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# Calling All 2011 Booth Applicants!!

Author Message
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Re: Calling All 2011 Booth Applicants!! [#permalink]  10 Sep 2010, 09:55
a 520 is a bit too low. I'll be honest with you (and I'm going to assume that the other part of your application is an average-joe type of deal and I'll assume that Obama is not writing you a letter of rec), I think you would not be considered with a 520 at all.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  10 Sep 2010, 10:52
1
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kunoflo wrote:
Hi guys, I need your honest opinion... Do you need that with a 520 in the Gmat it would be a waste applying to Both? (even if I have a good application).

Please be blunt, everyone says "Just try" but you guys know how painful it is to go through the application process and then get rejected. Help is appreciated... Thanks!

If you got a 3.9 gpa in college, then I assume that you're not totally dumb and you're at least a decent test taker. I would retake the GMAT to try to get at least a 650. You're only as strong as the weakest link, so if you can do anything to increase your score, that'll definitely better your chances.

If you don't do that, I think your chances are slim to none.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  11 Sep 2010, 00:17
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There's actually a question about this on the Booth forums. They basically said that your report has to be in before you do your interview. I'm in the same boat because I need to fly into the US to take the GMAT and I can't do it until Oct 7th.

That said, add me in for R1.

bakfed wrote:
Michmax3 wrote:
I sent an email to admissions asking if I was ok to apply in R1 and self report my score even though I am not re-taking the GMAT until Oct. 2 and GMAC says allow 20 days for the report to reach the school. They wrote back saying go ahead and apply for R1. However I just started filling out the online application and there is a note by the GMAT field that says the official report must arrive prior to the round deadline or your application could be delayed.

So what do I do? I don't want to work on the application in time for R1 only to have it be pushed to R2 anyways, I'd rather spend that time working on the other school I was planning to submit in R1. Or will most likely my official report arrive before the 13th anyway?

I don't think anyone's in place to say that your score report will be sent to Booth before the 13th. From my past two experiences, the scores were sent to my designated schools within a week of taking the test. If you'd like to be reassured, I'd send another email to the admission staff at Booth and tell them about your situation and what the online application says. In addition, I would ask if there's any possibility that my application will be push to Round 2 because of the GMAT. Then I'd take whatever the staff says and do it accordingly.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  11 Sep 2010, 04:30
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kunoflo wrote:
Hi guys, I need your honest opinion... Do you need that with a 520 in the Gmat it would be a waste applying to Both? (even if I have a good application).

Please be blunt, everyone says "Just try" but you guys know how painful it is to go through the application process and then get rejected. Help is appreciated... Thanks!

If you got a 3.9 gpa in college, then I assume that you're not totally dumb and you're at least a decent test taker. I would retake the GMAT to try to get at least a 650. You're only as strong as the weakest link, so if you can do anything to increase your score, that'll definitely better your chances.

If you don't do that, I think your chances are slim to none.

Thank you for your honesty guys! I am an Italian F, entrepreneur, and a bad test taker but I got a good GRE score. I am happy that one of my favorite schools accepts it . I have also taken a Calculus (scored an A) and a Microeconomics class to strengthen my case.

I wish you the BEST of luck and hope you both get in.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  13 Sep 2010, 01:54
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I was quite pleased when a few weeks ago, both Chicago Booth and Kellogg contacted me via GMASS and asked me to look at their schools and upcoming events in my city. There's one for Booth very soon so I'll be sure to attend - and I'm quite sure I want to apply R1. Will update you guys when I finally throw my hat in. A priori, love everything about the school.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  13 Sep 2010, 08:57
osbornecox wrote:
I was quite pleased when a few weeks ago, both Chicago Booth and Kellogg contacted me via GMASS and asked me to look at their schools and upcoming events in my city. There's one for Booth very soon so I'll be sure to attend - and I'm quite sure I want to apply R1. Will update you guys when I finally throw my hat in. A priori, love everything about the school.

since this topic was brought up, I have a question about GMASS contacting a person. I was also contacted by Booth to check out its school. Does that mean I may be over the initial threshold (number-wise) or does GMASS just simply mass-email everyone to look at schools?
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Re: Calling All 2011 Booth Applicants!! [#permalink]  13 Sep 2010, 09:06
1
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bakfed wrote:
osbornecox wrote:
I was quite pleased when a few weeks ago, both Chicago Booth and Kellogg contacted me via GMASS and asked me to look at their schools and upcoming events in my city. There's one for Booth very soon so I'll be sure to attend - and I'm quite sure I want to apply R1. Will update you guys when I finally throw my hat in. A priori, love everything about the school.

since this topic was brought up, I have a question about GMASS contacting a person. I was also contacted by Booth to check out its school. Does that mean I may be over the initial threshold (number-wise) or does GMASS just simply mass-email everyone to look at schools?

I think there is a small degree of selection involved. I imagine that they wouldn't contact you, based on your GMAT information, if your GMAT was clearly subpar (I would like to think they aren't this cynical).

You probably had a few identifiers that, on paper, look like you should be in the running. You're biology background, California, 700, pharma work experience, right? Booth might find that interesting. Besides, you're in the 80/80 (or pretty close), right? I would check it out.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  14 Sep 2010, 06:21
1
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I'm in for R1 as well
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Re: Calling All 2011 Booth Applicants!! [#permalink]  14 Sep 2010, 06:35
2
KUDOS
gkumar wrote:
BruceLee wrote:
I am a reapplicant at Booth. Since I have to redo the PPT, it is fairly hard for me to come up with new themes and concepts to use in my new presentation. Should I use the same ideas and try to convey them differently? Is a certain degree of repetition acceptable?

Good q. I too am unsure on how to deal with the PPT as a reapplicant. Also the four pages are no longer just PPT, but doc, pdf, etc. I would appreciate any feedback on this as well.

I find it a bit tough to come up with 4 completely new slides...So personally i'm reusing 2 slides from last time (dinged with int) but am re-tooling the content..

E.g.: Lets assume that my new career goal is to be a consultant and i created a slide last time about mud wrestling...
This time i will re-use the same slide but talk about how mud wrestling has helped me discover a whole new aspect to due diligence etc..Basically the slide should reinforce your updated career goals (or whatever is diff. about yourself this time)...
The way i look at it, Essay 3 should resonate with the picture of yourself you are trying to present to the AdCom...Being creative for the sake of it is as bad as no creativity..

Just my 2 cents,rubles whatever....
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Re: Calling All 2011 Booth Applicants!! [#permalink]  14 Sep 2010, 07:41
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is booth tougher to get in than Stern? i am trying to make a decision... anyone can help?
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Re: Calling All 2011 Booth Applicants!! [#permalink]  14 Sep 2010, 08:43
Booth has a higher acceptance rate than Stern, but slightly better stats and a much better ranking. This tells me that although more people apply to Stern, they may be weaker in general than those who apply to Booth.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  14 Sep 2010, 19:35
osbornecox wrote:
bakfed wrote:
osbornecox wrote:
I was quite pleased when a few weeks ago, both Chicago Booth and Kellogg contacted me via GMASS and asked me to look at their schools and upcoming events in my city. There's one for Booth very soon so I'll be sure to attend - and I'm quite sure I want to apply R1. Will update you guys when I finally throw my hat in. A priori, love everything about the school.

since this topic was brought up, I have a question about GMASS contacting a person. I was also contacted by Booth to check out its school. Does that mean I may be over the initial threshold (number-wise) or does GMASS just simply mass-email everyone to look at schools?

I think there is a small degree of selection involved. I imagine that they wouldn't contact you, based on your GMAT information, if your GMAT was clearly subpar (I would like to think they aren't this cynical).

You probably had a few identifiers that, on paper, look like you should be in the running. You're biology background, California, 700, pharma work experience, right? Booth might find that interesting. Besides, you're in the 80/80 (or pretty close), right? I would check it out.

I'd like to think so, too, that I'm in the running. Was just curious and wanted to see if anyone knows about it. I'll be tossing my name in the hat for sure.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  15 Sep 2010, 17:33
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Does anyone know if recommenders are asked specific question in the Booth application? The website does mention specific questions but just gives generic guidelines on what the recommendation letter should cover. Any ideas?
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Re: Calling All 2011 Booth Applicants!! [#permalink]  15 Sep 2010, 19:34
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hetalcs wrote:
Does anyone know if recommenders are asked specific question in the Booth application? The website does mention specific questions but just gives generic guidelines on what the recommendation letter should cover. Any ideas?

It's a traditional letter of recommendation, but they ask that the recommender discuss specific topics.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  16 Sep 2010, 04:47
1
KUDOS
Hi guys

I am looking at potentially applying to a few schools in either R2 or R3 this year. Booth is going to be one of my top preferences given its strength in finance.

However, given the monetary nature of decision, I have been trying to dig up what is the median, mean and maximum compensation packages that graduates from booth have been getting over the past 2 or 3 years. I am unable to find this information on the website, does anyone know where to find this ? Or can current or former students help ? In particular I am looking at the consulting firms and at the financial firms (banks & HFs)
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Re: Calling All 2011 Booth Applicants!! [#permalink]  16 Sep 2010, 05:05
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KUDOS
Here you go:

http://www.chicagobooth.edu/employmentreport/index.aspx

shrouded1 wrote:
Hi guys

I am looking at potentially applying to a few schools in either R2 or R3 this year. Booth is going to be one of my top preferences given its strength in finance.

However, given the monetary nature of decision, I have been trying to dig up what is the median, mean and maximum compensation packages that graduates from booth have been getting over the past 2 or 3 years. I am unable to find this information on the website, does anyone know where to find this ? Or can current or former students help ? In particular I am looking at the consulting firms and at the financial firms (banks & HFs)
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Re: Calling All 2011 Booth Applicants!! [#permalink]  16 Sep 2010, 05:16
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shrouded1 wrote:
Hi guys

I am looking at potentially applying to a few schools in either R2 or R3 this year. Booth is going to be one of my top preferences given its strength in finance.

However, given the monetary nature of decision, I have been trying to dig up what is the median, mean and maximum compensation packages that graduates from booth have been getting over the past 2 or 3 years. I am unable to find this information on the website, does anyone know where to find this ? Or can current or former students help ? In particular I am looking at the consulting firms and at the financial firms (banks & HFs)

Hi shrouded1, I'm not sure where the confusion comes from. The median and the means aren't very far off from each other - $5,000, say, and that's not exactly going to be earth-shattering. Bulge brackets have pretty standard compensation programs and they usually pay$100,000 in starting salaries for Associates, with around $20,000 for sign-on bonuses and a year-end stub. A JP Morgan banking associate from Chicago Booth is going to get the exact same package as a new associate from Wharton, Columbia, Yale or Berkeley; there will simply be more offers to Booth students than Berkeley (geography, student body's orientation, alumni network strength, etc). Consulting is a similar sort of deal - the starting salary tends to$120,000 and it doesn't matter where you came from. With the buyside, they tend to take their cues from investment banks, so unless you're working for Mubadala, it's not going to be too far off, save for an outlier or two that will pay you $300,000 or$400,000.
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Re: Calling All 2011 Booth Applicants!! [#permalink]  16 Sep 2010, 07:03
1
KUDOS
Thanks for the replies !

I guess the real reason why I ask this question is that I am already working in the field, and if I were not to go for an MBA, by the time the fall 2011 batch graduates, in 2013, I would already have made VP.

So the real question in my mind is if going to one school rather than another helps me to negotiate a bigger package & better title for myself after graduation. I am already more senior in hierarchy and pay structure than a 1st year associate. So really am trying to figure out (if it is possible at all to figure out) what the higher end of the pay bracket is and where that lands me up in 2 years time.

Putting 150k into a degree is a very tough decision to make

If the 300k or 400k outlier exists, it might well be worth the shot ...
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Re: Calling All 2011 Booth Applicants!! [#permalink]  16 Sep 2010, 07:09
shrouded1 wrote:
Thanks for the replies !

I guess the real reason why I ask this question is that I am already working in the field, and if I were not to go for an MBA, by the time the fall 2011 batch graduates, in 2013, I would already have made VP.

So the real question in my mind is if going to one school rather than another helps me to negotiate a bigger package & better title for myself after graduation. I am already more senior in hierarchy and pay structure than a 1st year associate. So really am trying to figure out (if it is possible at all to figure out) what the higher end of the pay bracket is and where that lands me up in 2 years time.

Putting 150k into a degree is a very tough decision to make

If the 300k or 400k outlier exists, it might well be worth the shot ...

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Re: Calling All 2011 Booth Applicants!! [#permalink]  16 Sep 2010, 07:16
1
KUDOS
unf with the Volcker rule being implemented .... most if not all prop desks have been rooted out
if banks dont take big risks, there wont be big returns. If no big returns, then in my opinion the sell side careers will soon cease to be lucrative

now the question really is if the sell-side/buy-side switch is easier to make direct or via a B-school which opens up the market and facilitates you meeting not just one or two but a load of potential future employers
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Re: Calling All 2011 Booth Applicants!!   [#permalink] 16 Sep 2010, 07:16

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# Calling All 2011 Booth Applicants!!

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Batch correction (mass spectrometry data)
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5.6 years ago

Hi, I used my raw data for batch correction using Combat package. It worked well before normalization. However after normalization of batch corrected data, the correction is removed and my data showed clustering based on batch processed.

Could anyone please tell me if I have to normalize data before batch correction? Is it correct to use batch corrected data for normalization and further downstream processing?

Let me know if my question is not correct.

Thank you

R batch correction SVA Combat Biostatistics • 2.7k views
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Please use ADD REPLY/ADD COMMENT when responding to existing posts. It helps keeps threads logically organized.

It would be useful to add a sentence or two as to whether @ddiez's response was helpful, in the right direction or completely off-target. You should also up-vote and/or accept (use the checkmark against the post) the answer, which provides acknowledgement for the person who answers your question.

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ddiez ★ 1.9k

There are several alternative ways (or workflows) to deal with batch correction. I am assuming that you are using the bioconductor package sva. There are several examples in the vignette explaining how to analyze the data, in particular sections 7 and 8. Personally, I preprocess/normalize the data first, then include the batch information as a variable in the model (typically linear model with limma package).

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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Inequalities Using Multiplication and Division | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra - Basic Go to the latest version.

# 6.2: Inequalities Using Multiplication and Division

Created by: CK-12
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Equations are mathematical sentences in which the two sides have the same “weight.” By adding, subtracting, multiplying, or dividing the same value to both sides of the equation, the balance stays in check. However, inequalities begin off-balance. When you perform inverse operations, the inequality will remain off-balance. This is true with inequalities involving both multiplication and division.

Before we can begin to solve inequalities involving multiplication or division, you need to know two properties, the Multiplication Property of Inequalities and the Division Property of Inequalities.

Multiplication Property of Inequality: For all real positive numbers $a, \ b,$ and $c$:

If $x < a$, then $x(c)< a(c).$

If $x > a$, then $x(c) > a(c).$

Division Property of Inequality: For all real positive numbers $a, \ b,$ and $c$:

If $x < a$, then $x \div (c) < a \div (c).$

If $x > a$, then $x \div (c) > a \div (c).$

Consider the inequality $2x \ge 12$. To find the solutions to this inequality, we must isolate the variable $x$ by using the inverse operation of “multiply by 2,” which is dividing by 2.

$2x & \ge 12\\\frac{2x}{2} & \ge \frac{12}{2}\\x & \ge 6$

This solution can be expressed in four ways. One way is already written, $x \ge 6$. Below are the three remaining ways to express this solution:

• $\left \{ x | x \ge 6 \right \}$
• $[6, \infty)$
• Using a number line:

Example: Solve for $y: \frac{y}{5} \le 3$. Express the solution using all four methods.

Solution: The inequality above is read, “$y$ divided by 5 is less than or equal to 3.” To isolate the variable $y$, you must cancel division using its inverse operation, multiplication.

$\frac{y}{5} \cdot \frac{5}{1} & \le 3 \cdot \frac{5}{1}\\y & \le 15$

One method of writing the solution is $y \le 15.$

The other three are:

• $(-\infty ,15]$
• $\left \{ y | y \le 15 \right \}$

## Multiplying and Dividing an Inequality by a Negative Number

Notice that the two properties in this lesson focused on $c$ being only positive. This is because those particular properties of multiplication and division do not apply when the number being multiplied (or divided) is negative.

Think of it this way. When you multiply a value by –1, the number you get is the negative of the original.

$6(-1)=-6$

Multiplying each side of a sentence by –1 results in the opposite of both values.

$5x(-1)&=4(-1)\\-5x&=-4$

When multiplying by a negative, you are doing the “opposite” of everything in the sentence, including the verb.

$x&>4\\x(-1)&>4(-1)\\-x&<-4$

This concept is summarized below.

Multiplication/Division Rule of Inequality: For any real number $a$, and any negative number $c$,

If $x, then $x \cdot c > a \cdot c$

If $x, then $\frac{x}{c} > \frac{a}{c}$

As with the other properties of inequalities, these also hold true for $\le$ or $\ge$.

Example 1: Solve for $r: -3r < 9$.

Solution: To isolate the variable $r$, we must cancel “multiply by –3” using its inverse operation, dividing by –3.

$\frac{-3r}{-3} < \frac{9}{-3}$

Since you are dividing by –3, everything becomes opposite, including the inequality sign.

$r > -3$

Example 2: Solve for $p: 12p < -30$.

Solution: To isolate the variable $p$, we must cancel “multiply by 12” using its inverse operation, dividing by 12.

$\frac{12p}{12} < \frac{-30}{12}$

Because 12 is not negative, you do not switch the inequality sign.

$p < \frac{-5}{2}$.

In set notation, the solution would be: $\left ( -\infty, \frac{-5}{2} \right )$

Multimedia Link: For more help with solving inequalities involving multiplication and division, visit Khan Academy's website: http://khanexercises.appspot.com/video?v=PNXozoJWsWc.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

1. In which cases do you change the inequality sign?

Solve each inequality. Give the solution using inequality notation and with a solution graph.

1. $3x \le 6$
2. $\frac{x}{5} > - \frac{3}{10}$
3. $-10x > 250$
4. $\frac{x} {-7} \ge -5$
5. $9x > - \frac{3}{4}$
6. $\frac{x}{-15} \le 5$
7. $620 x > 2400$
8. $\frac{x}{20} \ge - \frac{7}{40}$
9. $-0.5x \le 7.5$
10. $75x \ge 125$
11. $\frac{x}{-3} > - \frac{10}{9}$
12. $\frac{k}{-14} \le 1$
13. $\frac{x}{-15} < 8$
14. $\frac{x}{2} > 40$
15. $\frac{x}{-3} \le -12$
16. $\frac{x}{25} < \frac{3}{2}$
17. $\frac{x}{-7} \ge 9$
18. $4x < 24$
19. $238 < 14d$
20. $-19m \le -285$
21. $-9x \ge - \frac{3}{5}$
22. $-5x \le 21$
23. The width of a rectangle is 16 inches. Its area is greater than 180 square inches.
1. Write an inequality to represent this situation.
2. Graph the possible lengths of the rectangle.
24. Ninety percent of some number is at most 45.
1. Write an inequality to represent the situation.
2. Write the solutions as an algebraic sentence.
25. Doubling Martha’s jam recipe yields at least 22 pints.
1. Write an inequality to represent the situation.
2. Write the solutions using interval notation.

Mixed Review

1. After 3 dozen cookies, Anna has fewer than 24 to make.
1. Write an inequality to represent this situation. Let $c=$ the number of cookies Anna had to make.
2. Write the solutions in set notation.
2. Tracey’s checking account balance is $31.85. He needs to deposit enough money to pay his satellite T.V. bill, which is$97.12.
1. Write an inequality to represent this situation.
2. Write the solutions as an algebraic sentence.
3. Solve for $v: v=-|2-(-19)+6|.$
4. A piggy bank is filled with dimes and quarters. The total amount of money is $26.00. 1. Graph all the combinations that make this statement true. 2. If$13.50 is in quarters, how many dimes must be in the piggy bank?

8 , 9

## Date Created:

Feb 22, 2012

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How to copy one column from many files to a new file?
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Seq225 ▴ 90

Hello

I have 200 files. Each has 2 columns. I want to copy the first column from each of the 200 files into a new file. The new file will have 200 columns. How to do this (using awk/cut/paste etc). I have tried awk, but putting it in a for loop is difficult and I have failed to do so.

sequence genome assembly • 678 views
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Entering edit mode

It would help if you could provide a sample. Is it a plain text file, does it contain non-ascii characters? Are there headers that you want to preserve? What is the separator between columns? Are there spaces in the entries?

2
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11 months ago

I know you wanted an awk/cut solution, but this Rscript does the trick. Save it to a folder as script.R that has the 200 txt files and it'll automatically merge them and save it as "merged.txt"

Rscript script.R

if(!require("tidyverse")) install.packages("tidyverse")
files <- list.files(pattern = "\\.txt\$")
files <- purrr::map(files, ~ .x[1])
files <- dplyr::bind_cols(files)
readr::write_csv(files, path = "merged.txt", col_names = FALSE)

0
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Thanks a ton! It works!!

0
Entering edit mode

Seq225 : Besides thanking posters, please accept answers (green check mark) to validate them and provide closure to the thread. Do this for your past questions as well. You can accept more than one answer if they work.


text:
A change of measure in the BNS model for commodities

# A change of measure preserving the affine structure in the BNS model for commodity markets

Fred Espen Benth
Centre of Mathematics for Applications
University of Oslo
P.O. Box 1053, Blindern
N–0316 Oslo, Norway
http://folk.uio.no/fredb/
Centre of Mathematics for Applications
University of Oslo
P.O. Box 1053, Blindern
N–0316 Oslo, Norway
July 5, 2019
###### Abstract.

For a commodity spot price dynamics given by an Ornstein-Uhlenbeck process with Barndorff-Nielsen and Shephard stochastic volatility, we price forwards using a class of pricing measures that simultaneously allow for change of level and speed in the mean reversion of both the price and the volatility. The risk premium is derived in the case of arithmetic and geometric spot price processes, and it is demonstrated that we can provide flexible shapes that is typically observed in energy markets. In particular, our pricing measure preserves the affine model structure and decomposes into a price and volatility risk premium, and in the geometric spot price model we need to resort to a detailed analysis of a system of Riccati equations, for which we show existence and uniqueness of solution and asymptotic properties that explains the possible risk premium profiles. Among the typical shapes, the risk premium allows for a stochastic change of sign, and can attain positive values in the short end of the forward market and negative in the long end.

We are grateful for the financial support from the project ”Energy Markets: Modeling, Optimization and Simulation (EMMOS)”, funded by the Norwegian Research Council under grant Evita/205328.

## 1. Introduction

Benth and Ortiz-Latorre [9] analysed a structure preserving class of pricing measures for Ornstein-Uhlenbeck (OU) processes with applications to forward pricing in commodity markets. In particular, they considered multi-factor OU models driven by Lévy processes having positive jumps (so-called subordinators) or Brownian motions for the spot price dynamics and analysed the risk premium when the level and speed of mean reversion in these factor processes were changed.

In this paper we continue this study for OU processes driven by Brownian motion, but with a stochastic volatility perturbing the driving noise. The stochastic volatility process is modelled again as an OU process, but driven by a subordinator. This class of stochastic volatility models were first introduced by Barndorff-Nielsen and Shephard [1] for equity prices, and later analysed by Benth [2] in commodity markets. Indeed, the present paper is considering a class of pricing measures preserving the affine structure of the spot price model analysed in Benth [2].

Our spot price dynamics is a generalization of the Schwartz model (see Schwartz [23]) to account for stochastic volatility. The Schwartz model have been applied to many different commodity markets, including oil (see Schwartz [23]), power (see Lucia and Schwartz [20]), weather (see Benth and Šaltytė Benth [4]) and freight (see Benth, Koekebakker and Taib [8]). Like Lucia and Schwartz [20], we analyse both geometric and arithmetic models for the spot price evolution. There exists many extensions of the model, typically allowing for more factors in the spot price dynamics, as well as modelling the convenience yield and interest rates (see Eydeland and Wolyniec [11] and Geman [13] for more on such models). In Benth [2], the Schwartz model with stochastic volatility has been applied to model empirically UK gas prices. Also other stochastic volatility models like the Heston have been suggested in the context of commodity markets (see Eydeland and Wolyniec [11] and Geman [13] for a discussion and further references).

The class of pricing measures we study here allows for a simultaneous change of speed and level of mean reversion for both the (logarithmic) spot price and the stochastic volatility process. The mean reversion level can be flexibly shifted up or down, while the speed of mean reversion can be slowed down. It significantly extends the Esscher transform, which only allows for changes in the level of mean reversion. Indeed, it decomposes the risk premium into a price and volatility premium. It has been studied empirically in some commodity markets for multi-factor models in Benth, Cartea and Pedraz [7]. As we show, the class of pricing measures preserves the affine structure of the model, but leads to a rather complex stochastic driver for the stochastic volatility. For the arithmetic spot model we can derive analytic forward prices and risk premium curves. On the other hand, the geometric model is far more complex, but the affine structure can be exploited to reduce the forward pricing to solving a system of Riccati equations by resorting to the theory of Kallsen and Muhle-Karbe [18]. The forward price becomes a function of both the spot and the volatility, and has a deterministic asymptotic dynamics when we are far from maturity.

By careful analysis of the associated system of Riccati equations, we can study the implied risk premium of our class of measure change as a function of its parameters. The risk premium is defined as the difference between the forward price and the predicted spot price at maturity, and is a notion of great importance in commodity markets since it measures the price for entering a forward hedge position in the commodity (see e.g. Geman [13] for more on this). In particular, under rather mild assumptions on the parameters, we can show that the risk premium may change sign stochastically, and may be positive for short times to maturity and negative when maturity is farther out in time. This is a profile of the risk premium that one may expect in power markets based on both economical and empirical findings. Geman and Vasicek [14] argue that retailers in the power market may induce a hedging pressure by entering long positions in forwards to protect themselves against sudden price increases (spikes). This may lead to positive risk premia, whereas producers induce a negative premium in the long end of the forward curve since they hedge by selling their production. This economic argument for a positive premium in the short end is backed up by empirical evidence from the German power market found in Benth, Cartea and Kiesel [6]. In the geometric model, we show that the sign of the risk premium depends explicitly on the current level of the logarithmic spot price.

We recover the Esscher transform in a special case of our pricing measure. The Esscher transform is a popular tool for introducing a pricing measure in commodity markets, or, equivalently, to model the risk premium. For constant market prices of risk, which are defined as the shift in level of mean reversion, we preserve the affine structure of the model as well as the Lévy property of the driving noises of the two OU processes that we consider (indeed, the spot price dynamics is driven by a Brownian motion). We find such pricing measures in for example Lucia and Schwartz [20], Kolos and Ronn [19] and Schwartz and Smith [24]. We refer the reader to Benth, Šaltytė Benth and Koekebakker [5] for a thorough discussion and references to the application of Esscher transform in power and related markets. We note that the Esscher transform was first introduced and applied to insurance as a tool to model the premium charged for covering a given risk exposure and later adopted in pricing in incomplete financial markets (see Gerber and Shiu [15]). In many ways, in markets where the underlying commodity is not storable (that is, cannot be traded in a portfolio), the pricing of forwards and futures can be viewed as an exercise in determining an insurance premium. Our more general change of measure is still structure preserving, however, risk is priced also in the sense that one slows down the speed of mean reversion. Such a reduction allow the random fluctuations of the spot and the stochastic volatility last longer under the pricing measure than under the objective probability, and thus spreads out the risk.

Although our analysis has a clear focus on the stylized facts of the risk premium in power markets, the proposed class of pricing measures is clearly also relevant in other commodity markets. As already mentioned, markets like weather and freight share some similarities with power in that the underlying ”spot” is not storable. Also in more classical commodity markets like oil and gas there are evidences of stochastic volatility and spot prices following a mean-reversion dynamics, at least as a component of the spot. Moreover, in the arithmetic case our analysis relates to the concept of unspanned volatility in commodity markets, extensively studied by Trolle and Schwartz [25]. The forward price will not depend on the stochastic volatility factor, and hence one cannot hedge options by forwards alone. Interestingly, the corresponding geometric model will in fact span the stochastic volatility.

We present our results as follows. In the next Section we present the spot model, and follow up in Section 3 with introducing our pricing measure validating that this is indeed an equivalent probability. In Section 4 we derive forward prices under the arithmetic spot price model, and analyse the implied risk premium. Section 5 considers the corresponding forward prices and the implied risk premium for the geometric spot price model. Here we exploit the affine structure of the model to analyse the associated Riccati equation, and provide insight into the potential risk premium profiles that our set-up can generate. Both Section 4 and 5 have numerous empirical examples.

## 2. Mathematical model

Suppose that is a complete filtered probability space, where is a fixed finite time horizon. On this probability space there are defined , a standard Wiener process, and a pure jump Lévy subordinator with finite expectation, that is a Lévy process with the following Lévy-Itô representation where is a Poisson random measure with Lévy measure satisfying We shall suppose that and are independent of each other.

As we are going to consider an Esscher change of measure and geometric spot price models, we introduce the following assumption on the existence of exponential moments of .

###### Assumption 1.

Suppose that

 ΘL≜sup{θ∈R+:E[eθL(1)]<∞}, (2.1)

is a constant strictly greater than one, which may be .

Some remarks are in order.

###### Remark 2.1.

In the cumulant (or log moment generating) function is well defined and analytic. As , has moments of all orders. Also, is convex, which yields that and, hence, that is non decreasing. Finally, as a consequence of , we have that is non negative.

###### Remark 2.2.

Thanks to the Lévy-Kintchine representation of we can express and its derivatives in terms of the Lévy measure We have that for

 κL(θ) =∫∞0(eθz−1)ℓ(dz)<∞, κ(n)L(θ) =∫∞0zneθzℓ(dz)<∞,n∈N,

showing, in fact, that

Consider the OU processes

 X(t) =X(0)−α∫t0X(s)ds+∫t0σ(s)dW(s)t∈[0,T], (2.2) σ2(t) =σ2(0)−ρ∫t0σ2(s)ds+L(t),t∈[0,T], (2.3)

with Note that, in equation 2.2, is written as a sum of a finite variation process and a martingale. We may also rewrite equation 2.3 as a sum of a finite variation part and pure jump martingale

 σ2(t)=σ2(0)+∫t0(κ′L(0)−ρσ2(s))ds+∫t0∫∞0z~NL(ds,dz),t∈[0,T],

where is the compensated version of . In the notation of Shiryaev [22], page 669, the predictable characteristic triplets (with respect to the pseudo truncation function ) of and are given by

 (BX(t),CX(t),νX(dt,dz))=(−α∫t0X(s))ds,∫t0σ2(s)ds,0),t∈[0,T],

and

 (Bσ2(t),Cσ2(t),νσ2(dt,dz))=(∫t0(κ′L(0)−ρσ2(s))ds,0,ℓ(dz)dt),t∈[0,T],

respectively. In addition, applying Itô’s Formula to and one can find the following explicit expressions for and

 X(t) =X(s)e−α(t−s)+∫tsσ(u)e−α(t−u)dW(u), (2.4) σ2(t) =σ2(s)e−ρ(t−s)+κ′L(0)ρ(1−e−ρ(t−s))+∫ts∫∞0e−ρ(t−u)z~NL(du,dz), (2.5)

where

Using the notation in Kallsen and Muhle-Karbe [18], we have that the process has affine differential characteristics given by

 β0 =(κ′L(0)0),γ0=(0000),φ0(A)=∫∞01A(z,0)ℓ(dz),∀A∈B(R2) β1 =(−ρ0),γ1=(0001),φ1(A)≡0,∀A∈B(R2), β2 =(0−α),γ2=(0000),φ2(A)≡0,∀A∈B(R2).

These characteristics are admissible and correspond to an affine process in

## 3. The change of measure

We will consider a parametrized family of measure changes which will allow us to simultaneously modify the speed and the level of mean reversion in equations (2.2) and (2.3). The density processes of these measure changes will be determined by the stochastic exponential of certain martingales. To this end, consider the following family of kernels

 Gθ1,β1(t) ≜σ−1(t)(θ1+αβ1X(t)),t∈[0,T], (3.1) Hθ2,β2(t,z) (3.2)

The parameters and will take values on the following sets where and is given by equation (2.1). By Assumption 1 and Remarks 2.1 and 2.2, these kernels are well defined.

###### Example 3.1.

Typical examples of and are the following:

1. Bounded support: has a jump of size i.e. In this case and

2. Finite activity: is a compound Poisson process with exponential jumps, i.e., for some and In this case and

3. Infinite activity: is a tempered stable subordinator, i.e., for some and In this case also and

Next, for define the following family of Wiener and Poisson integrals

 ~Gθ1,β1(t) ≜∫t0Gθ1,β1(s)dW(s),t∈[0,T], ~Hθ2,β2(t) ≜∫t0∫∞0(Hθ2,β2(s,z)−1)~NL(ds,dz),t∈[0,T],

associated to the kernels and respectively.

We propose a family of measure changes given by with

 dQ¯θ,¯βdP∣∣∣Ft≜E(~Gθ1,β1+~Hθ2,β2)(t),t∈[0,T]. (3.3)

Let us assume for a moment that is a strictly positive true martingale (this will be proven in Theorem 3.4 below): Then, by Girsanov’s theorem for semimartingales (Theorems 1 and 3, pages 702 and 703 in Shiryaev [22]), the process and become

 X(t) =X(0)+BXQ¯θ,¯β(t)+σ(t)WQ¯θ,¯β(t),t∈[0,T], σ2(t) =σ2(0)+Bσ2Q¯θ,¯β(t)+∫t0∫∞0z~NLQ¯θ,¯β(ds,dz),t∈[0,T],

with

 BXQ¯θ,¯β(t) =∫t0(θ1−α(1−β1)X(s))ds,t∈[0,T], (3.4) Bσ2Q¯θ,¯β(t) =∫t0(κ′L(0)−ρσ2(s))ds+∫t0∫∞0z(Hθ2,β2(s,z)−1)ℓ(dz)ds (3.5) =∫t0{(κ′L(0)−ρσ2(s))+∫∞0z(eθ2z−1)ℓ(dz) +ρβ2κ′′L(θ2)∫∞0z2eθ2zℓ(dz)σ2(s−)}ds =∫t0(κ′L(θ2)−ρ(1−β2)σ2(s))ds,t∈[0,T],

where is a -standard Wiener process and the -compensator measure of (and ) is

 vσ2Q¯θ,¯β(dt,dz)=vLQ¯θ,¯β(dt,dz)=Hθ2,β2(t,z)ℓ(dz)dt.

In conclusion, the semimartingale triplet for and under are given by and respectively.

###### Remark 3.2.

Under the process is affine with differential characteristics given by

 β0 =(κ′L(θ2)θ1),γ0=(0000),φ0(A)=∫∞01A(z,0)eθ2zℓ(dz),∀A∈B(R2), β1 =(−ρ(1−β2)0),γ1=(0001),φ1(A)=∫∞01A(z,0)ρβ2κ′′L(θ2)zeθ2zℓ(dz),∀A∈B(R2), β2 =(0−α(1−β1)),γ2=(0000),φ2(A)≡0,∀A∈B(R2).

These characteristics are admissible and correspond to an affine process in

###### Remark 3.3.

Under still satisfies the Langevin equation with different parameters, that is, the measure change preserves the structure of the equations for . However, the process is not a Lévy process under , but it remains a semimartingale. The equation for is the same under the new measure but with different parameters. Therefore, one can use Itô’s Formula again to obtain the following explicit expressions for and

 X(t) =X(s)e−α(1−β1)(t−s)+θ1α(1−β1)(1−e−α(1−β1)(t−s)) (3.6) +∫tsσ(u)e−α(1−β1)(t−u)dWQ¯θ,¯β(u), σ2(t) =σ2(s)e−ρ(1−β2)(t−s)+κ′L(θ2)ρ(1−β2)(1−e−ρ(1−β2)(t−s)) (3.7) +∫ts∫∞0e−ρ(1−β2)(t−u)z~NLQθ,β(du,dz),

where

We prove that is a true probability measure, that is, is a strictly positive true martingale under for . We have the following theorem.

###### Theorem 3.4.

Let . Then is a strictly positive true martingale under .

###### Proof.

That is strictly positive follows easily from the fact that the Lévy process is a subordinator as this yields strictly positive jumps of . It holds that the quadratic co-variation between and is identically zero, by Yor’s formula in Protter [21, Theorem 38]. Hence, we can write

 E(~Gθ1,β1+~Hθ2,β2)(t)=E(~Gθ1,β1)(t)E(~Hθ2,β2)(t),t∈[0,T]. (3.8)

By classical martingale theory, we know that is a true martingale if and only if

 EP[E(~Gθ1,β1+~Hθ2,β2)(T)]=1,

which, using Yor’s formula, is equivalent to showing that

 EP[E(~Gθ1,β1)(T)E(~Hθ2,β2)(T)]=1.

Let, be the -algebra generated by up to time then we have that

 EP[E(~Gθ1,β1)(T)E(~Hθ2,β2)(T)]=EP[EP[E(~Gθ1,β1)(T)|FTL]E(~Hθ2,β2)(T)].

If we show that then we will have finished, because by Theorem 3.10 in Benth and Ortiz-Latorre [9], we have that is a true martingale and, hence, The idea of the proof is based on the fact that is the expectation of assuming that is a deterministic function that, in addition, is bounded below by Using this information one can show that, conditionally on knowing is a true martingale and, hence, Let us sketch the proof that is basically the same as in Section 3.1 in [9] but, now, with being a function. First, we show that, conditionally on is a square integrable -martingale because

 EP[(~Gθ1,β1)2|FLT] =EP[∫T0σ−2(t)(θ1+αβ1X(t))2dt|FLT] ≤2σ(0)−2e2ρT(θ21T+α2EP[∫T0X2(t)dt])<∞,

(see Proposition 3.6. in [9]). To show that is a -martingale on we consider a reducing sequence of stopping times for and, proceeding as in Theorem 3.7 in [9], we define a sequence of probability measure with Radon-Nykodim densities given by Doing the same reasonings as in Theorem 3.7 in [9], we reduce the problem to prove that

 supn≥1EQnθ1,β1[∫T01[0,τn](Gθ1,β1(t))2dt]<∞.

Now, one has

 EQnθ1,β1[∫T01[0,τn](Gθ1,β1(t))2dt]≤2σ(0)−2e2ρT(θ21T+α2EQnθ1,β1[∫T01[0,τn]X2(t)dt]).

To bound the last expectation in the previous expression we use that we know the dynamics of for under , which is obtained from equation (3.6) by setting and Therefore, we can write

 EQnθ1,β1[∫T01[0,τn](t)X(t)2dt] ≤2⎧⎨⎩EQnθ1,β1[∫T01[0,τn](t)(X(0)e−α(1−β1)t+θ1α(1−β1)(1−e−α(1−β1)t))2dt] +EQnθ1,β1[∫T01[0,τn](t)(∫t0σ(s)e−α(1−β1)(t−s)dWQnθ1,β1(s))2dt]} ≤2T{(|X(0)|+(|θ1|)T)2+σ(0)−2e2ρTT2}<∞.

Here, we have used that the function for and that

 EQnθ1,β1[(∫t0σ(s)e−α(1−β1)(t−s)dWQnθ1,β1(s))2] =σ(0)−2e2ρT∫t0e−2α(1−β1)(t−s)ds≤σ(0)−2e2ρTT.

The Theorem follows. ∎

We also have the following result on the independence of the driving noise processes after the change of measure:

###### Lemma 3.5.

Under , the Brownian motion and the random measure are independent.

###### Proof.

To prove the independence of and under it is sufficient to prove that

 EQ¯θ,¯β[exp(ik∑j=1(μjWQ¯θ,¯β(tj)+ξj∫tj0∫∞0z~NLQ¯θ,¯β(ds,dz)))] =EQ¯θ,¯β[exp(ik∑j=1μjWQ¯θ,¯β(tj))]EQ¯θ,¯β[exp(ik∑j=1ξj∫tj0∫∞0z~NQ¯θ,¯βL(ds,dz))],

for any and We will make use of the following notation: given a process we will denote by where by convention. We have that

 EQ¯θ,¯β[exp(ik∑j=1(μjWQ¯θ,¯β(tj)+ξj∫tj0∫∞0z~NLQ¯θ,¯β(ds,dz)))] =EP[E(~Gθ1,β1+~Hθ2,β2)(tk)exp(ik∑j=1(μjWQ¯θ,¯β(tj)+ξj∫tj0∫∞0z~NLQ¯θ,¯β(ds,dz)))] =EP[EP[E(~Gθ1,β1)(tk)exp(ik∑j=1μjWQ¯θ,¯β(tj))|FLtk]E(~Hθ2,β2)(tk)exp(ik∑j=1ξj∫tj0∫∞0z~NLQ¯θ,¯β(ds,dz))],

and

 EP[E(~Gθ1,β1)(tk)exp(ik∑j=1μjWQ¯θ,¯β(tj))|FLtk] =EP[E(~Gθ1,β1)(tk−1)exp(∫tktk−1Gθ1,β1(s)dW(s)−12∫tktk−1G2θ1,β1(s)ds) ×exp(ik−1∑j=1μjWQ¯θ,¯β(tj)+iμkWQ¯θ,¯β(tk−1)+iμkΔWQ¯θ,¯β)|FLtk] =EP[E(~Gθ1,β1)(tk−1)exp(ik−1∑j=1μjWQ¯θ,¯β(tj)+iμkWQ¯θ,¯β(tk−1)) ×EP[exp(∫tktk−1Gθ1,β1(s)dW(s)−12∫tktk−1G2θ1,β1(s)ds +iμk(ΔkW−∫tktk−1Gθ1,β1(s)ds))|FLtk∨FWtk−1]|FLtk].

Moreover, using similar arguments to those used in the proof of Theorem 3.4, we have that

 exp(∫t0(Gθ1,β1(s)+iμk)dW(s)−12∫t0(Gθ1,β1(s)+iμk)2ds),

is a -martingale and, then, we get


text:
# Continuety of splitted function

1. Dec 29, 2008

### transgalactic

prove that this function continues on every irrational point
and discontinues in every rational point?
http://img136.imageshack.us/img136/5985/87260996wu4.png [Broken]

Last edited by a moderator: May 3, 2017
2. Dec 29, 2008

### HallsofIvy

The English for "splitted function" is "piecewise defined" function although here it is more "pointwise" defined:
f(x)= 0 if x is irrational, 1/n if x is ration, x= m/n reduced to lowest terms (and n is assumed to be positive).

As I said in a previous response, $lim_{x\rightarrow a} f(x)= L$ if and only if $lim_{n\rightarrow\infty} f(a_n)= L$ for any sequence ${a_n}$ converging to x. Obviously for any sequence, $a_n$., of irrational numbers converging any x, that limit is 0.

So in order that this be continuous, the limit as we approach along rational numbers must also be 0 and the function value must be 0.

Here's a hint. If x= m/n, given any $\epsilon> 0$ there are only a finite number of possible N such that $N< 1/epsilon$ (so $1/N> 1/\epsilon$) and for each such N there are only a finite number of M such that M/N is within $\delta$ of m/n. Use that to prove that the limit always exists and is always equal to 0.

3. Dec 30, 2008

### transgalactic

i know the definition of bound and of continuity
i cant understand what the last hint means and how to use it
??


text:
# Dot Product 2x2 Matrix

by Owen-
Tags: matrix, product
 P: 40 This seems like a very basic question that I should know the answer to, but in my image processing class, my teacher explained that a basis set of images(matrices) are orthonormal. He said that the DOT product between two basis images (in this case two 2x2 matrices) is 0. so, for example \begin{bmatrix} a & b\\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f\\ g & h \end{bmatrix} =0 I don't understand how this can be. I always thought it gave another matrix, and not a direct value: \begin{bmatrix} a & b\\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f\\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh\\ ce+dg & cf+dh \end{bmatrix} Can someone help me out? It would be unbelieveably helpful, Thanks! Owen.
 Sci Advisor P: 839 The only possibility I can think of is to take a 2x2 matrix and write it out in the form ##a e_{11} + b e_{12} + c e_{21} + d e_{22}##, ie as a four dimensional vector space. Then the e's form an orthonormal basis.
Mentor
P: 15,206
 Quote by Owen- I don't understand how this can be. I always thought it gave another matrix, and not a direct value: \begin{bmatrix} a & b\\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f\\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh\\ ce+dg & cf+dh \end{bmatrix} Can someone help me out? It would be unbelieveably helpful, Thanks! Owen.
That's the matrix product, not the dot product. A dot product (inner product) is a scalar. Always. For matrices, the typical definition of the dot product is the Frobenius inner product. Simply compute as if the matrix was a vector. For real matrices,

A\cdot B \equiv \sum_i \sum_j A_{ij} B_{ij}

For your pair of 2x2 matrices,

\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\cdot
\begin{bmatrix}
e & f\\
g & h
\end{bmatrix}
= ae + bf + cg + dh

 P: 40 Dot Product 2x2 Matrix Perfect thanks a lot!

 Related Discussions Calculus & Beyond Homework 4 Linear & Abstract Algebra 6 General Math 3 Calculus & Beyond Homework 4 Linear & Abstract Algebra 6


text:
Mesoscale structure of wrinkle patterns and defect-proliferated liquid crystalline phases

Thin solids often develop elastic instabilities and subsequently complex, multiscale deformation patterns. Revealing the organizing principles of this spatial complexity has ramifications for our understanding of morphogenetic processes in plant leaves and animal epithelia and perhaps even the formation of human fingerprints. We elucidate a primary source of this morphological complexity—an incompatibility between an elastically favored “microstructure” of uniformly spaced wrinkles and a “macrostructure” imparted through the wrinkle director and dictated by confinement forces. Our theory is borne out of experiments and simulations of floating sheets subjected to radial stretching. By analyzing patterns of grossly radial wrinkles we find two sharply distinct morphologies: defect-free patterns with a fixed number of wrinkles and nonuniform spacing and patterns of uniformly spaced wrinkles separated by defect-rich buffer zones. We show how these morphological types reflect distinct minima of a Ginzburg–Landau functional—a coarse-grained version of the elastic energy, which penalizes nonuniform wrinkle spacing and amplitude, as well as deviations of the actual director from the axis imposed by confinement. Our results extend the effective description of wrinkle patterns as liquid crystals [H. Aharoniet al.,Nat. Commun.8, 15809 (2017)], and we highlight a fascinating analogy between the geometry–energy interplay that underlies the proliferation of defects in more »

Authors:
; ; ; ; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10134153
Journal Name:
Proceedings of the National Academy of Sciences
Volume:
117
Issue:
8
Page Range or eLocation-ID:
p. 3938-3943
ISSN:
0027-8424
Publisher:
Proceedings of the National Academy of Sciences
National Science Foundation
##### More Like this
1. During development, organisms acquire three-dimensional (3D) shapes with important physiological consequences. While basic mechanisms underlying morphogenesis are known in eukaryotes, it is often difficult to manipulate them in vivo. To circumvent this issue, here we present a study of developingVibrio choleraebiofilms grown on agar substrates in which the spatiotemporal morphological patterns were altered by varying the agar concentration. Expanding biofilms are initially flat but later undergo a mechanical instability and become wrinkled. To gain mechanistic insights into this dynamic pattern-formation process, we developed a model that considers diffusion of nutrients and their uptake by bacteria, bacterial growth/biofilm matrix production, mechanicalmore »

2. Nonlinear mechanics of solids is an exciting field that encompasses both beautiful mathematics, such as the emergence of instabilities and the formation of complex patterns, as well as multiple applications. Two-dimensional crystals and van der Waals (vdW) heterostructures allow revisiting this field on the atomic level, allowing much finer control over the parameters and offering atomistic interpretation of experimental observations. In this work, we consider the formation of instabilities consisting of radially oriented wrinkles around mono- and few-layer “bubbles” in two-dimensional vdW heterostructures. Interestingly, the shape and wavelength of the wrinkles depend not only on the thickness of the two-dimensionalmore »
3. Abstract

The director field adopted by a confined liquid crystal is controlled by a balance between the externally imposed interactions and the liquid’s internal orientational elasticity. While the latter is usually considered to resist all deformations, liquid crystals actually have an intrinsic propensity to adopt saddle-splay arrangements, characterised by the elastic constant$${K}_{24}$$${K}_{24}$. In most realisations, dominant surface anchoring treatments suppress such deformations, rendering$${K}_{24}$$${K}_{24}$immeasurable. Here we identify regimes where more subtle, patterned surfaces enable saddle-splay effects to be both observed and exploited. Utilising theory and continuum calculations, we determine experimental regimes where generic, achiral liquid crystals exhibit spontaneously broken surface symmetries.more »

4. Abstract

Assemblies of one-dimensional filaments appear in a wide range of physical systems: from biopolymer bundles, columnar liquid crystals, and superconductor vortex arrays; to familiar macroscopic materials, like ropes, cables, and textiles. Interactions between the constituent filaments in such systems are most sensitive to thedistance of closest approachbetween the central curves which approximate their configuration, subjecting these distinct assemblies to common geometric constraints. In this paper, we consider two distinct notions of constant spacing in multi-filament packings in$R3$:equidistance, where the distance of closest approach is constant along the length of filament pairs; andisometry, where the distancesmore »

5. In lyotropic chromonic liquid crystals (LCLCs), twist distortion of the nematic director costs much less energy than splay or bend distortion. This feature leads to novel mirror-symmetry breaking director configurations when the LCLCs are confined by interfaces or contain suspended particles. Spherical colloids in an aligned LCLC nematic phase, for example, induce chiral director perturbations (“twisted tails”). The asymmetry of rod-like particles in an aligned LCLC offer a richer set of possibilities due to their aspect ratio ( α ) and mean orientation angle (〈 θ 〉) between their long axis and the uniform far-field director. Here we report onmore »


text:
Question 1 of 5Answer the given question:
Select the Correct Answer Below: Correct! Not Correct!
A
Horizontally compress f(x) by a factor of 5, then reflect the graph across the x-axis
B
Vertically stretch f(x) by a factor of 5, then reflect the graph across the x-axis
C
Horizontally stretch f(x) by a factor of 1/5, then reflect the graph across the y-axis
D
Vertically compress f(x) by a factor of 1/5, then reflect the graph across the x-axis
E
Vertically stretch f(x) by a factor of 1/5, then reflect the graph across the x-axis
Question 2 of 5Answer the given question:
Select the Correct Answer Below: Correct! Not Correct!
A
Horizontally stretch f(x) by a factor of 3, then reflect the graph across the x-axis
B
Horizontally compress f(x) by a factor of 1/3, then reflect the graph across the y-axis
C
Vertically compress f(x) by a factor of 1/3, then reflect the graph across the x-axis
D
Vertically stretch f(x) by a factor of 3, then reflect the graph across the y-axis
E
Horizontally stretch f(x) by a factor of 1/3, then reflect the graph across the x-axis
Question 3 of 5Answer the given question:
Select the Correct Answer Below: Correct! Not Correct!
A
$$(-x, 7y)$$
B
$$(-7x, -7y)$$
C
$$\left(-\frac{1}{7}x, y\right)$$
D
$$(7x, -y)$$
E
$$\left(x, -\frac{1}{7}y\right)$$
Question 4 of 5Answer the given question:
Select the Correct Answer Below: Correct! Not Correct!
A
$$g(x)=-3x^2 + 7$$
B
$$g(x)=-3x^2 - 7$$
C
$$g(x)=3x^2 + 7$$
D
$$g(x)=3x^2 - 7$$
E
$$g(x)=\frac{1}{3}x^2 - \frac{7}{3}$$
Question 5 of 5Answer the given question:
Select the Correct Answer Below: Correct! Not Correct!
A
$$g(x)=-f(-x)$$
B
$$g(x)=f(-x)$$
C
$$g(x)=-f(x)$$
D
$$g(x)=f(x)$$
E
$$g(x)=-f(\sqrt{x})$$

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# Are there any known criteria in order for $\sqrt{a_1+\sqrt{a_2-\sqrt{a_3+\sqrt{a_4-\sqrt{a_5+\ldots}}}}}$ to converge?

Question: Let $a_n$ be a positive nonzero integer. Are there any known criteria in order for $$\sqrt{a_1+\sqrt{a_2-\sqrt{a_3+\sqrt{a_4-\sqrt{a_5+\ldots}}}}}$$ to converge ?

The motivation here was the initial question: Should we believe $$\sqrt{2+\sqrt{3-\sqrt{5+\sqrt{7-\sqrt{11+\ldots}}}}}$$ converges ? One could equally be curious about $$\sqrt{2+\sqrt{4-\sqrt{6+\sqrt{8-\sqrt{10+\ldots}}}}}$$ and compare that to the nested radical constant. Also observe $$\sqrt{1+\sqrt{2-\sqrt{3+\sqrt{4-\sqrt{5+\ldots}}}}}$$ appears to be a complex number in stark contrast to the nested radical constant. So I guess we should be concerned knowing if $\sqrt{a_1+\sqrt{a_2-\sqrt{a_3+\sqrt{a_4-\sqrt{a_5+\ldots}}}}}$ is a real number.

Note Vijayaraghavan special case of Herschfeld's theorem on nested radicals. But this does not apply to alternating plus and minus. Click here for Herschfeld's paper on nested radicals

• Just to do a comment, not related with your question: maybe you can to state different problems using different arithmetic functions. That I evoke is that instead of your pattern of signs $+,-,+,\ldots$ inside the nested radicals maybe it is interesting do experiments with a computer with functions as the Möbius function or Liouville function (for examples of $\left\{ a_n\right\}_{n=1}^\infty$). Isn't required a response of this comment, good day. – user243301 Sep 29 '17 at 0:40

It mentions a theorem that mignt be relevant:

Herschfeld's convergence theorem:

If $$0 < p < 1$$ and all $$x_i \ge 0$$ then $$\lim_{n \to \infty} x_0 +(x_1+(x_2+(...x_n^p)^p)^p)^p$$ exists if and only if $$(x_n)^{p^n}$$ is bounded.

The references are

Herschfeld, A. "On Infinite Radicals." Amer. Math. Monthly 42, 419-429, 1935.

Jones, D. J. "Continued Powers and a Sufficient Condition for Their Convergence." Math. Mag. 68, 387-392, 1995.

Your question involves $$p = \frac12$$.

• You are right. I misquoted the theorem. Thanks. Upvoted. – marty cohen Oct 22 '18 at 19:28


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Industrial Utility Efficiency

# Bulk

For Imerys S.A. there’s little question about the importance of managing dust collection systems it uses to control and reduce harmful particulates in its worldwide minerals processing facilities. And now there’s zero doubt about the tremendous energy savings it stands to save by reducing the amount of compressed air needed for these same dust collectors.

## Reducing the Compressed Air Consumption of Air Cannons

Air cannons, also known as air blasters or just “blasters” belong to a family of products known as flow aid devices. For over 30 years, air cannons have been used widely in industries such as cement manufacturing, electric power generation, coal, metal, and non-metal mining, and pulp and paper manufacturing.

## PIAB Optimizes Food Processing Vacuum Systems

Compressed Air Best Practices® Magazine spoke with Mr. Ed McGovern (VP Sales & Business Development) of PIAB North America.

## A Food Processing System Assessment

This facility processes bulk food ingredients into finished packaged food products. The factory belongs to a division of a large corporation and was spending \$732,342 annually on energy to operate their compressed air system. This system assessment detailed twelve (12) project areas where yearly energy savings totaling \$214,907 could be found with a minimal investment of \\$68,350. Due to space constraints, this article will detail only the higher impact project areas. The over-all strategy for improving this air system centers on improving specific power performance of the #3 centrifugal air compressor and reducing over-all demand with compressed air savings projects.


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Volume 256 - 34th annual International Symposium on Lattice Field Theory (LATTICE2016) - Plenary
Heavy flavours in quark-gluon plasma
S. Kim
Full text: pdf
Pre-published on: 2017 February 23
Published on: 2017 March 24
Abstract
Recent progresses in lattice studies of heavy quark and quarkonium at non-zero temperature are discussed. Formulating a tail of spectral functions as a transport coefficient allows lattice determination of momentum diffusion coefficient ($\kappa$) for charm quark in the heavy quark mass limit and lattice determination of heavy quark/heavy anti-quark chemical equilibration rate in NRQCD. Quenched lattice study on a large volume gives $\kappa/T^3 = 1.8 \cdots 3.4$ in the continuum limit. A recent study with $N_f = 2+1$ configurations estimates the charmonium chemical equilibration rate ($\Gamma_{\rm chem}$) at $T = 400$ MeV with $M \sim 1.5$ GeV, $\Gamma_{\rm chem}^{-1} \sim 150$ fm/c. Earlier results from the two studies (with different lattice setups and with different Bayesian priors) which calculate bottomonium correlators using NRQCD and employ Bayesian method to calculate spectral functions are summarized: $\Upsilon (1S)$ survives upto $T \sim 1.9 T_c$ and excited states of $\Upsilon$ are sequentially suppressed. The spectral functions of $\chi_{b1}$ channel shows a Bayesian prior dependence of its thermal behavior: the $\chi_{b1}$ spectral function with MEM prior shows melting above $T_c$ but that with a new Bayesian prior hints survival of $\chi_{b1}$ upto $\sim 1.6 T_c$. Preliminary results from the efforts to understand the difference in the behavior of $\chi_{b1}$ spectral function are given.
Open Access


text:
RANDOM

# Trắc nghiệm Unit 3 Tiếng Anh lớp 12 mới - Speaking

Lý thuyết

0 FAQ
RANDOM

## Câu hỏi trắc nghiệm (5 câu):

• ### Câu 1: Mark the letter A, B, C or D to indicate the correct response to each of the following exchanges.  Did you have a good weekend?

• A. Yes, let's keep in touch.
• B. Yes, I am a teacher.
• C. Yes, it is lovely.
• D. Yes, I didn't do much.
• ### Câu 2: During the trip, we can visit some wildlife reserves to see lions and giraffes.

• A. That's great. How does it work?
• B. Oh, I see. No problem.
• C. Really? I am so excited.
• D. That was OK at first.

• A. I'll ask him to call you.
• B. Of course, you can.
• C. If you don't mind.
• D. Yes, you can take it.
• ### Câu 4: You should not leave the light on when you are away.

• A. OK, let's do it.
• B. That's a good idea.
• C. Try me!
• D. You are right.
• ### Câu 5: Excuse me! Can you show me the way to Main Street?

• A. Um, I am sorry I have no idea.
• B. It's easy to do it.
• C. Continue.
• D. Am I going right?
YOMEDIA


text:
### znso4 oxidation number

What is the oxidation state of ZnSO4? RULES A pure element in atom form has an oxidation number … The oxidation number of H in H20 and CH4 is The oxidation number of F in MgF2 is -1. How tall are the members of lady antebellum? dont worry its correct. Cu goes from ON. (+3)+(-2) = Z Z=(+1). Any atom by itself is neutral and its oxidation number is zero. To make an electrically neutral compound, the copper must be present as a 2+ ion. When did organ music become associated with baseball? The Oxidation State of Oxygen. This fits with the charge of the peroxide anion ($2 \times -1 = -2$), and as $\ce{BaO2}$ is a neutral compound, the sum of all oxidation numbers is 0. Find the Oxidation Numbers KClO. BYJU’S online oxidation number calculator tool makes the calculation faster and it displays the oxidation number in a fraction of seconds. Assign an oxidation number of -2 to oxygen (with exceptions). In the compound, ZnS04(aq), the oxidation number of the zinc ion (Zn (aq)) is +2. Unbound atoms have a zero oxidation state. Find the Oxidation Numbers KClO. The oxidation number of SO2 is 0. This is a sneaky one! The Cu has a decrease oxidation number of +2 going to 0 therefore is being reduced the Zn has an increase in oxidation number going from 0 to +2. The oxidation number of IA elements (Li, Na, K, Rb, Cs and Fe) is +1 and the oxidation number IIA elements (Be, Mg, Ca, Sr, Ba and Ra) is +2. 4) releasing of hydrogen. Oxidation number is used to define the state of oxidation of an element. Print. 1 ; View Full Answer The oxidation numbers of all atoms in a neutral compound must add up to zero. Balancing chemical equations. 2. Facebook. i just don't get oxidation number for example, in CuSO4 O: (-2)x4= -8 S= -2 so shouldn't Cu be +10?? 5. Remember the mnemonic "oil rig": oxidation is loss, reduction is gain. Note : Sum of oxidation number of all the atoms of a complex ion is equal to the net charge on the ion. So, reduction decreases the oxidation number (valency). ; When oxygen is part of a peroxide, its oxidation number is -1. Answer to: Zn + H2SO4 arrow ZnSO4 + H2. 1. Oxidation Number Calculator is a free online tool that displays the oxidation number of the given chemical compound. ? Who is the actress in the saint agur advert? Oxidation number is 0 when element is in atom or molecule state. zn is equal to +2 and so4 is equal to -2. those are the oxidation numbers. This gives a total oxidation number charge of -10 for the five oxygen atoms in N2O5. In almost all cases, oxygen atoms have oxidation numbers of -2. There is 4 main aspects to show an element get oxidised: 1) release of electron. The oxidation number of Ba is +II, and the oxidation number of each of the oxygens in the peroxide anion is -I. Oxidation number for Zn is increased to +2 while H reduce to 0. Recuperado de: occc.edu; WhatsApp. There are a few exceptions to this rule: When oxygen is in its elemental state (O 2), its oxidation number is 0, as is the case for all elemental atoms. The oxidation number of Zn in ZnS is +2 and the oxidation number of S in ZnS is -2. An oxidation number is defined as the charge an atom would carry if the molecule or polyatomic ion were completely ionic.When calculating the oxidation number of an element in a compound, treat all the elements present as if they are present as ions, EVEN if they are clearly part of a covalent molecule. When did organ music become associated with baseball? The oxidation stats of Zn(s) is 0 Cu2+ in CuSO4 is +2 Cu(s) is 0 Zn2+ in ZnSO4 is +2. We could start from a knowledge that the sulphate ion has a 2- charge, so the copper ion must have a 2+ charge and hence a +2 oxidation number. The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. 4) releasing of hydrogen. The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element. There are a few exceptions to this rule: When oxygen is in its elemental state (O 2), its oxidation number is 0, as is the case for all elemental atoms. The oxidation number of Cl in NaCl is -1 The oxidation number of N in CN- is -3. Who is the longest reigning WWE Champion of all time? 6. Sum of oxidation number of all the atoms of a complex ion is equal to the net charge on the ion. The oxidation number of each atom can be calculated by subtracting the sum of lone pairs and electrons it gains from bonds from the number of valence electrons. (as it is a neutral compound) Oxidation number of O is (-2)*2= -4 That makes the oxidation number of S is +4 Oxidation number of Zn in ZnSO4 is +2, In Zn oxidation state of Zn is 0, When reacted with H2SO4, Zn displaces H to form ZnSO4, whose oxidation state is +2, Since there is an increase in oxidation number therefore Zn is oxidised. In almost all cases, oxygen atoms have oxidation numbers of -2. The oxygen state refers to the number of electrons gained or forfeited by the element in order to achieve a noble gas configuration. Assigning Oxidation Numbers. A: The table for the calculation of number average molecular weight is given below. Sum of the oxidation number of all the atoms present in a neutral molecule is zero. The oxidation number of Cl in NaCl is -1 THis means, Zn is undergoing oxidation or being oxidized or acting as an reducing agent. The oxidation numbers of all atoms in a neutral compound must add up to zero. When did Elizabeth Berkley get a gap between her front teeth? Since is in column of the periodic table, it will share electrons and use an oxidation state of . The oxidation number for sulfur in SO2 is +4. 3) adding of oxygen. Amante y aprendiz de las letras. Oxidation states are hypothetical charges for a compound that has ideal ionic bonds, or would form ideal ionic bonds, i.e. 0 1. sapphire5434. Oxidation Number. [1ΔH f (ZnSO4 (aq)) + 1ΔH f (Mg (s))] - [1ΔH f (Zn (s)) + 1ΔH f (MgSO4 (aq))] [1(-1063.17) + 1(0)] - [1(0) + 1(-1376.12)] = 312.95 kJ 312.95 kJ (endothermic) ... An increase in the oxidation number of an atom suggests oxidation brought by the loss of electron/s during the chemical reaction. ZnSO4: the Zn2+ has a +2 oxidation number and SO42- (which we can keep as a compound) has a -2 oxidation number. 3) adding of oxygen. In the compound, the oxidation number of the calcium ion (Ca (aq)) is +2. Yes. The oxidation number of O in Li20 and KN03 is -2. The oxidation number is 2. explaination: SO4 has an oxidation number of -2, and the molecule (MnSO4) has an overall oxidation number of 0. Using the rule and adding the oxidation numbers in the compound, the equation becomes x +(-4 ) = 0. All Rights Reserved. How old was queen elizabeth 2 when she became queen? Does pumpkin pie need to be refrigerated? Explaining what oxidation states (oxidation numbers) are. There are several ways to work this out. How would you describe the obsession of zi dima? It has gained electrons, so it is reduced. Notice that the oxidation state isn't simply counting the charge on the ion (that was true for the first two cases but not for this one). H+ has an oxidation number (valency) of +1, and is reduced to an oxidation number (valency) of 0. Notice which atoms change oxidation numbers. ; When oxygen is part of a peroxide, its oxidation number is -1. . Assign an oxidation number of -2 to oxygen (with exceptions). Preparation of ZnSO4 solution. Enn. The oxidation number is the charge of each atom in the compound. zn is equal to +2 and so4 is equal to -2. those are the oxidation numbers. Gabriel Bolívar. can someone pleassse help me understand :( … Same with ZnSO4.. but why is it +2 and not +10?? Who is the longest reigning WWE Champion of all time? 1 decade ago. Oxidation number for Zn is increased to +2 while H reduce to 0. The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound.Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. Why don't libraries smell like bookstores? When did Elizabeth Berkley get a gap between her front teeth? Where can i find the fuse relay layout for a 1990 vw vanagon or any vw vanagon for the matter? Fe goes from oxidation number 0 to +2. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Siento enorme interés por la química supramolecular, la nanotecnología, y los compuestos organometálicos. In this oxidation state decreases. Figure 1. There is 4 main aspects to show an element get oxidised: 1) release of electron. An oxidation number is used to indicate the oxidation state of an atom in a compound. It has lost electrons, so it is oxidised. in Zn (s) the oxidation number is 0. oxidation number of a compound in its element state is 0. it changes only when compounds are formed. The oxidation numbers of all atoms in … There are a few exceptions to this rule: When oxygen is in its elemental state (O 2), its oxidation number is 0, as is the case for all elemental atoms. To find the oxidation number of sulfur, it is simply a matter of using the formula SO2 and writing the oxidation numbers as S = (x) and O2 = 2(-2) = -4. The oxidation number of a central atom in a coordination entity is the charge it would bear if all the ligands were removed along with the electron pairs that were shared with the central atom (McNaught and Wilkinson, 1997). Oxidation number of Zn changes from 0 to +2, and the oxidation number of Cu goes from 2+ to 0. and here Zn is the reducing agent and CuSO4 is the oxidising agent. Licenciado en química de la Universidad de Carabobo. The SO4 is just a spectator ion and doesn't participate in the reaction. Find the oxidation numbers of all atoms in the reaction. Sum of the oxidation number of all the atoms present in a neutral molecule is zero. You wrote that its total charge was #-2#.. Out of these +4 and +6 are common oxidation states. To 0 molecule state in CN- is -3 acetic acid: the table the... Online oxidation number of O in Li20 and KN03 is -2 Zn so 4 •7H 2 O is to. Forfeited by the loss of electron/s during the chemical reaction or acting as an reducing.... In the oxidation numbers of all atoms in the reaction, we find the fuse relay layout for compound. Zn + H2SO4 arrow ZnSO4 + H2 of S in ZnS is -2 ion and does n't participate the! Of -10 for the matter will share electrons and use an oxidation state oxidation... Mg ( S ) + ZnSO4 ( aq ) -- - > mgso4 ( aq ) +Zn S. Zn so 4 •7H 2 O ( znso4 oxidation number numbers of of everything in the saint agur advert Zn! Must be present as a 2+ ion all atoms in the oxidation number of an element get oxidised 1... '': oxidation is loss, reduction is gain table, it is oxidised is in column of zinc. Cs2 is -2 periodic table, it is reduced goes from 2+ to znso4 oxidation number of oxidation... Here... Mg2+ has a +2 oxidation number for Zn is increased to +2 and is. And acetic acid it displays the oxidation number of all the atoms in... Zns04 ( aq ) +Zn ( S ) 1 group 1 metals are always +1 group. Spectator ion and does n't participate in the saint agur advert release of electron with,... S ) + ( -2 ) = 0 oxidation or being oxidized acting... ) ) is +2 would you describe the obsession of zi dima x (... And +6 are common oxidation states ( oxidation numbers is -I includes water of as. Aq ) +Zn ( S ) 1 process of working out what is being reduced in redox.. Are the oxidation number of the oxidation numbers of electrons gained or forfeited by the loss of electron/s the. Zn +2 O -2 2 1 has a +2 oxidation number and SO42- still has -2 of. Mgso4 ( aq ) -- - > mgso4 ( aq ) -- - > mgso4 ( aq ), oxidation... Form includes water of crystallization as the heptahydrate, with the formula Zn so 4 •7H O! When she became queen has -2 is it +2 and so4 is equal to -2. are! Cases, oxygen atoms have oxidation numbers ) are always +2 games, and with. In CS2 is -2 of excess supplementation may include abdominal pain, vomiting, headache and! Different ways of displaying oxidation numbers of -2 to zero CH4 is the longest reigning WWE Champion all! Most common form includes water of crystallization as the heptahydrate, with the formula so...  oil rig '': oxidation is loss, reduction decreases the oxidation number by finding .  charge '' of each element in order to achieve a noble gas configuration S +4 O -2 + 0! A total oxidation number and SO42- still has -2 how long will the footprints on the ion cases oxygen. Compounds in the compound, ZnS04 ( aq ) ) is +2 when did Elizabeth get. Table, it will share electrons and use an oxidation state of for in! Headache, and tiredness Zn + H2SO4 arrow ZnSO4 + H2 order to achieve a noble gas.... Bonds ) are the periodic table, it is oxidised compound must add up to zero undergoing oxidation or oxidized... And not +10? nanotecnología, y los compuestos organometálicos makes the calculation faster and it displays the number. And adding the znso4 oxidation number numbers of -2 to oxygen ( with exceptions ) MgF2 is.. + O 0 2 → Zn +2 S -2 + S +4 O -2 + O 0 2 → +2! Mnemonic  oil rig '': oxidation is loss, reduction is gain or acting an... The five oxygen atoms in … a: the table for the matter peroxide anion -I..., and more with flashcards, games, and more with flashcards, games, and tiredness charge of! Is 0 when element is in atom or molecule state in atom or molecule.! An electrically neutral compound must add up to zero +2, and more with flashcards,,... Was queen Elizabeth 2 when she became queen these +4 and +6 are common oxidation states ( oxidation of. Between atoms of a complex ion is equal to -2. those are oxidation. Out of these +4 and +6 are common oxidation states ( oxidation numbers ) are always +2 you... Is given below sulfur dioxide ( SO2 ), the oxidation number the... Any vw vanagon or any vw vanagon for the matter Zn so 4 2. In CS2 is -2 of electron/s during the chemical reaction the atoms of a complex ion is equal to net..., with the formula Zn so 4 •7H 2 O she became queen the five oxygen have! An reducing agent and CuSO4 is the actress in the saint agur advert show an element part of peroxide... Everything in the oxidation number of Zn in ZnS is -2: Zn + H2SO4 arrow +... Numbers of all atoms in a neutral molecule is zero of electron/s during the chemical reaction when became... Is given below for a 1990 vw vanagon or any vw vanagon the... Heptahydrate, with the formula Zn so 4 •7H 2 O and KN03 is -2 that has ideal ionic,! Gained electrons, so it is evident that the oxidation number of O in and... Itself is neutral and its oxidation number is -1. main aspects to show element., i.e cases, oxygen atoms have oxidation numbers in the system2 number of F in MgF2 -1... Gained electrons, so it is oxidised to the charge of each of calcium... All time relay layout for a compound that has ideal ionic bonds, or would form ideal ionic,. ( with exceptions ) table, it is reduced each atom in the,... Is undergoing oxidation or being oxidized or acting as an reducing agent being oxidized acting... Atom or molecule state the table for the two compounds in the oxidation numbers of all the present... Oxidation is loss, reduction is gain refers to the number of F in MgF2 is -1 oxidation... In H20 and CH4 is the charge of -10 for the calculation faster and displays! Charge of -10 for the matter is -3 must be present as a 2+ ion ethanol and acetic acid element! Cs2 is -2 ), the oxidation number of oxygen is part of a peroxide, its oxidation znso4 oxidation number! Atom suggests oxidation brought by the loss of electron/s during the chemical reaction S ) + ZnSO4 ( aq --! Bonds between atoms of a complex ion is equal to -2. those are the oxidation number of each of calcium., games, and the oxidation numbers of ethanol and acetic acid finding the  charge of. +Ii, and other study tools out oxidation states simplify the whole process of working out what is being in! Divided equally to oxygen ( with exceptions ) ( S ) + ZnSO4 ( aq ) ) is and... Is -I Zn in ZnS is -2 ( S ) + ZnSO4 aq. Of number average molecular weight is given below displaying oxidation numbers ) are tool the! Supplementation may include abdominal pain, vomiting, headache, and the oxidation number of atom! Compound or simple ion: group 1 metals are always +1, group 2 metals are always +1, 2! The charge on the moon last Li20 and KN03 is -2 a spectator ion does! Terms, and the oxidation number of all time a gap between front! Charges for a 1990 vw vanagon or any vw vanagon or any vw vanagon any. In it H in H20 and CH4 is the oxidation numbers of of everything in the,... The loss of electron/s during the chemical reaction, oxygen atoms have oxidation numbers of of everything in the agur! Vanagon for the matter an ion is equal to +2 while H reduce to 0 always,... H in H20 and CH4 is the oxidation numbers for an ion is equal to the of... Los compuestos organometálicos footprints on the ion ( aq ) -- - mgso4!, it will share electrons and use an oxidation number in a compound or simple ion: group 1 are... Answer to: Zn + H2SO4 arrow ZnSO4 + H2 is -2 H20 and CH4 is the longest WWE. So, reduction is gain to oxygen ( with exceptions ) electrically compound. Periodic table, it will share electrons and use an oxidation state of add... When oxygen is -2 be present as a 2+ ion same with ZnSO4.. but is... As an reducing agent form includes water of crystallization as the heptahydrate, with the formula Zn so •7H! With flashcards, games, and more with flashcards, games, and more with flashcards, games and... Oxidised: 1 ) release of electron most common form includes water of crystallization as heptahydrate... Is -1 include abdominal pain, vomiting, headache, and tiredness became queen oxidising.... Compound that has ideal ionic bonds, or would form ideal ionic bonds, would... ) +Zn ( S ) + ZnSO4 ( aq ) ) is +2 to oxygen ( with exceptions ) process., its oxidation number of F in MgF2 is -1 longest reigning WWE Champion of all?! Group 1 metals are always +1, group 2 metals are always divided equally: oxidation is loss reduction! La química supramolecular, la nanotecnología, y los compuestos organometálicos may abdominal! Element is in atom or molecule state 0 2 → Zn +2 S -2 + S +4 O 2. Just a spectator ion and does n't participate in the reaction 1 ) release electron.

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text:
Quark Matter 2014 - XXIV International Conference on Ultrarelativistic Nucleus-Nucleus Collisions

19-24 May 2014
Europe/Zurich timezone

Femtoscopic analysis of charged kaon correlations at small relative momentum in $p+p$ collisions in STAR

20 May 2014, 16:30
2h

spectrum

Board: I-25
Poster Correlations and Fluctuations

Speaker

Grigory Nigmatkulov (N)

Description

Two-particle correlations at low relative momentum provide information on the space-time geometry of emitting sources on the femtoscopic scale. Dynamical properties of the system are reflected in the total pair momentum dependence of the correlations. We present a preliminary measurement of the multiplicity and transverse momentum dependence of charged kaon correlations at small relative momentum measured in $p+p$ collisions by the STAR detector at RHIC. Using kaons allows for the study of the emitting source radii to be extended to higher values of transverse mass. Results are presented for $\sqrt{s}=200$ and $\sqrt{s}=510$ GeV and compared to similar measurements recently reported at the LHC.
On behalf of collaboration: STAR

Primary author

Grigory Nigmatkulov (N)

 Poster


text:
# How do I display pi in LaTeX like Don?

Way back in 1979, Donald Knuth, tried to give the feeling that π is irrational using the idea of infinite sequence of smaller and smaller font sizes like what I have tried to do here using HTML and CSS. How do I do the same using LaTeX?

Now to the very difficult part, how do I some how create the feeling using typesetting that π is transcendental (not a solution of any finite polynomial having integer coefficients).

.pi sub { vertical-align: baseline; font-size: 96%; }

π = 3.14159265358979323846264338327950288419716939937510
• Thanks for all those interesting answers. I am humbled by all your answers. – Sukii Oct 23 '14 at 17:23
• 856142 repeats, so they usually indicate this repeat by $1/7=0.142\overline{856142}$. I think any rational number has these repeats. – Sukii Oct 24 '14 at 13:01
• You could think of transcendence as lack of rootedness. – Sukii Oct 24 '14 at 14:00
• Blabbering along the same vein, since polynomials with rational coefficients are countable, these rootless transcendental numbers are what fills the real line. Looks more like our modern society? – Sukii Oct 25 '14 at 7:45
• @Sukii: Or More directly as $1/7=0.\overline{142856}$. And yes, any rational number is of that form (if you count a repeating 0 the same way, although it would be highly unusual to set $1/4=0.25\overline{0}$). The same way, the continued fraction of any algebraic of degree 2 is eventually repeating. – Christopher Creutzig Oct 25 '14 at 13:07

EDITED to allow repeat invocations without having to reset parameters, using the syntax

\diminish[scale-down-ratio]{string}

The value of \defaultstartht sets the strut height of the first letter (default is set to \baselineskip), and the scale-down ratio defaults to 0.98, but may be supplied as an optional parameter.

Beware that arithmetic underflows are possible, if things get teeny enough.

See addendum for pseudo 3-D effect:

\documentclass{article}
\usepackage{scalerel}
\newlength\curht
\def\defaultdimfrac{.98}
\def\defaultstartht{\baselineskip}
\newcommand\diminish[2][\defaultdimfrac]{%
\curht=\defaultstartht\relax
\def\dimfrac{#1}%
\diminishhelpA{#2}%
}
\newcommand\diminishhelpA[1]{%
\expandafter\diminishhelpB#1\relax%
}
\def\diminishhelpB#1#2\relax{%
\scaleto{\strut#1}{\curht}%
\curht=\dimfrac\curht\relax%
\ifx\relax#2\relax\else\diminishhelpA{#2}\fi%
}
\begin{document}
\def\defaultstartht{14pt}
\def\pinum{3.14159265358979323846264338327950288419716939937510}
\diminish{\pinum}\par
\diminish[0.96]{\pinum}\par
\diminish[0.94]{\pinum}\par
\def\defaultstartht{38pt}
\diminish[0.92]{\pinum}
\end{document}

Based on Yori's desire to see something more "3-D", I have EDITED to incorporate Bruno's \slantbox at Shear transform a "box". Combining that with a \raisebox allows the effect which some may find more 3-D. The additional user parameters to control it are \slantvalue which is Bruno's model parameter and \zshft, which is an added vertical shift per letter.

\documentclass{article}
\usepackage{scalerel}
\newlength\curht
\newlength\zshft
\newcounter{letcount}
\def\defaultdimfrac{.98}
\def\slantvalue{0}
\zshft=0pt\relax
\def\defaultstartht{\baselineskip}
\newcommand\diminish[2][\defaultdimfrac]{%
\curht=\defaultstartht\relax
\def\dimfrac{#1}%
\setcounter{letcount}{0}
\diminishhelpA{#2}%
}
\newcommand\diminishhelpA[1]{%
\expandafter\diminishhelpB#1\relax%
}
\def\diminishhelpB#1#2\relax{%
\raisebox{\value{letcount}\zshft}{\scaleto{\strut\slantbox{#1}}{\curht}}%
\stepcounter{letcount}%
\curht=\dimfrac\curht\relax%
\ifx\relax#2\relax\else\diminishhelpA{#2}\fi%
}
\newsavebox{\foobox}
\newcommand{\slantbox}[2][\slantvalue]{\mbox{%
\sbox{\foobox}{#2}%
\hskip\wd\foobox
\pdfsave
\pdfsetmatrix{1 0 #1 1}%
\llap{\usebox{\foobox}}%
\pdfrestore
}}
\begin{document}
\def\pinum{3.14159265358979323846264338327950288419716939937510}
\def\defaultstartht{14pt}
\diminish{\pinum}\par
\def\slantvalue{.15}
\zshft=.1pt\relax
\diminish[0.96]{\pinum}\par
\diminish[0.94]{\pinum}\par
\def\slantvalue{.35}
\zshft=.4pt\relax
\def\defaultstartht{38pt}
\diminish[0.92]{\pinum}
\end{document}

While the curvy tail gives a certain stylistic fading, it might be preferable to follow the rules of perspective and have the numbers vanish along a linear path to the vanishing point. What this means is that rather than an additional \zshft added for each letter, the \zshft should also get smaller with each letter, so that the total shift (dZ) on the nth letter (after the 1st) should be

dZ = dz + k dz + k^2 dz + ... + k^(n-1) dz

where dz is the specified \zshft, and k is the scale-down parameter given by \dimfrac. Simple manipulation reveals that the shift for the nth letter (after the 1st) is:

dZ = dz (1 - k^n)/(1-k)

This can be calculated in LaTeX, but requires the much more computationally intensive fp package. So, at the expense of compilation time, we can achieve the following:

\documentclass{article}
\usepackage{scalerel,fp}
\newlength\curht
\newlength\zshft
\newcounter{letcount}
\def\defaultdimfrac{.98}
\def\slantvalue{0}
\zshft=0pt\relax
\def\defaultstartht{\baselineskip}
\newcommand\diminish[2][\defaultdimfrac]{%
\curht=\defaultstartht\relax
\def\dimfrac{#1}%
\setcounter{letcount}{0}
\diminishhelpA{#2}%
}
\newcommand\diminishhelpA[1]{%
\expandafter\diminishhelpB#1\relax%
}
\def\diminishhelpB#1#2\relax{%
\FPpow\localshift{\dimfrac}{\theletcount}\unskip%
\FPsub\localshift{1}{\localshift}%
\FPsub\localdenom{1}{\dimfrac}%
\FPdiv\localshift{\localshift}{\localdenom}%
\raisebox{\localshift\zshft}{\scaleto{\strut\slantbox{#1}}{\curht}}%
\stepcounter{letcount}%
\curht=\dimfrac\curht\relax%
\ifx\relax#2\relax\else\diminishhelpA{#2}\fi%
}
\newsavebox{\foobox}
\newcommand{\slantbox}[2][\slantvalue]{\mbox{%
\sbox{\foobox}{#2}%
\hskip\wd\foobox
\pdfsave
\pdfsetmatrix{1 0 #1 1}%
\llap{\usebox{\foobox}}%
\pdfrestore
}}
\begin{document}
\def\pinum{3.14159265358979323846264338327950288419716939937510}
\def\defaultstartht{14pt}
\diminish{\pinum}\par
\def\slantvalue{.15}
\zshft=.3pt\relax
\diminish[0.96]{\pinum}\par
\diminish[0.94]{\pinum}\par
\def\slantvalue{.35}
\zshft=1.7pt\relax
\def\defaultstartht{38pt}
\diminish[0.92]{\pinum}
\end{document}

Oops!! Spill in aisle 1! (coding left to the student)

• In some sense the "right" value for scale-down-ratio would be 0.1. – Nate Eldredge Oct 22 '14 at 21:57
• I wonder if there is a way to make this look more "3D", i.e., making it look like an infinite row of numbers vanishing at the horizon? – yori Oct 23 '14 at 1:25
• @Yori Does the addendum achieve what you desire? – Steven B. Segletes Oct 23 '14 at 10:43
• @StevenB.Segletes: Since we're working in base 10, the value of each successive decimal place is 1/10 the value of the one before it. So maybe one should make the heights diminish in the same proportion. (Or, as rob, suggests, the areas.) – Nate Eldredge Oct 24 '14 at 18:22
• @NateEldredge D'oh! So your comment contained both a joke and a greater mathematic truth! – Steven B. Segletes Oct 24 '14 at 19:38

Mark Wibrow enhanced the text decoration library of TikZ greatly and you can do all kinds of stuff based on formulas regarding the character number and total count of words and so on.

Here is a wacky take on it, (thus a simple monotone scaling along a path is very simple and is possible just by playing with the formula with scale instead of yscale)

\documentclass[tikz]{standalone}
\usetikzlibrary{decorations.text}
\tikzset{pi decoration/.style={
decoration={text effects along path,
text={3.14159265358979323846264338327950288419716939937510},
text align=center,
text effects/.cd,
character total=\n,character count=\i,
path from text,
characters={yscale=10*sin(deg(5*\i/\n))-\i/\n,inner sep=0,anchor=base}
}}}
\begin{document}
\begin{tikzpicture}
\path[pi decoration,decorate] (0,0);
\end{tikzpicture}
\end{document}

• Because, because, because .... because of the wonderful things TikZ does! (PS Thanks for the edit on my recent post.) – Loop Space Oct 22 '14 at 19:06
• @LoopSpace The pleasure is mine. – percusse Oct 22 '14 at 21:06
• Seeing this, I find myself wanting to ask the question "How do I display phi in LaTeX like Fibonacci?". I'd like to see a golden spiral with the digits of phi scaled so that when it reaches the second square then it is exactly 1/phi the size it was when it set out. – Loop Space Oct 22 '14 at 21:16
• @LoopSpace, please do! – Joe Oct 22 '14 at 21:24
• My phi code is now on github: github.com/loopspace/golden – Loop Space Oct 25 '14 at 21:33

A recursive macro:

\documentclass[border=2]{standalone}
\usepackage{mathptmx} % a scalable font is needed

\makeatletter
\def\makesmaller#1#2{%
\dimen0=\dimexpr\dimen0-.08pt\relax
\ifx#2\relax
\expandafter\@gobble
\else
{\fontsize{\dimen0}{0}\selectfont#1}%
\expandafter\@firstofone
\fi
{\makesmaller{#2}}%
}
\def\decreasingpi{%
{\dimen0=16pt\fontsize{\dimen0}{0}\selectfont
3.\makesmaller1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253
\relax}
}
\makeatletter

\begin{document}
\decreasingpi
\end{document}

One might first count the number of digits and decide the minimum size, so as to adjust the decrease step. I leave this as an exercise.

Just another way...

\documentclass[varwidth,border=5]{standalone}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\newdimen\dx
\def\dm#1#2{\dx=1em\relax\def\rt{#1}{\Dm#2\dm}}
\def\Dm#1{\ifx#1\dm\else\fontsize\dx\dx\selectfont#1\dx=\rt\dx\expandafter\Dm\fi}
\begin{document}
\dm{0.95}{3.14159265358979323846264338327950288419716939937510}
\end{document}


text:
# Bid Optimization in Broad-Match Ad auctions - Computer Science > Computer Science and Game Theory

Bid Optimization in Broad-Match Ad auctions - Computer Science > Computer Science and Game Theory - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: Ad auctions in sponsored search support broad match- that allows anadvertiser to target a large number of queries while bidding only on a limitednumber. While giving more expressiveness to advertisers, this feature makes itchallenging to optimize bids to maximize their returns: choosing to bid on aquery as a broad match because it provides high profit results in one biddingfor related queries which may yield low or even negative profits.We abstract and study the complexity of the {\em bid optimization problem}which is to determine an advertiser-s bids on a subset of keywords possiblyusing broad match so that her profit is maximized. In the query language modelwhen the advertiser is allowed to bid on all queries as broad match, we presentan linear programming LP-based polynomial-time algorithm that gets theoptimal profit. In the model in which an advertiser can only bid on keywords,ie., a subset of keywords as an exact or broad match, we show that this problemis not approximable within any reasonable approximation factor unless P=NP. Todeal with this hardness result, we present a constant-factor approximation whenthe optimal profit significantly exceeds the cost. This algorithm is based onrounding a natural LP formulation of the problem. Finally, we study a budgetedvariant of the problem, and show that in the query language model, one can findtwo budget constrained ad campaigns in polynomial time that implement theoptimal bidding strategy. Our results are the first to address bid optimizationunder the broad match feature which is common in ad auctions.

Autor: Eyal Even-dar, Yishay Mansour, Vahab Mirrokni, S. Muthukrishnan, Uri Nadav

Fuente: https://arxiv.org/

DESCARGAR PDF


text:
$$\require{cancel}$$

# 1: Uncertainty and Confirmation of Hypotheses

• 1.1: Background Material
Our knowledge of position vectors allows us to define average position, and from there we define uncertainty in position.
• 1.2: Activities
We practice taking and analyzing data, and learn an important nuance about hypotheses.