text:
A Note on the Monotone Product of Nuclear $C^{*}$-Algebras
Received:April 02, 2007  Revised:November 22, 2007
Key Words: monotone product   GNS representations   nuclear $C^{*}$-algebras.
Fund Project:the Youth Foundation of Sichuan Education Department (No.2003B017); the Doctoral Foundation of Chongqing Normal University (No.08XLB013).
 Author Name Affiliation WU Wen Ming College of Mathematics and Computer Science, Chongqing Normal University, Chongqing 400047, China Department of Mathematical Science, Tsinghua University, Beijing 100084, China ZHAO Yong School of Mathematics and Information, China-West Normal University, Sichuan 637002, China YANG Fang College of Mathematics and Computer Science, Chongqing Normal University, Chongqing 400047, China
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Abstract:
Given two nuclear $C^{*}$-algebras ${\cal A}_{1}$ and ${\cal A}_{2}$ with states $\varphi_{1}$ and $\varphi_{2}$, we show that the monotone product $C^{*}$-algebra ${\cal A}_{1}\rhd{\cal A}_{2}$ is still nuclear. Furthermore, if both the states $\varphi_{1}$ and $\varphi_{2}$ are faithful, then the monotone product ${\cal A}_{1}\rhd{\cal A}_{2}$ is nuclear if and only if the $C^{*}$-algebras ${\cal A}_{1}$ and ${\cal A}_{2}$ both are nuclear.
Citation:
DOI:10.3770/j.issn:1000-341X.2009.03.013
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# Phi coefficient

34,201pages on
this wiki

It has been suggested that this article or section be merged with Matthews correlation coefficient. (Discuss)

In statistics, the phi coefficient φ or rφ is a measure of association for two binary variables. This measure is similar to the Pearson correlation coefficient in its interpretation. In fact, a Pearson correlation coefficient estimated for two binary variables will return the phi coefficient.[1] The Phi coefficient is also related to the chi-square statistic for a 2×2 contingency table (see Pearson's_chi-square_test)[2]

$\phi = \sqrt{\frac{\chi^2}{n}}$

where n is the total number of observations. Two binary variables are considered positively associated if most of the data falls along the diagonal cells. In contrast, two binary variables are considered negatively associated if most of the data falls off the diagonal. If we have a 2×2 table for two random variables x and y

 y = 1 y = 0 total x = 1 $n_{11}$ $n_{10}$ $n_{1\bullet}$ x = 0 $n_{01}$ $n_{00}$ $n_{0\bullet}$ total $n_{\bullet1}$ $n_{\bullet0}$ $n$

where n11, n10, n01, n00, are non-negative "cell cell counts" that sum to n, the total number of observations. The phi coefficient that describes the association of x and y is

$\phi = \frac{n_{11}n_{00}-n_{10}n_{01}}{\sqrt{n_{1\bullet}n_{0\bullet}n_{\bullet0}n_{\bullet1}}}$

## Maximum values Edit

Although computationally the Pearson correlation coefficient reduces to the phi coefficient in the 2×2 case, the interpretation of a Pearson correlation coefficient and phi coefficient must be taken cautiously. The Pearson correlation coefficient ranges from −1 to +1, where ±1 indicates perfect agreement or disagreement, and 0 indicates no relationship. The phi coefficient has a maximum value that is determined by the distribution of the two variables. If both have a 50/50 split, the range of phi will range from −1 to +1. See Davenport El-Sanhury (1991) [3] for a thorough discussion.


text:
# Please consider using book references that will be understood internationally

This originally was an off-topic comment elsewhere. I think it's important enough to warrant it's own post.

As this is an international site, our language is English. However, many people play games in their native language, using translated books. Posting here means we can translate language from English to our native language and back quite well, but book layout does not translate at all. References to books should ideally mention texts that we can search for to find the mentioned sentences or paragraphs in our native language books. Because layout and maybe even cover pictures may change with editions and translations.

The very first sentence on pg. 73 of the PHB proves you wrong!

This is a reference, non-English players have a lot of problems finding. Page 73 in the English book is somewhere between page 60 and 90 in my book. And a first sentence on a page will probably not be the first sentence on any of my pages. So I'd have to search 30 pages, sentence by sentence to check if one might prove me wrong.

In the players handbook, in the Chapter "Combat" under the heading "Attack Actions", the first sentence of the second paragraph proves you wrong (PHB pg. 73).

Now this is way easier. I have the books name, the chapter and paragraph. I can find all of that by translating words and skimming over the table of contents. Maybe the chapter starts at page 68 and have have to flip over to page 69 to see the heading mentioned. But it is spot on.

So please, if you reference books, think of the non-English speakers. We don't expect a reference into our books, but it's way easier for us to find it, if you provide context from your books instead of page numbers.

Because context translates, page numbers do not.

They even translate well enough the other way. I cannot quote verbatim or cite page numbers in any meaningful way (at least not to a non-German in my case) but I can give a description and Chapter heading. If I guess the chapter is named "Combat" and it happens to be named "Battle", then my guess is still better than nothing.

Please note that this is a suggestion how to improve your posts. I'm not looking to make this mandatory or downvote anybody if he does not. This will take a post from good to great, it's in no way required that you do this.

• Even the long form doesn't necessarily help. As an example, German editions of Shadowrun also change the structure of the text.... – Mala Feb 26 '16 at 17:56
• @Mala Yes, but with the Chapter and Heading, at least you know where to look. – nvoigt Feb 26 '16 at 18:14
• No, that's what I mean. They re-structured it. They didn't just translate, the original chapters and subsections were not mapped 1:1 but rewritten to be easier to understand. – Mala Feb 26 '16 at 18:17
• @Mala hm, I did not notice that, but I don't own the English Shadowrun books to compare. With most Games, the Chapters and heading stay the same, just get a different layout based on picture selection and word count in different languages. – nvoigt Feb 26 '16 at 18:19
• Well, they even called it 4.01d (to show that the changes, etc). Anyway, it was just an example, there are others... The point is that even the additional information (which is clearly harder to write and makes texts harder to read) does not always solve the problem - so it's not an ideal approach for a convention in my eyes. – Mala Feb 26 '16 at 18:23
• @Mala: it's certainly not worse, though, right? It seems to me that a little reminder to focus on content rather than (or as much as, or in addition to) probably isn't bad advice. – nitsua60 Feb 26 '16 at 19:36
• @nitsua60 I find the suggestion adds noise to an answer. HeyICanChan's answer demonstrates this nicely, so in my eyes, it might be an improvement as a footnote, but not good inline in the text. – Mala Feb 26 '16 at 21:47
• @Mala I don't know, I find HeyICanChan's answer to be overwrought. I'll likely be adding header to any of my references from now on, and others can do as they wish. (Obviously.) So my version of HeyICanChan's quote would have read "quote quote" (DMG p.136 "Identifying A Magic Item") – nitsua60 Feb 26 '16 at 21:53
• @nitsua60 You might want to propose that compromise as an answer. (However, based on nvoigt's sample in this question, I'm not sure how much help that limited amount of context will provide.) – Hey I Can Chan Feb 26 '16 at 22:49
• Any amount of context is better than no context. – nvoigt Feb 27 '16 at 11:49
• This is sometimes useful for games that have one (game) edition but many (literary) editions. For example, Stars Without Numbers has a free no-art edition that is the same rules as the full for-pay edition, but with different pagination. Giving chapter/section/subsection information is helpful there too. – SevenSidedDie Feb 28 '16 at 7:02
• @SevenSidedDie There's a difference between a question or answer identifying the text it uses (like here)—which is common courtesy—and asking that a question or answer pinpoint where text is located in that edition without using page numbers—which is a heavy burden on an already conscientious querent or respondent. – Hey I Can Chan Feb 28 '16 at 11:01
• @HeyICanChan I agree there's a difference, and it is a burden, yes. It's something that it's good to be aware of, and does come up naturally. I'm not sure how well it can be applied across the entire site and everyone who might answer though, yeah. Especially for popular games like D&D 5e, the expected volume of answers and variety of answerers makes it nearly impossible to mandate. – SevenSidedDie Feb 28 '16 at 20:42
• @SevenSidedDie Even on a small scale, what's being asked for here is greater than what's asked for by a college professor or a professional journal. If a prof told me Cite your quotations so a reader of a foreign-language edition of the text you quoted can use context to find the text you quoted in his foreign-language edition, I'd think he was joking. – Hey I Can Chan Feb 29 '16 at 8:53
• Does this happen frequently? I don't see many people making mention of page numbers being inaccurate because of language differences. Would it help if question askers indicated that they were non-english native for this purpose? Would even a tag be necessary? – Premier Bromanov Mar 1 '16 at 16:27

Instead trying to accommodate non-English texts in general, I suggest a user of a non-English edition Comment how the desired information can't be located in that non-English edition and request from the querent or respondent more context.

That's because, while I think it's a great idea to accommodate non-English speakers, asking folks to use a text's headers, subheads, and sentence numbers instead of or in addition to page numbers is, I think, a burden too great and may possibly be even more confusing.

For example, consider this:

"Activating a spell trigger item is a standard action and does not provoke attacks of opportunity." (DMG 213)

And compare it to this:

"Activating a spell trigger item is a standard action and does not provoke attacks of opportunity." (DMG 213 or Chapter 7: Magic Items on Handling Magic Items on Using Items in the paragraph Spell Trigger in the final sentence)

While the second example does locate the text precisely, it assumes that, when translated, all of those terms also match up, and there's no way to be sure that they do. Further, there's not even a standardized international method of indicating page numbers, so using an even more complicated non-standardized long-form of text location will be messy and difficult for English and non-English readers alike.

It does suck when someone else's page numbers don't sync up to one's own, and you have my sympathy, but I think asking for clarification on a case-by-case basis is a better solution.

(Just to be sure, a search for first sentence brings up about 50 results, and many seem to reference a quotation already presented, so rummaging through a text fairly to find that quotation should be fairly quick. And I totally know that's an imperfect measure.)

• Do you really think that requesting more context will help after a year or so? After all, we strive to be the resource for people coming from Google, we aren't always onto the hot topics only. – nvoigt Feb 26 '16 at 18:15
• Both your examples are comments. I don't expect that from comments, they aren't meant to be permanent anyway. But I do expect that from a full blown answer. An English book's page number is about as useless to me as if you said "it's about 3mm of pages into the book". – nvoigt Feb 26 '16 at 18:17
• @nvoigt A diligent author should be paying attention to comments on questions and answers and update those to reflect greater interest. I mean, yes, it sucks if the author isn't diligent, but you can downvote that answer to express your displeasure if it goes unchanged, and make it clear in a comment that you'll upvote it when it's updated to reflect your needs. That answer is by definition not useful to you until it's updated, after all. – Hey I Can Chan Feb 26 '16 at 18:21
• @nvoigt I'm not sure how my examples are Comments. I was using examples that could have come from any of my own answers or questions as well as comments. – Hey I Can Chan Feb 26 '16 at 18:22
• I certainly won't downvote an answer because it's "good" instead of "picture perfect to my personal taste". That would be the sort of dick move I'm arguing against in almost every post here. If some people want to improve their posts in the face of the fact that not everyone is reading the English books, he or she can read my post and act accordingly. It will make me more inclined to upvote it. – nvoigt Feb 27 '16 at 11:47
• @nvoigt I suggested commenting first then using the only remaining tool — a downvote — after having been ignored or refused; seriously, I, too, wish there were options more granular than nothing, up, down, and discuss, but that's not the site's format. And, just to be clear, to understand my reluctance to accommodate non-English editions in the manner you suggest, imagine, for example, my answer here but with the addition of contextual textual locators like those you mention. – Hey I Can Chan Feb 27 '16 at 15:49
• I'm just saying that this is a means to make answers better, I will comment to ask for improvement, I will not downvote an other wise good answer just because I personally think it could be better. That's not what a downvote is for from my perspective. – nvoigt Feb 28 '16 at 10:36
• I just edited in the page number in my second example. I'm not sure if that was what put you off, but I never meant for that information to get lost. I just forgot it in my post. I want more information, not less. – nvoigt Feb 28 '16 at 11:09
• @nvoigt It's cool. My second example had already followed what I assumed you meant. (Not including page numbers at all makes me cringe; I don't think I'm capable of excluding page numbers completely unless there's a corresponding link I can use instead.) – Hey I Can Chan Feb 28 '16 at 11:14
• @nvoigt RE: The uselessness of It's about 3mm of pages into the book. Interestingly, an answer to the question Angelo Fuchs posed on History.SE proposes using % into the chapter, which I think would probably have about the same degree of usefulness as measuring page depth in mm. – Hey I Can Chan Mar 2 '16 at 17:51

I'd consider this to be different by context.

In cases where you cite something easily searchable (a spell, a weapon, a creature) you must cite the book, but usually not the page (or anything else). That is everybody, English speaking or not, can look up 'owl-bear' in the index and translate it to their relevant language easily.

In cases where you cite something that is buried in large text sections it becomes more complicated.

I suggest you keep it as short as possible while still being searchable. It doesn't need to be understood by a machine, but by a human.

Activating a spell trigger item is a standard action and does not provoke attacks of opportunity." (Chapter 7: Using Items, Spell Trigger paragraph)

Note: It might also be that your citation is lost due translation and just not there in another language (e.G. the Cyberpunk 2020 basic rules differ in the area of martial arts between the German and the English version).

On a personal note: I usually own German books and I pirated the English ones to be able to look up page numbers, so I can translate them and then look it up on the dead-tree. So, I totally agree that this is an issue.

• Because there's no standardized method of presenting the context clues, how much good will this actually do? I mean, in your example, you've quotation marks around Spell Trigger which aren't present in the English original and the quotation marks are used in a nonstandard way (in English anyway). Searching an English DMG for "Spell Trigger", for instance, yields 0 results. I can get behind a meta-movement to develop a standard, but telling everyone Fake it for the greater good! will just make a mess. – Hey I Can Chan Feb 29 '16 at 9:00
• @HeyICanChan The " aren't there in the German book either, I just used them to mark the name. Lets work on a meta-movement to develop a standard. I'll remove the " from my quote as they are misleading. – Angelo Fuchs Feb 29 '16 at 9:45
• @HeyICanChan Also, they still will do a lot of good. "DMG p. 213" is 100% useless to someone who does not happen to own the same copy as you do. "Chatper 7: Using Items" is much more helpful. "Spell Trigger paragraph" makes it more clear. As I mentioned: They don't need to be machine readable (e.G. Full text search compatible) they need to be human readable (e.G. opening the book at Chapter 7, look for Using Items). If my copy now happens to have Chapter 7 being about creation of monsters because the translators thought it would be better, I still know what to look for in the index. – Angelo Fuchs Feb 29 '16 at 9:49
• @HeyICanChan compare this answer of mine I can't quote the English page number (because I don't have the English book), but I could quote the chapter and section. And if you look into the edit history, you note that even the words I used most likely never were full text searchable anyway because I needed to translate them from German (and did so poorly). – Angelo Fuchs Feb 29 '16 at 9:52
• @HeyICanChan If this meta were successful I will look up Chapter and Section and edit my answer accordingly. – Angelo Fuchs Feb 29 '16 at 9:53
• Fair enough. I look forward to participating in the development of an international textual citation standard. (That sounds sarcastic, but it's really not. As a conscientious contributor, I want my questions and answers to be applicable internationally.) – Hey I Can Chan Feb 29 '16 at 9:57
• @HeyICanChan I asked a question on History.SE: history.stackexchange.com/questions/27752/… to figure out how the "pros" do it. Seems overly complicated for our hobby, but sure a good starting point for a standard. – Angelo Fuchs Feb 29 '16 at 17:12


text:
# 75^40=9^20*5^x

can anyone tell me how to solve this.

Note by Daniel Kua
3 years, 9 months ago

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2^{34} $$2^{34}$$
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X=80

- 3 years, 6 months ago

x=80

- 3 years, 8 months ago

Which one do you mean I or II :

• $$75^{40} = 9^{20 \times 5^{x}} . . . . . I$$

• $$75^{40} = 9^{20} \times 5^{x} . . . . .II$$

- 3 years, 9 months ago

the 2nd one

- 3 years, 9 months ago

Sorry for the late response,

$$75^{40} = 9^{20} \times 5^{x}$$

$$75^{40} = (3^2)^{20} \times 5^{x}$$

$$75^{40} = 3^{40} \times 5^{x}$$

$$\frac{75^{40}}{3^{40}} = 5^{x}$$

$$(\frac{75}{3})^{40} = 5^{x}$$

$$25^{40} = 5^{x}$$

$$(5^2)^{40} = 5^{x}$$

$$5^{80} = 5^{x}$$

$$\boxed{x = 80}$$

- 3 years, 9 months ago

$$75^{40}= (3*25)^{40}$$

$$= 3^{40} * 25^{40}$$

$$= 3^{2*20} * (5^2)^{40}$$

$$= (3^2)^{20} * 5^{2*40}$$

$$= 9^{20}*5^{80}$$

Therefore $$x = 80$$. I only used 2 formulae, $$(ab)^n = a^n * b^n$$ and $$(a^n)^m = a^{m*n}$$

Also, to write all the signs, read this

- 3 years, 9 months ago

3^40 is not equal to 3^20+2

- 3 years, 8 months ago

Its $$3^{20 \times 2}$$. I used the $$*$$ symbol.

- 3 years, 8 months ago

Of course not. Look at this exponent's formulas :

$$a^x \times a^y = a^{x+y}$$

$$(a^x)^y = a^{xy}$$

So,

$$3^{40} \neq 3^{20+2}$$ but

$$3^{40} = 3^{2 \times 20} = (3^2)^{20}$$

Got it?

- 3 years, 8 months ago

how do u type multiple again? tell me pls.

- 3 years, 9 months ago

You can see how to type in math formatting by click on "formatting guide" at the bottom-left corner of your comment box.

To type multiple, just type " \times " in math format (remember, in math format! See how to start a math format in the formatting guide.

The more easy ways to know the format of symbols, just point your cursor at the symbols, the format then will appears.

- 3 years, 9 months ago

no.2

- 3 years, 9 months ago


text:
# Prove that the square of an odd integer

1. Sep 9, 2004

### Yodums

I'm not a very logical person, and I would hardly consider math a strength so, I'm stuck with these proofs:

1. Prove that every positive integer, ending in 5 creates a number that when squared, ends in 25

2. Prove that if n is an even positive integer, then n^3 - 4n is always divisible by 48.

3. Prove that the square of an odd integer is always of the form 8k + 1 , where k is an integer.

Thank you.

2. Sep 9, 2004

### shmoe

In all these problems you know something about the form of the integers being considered, use this information.

For example in Q2, n is even so you can write it as n=2m for some positive integer m. Substitute this into n^3-4n and try to show it's divisible by 48, irregardless of the value of m (do you know modular arithmetic yet?).

Same deal will work for Q3, except your number is odd here.

For Q1, can you think of a simple general form for numbers ending in 5? I can't think of a good hint off hand that doesn't give too much away...

3. Sep 9, 2004

### Yodums

I've never studied the modular arithmetic yet. Unless you are referring to a arithmetic sequence?

Anyways, working ahead on Q2, I've gotten from:

n^3 - 4n

n=2m

So:

2m^3 - 4 (2m)

2m^3 - 8m

2m (m^2 - 4)

2m (m + 2) (m - 2)

Not sure where to go from there?

4. Sep 9, 2004

### Tide

That should be $8m^3-8m$ which, by the way can be expressed as $8m(m-1)(m+1)$. :-)

5. Sep 9, 2004

### musky_ox

I really have no clue what your doing but doesnt that factor to 8m(m-1)(m+1)?

I dont understand what we are trying to do here...

n=2
(2)^3-4(2) = 0
is 0 divisible by 48??? I guess so, but then again.. everything is divisible by everything, its just not guaranteed to give a whole number as an answer.

6. Sep 9, 2004

### shmoe

Careful with those substitutions, brackets are easy to miss:)

Now, you can clearly see that 8 divides $8m(m-1)(m+1)$, because of the 8 in front. What can you say about the $m(m-1)(m+1)$ part? Is it even? odd? divisibly by what? Try subbing in a few values of m (say 1 to 5) and see if anything jumps at you.

ps. Modular arithmetic is different from arithmetic sequences, and can give a handy way of dealing with divisibility kind of problems. It's not necessary here though, so don't worry about it.

pps. 0 is divisible by everything, in the integer sense of "divisible". we say a is divisible by b if there exist c where a=bc (a,b, c all integers). In the case of a=0 and b=anything, take c=0.

7. Sep 9, 2004

### Tide

schmoe: BTW - you don't need modulo arithmetic.

musky ox: Divisibility in math has a specific meaning - namely an integer is divisible by another integer if there is no remainder.

8. Sep 9, 2004

### Alkatran

For the 5 proof, this is how I would go about it (I do this for ending in 5, and NOT ending in 25, to give you something to expand on):

We have a number ending with 5 which has another digit in it, x is the other digit, and y it's position (with 0 being the position of the 5, so y > 0, and x and y are contained within Z). This way we can show if a digit is in any position to the left of the 5, it won't affect the final result.

(x*10^y+5)^2 = (x*10y)^2 + 2*x*10^y*5 + 5^2 = 100*(xy)^2 + 10*x*10^y + 25

Now, we only want the last digit so we mod by 25, now:
(x + y) mod z = ((x mod z) + (y mod z)) mod z
and:
(x*y) mod z = ((x mod z)(y mod z)) mod z
so:

*by the way, this equation really spreads out, and I don't know latex. The underlined abd bolded portions signify work being done.

(100*(xy)^2 + 2x*10^2 + 25) mod 10
= [([100*(xy)^2] mod 10) + ([10*x*10^y] mod 10) + (25 mod 10)] mod 10
= ([([100 mod 10]*[(xy)^2 mod 10]) mod 10] + ([10 mod 10] * [(x*10^y) mod 10] mod 10) + 5) mod 10
= ([(0 * [(xy)^2 mod 10]) mod 10] + (0 * [(x*10^y) mod 10] mod 10) + 5) mod 10
= (0 + 0 + 5) = 5

By the way.. I just guessed at the mod rules, and if they don't stand the entire proof is false.

Last edited: Sep 9, 2004
9. Sep 9, 2004

### CTS

1.)
(10k + 5)^2 = 100k^2 + 100k + 25 = 100(k^2 + k) + 25

2.)
(2n)^3 -4*(2n) = 8n^3 - 8n = 8n(n-1)(n+1) = 8(n-1)n(n+1)
One of (n-1), n, (n+1) must be divisible by three, and at least one of them must be even, therefore 48|(2n)^3 -4*(2n)

3.)
(2n + 1)^2 = 4n^2 + 4n + 1 = 4n(n + 1) + 1
Either n or (n+1) is even, therefore (2n + 1)^2 = 8k+1

10. Sep 10, 2004

### HallsofIvy

Staff Emeritus
A lot of really complicated responses and then CTS gives the obvious way to do them! (Just before I could!)

11. Sep 10, 2004

### Alkatran

I considered going down that route, but then I realised that someone could argue that the digit wasn't in the "ten" position, and thus the proof didn't cover it. Much shorter than my work though!

12. Sep 10, 2004

### hedlund

3: (2m+1)^2 = 4m^2+4m+1
4m^2+4m+1 = 4m(m+1)+1
Now we have to cases, if m is even let m=2n then we get
8n(2n+1)+1 which is of the form 8n*(2n+1)+1. If m is odd we have m=2u+1 which yields
4*(2u+1)*(2u+2) + 1 = 8*(2u+1)*(u+1)+1 which is of the form 8k+1.

13. Sep 10, 2004

### shmoe

I may have been more complicated, but my hope was to kick Yodlums down the path to understand how to actually think about solving these sort of problems. I figured that would be much more instructive than to just hand out the solutions (teach a man to fish and all that jazz). Oh well.

Tide-I know modular arithmetic isn't necessary (see my post above yours) but it's often taught close by divisibility. If that was the case here, it would have made sense to use that language where appropriate.

14. Sep 10, 2004

### Tide

Schmoe - yes I reread your post and realize I had read it too fast the first time! Thanks for the clarification.

15. Sep 12, 2004

### decibel

The first one i thought was pretty simple, i'm taking the same course as you right now (i'm guessing), and probably have the same textbook too (Goemetry and Discrete Mathematics)...this is how i did #1 for my homework :zzz:

1. Prove that every positive integer, ending in 5 creates a number that when squared, ends in 25

Okay, listen to what the question is asking you to prove, it says prove that for every POSITIVE INTEGER ending in 5 and then SQUARED, generates a number that ends in 25. Think about a formula that would generate a number ending in 5 for whatever positive integer you input, i used this..

Let n = 5(2x+1) (every odd number multiplied by 5 will end in 5 )
= 10x+5 (expand)

We now have a number that ends in 5 no matter what we plug in (positive integers)...now it has to be squared doesent it?...cause remember what the original question said...so we simply square our "formula".

Let P(n) = n^2
=(10x+5)^2
= 100x^2 + 100x +25

Now this is where you have to think, the 100's in the equation are gonna make numbers ending in 00 no matter what so you can conclude that...

= 00 + 00 + 25
= 25

Theres number one, i realize that other people already posted their answer to this, but i decided to show it in a more simple manner.

16. Sep 12, 2004

### HallsofIvy

Staff Emeritus
Me too, shmoe, I was really referring to the posts that answered in terms of "mod" which was clearly too technical for the original question.


text:
# Unit of Density

Before we discuss the unit of density, let’s quickly go through what density is. Well, the density of a substance is basically the measure of how much stuff an object has in a unit volume. It is represented by the symbol ρ or it can also simply be represented by letter D.

## What is Density?

The density of material shows the denseness of that material in a specific given area. A material’s density is defined as its mass per unit volume. Density is essentially a measurement of how tightly matter is packed together. It is a unique physical property for a particular object. The principle of density was discovered by the Greek scientist Archimedes.  It is easy to calculate density if you know the formula and understand its related units.

## How is Density Calculated?

Mathematically, the density of an object is expressed as follows:

$Density= \frac{Mass}{Volume}$ $\rho = \frac{m}{v}$

Where,

• ρ is the density
• m is the mass
• V is the volume

## Unit of Density

• Though SI unit of density is kg/m³, for convenience we use g/cm³ for solids, g/ml for liquids and g/L for gases.
• Density can be explained as the relationship between the mass of the substance and volume it takes up.
• In a qualitative term, it shows how much heavy an object is at constant volume.
• Different substances have different density, which means for the same volume of different substances weigh differently.

## SI Unit of Density

Each substance has a specific density. Generally, the density of water (which is approximately about 1 gram/cubic centimetre) is taken as the standard value for calculating the density of substances. However, the SI unit of Density is measured using kilogram per cubic metre (kg/m3).

### Other Density Units

Talking about other density units, metric tons and litre are also used even though they are not part of the SI. Some of the other units include:

• gram per millilitre (g/mL)
• metric ton per cubic metre (t/m3)
• kilogram per litre (kg/L)
• megagram (metric ton) per cubic metre (Mg/m3)
• gram per cubic centimetre (g/cm3)
1 g/cm3 = 1000 kg/m3
• kilogram per cubic decimetre (kg/dm3)

In addition to this, in the cgs system density is measured in gram per cubic centimetre (g/cm3).

You may also want to check out these topics given below!

## Applications of Density in Real Life

Many applications of density are there in our real-life like a few examples are in pipe design, shipbuilding, helium balloons, weight distribution in the aeroplane and the fact that ice floats on water.

• The knowledge of densities of two substances helps you in separation techniques. For example, separation of oil from water. Leakage of an oil tank in the ocean then oil drops starts to float on the water due to their less density in the water.
• Another well-known application of density is determining whether an object will float on water or not. Floating of ships and diving of submarines are due to their density difference.

## Solved Examples

### Question 1:

Take two boxes which have the same volume. Fill the first box with x balls and the second box with 6x balls. If the mass of each ball is the same, which box would weigh more?

Solution:

The box that has more balls has more mass per unit of volume.

Here the first box contains x number of balls and the second box contains 6x number of balls. Since the number of balls in the second box is 6 times the first box, the second box would weigh more. This property of matter is called density.

### Question 2:

Calculate the density of water if it has a mass of 1160 Kg and volume of 1m3?

Solution:

Given,

Mass = 1160 Kg

Volume = 1m3

Density is given by the formula:

Density = Mass/Volume

ρ = 1160/1 = 1160 kg/m3

### Questions 3:

If you find a shiny rock, a carbon allotrope with a volume of 0.042cm³ and a mass of 0.14 g, is it graphite or diamond? The density of graphite is 2.266 g/cm3 and the density of diamond is 3.51g/cm3.

Solution:

Given,

Volume of the shiny rock =0.042cm³

Mass of the shiny rock = 0.15 g

Density of graphite = 2.266 g/cm3

Density of diamond = 3.51g/cm3

Use the density equation to solve for m, mass of graphite and for the mass of a diamond.

ρ = m/V

m = ρV

m = 2.266 g/cm³ x 0.042cm³ = 0.0951g for graphite

m = 3.51g/cm3 x 0.042cm³ = 0.1474 g for diamond

The mass of the shiny rock you found is identical with the mass of diamond.


text:
# [tex-live] that nasty \undefined in supp-pdf.mkii again

Joseph Wright joseph.wright at morningstar2.co.uk
Tue Oct 24 08:32:01 CEST 2017

On 24/10/2017 04:14, Norbert Preining wrote:
> Dear all,
>
> back in 2010 (!!!) https://tug.org/pipermail/tex-live/2010-March/025102.html
> I complained about the fact that a minimal tex document
>   \documentclass{article}
>   \usepackage{graphicx}
>   \newcommand{\undefined}{\textbf{undefined}}
>   \begin{document}
>   This is an \undefined\ command.
>   \end{document}
> does not work anymore. This was due to the auto-loading of supp-pdf.mkii
> that was introduced in TL2009 (or so).
>
> The reason is that supp-pdf.mkii uses
>   \ifx\writestatus\undefined
> to check for an undefined commands.
>
> I argued that this is wrong to do it this way, especially since this
> is loaded now in LaTeX. See
> https://tug.org/pipermail/tex-live/2010-March/025107.html
>
> There was the conclusion that pdftex.def should deal with this and
> Heiko send a fixed pftex.def:
> https://tug.org/pipermail/tex-live/2010-March/025126.html
>
> I tested that and it worked.
>
> But alas here we are in 2017 and a colleague just asked me the very same
> question, because he writes a logic text book and uses \undefined as
> a macro.
>
> Can we somehow work that out again?
>
> All the best
>
> Norbert

I have to say I'm surprised anyone would use \undefined as a macro, but
I guess as it's a 'user' name (cf. LaTeX's \@ifundefined) then there's
always a chance.

Heiko's earlier mail doesn't say what fix he proposed, and the old FTP
site is no longer available. It seems most likely that the approach of
ensuring \undefined is really not defined is likely to be best,
something like altering the current

\InputIfFileExists{supp-pdf.mkii}{}{}%

to

\let\GPT at saved@undefined\undefined
\let\undefined\GPT at undefined
\InputIfFileExists{supp-pdf.mkii}{}{}%
\let\undefined\GPT at saved@undefined

I guess will do it. I'll discuss with David C.

Joseph


text:
# Storing results of a calculation as a list

Clear[Pressure, EnergyDensity, fromPressureToEnergy,
fromEnergyToPressure, Ae, Be, Ap, Bp, Er, Mr, Pr, r, h, g1, f1, k1,
m1, f2, g2, k2, m2, f3, g3, k3, m3, f4, m4, g4, k4, M, R];

Ae = Import["D:\\Energy density.xlsx", {"Data", 1, All, 1}];
Be = Ae*(10^-6/1.1154907);
EnergyDensity = Be/(0.08969);
Ap = Import["D:\\Pressure.xlsx", {"Data", 1, All, 1}];
Bp = Ap*(10^-6/1.1154907);
Pressure = Bp/(0.08969);
fromPressureToEnergy =
Interpolation[Most@Transpose[{Pressure, EnergyDensity}],
InterpolationOrder -> 1];
fromEnergyToPressure =
Interpolation[Most@Transpose[{EnergyDensity, Pressure}],
InterpolationOrder -> 1];
Er = (1637.99731*(10^-6/1.1154907)/0.08969); (*Initial value*)
h = 0.0001;
Mr = 10^-33;
Pr = fromEnergyToPressure[Er];
r = h;

While[Pr > 0,

g1 = 4*\[Pi]*r^2*Er*.08969;
f1 = -(1.47*Mr*
Er*(1 + Pr/Er)*(1 + 4*\[Pi]*r^3*Pr*.08969/Mr))/(r^2*(1 -
2*Mr*1.47/r));
k1 = h*f1;
m1 = h*g1;

f2 = -(1.47*Mr*
Er*(1 + (Pr + k1/2)/Er)*(1 +
4*\[Pi]*(r + h/2)^3*(Pr + k1/2)*.08969/Mr))/((r + h/2)^2*(1 -
2*Mr*1.47/(r + h/2)));
g2 = 4*\[Pi]*(r + h/2)^2*(Er + m1/2)*.08969;
k2 = h*f2;
m2 = h*g2;

f3 = -(1.47*Mr*
Er*(1 + (Pr + k2/2)/Er)*(1 +
4*\[Pi]*(r + h/2)^3*(Pr + k2/2)*.08969/Mr))/((r + h/2)^2*(1 -
2*Mr*1.47/(r + h/2)));
g3 = 4*\[Pi]*(r + h/2)^2*(Er + m2/2)*.08969;
k3 = h*f3;
m3 = h*g3;

f4 = -(1.47*Mr*
Er*(1 + (Pr + k3)/Er)*(1 +
4*\[Pi]*(r + h)^3*(Pr + k3)*.08969/Mr))/((r + h)^2*(1 -
2*Mr*1.47/(r + h)));
g4 = 4*\[Pi]*(r + h)^2*(Er + m3)*.08969;
k4 = h*f4;
m4 = h*g4;

Pr = Pr + 1/6*(k1 + 2*k2 + 2*k3 + k4);
Mr = Mr + 1/6*(m1 + 2*m2 + 2*m3 + m4);

Er = fromPressureToEnergy[Pr];
r = r + h;
]

M = Mr; (*result*)
R = r;(*result*)

Dear Friends in the code above I have an initial value for (Er) which is (1637.99731*(10^-6/1.1154907)/0.08969) and the final results are (M) and (R).

My question is that how I can solve this code for various (Er) which varies from the initial value above to (135*(10^-6/1.1154907)/0.08969) and for each of the (Er) I want to save M and R. Therefore, finally I could have something like a list that consists of different (M) and (R) that related to various (Er) which are considered as the initial value.

In fact, my code only works for one (Er) and as a result one (M) and (R) for the chosen (Er). What I need to do is solving my code for various (Er) regarding the mentioned interval.

• Wrap the whole lot into a function that takes those parameters (except anything constant, leave outside) and then call it in a Table with different Er arguments. Return a list or association from the function with your values of M,R. – flinty Jul 24 '20 at 15:09
• There are many ways to do this. Map, Table and Sow + Reap come to mind. See, e.g., mathematica.stackexchange.com/a/181603/43522 – Sjoerd Smit Jul 24 '20 at 15:48


text:
## Seminar in Probability Theory

The Seminar in Probability Theory takes place during the semester, normally on Wednesday at 11:00.

### Program HS 2018

Date/Time Speaker Title Location
6 September 2018 Lisa Hartung
New York University
The Ginibre ensemble and Gaussian multiplicative chaos >
It was proven by Rider and Virag that the logarithm of the characteristic polynomial of the Ginibre ensemble converges to a logarithmically correlated random field. In this talk we will see how this connection can be established on the level if powers of the characteristic polynomial by proving convergence to Gaussian multiplicative chaos. We consider the range of powers in the $$L^2$$ phase.
(Joint work in progress with Paul Bourgade and Guillaume Dubach).
Spiegelgasse 1 Room 00.003
19 September 2018 Alexander Drewitz
Universität Köln
Ubiquity of phases in some percolation models with long-range correlations >
We consider two fundamental percolation models with long-range correlations: The Gaussian free field and (the vacant set) of Random Interlacements. Both models have been the subject of intensive research during the last years and decades, on $$\mathbb Z^d$$ as well as on some more general graphs. We investigate some structural percolative properties around their critical parameters, in particular the ubiquity of the infinite components of complementary phases.
This talk is based on joint works with A. Prévost (Köln) and P.-F. Rodriguez (Bures-sur-Yvette).
Spiegelgasse 1 Room 00.003
31 October 2018 Anton Klimovsky
Universität Duisburg-Essen
High-dimensional Gaussian fields with isotropic increments seen through spin glasses >
Finding the (space-height) distribution of the (local) extrema of high-dimensional strongly correlated random fields is a notorious hard problem with many applications. Following Fyodorov and Sommers (2007), we focus on the Gaussian fields with isotropic increments and take the viewpoint of statistical physics. By exploiting various probabilistic symmetries, we rigorously derive the Fyodorov-Sommers formula for the log-partition function in the high-dimensional limit. The formula suggests a rich picture for the distribution of the local extrema akin to the celebrated spherical Sherrington-Kirkpatrick model with mixed p-spin interactions.
Spiegelgasse 1 Room 00.003
7 November 2018 Dominik Schröder
IST Austria
Cusp Universality for Wigner-type Random Matrices >
For Wigner-type matrices, i.e. Hermitian random matrices with independent, not necessarily identically distributed entries above the diagonal, we show that at any cusp singularity of the limiting eigenvalue distribution the local eigenvalue statistics are universal and form a Pearcey process. Since the density of states typically exhibits only square root or cubic root cusp singularities, our work complements previous results on the bulk and edge universality and it thus completes the resolution of the Wigner-Dyson-Mehta universality conjecture for the last remaining universality type.
Spiegelgasse 1 Room 00.003
14 November 2018 Marius Schmidt
Universität Basel
Oriented first passage percolation on the hypercube >
Consider the hypercube as a graph with vertex set $${0,1}^N$$ and edges between two vertices if they are only one coordinate flip apart. Choosing independent standard exponentially distributed lengths for all edges and asking how long the shortest directed paths from $$(0,..,0)$$ to $$(1,..,1)$$ is defines oriented first passage percolation on the hypercube. We will discuss the conceptual steps needed to answer this question to the precision of extremal process following the two paper series "Oriented first passage percolation in the mean field limit" by Nicola Kistler, Adrien Schertzer and Marius A. Schmidt: arXiv:1804.03117 [math.PR] and arXiv:1808.04598 [math.PR].
Spiegelgasse 1 Room 00.003
21 November 2018 Antti Knowles
University of Geneva
Local law and eigenvector delocalization for supercritical Erdos-Renyi graphs >
We consider the adjacency matrix of the Erdos-Renyi graph $$G(N,p)$$ in the supercritical regime $$pN > C \log N$$ for some universal constant C. We show that the eigenvalue density is with high probability well approximated by the semicircle law on all spectral scales larger than the typical eigenvalue spacing. We also show that all eigenvectors are completely delocalized with high probability. Both results are optimal in the sense that they are known to be false for $$pN < \log N$$. A key ingredient of the proof is a new family of large deviation estimates for multilinear forms of sparse vectors. Joint work with Yukun He and Matteo Marcozzi.
Spiegelgasse 1 Room 00.003
28 November 2018 Gaultier Lambert
University of Zurich
How much can the eigenvalue of a random matrix fluctuate? >
The goal of this talk is to explain how much the eigenvalues of large Hermitian random matrices deviate from certain deterministic locations. These are known as “rigidity estimates” in the literature and they play a crucial role in the proof of universality. I will review some of the current results on eigenvalues’ fluctuations and present a new approach which relies on the theory of Gaussian Multiplicative Chaos and leads to optimal rigidity estimates for the Gaussian Unitary Ensemble. I will also mention how it is also deduce a central limit theorem from our proof. This is joint work with Tom Claeys, Benjamin Fahs and Christian Webb.
Spiegelgasse 1 Room 00.003
12 December 2018 Ioan Manulescu
University of Fribourg
Uniform Lipschitz functions on the triangular lattice have logarithmic variations >
Uniform integer-valued Lipschitz functions on a finite domain of the triangular lattice are shown to have variations of logarithmic order in the radius of the domain. The level lines of such functions form a loop $$O(2)$$ model on the edges of the hexagonal lattice with edge-weight one. An infinite-volume Gibbs measure for the loop $$O(2)$$ model is constructed as a thermodynamic limit and is shown to be unique. It contains only finite loops and has properties indicative of scale-invariance: macroscopic loops appearing at every scale. The existence of the infinite-volume measure carries over to height functions pinned at 0; the uniqueness of the Gibbs measure does not. The proof is based on a representation of the loop $$O(2)$$ model via a pair of spin configurations that are shown to satisfy the FKG inequality. We prove RSW-type estimates for a certain connectivity notion in the aforementioned spin model. Based on joint work with Alexander Glazman.
Spiegelgasse 1 Room 00.003


text:
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# Graphing Absolute Value Functions

## Graphing Absolute Value Functions 1.11 - Solution

In order to graph the given absolute value function, we can make a table of values.

$x$ $f(x)=\text{-}\dfrac{3}{5}|x|$ Simplified Absolute Value $f(x)$
${\color{#0000FF}{\text{-}5}}$ $f({\color{#0000FF}{\text{-}5}})=\text{-}\dfrac{3}{5}|{\color{#0000FF}{\text{-}5}}|$ $f({\color{#0000FF}{\text{-}5}})=\text{-}\dfrac{3}{5}(5)$ $\text{-}3$
${\color{#0000FF}{\text{-}3}}$ $f({\color{#0000FF}{\text{-}3}})=\text{-}\dfrac{3}{5}|{\color{#0000FF}{\text{-}3}}|$ $f({\color{#0000FF}{\text{-}3}})=\text{-}\dfrac{3}{5}(3)$ $\text{-} 1.8$
${\color{#0000FF}{0}}$ $f({\color{#0000FF}{0}})=\text{-}\dfrac{3}{5}|{\color{#0000FF}{0}}|$ $f({\color{#0000FF}{0}})=\text{-}\dfrac{3}{5}(0)$ $0$
${\color{#0000FF}{3}}$ $f({\color{#0000FF}{3}})=\text{-}\dfrac{3}{5}|{\color{#0000FF}{3}}|$ $f({\color{#0000FF}{3}})=\text{-}\dfrac{3}{5}(3)$ $\text{-} 1.8$
${\color{#0000FF}{5}}$ $f({\color{#0000FF}{5}})=\text{-}\dfrac{3}{5}|{\color{#0000FF}{5}}|$ $f({\color{#0000FF}{5}})=\text{-}\dfrac{3}{5}(5)$ $\text{-}3$

Now we can graph $f(x)$ by plotting the points and connecting them with straight lines on either side of the vertex.


text:
# Showing a product of two Lebesgue integrals is $\geq 1$ if the product of the integrands is $\geq 1$

Let $$\mu$$ be a probability measure on a set $$X$$, i.e. $$\mu(X)=1$$, and let $$f$$ and $$g$$ be positive measurable functions on $$X$$. Show that if $$fg\geq1$$, then the integral of $$f$$ times the integral of $$g$$ is greater than or equal to $$1$$.

I'm not sure how to approach this. Since $$f$$ and $$g$$ are positive, the integral of $$f$$ and the integral of $$g$$ are equal to the $$L_1$$ norm of $$f$$ and the $$L_1$$ norm of $$g$$, but I'm not sure why $$fg\geq 1$$ implies that the product of their $$L_1$$ norms is greater than or equal to $$1$$. Considering that $$fg$$ is involved, I was thinking of using Holder's inequality, but I'm not sure what $$p$$ and $$q$$ to use.

This relies more on an arithmetic trick than on analysis. Since $$fg \ge 1$$ you have also that $$\sqrt{fg} \ge 1$$ so that $$1 \le \int_X \sqrt{fg} \le \left(\int_X f \right)^{1/2} \left( \int_X g \right)^{1/2}$$ by Holder's inequality. Now square both sides.
• I don't understand your application of Holder's inequality. What are your $p$ and $q$? – Keshav Srinivasan Nov 8 '18 at 20:31
• $2$ and $2$. $\mbox{}$ – Umberto P. Nov 8 '18 at 20:32
• And how do we know that $\sqrt f$ and $\sqrt g$ are in $L_2$? – Keshav Srinivasan Nov 8 '18 at 20:35
• You don't. Since $f$ and $g$ are positive the integrals are all defined, even if possibly infinite. – Umberto P. Nov 8 '18 at 20:39


text:
# Composition of functions

If we have two or more functions that are contained within each other, we call it compound functions. Its general expression is as follows:

The value of the compound function in x is equal to the function f evaluated in g (x).

## Definition

Given two functions f and g, the composite function of f with g is defined as:

his domain is the set of all x (variable) that are in the domain of g, such that g (x) belongs to the domain of f. That is to say:

Exercise 1: Sean:

y

Determine:

We know that:

We first substitute the value of g (x):

Now, to evaluate f, we substitute x = g (x):

We simplify:

We solve the binomial:

To determine the domain we must remember that:

We will first look for the domain of g (x); Since the function has a root and the variable is inside it, it cannot be negative:

Then the domain of g will be all values greater than or equal to four, that is:

For the domain of f: since its function is a polynomial, it will be all real numbers, that is:

Since we are looking for the domain of the compound function, we know that this domain depends directly on g (x), therefore:


text:
# Determine square roots of rational numbers

#### All in One Place

Everything you need for JC, LC, and college level maths and science classes.

#### Learn with Ease

We’ve mastered the national curriculum so that you can revise with confidence.

#### Instant Help

0/15
##### Examples
###### Lessons
1. Use the graph below to determine a rational number with a square root between 4 and 5.

1. Use the side lengths below to estimate and calculate the area of each square.
1. 5.2 cm
2. 0.027 km
2. Find out if each of the following rational numbers is a perfect square.
1. $\frac{{81}}{{16}}$
2. 0.1
3. 0.01
4. $\frac{5}{{14}}$
3. Evaluate.
1. $\sqrt {361}$
2. $\sqrt {2209}$
3. $\sqrt {0.0169}$
4. $\sqrt {5.76}$
4. Calculate.
1. $\sqrt {56}$, to the nearest tenth
2. $\sqrt {3.7}$, to the nearest hundredth
3. $\sqrt {0.96}$ , to the nearest hundredth
4. $\sqrt {0.066}$ , to the nearest hundredth
0%
##### Practice
###### Free to Join!
StudyPug is a learning help platform covering math and science from grade 4 all the way to second year university. Our video tutorials, unlimited practice problems, and step-by-step explanations provide you or your child with all the help you need to master concepts. On top of that, it's fun - with achievements, customizable avatars, and awards to keep you motivated.
• #### Easily See Your Progress

We track the progress you've made on a topic so you know what you've done. From the course view you can easily see what topics have what and the progress you've made on them. Fill the rings to completely master that section or mouse over the icon to see more details.
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###### Topic Notes
In this section, we will look at how to evaluate a rational number by using square roots. We will also work on questions determining whether a rational number is a perfect square. We will also evaluate the square roots of rational numbers.


text:
# A Comprehensive guide to Bayesian Convolutional Neural Network with Variational Inference

8 Jan 2019

Artificial Neural Networks are connectionist systems that perform a given task by learning on examples without having prior knowledge about the task. This is done by finding an optimal point estimate for the weights in every node... (read more)

PDF Abstract


text:
# How to set dropdown values to null using ng-change angular js?

• I have one dropdown lets say Drop-down A, based on option selected from that, new dropdowns will be populated. For example,on selection of option 1 from Drop-down A, 2 new dropdowns will be showed. similarly on selection of other options new dropdowns will be populated and other dropdowns will be removed.
• Now when I select options from newly populated dropdowns,I can get their value.
• Problem comes when I select option 1 from Drop-down A and then select values from newly generated dropdowns, but now I wish to change Drop-down A slection to option 2. Now based on current selection, new dropdowns will be generated.
So if now I post the values, my controller still shows, previous option 1 values.

So how can I change unselected dropdown values to null?


text:
# Analyticity of critical exponents of the $O(N)$ models from nonperturbative renormalization

### Submission summary

 As Contributors: Andrzej Chlebicki · Pawel Jakubczyk Arxiv Link: https://arxiv.org/abs/2012.00782v3 (pdf) Date submitted: 2020-12-10 09:57 Submitted by: Chlebicki, Andrzej Submitted to: SciPost Physics Academic field: Physics Specialties: High-Energy Physics - Theory Statistical and Soft Matter Physics Approach: Theoretical

### Abstract

We employ the functional renormalization group framework at second order in the derivative expansion to study the $O(N)$ models continuously varying the number of field components $N$ and the spatial dimensionality $d$. We in particular address the Cardy-Hamber prediction concerning nonanalytical behavior of the critical exponents $\nu$ and $\eta$ across a line in the $(d,N)$ plane, which passes through the point $(2,2)$. By direct numerical evaluation of $\eta(d,N)$ and $\nu^{-1}(d,N)$ we find no evidence of discontinuous or singular first and second derivatives of these functions for $d>2$. The computed derivatives of $\eta(d,N)$ and $\nu^{-1}(d,N)$ become increasingly large for $d\to 2$ and $N\to 2$ and it is only in this limit that $\eta(d,N)$ and $\nu^{-1}(d,N)$ as obtained by us are evidently nonanalytical. We provide a discussion of the evolution of the obtained picture upon varying $d$ and $N$ between $(d,N)=(2,2)$ and other, earlier studied cases, such as $d\to 3$ or $N\to \infty$.

###### Current status:
Has been resubmitted

### Submission & Refereeing History

Resubmission 2012.00782v4 on 1 April 2021

Submission 2012.00782v3 on 10 December 2020

## Reports on this Submission

### Strengths

Applies FRG approach to a basic question about RG flows in an important model
Explains motivation clearly
Raises interesting questions

### Weaknesses

Approximate nature of method prevents definitive answer to the question motivating the study
Data shown is limited to plots of two exponents

### Report

This paper applies the functional RG to the O(N) model, aiming to determine the structure of fixed points in the (N, dimension) plane. The motivation comes from work by Cardy and Hamber arguing for a line in this plane where the critical exponents are non-analytic.

Revisiting this prediction using these FRG tools is a worthwhile idea, as the Cardy-Hamber idea has not been sufficiently tested and extended. This paper is a useful contribution. It is well motivated in the introduction, and clearly written (though concise about some technical details). However, the insight provided is limited, as the paper does not give a clear answer to the basic motivating question about the correctness of the Cardy-Hamber prediction. This is mostly because of the approximate nature of the FRG approach (which also involves some ad hoc tuning as implemented) and the difficulty in quantifying its error. The authors state that they do not see the predicted nonanalyticity when the dimension is larger than 2, but acknowledge that it could be that this nonanalyticity is smoothed in the approximate calculation.

In the absence of a definitive conclusion on this issue, the paper gives approximate information about exponents in the (N, dimension) plane and shows that the FRG is capturing quite a lot of the basic physics. Reopening questions about the CH prediction, and about the accuracy of the FRG approach in various limits, may in itself be a valuable contribution.

It is hard for me to judge whether this is sufficient for publication in SciPost. My impression is that this journal is intended to be significantly more selective than a standard Physical Review journal. If so, the answer is probably no. The case might be strengthened if the authors could push their data a little further, for example to provide information about the subleading exponent, which should vanish on the CH line, or to give interesting qualitative information about the shape of the effective potential.

- Irrelevant exponent.

The authors search for the CH line by looking for divergent derivatives of exponents. Taking the CH equations at face value would give a step discontinuity in the derivative of 1/nu as the CH line is crossed, with 1/nu and its derivatives being finite on either side of the step. This is to be compared with Fig 7. From this plot it seems hard to rule out the possibility that the locations marked by stars are such nonanalyticities, rounded by errors caused by the approximations used. For example, the magnitude of the errors in Fig 5 is not all that small.

As an alternative way to look for the CH line, is it possible to examine the irrelevant (negative) RG eigenvalue numerically, as done in some FRG works? This should vanish as the CH line is approached.

- Qualitative information about the effective potential? The figures that the authors show focus on exponents. Does the shape of the effective potential show any interesting change as, say, d is varied at fixed N, going from one side of the CH line to the other?

- Sources of error. Is the truncation of the derivative expansion the only source of error, or are there others? Can we assume that the numerical solution of the truncated equations is essentially exact? For readers unfamiliar with the method, it might be useful to have a concise statement of where possible error can come from.

The method is not exact for the power law phase at N=2, d=2, despite the fact that this fixed line has an exact representation using a Lagrangian with two derivatives. Does this indicate that there are additional sources of error, beyond the truncation?

- Another concern is that the method has to be tuned in a somewhat ad hoc way to reproduce the expected behavior at N=2, d=2. If this is necessary at this point, why should the method be expected to accurately reproduce the structure of the flows at other points?

- The authors mention that the equations cannot be solved when N>2 if the dimension is too close to 2. Could the authors comment in the text on what goes wrong?

- Can the authors comment in the manuscript on the level of difficulty in going to the next order in the derivative expansion?

- p11 “These results are compared to the predictions of the epsilon expansion”. Does this mean the direct result from the “traditional” 2+epsilon expansion, or does it mean some kind of resummation taken from the reference cited [38]? This resummation may assume that the exponents are analytic between d=2 and d=4.

- p6: The statement about “exactly controlled limits” being recovered should be clarified. It could give the impression that the method is exact for d=2, N<=2 which as the authors explain is not the case.

• validity: -
• significance: -
• originality: -
• clarity: -
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Author Andrzej Chlebicki on 2021-04-01
(in reply to Report 2 on 2021-03-01)
Category:
correction

Dear Referee, thank you very much for carefully reviewing our manuscript and preparing the report. We extended the paper in the two directions proposed by you. The obtained results allowed us to strengthen our conclusion and claim that the existence of the nonanalyticity line of C-H is excluded except (perhaps) for the close vicinity of d=2, where our truncation is not sufficiently accurate.

We first briefly comment on the 'weaknesses' listed in your review. - The method implemented by us is indeed approximate, but accurate. When extended to the 4-th order (DE4) (which was achieved only last year in Refs. 21 and 22) it also provides ways of controlling the error bars. Note that in d=3 the DE4 yields estimates of the critical exponents of accuracy comparable to Monte Carlo. In addition, these predictions are compatible with the values from conformal bootstrap. The amended version of the paper gives a definite answer to the question motivating the study except for the close vicinity of d=2. - In the amended version of the manuscript we present results illustrating the evolution of the profiles of the functional fixed points when varying d and N across the crossover as well as a calculation of the irrelevant exponent. We are grateful for the suggestion concerning enriching the paper in these directions.

We now comment on the points from your report: 1. In the present version of the manuscript we provide an analysis of the irrelevant exponent lowering dimensionality starting from d=3. Down to d≈2.2, we find no indication of its vanishing at any value of N. Our truncation is not sufficient to properly resolve the limit d->2 concerning this respect. However, for a broad vicinity of d=3 it gives a fully reliable negative answer as to the possibility of a collision with another fixed point as anticipated by the C-H study. Recovering the limit d->2 is a topic of our ongoing research, which however must go beyond the present approximation level. The corresponding analysis is contained in the new Sec. 5.3. 2. Both the effective potential and the longitudinal stiffness exhibit interesting and rapid (however smooth) changes upon crossing the C-H (crossover) line. The corresponding analysis is contained in the new Sec. 6. 3. For all the practical purposes of the present paper one may treat the truncation of the derivative expansion as the only really relevant source of errors. We verified this by confronting the results obtained by the stability matrix analysis against those obtained from direct integration of the flow. We also extensively tested the dependencies of our results on the grid size, discretization, way of implementing the discrete derivatives, integration accuracy, etc. We comment on this in the new version of the manuscript (p.6, "We have extensively tested..."). 4. The algebraic phase in d=N=2 may be exactly recovered in our framework by freezing the longitudinal mode (see Ref.29). Inclusion of this mode is however necessary to capture the transition. The finite (even though enormously large) correlation length in the low-T phase is due to truncation (not errors of other sources). 5. Upon tuning the regulator in the way implemented by us, the KT singularity is well captured. In particular, this yields the value of \eta, the universal stiffness jump, and the essential singularity of the correlation length in the high-temperature phase. For a non-tuned ('arbitrary') cutoff, our framework yields the KT transition in the form of an extremely sharp crossover (See Refs 28,29). In the low-T phase, the correlation length is 'astronomically' large, and in a realistic experiment or simulation, the two pictures are not distinguishable. This behavior drastically changes when going away from d=N=2 for all the tested regulators. We commented on this in the amended version of the manuscript (see p. 8 "We also point out..."). 6. The numerical problems encountered for d close to 2 and N>2 are related to very large derivatives of the effective potential and the longitudinal stiffness z(\rho) as well as the presence of a pole of the propagator. We commented on this in the new version of the paper (Sec.6, two paragraphs beginning with "The analysis of the ..."). 7. The effort of going to order DE4 is tremendous and this was achieved only last year (Ref. 20). Our motivation for such a study in the present context is in fact quite limited. We are convinced that the picture obtained at order DE2 and DE4 would yield the same reliable conclusions when the close vicinity of d=2 is excluded. Note in particular that the values obtained for the irrelevant exponent (Fig. 10) at order DE2 and DE4 are close from the point of view of our requirements. On the other hand, we do not expect that the limit d->2^+ would be exactly resolved at order DE4 (although improvement may certainly be anticipated). We commented on this in the amended manuscript (see p.7 "An extension of this study to the fourth-order...") 8. We used a direct result from the "traditional" 2+epsilon expansion. We changed the citation, so to cite the original source (Ref 40). 9. We removed the sentence in question.

We would like to thank you once again for your helpful report. We are particularly grateful for your suggestions for amendments, which directed us in a way allowing for strengthening the major conclusion of the paper.

Sincerely yours, Andrzej Chlebicki Pawel Jakubczyk

### Weaknesses

Not sure the model in this paper has a chance to capture the Cardy-Hamber mechanism

### Report

Cardy and Hamber back in 1980 argued that vortices may affect the transition in the $O(n)$ model for $d>2$. These are the same vortices which cause the BKT transition for $n=2$, $d=2$. Cardy and Hamber argued that for $n>2$ and $d$ very close to 2 the vortices are irrelevant, so that the standard $2+\epsilon$ expansion a la Brezin and Zinn-Justin (which does not take vortices into account) is applicable, while for larger $d$ vortices become relevant and change the universality class of the transition. Along the line in the $(n,d)$ plane where vortices become relevant, the critical exponents should be continuous but not analytic.

One caveat of the Cardy-Hamber paper is that they do not present a first-principle derivation of their RG equations (Ref. [10] in Cardy-Hamber promises a derivation but I’m not sure it appeared in print.) So a better understanding of their mechanism, and of the difference in the transition in systems with allowed and suppressed vertices, would be welcome. I’m not aware of much work in this direction. One relevant paper is Ref. https://arxiv.org/abs/cond-mat/0311222 which discussed the phase transition in the O(3) model in $d=3$ with suppressed vortices. On the other hand, there was much related work about phase transitions in the 3d gauge theories coupled to matter, with monopoles playing the role of vortices. In this case it is known that depending on the number of the matter species, monopoles may be relevant or irrelevant. These latter models are sometimes called $CP^{N-1}$ models because of their microscopic description, and the above-mentioned $O(3)$ model with suppressed vortices is also called “non-compact $CP^{N-1}$”.

The paper under review studies the phase transition in the $O(n)$ model for various $d$ and $n$ using the FRG. They do not observe any non-analyticity line, and interpret this as evidence against Cardy-Hamber. I am not sure however I agree with this interpretation. Indeed, to test the Cardy-Hamber mechanism, one should to begin with have a model which allows for vortices. On the contrary, their model is a continuous scalar field theory with $O(n)$ symmetry. Since the field is continuous, the vortices are totally suppressed for any $n$ and $d$. Therefore, their model has nothing to do with the Cardy-Hamber mechanism, analyticity being totally expected given the built-in suppression of vortices. If my interpretation is correct, the paper might be misleading.

• validity: -
• significance: -
• originality: -
• clarity: -
• formatting: -
• grammar: -

Author Andrzej Chlebicki on 2021-04-01
(in reply to Report 1 by Slava Rychkov on 2021-01-17)
Category:


text:
# Latexmk

If you use cross-references, you often have to run LaTeX multiple times. If you use BibTeX for your bibliography or if you want to have a glossary you even need to run external programs in-between.

To avoid all this hassle, you should simply use Latexmk.

Latexmk is a Perl script which you just have to run once and it takes care of everything else for you.

Follow the instruction given below for a first test drive with Latexmk using the CLI. See the online reference of Latexmk for further information.

Replace FILE with the name of the main file. To build the template use myFile.tex in place of FILE

Note

The BFH

## CLI Usage

• In the simplest case you just have to type

$latexmk  This will run LaTeX on all .tex files in the current directory using the output format specified by the configuration files. • If you want to compile only one specific .tex file in the current directory, just provide the file name: $ latexmk FILE

• If you want to preview the resulting output file(s), just use

$latexmk -pv  • And now the Killer Feature: If you want Latexmk to continuously check all input files for changes and re-compile the whole thing if needed and always display the result, type $ latexmk -pvc

Then, whenever you change something in any of your source files and save your changes, the preview is automatically updated.

Warning

This doesn’t work with all viewers, especially not with Adobe Reader. See the section about configuration files below for setting a suitable viewer application.

• Of course, options can be combined. See the example given below.

$latexmk -outdir=_build -pv FILE  ## Cleaning Up • After running LaTeX, the current directory is contaminated with a myriad of temporary files; you can get rid of them with $ latexmk -c

• Previous command doesn’t delete the final .pdf/.ps/.dvi files. If you want to delete those too, use

$latexmk -C  ## Configuration File To customize some latexmk build setting such as the default viewer for the preview, edit the file .latexmkrc. A snippet with some good pre-configurations for latexmk is given below. Add the configuration file to the top-level directory in your documentation folder. Note The snippet below is in the so called heredoc format. On a GNU/Linux Shell it creates the .latexmkrc configuration file with the content between the first and last line of that snippet. If one is not able to use command line, one can copy the snippet without the first cat << "EOF" > .latexmkrc and the last EOF line, into a file called .latexmkrc cat << "EOF" > .latexmkrc ## ## Default PDF viewer -- system dependent setting ## See the documentation -- https://mirror.foobar.to/CTAN/support/latexmk/latexmk.pdf ## Windows #$pdf_previewer = 'start acroread %O %S';
## OS X
#$pdf_previewer = 'open %S'; ## GNU/Linux (Debian/Ubuntu like) #$pdf_previewer = 'start xdg-open %S';

##
## EPS to PDF conversion hook
##
@cus_dep_list = (@cus_dep_list, "eps pdf 0 eps2pdf");
sub eps2pdf {
system("epstopdf $_[0].eps"); } ## ## GLO to GLS conversion hook ## add_cus_dep( 'slo', 'sls', 0, 'makeglossaries' ); add_cus_dep( 'acn', 'acr', 0, 'makeglossaries' ); add_cus_dep( 'glo', 'gls', 0, 'makeglossaries' );$clean_ext .= "slo sls slg acr acn alg glo gls glg";
sub makeglossaries {
my ($base_name,$path) = fileparse( $_[0] ); pushd$path;
my $return = system "makeglossaries",$base_name;
popd;
return $return; } ## ## To enable shell-escape for all *latex commands ## Used i.e. for svg package invoking inkscape ## set_tex_cmds( '--shell-escape %O %S' ); ## ## Set default TeX file ## #@default_files = ('reportTitle.tex'); ## ## Latexmk build properties ##$pdf_mode = 4;
$postscript_mode =$dvi_mode = 0;
$bibtex_use = 1; ## ## Build directory ##$out_dir = '_build';

##
## Post process hooks (Linux, OS X only; For Windows install CLI tools)
##
## Copy PDF to a sub directory named "_output"
#$success_cmd = 'mkdir -p _output && cp _build/*.pdf _output/'; ## Copy PDF to a sub directory named "_output" and create a link from top level to the PDF file #$success_cmd = 'mkdir -p _output && cp _build/*.pdf _output/ && ln -s _output/%R.pdf';

##
## List of file extensions to clean up
##
$clean_ext .= '%R.aux %R.dvi %R.log %R.out tex~';$clean_full_ext = 'bbl synctex.gz _build';
EOF


text:
# Product metric

Product metric

In mathematics, the product metric is a definition of metric on the Cartesian product of two metric spaces.

Definition

Let $\left(X, d_\left\{X\right\}\right)$ and $\left(Y, d_\left\{Y\right\}\right)$ be metric spaces and let $1 leq p leq + infty$. Define the $p$-product metric $d_\left\{p\right\}$ on $X imes Y$ by

:$d_\left\{p\right\} left\left( \left(x_\left\{1\right\}, y_\left\{1\right\}\right) , \left(x_\left\{2\right\}, y_\left\{2\right\}\right) ight\right) := left\left( d_\left\{X\right\} \left(x_\left\{1\right\}, x_\left\{2\right\}\right)^\left\{p\right\} + d_\left\{Y\right\} \left(y_\left\{1\right\}, y_\left\{2\right\}\right)^\left\{p\right\} ight\right)^\left\{1/p\right\}$ for $1 leq p < infty;$

:$d_\left\{infty\right\} left\left( \left(x_\left\{1\right\}, y_\left\{1\right\}\right) , \left(x_\left\{2\right\}, y_\left\{2\right\}\right) ight\right) := max left\left\{ d_\left\{X\right\} \left(x_\left\{1\right\}, x_\left\{2\right\}\right), d_\left\{Y\right\} \left(y_\left\{1\right\}, y_\left\{2\right\}\right) ight\right\}.$

for $x_\left\{1\right\}, x_\left\{2\right\} in X$, $y_\left\{1\right\}, y_\left\{2\right\} in Y$.

Wikimedia Foundation. 2010.


text:
SMS scnews item created by Oded Yacobi at Thu 17 May 2018 0704
Type: Seminar
Distribution: World
Expiry: 24 May 2018
Calendar1: 23 May 2018 1600-1700
CalLoc1: Carslaw 830
CalTitle1: Hitchin System and Ramifications
Auth: oyacobi@n114-73-194-245.sbr1.nsw.optusnet.com.au (oyacobi) in SMS-WASM

# Seminar: Masoud Kamgarpour -- Hitchin system and ramifications

Masoud Kamgarpour (Queensland) is giving a representation theory seminar on Wednesday
4-5pm, in Carslaw 830.

Title: Hitchin System and ramifications

Abstract: The Hitchin system is a remarkable mathematical object sitting at the
crossroad between geometry, representation, number theory, and physics.  The first half
of the talk will be a general introduction to this area of mathematics.  In the second
half, I will discuss some recent advances in understanding what happens to the Hitchin
system if one allows for â€œramificationâ€ at finitely many points.

If you are registered you may mark the scnews item as read.
School members may try to .


text:
# e'ing a matrix and then finding eigenvalues and a ivp to converge to the origion

The question is as follow's consider the Matrix A

$$A = \begin{pmatrix} \lambda & 0 &0\\ 1 & \lambda&0 \\ 0&1& \lambda \end{pmatrix},$$

Compute $e^{tA}$, and use it to solver the initial value problem

$X^{'}$ = AX,

$X(0)$=$X_{0}$=$(x_{0},y_{0},z_{0})$

Question (ii)

Under what conditions on $\lambda$ and $X_{0}$ does the solution $X(t)$ of the initial value problem converges to the origin as t tends to $\infty$ Justify.

For part 1 i know what it says, it wants me to write $e^{tA}$ where A is actually my matrix i was sick for class and the notes i have are barely legible and as per usual my textbook doesn't have any worked out example's only a proof.

So if someone could show me how to set this up and/or point me in the direction of some dummy level examples with a nice explanation i would appreciate that.

For part 2 i think id like to construct a stable star node but any value where all the eigenvalues $\lambda <0$ should converge to (0,0,0) in infinite time ?

also why does the IVP matter in this case everything should move to (0,0,0) in eithier finite or infinite time if all the eigenvalues are negative?

-
To compute the powers of $A$, it may help to write $A=\lambda I_3 +N$ where $N$ is nilpotent and commutes, of course, with $\lambda I_3$. Once you have that, you'll get $\exp (tA)$. –  1015 Mar 4 '13 at 1:03
Whats nilpotent mean? –  Faust7 Mar 4 '13 at 1:12
$M$ is nilpotent $\Leftrightarrow \exists k\in\mathbb{N}, M^k=0_n$ –  xavierm02 Mar 4 '13 at 1:15
It makes it really easy to calculate the exponential because you have a finite number of terms to add since $\sum\limits_{k=0}^{+\infty}\cfrac{M^k}{k!}=\sum\limits_{k=0}^{n-1}\cfrac{M^k}{k!}+\sum\limits_{k=n}^{+\infty}\cfrac{M^k}{k!}=\sum\limits_{k=0}^{n-1}\cfrac{M^k}{k!}+\sum\limits_{k=n}^{+\infty}\cfrac{0_n}{k!}$$=\sum\limits_{k=0}^{n-1}\cfrac{M^k}{k!}$ –  xavierm02 Mar 4 '13 at 1:17
ah, ic and $0_{n}$ is the 0 matrix where n is the number of times we multiply it by itself to get the 0 matrix. –  Faust7 Mar 4 '13 at 1:19

$A = \begin{pmatrix} \lambda & 0 &0\\ 1 & \lambda&0 \\ 0&1& \lambda \end{pmatrix} = \lambda\begin{pmatrix} 1 & 0 &0\\ 0 & 1&0 \\ 0&0& 1 \end{pmatrix}+ \begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix} = \lambda I_3 + N$

$e^A = e^{\lambda I_3 + N}$

Since the two matrices commute, ie $(\lambda I_3) N = N(\lambda I_3)$

$e^A = e^{\lambda I_3}e^{N}$

$e^{\lambda I_3}=\sum\limits_{n=0}^{+\infty}\cfrac{\left(\lambda I_3\right)^n}{n!}=\sum\limits_{n=0}^{+\infty}\cfrac{\lambda^n I_3^n}{n!}=\sum\limits_{n=0}^{+\infty}\cfrac{\lambda^n I_3}{n!}=\left(\sum\limits_{n=0}^{+\infty}\cfrac{\lambda^n }{n!}\right)I_3 = e^\lambda I_3$

$N^2=\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix}\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix} = \begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 1&0& 0 \end{pmatrix}$

$N^3=N N^2 =\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix}\begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 1&0& 0 \end{pmatrix} = \begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 0&0& 0 \end{pmatrix} = 0_3$

$e^N = \sum\limits_{n=0}^{+\infty}\cfrac{\left(\lambda N\right)^n}{n!}=\sum\limits_{n=0}^{2}\cfrac{\left(\lambda N\right)^n}{n!}+\sum\limits_{n=3}^{+\infty}\cfrac{\left(\lambda N\right)^n}{n!}=\sum\limits_{n=0}^{2}\cfrac{\left(\lambda N\right)^n}{n!} = \cfrac{I_3}{0!}+\cfrac{N}{1!}+\cfrac{N^2}{2!}=\begin{pmatrix} 1 & 0 &0\\ 0 & 1&0 \\ 0&0& 1 \end{pmatrix}+\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix}+\cfrac{1}{2}\begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 1&0& 0 \end{pmatrix}=\cfrac{1}{2}\begin{pmatrix} 2 & 0 &0\\ 2 & 2&0 \\ 1&2& 2 \end{pmatrix}$

$e^A = e^{\lambda I_3}e^{N}=\cfrac{e^{\lambda}}{2}\begin{pmatrix} 2 & 0 &0\\ 2 & 2&0 \\ 1&2& 2 \end{pmatrix}$

$e^{tA}=e^{t\lambda I_3 + tN}=e^{t\lambda I_3}e^{tN}=\cfrac{e^{\lambda t}}{2}\begin{pmatrix} 2 & 0 &0\\ 2t & 2&0 \\ t^2&2t& 2 \end{pmatrix}$

$X(0)=X_0=\begin{pmatrix} x_0\\ y_0 \\ z_0 \end{pmatrix}$

$X'=AX\tag{E}$

The unique solution is $X(t)=e^{tA}X_0=\cfrac{e^{\lambda t}}{2}\begin{pmatrix} 2 & 0 &0\\ 2t & 2&0 \\ t^2&2t& 2 \end{pmatrix}\begin{pmatrix} x_0\\ y_0 \\ z_0 \end{pmatrix}=\cfrac{e^{\lambda t}}{2}\begin{pmatrix} 2x_0\\ 2tx_0+2y_0 \\ t^2 x_0 + 2t y_0 + 2z_0 \end{pmatrix}=\begin{pmatrix} e^{\lambda t}x_0\\ e^{\lambda t}\left(tx_0+y_0\right) \\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right) \end{pmatrix}$

We want $\begin{pmatrix} e^{\lambda t}x_0\\ e^{\lambda t}\left(tx_0+y_0\right) \\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right) \end{pmatrix}\underset{t\to +\infty}{\longrightarrow} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

Since the space is finite dimensional ($3 < +\infty$), convergence is the same thing as convergence coordinate wise.

So $\begin{pmatrix} e^{\lambda t}x_0\\ e^{\lambda t}\left(tx_0+y_0\right) \\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right) \end{pmatrix}\underset{t\to +\infty}{\longrightarrow} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\Leftrightarrow \left\{ \begin{array}{ll} e^{\lambda t}x_0 \underset{t\to +\infty}{\longrightarrow}0 \\ e^{\lambda t}\left(tx_0+y_0\right)\underset{t\to +\infty}{\longrightarrow}0\\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right)\underset{t\to +\infty}{\longrightarrow}0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{ll} \lambda < 0 \mbox{ or } x_0 = 0\\ \lambda < 0 \mbox{ or } x_0=y_0 = 0\\ \lambda < 0 \mbox{ or } x_0=y_0=z_0 = 0 \end{array} \right.\Leftrightarrow \lambda < 0 \mbox{ or } x_0=y_0=z_0 = 0$

-
Wow that's extremely straight forward... –  Faust7 Mar 4 '13 at 1:12
Well usually the tricky part is to decompose the matrix into a diagonal matrix (because you know that $e^{diag(a_1,\dots,a_n)}=diag(e^{a_1},\dots,e^{a_n})$) and a nilpotent matrix (because then you only have a finite number of terms to add). And in your case, the matrix was made to make this decomposition easy to find. –  xavierm02 Mar 4 '13 at 1:21
That doesn't surprise me the textbook talks about this idea for about, 3 pages. any hints for the IVP n why i can't just pick anything? –  Faust7 Mar 4 '13 at 1:26
@Faust7: I edited. –  xavierm02 Mar 4 '13 at 1:55


text:
# 12.6: Axiom III

Note that the first part of Axiom III follows directly from the definition of the h-angle measure defined on page . It remains to show that $$\measuredangle_h$$ satisfies the conditions Axiom IIIa, Axiom IIIb, and Axiom IIIc.

The following two claims say that $$\measuredangle_h$$ satisfies IIIa and IIIb.

Claim $$\PageIndex{1}$$

Given an h-half-line $$[O P)_h$$ and $$\alpha\in(-\pi,\pi]$$, there is a unique h-half-line $$[O Q)_h$$ such that $$\measuredangle_h P O Q= \alpha$$.

Claim $$\PageIndex{2}$$

For any h-points $$P$$, $$Q$$, and $$R$$ distinct from an h-point $$O$$, we have

$$\measuredangle_h P O Q+\measuredangle_h Q O R \equiv\measuredangle_h P O R.$$

Proof of $$\PageIndex{1}$$ and $$\PageIndex{2}$$

Applying the main observation, we may assume that $$O$$ is the center of the absolute. In this case, for any h-point $$P \ne O$$, the h-half-line $$[OP)_h$$ is the intersection of the Euclidean half-line $$[OP)$$ with h-plane. Hence the claim $$\PageIndex{1}$$ and Claim $$\PageIndex{2}$$ follow from the axioms Axiom IIIa and Axiom IIIb of the Euclidean plane.

Claim $$\PageIndex{3}$$

The function

$$\measuredangle_h\:(P,Q,R)\mapsto\measuredangle_h P Q R$$

is continuous at any triple of points $$(P,Q,R)$$ such that $$Q\ne P$$, $$Q\ne R$$, and $$\measuredangle_h P Q R\ne\pi$$.

Proof

Suppose that $$O$$ denotes the center of the absolute. We can assume that $$Q$$ is distinct from $$O$$.

Suppose that $$Z$$ denotes the inverse of $$Q$$ in the absolute; suppose that $$\Gamma$$ denotes the circle perpendicular to the absolute and centered at $$Z$$. According to Lemma 12.3.1, the point $$O$$ is the inverse of $$Q$$ in $$\Gamma$$.

Let $$P'$$ and $$R'$$ denote the inversions in $$\Gamma$$ of the points $$P$$ and $$R$$ respectively. Note that the point $$P'$$ is completely determined by the points $$Q$$ and $$P$$. Moreover, the map $$(Q,P)\mapsto P'$$ is continuous at any pair of points $$(Q,P)$$ such that $$Q\ne O$$. The same is true for the map $$(Q,R)\mapsto R'$$

According to the main observation

$$\measuredangle_h P Q R\equiv -\measuredangle_h P' O R'.$$

Since $$\measuredangle_h P' O R'=\measuredangle P' O R'$$ and the maps $$(Q,P)\mapsto P'$$, $$(Q,R)\mapsto R'$$ are continuous, the claim follows from the corresponding axiom of the Euclidean plane.


text:
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I need some new R skills, and there is no better motivation to learn XML in R than one of the scariest financial datasets out there—the US Department of the Treasury Office of the Comptroller of the Currency (OCC) Quarterly Derivatives Report. I’ll move to the even scarier global BIS derivatives data in a later post to avoid inducing nightmares and sleepless nights.

The massive damage caused in 2008-2009 by the sliver of derivatives called credit default swaps seems like a faint warning siren when we see that they only represent < 7% of total derivatives exposure.  Interest rate and currency derivatives, also where I think the next disaster occurs, are more than 10 times larger than these credit contracts at $226 Trillion. While some take comfort in “Bilateral Netting: A legally enforceable arrangement between a bank and a counterparty that creates a single legal obligation covering all included individual contracts. This means that a bank‟s receivable or payable, in the event of the default or insolvency of one of the parties, would be the net sum of all positive and negative fair values of contracts included in the bilateral netting arrangement.” bilateral netting does not work so well when your counterparty fails like Lehman Brothers.  From TimelyPortfolio I know there must be a better way to read Microsoft Excel XML spreadsheets in R, but these were my two most elegant methods. Please let me know how I could have done better. R code (click to download): #read xml derivatives data from the #US Treasury OCC Quarterly Derivatives Report #2 methods of as far as I can tell #the first published example of how to read #Microsoft Excel xml workbooks require(XML) require(ggplot2) url = "http://www.occ.treas.gov/topics/capital-markets/financial-markets/trading/derivatives/dq410-xml.xml" #url = "dq410-xml.xml" doc = xmlInternalTreeParse("dq410-xml.xml") #define namespaces #figuring this out took hours #but using getNodeSet was much cleaner than the #next method namespaces = c(o="urn:schemas-microsoft-com:office:office", x="urn:schemas-microsoft-com:office:excel", ss="urn:schemas-microsoft-com:office:spreadsheet", html="http://www.w3.org/TR/REC-html40") ns <- getNodeSet(doc,"/ss:Workbook/ss:Worksheet[@ss:Name='Table 3']/ss:Table/ss:Row",namespaces)[[44]] pct <- df <- as.data.frame(as.numeric(xmlSApply(ns, xmlValue))[5:8]) ns <- getNodeSet(doc,"/ss:Workbook/ss:Worksheet[@ss:Name='Table 3']/ss:Table/ss:Row",namespaces)[[8]] lab <- as.character(xmlSApply(ns, xmlValue)[8:11]) df <- cbind(lab,df) #jpeg(filename="derivatives by type.jpg",quality=100, # width=6.25, height = 8, units="in",res=96) barplot(df[,2],names.arg = df[,1],main="US Bank Derivatives by Type Q4 2010",ylab="% of Total") mtext("Source: US Dept of Treasury OCC Quarterly Derivatives Report", side=1,line=3,cex=0.8,adj=0) #dev.off() #here is the original very hacked brute force method doc = xmlRoot(xmlTreeParse(url)) #set xmlNodeList to Table 3 #hacked with brute force table3 <- doc[6]$Worksheet
df <- as.data.frame(cbind(
as.numeric(
xmlApply(table3[[1]],function(x) xmlApply(x,xmlValue))[53]$Row[5:8])), stringsAsFactors=FALSE) df <- cbind( as.character(xmlApply(table3[[1]],function(x) xmlApply(x,xmlValue))[17]$Row[c(8:11)]),
df)
colnames(df) <- c("DerivativeType","PctOfTotal")
barplot(df[,2],names.arg = df[,1],main="US Bank Derivatives by Type
Q4 2010",ylab="% of Total")
mtext("Source: US Dept of Treasury OCC Quarterly Derivatives Report",
side=1,line=3,cex=0.8,adj=0)


text:
# tf.contrib.layers.layer_norm

tf.contrib.layers.layer_norm(
inputs,
center=True,
scale=True,
activation_fn=None,
reuse=None,
variables_collections=None,
outputs_collections=None,
trainable=True,
begin_norm_axis=1,
begin_params_axis=-1,
scope=None
)

Adds a Layer Normalization layer.

Based on the paper:

"Layer Normalization"

Jimmy Lei Ba, Jamie Ryan Kiros, Geoffrey E. Hinton

https://arxiv.org/abs/1607.06450.

Can be used as a normalizer function for conv2d and fully_connected.

Given a tensor inputs of rank R, moments are calculated and normalization is performed over axes begin_norm_axis ... R - 1. Scaling and centering, if requested, is performed over axes begin_params_axis .. R - 1.

By default, begin_norm_axis = 1 and begin_params_axis = -1, meaning that normalization is performed over all but the first axis (the HWC if inputs is NHWC), while the beta and gamma trainable parameters are calculated for the rightmost axis (the C if inputs is NHWC). Scaling and recentering is performed via broadcast of the beta and gamma parameters with the normalized tensor.

The shapes of beta and gamma are inputs.shape[begin_params_axis:], and this part of the inputs' shape must be fully defined.

#### Args:

• inputs: A tensor having rank R. The normalization is performed over axes begin_norm_axis ... R - 1 and centering and scaling parameters are calculated over begin_params_axis ... R - 1.
• center: If True, add offset of beta to normalized tensor. If False, beta is ignored.
• scale: If True, multiply by gamma. If False, gamma is not used. When the next layer is linear (also e.g. nn.relu), this can be disabled since the scaling can be done by the next layer.
• activation_fn: Activation function, default set to None to skip it and maintain a linear activation.
• reuse: Whether or not the layer and its variables should be reused. To be able to reuse the layer scope must be given.
• variables_collections: Optional collections for the variables.
• outputs_collections: Collections to add the outputs.
• trainable: If True also add variables to the graph collection GraphKeys.TRAINABLE_VARIABLES (see tf.Variable).
• begin_norm_axis: The first normalization dimension: normalization will be performed along dimensions begin_norm_axis : rank(inputs)
• begin_params_axis: The first parameter (beta, gamma) dimension: scale and centering parameters will have dimensions begin_params_axis : rank(inputs) and will be broadcast with the normalized inputs accordingly.
• scope: Optional scope for variable_scope.

#### Returns:

A Tensor representing the output of the operation, having the same shape and dtype as inputs.

#### Raises:

• ValueError: If the rank of inputs is not known at graph build time, or if inputs.shape[begin_params_axis:] is not fully defined at graph build time.


text:
If you have ever watched a commercial airplane roll back from an airport gate, you may have noticed a crewmember on the ground giving signals to the pilot in the cockpit, many feet above. Have you ever thought that the angle of sight for the ground crewmember is the same angle of sight for the pilot? The pilot's viewing angle is called the angle of depression, while the ground crewmember's viewing angle is the angle of elevation.

## Angle of Depression

The Airbus A380 commercial airliner places the pilot's cockpit window an astounding $7.13$ meters above the ground. From the height of a two-story building, the pilot looks down to the airport tarmac and must be able to see airport crewmembers giveing visual signals.

If the pilot looks straight out, that line of sight is horizontal (focused on the distant horizon). The pilot might see the airport's runway, distant buildings, or even mountains in her field of view. If the pilot looks down at the ground, she will see the ground crew. The angle of depression is the angle that is formed between the horizontal and the downward looking angle. It is always a downward view, an angle below the horizon.

A line of sight is a straight line from an observer's eyes to the object being observed. You can look up, look straight ahead, or look down.
A horizontal line is a line viewed straight in front of you, with your eyes neither looking up or looking down. You are looking to the horizon.

## Angle of Elevation

For the ground crewmember, looking down would not be much help in communicating with the pilot in her cockpit seat. The crewmember will look up, above the horizontal line, to see the pilot. The ground crewmember's angle of vision if an angle of elevation, the angle above the horizon.

For both people's viewpoints -- the pilot's eyes in her cockpit seat $7.13$ meters (around 23') above the tarmac, and the ground crewmember's eyes roughly $1.7$ meters (about 5'-6") above the tarmac, both angles are the same. Look at this sketch:

Notice the angle of depression is the same as the angle of elevation. Since the horizontal lines are parallel, this is a case of alternate interior angles created by a transversal (the line of sight) cutting across two parallel lines (the horizontals).

## How to Find the Angle of Elevation & Depression

Animals and artists, pilots and pelicans all instinctively use the angle of depression. From a park ranger in a fire tower to that pilot taxiing or landing an A380 airliner, the angle of depression provides important information about distance and height.

Its opposite, the angle of elevation, is what humans do when they want to look up at tall mountains, nests in trees, or towering buildings.

### Formulas for Angle of Elevation & Depression

Recall the three trigonometric ratios (or identities), sine, cosine, and tangent. They each are the mathematical relationship between angles and sides and the hypotenuse of a right triangle:

If you know the tangent of an unknown angle (using the tangent formula), you can use the inverse of tangent, arctangent, to find the actual angle.

### Finding The Angles of Elevation & Depression

Let's look again at our pilot and ground crewmember. We can create a right triangle with the pilot at the top of the short side, the crewmember at the far end of the long side, and a hypotenuse (the line of sight) stretching between them. The long side is the ground crewmember's horizon line. The short side is the right angle straight up to the pilot.

The height from the crewmember's horizon line to the pilot's eyes is our opposite side, and the adjacent side is the distance from that side to the ground crewmember. We know the height of the opposite side, from basic subtraction of the pilot's and the ground crewmember's eye positions above the tarmac:

The ground crewmember stands far enough away from the aircraft to see the pilot and be seen by her. Let's call that 25 meters, the length of the triangle's long side. We know the opposite side and adjacent side measurements, so we can find the tangent of the angle of depression for the pilot and angle of elevation for the crewmember (they are the same number, remember).

Setting up to find the tangent allows us to find the inverse, the angle itself, using arctangent:

Inverse of tangent to find the angle:

So the pilot is having to look down only $12.25°$ below horizontal, and the ground crewmember is looking up only $12.25°$ above horizontal, to see each other.

If you know the angle of depression (or elevation), you can calculate distances (either height or length) using tangent.

If you know the heights and lengths of the two legs of the right triangle, you can calculate the angle of depression (or elevation) using the inverse of tangent, arctangent.

## Example Problems

You dropped an egg on the kitchen floor. You are 1.5 meters above the floor, and 3.75 meters from the egg. To see the egg, what is your angle of depression?

You are looking down the Grand Canyon at an angle of depression of $45°$ and see a lovely rock formation on the canyon floor, which is one mile straight down. How far away from you is the rock formation?

So, the formation is one mile away from the canyon wall.

## Lesson Summary

Now that you have studied this lesson, you are able to recall and define angle of depression and angle of elevation, relate the angle of depression to the angle of elevation, cite examples from everyday life of the angle of depression and the angle of elevation, and use tangent and arctangent to calculate an angle of depression or an angle of elevation.

## What you learned:

After working your way through this lesson and video, you will learn to:

• Recall and define angle of depression and angle of elevation
• Relate the angle of depression to the angle of elevation
• Cite examples from everyday life of the angle of depression and the angle of elevation
• Use tangent and arctangent (tangent's inverse) to calculate an angle of depression or an angle of elevation
Instructor: Malcolm M.
Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

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text:
thatgirl666 2 years ago Solve? 4/5y+2=6

1. ash2326

$\frac 4 5 y+2=6$ Subtract both sides by 2, what would you get? $\frac 4 5 y +2- 2= 6-2$

2. thatgirl666

4/5 y= 4?

3. ash2326

Yes, good :D Now multiply both sides by 5

4. thatgirl666

how do i multiply a fraction by 5?

5. ash2326

$\frac 4 5 y= 4$ $(\frac 4 5 \times 5 ) y= 4\times 5$ $(\frac 4 {\cancel 5} \times \cancel 5 ) y= 4\times 5$ $4y=4\times 5$ Do you get this?

6. ash2326

@thatgirl666 ????

7. thatgirl666

Not really so the answer is 20?

8. thatgirl666

Nevermind. i then divide by 4 and it equals 5

9. ash2326

Yeah, let me explain you the multiplication of fractions

10. thatgirl666

Ok

11. ash2326

Suppose you have the fraction $\frac a b$ You need to multiply by a no. say c $\frac ab \times c=\frac{a\times b}c$ If you had to multiply by 1/c $\frac ab \times \frac 1 c=\frac a {b\times c}$ here $a \ne b \ne c$ Do you get this?

12. thatgirl666

Yes i do now Thank you

13. ash2326

If you multiply a/b by b, what would you get?

14. ash2326

$\frac a b \times b= ???$

15. thatgirl666

c

16. ash2326

We don't even have c in the expression $\frac 2 4 \times 4=???$

17. thatgirl666

*Question Closed*

18. ash2326

It's a simple thing, you should learn it


text:
# Find the Derivative – d/dx f(x) = natural log of e^(9x)+4

Differentiate using the chain rule, which states that is where and .
To apply the Chain Rule, set as .
The derivative of with respect to is .
Replace all occurrences of with .
By the Sum Rule, the derivative of with respect to is .
Differentiate using the chain rule, which states that is where and .
To apply the Chain Rule, set as .
Differentiate using the Exponential Rule which states that is where =.
Replace all occurrences of with .
Differentiate.
Since is constant with respect to , the derivative of with respect to is .
Move to the left of .
Differentiate using the Power Rule which states that is where .
Multiply by .
Since is constant with respect to , the derivative of with respect to is .
Combine fractions.
Combine and .
Combine and .
Find the Derivative – d/dx f(x) = natural log of e^(9x)+4

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