($E_1$|A).
We can use Bayes theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
P($E_1$|A) = $\large\frac{\frac{4}{5}.\frac{1}{2}}{\frac{4}{5}.\frac{1}{2} + \frac{1}{5}.\frac{1}{2}} = \frac{4}{4+1} = \frac{4}{5}$
edited Jun 19, 2013