url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
https://www.snapxam.com/problems/47405623/integral-of-1-x-1-0-x-2-dx-from-2-to-5
|
# Step-by-step Solution
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2
e
π
ln
log
log
lim
d/dx
Dx
|◻|
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
## Step-by-step explanation
Problem to solve:
$\int_{2}^{5}\frac{1}{\left(x-1\right)\left(x+2\right)}dx$
Learn how to solve definite integrals problems step by step online.
$\frac{1}{\left(x-1\right)\left(x+2\right)}=\frac{A}{x-1}+\frac{B}{x+2}$
Learn how to solve definite integrals problems step by step online. Integrate 1/((x-1)(x+2)) from 2 to 5. Rewrite the fraction \frac{1}{\left(x-1\right)\left(x+2\right)} in 2 simpler fractions using partial fraction decomposition. Find the values for the unknown coefficients: A, B. The first step is to multiply both sides of the equation from the previous step by \left(x-1\right)\left(x+2\right). Multiplying polynomials. Simplifying.
$\frac{62}{225}$$\,\,\left(\approx 0.2755595243948227\right)$
$\int_{2}^{5}\frac{1}{\left(x-1\right)\left(x+2\right)}dx$
|
2020-10-24 20:03:37
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9392554759979248, "perplexity": 1268.23480119844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107884755.46/warc/CC-MAIN-20201024194049-20201024224049-00275.warc.gz"}
|
http://mathhelpforum.com/advanced-applied-math/193352-determining-step-size-using-golden-ratio.html
|
# Thread: Determining step size using golden ratio
1. ## Determining step size using golden ratio
I would like to calculate the minimum of the function $g(x,y) = (y-x^2)^2+(1.1 -x)^2$ using the steepest descent method. I am, however, stuck at determining the step size. I want to calculate the step size using golden ratio, however, as I have only use the method in 1 dimension, I don't know how to use it here. Here is what I did so far:
Step 1: Choose starting points. These were given already as $(0.5, 0.5)$
Step 2: Calculate the gradient $\nabla$. I did this and got $x (2-4y)+4x^3-2.2, 2 (y-x^2)$. Substituting for the starting points $0.5, 0.5$, we get $-\nabla = 1.7, -0.5$
Step 3: Calculate $g(x+ \alpha -\nabla)$.
I am stuck at step 3. I don't know how I can get $\alpha$. If there is a possibility to use the golden ratio to determine alpha here, I would welcome. Any help in explaining it is highly appreciated. Thanks in advance!
2. ## Re: Determining step size using golden ratio
If you know what to do in 1-dimension, it's easy to apply the algorithm in 2-dimensions. Just considers the 1-d function :
$f(\alpha) = g(x - \alpha \nabla_x, y - \alpha \nabla_y)$
And apply the method on $f(\alpha)$.
|
2013-12-18 12:46:26
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7531289458274841, "perplexity": 241.433849186607}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345758566/warc/CC-MAIN-20131218054918-00022-ip-10-33-133-15.ec2.internal.warc.gz"}
|
http://ndp.jct.ac.il/tutorials/infitut1/node123.html
|
# Taylor series.
Definition 10.5.1 Let be a function having derivatives of any order at . The power series
is called the Taylor series of the function at .
If , this series is called the Mac-Laurin series of .
Example 10.5.2 (A couple of MacLaurin series)
• , for .
Theorem 10.5.3 Let and be two functions having Taylor series expansions and respectively, converging on the interval . Then has as its Taylor series expansion the series and it converges on .
Theorem 10.5.4 Let and be two functions having Taylor series expansions and respectively, converging on the interval . Then has as its Taylor series expansion the product of these series and it converges on .
Example 10.5.5 On the one side we have:
On the other side, we have:
Thus, we have:
|
2017-05-23 07:04:56
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9964730143547058, "perplexity": 889.4943271433701}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607591.53/warc/CC-MAIN-20170523064440-20170523084440-00287.warc.gz"}
|
http://theagora.info/imgs/h/h-x-y.html
|
# GALLERY: H X Y
The joint Shannon entropy (in bits ) of two discrete random variables X {\displaystyle X} and Y {\displaystyle Y} with images X {\displaystyle {\mathcal {X}}} and Y {\displaystyle {\mathcal {Y}}} is defined as [2] : 16
|
2019-01-20 09:58:53
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999321460723877, "perplexity": 6629.176459171883}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583705091.62/warc/CC-MAIN-20190120082608-20190120104608-00469.warc.gz"}
|
https://math.meta.stackexchange.com/questions/6362/reopened-question/6364
|
# Reopened question
What was the rationale for reopening this question this question? (The initial comments refer to the link that is struck out. But the subject of my query is the other link)
My understanding was that this question was:
• voted closed,
• voted deleted,
• undeleted at the request of the OP
• voted closed in reaction to the apparent disregard for the previous votes,
• reopened, with no visible reason given.
A recent meta discussion leads me to believe the final action was not done in good faith. But before I form opinions on the incident, I would like to get my facts straight.
Has there been a vote to re-open? Has there somehow been a (subsequently deleted) discussion on the topic of re-opening it during the fairly brief course of events?
Or has action truly been taken with utter disregard for the previous votes? Do I have the wrong post and this was never actually voted closed / deleted? (was it this then? Or another one entirely?) Is there some other case I haven't thought of?
• The (first struck-out) post was closed as "nonconstructive", presumably for the many nonconstructive comments discussing downvotes. These were purged, so there no longer remains anything nonconstructive. What remains is an interesting number-theoretical question, which, if answered, would add great value to the site. – Bill Dubuque Oct 16 '12 at 4:37
• @Bill: That is a flat out lie. The original voter cited the discussion at meta.math.stackexchange.com/questions/6354/… as rationale and I concurred. (Unless I have somehow thoroughly mixed up two posts) (P.S. sorry for accidentally deleting my first draft after you commented) – user14972 Oct 16 '12 at 4:41
• Which conveniently excludes the comment history you deleted. – user14972 Oct 16 '12 at 4:59
• Dear Hurkyl, since I don't have enough reputation, I can't see deleted comments, so I don't know what could have been said in the deleted comments, but to be honest, I don't see anything wrong with the question at all. In fact, I'm quite shocked that all this fuss has being raised about Bill reopening the question. To me it seems like a well posed question and actually, I would instead ask, what was the rationale for closing the question and downvoting it in the first place? – Adrián Barquero Oct 16 '12 at 5:04
• @Hurkyl Please get the facts straight. There were no votes to delete. The post was deleted automatically by the SE software because it met some criteria devised by the SE designers to automatically attempt to infer which posts are so worthless that they should be automatically deleted without any human intervention. That failed quite badly here. – Bill Dubuque Oct 16 '12 at 5:12
• @BillDubuque, while many such questions «haven't even been closed, let alone deleted», the one you undeleted was. This post is about that, precisely: your overriding other users votes to close and to delete. The contrast to your «Let the community make decisions vs. use superpowers» at math.stackexchange.com/election/3 is probably part of the discussion, too, btw. – Mariano Suárez-Álvarez Oct 16 '12 at 5:39
• Are you kidding me? You wrote in a comment above « I undeleted a few posts that I thought were automatically deleted by the SE software. I had missed the fact that one of them was actually deleted by 20k users» Honestly, I find this absolutely pointless. – Mariano Suárez-Álvarez Oct 16 '12 at 6:00
• @Bill: You undeleted a question that was deleted by three users; and you did reopened a post that was closed by five users. You alone. If you think something done by a jury of members is incorrect, you should post a meta thread calling for a reconsideration. Matt E could have voiced his opinion, and the posts were likely to be undeleted, reopened, etc. I guess that all users are equally important here, but some users are more equal than others. – Asaf Karagila Oct 16 '12 at 6:02
• Oh well. Do as you see fit, Bill, as apparently you are right, as usual. For the record, your unilateral use of your superpowers to override other users' votes on various instances has my fervent opposition. I walk out of this discussion, and let other users figure out if they are happy with this situation. – Mariano Suárez-Álvarez Oct 16 '12 at 6:09
• @AsafKaragila You can always write to team@stackexchange.com, the SE community team is the correct address if you have a complaint about moderator actions. – Mad Scientist Oct 16 '12 at 6:55
• I cannot do anything about it, except make it clear that I oppose certain actions. – Mariano Suárez-Álvarez Oct 16 '12 at 6:55
• @AsafKaragila: I certainly agree with you, but I think it's unfortunate that we're in a situation where one moderator's opinion over-rules all the other moderators. – Noah Snyder Oct 16 '12 at 14:46
• @NoahSnyder, well, i was implicitly saying that there is nothing I can sensibly do. I am not going to start doing stupid things to cope with Bill's actions. – Mariano Suárez-Álvarez Oct 16 '12 at 17:00
• And also the three votes to delete; and then another five votes to close a question so its re-deletion could be re-considered. – Asaf Karagila Oct 16 '12 at 20:17
• Your new mantra about confusion is not helping... :-/ – Mariano Suárez-Álvarez Oct 20 '12 at 4:01
Let me give a few points in the history of this meta thread:
1. MK asks three questions; two had a negative sum of votes and automatically deleted, another was closed by five non-moderator users, and then deleted by another three.
2. MK asks "Why was my question deleted?" and Mariano points out that it was an automatic process.
3. Bill takes upon himself to undo the software's action, he undeleted the originally discussed question.
4. MK thanks Bill, and asks for another automatically deleted question to be restored as well.
5. Bill complies. Bill also restores, without being asked to, the third question.
6. When asked whether or not he believes that overriding five closure votes and another three is an act reasonable by a moderator who gives the impression that the community should manage itself, Bill insists that automatic deletion is wrong to begin with.
7. Bill refuses to continue the discussion, Hurkyl gives what I believe a good reason for the restored questions to be closed, and pend their deletion.
8. The first question reopened by Bill, again to the contrary of five members.
What is unclear to me is why a moderator who repeatedly claim that moderators should not act based on their personal opinion, acts on his personal opinion, and repeatedly denies users of their vote. I will not get into a discussion why the question was reopened or undeleted, and whether or not it is right. I just wanted to point out the irony of these actions.
• As you are well aware, there is a long history of wrongful downvoting and closure of MK's questions. The matters prompting such behavior have only recently been solved. These questions were on the tail end of such resolution. It is better to let these matters rest. I have nothing more to say on the matter other than what I've already said here and in the parallel meta thread. – Bill Dubuque Oct 16 '12 at 5:45
• Let us please avoid drama-charged terms like «wrongful»... This is not an Austen novel. – Mariano Suárez-Álvarez Oct 16 '12 at 5:46
• @Mariano Truth $\ne$ drama. – Bill Dubuque Oct 16 '12 at 5:47
• @Bill: If that had been given as a rationale, the discussion would have gone quite differently (and possibly more quietly)! – user14972 Oct 16 '12 at 5:48
• @Bill: You only provided one reason for these actions, you find them right. Not the community, not three or fives members. One member. I don't recall this community being ran by you. You are only a moderator. – Asaf Karagila Oct 16 '12 at 5:49
• @Bill: One man's truth is another's propaganda. While there may be some kernel of truth in what you assert, there is still some dramatic, even accusatory connotations with the word "wrongful". But you've long demonstrated your firm belief that anyone whose value judgment is not the same as your own is conveniently "wrong", so I wouldn't expect any better of you. So what would you say to a younger and wiser Bill Dubuque who disavowed any and all unilateral moderator action overriding community decisions? – Willie Wong Oct 16 '12 at 6:32
• @Willie To set the record straight, I never "disavowed any and all unilateral actions". Rather, I argued against a longstanding pattern of unilateral closures and deletions that were frequently made by certain moderators (both past and current), often without community input and often disregarding the charter of MSE. Thankfully we are past most of that. And we are almost past all of the problems that sparked this thread. Hopefully we can move on and forgive those who made mistakes, since I don't believe anyone had bad intentions. – Bill Dubuque Oct 16 '12 at 6:41
• @Bill: You ran on a platform of Let the community make decisions vs. use superpowers., that is a fairly clear message to me. I'm not a native speaker though, so maybe I'm not qualified to read this sentence... If you want to be forgiven for your actions, you can set things right: Delete the question again and ask on meta for it to be reopened by the community. – Michael Greinecker Oct 16 '12 at 8:34
• Dear Asaf, One thing I am confused by: do you think that the number theory questions of MK should stay deleted, because an automatic software process deleted them? (I can appreciate that this might be a reasonable person's position, although I don't share it; but I'm not sure if it's actually your position or not. One point of confusion for me is that the question this post was discussing seemed to change over time, and so it's not always obvious what people are actually objecting to.) Regards, – Matt E Oct 16 '12 at 17:32
• @Matt: No. I think that not every question should be automatically undeleted. There are enough 10k users now that Bill should have posted a link in an answer, and asking members to actively undelete the post[s]. I think that Hurkyl's reasons given in the MK thread comments were excellent. The question was deleted, which is enough not to initiate automatic undeletion upon request. When Bill did that, and especially his unwillingness to admit a mistake [when I pointed out the third undeletion was unfair] made things far far worse. – Asaf Karagila Oct 16 '12 at 17:48
• @Asaf It may be worth discussing a general policy on undeletion of automatically deleted posts. It is conceivable that this could happen again, and having a policy in place would help to avoid a repeat of what happened here. – Bill Dubuque Oct 16 '12 at 18:45
• @Bill: Yes, a general policy would be helpful. But for someone who wishes to let the community govern itself you seem awfully bureaucratic with all these policies. – Asaf Karagila Oct 16 '12 at 18:51
• The problem with policies is that they enshrine what the current community thinks (well, I say current community - what I mean is the 10-20 members who will read the policy and vote on it) without regard to the future community. Is there also going to be a policy of regularly reviewing old policies to ensure that the current set of policies are up to date? (I am being ironic, but only about 50% ironic). – Chris Taylor Oct 17 '12 at 23:03
• @Chris: There is actually a policy about being ironic, and it states that you clearly have to be at least 57% ironic. (I am being ironic, or rather ridiculous, 100% :-)) – Asaf Karagila Oct 17 '12 at 23:05
I don't know the particular open/close history of the question, but I can't see any reason why it should be closed (and it is strange that it should have a negative vote total). It is a perfectly good question in number theory (in fact a rather famous one!), and I would certainly vote to reopen it if it were closed.
Added: It seems that the question of MK that my answer responds to (and which was the one originally linked in this post) is in fact not the one that the post intended to refer to. This is just one reflection of the confusion that seems to be surrounding this issue.
Since I am adding to this answer, let me add a further remark: as far as I can tell, there are three MK questions involved in this discussion, of which two were substantial number theory questions and of which the third (which this post was intended to refer to) was about notation. The third was closed as non-constructive, quite possibly with solid justification. It seems to have been undeleted by error. On the other hand, I agree with Bill that the two number theory questions are reasonable, and should not have been deleted. I understand that they were deleted for reasons related to the automatic functioning of the software, but this seems to me to be akin to a case when an OP's question or answer becomes CW because of multiple edits. Certainly there have been cases when that CW status has been removed by a moderator because they deem it to be inappropriate, despite the software's automatic process; similarly, in this case it would seem to be silly to have solid number theory questions deleted just because of an automatic process.
• I not really concerned about the actual status of the question, whether deleted or opened, but about the process by which it got there. – user14972 Oct 16 '12 at 5:05
• The two SE-automatic deletions that I reversed at the OP's request are this one and this one. I would be quite surprised if anyone thinks that these questions deserve to be automatically deleted by the SE heuristics on worthless questions, without any human intervention whatsoever. – Bill Dubuque Oct 16 '12 at 7:05
• Dear Bill: there was human intervention. The software only deletes questions which have negative vote count (and which have not been answered) – Mariano Suárez-Álvarez Oct 16 '12 at 21:26
• No, there was no human intervention on the two questions that I linked to. They were deleted automatically by the SE software. – Bill Dubuque Oct 16 '12 at 22:00
• ...because they had negative vote totals, which was the 'human intervention' that Mariano was referring to. – Chris Taylor Oct 16 '12 at 22:41
• @Chris But those votes were downvotes, not votes to delete. None of our users made an explicit decision to delete either of those questions. That decision was made by heuristics in some algorithm devised by SE. I agree wholeheartedly with Matt E. that there is no reason that either of those questions should have been closed (let alone deleted), and that it is strange that they had negative vote totals. I'm happy to see that the first post has since received some upvotes. The second post still needs an upvote to protect it from the SE grim reaper. – Bill Dubuque Oct 17 '12 at 20:47
• @Bill I realize that they were downvotes, not votes to delete. I was explaining a comment of Mariano's that you seem to have misunderstood. – Chris Taylor Oct 17 '12 at 22:58
• @Chris I understood Mariano's comment. But, as I explained, I don't think it is pertinent. – Bill Dubuque Oct 18 '12 at 0:55
Since Bill hasn't seen fit to answer this question, I will.
Two of Makoto Kato's questions were automatically deleted by the system. MK requested undeletion, and Bill complied. While doing so, he also undeleted this question, incorrectly believing that it had also been automatically deleted. While Bill feels personally that such questions should not be deleted, this particular undeletion was clearly an oversight, and not an attempt to override the wishes of the community.
Given that it was closed (correctly) as Not Constructive, and deleted after a sufficient time for improvement had passed without any edits or efforts to re-open it being made, I've re-deleted it.
Moderators are human. They make mistakes. Fortunately, those mistakes can be corrected without ridiculous amounts of drama.
• As far as I can see, MK did not ask for the questions to be undeleted. – Mariano Suárez-Álvarez Oct 16 '12 at 16:55
• @Mariano: no; he identified two other questions (which were also undeleted, but which had been automatically deleted and had not been closed). So far as I can discern, this particular question was incorrectly identified and incorrectly undeleted. It is regrettable that this mistake was not corrected sooner. – Shog9 Oct 16 '12 at 17:01
• @Shog: As I wrote to Matt in a comment to my answer, what exacerbated the drama was the fact that Bill were adamant that he did not make any mistakes, and that he is correct. Telling users that he is unwilling to discuss anything further, and then deleting the comments and reopening a closed question, seems to me as acts of tyranny. – Asaf Karagila Oct 16 '12 at 17:51
• @Asaf (1) I've already mentioned a couple times that the 3rd question was mistakenly believed to also have been automatically deleted. That was simply an oversight. I did not immediately reverse that since then most readers would not have been able to read and discuss it here (and see that it was not reopened). (2) The reason that I did not intend to address it any further here is due to the fact that this thread is so riddled with misinformation that it seems hopeless to try to get it steered back on course. Instead, I had planned on opening a new meta thread to address these matters. – Bill Dubuque Oct 16 '12 at 18:29
• @Bill: The facts are that I pointed that out several times on the comments to MK's meta thread. You kept insisting otherwise and declared that you will not discuss your actions unless we have a "reasonable" reason why the question should be deleted again. This is not how a moderator should behave. It reminds me a lot more of the times I would argue with my commander during my army service and he ended up pulling rank and telling me "I just said so, I don't care what you think. Do it my way.", rather than a conversation with a moderator claiming to be open to suggestions from users. – Asaf Karagila Oct 16 '12 at 18:33
• @Asaf I did not immediately realize which question you were referring to (nor did the OP here). Originally I thought you were referring to one of the other two automatically deleted questions. Only later did I realize what happened with the 3rd question. Both threads are permeated by misunderstandings throughout. That is why I thought it would be wise to open a new thread. – Bill Dubuque Oct 16 '12 at 18:37
• @Bill: Yes, there has been some confusion. Undeleting on your own accord, with the only reason "I believe this is a good question" is another problem, which Hurkyl addressed. The fact that you reopened a closed question is yet another. Once the question was closed the community could have decided whether or not it wants it reopened or deleted. Your action makes it impossible to let the community take its course. What is interesting is that you rather declare that you essentially remove yourself from discussion than taking a look at what you have done. – Asaf Karagila Oct 16 '12 at 18:49
• @Shog9 Do questions that are automatically deleted behave any differently from questions deleted by 20K users or moderators? In particular, (how) can the community undelete such automatic deletions? Iirc, content deleted by moderators cannot be undeleted by the community, correct? Knowing such is useful when discussing policy on such matters. – Bill Dubuque Oct 16 '12 at 18:56
• @Bill: trusted users should be able to undelete automatically-deleted posts. Only moderator-deleted posts are protected from this (this includes spam- or offensive-deleted posts, which are additionally locked to prevent ANY modification). For the record, moderators are able to undelete any deleted post, an ability they are trusted to use judiciously. – Shog9 Oct 16 '12 at 19:09
• @Shog9 Thanks, that's helpful. Readers may recall that trusted users are those users with 20K+ rep, and they have the ability to vote to delete posts with a score of -1 or lower (it takes 3 such votes to delete a post). We currently have 48 trusted users (not all of which are active on meta). – Bill Dubuque Oct 16 '12 at 19:22
• @Bill: There are eight [non-moderator] trusted users on the front page of the meta users list ordered by participation. I am certain that there are several others which are avid readers, but don't participate often. – Asaf Karagila Oct 16 '12 at 19:32
• @Shog9: I’ve no problem with the accidental error; it’s Bill’s evasive responses to Mariano and Asaf that I find distressing. – Brian M. Scott Oct 17 '12 at 12:31
• @Bell: I did read all of the comments, here and in this thread as well. I think that you’re being disingenuous. – Brian M. Scott Oct 17 '12 at 21:07
• @Bill: I am not confused. I made no false inferences. And if you’re not being deliberately disingenuous $-$ and I’m quite prepared to believe that you’re not, just as I believe that you actually do have the best interests of MSE at heart $-$ then you have a somewhat distorted view of your own words in these threads. – Brian M. Scott Oct 17 '12 at 21:19
• Okay, @Bill, let's run a discussion by your rules. When might we see this meta thread you speak of for "address(ing) both the specific and general issues", that is "free of all the misunderstandings here"? – J. M. is a poor mathematician Oct 18 '12 at 8:02
It seems quite clear to me that what Bill actually objects to is closing questions, not the means by which they're closed (community, moderator, etc.).
• This could not be further from the truth. Again, please refrain from speaking for others. Moreover, the issue here is not closure but, rather, the much more extreme action of deletion. – Bill Dubuque Oct 16 '12 at 14:38
• @Bill Noah wrote "It seems [...]", so don't worry that he is speaking for you. – Michael Greinecker Oct 16 '12 at 16:31
I believe the actual subject of my question is this question which, while undeleted, was not actually reopened.
• Since that's the question that you intend to refer to, in order to prevent this thread from further spiralling out of control due to further confusion, I struck out the final two bullet items in your question, since they do not apply to this question. The only recent action on that question was that it was undeleted (because I mistakenly thought that it was one of the questions automatically deleted by SE). – Bill Dubuque Oct 16 '12 at 6:55
• That's fair. I still haven't decided exactly how I wanted to modify my question since I want to consider both the question I originally linked and the one I intended to refer to, and that's as good as any for now. – user14972 Oct 16 '12 at 18:37
|
2019-11-13 22:12:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4988727569580078, "perplexity": 1331.140587175453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496667442.36/warc/CC-MAIN-20191113215021-20191114003021-00435.warc.gz"}
|
https://brilliant.org/discussions/thread/just-think-about-it-its-urgent/
|
Can 2 raise to the power something be equal to 0? If yes then at what value?
Note by Prabhav Bansal
3 years, 1 month ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
Can we say 0.0000000000000000000000000000000000000000000001 equal to zero.
- 3 years, 1 month ago
This would only approach 1. Any negative power would be less than 1.
So technically, $$2^{-\infty} = 0$$. But then again, $$-\infty$$ isn't a number.
- 3 years, 1 month ago
But still if we say that 2^-100000000000 it is zero.
- 3 years, 1 month ago
No, it's still not absolute zero.
- 3 years, 1 month ago
So, finally we conclude that anything raise to the power any integer is not zero.
- 3 years, 1 month ago
0 raised to the power of any positive real is 0.
You need to qualify what "anything" refers to.
Staff - 3 years, 1 month ago
No, that's the law of logarithm.
- 3 years, 1 month ago
No.
- 3 years, 1 month ago
No. Because 0 is not a number.
- 3 years, 1 month ago
Because 0 is not a number.
Are you sure? xP
I'm guessing you mean infinity?
- 3 years, 1 month ago
|
2018-10-18 23:35:12
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9979054927825928, "perplexity": 9783.815546637969}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512015.74/warc/CC-MAIN-20181018214747-20181019000247-00436.warc.gz"}
|
https://www.ncatlab.org/nlab/show/Vincent+Braunack-Mayer
|
# nLab Vincent Braunack-Mayer
(formerly V. Schlegel)
## Selected writings
On dg-algebraic models for rational parameterized stable homotopy theory:
• Combinatorial parametrised spectra,
On rational parameterized stable homotopy theory applied to the mathematical analysis of the duality between M-theory and type IIA string theory:
category: people
Last revised on April 29, 2021 at 06:30:43. See the history of this page for a list of all contributions to it.
|
2021-06-20 04:32:31
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2621985971927643, "perplexity": 2469.7211174066233}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00617.warc.gz"}
|
http://pub.acta.hu/acta/showCustomerArticle.action?id=2103&dataObjectType=article&returnAction=showCustomerVolume&sessionDataSetId=5fb2b7db720de100&style=
|
ACTA issues
## Continuity theorems and ratio-Tauberian theorems for integral transforms
Abstract. We consider continuity theorems for multiplicative integral transforms $\hat{f}$ of functions $f\colon(0, \infty ) \to{\msbm R},$ i.e., we relate the limit behaviours $f_n \to f$ with $\hat{f_n} \to\hat {f}.$ These continuity theorems based on sequential compactness arguments lead to Abelian-- and ratio--Tauberian theorems for these integral transforms. AMS Subject Classification (1991): 44A05, 40E05 Keyword(s): Integral transform, continuity theorems, Abelian theorems, Tauberian theorems, dominated variation, regular variation Received January 29, 1999. (Registered under 2707/2009.)
|
2023-03-29 10:16:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9538500308990479, "perplexity": 3807.6723050155015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00138.warc.gz"}
|
https://electronics.stackexchange.com/posts/46855/revisions
|
Post Locked by Community
Post Migrated Away to dsp.stackexchange.com by Kortuk
Post Closed as "off topic" by Kortuk
4 A better explaining title
# FFT of audio, 2^8 samplingspacket sampling and others
Hi I am just toying around with compression now that we are learning about fourierFourier transforms in my class, compression (compression is not a part of the class but I became a bit curious). Trying to do this with a wav file and I have some questions.
Assume I don't know the sampling rate of the signal (.wav) how can I figure out what the fftFFT frequency range is in Hz? I know that the bins are multiples of the samplingratesampling rate (Fs$$\F_s\$$) divided by N, or they are factors of Fs, eg 0.1*Fs1*$$\F_s\$$, 0.2*Fs2*$$\F_s\$$ etc... but how does this help me find the sampling frequency in Hz, so I can actually see which frequencies are doing what.?
Is there any clever trick in storing the coefficients of a real signal (wav)? They are always complex unless the signal is a pure cosine but what audio is? If I want to store the full range of coefficients I need at least as much storage as the .wav file (exploiting symmetry at the midpoint and having 2 units (real, imag) for every real input). This is of particular importance, if the coefficients are small they can be stored with less bits than the bits needed to represent the .wav datapoints.
Since for the case of functions you can split them in subintervalls and take the fourierFourier transform and depending on the result get some easier or more compact expression I was wondering if for the sake of compression I could split the audio-signal?
If I have a .wav file with N datapoints then split into ten N/10 segments and then fft on each segment. But this then means that my sampling frequency is essentially reduced to N/10 Fs$$\F_s\$$? This seems intuitive, if I "zoom" into a rapidly osciallitaing function then in that window it will look very slow and appear to have a low frequency.
Thanks for any input!
Hi I am just toying around with compression now that we are learning about fourier transforms in my class, compression is not a part of the class but I became a bit curious. Trying to do this with a wav file and I have some questions.
Assume I don't know the sampling rate of the signal (.wav) how can I figure out what the fft frequency range is in Hz? I know that the bins are multiples of the samplingrate (Fs) divided by N, or they are factors of Fs, eg 0.1*Fs, 0.2*Fs etc but how does this help me find the sampling frequency in Hz so I can actually see which frequencies are doing what.
Is there any clever trick in storing the coefficients of a real signal (wav)? They are always complex unless the signal is a pure cosine but what audio is? If I want to store the full range of coefficients I need at least as much storage as the .wav file (exploiting symmetry at the midpoint and having 2 units (real, imag) for every real input). This is of particular importance, if the coefficients are small they can be stored with less bits than the bits needed to represent the .wav datapoints.
Since for the case of functions you can split them in subintervalls and take the fourier transform and depending on the result get some easier or more compact expression I was wondering if for the sake of compression I could split the audio-signal? If I have a .wav file with N datapoints then split into ten N/10 segments and then fft on each segment. But this then means that my sampling frequency is essentially reduced to N/10 Fs? This seems intuitive, if I "zoom" into a rapidly osciallitaing function then in that window it will look very slow and appear to have a low frequency.
Thanks for any input!
Hi I am just toying around with compression now that we are learning about Fourier transforms in my class (compression is not a part of the class but I became a bit curious). Trying to do this with a wav file and I have some questions.
Assume I don't know the sampling rate of the signal (.wav) how can I figure out what the FFT frequency range is in Hz? I know that the bins are multiples of the sampling rate ($$\F_s\$$) divided by N, or they are factors of Fs, eg 0.1*$$\F_s\$$, 0.2*$$\F_s\$$ etc... but how does this help me find the sampling frequency in Hz, so I can actually see which frequencies are doing what?
Is there any clever trick in storing the coefficients of a real signal (wav)? They are always complex unless the signal is a pure cosine but what audio is? If I want to store the full range of coefficients I need at least as much storage as the .wav file (exploiting symmetry at the midpoint and having 2 units (real, imag) for every real input). This is of particular importance, if the coefficients are small they can be stored with less bits than the bits needed to represent the .wav datapoints.
Since for the case of functions you can split them in subintervalls and take the Fourier transform and depending on the result get some easier or more compact expression I was wondering if for the sake of compression I could split the audio-signal?
If I have a .wav file with N datapoints then split into ten N/10 segments and then fft on each segment. But this then means that my sampling frequency is essentially reduced to N/10 $$\F_s\$$? This seems intuitive, if I "zoom" into a rapidly osciallitaing function then in that window it will look very slow and appear to have a low frequency.
2 added 153 characters in body
Hi I am just toying around with compression now that we are learning about fourier transforms in my class, compression is not a part of the class but I became a bit curious. Trying to do this with a wav file and I have some questions.
Assume I don't know the sampling rate of the signal (.wav) how can I figure out what the fft frequency range is in Hz? I know that the bins are multiples of the samplingrate (Fs) divided by N, or they are factors of Fs, eg 0.1*Fs, 0.2*Fs etc but how does this help me find the sampling frequency in Hz so I can actually see which frequencies are doing what.
Is there any clever trick in storing the coefficients of a real signal (wav)? They are always complex unless the signal is a pure cosine but what audio is? If I want to store the full range of coefficients I need at least as much storage as the .wav file (exploiting symmetry at the midpoint and having 2 units (real, imag) for every real input). This is of particular importance, if the coefficients are small they can be stored with less bits than the bits needed to represent the .wav datapoints.
Since for the case of functions you can split them in subintervalls and take the fourier transform and depending on the result get some easier or more compact expression I was wondering if for the sake of compression I could split the audio-signal? If I have a .wav file with N datapoints then split into ten N/10 segments and then fft on each segment. But this then means that my sampling frequency is essentially reduced to N/10 Fs? This seems intuitive, if I "zoom" into a rapidly osciallitaing function then in that window it will look very slow and appear to have a low frequency.
Thanks for any input!
Hi I am just toying around with compression now that we are learning about fourier transforms in my class, compression is not a part of the class but I became a bit curious. Trying to do this with a wav file and I have some questions.
Assume I don't know the sampling rate of the signal (.wav) how can I figure out what the fft frequency range is in Hz? I know that the bins are multiples of the samplingrate (Fs) divided by N, or they are factors of Fs, eg 0.1*Fs, 0.2*Fs etc but how does this help me find the sampling frequency in Hz so I can actually see which frequencies are doing what.
Is there any clever trick in storing the coefficients of a real signal (wav)? They are always complex unless the signal is a pure cosine but what audio is? If I want to store the full range of coefficients I need at least as much storage as the .wav file (exploiting symmetry at the midpoint and having 2 units (real, imag) for every real input).
Since for the case of functions you can split them in subintervalls and take the fourier transform and depending on the result get some easier or more compact expression I was wondering if for the sake of compression I could split the audio-signal? If I have a .wav file with N datapoints then split into ten N/10 segments and then fft on each segment. But this then means that my sampling frequency is essentially reduced to N/10 Fs? This seems intuitive, if I "zoom" into a rapidly osciallitaing function then in that window it will look very slow and appear to have a low frequency.
Thanks for any input!
Hi I am just toying around with compression now that we are learning about fourier transforms in my class, compression is not a part of the class but I became a bit curious. Trying to do this with a wav file and I have some questions.
Assume I don't know the sampling rate of the signal (.wav) how can I figure out what the fft frequency range is in Hz? I know that the bins are multiples of the samplingrate (Fs) divided by N, or they are factors of Fs, eg 0.1*Fs, 0.2*Fs etc but how does this help me find the sampling frequency in Hz so I can actually see which frequencies are doing what.
Is there any clever trick in storing the coefficients of a real signal (wav)? They are always complex unless the signal is a pure cosine but what audio is? If I want to store the full range of coefficients I need at least as much storage as the .wav file (exploiting symmetry at the midpoint and having 2 units (real, imag) for every real input). This is of particular importance, if the coefficients are small they can be stored with less bits than the bits needed to represent the .wav datapoints.
Since for the case of functions you can split them in subintervalls and take the fourier transform and depending on the result get some easier or more compact expression I was wondering if for the sake of compression I could split the audio-signal? If I have a .wav file with N datapoints then split into ten N/10 segments and then fft on each segment. But this then means that my sampling frequency is essentially reduced to N/10 Fs? This seems intuitive, if I "zoom" into a rapidly osciallitaing function then in that window it will look very slow and appear to have a low frequency.
Thanks for any input!
1
|
2019-08-24 22:23:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7398927211761475, "perplexity": 570.701461859203}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321786.95/warc/CC-MAIN-20190824214845-20190825000845-00107.warc.gz"}
|
https://mattermodeling.stackexchange.com/questions/174/what-are-the-situations-problems-where-born-oppenheimer-approximation-is-invalid
|
# What are the situations/problems where Born-Oppenheimer approximation is invalid?
We use the Born-Oppenheimer approximation in both Hartree-Fock method and DFT. What are the problems where we cannot use this approximations.
## 5 Answers
Adding to the answer given by Cody Aldaz, there are many situations in chemistry when the Born-Oppenheimer approximation (BOA) breaks down, conical intersection is just one of them! In fact, the electronic states do not necessarily have to cross/intersect each other for BOA to be invalid. A classic example of this is given by what is famously known as "avoided crossing". The general term for phenomena when the BOA breaks down is referred to as "non-adiabatic processes". The entire field of photochemistry is filled with examples of non-adiabatic processes. Some examples include internal conversion and singlet fission. The former is even responsible for our vision!
From a quantum chemistry point of view, this is very common when one computes excited states (in order to model the above mentioned photochemical processes). You mentioned about Hartree-Fock and DFT. Well, they are single-reference methods to compute ground state energies. For excited state problems, when BOA is no longer valid, one will have to resort to multi-reference methods such as MRCI or EOM-CC. In addition to computing the excitation energies, these methods then allow to compute the "coupling" between different electronic states, precisely the terms that have been neglected in the derivation of BOA.
In short, most excited state problems such as vibronic spectroscopy, charge transfer, photochemistry, singlet fission, and many others will require going beyond Born-Oppenheimer approximation!
• By EOM-CC, do you mean MR-EOM? – Nike Dattani May 11 '20 at 21:56
## TL;DR conical intersections, and polarons. Or any other case when the velocity of the nuclei is faster than the electrons can respond nearly instantaneous
The long answer requires a lot of mathematics!
## The Mathematical derivation
The total nuclear and electronic Hamiltonian can be written as
$$\hat{H} = \hat{T}_N + \hat{T}_e +\hat{V}_{ee} + \hat{V}_{NN} + \hat{V}_{eN}$$
$$(\hat{T}_N + \hat{T}_e +\hat{V}_{ee} + \hat{V}_{NN} + \hat{V}_{eN})\Psi = E\Psi$$
where N are the nuclei and e are the electrons.
The BO approximation starts by asking what happens if we we freeze the nuclei at position $$\mathbf{R}_0$$ and take $$\Phi(\mathbf{r})\chi(\mathbf{R}_0)$$ as an ansatz of the total $$\Psi$$
This is mathematically much easier to solve because nuclear and electronic wavefunctions can be solved seperately. As we will see below, this is a good approximation in most situations.
First separating the electronic component we have
$$(\hat{T}_e +\hat{V}_{ee} + \hat{V}_{NN} + \hat{V}_{eN})\Phi_k(\mathbf{r};\mathbf{R}_0) = E_{el,k}\Phi_k(\mathbf{r};\mathbf{R}_0)$$
Combining this with the total Hamiltonian and the ansatz
$$[\hat{T}_{N} + E_{el,k}]\Phi_k(\mathbf{r};\mathbf{R}_0)\chi(R) = E\Phi_k(\mathbf{r};\mathbf{R}_0)\chi(R)$$
Rearranging,
$$\frac{\hat{T}_{N}\Phi_k(\mathbf{r};\mathbf{R}_0)\chi(R)}{\Phi_k(\mathbf{r};\mathbf{R}_0)} +E_{el,k}\chi(R) = E\chi(R)$$
Now here is the technical BO approximation::
$$\hat{T}_{N}\Phi_k(\mathbf{r};\mathbf{R}_0)\chi(R)\approx \Phi_k(\mathbf{r};\mathbf{R}_0)\hat{T}_{N}\chi(R)$$
Essentially this says that the kinetic energy of the nuclei doesn't affect the electronic states. Therefore, the nuclear wavefunction can be solved separately
$$[\hat{T}_{N} + E_{el,k}]\chi(R) = E\chi(R)$$
Thus under the BO approx., $$\chi(\mathbf{R})$$ Is solved by solving a nuclear Schrodinger equation where the potential is given by $$E_{el}$$
## But why is the B.O. approximation justified?
The nuclear kinetic energy operators are derivatives with respect to the nuclei $$\hat{T}_{N}\Phi_k(\mathbf{r};\mathbf{R}_0)\chi(R)=-\sum_{l=1}^{N_a}\frac{1}{2M_l}\nabla_l^2[\Phi_k(\mathbf{r};\mathbf{R}_0)\chi(R)]$$
Now doing some painful differentiation this equals
$$-\sum_{l=1}^{N_a}\frac{1}{2M_l} \Phi_k(\mathbf{r};\mathbf{R}_0) \nabla_l^2\chi(\mathbf{r}) - \sum_{l=1}^{N_a}\frac{1}{2M_l}\chi(\mathbf{r})\nabla_l^2\Phi_k(\mathbf{r};\mathbf{R}_0) - \sum_{l=1}^{N_a}\frac{1}{M_l}[\nabla_l\chi(\mathbf{R})][\nabla_l \Phi_k(\mathbf{r};\mathbf{R}_0) ]$$
The BO approx. amounts to assuming that the last two terms on the right hand side are negligible because
$$\frac{1}{M_l}[\nabla_l\chi(\mathbf{R})][\nabla_l \Phi_k(\mathbf{r};\mathbf{R}_0)]\approx \frac{1}{2M_l}\chi(\mathbf{R})\nabla_l^2 \Phi_k(\mathbf{r};\mathbf{R}_0) ≪ \frac{1}{2M_e}\chi(\mathbf{R})\nabla_e^2 \Phi_k(\mathbf{r};\mathbf{R}_0)$$
which is in turn because $$\nabla_r \Phi_k (r;R)≈∇_e Φ_k (r;R)$$ and $$m_e≪M_l$$
In other words, because the mass of an electron is much less than the mass of a nuclei it moves slower and the nuclei can be considered stationary with respect to the motion of the electron. Whew!
However, this reveals something VERY interesting. As we can see, it tells us that we can expect breakdown of the BO approximation any time the nuclear derivative of the electronic wavefunction changes rapidly.
For example in a conical intersection!!!
As you can see the electronic wavefunction changes rapidly with nuclear coordinates and it actually has a discontinuity.
Polarons are another example which have breakdown of the BO approximation and that's simply because the nuclei and electron are so far apart that the electrons cannot respond fast enough to the nuclei.
In addition to the classic examples of where non-adiabatic effects are important, the Born-Oppenheimer approximation cannot be taken for granted in the electronic structure computations of small chemical systems, where high-accuracy is needed (e.g. isotope dependence of molecular properties, high-resolution rovibrational spectroscopy, and quantum nuclear dynamics).
In some cases, a (relatively) easy fix is possible - DBOC (diagonal Born-Oppenheimer correction), which is to take into account first-order correction only. Otherwise, non-adiabatic effects need to be taken into account (e.g. here is such a calculation for water).
The Born-Oppenheimer approximation comprises two different approximations:
1. Adiabatic separation of electron and nuclear coordinates
2. Semi-classical approximation for nuclei
Previous answers have already addressed (1). I would also like to add that although it neglects electron-phonon coupling, this can be added back in using density functional perturbation theory (DFPT).
The semi-classical approximation (2) can also cause some issues, particularly for light nuclei where quantum delocalisation, zero-point energy etc. are an issue. These quantum effects can be added back, for example using path-integral dynamics (esp. suitable for bosons), DFPT or vibrational SCF, or the light nuclei can be treated in an explicit quantum mechanical treatment (e.g. the double adiabatic separation). As well as systems with light nuclei, there are also experimental techniques where these issues are vital, e.g. muon spectroscopy, where the muons can be treated as very low-mass nuclei.
The first answer is most relevant to materials modeling, but I also want to chip in that Born-Oppenheimer is more than just a kinetic assumption, it also assumes nuclei are point charges. This is especially important for actinides and super heavy elements where nuclei become not only much larger relative to the electron cloud but also more ovoid with unequal charge distribution in space. This can play a role in chemical properties related to spin effects as the dipole moment of the nuclei are much different than what Born-Oppenheimer would tell you.
• Is it really the "Born-Oppenheimer" approximation that assumes that the nuclei are point charges? Or would that be the "point-nucleus" approximation? Is it really incompatible to have a finite-sized nucleus and still be Born-Oppenheimer, as long as the wavefunction is a product of the nuclear and electronic wavefunctions? – Nike Dattani May 4 '20 at 2:25
• My understanding is that the Born-Oppenheimer approximation is that the point-nucleus is a part of it. I am by no means an expert. – Raz Jugovic May 4 '20 at 18:01
|
2021-06-24 12:45:08
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.810215175151825, "perplexity": 774.6246977827992}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00136.warc.gz"}
|
https://ask.bc-pf.org/t/bubble-connected-by-straws/5938
|
# Bubble connected by straws
Please provide a solution. I am not able to understand the problem. What does final state mean? @Alisher @Miras @Ersultan
2 лайка
It is known that the internal energy of a bubble equals U=\sigma A, where A is the surface area. In fact, this is only an approximate relation which assumes that \sigma doesn’t depend on temperature T, and, in general, this relation holds in terms of Helmholtz free energy F\equiv U-TS=\sigma A. Now, using the 1st law of thermodynamics for a bubble \delta Q = dU -\sigma dA, you may prove that
U=\left(\sigma-T\frac{d\sigma}{dT}\right)A.
6 лайков
Anyone wants to help me…
oh thanks, let me read about it… and for the 2nd and 3rd parts… how will we go around solving it…? Entropy change is confusing especially, never saw it for bubbles
1 лайк
“A thousand-mile journey begins with the first step.”
6 лайков
ok well… in all seriousness, what does Final state mean in problem2? what answer are you getting for problem 2?
Honestly, I have no clue how to solve that. I supposed they meant that under perturbations the temperature of the system will change, thereby changing the radii of the bubbles. However, after writing equations for energy and pressure, I got an equation with the only solution R=R_0. Probably after perturbations system will be asymmetrical, but there won’t be enough equations to find the final state in this case.
3 лайка
oh yeah exactly, I am struggling on this for around 2 days now… still no progress. The answers given are:
1. Proof
2. R_f \approx 2^{1 / 3} R_0+\frac{4 \sigma}{3 p_0}\left(2^{1 / 3}-1\right)
P_f=P_0+\frac{4 \sigma}{2^{1 / 3} R_0}
1. \Delta S = \frac{8 \pi R_0^2}{T}\left(1-2^{-1 / 3}\right)\left(\frac{4}{3} \sigma-b T\right)
3 лайка
@Ersultan @Alisher Any progress sir?
|
2023-02-09 05:23:55
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9328434467315674, "perplexity": 1622.8315317808826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00164.warc.gz"}
|
https://www.physicsforums.com/threads/help-please-difference-with-clarification.571555/
|
# Help please! difference with clarification
1. Jan 28, 2012
### dpa
what is difference between
B.what are Flatness and horizon problems with standard cosmology model tried to overcome by string theory model.
2. Jan 30, 2012
### bapowell
A. The Hubble radius is the distance at which objects recede from us at light speed. This comes from Hubble's Law: $v = Hr$: when $r=r_H=c/H$, the recession velocity is c. I'm not sure what you mean by spatial radius. Perhaps you are referring to the curvature scale, also called the radius of curvature. This is simply the inverse of the Gaussian curvature, K, of the spatial geometry: $R \sim 1/K$. The curvature scale is infinite for flat geometries.
B. The horizon and flatness problems are well known and probably best looked up on Wikipedia or similar. String theory does not directly address these problems. The inflationary universe was developed to solve these (and other) problems.
|
2018-06-25 10:43:13
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4638688266277313, "perplexity": 618.0750805863734}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867644.88/warc/CC-MAIN-20180625092128-20180625112128-00321.warc.gz"}
|
http://en.mimi.hu/astronomy/number_density.html
|
Home »
# Number Density
Number density: The distribution with equivalent with at fixed redshift is close to a power law in column density.
Number Density
Number of particles per cm3 (cf. column density).
Numbers (Dirac): ...
The Number density is the number of a particular object or species per unit volume and might be used when describing the number of electrons per cubic centimetre in a plasma or the number of stars in the core of a globular cluster.
The number density $N_{\rm n}$ of atoms in the $n$th electronic energy level is given by the Saha equation, a generalization of the Boltzmann equation. For a derivation of the Saha equation, see Rybicki & Lightman Eq. 9.47).
n_s is the number density of species s (s = e,Z for electrons, ion species Z), T_e is the electron temperature in energy units, and symbols not defined here are physical constants. Note that the third bracketed factor on the right-hand side determines its units.
where is the number density and is the lifetime.
References
Harwit, M. New York: Springer-Verlag, p. 308, 1988.
CHARGE CARRIER NUMBER DENSITY (n) - Number density (number per volume) ofcharge carriersin a material. In conductors, electrons are free to move, so n for a conductor equals the number density of conduction electrons in the material.
where ni is the ion number density and mi is the ion mass.
Relativistic case
The general Alfvén wave velocity is defined by [1] ...
Number density has units cm-3 (particles per cubic centimeter). Differential Rotation The change in solar rotation rate with latitude. Low latitudes rotate at a faster angular rate (approx. 14 degrees per day) than do high latitudes (approx. 12 degrees per day).
Number density has units cm-3 (particles per cubic centimeter).
Electromagnetic Radiation Radiation that travels through vacuous space at the speed of light and propagates by the interplay of oscillating electric and magnetic fields. This radiation has a wavelength and a frequency.
Where is the "contact surface" (the cometary ionosphere boundary) and what are the number density and chemical composition of the ions in the cometary ionosphere?
If the neutrinos are not massless, then they could have a larger mass density now consisting of their number density times their rest mass. Each neutrino species has a number density of n = (3/4)(4\pi)\Gamma(3)\zeta(3) (kTn/hc)3 where \Gamma is the gamma function (\Gamma(n+1) = n! ...
An image illustrating the number density of galaxies estimated to be four billion light years from the Earth. Bright areas indicate high-density regions. The brightest region in the centre corresponds to the main body of the CL0939 cluster.
Where N(v) is the number density of particles with velocities in the range v to v + dv, N is the total number density of the particles, m is the particle mass, T the temperature and k is boltzmann's constant.
(b) Within the confines of the HR diagram and/or the corresponding Color-Magnitude diagram it is also possible to give representation to the number density of stars at any particular luminosity and color/spectral-type by the use of contours or grey scale plotting.
On the basis of the numbers of cosmic rays hitting Earth, astronomers estimate the number density of cosmic rays in interstellar space to
be on the order of 10-3 particle/m3. That makes them rare, although still more numerous than interstellar dust particles.
Outside the galaxy, in intergalactic space, the number density of particles is thought to fall off to about one atom or molecule per cubic meter (10-6 / cm3).
Motions Within the Solar System ...
The amount of mass or number of particles per unit volume. In cgs units mass density has units of gm cm-3. Number density has units cm-3 (particles per cubic centimeter).
Disk
The visible surface of the Sun (or any heavenly body) projected against the sky.
This is that old PERVNERT that we all learned when we were in Chemistry in High School. Pressure times Volume is equal to the number density times the constant times the temperature.
Fraser: I never learned that.
Pamela: You didn't?
Variations in brightness are due to a combination of differences in ring particle number density and light scattering properties. The B-ring is the most opaque of Saturn's rings. The Cassini division is located at the upper left.
You can have a million-degree plasma that wouldn't have the power to boil a teaspoon of water, if the number density is low enough. Even if the CME is large, the density is really small.
This effect is called pressure broadening - the greater the atmospheric density and pressure, the greater the broadening of the spectral lines. So, by careful analysis of the star's spectrum, an astronomer can determine the number density of atoms of a particular element in the star's ...
|
2015-03-31 11:21:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.754170835018158, "perplexity": 893.8404617018725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131300472.97/warc/CC-MAIN-20150323172140-00044-ip-10-168-14-71.ec2.internal.warc.gz"}
|
http://math.stackexchange.com/questions/201615/on-the-existence-of-a-universal-field-without-algebraic-elements
|
# On the existence of a “universal” field without algebraic elements
Let $\mathfrak{M}$ be an infinite cardinal. Consider all fields $F$ which have the following properties:
(1) $F$ contains $\mathbb{Q}$.
(2) $F$ has cardinality $\leqslant \mathfrak{M}$.
(3) All elements of $F \setminus \mathbb{Q}$ are transcendental over $\mathbb{Q}$.
(Such a field need not be a purely transcendental extension of $\mathbb{Q}$.)
Does there exist a field that satisfies (1)-(3) and contains an isomorphic copy of any field which has properties (1)-(3)?
-
what about taking $\mathfrak M$ number of transcendentals $(\xi_i)_{i<\mathfrak M}$ over $\mathbb Q$, and consider the algebraic closure of $\mathbb Q(\xi_i)_i$?
Edit: Instead of the algebraic closure, consider only (all the roots of) all irreducible polynomials that has at least one transcendental over $\mathbb Q$ among its coefficients..
This contains the entire algebraic closure of $\mathbb{Q}$. – Chris Eagle Sep 24 '12 at 11:33
I assume you mean irreducible monics, since otherwise $\xi x^2-2\xi$ is such an irreducible with $\sqrt{2}$ as a root. In that case, what you describe isn't even a field: it contains $\xi$ (root of $x-\xi$) and $\xi+\sqrt{2}$ (root of $x^2-2\xi x+\xi^2-2$) but not $\sqrt{2}$. – Chris Eagle Sep 24 '12 at 11:49
|
2016-07-26 12:34:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9492368698120117, "perplexity": 271.6474609571056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00056-ip-10-185-27-174.ec2.internal.warc.gz"}
|
https://www.clutchprep.com/physics/practice-problems/146672/a-proton-with-an-initial-speed-of-800000-m-s-is-brought-to-rest-by-an-electric-f-1
|
Relationships Between Force, Field, Energy, Potential Video Lessons
Concept
# Problem: A proton with an initial speed of 800000 m/s is brought to rest by an electric field.Part A. Did the proton move into a region of higher potential or lower potential?Part B. What was the potential difference that stopped the proton?ΔV = ________ VPart C - What was the initial kinetic energy of the proton, in electron volts?Ki =_________ eV
###### FREE Expert Solution
We are required to identify the effects of change in potential on the motion of a positively charged particle and perform calculations involving energy.
Law of conservation of energy:
$\overline{){{\mathbf{K}}}_{{\mathbf{0}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{0}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$, where Wnc is the work done by non-conservative forces such as friction.
Kinetic energy:
$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}}$
86% (500 ratings)
###### Problem Details
A proton with an initial speed of 800000 m/s is brought to rest by an electric field.
Part A. Did the proton move into a region of higher potential or lower potential?
Part B. What was the potential difference that stopped the proton?
ΔV = ________ V
Part C - What was the initial kinetic energy of the proton, in electron volts?
Ki =_________ eV
|
2020-12-02 18:50:26
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.25434890389442444, "perplexity": 1338.1300569549226}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141715252.96/warc/CC-MAIN-20201202175113-20201202205113-00039.warc.gz"}
|
https://zbmath.org/?q=an:1111.60020
|
## Sectorial local non-determinism and the geometry of the Brownian sheet.(English)Zbl 1111.60020
Summary: We prove the following results about the images and multiple points of an $$N$$-parameter, $$d$$-dimensional Brownian sheet $$B=\{B(t)\}_{t\in \mathbb{R}_+^N}$$:
(1) If $$\dim_HF\leq d/2$$, then $$B(F)$$ is almost surely a Salem set.
(2) If $$N\leq d/2$$, then with probability one $\dim_HB(F) =2\dim_HF\quad\text{for all Borel sets }F\subset\mathbb{R}_+^N,$ where “$$\dim_H$$” could be everywhere replaced by the “Hausdorff”, “packing”, “upper Minkowski”, or “lower Minkowski dimension”.
### MSC:
60G15 Gaussian processes 28A80 Fractals 60G17 Sample path properties 82B41 Random walks, random surfaces, lattice animals, etc. in equilibrium statistical mechanics
Full Text:
|
2022-09-26 22:21:13
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.553017795085907, "perplexity": 2861.492740970351}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334942.88/warc/CC-MAIN-20220926211042-20220927001042-00257.warc.gz"}
|
https://www.physicsforums.com/threads/thermodynamic-question-first-law.806466/
|
# Thermodynamic question: first law
Tags:
1. Apr 3, 2015
### Superhawkkodaka
Estimate the kilowatt pump you would need to pump 450 L/min of liquid water at 20c through a rise in elevation of 100m. Assume the water is pumped adiabatically and with negligible changes changes in temperature and kinetic energy.
answer given to us is -7.35kw..
Question.. what formula should i use?.. im lost in this one.. please be specific guys.. i've used all i know and still cant get the answer..
2. Apr 3, 2015
### CWatters
Hints:
How much energy would it take to raise 1L (=1kg) up by 100m?
What is the equation that relates power, energy and time?
3. Apr 3, 2015
### Staff: Mentor
You are supposed to use the open system steady flow version of the first law:
$mΔh+mgΔz=Q-W_s$
where Ws is the rate of doing shaft work, m is the mass flow rate, and h is the specific enthalpy.
Chet
4. Apr 3, 2015
### Superhawkkodaka
weew thanks!!!
|
2017-11-22 17:41:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3080671727657318, "perplexity": 3372.8120750511894}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806615.74/warc/CC-MAIN-20171122160645-20171122180645-00362.warc.gz"}
|
http://math.stackexchange.com/questions/356713/how-can-i-determine-the-values-of-a-third-point-to-make-all-3-collinear-in-mat/356718
|
# How can I determine the values of a third point to make all 3 collinear in $\mathbb{R}^3$
I am given the points $(1,2,3), (2,4,1)$ and $(x,y,2)$ and I am being asked to find some values for $x$ and $y$ such that these 3 points will be collinear in $\mathbb{R}^3$
Most of this topic is covered with videos and tutorials in $\mathbb{R}^2$ and my teacher did not cover this topic at all, and it is not discussed in the book.
So far I have found out that the points are said to be collinear when the two shorter distances summed equals the longer distance between two points.
I started with the equation $\sqrt{(x-1)^2 + (y-2)^2 + 1} + \sqrt{(5-x)^2 + (3-y)^2 + 1} = \sqrt{21}$
But with two variables and one equation this does not appear to be the correct approach. I also feel it is incorrect as the $\sqrt{21}$ comes from the distance between the two known points, but not necessarily the longest distance as either the first or second point may be the point in-between.
I tried writing a short program which is checking for points between the two known points, but it is not returning any results and I would rather not expand the program to find the solution as this technique will not be available on a test of course. This implies that one of the two known points is somewhere in the middle.
-
If you know determinants, there is a characterization of collinearity by the vanishing of the dterminant of all coordinates. See here, next to the end, for the $\mathbb{R}^3$ case. – 1015 Apr 10 '13 at 4:03
A simpler approach is to remember the parametric equation for a line in $\Bbb R^3$. It is (some point)+t(direction vector). The line through $(1,2,3)$ and $(2,4,1)$ is then $(1,2,3)+t(2-1,4-2,1-3)=(1,2,3)+t(1,2,-2)$. Now you need to find $x$ and $y$ such that $(x,y,2)$ is of this form. The third coordinate will let you find $t \ldots$
Added: you are correct with $t=1/2$. You missed adding in $(1,2,3)$, so the point is $(1,2,3)+(1/2,1,-1)=(3/2,3,2)$ and you can read off $x$ and $y$. The idea is that you describe a line by adding a point on the line to a vector of any length in the direction the line points. As a simple example, the line through $(1,0,0)$ and parallel to the $y$ axis is all the points $(1,t,0)$. The above would express it as $(1,0,0)+t(0,1,0)$
I think this is a good answer, but I am lacking so much basic knowledge that it is hard to understand exactly what I am doing. My instinct tells me based on what you say that $3 + t(-2) = 2$ so $t = 1/2$ and $(1/2, 1, -1)$ satisfies the equation? – Leonardo Apr 10 '13 at 4:18
|
2015-11-26 16:20:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6251871585845947, "perplexity": 93.18926519710595}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398447758.91/warc/CC-MAIN-20151124205407-00057-ip-10-71-132-137.ec2.internal.warc.gz"}
|
https://www.genealogy.math.ndsu.nodak.edu/id.php?id=207531
|
## John Bowman
Dissertation: FINITE-DIMENSIONAL MODULES FOR THE QUANTUM AFFINE ALGEBRA $U_q(\mathfrak{g})$ AND ITS BOREL SUBALGEBRA
Mathematics Subject Classification: 17—Nonassociative rings and algebras
No students known.
If you have additional information or corrections regarding this mathematician, please use the update form. To submit students of this mathematician, please use the new data form, noting this mathematician's MGP ID of 207531 for the advisor ID.
|
2019-07-21 00:57:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2326076179742813, "perplexity": 2799.38628650365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00448.warc.gz"}
|
http://www.lmfdb.org/GaloisGroup/12T272
|
# Properties
Label 12T272 Order $$7920$$ n $$12$$ Cyclic No Abelian No Solvable No Primitive Yes $p$-group No Group: $M_{11}$
# Related objects
## Group action invariants
Degree $n$ : $12$ Transitive number $t$ : $272$ Group : $M_{11}$ CHM label : $M_{11}(12)$ Parity: $1$ Primitive: Yes Nilpotency class: $-1$ (not nilpotent) Generators: (1,7,3,10,5,9,6,12)(2,11,8,4), (1,6,3,9)(2,7,12,10,4,5,11,8) $|\Aut(F/K)|$: $1$
## Low degree resolvents
None
Resolvents shown for degrees $\leq 47$
Degree 2: None
Degree 3: None
Degree 4: None
Degree 6: None
## Low degree siblings
11T6, 22T22
Siblings are shown with degree $\leq 47$
A number field with this Galois group has no arithmetically equivalent fields.
## Conjugacy classes
Cycle Type Size Order Representative $1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$ $1$ $1$ $()$ $5, 5, 1, 1$ $1584$ $5$ $( 1,12,10, 2, 9)( 3, 4, 7, 8, 5)$ $11, 1$ $720$ $11$ $( 1,12,11, 8, 7, 9, 4, 6, 2, 3,10)$ $11, 1$ $720$ $11$ $( 1,10, 3, 2, 6, 4, 9, 7, 8,11,12)$ $2, 2, 2, 2, 1, 1, 1, 1$ $165$ $2$ $( 1, 2)( 3,12)( 5, 9)( 8,10)$ $4, 4, 2, 2$ $990$ $4$ $( 1, 3, 2,12)( 4,11)( 5,10, 9, 8)( 6, 7)$ $8, 4$ $990$ $8$ $( 1,10,12, 5, 2, 8, 3, 9)( 4, 7,11, 6)$ $8, 4$ $990$ $8$ $( 1, 9, 3, 8, 2, 5,12,10)( 4, 6,11, 7)$ $3, 3, 3, 1, 1, 1$ $440$ $3$ $( 1, 4, 2)( 3,11,12)( 5, 9, 7)$ $6, 3, 2, 1$ $1320$ $6$ $( 1, 2, 4)( 3, 5,11, 9,12, 7)( 8,10)$
## Group invariants
Order: $7920=2^{4} \cdot 3^{2} \cdot 5 \cdot 11$ Cyclic: No Abelian: No Solvable: No GAP id: Data not available
Character table: 2 4 . . . 4 1 1 3 3 3 3 2 . . . 1 2 1 . . . 5 1 . . 1 . . . . . . 11 1 1 1 . . . . . . . 1a 11a 11b 5a 2a 3a 6a 4a 8a 8b 2P 1a 11b 11a 5a 1a 3a 3a 2a 4a 4a 3P 1a 11a 11b 5a 2a 1a 2a 4a 8a 8b 5P 1a 11a 11b 1a 2a 3a 6a 4a 8b 8a 7P 1a 11b 11a 5a 2a 3a 6a 4a 8b 8a 11P 1a 1a 1a 5a 2a 3a 6a 4a 8a 8b X.1 1 1 1 1 1 1 1 1 1 1 X.2 10 -1 -1 . 2 1 -1 2 . . X.3 10 -1 -1 . -2 1 1 . B -B X.4 10 -1 -1 . -2 1 1 . -B B X.5 11 . . 1 3 2 . -1 -1 -1 X.6 16 A /A 1 . -2 . . . . X.7 16 /A A 1 . -2 . . . . X.8 44 . . -1 4 -1 1 . . . X.9 45 1 1 . -3 . . 1 -1 -1 X.10 55 . . . -1 1 -1 -1 1 1 A = E(11)^2+E(11)^6+E(11)^7+E(11)^8+E(11)^10 = (-1-Sqrt(-11))/2 = -1-b11 B = -E(8)-E(8)^3 = -Sqrt(-2) = -i2
|
2020-09-28 10:07:05
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6930928826332092, "perplexity": 596.5862611675142}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00079.warc.gz"}
|
https://socratic.org/questions/the-half-life-of-plutonium-239-is-24-110-years-of-an-original-mass-of-100-g-how-
|
# The half-life of plutonium-239 is 24,100 years. Of an original mass of 100g, how much plutonium-239 remains after 96,440 years?
Nov 10, 2015
$6.25 g$
#### Explanation:
Note: I edited the question to use the half-life of plutonium-239 as the correct value of $24 , 100$ years, not the value in the original question of $24 , 110$ years, which I assume was a typo.
Use the equation
${M}_{r} = {M}_{i} \times {\left(\frac{1}{2}\right)}^{n}$
Where $n$ is the number of half-lives
${M}_{r}$ is the mass remaining after $n$ half-lives
and ${M}_{i}$ is the initial mass of the sample
to find $n$, the number of half-lives, divide the total time ($96 , 400$) by the time of the half-life ($24 , 100$)
$n = \frac{96 , 400}{24 , 100} = 4$
So:
${M}_{r} = 100 g \times {\left(\frac{1}{2}\right)}^{4} = 6.25 g$
The mass remaining is $6.25 g$
Nov 10, 2015
$\text{6.25 g}$
#### Explanation:
The nuclear half-life of a nuclide tells you how much time must pass before half of an initial sample of said nuclide undergoes radioactive decay.
Simply put, an initial sample of radioactive isotope is halved with every passing of a half-life.
This means that for a ${A}_{0}$ sample of a radioactive isotope, you can say that you'll be left with
• ${A}_{0} \cdot \frac{1}{2} \to$ after one half-life
• ${A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / 4 \to$ after two half-lives
• ${A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / 8 \to$ after three half-lives
• ${A}_{0} / 8 \cdot \frac{1}{2} = {A}_{0} / 16 \to$ after four half-lives
$\vdots$
and so on.
Notice that you can express the amount of the sample that remains after $n$ half-lives like this
$\textcolor{b l u e}{{A}_{n} = {A}_{0} \cdot \frac{1}{2} ^ n}$
Now, notice that the time given to you is actually a multiple of the half-life
$n = \frac{96440}{24110} = 4$
This means that four half-lives of plutonium-239 will pass in $\text{96,440}$, and so the remaining sample after this period of time will be
$m = \text{100 g" * 1/2^4 = "6.25 g}$
I'll leave the answer rounded to three sig figs, despite the fact that you only have one sig fig for the mass of the sample.
|
2019-07-20 02:56:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 26, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9383370280265808, "perplexity": 847.0520336898883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526408.59/warc/CC-MAIN-20190720024812-20190720050812-00071.warc.gz"}
|
http://problemasteoremas.wordpress.com/2010/01/01/1-%C2%BA-problema-de-2010-um-integral-de-stieltjes-2010-problem-1-a-stieltjes-integral/
|
## 1.º Problema de 2010: um integral de Stieltjes :: 2010 Problem #1 – A Stieltjes Integral
Prove que/prove that
$\zeta \left( 2\right) =\dfrac{p}{q}\displaystyle\int_{-1}^{\sqrt{3}}\arctan (x)\,d\left( \arctan (x)\right)$,
where/em que $(p,q)\in\mathbb{Z}^{2}$.
## Sobre Américo Tavares
eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Análise Matemática, Calculus, Cálculo, Integrais, Matemática, Math, Problemas com as tags , , , , . ligação permanente.
### 6 respostas a 1.º Problema de 2010: um integral de Stieltjes :: 2010 Problem #1 – A Stieltjes Integral
1. Wouldn’t there be a problem if $q = 0$?
• According to my evaluation $q\ne 0$
2. Also, I am having a little difficulty in understanding the right hand side expression. Isn’t there supposed to be some summation sign? I mean, $p/q$ should be summed over $\mathbb{Z}^2$, is that right (of course, not allowing $q = 0$)?
• What I mean is this: prove that the given integral can be expressed as a racional multiple of $\zeta(2)$, i. e. $\left( p,q\right) \in\mathbb{Z}^{2}$.
I edited the problem statement a little bit.
3. Ah, I see now. Thanks for the clarification. But, I must say that the integral in its present form is somewhat easy to calculate.
• Then you can move to the last one. That is an important intermediate result. One gets $\zeta(3)$ expressed as another series that converges faster to $\zeta(3)$ (see e.g. van der Poorten’s paper A proof Euler missed…, Apéry’s proof of the irrationality of $\zeta(3)$, An informal report). You find it online. Go to the page Consulta de publicações.
PS. I think that most of my readers doesn’t even know what a Stiltjes integral is.
|
2013-05-25 10:03:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9440450072288513, "perplexity": 1665.582131100091}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705926946/warc/CC-MAIN-20130516120526-00048-ip-10-60-113-184.ec2.internal.warc.gz"}
|
https://noa.gwlb.de/receive/cop_mods_00000564
|
# The remote sensing of radiative forcing by light-absorbing particles (LAPs) in seasonal snow over northeastern China
Light-absorbing particles (LAPs) deposited on snow can decrease snow albedo and affect climate through snow-albedo radiative forcing. In this study, we use MODIS observations combined with a snow-albedo model (SNICAR – Snow, Ice, and Aerosol Radiative) and a radiative transfer model (SBDART – Santa Barbara DISORT Atmospheric Radiative Transfer) to retrieve the instantaneous spectrally integrated radiative forcing at the surface by LAPs in snow (${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$) under clear-sky conditions at the time of MODIS Aqua overpass across northeastern China (NEC) in January–February from 2003 to 2017. ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$ presents distinct spatial variability, with the minimum (22.3 W m−2) in western NEC and the maximum (64.6 W m−2) near industrial areas in central NEC. The regional mean ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$ is $\sim \mathrm{45.1}±\mathrm{6.8}$ W m−2 in NEC. The positive (negative) uncertainties of retrieved ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$ due to atmospheric correction range from 14 % to 57 % (−14 % to −47 %), and the uncertainty value basically decreases with the increased ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$. We attribute the variations of radiative forcing based on remote sensing and find that the spatial variance of ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$ in NEC is 74.6 % due to LAPs and 21.2 % and 4.2 % due to snow grain size and solar zenith angle. Furthermore, based on multiple linear regression, the BC dry and wet deposition and snowfall could explain 84 % of the spatial variance of LAP contents, which confirms the reasonability of the spatial patterns of retrieved ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$ in NEC. We validate ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$ using in situ radiative forcing estimates. We find that the biases in ${\mathrm{RF}}_{\mathrm{MODIS}}^{\mathrm{LAPs}}$ are negatively correlated with LAP concentrations and range from ∼5 % to ∼350 % in NEC.
### Zitieren
Zitierform:
Pu, Wei / Cui, Jiecan / Shi, Tenglong / et al: The remote sensing of radiative forcing by light-absorbing particles (LAPs) in seasonal snow over northeastern China. 2019. Copernicus Publications.
### Rechte
Rechteinhaber: Wei Pu et al.
Nutzung und Vervielfältigung:
|
2019-12-06 04:05:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7294653654098511, "perplexity": 5415.235176407048}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540484477.5/warc/CC-MAIN-20191206023204-20191206051204-00086.warc.gz"}
|
https://www.bartleby.com/questions-and-answers/as-requested-i-include-full-question.-only-need-answer-part-d-httpswww.bartleby.comquestionsandanswe/28789371-9c85-4390-9ec5-78421f1ca2c4
|
# As requested, I include full question. only need answer part d) https://www.bartleby.com/questions-and-answers/q1.-consider-an-allequity-firm-that-is-contemplating-going-into-debt.-the-market-value-of-equity-is-/c24b4703-bc9d-4a3a-8d7f-8964bef82326 Consider an all-equity firm that is contemplating going into debt. The market value of equity is calculated as Free Cash Flow/required rate of return. Current ProposedAssets $10,000$18,000Debt $0$8,000Equity $10,000$10,000Debt/Equity ratio 0.00 1.00Interest rate n/a 7%Shares outstanding 500 500Share price $20$20 (a) If the required rate of return on unlevered equity is 10%, fill out the following table for the company before the debt is issued: Recession Expected ExpansionEBIT $500$1,000 $1,500Interest 0 0 0Net incomeEPSROAROE (b) If the company adds the proposed amount of debt and EBIT is expected to expand proportionally, fill out the table in (a) after the debt is issued. (c) If an investor is not happy with the debt the company added, show the steps the investor can take to do homemade “un-leverage” and earn the same ROA and ROE as in part (a). Set up a table to show that the unleveraged works.(d) If there are no taxes, calculate the WACC before and after the debt is added (assume the 7% payment on the debt = YTM). (e ) If the company stock price goes up by 2% from announcing it is adding debt to expand the business, what effect does this have on the WACC? Question As requested, I include full question. only need answer part d) https://www.bartleby.com/questions-and-answers/q1.-consider-an-allequity-firm-that-is-contemplating-going-into-debt.-the-market-value-of-equity-is-/c24b4703-bc9d-4a3a-8d7f-8964bef82326 Consider an all-equity firm that is contemplating going into debt. The market value of equity is calculated as Free Cash Flow/required rate of return. Current Proposed Assets$10,000 $18,000 Debt$0 $8,000 Equity$10,000 $10,000 Debt/Equity ratio 0.00 1.00 Interest rate n/a 7% Shares outstanding 500 500 Share price$20 $20 (a) If the required rate of return on unlevered equity is 10%, fill out the following table for the company before the debt is issued: Recession Expected Expansion EBIT$500 $1,000$1,500
Interest 0 0 0
Net income
EPS
ROA
ROE
(b) If the company adds the proposed amount of debt and EBIT is expected to expand proportionally, fill out the table in (a) after the debt is issued.
(c) If an investor is not happy with the debt the company added, show the steps the investor can take to do homemade “un-leverage” and earn the same ROA and ROE as in part (a). Set up a table to show that the unleveraged works.
(d) If there are no taxes, calculate the WACC before and after the debt is added (assume the 7% payment on the debt = YTM).
(e ) If the company stock price goes up by 2% from announcing it is adding debt to expand the business, what effect does this have on the WACC?
|
2021-01-18 20:49:49
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21859824657440186, "perplexity": 3760.9667861015423}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00577.warc.gz"}
|
https://search.r-project.org/CRAN/refmans/Anthropometry/html/plotPrototypes.html
|
plotPrototypes {Anthropometry} R Documentation
## Prototypes representation
### Description
This function represents the scatter plots of bust circumference against other selected variable (chest, hip, neck to ground or waist) jointly with the prototypes obtained for each bust class provided by either trimowa or hipamAnthropom. In addition, the prototypes defined by the European standard on sizing systems. Size designation of clothes. Part 3: Measurements and intervals can be also displayed.
### Usage
plotPrototypes(data,prototypes,nsizes,bustVariable,variable,col,xlim,ylim,
main,EN)
### Arguments
data Data frame. It should contain the chest, neck to ground, waist, hip and bust measurements of the individuals. In order to be able to represent them, the name of the columns of the database must be 'chest', 'necktoground', 'waist', 'hip' and 'bust' respectively, see sampleSpanishSurvey. Each row corresponds to an observation, and each column corresponds to a variable. All variables are numeric. prototypes Prototypes (medoids) i.e., typical persons within the sample, obtained with trimowa or hipamAnthropom. nsizes Number of subsets (classes), into the database is segmented. In our approach, the whole anthropometric Spanish survey is segmented into twelve bust segments, according to the European standard on sizing systems. Size designation of clothes. Part 3: Measurements and intervals. bustVariable Bust variable. variable Anthropometric variable to be plotted. It can be 'chest', 'necktoground', 'waist' and 'hip'. col A specification for the medoids color in each bust class. xlim Axis lenght of the x axis according to the range of the bust variable. ylim Axis lenght of the y axis according to the range of the selected variable among chest, hip, neck to ground and waist. main Main title of the plot. EN A logical value. If TRUE, the prototypes defined by the European standard for each variable are represented. See section Details for more details.
### Details
Ir order to check the goodness of trimowa, the sizes defined by the prototypes can be compared with those defined by the European standard to sizing system. This standard establishes 12 sizes according to the combinations of the bust, waist and hip measurements and does not fix neither chest nor height standard measurements. We can approximate the chest measurements through a linear regression analysis, taking the bust measurements detailed in the standard as independent variable. Besides, we take as neck to ground measurements for the standard sizing system, the values 132, 136 and 140 cm because those are the most repeated values and they are those which best cover our data set. See Ibanez et al. (2012) for a complete explanation.
### Value
A device with the desired plot.
### Note
As mentioned, this function is especially defined for the sizes established by the European standard on sizing systems. Part 3: Measurements and intervals. In order to use this function with other standard, this function must be adapted.
Guillermo Vinue
### References
Ibanez, M. V., Vinue, G., Alemany, S., Simo, A., Epifanio, I., Domingo, J., and Ayala, G., (2012). Apparel sizing using trimmed PAM and OWA operators, Expert Systems with Applications 39, 10512–10520.
Vinue, G., Leon, T., Alemany, S., and Ayala, G., (2014). Looking for representative fit models for apparel sizing, Decision Support Systems 57, 22–33.
European Committee for Standardization. Size designation of clothes. Part 3: Measurements and intervals. (2005).
sampleSpanishSurvey, weightsMixtureUB, trimowa, getDistMatrix, trimmedoid, hipamAnthropom
### Examples
#TRIMOWA ALGORITHM:
dataTrimowa <- sampleSpanishSurvey
numVar <- dim(dataTrimowa)[2]
bust <- dataTrimowa$bust bustSizes <- bustSizesStandard(seq(74, 102, 4), seq(107, 131, 6)) orness <- 0.7 weightsTrimowa <- weightsMixtureUB(orness, numVar) numClust <- 3 ; alpha <- 0.01 ; niter <- 10 ; algSteps <- 7 ah <- c(23, 28, 20, 25, 25) #For reproducing results, seed for randomness: #suppressWarnings(RNGversion("3.5.0")) #set.seed(2014) numSizes <- 2 res_trimowa <- computSizesTrimowa(dataTrimowa, bust, bustSizes$bustCirc, numSizes,
weightsTrimowa, numClust, alpha, niter, algSteps,
ah, FALSE)
prototypes <- anthrCases(res_trimowa, numSizes)
bustVariable <- "bust"
xlim <- c(72, 132)
color <- c("black", "red", "green", "blue", "cyan", "brown", "gray",
"deeppink3", "orange", "springgreen4", "khaki3", "steelblue1")
variable <- "chest"
range(dataTrimowa[,variable])
#[1] 76.7755 135.8580
ylim <- c(70,140)
title <- "Prototypes \n bust vs chest"
plotPrototypes(dataTrimowa, prototypes, numSizes, bustVariable,
variable, color, xlim, ylim, title, FALSE)
plotPrototypes(dataTrimowa, prototypes, numSizes, bustVariable,
variable, color, xlim, ylim, title, TRUE)
#For other plots and an example for the hipam algorithm,
#see www.uv.es/vivigui/softw/more_examples.R
[Package Anthropometry version 1.17 Index]
|
2022-12-04 15:22:18
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35297277569770813, "perplexity": 11922.35840207378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00417.warc.gz"}
|
https://www.shaalaa.com/question-bank-solutions/a-potted-plant-was-taken-order-prove-factor-necessary-photosynthesis-potted-plant-was-kept-dark-24-hours-concept-photosynthesis_18544
|
Share
# A Potted Plant Was Taken in Order to Prove a Factor Necessary for Photosynthesis. the Potted Plant Was Kept in the Dark for 24 Hours. - ICSE Class 10 - Biology
#### Question
A potted plant was taken in order to prove a factor necessary for photosynthesis. The potted plant was kept in the dark for 24 hours. One of the leaves was covered with black paper in the centre. The potted plant was then placed in sunlight for a few hours.
(i) What aspect of photosynthesis was being tested?
(ii) Why was the plant placed in the dark before beginning the experiment?
(iii) During the starch test, why was the leaf
(1) boiled in water
(2) boiled in methylated spirit
(iv) Write a balanced chemical equation to represent the process of photosynthesis.
(v) Draw a neat diagram of a chloroplast and label its parts.
#### Solution
(i) Light is required for photosynthesis.
(ii) Before beginning the experiment, the plant was kept in dark in order to destarch it, i.e. to remove the pre-existing starch from the storage organs.
(iii)
(1) The leaf was boiled in water to destroy enzymes so that further chemical changes do not take place in the leaf.
(2) The leaf was boiled in methylated spirit to dissolve chlorophyll.
(iv) Chemical equation for the process of photosynthesis:
6CO2+12H2O"Light"/"Chlorophyll"C6H12O6+6O2+6H2O
(v)Chloroplast:
Is there an error in this question or solution?
#### APPEARS IN
Solution A Potted Plant Was Taken in Order to Prove a Factor Necessary for Photosynthesis. the Potted Plant Was Kept in the Dark for 24 Hours. Concept: Concept of Photosynthesis.
S
|
2019-12-14 08:00:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.709431529045105, "perplexity": 2951.37411911008}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540585566.60/warc/CC-MAIN-20191214070158-20191214094158-00469.warc.gz"}
|
http://www.pandora.com/red/end-of-silence/already-over-pt-2
|
It is taking longer than expected to fetch the next song to play. The music should be playing soon. If you get tired of waiting, you can try reloading your browser.
Please check our Help page for information about troubleshooting Pandora on your browser.
Please ensure you are using the latest Flash Player.
If you are unable or do not wish to upgrade your Flash Player,
please try a different browser.
Please check our Help page for information about troubleshooting Pandora on your browser.
Your Pandora One subscription will expire shortly.
Your Pandora One trial will expire shortly.
Your Pandora One trial subscription will expire shortly. Upgrade to continue unlimited, ad-free listening.
You've listened to hours of Pandora this month. Consider upgrading to Pandora One.
Hi . Pandora is using Facebook to personalize your experience.
|
-0:00
0:00
Change Skin
We created Pandora to put the Music Genome Project directly in your hands
It’s a new kind of radio –
stations that play only music you like
Now Playing
Music Feed
My Profile
Create a Station
# Already Over, Pt. 2
by Red
## Features of This Track
a subtle use of vocal harmony
acoustic sonority
intricate melodic phrasing
thru composed melodic style
acoustic rhythm piano
minor key tonality
melodic string accompaniment
a good dose of acoustic guitar pickin'
acoustic rhythm guitars
solo strings
These are just a few of the hundreds of attributes cataloged for this track by the Music Genome Project.
## Similar Tracks
Report as inappropriate
this song is making me cry
Report as inappropriate
Beautiful ��
Report as inappropriate
Everything Red makes is just, amazing. This song and every song they make is amazing.
Report as inappropriate
I got chills. His voice is amazing wether it's his yelling in songs or just clean vocals like this. AMAZING.
Report as inappropriate
Amazing god is so awesome
Already over now, give it all to you reaching as I fall god so grab my hand and take me far
Report as inappropriate
I love already over the original but part 2 isn't so bad. The only thing is its a bit too slow paced for me but other than that i love it!
Report as inappropriate
Butifule we r his forever we r his bride
Report as inappropriate
<3 this song
Report as inappropriate
Already over part 2 ALL THE WAY
Report as inappropriate
Love this song and his voice!!!
Report as inappropriate
Wow.
Report as inappropriate
I love this song soooooo much it's just beautiful! I love it!!!
Report as inappropriate
cfosholdt
Simply Beautiful!
Report as inappropriate
I love the first a little more, but this one is really good too.
Report as inappropriate
WHAT!?! THERE'S A PART TWO!!! LUV IT! :)
Report as inappropriate
haha, same here
Report as inappropriate
I looooove this song!!!!!!!! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
!!!!!!!!!!!! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
We're sorry, but a browser plugin or firewall may be preventing Pandora from loading.
In order to use Pandora internet radio, please upgrade to a more current browser.
|
2015-08-28 23:10:19
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9972754716873169, "perplexity": 4673.843085279324}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064019.39/warc/CC-MAIN-20150827025424-00087-ip-10-171-96-226.ec2.internal.warc.gz"}
|
https://www.aimsciences.org/article/doi/10.3934/dcdsb.2020298
|
# American Institute of Mathematical Sciences
• Previous Article
Modeling multiple taxis: Tumor invasion with phenotypic heterogeneity, haptotaxis, and unilateral interspecies repellence
• DCDS-B Home
• This Issue
• Next Article
On a singularly perturbed semi-linear problem with Robin boundary conditions
January 2021, 26(1): 415-441. doi: 10.3934/dcdsb.2020298
## The dynamics of a two host-two virus system in a chemostat environment
1 Department of Mathematics, National Tsing Hua University, Hsinchu 300, Taiwan 2 Department of Mathematics, University of Nebraska-Lincoln, Lincoln, NE 68588, USA
* Corresponding author: Sze-Bi Hsu
Received June 2020 Revised August 2020 Published October 2020
Fund Project: The first author is supported by the grant MOST 108-2115-M-007-007. The second author is supported by NSF Grant DMS 1411703
The coevolution or coexistence of multiple viruses with multiple hosts has been an important issue in viral ecology. This paper is to study the mathematical properties of the solutions of a chemostat model for two host species and two virus species. By virtue of the global dynamics of its submodels and the theories of uniform persistence and Hopf bifurcation, we derive sufficient conditions for the coexistence of two hosts with two viruses and coexistence of two hosts with one virus, as well as occurrence of Hopf bifurcation.
Citation: Sze-Bi Hsu, Yu Jin. The dynamics of a two host-two virus system in a chemostat environment. Discrete & Continuous Dynamical Systems - B, 2021, 26 (1) : 415-441. doi: 10.3934/dcdsb.2020298
##### References:
show all references
##### References:
The projection of the phase diagram of model (1) onto the $N_1N_2$ plane. Left: $\beta = 11.5$; right: $\beta = 20$
The time series of model (1). Left: $\beta = 11.5$; right: $\beta = 20$
The conditions for existence and local stability of equilibria of (4). Here, an equilibrium exists means it is nonnegative for $E_1^{nnv}$-$E_4^{nnv}$ and positive for $E_5^{nnv}$
Equilibrium Existence condition Stability condition $E_0^{nnv}=(0,0,0)$ $r_1 <\omega$, $r_2 <\omega$ $E_1^{nnv}=(\tilde{N}_1,0,0)$ $r_1 > \omega$ $r_1 >r_2$, $N_1^\ast >\tilde{N}_1$ $E_2^{nnv}=(0,\tilde{N}_2,0)$ $r_2 > \omega$ $r_1 \tilde{N}_2$ $E_3^{nnv}=(N_1^\ast,0,\tilde{V}^\ast)$ $N_1^\ast <\tilde{N}_1$ ($r_1 >\omega$ required) $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)(N_1^\ast-\eta) >0$ $E_4^{nnv}=(0,N_2^\ast,V^\ast)$ $N_2^\ast <\tilde{N}_2$ ($r_2 >\omega$ required) $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)(N_2^\ast-\eta) <0$ $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)(r_1-r_2) >0$ $E_5^{nnv}=(N_1^c,N_2^c,V^c)$ $(N_1^\ast-\eta)\left(\frac{\phi_1}{\phi_2}-\frac{\beta_2}{\beta_1}\right) <0$ $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)\left(\frac{\phi_1}{\phi_2}-\frac{\beta_2}{\beta_1}\right) >0$ $(N_2^\ast-\eta)\left(\frac{\phi_1}{\phi_2}-\frac{\beta_2}{\beta_1}\right) >0$
Equilibrium Existence condition Stability condition $E_0^{nnv}=(0,0,0)$ $r_1 <\omega$, $r_2 <\omega$ $E_1^{nnv}=(\tilde{N}_1,0,0)$ $r_1 > \omega$ $r_1 >r_2$, $N_1^\ast >\tilde{N}_1$ $E_2^{nnv}=(0,\tilde{N}_2,0)$ $r_2 > \omega$ $r_1 \tilde{N}_2$ $E_3^{nnv}=(N_1^\ast,0,\tilde{V}^\ast)$ $N_1^\ast <\tilde{N}_1$ ($r_1 >\omega$ required) $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)(N_1^\ast-\eta) >0$ $E_4^{nnv}=(0,N_2^\ast,V^\ast)$ $N_2^\ast <\tilde{N}_2$ ($r_2 >\omega$ required) $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)(N_2^\ast-\eta) <0$ $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)(r_1-r_2) >0$ $E_5^{nnv}=(N_1^c,N_2^c,V^c)$ $(N_1^\ast-\eta)\left(\frac{\phi_1}{\phi_2}-\frac{\beta_2}{\beta_1}\right) <0$ $\left(\frac{\phi_1}{\phi_2}-\frac{r_1}{r_2}\right)\left(\frac{\phi_1}{\phi_2}-\frac{\beta_2}{\beta_1}\right) >0$ $(N_2^\ast-\eta)\left(\frac{\phi_1}{\phi_2}-\frac{\beta_2}{\beta_1}\right) >0$
Global or local dynamics of (4). $E_0^{nnv}$-$E_5^{nnv}$ are defined in (5). Conditions for $E_5^{nnv}$ to be positive or not may not be all listed. "-" represents that some compartments of the equilibrium are negative. "U" represents "unstable"; "GAS" represents "globally asymptotically stable", "S" represents "locally asymptotically stable"
Condition $E_0^{nnv}$ $E_1^{nnv}$ $E_2^{nnv}$ $E_3^{nnv}$ $E_4^{nnv}$ $E_5^{nnv}$ (a) $r_1 <\omega$, $r_2 <\omega$ GAS - - - - - (b) $r_2 <\omega \tilde{N}_1$ U GAS - - - - (c) $r_2 <\omega \tilde{N}_2$ U - GAS - - - (e) $r_1 <\omega \omega, N_1^\ast <\tilde{N}_1, N_2^\ast <\tilde{N}_2 \\ (\phi_1r_2-\phi_2r_1)(N_1^\ast-\eta) <0 \\ (\phi_1r_2-\phi_2r_1)(N_2^\ast-\eta) <0 \end{array}$ U U U U GAS - (g) $\begin{array}{l} r_1,r_2 >\omega, , N_1^\ast <\tilde{N}_1, N_2^\ast <\tilde{N}_2 \\ (\phi_1r_2-\phi_2r_1)(N_1^\ast-\eta) >0 \\ (\phi_1r_2-\phi_2r_1)(N_2^\ast-\eta) >0 \end{array}$ U U U GAS U - (h) $r_1 >r_2 >\omega$, $N_1^\ast >\tilde{N}_1$, $N_2^\ast <\tilde{N}_2$ U GAS U - U - (i) $\omega \tilde{N}_2$ U U GAS U - - (j) $r_1 >r_2 >\omega$, $N_1^\ast <\tilde{N}_1$, $N_2^\ast >\tilde{N}_2$ U U U GAS - - (k) $\omega \tilde{N}_1$, $N_2^\ast <\tilde{N}_2$ U U U - GAS - (l) $r_1 >r_2 >\omega$, $N_1^\ast >\tilde{N}_1$, $N_2^\ast >\tilde{N}_2$ U GAS U - - - (m) $\omega \tilde{N}_1$, $N_2^\ast >\tilde{N}_2$ U U GAS - - - (n) $\begin{array}{l} \mbox{(a) }r_1 >r_2 >\omega, \phi_1 r_2 >\phi_2r_1, \\ \tilde{N}_1 >N_1^\ast >\eta >N_2^\ast, \tilde{N}_2 >N_2^\ast; \\ \mbox{or (b) } \omega r_2 >\omega, \phi_1 r_2 >\phi_2r_1, \\ N_1^\ast <\eta \tilde{N}_1 >N_1^\ast >\eta >N_2^\ast \end{array}$ U U U U U S (p) $r_1 >r_2 >\omega$, $\phi_1 r_2 >\phi_2r_1$, U S U - S U $N_1^\ast >\tilde{N}_1 >\tilde{N}_2 >\eta >N_2^\ast$ (q) $\begin{array}{l} \omega r_2 >\omega, \phi_1 r_2 >\phi_2r_1, \\ N_1^\ast <\eta <\tilde{N}_2 <\tilde{N}_1, \tilde{N}_2 \tilde{N}_1, \tilde{N}_2 >\tilde{N}_1 >\eta >N_2^\ast \end{array}$ U U U - U S
Condition $E_0^{nnv}$ $E_1^{nnv}$ $E_2^{nnv}$ $E_3^{nnv}$ $E_4^{nnv}$ $E_5^{nnv}$ (a) $r_1 <\omega$, $r_2 <\omega$ GAS - - - - - (b) $r_2 <\omega \tilde{N}_1$ U GAS - - - - (c) $r_2 <\omega \tilde{N}_2$ U - GAS - - - (e) $r_1 <\omega \omega, N_1^\ast <\tilde{N}_1, N_2^\ast <\tilde{N}_2 \\ (\phi_1r_2-\phi_2r_1)(N_1^\ast-\eta) <0 \\ (\phi_1r_2-\phi_2r_1)(N_2^\ast-\eta) <0 \end{array}$ U U U U GAS - (g) $\begin{array}{l} r_1,r_2 >\omega, , N_1^\ast <\tilde{N}_1, N_2^\ast <\tilde{N}_2 \\ (\phi_1r_2-\phi_2r_1)(N_1^\ast-\eta) >0 \\ (\phi_1r_2-\phi_2r_1)(N_2^\ast-\eta) >0 \end{array}$ U U U GAS U - (h) $r_1 >r_2 >\omega$, $N_1^\ast >\tilde{N}_1$, $N_2^\ast <\tilde{N}_2$ U GAS U - U - (i) $\omega \tilde{N}_2$ U U GAS U - - (j) $r_1 >r_2 >\omega$, $N_1^\ast <\tilde{N}_1$, $N_2^\ast >\tilde{N}_2$ U U U GAS - - (k) $\omega \tilde{N}_1$, $N_2^\ast <\tilde{N}_2$ U U U - GAS - (l) $r_1 >r_2 >\omega$, $N_1^\ast >\tilde{N}_1$, $N_2^\ast >\tilde{N}_2$ U GAS U - - - (m) $\omega \tilde{N}_1$, $N_2^\ast >\tilde{N}_2$ U U GAS - - - (n) $\begin{array}{l} \mbox{(a) }r_1 >r_2 >\omega, \phi_1 r_2 >\phi_2r_1, \\ \tilde{N}_1 >N_1^\ast >\eta >N_2^\ast, \tilde{N}_2 >N_2^\ast; \\ \mbox{or (b) } \omega r_2 >\omega, \phi_1 r_2 >\phi_2r_1, \\ N_1^\ast <\eta \tilde{N}_1 >N_1^\ast >\eta >N_2^\ast \end{array}$ U U U U U S (p) $r_1 >r_2 >\omega$, $\phi_1 r_2 >\phi_2r_1$, U S U - S U $N_1^\ast >\tilde{N}_1 >\tilde{N}_2 >\eta >N_2^\ast$ (q) $\begin{array}{l} \omega r_2 >\omega, \phi_1 r_2 >\phi_2r_1, \\ N_1^\ast <\eta <\tilde{N}_2 <\tilde{N}_1, \tilde{N}_2 \tilde{N}_1, \tilde{N}_2 >\tilde{N}_1 >\eta >N_2^\ast \end{array}$ U U U - U S
The conditions for existence and stability of equilibria of model (1). Here, an equilibrium exists means it is nonnegative for $E_1$-$E_8$ and positive for $E_9$. The notations are defined in (9) and (10)
Equilibrium Existence condition Stability condition $E_0=(0,0,0,0)$ $r_1 <\omega$, $r_2 <\omega$ $E_1=(\tilde{N}_1,0,0,0)$ $r_1 > \omega$ $r_1 >r_2$, $N_{1,1}^\ast >\tilde{N}_1$, $N_{1,2}^\ast >\tilde{N}_1$ $E_2=(0,\tilde{N}_2,0,0)$ $r_2 > \omega$ $r_1 \tilde{N}_2$, $N_{2,2}^\ast >\tilde{N}_2$ $E_3=(N_{1,1}^\ast,0,\frac{r_1(\tilde{N}_1-N_{1,1}^\ast)}{K\phi_{11}},0)$ $N_{1,1}^\ast <\tilde{N}_1$ $\begin{array}{l} B\Phi_3 >0 \\ \Phi R_1\cdot (N_{1,1}^\ast-\eta_1) >0 \end{array}$ $E_4=(N_{1,2}^\ast,0,0,\frac{r_1(\tilde{N}_1-N_{1,2}^\ast)}{K\phi_{12}})$ $N_{1,2}^\ast <\tilde{N}_1$ $\begin{array}{l} B\Phi_3 <0 \\ \Phi R_2\cdot (N_{1,2}^\ast-\eta_2) >0 \end{array}$ $E_5=(0,N_{2,1}^\ast,\frac{r_2(\tilde{N}_2-N_{2,1}^\ast)}{K\phi_{21}},0)$ $N_{2,1}^\ast <\tilde{N}_2$ $\begin{array}{l} B\Phi_4 >0 \\ \Phi R_1\cdot (N_{2,1}^\ast-\eta_1) <0 \end{array}$ $E_6=(0,N_{2,2}^\ast,0,\frac{r_2(\tilde{N}_2-N_{2,2}^\ast)}{K\phi_{22}})$ $N_{2,2}^\ast <\tilde{N}_2$ $\begin{array}{l} B\Phi_4 <0 \\ \Phi R_2\cdot(N_{2,2}^\ast-\eta_2) <0 \end{array}$ $E_7=(N_1^c,N_2^c,V_1^c,0)$ $\begin{array}{l} (N_{2,1}^\ast-\eta_1)\cdot B\Phi_1 >0 \\ (N_{1,1}^\ast-\eta_1)\cdot B\Phi_1 <0 \\ (r_1-r_2)\Phi R_1 >0 \end{array}$ $\begin{array}{l} NN <0 \\ \Phi R_1\cdot B\Phi_1 >0 \end{array}$ $E_8=(\hat{N}_1^c,\hat{N}_2^c,0,\hat{V}_2^c)$ $\begin{array}{l} (N_{2,2}^\ast-\eta_2)\cdot B\Phi_2 >0 \\ (N_{1,2}^\ast-\eta_2)\cdot B\Phi_2 <0 \\ (r_1-r_2)\Phi R_2 >0 \end{array}$ $\begin{array}{l} NN_h <0 \\ \Phi R_2\cdot B\Phi_2 >0 \end{array}$ $E_9=(N_1^p,N_2^p,V_1^p,V_2^p)$ $\begin{array}{l}B\Phi\cdot B\Phi_3 <0 \\ B\Phi\cdot B\Phi_4 >0 \\ \Phi R_1\cdot B\Phi_1\cdot NN\cdot B\Phi \cdot \Phi\Phi >0 \\ \Phi R_2\cdot B\Phi_2\cdot NN_h\cdot B\Phi \cdot \Phi\Phi >0 \end{array}$ $\begin{array}{l} B\Phi \cdot \Phi\Phi >0 \\ (14) \end{array}$
Equilibrium Existence condition Stability condition $E_0=(0,0,0,0)$ $r_1 <\omega$, $r_2 <\omega$ $E_1=(\tilde{N}_1,0,0,0)$ $r_1 > \omega$ $r_1 >r_2$, $N_{1,1}^\ast >\tilde{N}_1$, $N_{1,2}^\ast >\tilde{N}_1$ $E_2=(0,\tilde{N}_2,0,0)$ $r_2 > \omega$ $r_1 \tilde{N}_2$, $N_{2,2}^\ast >\tilde{N}_2$ $E_3=(N_{1,1}^\ast,0,\frac{r_1(\tilde{N}_1-N_{1,1}^\ast)}{K\phi_{11}},0)$ $N_{1,1}^\ast <\tilde{N}_1$ $\begin{array}{l} B\Phi_3 >0 \\ \Phi R_1\cdot (N_{1,1}^\ast-\eta_1) >0 \end{array}$ $E_4=(N_{1,2}^\ast,0,0,\frac{r_1(\tilde{N}_1-N_{1,2}^\ast)}{K\phi_{12}})$ $N_{1,2}^\ast <\tilde{N}_1$ $\begin{array}{l} B\Phi_3 <0 \\ \Phi R_2\cdot (N_{1,2}^\ast-\eta_2) >0 \end{array}$ $E_5=(0,N_{2,1}^\ast,\frac{r_2(\tilde{N}_2-N_{2,1}^\ast)}{K\phi_{21}},0)$ $N_{2,1}^\ast <\tilde{N}_2$ $\begin{array}{l} B\Phi_4 >0 \\ \Phi R_1\cdot (N_{2,1}^\ast-\eta_1) <0 \end{array}$ $E_6=(0,N_{2,2}^\ast,0,\frac{r_2(\tilde{N}_2-N_{2,2}^\ast)}{K\phi_{22}})$ $N_{2,2}^\ast <\tilde{N}_2$ $\begin{array}{l} B\Phi_4 <0 \\ \Phi R_2\cdot(N_{2,2}^\ast-\eta_2) <0 \end{array}$ $E_7=(N_1^c,N_2^c,V_1^c,0)$ $\begin{array}{l} (N_{2,1}^\ast-\eta_1)\cdot B\Phi_1 >0 \\ (N_{1,1}^\ast-\eta_1)\cdot B\Phi_1 <0 \\ (r_1-r_2)\Phi R_1 >0 \end{array}$ $\begin{array}{l} NN <0 \\ \Phi R_1\cdot B\Phi_1 >0 \end{array}$ $E_8=(\hat{N}_1^c,\hat{N}_2^c,0,\hat{V}_2^c)$ $\begin{array}{l} (N_{2,2}^\ast-\eta_2)\cdot B\Phi_2 >0 \\ (N_{1,2}^\ast-\eta_2)\cdot B\Phi_2 <0 \\ (r_1-r_2)\Phi R_2 >0 \end{array}$ $\begin{array}{l} NN_h <0 \\ \Phi R_2\cdot B\Phi_2 >0 \end{array}$ $E_9=(N_1^p,N_2^p,V_1^p,V_2^p)$ $\begin{array}{l}B\Phi\cdot B\Phi_3 <0 \\ B\Phi\cdot B\Phi_4 >0 \\ \Phi R_1\cdot B\Phi_1\cdot NN\cdot B\Phi \cdot \Phi\Phi >0 \\ \Phi R_2\cdot B\Phi_2\cdot NN_h\cdot B\Phi \cdot \Phi\Phi >0 \end{array}$ $\begin{array}{l} B\Phi \cdot \Phi\Phi >0 \\ (14) \end{array}$
[1] Tao Feng, Zhipeng Qiu, Xinzhu Meng. Dynamics of a stochastic hepatitis C virus system with host immunity. Discrete & Continuous Dynamical Systems - B, 2019, 24 (12) : 6367-6385. doi: 10.3934/dcdsb.2019143 [2] Hossein Mohebbi, Azim Aminataei, Cameron J. Browne, Mohammad Reza Razvan. Hopf bifurcation of an age-structured virus infection model. Discrete & Continuous Dynamical Systems - B, 2018, 23 (2) : 861-885. doi: 10.3934/dcdsb.2018046 [3] Cameron J. Browne, Sergei S. Pilyugin. Global analysis of age-structured within-host virus model. Discrete & Continuous Dynamical Systems - B, 2013, 18 (8) : 1999-2017. doi: 10.3934/dcdsb.2013.18.1999 [4] Shikun Wang. Dynamics of a chemostat system with two patches. Discrete & Continuous Dynamical Systems - B, 2019, 24 (11) : 6261-6278. doi: 10.3934/dcdsb.2019138 [5] Bernold Fiedler. Global Hopf bifurcation in networks with fast feedback cycles. Discrete & Continuous Dynamical Systems - S, 2021, 14 (1) : 177-203. doi: 10.3934/dcdss.2020344 [6] Hui li, Manjun Ma. Global dynamics of a virus infection model with repulsive effect. Discrete & Continuous Dynamical Systems - B, 2019, 24 (9) : 4783-4797. doi: 10.3934/dcdsb.2019030 [7] Tin Phan, Bruce Pell, Amy E. Kendig, Elizabeth T. Borer, Yang Kuang. Rich dynamics of a simple delay host-pathogen model of cell-to-cell infection for plant virus. Discrete & Continuous Dynamical Systems - B, 2021, 26 (1) : 515-539. doi: 10.3934/dcdsb.2020261 [8] Zhikun She, Xin Jiang. Threshold dynamics of a general delayed within-host viral infection model with humoral immunity and two modes of virus transmission. Discrete & Continuous Dynamical Systems - B, 2021, 26 (7) : 3835-3861. doi: 10.3934/dcdsb.2020259 [9] Hossein Pourbashash, Sergei S. Pilyugin, Patrick De Leenheer, Connell McCluskey. Global analysis of within host virus models with cell-to-cell viral transmission. Discrete & Continuous Dynamical Systems - B, 2014, 19 (10) : 3341-3357. doi: 10.3934/dcdsb.2014.19.3341 [10] Zhihua Liu, Hui Tang, Pierre Magal. Hopf bifurcation for a spatially and age structured population dynamics model. Discrete & Continuous Dynamical Systems - B, 2015, 20 (6) : 1735-1757. doi: 10.3934/dcdsb.2015.20.1735 [11] Stephen Pankavich, Nathan Neri, Deborah Shutt. Bistable dynamics and Hopf bifurcation in a refined model of early stage HIV infection. Discrete & Continuous Dynamical Systems - B, 2020, 25 (8) : 2867-2893. doi: 10.3934/dcdsb.2020044 [12] Fabien Crauste. Global Asymptotic Stability and Hopf Bifurcation for a Blood Cell Production Model. Mathematical Biosciences & Engineering, 2006, 3 (2) : 325-346. doi: 10.3934/mbe.2006.3.325 [13] Pengmiao Hao, Xuechen Wang, Junjie Wei. Global Hopf bifurcation of a population model with stage structure and strong Allee effect. Discrete & Continuous Dynamical Systems - S, 2017, 10 (5) : 973-993. doi: 10.3934/dcdss.2017051 [14] Xiaoling Zou, Dejun Fan, Ke Wang. Stationary distribution and stochastic Hopf bifurcation for a predator-prey system with noises. Discrete & Continuous Dynamical Systems - B, 2013, 18 (5) : 1507-1519. doi: 10.3934/dcdsb.2013.18.1507 [15] Tewfik Sari, Frederic Mazenc. Global dynamics of the chemostat with different removal rates and variable yields. Mathematical Biosciences & Engineering, 2011, 8 (3) : 827-840. doi: 10.3934/mbe.2011.8.827 [16] Yan-Xia Dang, Zhi-Peng Qiu, Xue-Zhi Li, Maia Martcheva. Global dynamics of a vector-host epidemic model with age of infection. Mathematical Biosciences & Engineering, 2017, 14 (5&6) : 1159-1186. doi: 10.3934/mbe.2017060 [17] Cuicui Jiang, Kaifa Wang, Lijuan Song. Global dynamics of a delay virus model with recruitment and saturation effects of immune responses. Mathematical Biosciences & Engineering, 2017, 14 (5&6) : 1233-1246. doi: 10.3934/mbe.2017063 [18] Cuicui Jiang, Wendi Wang. Complete classification of global dynamics of a virus model with immune responses. Discrete & Continuous Dynamical Systems - B, 2014, 19 (4) : 1087-1103. doi: 10.3934/dcdsb.2014.19.1087 [19] Feng-Bin Wang, Junping Shi, Xingfu Zou. Dynamics of a host-pathogen system on a bounded spatial domain. Communications on Pure & Applied Analysis, 2015, 14 (6) : 2535-2560. doi: 10.3934/cpaa.2015.14.2535 [20] Xiulan Lai, Xingfu Zou. A reaction diffusion system modeling virus dynamics and CTLs response with chemotaxis. Discrete & Continuous Dynamical Systems - B, 2016, 21 (8) : 2567-2585. doi: 10.3934/dcdsb.2016061
2019 Impact Factor: 1.27
|
2021-05-06 17:59:00
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.653383731842041, "perplexity": 2671.8022760203685}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988759.29/warc/CC-MAIN-20210506175146-20210506205146-00588.warc.gz"}
|
https://codereview.stackexchange.com/questions/154334/magic-trick-google-code-jam-2014-qr-problem-a-in-python/154339
|
# Magic Trick (Google Code Jam 2014 QR Problem A) in Python
I'm learning Python by solving previous GCJ problems. This problem is quite simple to solve, but I'd like my code to be reviewed for advance (more Pythonic expression and more efficient algorithm).
The magician starts by arranging 16 cards in a square grid: 4 rows of cards, with 4 cards in each row. Each card has a different number from 1 to 16 written on the side that is showing. Next, the magician asks a volunteer to choose a card, and to tell him which row that card is in.
Finally, the magician arranges the 16 cards in a square grid again, possibly in a different order. Once again, he asks the volunteer which row her card is in. With only the answers to these two questions, the magician then correctly determines which card the volunteer chose. Amazing, right?
You decide to write a program to help you understand the magician's technique. The program will be given the two arrangements of the cards, and the volunteer's answers to the two questions: the row number of the selected card in the first arrangement, and the row number of the selected card in the second arrangement. The rows are numbered 1 to 4 from top to bottom.
Your program should determine which card the volunteer chose; or if there is more than one card the volunteer might have chosen (the magician did a bad job); or if there's no card consistent with the volunteer's answers (the volunteer cheated).
Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line containing an integer: the answer to the first question. The next 4 lines represent the first arrangement of the cards: each contains 4 integers, separated by a single space. The next line contains the answer to the second question, and the following four lines contain the second arrangement in the same format.
Output
For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1).
If there is a single card the volunteer could have chosen, y should be the number on the card. If there are multiple cards the volunteer could have chosen, y should be "Bad magician!", without the quotes. If there are no cards consistent with the volunteer's answers, y should be "Volunteer cheated!", without the quotes. The text needs to be exactly right, so consider copying/pasting it from here.
Sample
Input
3
2
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
3
1 2 5 4
3 11 6 15
9 10 7 12
13 14 8 16
2
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
2
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
2
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
3
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output
Case #1: 7
Case #3: Volunteer cheated!
And here's my code:
import sys
from itertools import repeat
def main():
submit = False
filename = "A-small-practice"
sys.stdin = open(filename + ".in")
if submit:
sys.stdout = open(filename + ".out", "w")
test_cases = int(input())
main2(test_cases)
def main2(test_cases):
for cn in range(1, test_cases + 1):
print "Case #" + str(cn) + ":",
choices = {}
# Each test case has 2 answers.
for i in repeat(None, 2):
# Volunteer's answer (row # which contains his choice)
row = int(input())
# Cards are in 4 rows.
for r in range(1, 5):
nums = raw_input()
if r == row:
# A row has 4 cards.
for n in nums.split():
choices[n] = choices.get(n, 0) + 1
# The chosen card(s) would be appeared twice.
# So, pick items in choices whose value(appearance) is 2, and filter
# others.
choices = {k: v for k, v in choices.iteritems() if v == 2}
# Only one choice should exist after filtering.
if len(choices) == 1:
print choices.keys()[0]
elif len(choices) > 1:
else:
print "Volunteer cheated!"
main()
• Naming: main is reserved so that's ok, but main2 makes no sense. The name should be a hint about what the function does, I'd go for guess_card.
• Since there's no mention of a file input, I'd say go for standard input. You can always read a file and redirect the input if that's needed. The same goes for standard output.
• You're using both input and raw_input, why is that? If you want to do input validation, you should do it properly, trying to parse it and eventually catching exceptions. But if you go down that road, then you should also not assume that the rest of the input is correct. I'd say that for this specific program, you can safely say that the syntax of the input will be fine.
• Comments used like this are pretty much useless, they should tell why things are happening the way they are, not re-describe the code
• Your second function is not modular, it will take all cases and do everything. I think you should make that function guess one of the cases and loop around it instead of inside.
• There's really no need to do anything differently from what the "magician" is doing, no need to import itertools, just keep in mind that user answers are a 1-based index, while arrays are 0-based.
Here's my take on it:
def guess_card(user_rows, card_rows):
possible_cards = card_rows[user_rows[0] - 1]
for card in possible_cards:
if card in card_rows[user_rows[1] + 3]:
return 'Volunteer cheated!'
def main():
n_test_cases = int(raw_input())
while n_test_cases:
n_test_cases -= 1
user_rows = []
card_rows = []
user_rows.append(int(raw_input()))
for i in range(0, 4):
card_rows.append(raw_input().split(' '))
user_rows.append(int(raw_input()))
for i in range(4, 8):
card_rows.append(raw_input().split(' '))
if __name__ == "__main__":
main()
If you're wondering about the if __name__ == "__main__": line, you can have a look here.
• In guess_card you could calculate card_rows[user_rows[1] + 3] at the beginning of the function (and possibly make it a set. In main you could use for _ ni range(n_test_cases) instead of the while loop. – Graipher Feb 3 '17 at 11:45
## Use with for file objects
And generally anything that connects with the outside world. with statements automatically handle clean-up of these pesky objects at the end of the block.
## Use list and dictionary objects where possible
Most of the built-in Python functions apply to these objects. Using them allows you to employ more of the handy Python built-ins.
On that note:
## Some points about Python-specific techniques:
• An input file (as stated in the assignment) is easily converted to a list using a list comprehension.
As in:
file_lines = [i[:-1] for i in open("input.txt")]
You can also use this alongside the with function. Turning the file handler object immediately into a list means you can start using those amazing Python abilities ASAP.
• Use set objects to find matches
The magician's trick works by finding cards that are common to both rows, before and after moving the cards. Let's call them a and b. Once you have both rows as sets, it's easy to find the common cards:
a.intersection(b)
• use slices and split to handle string input data.
Good work for doing this eg. in for n in nums.split. In fact, most of this assignment is simple string manipulation, which doesn't really require a lot of dressing-up in your code.
I've added my short, simple take on the assignment below:
message = ["Bad magician!",0] + ["Volunteer cheated!"]*3
file_lines = [i[:-1] for i in open("input.txt")]
for i in range(int(file_lines[0])):
offset = (i * 10) + 1
a = set(file_lines[offset + 0 + int(file_lines[offset + 0])].split(" "))
b = set(file_lines[offset + 5 + int(file_lines[offset + 5])].split(" "))
message[1] = a.intersection(b)
print("Case #" + str(i + 1) + ": " + str(message[len(message[1])]))
(Note: It's likely an object-oriented version would be most re-useable and adabtable to other projects. If you'd like ideas about that ask me or anyone else here.)
• Thanks for the tips. BTW the output of your code shows a guess number as Case #1: set(['7']), not Case #1: 7, so I edited a little. Can I edit this code more simpler? – philipjkim Feb 7 '17 at 5:31
• You can add [0] after msg[len(msg[1])] in the last line, and add an extra set of square brackets around "Bad magician!" and "Volunteer cheated!". This makes it so the output will be sliced properly regardless of whether it is the card number or the text. It works, but it's not exactly pretty. – Ryan Mills Feb 7 '17 at 6:23
• I'm sure there's a way to do this exercise and make it short AND sweet. We're not there yet. – Ryan Mills Feb 7 '17 at 6:24
|
2020-10-23 02:51:26
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36721813678741455, "perplexity": 1264.0191240621918}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00186.warc.gz"}
|
https://learn.careers360.com/ncert/question-a-field-is-in-the-shape-of-a-trapezium-whose-parallel-sides-are-25-m-and-10-m-the-non-parallel-sides-are-14-m-and-13-m-find-the-area-of-the-field/
|
# Q9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
The trapezium field is shown below in figure:
Drawing line CF parallel to AD and a line perpendicular to AB, we obtain
$CF || AD$ ............................ $\left [ \because\ by\ construction \right ]$
$CD || AF$ ............................ $\left [ \because ABCD\ is\ a\ trapezium \right ]$
So, $AD = CF = 13\ m$ and $CD = AF = 10\ m$
$\left ( \because Opposite\ sides\ of\ a\ parallelogram \right )$
Therefore, $BF = AB-AF = 25-10 = 15\ m$
Now, the sides of the triangle;
$a = 13\ m,\ b =14\ m\ and\ c = 15\ m.$
So, the semi-perimeter of the triangle will be:
$s= \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21\ m$
Therefore, the area of the triangle can be found by using Heron's Formula:
$Area = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{21(21-13)(21-14)(21-15)}$
$= \sqrt{21(8)(7)(6)}$
$= \sqrt{7056} = 84\ m^2.$
Also, the area of the triangle is given by,
$Area = \frac{1}{2}\times BF\times CG$
$\Rightarrow \frac{1}{2}\times BF\times CG = 84\ m^2$
$\Rightarrow \frac{1}{2}\times 15\times CG = 84\ m^2$
Or,
$\Rightarrow CG = \frac{84\times2}{15} = 11.2\ m$
Therefore, the area of the trapezium ABCD is:
$= \frac{1}{2} \times (AB+CD)\times CG$
$= \frac{1}{2} \times (25+10)\times 11.2$
$= 35\times5.6$
$= 196\ m^2$
Hence, the area of the trapezium field is $196\ m^2.$
## Related Chapters
### Preparation Products
##### JEE Main Rank Booster 2021
This course will help student to be better prepared and study in the right direction for JEE Main..
₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021
This course will help student to be better prepared and study in the right direction for NEET..
₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)
An exhaustive E-learning program for the complete preparation of JEE Main..
₹ 4999/-
##### Knockout NEET May 2021
An exhaustive E-learning program for the complete preparation of NEET..
₹ 22999/- ₹ 14999/-
|
2020-10-20 22:41:13
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8465810418128967, "perplexity": 3728.0160483750583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107874340.10/warc/CC-MAIN-20201020221156-20201021011156-00581.warc.gz"}
|
https://merlinblog.xyz/music/tolerate-it.html
|
Perusal
I take your indiscretions all in good fun
ndiscretion
n. 不慎重的举止,轻率的言行
If you talk about someone's indiscretion, you mean that they have done or said something that is risky, careless, or likely to upset people.
①And you, in turn can forgive me for my indiscretions.
②Indiscretion broke both marriage and career.
I sit and listen I polish plates until they gleam and glisten
gleam and glisten
gleam, glare, glisten, glimmer
Glare implies a bright light.
Gleam implies no such brightness.
Glisten implies that something is wet or seems wet. And 'glimmer' implies a dim and wavering light.
Where's that man who'd throw blankets over my barbed wire?
barbed
①A barbed remark or joke seems polite or humorous, but contains a cleverly hidden criticism.（延伸义）
②a barbed hook or arrow has one or more sharp curved points on it.
①That's all barbed wire around there.
②Back then, the movie drew acclaim for its imaginative storyline and barbed commentary.
I made you my temple, my mural, my sky
mural
n. 壁画；（美）壁饰
A mural is a picture painted on a wall.
①One of the walls enclosing the park is decorated with a huge mural showing Hollywood stars.
②The murals have words like " Let Us Breathe" and " God is Love" .
Drawing hearts in the byline
byline
n. 标题下署名之行；铁路支线；副业
vt. 署名
A byline is a line at the top of an article in a newspaper or magazine giving the author's name.
①One day, I'm gonna see your byline in some magazine or newspaper, and I'm gonna know I was lucky enough to have loved that girl.
②I didn't see your name on the byline.
Took this dagger in me and removed it
dagger
Threw out our cloaks and our daggers because it's morning now
used to describe an exciting story involving secrets and mystery, often about spies, or something that makes you think of this.
I'm tired of all these cloak-and-dagger (= secretive) meetings - let's discuss the issues openly.
GsouCloud·全专线机场
MZFast机场|高性价比
N3RO|专业带机场|专属优惠
VeePN骨折优惠码SABRINA1551
Last modification：January 13th, 2021 at 11:15 am
If you think my article is useful to you, please feel free to appreciate
|
2021-05-15 17:24:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20714841783046722, "perplexity": 11711.045422720628}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00218.warc.gz"}
|
http://math.stackexchange.com/questions/236694/integer-solutions-to-linear-equation-triangle-with-set-perimeter
|
# Integer solutions to linear equation – Triangle with set perimeter
We have a triangle with the sides a, b and c where:
/ \
/ \
a / \ b
/ \
/ \
/________________\
c
$\ P = a+b+c$
$\ a+b > c$ else it's not a triangle
$\ a,b,c < P$ no side can be negative
P is known and I'm trying to find all integer solutions of (a,b,c).
By plotting $\ c = P - a - b$ we get a plane where a triangle shows all positive solutions. The solutions where $\ a+b > c$ valid is the the bottom part of this triangle. More I don't know.
EDIT
I'm looking for all solutions, not only the number of solutions. And if $\ P=12$ I think (3,4,5) and (4,3,5) should be treated like different solutions.
Edit 2
I made an algorithm with C. With help of solutions for a range of vaules for P (See picture) I came up with the following rules:
(note: calling the sides x, y and z now, and now I treat (3,4,5) to be equal with (5,3,4))
Z will vary from (1) to (2)
(1) $Z_0=floor(P/2)$
(2) $Z_m=ceil(P/3)$
For a certain Z will X vary from (3) to (4)
(3) $X_0=Z$
(4) $X_m=floor((P-Z)/2)$
-
I get that you are looking for the count for a given $P$ of the number of solutions in positive integers of $x+y+z=P$ for which $x+y>z$ and $y+z>x$ and $z+x>y$. This is likely not too hard to get a formula for, but it makes a difference whether you are counting the ordered triples $(x,y,z)$ or simply the multisets $\{x,y,z\}$ for your solution count. For example if $P=12$ are you counting $(3,4,5)$ as different from $(4,3,5)$ or not? – coffeemath Nov 13 '12 at 21:24
Are you trying to find all the integer solutions, or will you settle for the number of integer solutions? The latter is given at this link.
EDIT: I'll indicate a systematic way to list all the solutions for $P=101$, and leave it to you to code it:
50 50 1; 50 1 50; 1 50 50.
50 49 2; 50 2 49; 49 50 2; 49 2 50; 2 50 49; 2 49 50.
50 48 3; etc.
.
.
.
49 49 3; 49 3 49; 3 49 49.
49 48 4; 49 4 48; 48 49 4; 48 4 49; 4 49 48; 4 48 49.
.
.
.
34 34 33; 34 33 34; 33 34 34.
-
Thanks! But I'm looking for all solutions, not only the count. – Johannes Lund Nov 14 '12 at 8:13
If you're looking to list all the solutions, then it's more of a programming problem than a mathematics problem. Suggest you visit the coding site. – Gerry Myerson Nov 14 '12 at 10:55
|
2016-05-07 01:15:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5513628125190735, "perplexity": 382.15056106870196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461864953696.93/warc/CC-MAIN-20160428173553-00180-ip-10-239-7-51.ec2.internal.warc.gz"}
|
https://nbi.ku.dk/english/staff/?pure=en%2Fpublications%2Foneloop-partition-functions-in-deformed-mathcaln4-sym-theory(09be457d-f8ba-4f94-a041-5e93901128a0).html
|
## One-Loop Partition Functions in Deformed $\mathcal{N}=4$ SYM Theory
Research output: Contribution to journalJournal articleResearchpeer-review
### Documents
We study the thermodynamic behaviour of the real $\beta$- and $\gamma_i$-deformation of $\mathcal{N}=4$ Super Yang-Mills theory on $\mathbb{R}\times S^3$ in the planar limit. These theories were shown to be the most general asymptotically integrable supersymmetric and non-supersymmetric field-theory deformations of $\mathcal{N}=4$ Super Yang-Mills theory, respectively. We calculate the first loop correction to their partition functions using an extension of the dilatation-operator and P\'{o}lya-counting approach. In particular, we account for the one-loop finite-size effects which occur for operators of length one and two. Remarkably, we find that the $\mathcal{O}(\lambda)$ correction to the Hagedorn temperature is independent of the deformation parameters, although the partition function depends on them in a non-trivial way.
Original language English 018 Journal of High Energy Physics (Online) 2015 03 32 1126-6708 https://doi.org/10.1007/JHEP03(2015)018 Published - 2015 Yes
### Research areas
• hep-th, hep-lat, hep-ph
ID: 227489112
|
2023-01-30 04:24:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39228111505508423, "perplexity": 1125.8605648670919}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00768.warc.gz"}
|
https://www.thestudentroom.co.uk/showthread.php?t=3224117
|
# Derivatives and integrals of Fourier series
Watch
Announcements
#1
I've got the series and the derivatives and integrals. How can I show what they sum to?
Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.
If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...
1
5 years ago
#2
(Original post by Goods)
I've got the series and the derivatives and integrals. How can I show what they sum to?
Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.
If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...
It is a fact that the Fourier series of the integral is the termwise integral of the Fourier series, and the Fourier series of the derivative is the termwise derivative of the Fourier series, assuming converges. All you need is uniform convergence of the series, and that condition turns out to be sufficient. More explicitly, you can use the Weierstrass M-test.
0
#3
(Original post by Smaug123)
It is a fact that the Fourier series of the integral is the termwise integral of the Fourier series, and the Fourier series of the derivative is the termwise derivative of the Fourier series, assuming converges. All you need is uniform convergence of the series, and that condition turns out to be sufficient. More explicitly, you can use the Weierstrass M-test.
are those an and bn of the original series?
can i quote as a standard result then as regards the term wise derivative?
0
5 years ago
#4
(Original post by Goods)
are those an and bn of the original series?
can i quote as a standard result then as regards the term wise derivative?
It is, yes. You probably want to say "by the Weierstrass M-test" or something just to dress it up a bit.
0
5 years ago
#5
(Original post by Goods)
I've got the series and the derivatives and integrals. How can I show what they sum to?
Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.
If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...
I have not done highly analytical stuff on fourier but I can tell you that
|x| differentiates to signum(x)
|x| integrates to 1/2 x2 signum(x)
both can be written as piecewise continuous functions.
the only thing you have to be careful is creating an a0 term when performing these operations, which does not apply to this example as |x| is even
0
5 years ago
#6
(Original post by Smaug123)
More explicitly, you can use the Weierstrass M-test.
Not sure how the M-test works here, the obvious choice for M_n is O(1/n) and then diverges.
Edit: in fact, I don't think the M-test *can* work here. If it does, then the sum converges uniformly, and so the result is continuous. But the derivative of |x| is not continuous...
0
5 years ago
#7
(Original post by DFranklin)
Not sure how the M-test works here, the obvious choice for M_n is O(1/n) and then diverges.
Edit: in fact, I don't think the M-test *can* work here. If it does, then the sum converges uniformly, and so the result is continuous. But the derivative of |x| is not continuous...
Ah. Curses.
0
5 years ago
#8
FWIW, from wording of the question I don't think proof is required. There is actually a relevant theorem: given a continuous and piecewise-smooth function f, differentiating the Fourier series for f gives the correct Fourier series for f', but I think the proof is quite technical.
0
#9
(Original post by DFranklin)
FWIW, from wording of the question I don't think proof is required. There is actually a relevant theorem: given a continuous and piecewise-smooth function f, differentiating the Fourier series for f gives the correct Fourier series for f', but I think the proof is quite technical.
is |x| continuous and piecewise-smooth? and what are the criteria for it to be so?
0
5 years ago
#10
(Original post by Goods)
is |x| continuous and piecewise-smooth? and what are the criteria for it to be so?
0
X
new posts
Back
to top
Latest
My Feed
### Oops, nobody has postedin the last few hours.
Why not re-start the conversation?
see more
### See more of what you like onThe Student Room
You can personalise what you see on TSR. Tell us a little about yourself to get started.
### Poll
Join the discussion
#### Current uni students - are you thinking of dropping out of university?
Yes, I'm seriously considering dropping out (49)
16.23%
I'm not sure (8)
2.65%
No, I'm going to stick it out for now (103)
34.11%
I have already dropped out (4)
1.32%
I'm not a current university student (138)
45.7%
|
2020-10-22 07:00:51
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9245184063911438, "perplexity": 790.345787145998}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00568.warc.gz"}
|
http://paws-public.wmflabs.org/paws-public/47830195/00%20-%20Intro%20to%20Jupyter.ipynb
|
# Introduction to Jupyter¶
Jupyter is a web version of the python shell. It's useful to write text like this, or can also write python code and execute them block by block. It can also be used for other languages like Julia, R, Java, Javascript, and so on (pretty much anything!). This notebook introduces the basic blocks that help us use the notebook.
It provides an interactive interface to be able to run blocks of code and also to write documentation to understand the code.
# 1. Cells¶
Jupyter's basic building block is a cell. There are majorly two different types of cells in jupyter:
• Code cells - Code cells have code which is executed by the kernel. Hence, the also have an output section right underneath the cell.
• Markdown cells - Markdown cells are parsed into HTML and using markdown stylization.
Focusing on cells: Cells can be selected by clicking on them. The currently focused cell is shown with a blue border. It is also possible to change the focused cell by using the arrow keys.
# 2. Editing cells¶
There are two modes that are available on Jupyter notebooks (Similar to vim or emacs):
• Edit mode - If a cell is in an edit mode, it is possible to type content into it. When a cell is in the edit mode, it has a **green border**.
• Command mode - If a cell is in the command mode, it can't be edited. But it's easier to move, copy, cut, run, etc. and perform operations on the cell itself. When a cell is in a command mode, it has a **blue border**.
### Creating a new cell¶
To create a new cell, you can either:
• Choose the "Insert" > "Insert cell below" or "Insert cell above" option in the menubar on the top.
• Type b to add a cell below the focused cell. Type a to add a cell above the focused cell. This needs to be done in the command mode.
### Deleting a cell¶
To delete a cell, you can either:
• Choose the "Edit" > "Delete a cell" option in the menubar on the top.
• Type dd to delete the cell in the command mode. (2 times d)
# 3. Running Code¶
Code cells allow you to enter and run code in them, unline markdown cells.
Run a code cell by:
1. Focus on the cell by clicking on it.
2. Then, press Shift-Enter or press the button in the toolbar above to run the cell.
# This creates a variable a. Hence, there's no output.
a = 10
print("The value of a is =", a)
There are two other keyboard shortcuts for running code:
• Alt-Enter runs the current cell and inserts a new cell below the one that was executed.
• Ctrl-Enter run the current cell and enters the edit mode of the cell to edit it.
You can check if a cell is running by looking on it's left. If it shows In [*]:, it means that the cell is still being run. Once the cell has completed running, it is assigned a number. For example: In [4]:. The number denotes the order in which the cells are run. Hence, if 4 is shown, the cell was the fourth to be run in the notebook.
Code cells show the output of the code below the cell. If there is no output (Like in a = 10) there is no output shown. If there is an error, it shows the error with helpful debugging information instead:
a =
# 4. The Kernel¶
All code is run using a kernel. The kernel is a separate process which links the notebook with the compiler/interpreter of the appropriate type (based on the language). The Kernel can be interrupted or restarted.
When you stop the execution of a kernel, python throws a KeyboardInterrupt exception.
Try running the following cell. It pauses and sleeps for 30 seconds. While it is running, hit the (stop) button in the toolbar above. It should give the KeyboardInterrupt exception:
import time
time.sleep(30)
## Restarting the kernels¶
The kernel maintains the state of a notebook's computations. You can reset this state by restarting the kernel. This is done by clicking on the in the toolbar above. On doing this, all your stored variables will be deleted from the session and will need to be redefined.
# 5. Output formatting¶
## Handling sys.stdout and sys.stderr¶
The stdout and stderr streams are displayed as text in the output area.
print("hi, stdout")
from __future__ import print_function
import sys
print('hi, stderr', file=sys.stderr)
## Output is asynchronous¶
All output is displayed asynchronously as it is generated in the Kernel. If you execute the next cell, you will see the output one piece at a time, not all at the end.
import time, sys
for i in range(5):
print(i)
time.sleep(0.5)
## Large outputs¶
Beyond a certain point, output will scroll automatically:
for i in range(500):
print(2**i)
|
2018-08-16 00:39:41
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22823914885520935, "perplexity": 2113.256036247483}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221210387.7/warc/CC-MAIN-20180815235729-20180816015729-00168.warc.gz"}
|
https://stats.stackexchange.com/questions/89717/multivariate-data-analyis-of-compositional-data
|
# Multivariate data analyis of compositional data
Suppose I have a multivariate, compositional dataset that depicts the concentration of different elements. However, the data are not available on a single scale; i.e., some are of form 0.00x while others are integers. Should I apply any kind of normalization / standardization technique before applying, or do I need to do anything with the data at all before apply transformations (isometric log ratio [ilr], centered log ratio [clr], etc.) and start the data analysis, imputation on missing value using robust methods, robust PCA and data clustering?
Some pointers for understanding compositional data analysis are also welcomed.
[Update]
For example: Consider two vectors:
[ 0.016, 71.2, 0.123, 1.74, 14.0, 0.002, 2310, 0.064, 0.29, 0.32,5.63, 96.5, 0.044,
4360, 1110, 585, 0.052, 62.9, 4.45, 1110, 1.50, 15.10, 783, 0.015,78.9, 5.61, 0.007,
0.022, 0.007, 0.53, 29.3 ]
[ 0.073, 245.0, 0.299, 2.77, 17.4, 0.039, 2460, 0.145, 0.85, 0.99, 20.40, 359.0 0.062,
4040, 1530, 148, 0.113, 217.0, 18.10, 1310, 4.61, 4.56, 880, 0.069, 230.0, 12.20,
0.028, 0.025, 0.013, 9.92, 34.1]
These two represent concentrations of different elements in soil samples collected from two different positions. If I were to analyse them using robust methods, should I preprocess them standardize / normalize in any form, or should I simply transform them into Aitchison geometry and start my analysis?
• This sentence: "But data is not available on a single scale i.e. some are of form 0.00x while others are integers" is a bit cryptic, especially if your data (as you write below) are concentrations of different elements. Why integers? Why ratios? What are the values? – amoeba says Reinstate Monica Mar 13 '14 at 14:22
|
2019-12-07 07:19:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8089509010314941, "perplexity": 3555.6026739201147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540496492.8/warc/CC-MAIN-20191207055244-20191207083244-00324.warc.gz"}
|
https://projecteuclid.org/euclid.aoap/1382447693
|
## The Annals of Applied Probability
### Second order reflected backward stochastic differential equations
#### Abstract
In this article, we build upon the work of Soner, Touzi and Zhang [Probab. Theory Related Fields 153 (2012) 149–190] to define a notion of a second order backward stochastic differential equation reflected on a lower càdlàg obstacle. We prove existence and uniqueness of the solution under a Lipschitz-type assumption on the generator, and we investigate some links between our reflected 2BSDEs and nonclassical optimal stopping problems. Finally, we show that reflected 2BSDEs provide a super-hedging price for American options in a market with volatility uncertainty.
#### Article information
Source
Ann. Appl. Probab., Volume 23, Number 6 (2013), 2420-2457.
Dates
First available in Project Euclid: 22 October 2013
https://projecteuclid.org/euclid.aoap/1382447693
Digital Object Identifier
doi:10.1214/12-AAP906
Mathematical Reviews number (MathSciNet)
MR3127940
Zentralblatt MATH identifier
1303.60049
#### Citation
Matoussi, Anis; Possamai, Dylan; Zhou, Chao. Second order reflected backward stochastic differential equations. Ann. Appl. Probab. 23 (2013), no. 6, 2420--2457. doi:10.1214/12-AAP906. https://projecteuclid.org/euclid.aoap/1382447693
#### References
• [1] Avellaneda, M., Lévy, A. and Paras, A. (1995). Pricing and hedging derivative securities in markets with uncertain volatilities. Appl. Math. Finance 2 73–88.
• [2] Bally, V., Caballero, M. E., Fernandez, B. and El Karoui, N. (2002). Reflected BSDEs, PDEs and variational inequalities. Preprint inria-00072133.
• [3] Bayraktar, E. and Huang, Y. J. (2011). On the multi-dimensional controller and stopper games. Preprint. Available at arXiv:1009.0932.
• [4] Bayraktar, E., Karatzas, I. and Yao, S. (2010). Optimal stopping for dynamic convex risk measures. Illinois J. Math. 54 1025–1067.
• [5] Bayraktar, E. and Yao, S. (2011). Optimal stopping for non-linear expectations—Part I. Stochastic Process. Appl. 121 185–211.
• [6] Bayraktar, E. and Yao, S. (2011). Optimal stopping for non-linear expectations—Part II. Stochastic Process. Appl. 121 212–264.
• [7] Bismut, J.-M. (1973). Conjugate convex functions in optimal stochastic control. J. Math. Anal. Appl. 44 384–404.
• [8] Chen, Z. and Peng, S. (2000). A general downcrossing inequality for $g$-martingales. Statist. Probab. Lett. 46 169–175.
• [9] Cheridito, P., Soner, H. M., Touzi, N. and Victoir, N. (2007). Second-order backward stochastic differential equations and fully nonlinear parabolic PDEs. Comm. Pure Appl. Math. 60 1081–1110.
• [10] Dellacherie, C. and Meyer, P.-A. (1975). Probabilités et Potentiel. Hermann, Paris.
• [11] Denis, L. and Martini, C. (2006). A theoretical framework for the pricing of contingent claims in the presence of model uncertainty. Ann. Appl. Probab. 16 827–852.
• [12] Ekren, I., Touzi, N. and Zhang, J. (2012). Optimal stopping under nonlinear expectation. Preprint. Available at arXiv:1209.6601.
• [13] El Karoui, N., Kapoudjian, C., Pardoux, E., Peng, S. and Quenez, M. C. (1997). Reflected solutions of backward SDE’s, and related obstacle problems for PDE’s. Ann. Probab. 25 702–737.
• [14] El Karoui, N., Pardoux, E. and Quenez, M. C. (1997). Reflected backward SDEs and American options. In Numerical Methods in Finance. Publications of the Newton Institute 13 215–231. Cambridge Univ. Press, Cambridge.
• [15] Fabre, E. (2012). Some contributions to stochastic control and backward stochastic differential equations in finance. Ph.D. thesis, Ecole Polytechnique.
• [16] Fahim, A., Touzi, N. and Warin, X. (2011). A probabilistic numerical method for fully nonlinear parabolic PDEs. Ann. Appl. Probab. 21 1322–1364.
• [17] Fremlin, D. H. (1984). Consequences of Martin’s Axiom. Cambridge Tracts in Mathematics 84. Cambridge Univ. Press, Cambridge.
• [18] Hamadène, S. (2002). Reflected BSDE’s with discontinuous barrier and application. Stoch. Stoch. Rep. 74 571–596.
• [19] Hamadène, S. and Popier, A. (2012). $L^{p}$-solutions for reflected backward stochastic differential equations. Stoch. Dyn. 12 1150016, 35.
• [20] Karandikar, R. L. (1995). On pathwise stochastic integration. Stochastic Process. Appl. 57 11–18.
• [21] Karatzas, I. and Sudderth, W. D. (1999). Control and stopping of a diffusion process on an interval. Ann. Appl. Probab. 9 188–196.
• [22] Karatzas, I. and Sudderth, W. D. (2001). The controller-and-stopper game for a linear diffusion. Ann. Probab. 29 1111–1127.
• [23] Karatzas, I. and Zamfirescu, I.-M. (2008). Martingale approach to stochastic differential games of control and stopping. Ann. Probab. 36 1495–1527.
• [24] Lepeltier, J. P., Matoussi, A. and Xu, M. (2005). Reflected backward stochastic differential equations under monotonicity and general increasing growth conditions. Adv. in Appl. Probab. 37 134–159.
• [25] Lepeltier, J. P. and Xu, M. (2005). Penalization method for reflected backward stochastic differential equations with one r.c.l.l. barrier. Statist. Probab. Lett. 75 58–66.
• [26] Matoussi, A. (1997). Reflected solutions of backward stochastic differential equations with continuous coefficient. Statist. Probab. Lett. 34 347–354.
• [27] Nutz, M. (2012). Pathwise construction of stochastic integrals. Electron. Commun. Probab. 17 1–7.
• [28] Pardoux, É. and Peng, S. G. (1990). Adapted solution of a backward stochastic differential equation. Systems Control Lett. 14 55–61.
• [29] Peng, S. (1999). Monotonic limit theorem of BSDE and nonlinear decomposition theorem of Doob–Meyer’s type. Probab. Theory Related Fields 113 473–499.
• [30] Peng, S. (2010). Nonlinear expectations and stochastic calculus under uncertainty. Preprint. Available at arXiv:1002.4546.
• [31] Peng, S. and Xu, M. (2005). The smallest $g$-supermartingale and reflected BSDE with single and double $L^{2}$ obstacles. Ann. Inst. Henri Poincaré Probab. Stat. 41 605–630.
• [32] Possamai, D. and Zhou, C. (2010). Second order backward stochastic differential equations with quadratic growth. Preprint. Available at arXiv:1201.1050.
• [33] Qian, Z. and Xu, M. (2011). Skorohod equation and reflected backward stochastic differential equations. Preprint. Available at arXiv:1103.2078.
• [34] Soner, H. M., Touzi, N. and Zhang, J. (2012). Wellposedness of second order backward SDEs. Probab. Theory Related Fields 153 149–190.
• [35] Soner, H. M., Touzi, N. and Zhang, J. (2013). Dual formulation of second order target problems. Ann. Appl. Probab. 23 308–347.
• [36] Stroock, D. W. and Varadhan, S. R. S. (1979). Multidimensional Diffusion Processes. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences] 233. Springer, Berlin.
• [37] Vorbrink, J. (2010). Financial markets with volatility uncertainty. Finance 1–39.
|
2019-07-23 20:31:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3714359998703003, "perplexity": 6096.278006296501}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529664.96/warc/CC-MAIN-20190723193455-20190723215455-00153.warc.gz"}
|
https://wizedu.com/questions/193/use-the-table-of-standard-reduction-potentials
|
In: Chemistry
# Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction:
Standard reduction potentials
Reduction half-reaction E∘ (V) Ag+(aq)+e−→Ag(s) 0.80 Cu2+(aq)+2e−→Cu(s) 0.34 Sn4+(aq)+4e−→Sn(s) 0.15 2H+(aq)+2e−→H2(g) 0 Ni2+(aq)+2e−→Ni(s) −0.26 Fe2+(aq)+2e−→Fe(s) −0.45 Zn2+(aq)+2e−→Zn(s) −0.76 Al3+(aq)+3e−→Al(s) −1.66 Mg2+(aq)+2e−→Mg(s) −2.37
Part A
Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction:
Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s)
Part B
Calculate the standard cell potential (E∘) for the reaction
X(s)+Y+(aq)→X+(aq)+Y(s)
if K = 3.04×10−3.
Express your answer to three significant figures and include the appropriate units.
## Solutions
##### Expert Solution
Concepts and reason The question is divided into 2 parts:
First part is about calculating the equilibrium constant using the standard cell potential. It can be done using the following formula:
$$\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=\frac{0.0591 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}$$
The second part is based on calculating the standard cell potential using the given equilibrium constant. It can also be done by the same formula given above.
Fundamentals
Standard cell potential $$\left(\mathrm{E}_{\mathrm{cell}}^{\circ}\right):$$ It is the cell potential of the overall net reaction. It is calculated as follows:
$$\mathrm{E}_{\mathrm{cell}}^{\circ}=\mathrm{E}_{\mathrm{cathode}}^{\circ}-\mathrm{E}_{\mathrm{anode}}^{\circ}$$
Here, $$\mathrm{E}_{\text {cathode }}^{\circ}$$ is the standard reduction potential of the reaction occurring at cathode; and $$\mathrm{E}_{\text {anode }}^{\circ}$$ is the standard reduction potential of the reaction occurring at anode. Relation between standard cell potential and equilibrium constant:
It is known that the relation of standard free energy change, $$\Delta \mathrm{G}^{\circ}$$ are given as follows:
$$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\mathrm{cell}}^{\mathrm{o}} \ldots \ldots(1)$$
And, $$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K} \ldots \ldots .(2)$$
Therefore, equating (1) and (2) as follows:
$$-\mathrm{nFE}_{\text {cell }}^{\mathrm{o}} =-\mathrm{RT} \ln \mathrm{K}$$
$$\mathrm{E}_{\text {cell }}^{\mathrm{o}} =\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{K}$$
At $$298 \mathrm{~K}$$, the equation can be rewritten as follows:
$$\mathrm{E}_{\text {cell }}^{\mathrm{o}}=\frac{0.0591 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}$$
Here, $$\mathrm{n}$$ is the number of electrons participating in the overall reaction; $$\mathrm{K}$$ is the equilibrium constant.
(A)
Consider the given reaction:
$$\mathrm{Fe}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ni}(\mathrm{s})$$
At cathode: $$\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{s}) \quad \mathrm{E}_{\text {cathode }}^{\mathrm{o}}=-0.26 \mathrm{~V}$$
At anode: $$\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \quad \mathrm{E}_{\text {anode }}^{\mathrm{o}}=-0.45 \mathrm{~V}$$
Calculate the standard cell potential as follows:
$$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ}$$
$$=-0.26 \mathrm{~V}-(-0.45 \mathrm{~V})$$
$$\quad=0.19 \mathrm{~V}$$
At cathode, reduction reaction occurs in which $$\mathrm{Ni}^{2+}$$ reduces to $$\mathrm{Ni}$$ and at anode, oxidation reaction occurs in which Fe oxidizes to $$\mathrm{Fe}^{2+}$$
Calculate equilibrium constant as follows:
$$\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=\frac{0.0592}{\mathrm{n}} \log \mathrm{K}$$
$$0.19 \mathrm{~V}=\frac{0.0592}{2} \log \mathrm{K}$$
$$\log \mathrm{K}=6.42$$
$$\mathrm{~K}=2.63 \times 10^{6}$$
Part A The equilibrium constant for the given reaction is found to be .
The standard cell potential for the given reaction is found to be $$0.19 \mathrm{~V}$$. The number of electrons, n, participating in the overall reaction is 2. These values are substituted in the formula, and the equilibrium constant is calculated.
(B) Consider the given reaction:
$$\mathrm{X}(\mathrm{s})+\mathrm{Y}^{+}(\mathrm{aq}) \rightarrow \mathrm{X}^{+}(\mathrm{aq})+\mathrm{Y}(\mathrm{s})$$
Calculate the standard cell potential as follows:
$$\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=\frac{0.0592 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}$$
$$=\frac{0.0592 \mathrm{~V}}{1} \log \left(3.04 \times 10^{-3}\right)$$
$$=-0.149 \mathrm{~V}$$
Part B The standard cell potential for the given reaction is
The equilibrium constant for the reaction is given as $$3.04 \times 10^{-3}$$. The number of electrons, n, participating in the overall reaction is $$1 .$$ These values are substituted in the formula, and standard reduction potential is calculated.
|
2021-02-27 15:54:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9118924140930176, "perplexity": 1156.346492860111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358976.37/warc/CC-MAIN-20210227144626-20210227174626-00622.warc.gz"}
|
http://math.stackexchange.com/questions/210996/latin-squares-proving-the-unique-number-of-sudoku-that-can-be-generated?answertab=oldest
|
# Latin Squares - Proving the Unique number of Sudoku that can be generated
I recently read that the Sudokus are just Latin Squares for $n = 9$. I know that proving the number of Latin Squares is considered difficult to generalize in terms of $n$. I would like to know if there is any method of proving the unique number of Sudokus taking a special case for $n = 9$.
The Wikipedia page on Latin Squares shows that for $n = 9 , number\ of\ possible\ Latin\ Squares = 5524751496156892842531225600$. I would like to know different possible methods used to arrive at the answer (e.g using Brute-force , or generating for the reduced Latin-Squares or any other combinatorial method).
PS: I have also read this question asked here and the paper referred by the answer. I am not able to understand why the method fails for $n>11$. Even the OEIS page does not have sequences for $n>11$
-
Sudokus are a proper subset of the Latin squares of order $9$, as they have the added restriction of the $3 \times 3$ boxes. Thus, all sudokus are Latin squares, but not all Latin squares are sudokus. As with Latin squares, there is no computer-free proof that these numbers are correct. Typically, these are checked by performing independent computations (possibly by slightly different methods).
That being said, most of the searching can be eliminated by identifying symmetries. Both Latin squares, in general, and sudokus have "symmetries", which can be exploited to give a significant reduction in the search space. In a Latin square, for instance, we can permute the rows arbitrarily to give another Latin square (so it'd be a waste of time to count all of these separately). I'm not particularly familiar with the methods used in enumerating sudokus, but a website by Jarvis (and the linked papers) offers much detail into his enumeration method.
There are $R_9=377597570964258816$ reduced Latin squares of order 9. I don't think any computer ever has counted from 1 to $R_9$, let alone play around with Latin squares at each step. Thus, it's safe to say a brute force enumeration is completely out of the question.
To date, the easiest way to find $R_9$ is to use Sade's method which I mention in my answer to the linked question. Sade's method is the only feasible way for order $10$ or greater. Unfortunately, Sade's method was published only in a very obscure paper and is hard to obtain. But I describe it in great detail in my survey paper (here).
Roughly speaking, Sade's method saves an enormous amount of time by identifying Latin rectangles that have the same number of completions, and clumping them together. We then count the number of ways of extending each equivalence class by one row, then identify which of these extended Latin rectangles admit the same number of completions, and so on recursively.
(Note: In an earlier version of this post, I claimed it was the only way for $9 \times 9$ squares, but there is actually another way in this case.)
For $9 \times 9$ Latin squares, it's possible to iterate through representatives of the 19270853541 main classes of Latin squares on a computer, and at each step calculate the size of the autoparatopism group $\mathrm{Par}(L)$. The total number of $9 \times 9$ Latin squares is thus $\sum_L \frac{6n!^3}{|\mathrm{Par}(L)|},$ where the sum is over the 19270853541 representatives. Generating these representatives can be done via the "canonical construction path" described in:
B. D. McKay, Isomorph-free exhaustive generation, J Algorithms, 26 (1998), 306-324.
See this paper for more details (including the relevant definitions: "main class" and "autoparatopism"):
B. D. McKay, A. Meynert, W. Myrvold, Small Latin squares, quasigroups, and loops. J. Combin. Des. 15 (2007), 98-119.
Sade's method for $12 \times 12$ Latin squares would work fine, if we had a sufficiently powerful computer with enough memory (and the budget to use it, and know-how to program it efficiently). I think it's safe to say we (as a species) could find this number in the next 100 years or so (I'm hoping to see $R_{12}$ before I die).
Note that the number of Latin squares grows quite fast. Most people think $n!$ grows quickly. Well, Smetaniuk showed that $L_{n+1} \geq (n+1)!\ L_n$ (where $L_n$ counts non-reduced Latin squares). The sheer number of Latin squares is the obstacle here.
B. Smetaniuk, A new construction of Latin squares. II. The number of Latin squares is strictly increasing, Ars Combin., 14 (1982), pp. 131-145.
-
$R_9 \approx 3e17$. That's no longer an "absurdly large" number for tasks that are parallelizable. The fastest supercomputer runs 20,000 Terraflops, ie $2e16$ floating point operations in one second. – Nick Alger Oct 11 '12 at 11:05
|
2015-04-18 07:26:32
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7520732283592224, "perplexity": 384.59117270159123}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246633972.52/warc/CC-MAIN-20150417045713-00053-ip-10-235-10-82.ec2.internal.warc.gz"}
|
https://emresahin.net/translating-ottoman-turkish-spelling-to-latin-alphabet-24600/
|
dervaze is a project I have started back in my Ph.D. work in 2015 to translate Ottoman Turkish to modern Turkish spelling and providing an OCR/ICR/handwriting recognition engine for Ottoman language.
The reason I had to stop was the lack of data, since without some considerable amount of data, statistical methods for both Natural Language Processing and Computer Vision fails. Producing and maintaining data seemed a much more important burden than having technical solutions, so I mostly gave up the idea that a working solution is obtainable with the classical OCR techniques. The research still waits me to finish.
I begin this series of explorations in Computer Vision, Machine Learning and related fields to document my achievements and provide some basis for further research. Time to time, I will provide results regarding the ideas here. Currently this is mostly a hobby/side project.
My current endeavour is to write a robust translation engine between Ottoman and Turkish in Dart.1
Instead of making a full fledged morphological analysis like TRMorph, Google Research Morphological Analyzer or Starlang Morphological Analyzer, our aim is to provide a surface level to surface level translation between Arabic and Turkish letters.
It is true that it's possible to use something like:
+-----------------+ +----------------+ +---------------+
| | | | | |
| | | | | |
| kelimeler | +-------> | kelime+PLU | +-----------> | کلمهلر |
| | | | | |
| | | | | |
+-----------------+ +----------------+ +---------------+
to translate between Turkish Latin and Ottoman, but there are two problems here:
1. There is no Ottoman morphological analyzer and although grammatically Turkish, surface level forms should be translated. Even Turkish Latin ones are relatively recent.
2. Translating the output of a Turkish Latin morphological analyzer to Ottoman seems a work more than writing a translation method itself.
For example for the query kelimeler TRMorph gives
kelime<N><pl>
kelime<N><0><V><cpl:pres><3p>
kelime<N><pl><0><V>
kelime<N><pl><0><V><cpl:pres><3p>
kelime<N><pl><0><V><cpl:pres><3s>
and although most of the items' surface forms are identical, we need to work through all suffixes and their different kinds of connections. Also, as the surface form of Ottoman Turkish has less information, Ottoman morphological analysis would yield much more results than the corresponding counterpart.
Because of these hindrances and trying to come up with a quick-and-dirty yet workable solution, I have made the following observations:
1. Our part-of-speech system should not need to be highly specialized. Actually we need to have only three classes: Nouns, verbs and proper nouns. Proper nouns are grammatically nouns but their ortography may require different rules. We will call these three word classes.
2. We can get away with a set of surface level rules to translate suffixes for each word class. These rules use attributes that can be derived from Turkish Latin ortography.
These rules are:
• Part of Speech: The root class we discussed above
• Last Vowel: The last vowel in a Turkish root that is required to find the actual suffix when vowel harmony is utilized.
• Last Consonant: The last consonant of the Turkish root. This is needed to find the palatalization/softening of certain consonants.
• Ends with Vowel: Whether the root ends with vowel like ata or not.
• Has Single Vowel: Whether the root has a single vowel. This is important in aorist inflection irregularities in verbs.
• Last Vowel Hard: Whether the last vowel is one of a, ı, o, u or not.
• Last Consonant Hard: Whether the last consonant is one of p, ç, t, k and undergoes softening when it receives a suffix starting with a vowel.
• Has Consonant Softening: Inverse of the last consonant hard rule, when a root ends with b, c, d, g and happens to receive a suffix.
From these rules which can be derived from the Turkish Latin forms of the words using regular expressions, we can write rules to translate Turkish to Ottoman and vice versa.
TBC.
1
Dart has this little benefit to be able to run both on mobile (in Android and iOS) and servers with support for the web, so write once, run on Linux, iOS and Android is feasible without much architecture jumping. I hope Flutter gains more traction and becomes the mainstream way of writing mobile applications.
|
2023-02-04 14:29:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3396390676498413, "perplexity": 2612.588444718544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00675.warc.gz"}
|
https://tsahota.github.io/NMproject/articles/a01_config.html
|
## Prerequisites
The following need to be installed and available:
• PsN >= 4.4.8
• NONMEM installed with valid license
• RStudio
## First time set up
### Set up code completion
It is highly recommended to set up code completion though as it eases the scripting process.
setup_code_completion()
This will ask for user consent as it will add entries to the user level snippets file.
### Verify installation
The following function checks for availability of PsN via system_nm(), this is a requirement for NMproject. There are also subsequent checks which not mandatory, but recommended for full functionality.
check_installation()
#> Test passed 🎉
Users working on a standalone machines (e.g. laptops and desktops) should expect these tests to pass and for NMproject to work out of the box (provided the pre-requisites are met).
## Custom configuration
More complex environments involving multiple servers and grids may require more configuration. The options shown here aim to facilitate that. You may need the help of an admin.
### NMTRAN checks
If working in a grid environment, it is highly recommended to perform NMTRAN checks prior to sending jobs off the grid. NMTRAN check provides immediate feedback of control file and dataset errors allowing you to address problems quickly just like if you were on a laptop/desktop environment. Without this, having to wait for a job to reach the grid and fail is a much slower (and more annoying) process.
NMproject tries to guess the location to the NMTRAN file in the NONMEM installation directory however if this fails and check_installation() cannot detect it, you will need to specify the NMTRAN command/file with the nm_tran_command() function. See the following for instructions and examples at the bottom of the help file:
nm_tran_command("/opt/NONMEM/nm75/tr/NMTRAN.exe")
To see more examples see ?nm_tran_command.
This setting will reset upon R session restart, if you want to make it permanent see Making configuration changes permanent.
### The system_nm() command
Under the hood, all calls to PsN proceed through the system_nm() command. This is a user configurable function. By default it’s the same as system() on unix systems and shell() on windows. However if the NONMEM server is remote and requires connection (e.g. via SSH), users may want to modify the behaviour of system_nm(). This is possible through setting the option() also named system_nm, e.g. the following allows runs the system() command on a remote machine named clustername:
options(system_nm=function(cmd,...) {
### Via a custom package
If you already have a package you maintain to deliver functions to users, this package can be used to configure set up options via the .onLoad hook. For example, the following will configure custom NMTRAN and code library settings on package load
.onLoad <- function(libname, pkgname){
## custom NMTRAN location
NMproject::nm_tran_command("/opt/NONMEM/nm75/tr/NMTRAN.exe")
## custom code library location
options(code_library_path = "/projects/myorganisation/code_library")
## message to users in interactive sessions
if (interactive()) usethis::ui_done("NMproject configured")
}
The advantage of a package is you can also bundle other functions in there for users to use.
|
2022-05-26 18:40:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3092429041862488, "perplexity": 4926.757914866391}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00483.warc.gz"}
|
https://www.semanticscholar.org/paper/BEER%3A-Fast-O(1%2FT)-Rate-for-Decentralized-Nonconvex-Zhao-Li/b2db41fe686fd81282e693fefd7646545f5ca2dd
|
• Corpus ID: 246430634
# BEER: Fast O(1/T) Rate for Decentralized Nonconvex Optimization with Communication Compression
@article{Zhao2022BEERFO,
title={BEER: Fast O(1/T) Rate for Decentralized Nonconvex Optimization with Communication Compression},
author={Haoyu Zhao and Boyue Li and Zhize Li and Peter Richt'arik and Yuejie Chi},
journal={ArXiv},
year={2022},
volume={abs/2201.13320}
}
• Published 31 January 2022
• Computer Science
• ArXiv
Communication efficiency has been widely recognized as the bottleneck for large-scale decentralized machine learning applications in multi-agent or federated environments. To tackle the communication bottleneck, there have been many efforts to design communication-compressed algorithms for decentralized nonconvex optimization, where the clients are only allowed to communicate a small amount of quantized information (aka bits) with their neighbors over a predefined graph topology. Despite significant…
7 Citations
## Figures and Tables from this paper
• Computer Science
• 2022
This work formalizes the multi-token semi-decentralized scheme, which subsumes the client-server and decentralized setups, and design a feature-distributed learning algorithm for this setup, which can be seen as a parallel Markov chain (block) coordinate descent algorithm.
• Computer Science
ArXiv
• 2022
This paper proposes a coreset framework by constructing coresets in a distributed fashion for communication-efficient VFL, and theoretically shows that using coresets can drastically alleviate the communication complexity, while nearly maintain the solution quality.
• Computer Science
ArXiv
• 2022
This work proposes and analyzes several stochastic gradient algorithms for finding stationary points or local minimum in nonconvex, possibly with nonsmooth regularizer, finite-sum and online optimization problems, and proposes an optimal algorithm, called SSRGD, based on SARAH, which can find an -approximate (first-order) stationary point by simply adding some random perturbations.
• Computer Science
ArXiv
• 2022
A framework called SoteriaFL is proposed, which accommodates a general family of local gradient estimators including popular stochastic variance-reduced gradient methods and the state-of-the-art shifted compression scheme, and is shown to achieve better communication complexity without sacrificing privacy nor utility than other private federated learning algorithms without communication compression.
• Computer Science
ArXiv
• 2022
A convergence lower bound for algorithms whether using unbiased or contractive compressors in unidirection or bidirection is established and an algorithm is proposed, NEOLITHIC, which almost reaches the lower bound (up to logarithm factors) under mild conditions.
• Computer Science
SIAM J. Math. Data Sci.
• 2022
A new algorithm, called DEcentralized STochastic REcurSive gradient methodS (DESTRESS) for nonconvex optimization, which matches the optimal incremental incremental oracle complexity of centralized algorithms for stationary points, while maintaining communication ef-ficiency.
## References
SHOWING 1-10 OF 64 REFERENCES
• Computer Science
IEEE Journal on Selected Areas in Information Theory
• 2021
Theoretical understanding is corroborated with experiments and the performance of the algorithm is compared with the state-of-the-art, showing that without sacrificing much on the accuracy, SQuARM-SGD converges at a similar rate while saving significantly in total communicated bits.
• Computer Science
ICLR
• 2020
The use of communication compression in the decentralized training context achieves linear speedup in the number of workers and supports higher compression than previous state-of-the art methods.
• Computer Science
• 2019
This paper proposes an elegant algorithmic design to employ error-compensated stochastic gradient descent for the decentralized scenario, named DeepSqueeze, and is the first time to apply the error-Compensated compression to the decentralized learning.
• Computer Science
NIPS
• 2017
Quantized SGD is proposed, a family of compression schemes for gradient updates which provides convergence guarantees and leads to significant reductions in end-to-end training time, and can be extended to stochastic variance-reduced techniques.
• Computer Science
ICML
• 2019
This work presents a novel gossip-based stochastic gradient descent algorithm, CHOCO-SGD, that converges at rate $\mathcal{O}\left(1/(nT) + 1/(T \delta^2 \omega)^2\right)$ for strongly convex objectives, where $T$ denotes the number of iterations and $\delta$ the eigengap of the connectivity matrix.
• Computer Science
NeurIPS
• 2018
This work analyzes Stochastic Gradient Descent with k-sparsification or compression (for instance top-k or random-k) and shows that this scheme converges at the same rate as vanilla SGD when equipped with error compensation.
This paper studies a D-PSGD algorithm and provides the first theoretical analysis that indicates a regime in which decentralized algorithms might outperform centralized algorithms for distributed stochastic gradient descent.
• Computer Science
NeurIPS
• 2021
It is proved that EF21 enjoys a fast O(1/T ) convergence rate for smooth nonconvex problems, beating the previous bound of O( 1/T ), which was shown under a strong bounded gradients assumption.
• Computer Science
IEEE Transactions on Automatic Control
• 2022
It is shown that C-GT inherits the advantages of gradient tracking-based algorithms and achieves linear convergence rate for strongly convex and smooth objective functions.
• Computer Science, Mathematics
SIAM J. Optim.
• 2022
Over infinite time horizon, it is established that all nodes in GT-SARAH asymptotically achieve consensus and converge to a first-order stationary point in the almost sure and mean-squared sense.
|
2023-02-02 12:32:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7004068493843079, "perplexity": 2702.7570405995352}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00516.warc.gz"}
|
https://stats.stackexchange.com/questions/539283/garch-use-the-normal-errors-formula-build-an-iterative-equation-for-predicting
|
# GARCH Use the Normal errors formula, build an iterative equation for predicting the variances in R
My arma model had residual from which I created GARCH model in R.My question is how using Normal errors formula, build an iterative equation for predicting the variances & figure a way to obtain the last ah and the last σh from the output.
I know variance formula in GARCH is given by Sigma(t)=omega+alphae(t-1)+Bsigma(t-1)
I am confused as to how to find value for e(t-1) and sigma(t-1)
I am sharing my R code below for reference:
#Generating data for 4 years
data1<-getSymbols("KO",from="2017-08-01", to="2021-08-01",auto.assign = F)
#Checking for Unit root/stationary
adf.test(data1$KO.Adjusted) logdata1<-na.omit(diff(log(data1$KO.Adjusted)))
#Order of my arma model is (4,1)
arma_model<-arima(logdata1,c(4,0,1))
#Checking for presence of residual
arma_model$residuals #Creating standard garch with arma order(4,1) and distribution as normal x=ugarchspec(variance.model=list(garchOrder=c(1,1)),mean.model=list(armaOrder=c(4,1)),distribution.model = "norm") I am confused how to proceed forward and how to calculate variance and last ah and the last σh. Kindly advise how to proceed forward. Explain using R. • Use ugarchforcast (for one step ahead prediction) and ugarchboot (for$h>1\$ prediction). Aug 6, 2021 at 0:08
• Also relate to stats.stackexchange.com/questions/201933/… for some formula Aug 6, 2021 at 0:10
• @Richard Hardy I saw your old post stats.stackexchange.com/questions/201933/… your input will help me out.Kindly advise how to calculate variance and last ah and the last σh in R?My code is above Aug 6, 2021 at 22:04
• I saw your post but am busy with other stuff. Sorry. Aug 7, 2021 at 5:55
|
2022-09-26 17:44:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5909269452095032, "perplexity": 3216.030264332958}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334912.28/warc/CC-MAIN-20220926144455-20220926174455-00429.warc.gz"}
|
https://math.stackexchange.com/questions/2731638/conditions-for-extending-convergence-in-distribution-of-a-subsequence-to-the-ent
|
# Conditions for extending convergence in distribution of a subsequence to the entire sequence
Suppose I have a sequence for which I can prove convergence in distribution of some infinite subsequence, e.g. let $\{X_n: n \geq 1\}$ be the original sequence and suppose I can show that
$$X_{n_k} \to X \textrm{ in distribution as } k \to \infty$$ for a subsequence $n_k$ satisfying $n_k \to \infty$ as $k \to \infty$.
Question: What is the weakest set of conditions on the entire sequence I can impose in order to be able to extend the convergence of the subsequence back up to the whole subsequence level?
I would imagine this would be true if we could bound the deviation of the elements between the elements of the subsequence, but I am just spitballing. For example, something like: for any $k$, $$\sup_{n_k \leq n \leq n_{k+1}} \left| X_n - X_{n_k} \right| \to 0 \textrm{ in probability as } k \to \infty$$
$\def\dto{\xrightarrow{\mathrm{d}}}\def\Pto{\xrightarrow{P}}\def\peq{\mathrel{\phantom{=}}{}}$A stronger proposition can be proved under your condition.
Suppose $\{X_t \mid t > 0\}$ is a sequence of random variables and $\{X_{t_n} \mid n \in \mathbb{N}_+\}$ is a subsequence such that $t_n \nearrow ∞$. If $X_{t_n} \dto X\ (n → ∞)$ and $\sup\limits_{t_n < t < t_{n + 1}} |X_t - X_{t_n}| \Pto 0\ (n → ∞)$, then $X_t \dto X\ (t → ∞)$.
Proof: Define $W_n = \sup\limits_{t_n < t < t_{n + 1}} |X_t - X_{t_n}|$, $C = \{x \in \mathbb{R} \mid F_X \text{ is continuous at }x\}$. Note that $\mathbb{R} \setminus C$ is at most countable, thus $C$ is dense in $\mathbb{R}$.
Consider fixed $x_0 \in C$. Take two sequences $\{y_m\}, \{z_m\} \subseteq C$ such that $y_m \nearrow x_0\ (m → ∞)$, $z_m \searrow x_0\ (m → ∞)$. Because for any $m \in \mathbb{N}_+$ and $t \geqslant t_1$, suppose $t_n \leqslant t < t_{n + 1}$, then\begin{align*} P(X_t \leqslant x_0) &= P(X_t \leqslant x_0) - P(X_{t_n} \leqslant y_m) + P(X_{t_n} \leqslant y_m)\\ &\geqslant -P(X_t > x_0,\ X_{t_n} \leqslant y_m) + P(X_{t_n} \leqslant y_m)\\ &\geqslant -P(|X_t - X_{t_n}| > x_0 - y_m) + P(X_{t_n} \leqslant y_m)\\ &\geqslant -P(W_n > x_0 - y_m) + P(X_{t_n} \leqslant y_m). \end{align*} Since $t_n \nearrow ∞$ and $W_n \Pto 0$, make $t → ∞$ to get$$\varliminf_{t → ∞} P(X_t \leqslant x_0) \geqslant -\lim_{n → ∞} P(W_n > x_0 - y_m) + \lim_{n → ∞} P(X_{t_n} \leqslant y_m) = P(X \leqslant y_m),$$ then make $m → ∞$ to get$$\varliminf_{t → ∞} P(X_t \leqslant x_0) \geqslant P(X \leqslant x_0).$$ Analogously,$$\varlimsup_{t → ∞} P(X_t \leqslant x_0) \leqslant P(X \leqslant x_0),$$ thus $\lim\limits_{n → ∞} P(X_t \leqslant x_0) = P(X \leqslant x_0)$. Therefore, $X_t \dto X\ (t → ∞)$.
|
2021-05-13 03:25:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9898125529289246, "perplexity": 429.68120266942987}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00432.warc.gz"}
|
http://pubman.mpdl.mpg.de/pubman/faces/viewItemOverviewPage.jsp?itemId=escidoc:2085240
|
de.mpg.escidoc.pubman.appbase.FacesBean
Deutsch
Hilfe Wegweiser Datenschutzhinweis Impressum Kontakt
# Datensatz
DATENSATZ AKTIONENEXPORT
Freigegeben
Zeitschriftenartikel
#### Neutrino physics with multi-ton scale liquid xenon detectors
##### MPG-Autoren
http://pubman.mpdl.mpg.de/cone/persons/resource/persons73111
Marrodán Undagoitia, Teresa
Division Prof. Dr. Manfred Lindner, MPI for Nuclear Physics, Max Planck Society;
##### Volltexte (frei zugänglich)
1309.7024.pdf
(Preprint), 774KB
##### Ergänzendes Material (frei zugänglich)
Es sind keine frei zugänglichen Ergänzenden Materialien verfügbar
##### Zitation
Baudis, L., Ferella, A., Kish, A., Manalaysay, A., Marrodán Undagoitia, T., & Schumann, M. (2014). Neutrino physics with multi-ton scale liquid xenon detectors. Journal of Cosmology and Astroparticles Physics, 2014(1): 044. doi:10.1088/1475-7516/2014/01/044.
We study the sensitivity of large-scale xenon detectors to low-energy solar neutrinos, to coherent neutrino-nucleus scattering and to neutrinoless double beta decay. As a concrete example, we consider the xenon part of the proposed DARWIN (Dark Matter WIMP Search with Noble Liquids) experiment. We perform detailed Monte Carlo simulations of the expected backgrounds, considering realistic energy resolutions and thresholds in the detector. In a low-energy window of 2-30 keV, where the sensitivity to solar pp and $^7$Be-neutrinos is highest, an integrated pp-neutrino rate of 5900 events can be reached in a fiducial mass of 14 tons of natural xenon, after 5 years of data. The pp-neutrino flux could thus be measured with a statistical uncertainty around 1%, reaching the precision of solar model predictions. These low-energy solar neutrinos will be the limiting background to the dark matter search channel for WIMP-nucleon cross sections below $\sim$2$\times$10$^{-48}$ cm$^2$ and WIMP masses around 50 GeV$\cdot$c$^{-2}$, for an assumed 99.5% rejection of electronic recoils due to elastic neutrino-electron scatters. Nuclear recoils from coherent scattering of solar neutrinos will limit the sensitivity to WIMP masses below $\sim$6 GeV$\cdot$c$^{-2}$ to cross sections above $\sim$4$\times$10$^{-45}$cm$^2$. DARWIN could reach a competitive half-life sensitivity of 5.6$\times$10$^{26}$ y to the neutrinoless double beta decay of $^{136}$Xe after 5 years of data, using 6 tons of natural xenon in the central detector region.
|
2018-07-23 10:17:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8592885732650757, "perplexity": 10878.671996741454}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676596204.93/warc/CC-MAIN-20180723090751-20180723110751-00554.warc.gz"}
|
http://kleine.mat.uniroma3.it/mp_arc-bin/mpa?yn=14-5
|
Below is the ascii version of the abstract for 14-5. The html version should be ready soon.
Amadeu Delshams, Marina Gonchenko, Pere Guti\'errez
Exponentially small lower bounds for the splitting of separatrices to whiskered tori with frequencies of constant type
(62K, LATeX 2e)
ABSTRACT. We study the splitting of invariant manifolds of whiskered tori with two frequencies in nearly-integrable Hamiltonian sys
tems, such that the hyperbolic part is given by a pendulum. We consider a 2-dimensional torus with a fast frequency vecto
r $\omega/\sqrt arepsilon$, with $\omega=(1,\Omega)$ where $\Omega$ is an irrational number of constant type, i.e. a num
ber whose continued fraction has bounded entries. Applying the Poincar\'e-Melnikov method, we find exponentially small lo
wer bounds for the maximal splitting distance between the stable and unstable invariant manifolds associated to the invar
iant torus, and we show that these bounds depend strongly on the arithmetic properties of the frequencies.
|
2017-11-20 21:05:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9229540824890137, "perplexity": 557.4257753390361}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00714.warc.gz"}
|
https://gilkalai.wordpress.com/2019/10/13/gerard-cornuejolss-bakers-eighteen-5000-dollars-conjectures/
|
Gérard Cornuéjols’s baker’s eighteen 5000 dollars conjectures
Gérard Cornuéjols
Gérard Cornuéjols‘s beautiful (and freely available) book from 2000 Optimization: Packing and Covering is about an important area of combinatorics which is lovely described in the preface to the book
The integer programming models known as set packing and set covering have a wide range of applications, such as pattern recognition, plant location and airline crew scheduling. Sometimes, due to the special structure of the constraint matrix, the natural linear programming relaxation yields an optimal solution that is integer, thus solving the problem. Sometimes, both the linear programming relaxation and its dual have integer optimal solutions. Under which conditions do such integrality properties hold? This question is of both theoretical and practical interest. Min-max theorems, polyhedral combinatorics and graph theory all come together in this rich area of discrete mathematics. In addition to min-max and polyhedral results, some of the deepest results in this area come in two flavors: “excluded minor” results and “decomposition” results. In these notes, we present several of these beautiful results. Three chapters cover min-max and polyhedral results. The next four cover excluded minor results. In the last three, we present decomposition results.
The last sentence of the preface gives this post some urgency
In particular, we state 18 conjectures. For each of these conjectures, we offer \$5000 as an incentive for the first correct solution or refutation before December 2020.
The book starts with Konig’s theorem, the first figure is the Petersen graph, and among the other mathematical heroes mentioned in the book are Edmonds, Johnson, Seymour, Lovász, Lehman, Camion, Tutte, and Truemper.
The title of this post refers to Baker’s dozen. In the 13th century Bakers who were found to have shortchanged customers could be liable to severe punishment, and to guard against the punishment of losing a hand to an axe, a baker would give 13 for the price of 12, to be certain of not being known as a cheat. (Wikipedia) In this post we mention a 19th problem for which Gerard offered 5000 dollars. (I am not sure if there is time limit for that problem. I am thankful to Maria Chudnovsky for telling me about the problem.)
What happened to the 18 problems in the list?
Perhaps the most difficult problem on the list was solved first: two of the problems in the list were about perfect graphs and were settled with the solution of the strong perfect graph conjecture by Chudnovsky, Robertson, Seymour, and Thomas. Three of the problems were about balanced bipartite graphs. They were solved by Chudnovsky and Seymour in 2006. Conjecture 4.14 in Chapter 4 was solved by Jonathan Wang (2010) 30,000 dollars were thus collected and 60,000 dollars are still offered (until Dec 2020).
Balanced bipartite graphs and the 19th problem.
Balanced bipartite graphs are sort of bipartite analogs of perfect graphs. They are bipartite graphs so that every induced cycle have length divisible by four. Gerard’s 19the prize money problem is also about balanced bipartite graphs.
Conjecture: Let G be balanced. Then there is $e \in E(G)$ such that $G\backslash e$ is a balanced graph.
In other words every balanced bipartite graph contains an edge which is not a unique chord in any cycle.
This conjecture is Conjecture 5.20 in
M. Conforti, G. Cornuejos, K. Vuskovic Balanced matrices
In that paper, this conjecture is attributed to:
M. Conforti and M.R Rao “Structural properties and decomposition of linear
balaced matrices”, Mathematical Programming 55 (1992) 129-168.
This entry was posted in Combinatorics, Computer Science and Optimization, Open problems and tagged , , , , . Bookmark the permalink.
3 Responses to Gérard Cornuéjols’s baker’s eighteen 5000 dollars conjectures
1. Gil Kalai says:
Let me share with you (with the writer’s permission) an email I received regarding the post:
Hi Prof. Kalai,
My name is Jourdain Lamperski, and I am PhD student at MIT. I hope all is well!
I just enjoyed reading your recent blog post on Gerard’s book. I have two remarks that you might find interesting:
1. The 19th problem to some extent appears as a conjecture in Gerard’s book (Conjecture 6.11, pg. 80), but it is stated in terms of balanced matrices.
2. Very coincidentally, two days before your blog post, I posted a solution to the 19th conjecture for the case of linear balanced graphs (the context in which it was originally conjectured in the cited 1992 paper by Conforti and Rao). Here is a link http://www.mit.edu/~jourdain/balanced.pdf if interested. I have actually already gone through the result with Gerard in person and he thought it was nice.
I thought the timing of our posts was unusually coincidental so I just thought I’d reach out!
Best,
Jourdain
|
2020-09-30 00:45:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6084772944450378, "perplexity": 1132.5590959782774}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00592.warc.gz"}
|
https://math.stackexchange.com/questions/843519/for-an-odd-prime-p-p-1-pmod-4-x21-0-pmod-p
|
For an odd prime $p$, $\;p =1 \pmod 4,\; \;x^2+1 = 0 \pmod p$
$$\text{ If } p \neq 2 \text{ is a prime, we know that: }$$ $$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$$
According to this,prove that: $$p \equiv 1 \pmod 4 \Rightarrow x^2+1\equiv 0 \pmod p \text{ has a solution}$$ $$\text{Which is the solution?}$$ Solve: $\displaystyle{ x^2+1 \equiv 0 \pmod{29} \text{ , } x^2+1 \equiv 0 \pmod{37}}$
Inversely,prove that $x^2+1 \equiv 0 \pmod p \text{ has a solution } \Rightarrow p \equiv 1 \pmod 4$
I have thought the following:
• $$\left(\left(\frac{p-1}{2}\right)!\right)^2-(-1)^{\frac{p+1}{2}} \equiv 0 \pmod p \Rightarrow \left(\left(\frac{p-1}{2}\right)!\right)^2+(-1)^{\frac{p+3}{2}} \equiv 0 \pmod p$$
As $p=4k+1$, $(-1)^{\frac{p+3}{2}}=(-1)^{\frac{4k+4}{2}}=1$
So,we have: $$\left(\left(\frac{p-1}{2}\right)!\right)^2+1 \equiv 0 \pmod p$$
So,we conclude that $x^2+1 \equiv 0 \pmod p \text{ has a solution , this one: } (\frac{p-1}{2})!$
Is it right or have I done something wrong?
• So,are the solutions of $x^2+1 \equiv 0 \pmod{29} \text{ and } x^2+1 \equiv 0 \pmod{37}$,these one: $\displaystyle{(14)! \text{ and } (18)! \text{ respectively}}$ ? Are these the only solutions??
• How can I show the inverse?
• You want to get your answer in the interval $(1,p-1).$ In fact, you can get $x \in (1,\frac{p-1}{2}).$ Remember, it is only the answer (mod $p$) that is important. The answer $18!$ is correct (mod $37$) but $18!$ is a ridiculously large integer, and you should try to simplify it. – Geoff Robinson Jun 22 '14 at 13:57
• @GeoffRobinson Do you mean that I have to write $18!$ like that:$1 \cdot 2 \cdots 18$ and then simplify the numbers?Also are $14! \text{ and } 18!,$ the only solutions of $x^2+1 \equiv 0 \pmod{29} \text{ and } x^2+1 \equiv 0 \pmod{37}$ respectively?? – evinda Jun 22 '14 at 14:22
• @GeoffRobinson Do you maybe know how I could prove the inverse? – evinda Jun 22 '14 at 19:48
• Hint for last question If $x^{2} = -1$, then $x$ has multiplicative order $4$ in the group of non-zero elements of the field under multiplication. Yes,, you should simplify the factorials, though from a theoretical point of view it is not essential. Second question, the answer are $\pm 14!$ and $\pm 18!$ respectively. – Geoff Robinson Jun 22 '14 at 22:11
• @GeoffRobinson I haven't get taught multiplicative orders...Isn't there an other way to show it? And why do we know that there are only these 2 solutions? – evinda Jun 22 '14 at 22:24
That seems correct to me. The last part comes from the fact that the equation $x^2 +1 \equiv 0$ (mod $p$) has at most 2 distinct solutions because it is a degree 2 polynomial and $p$ is prime. And if $x$ is a solution, then so is $-x$.
• And how to we conclude from this that $p=4k+1$ ? – evinda Jun 22 '14 at 13:50
• If $p \equiv -1$ (mod $4$), then if $x^2 \equiv -1$ (mod $p$), then $-1 \equiv (-1)^{\frac{p-1}{2}} \equiv x^{p-1} \equiv 1$ (mod $p$) by Fermat's Little Theorem - a contradiction – Mathmo123 Jun 22 '14 at 13:51
• You have already used the condition $p=4k+1$. If $p=4k+3$, you will get $\left(\frac{p-1}{2}\right)!^2\equiv 1\pmod p$, not really helpful. There are actually no solutions to $x^2\equiv -1\pmod p$ if $p=4k+3$, but you will probably use a bit different route to show that. – Ian Mateus Jun 22 '14 at 13:55
• I haven't really understood why the prime $p$ cannot be of the form $4k+3$..Could you explain it further to me???? – evinda Jun 22 '14 at 14:23
• @evinda If $p\equiv 3 \pmod 4$, there are no solutions to $x^2\equiv -1\pmod p$. We say that $-1$ is a quadratic nonresidue modulo $p$. I don't know how much number theory you have seen, but it shouldn't be too difficult to you to learn Euler's criterion and then show (using the Legendre symbol) $$\left(\frac{-1}{p}\right)\equiv (-1)^{(p-1)/2} \pmod p.$$ That is, $-1$ is a quadratic nonresidue modulo $p$ if $p\equiv 3\pmod 4$. – Ian Mateus Jul 1 '14 at 19:03
|
2019-08-22 10:07:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8727061152458191, "perplexity": 214.7512416274911}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00318.warc.gz"}
|
https://direct.mit.edu/jocn/article/31/3/431/28961/Visual-Simultaneity-Judgments-Activate-a-Bilateral
|
Abstract
In everyday life, we often make judgments regarding the sequence of events, for example, deciding whether a baseball runner's foot hit the plate before or after the ball hit the glove. Numerous studies have examined the functional correlates of temporal processing using variations of the temporal order judgment and simultaneity judgment (SJ) tasks. To perform temporal order judgment tasks, observers must bind temporal information with identity and/or spatial information relevant to the task itself. SJs, on the other hand, require observers to detect stimulus asynchrony but not the order of stimulus presentation and represent a purer measure of temporal processing. Some previous studies suggest that these temporal decisions rely primarily on right-hemisphere parietal structures, whereas others provide evidence that temporal perception depends on bilateral TPJ or inferior frontal regions (inferior frontal gyrus). Here, we report brain activity elicited by a visual SJ task. Our methods are unique given our use of two orthogonal control conditions, discrimination of spatial orientation and color, which were used to control for brain activation associated with the classic dorsal (“where/how”) and ventral (“what”) visual pathways. Our neuroimaging experiment shows that performing the SJ task selectively activated a bilateral network in the parietal (TPJ) and frontal (inferior frontal gyrus) cortices. We argue that SJ tasks are a purer measure of temporal perception because they do not require observers to process either identity or spatial information, both of which may activate separate cognitive networks.
INTRODUCTION
Human perception requires the integration of multiple sensory signals that are often presented in rapid sequential order or simultaneously. The ability to accurately perceive the temporal properties of sensory signals is critical to everyday human behavior, as this ability helps us reorient attention, prioritize responses, and act accordingly to our environment. For example, temporal information helps referees call sports games, dancers coordinate their actions, and musicians stay in sync. Additional evidence is necessary to understand the precise nature of brain networks underlying visual temporal information processing in healthy adults.
The significance of temporal processing may be seen when we observe deficits in individuals with neurological impairments such as visual extinction (Karnath & Zihl, 2003). These individuals only report a single item when two are presented simultaneously or report that the contralesional item appeared with an artificial delay. These individuals are often unaware of their impairments, exhibiting a potentially dangerous level of anosognosia (Vossel, Weiss, Eschenbeck, & Fink, 2013). The consequences of these impairments are varied and may include inaccurate duration judgments, poor temporal order judgments, and an inability to accurately engage attention to temporally sequenced stimuli (Rorden, Li, & Karnath, 2018). Here, we focus on visual order judgments made by neurologically healthy individuals performing a simultaneity task.
Neuroimaging research on visual temporal perception has been partly motivated by patient data with lesions to parietal or ventral frontal brain regions (Roberts, Lau, Chechlacz, & Humphreys, 2012). Patient studies provide us with valuable insight into which brain regions may be involved in various behaviors. Of these studies, most have focused on patients with the inability to interpret the order of sensory events or engage their attention to contemporaneous stimuli after injuries such as unilateral stroke (i.e., exaggerated attentional blink; Husain, Shapiro, Martin, & Kennard, 1997). Previous research has suggested that the TPJ may play a crucial role in integrating information across space and time (for a review, see Husain & Rorden, 2003). Battelli and colleagues boldly suggest that only the right hemisphere TPJ forms a dedicated “when” pathway (Battelli, Alvarez, Carlson, & Pascual-Leone, 2009; Battelli, Walsh, Pascual-Leone, & Cavanagh, 2008; VanRullen, Pascual-Leone, & Battelli, 2008), located between the heavily studied dorsal “where”/“how” pathway and the ventral “what” pathway (Creem & Proffitt, 2001; Goodale & Milner, 1992). However, there are conflicting reports as to which brain regions support visual temporal perceptions like order judgment or simultaneity. For example, seminal work by Coull and Nobre (1998) suggests that temporal tasks preferentially engage the left hemisphere, whereas others have produced results in favor of the right hemisphere being most involved (Agosta et al., 2017). Still, others argue that temporal perception engages a bilateral system in either parietal (and/)or inferior frontal regions (Davis, Christie, & Rorden, 2009; Lux, Marshall, Ritzl, Zilles, & Fink, 2003). It is important to note that the wide range of anatomical discrepancies might be explained by the various designs of behavioral tasks employed (Agosta et al., 2017).
We hypothesize that visual temporal information is encoded bilaterally in both inferior parietal and inferior frontal regions. This idea builds on previous work mentioned above but also assumes a more inclusive, bilateral network of brain regions supporting the perception of visual temporal information. The present article extends our knowledge by reporting on the brain correlates of visual processing during a simultaneity judgment (SJ) task, while explicitly controlling for extraneous perceptual processes in the spatial and identity processing domains. Parsing the contribution of many perceptual processes (e.g., spatial processing, identity processing, temporal processing) is a difficult task in many experiments. We chose the SJ task for its simplicity and because it allowed us to keep instructions and stimuli as similar as possible across all tasks.
Furthermore, our SJ experiment (including control tasks) was designed to be adaptive on a trial-by-trial basis. This crucial element is novel compared to similar experiments mentioned above. The adaptive nature of our task design maintains participants at their individual perceptual threshold throughout the entire experiment.
METHODS
Participants
The current experiment included 35 participants (28 women, mean age = 22 ± 4 years) who were recruited from the University of South Carolina (Columbia, South Carolina) and surrounding areas. All study procedures were reviewed and approved by the local institutional review board, and each participant provided written consent using documents approved by the institutional review board. All participants self-reported right-hand dominance, had normal or corrected-to-normal visual acuity, and were neurologically healthy by self-report. All participants were screened for MRI compatibility. Participants were paid $20 for their participation and were told that the top performer would receive an additional$80 to increase motivation.
We also included a previously collected data set consisting of 26 participants (16 women, mean age = 23 ± 5.84 years) who were recruited and consented in the same manner as above. Six of the participants in this data set were left-handed via self-report. Going forward, this data set of 26 participants is referred to as the out-of-sample data used in the machine learning classification analysis.
Stimuli and Procedure
Stimuli were created and presented using Psychtoolbox (Kleiner et al., 2007) on a Windows 7 computer connected to a projector with a long throw lens aimed at a screen inside the scanner room (1024 × 768 px). Stimuli were made isoluminant for each participant with the use of a visual flicker fusion thresholding procedure where they adjusted stimulus color intensities until flickering ceased. The lighting and conditions inside the MRI room were consistent throughout the entire study.
Figure 1.
Actual trial screenshots from the behavior tasks performed during MRI scanning (modified for print). Participants performed each of the tasks in a pseudorandom block design. Each trial lasted 1.8 sec, with 500-msec stimulus duration, 1100 msec for response, and 200-msec intertrial interval. Timing parameters were constant across tasks. Before each block, a word appeared on-screen for 1000 msec to indicate the task: time, angle, or color. A centrally located character (“T,” “A,” or “C”) provided a fixation point to reduce eye movement. This character remained on-screen during each block and did not provide any information relevant to the judgment task other than the current block type.
Figure 1.
Actual trial screenshots from the behavior tasks performed during MRI scanning (modified for print). Participants performed each of the tasks in a pseudorandom block design. Each trial lasted 1.8 sec, with 500-msec stimulus duration, 1100 msec for response, and 200-msec intertrial interval. Timing parameters were constant across tasks. Before each block, a word appeared on-screen for 1000 msec to indicate the task: time, angle, or color. A centrally located character (“T,” “A,” or “C”) provided a fixation point to reduce eye movement. This character remained on-screen during each block and did not provide any information relevant to the judgment task other than the current block type.
Stimuli were presented in a pseudorandomized block order so that one task would never repeat more than twice in a row. Each block lasted 29 sec and contained 16 individual trials each lasting 1.8 sec. Total stimulus duration from onset to offset was ∼500 msec (30 frames at 60-Hz presentation). Each block condition was shown seven times, totaling 21 stimulus blocks per fMRI run, with an accompanying rest block after each stimulus block that lasted 15 sec. For each participant, two fMRI runs were performed while in the scanner, with each block condition being presented 14 times across fMRI runs. Each participant performed a training session in the scanner during the T1 scan before the fMRI runs. This training session was used to build the starting logistic model for each participant used in the adaptive thresholding procedure.
Both of these experiments make use of an SJ task but differ in notable ways. The early experiment (with rectangles as stimuli) is referred to as our out-of-sample data. The two experiments do not contain any of the same participants. The data from this early experiment were only used to test the functional model developed from the later experiment via the machine learning classification procedures described later.
Our current version of the SJ paradigm we have developed was designed with great care to control for task difficulty across the different tasks. The early experiment was limited in that task difficulty was not as rigorously controlled via the PEST procedure. However, the early data set is still rich with information related to visual temporal information processing, and therefore we chose to use it as our out-of-sample data to predict task based on the functional activation model derived from the newer, current data set.
Imaging Protocol and Analysis
All imaging took place at the McCausland Center for Brain Imaging located at the Palmetto Health Richland Hospital (Columbia, SC) using a Siemens Prisma 3-T MRI system fitted with a 32-channel head coil. High-resolution structural images were obtained using a T1-weighted 3-D magnetized prepared rapid gradient echo scan with the following parameters: repetition time (TR) = 2400 msec, echo time (TE) = 2.24 msec, inversion time = 1060 msec, flip angle = 8°, field of view (FOV) = 167 × 240 × 256 mm, voxel size = 0.8 mm isotropic, bandwidth = 210 Hz/px, iPAT factor of 2, and acquisition time = 6:38 min. Two fMRI series, each with 1,360 volumes, were collected using x8 multiband (Xu et al., 2013), gradient-echo EPI, TR = 720 msec, TE = 37 msec, flip angle = 52°, FOV = 208 × 208 × 144 mm, slice thickness = 2.0 mm (72 slices, 2.0-mm isotropic voxels), echo spacing = 0.58 msec, and bandwidth = 2290 Hz/px. Single-band fMRI data were also acquired at the beginning of each run. These single-band images had the same acquisition settings as the task fMRI, with the exception that they do not use multiband speed enhancements. In addition, to correct for spatial distortions in the fMRI data, spin-echo images were acquired with the following parameters: TR = 7700 msec, TE = 58 msec, flip angle = 90°, FOV = 208 × 208 × 144 mm, slice thickness = 2.0 mm (72 slices, 2.0-mm isotropic voxels), multiband factor = 1, echo spacing = 0.59 msec, bandwidth = 2290 Hz/px, and acquisition time = 31 sec. fMRI data acquisition for the earlier data set was identical in every way except that 1,370 volumes were collected versus the 1,360 for the current experiment.
All neuroimaging data were analyzed using a combination of FMRIB Software Library (FSL; Jenkinson, Beckmann, Behrens, Woolrich, & Smith, 2012; Smith et al., 2004) and SPM routines. Specifically, the spin-echo images were used in FSL's topup and applytopup to compute and correct for spatial distortions in the fMRI data (Andersson, Skare, & Ashburner, 2003). Each participant's functional data (via single-band reference images) were then coregistered to their T1 anatomical image using the FSL implementation of the boundary-based registration method (Greve & Fischl, 2009). Motion correction was applied using the rigid body MCFLIRT routines in FSL (Jenkinson, Bannister, Brady, & Smith, 2002), and functional data were smoothed using a 5-mm Gaussian kernel. The last preprocessing step was linearly warping each participant's data to the 2-mm Montreal Neurological Institute template brain distributed with FSL using FLIRT with 12 degrees of freedom (Jenkinson & Smith, 2001).
Statistical analyses of the fMRI data were carried out using the GLM as implemented in SPM. Specifically, each participant's normalized functional data were modeled using a boxcar block design with block duration timing parameters specified in the task description section and a high-pass filter of 116 sec. Motion parameters (translation and rotation) were included in the GLM model as nuisance regressors. At the participant level, each participant's two fMRI runs were included in the model, in addition to another nuisance regressor to account for each fMRI run. At the participant level, each pairwise combination of task activation contrasts was modeled. At the group level, the same contrasts were modeled. Thresholded conjunction images were made to indicate significant voxels in each task compared with the others using the simple union of suprathreshold values (see Equations 13) described in Nichols, Brett, Andersson, Wager, and Poline (2005). See Figure 2 caption for thresholding information.
$SJconjunction=SJ>CL∪SJ>OR$
(1)
$CLconjunction=CL>SJ∪CL>OR$
(2)
$ORconjunction=OR>SJ∪OR>CL$
(3)
Figure 2.
(A) 3-D rendering of results from the group-level conjunction analysis shown overlaid on a standard anatomical template image. (B) Axial slices showing the same activation as in A. Red indicates voxels in which BOLD signal was significantly greater during the simultaneity task relative to the color task and significantly greater in the simultaneity task relative to the orientation task (conjunction analysis). These regions included TPJ (including SMG, pSTG, and MTG) and IFG (pars opercularis). Similarly, a conjunction analysis of BOLD signal during the orientation task (green) relative to the other two tasks revealed activation at sites in the LOC and SPL. Finally, the conjunction analysis for the color task (blue) revealed activity in the PCC and ANG. All images for the conjunction analyses were thresholded at the cluster level using p < .05, Bonferroni FWE corrected. Cluster extent thresholds for each group contrast were as follows: SJ > CL (k = 155), SJ > OR (k = 207), OR > SJ (k = 138), OR > CL (k = 126), CL > OR (k = 137), and CL > SJ (k = 136). MTG = middle temporal gyrus; LOC = lateral occipital cortex; SPL = superior parietal lobule; PCC = posterior cingulate cortex; ANG = angular gyrus.
Figure 2.
(A) 3-D rendering of results from the group-level conjunction analysis shown overlaid on a standard anatomical template image. (B) Axial slices showing the same activation as in A. Red indicates voxels in which BOLD signal was significantly greater during the simultaneity task relative to the color task and significantly greater in the simultaneity task relative to the orientation task (conjunction analysis). These regions included TPJ (including SMG, pSTG, and MTG) and IFG (pars opercularis). Similarly, a conjunction analysis of BOLD signal during the orientation task (green) relative to the other two tasks revealed activation at sites in the LOC and SPL. Finally, the conjunction analysis for the color task (blue) revealed activity in the PCC and ANG. All images for the conjunction analyses were thresholded at the cluster level using p < .05, Bonferroni FWE corrected. Cluster extent thresholds for each group contrast were as follows: SJ > CL (k = 155), SJ > OR (k = 207), OR > SJ (k = 138), OR > CL (k = 126), CL > OR (k = 137), and CL > SJ (k = 136). MTG = middle temporal gyrus; LOC = lateral occipital cortex; SPL = superior parietal lobule; PCC = posterior cingulate cortex; ANG = angular gyrus.
Support Vector Machine Analysis
We performed an additional multivariate analysis where we predicted the task from the BOLD signal measured at the ROIs defined by the SJ conjunction analysis. We carried out this multivariate analysis for the SJ versus CL and SJ versus OR contrasts separately. The network was defined as follows: From the SJ conjunction map, the center of mass coordinates of each cluster was computed and used to make spherical ROIs at each cluster location with a 10-mm radius. These ROIs were combined into a single image, which represented a visual temporal perception network and served to select the voxels (features) that were entered into support vector machine (SVM) classification. Within the task blocks, the BOLD signal for each volume was divided by the mean BOLD signal from the last half of the preceding resting block. This normalization procedure was used to normalize the task-related BOLD signal relative to the resting baseline (Schmah et al., 2010; McIntosh & Lobaugh, 2004). After this signal normalization, we computed the mean volume for the entire task block (Ku, Gretton, Macke, & Logothetis, 2008); the values for these averages within the spatial mask were used as observations for classification analysis.
To predict the task, we used SVM with linear kernel, as implemented in the LIBSVM package (Chang & Lin, 2011). The relative BOLD signal from the observations was scaled to [0–1] range: We computed the minimum and maximum values for each feature, subtracted the minimum value from the block averages, and divided by the range (maximum minus minimum). The value for the slack parameter C was selected to optimize the classification accuracy using 10-fold cross-validation on the training set (no data from the test set were used in this optimization procedure). The candidate values for the optimal C were 0.00001, 0.0001, 0.001, 0.01, 0.1, 10, 100, 250, 500, 750, 1,000, 1,500, 2,500, 5,000, and 10,000.
We performed two types of classification analysis: within participant and across participants. For within-participant analysis, prediction was performed using leave-one-block-per-condition-out procedure. To test our prediction, we selected one block average from the first task (SJ) and another block average from the second task (CL or OR) and used the remaining block averages to train the SVM model. This procedure was repeated exhaustively for each possible pairing of block averages from the two tasks. Participant-specific classification accuracy was computed as the average proportion of correctly predicted test cases. This value was then averaged across participants.
For the second type of analysis (across participants), we left out all the observations (block averages) for a particular participant to use for testing, and all block averages for the remaining participants were used for training. After training the SVM model, we predicted the task for each block average of the test participant (28 blocks per participant, 14 for each task). An additional step was performed during the scaling of features into [0–1] range: The minimum and maximum values (described above) were computed across the training participants, and the observations from the test participants were scaled into this range.
RESULTS
Behavioral
A one-way within-participant ANOVA was performed to determine an effect of Task on overall Accuracy. Before the ANOVA, a Fisher exact test was used to filter outlier participants who had significantly different hit–miss ratios for each pairwise combination of tasks. This filter preserved all participants. Task had no significant effect on accuracy measurements, F(2, 68) = 1.95, p = .15, indicating that the adaptive algorithm was able to maintain equal objective difficulty across tasks and within participants. The mean and standard deviation of accuracy measurements across tasks were 0.81 (0.04) for SJs, 0.82 (0.04) for color judgments, and 0.82 (0.04) for orientation judgments. Accuracy means include all task trials where there was a valid response (including catch trials).
Neuroimaging
The conjunction analysis revealed that the timing task (SJ) activated a ventral frontoparietal network including bilateral supramarginal gyrus (SMG), posterior superior temporal gyrus (pSTG), bilateral IFG, and bilateral temporo-occipital areas (but still connected to the larger TPJ cluster). The conjunction map for the OR task showed significant activity in bilateral inferior lateral occipital cortex, and right superior parietal lobule. Finally, the conjunction result for the CL task showed activation of bilateral posterior cingulate cortex and right angular gyrus. Detailed cluster information for all neuroimaging results are listed in Table 1, and Figure 2 shows all conjunction analysis results overlaid on surface- and volume-based brain renderings.
Table 1.
Center of Mass MNI Coordinates for SJ Activation Clusters
RegionxyzCluster Size
L IFG −50.54 8.18 12.73 320
L TPJ −54.32 −48.55 19 691
R IFG 50.22 17.63 8.21 531
R TPJ 56.46 −40.77 22.6 961
RegionxyzCluster Size
L IFG −50.54 8.18 12.73 320
L TPJ −54.32 −48.55 19 691
R IFG 50.22 17.63 8.21 531
R TPJ 56.46 −40.77 22.6 961
Here, we report the center of mass for each ROI that resulted from the conjunction procedure. To create a custom ROI mask, spheres with a 10-mm radius were generated at each of these points. The voxels from these ROIs were used as predictors in the SVM model. MNI = Montreal Neurological Institute.
SVM fMRI Classification
For the within-participant classification analysis, we achieved an accuracy of 99.4% for the SJ versus CL contrast and an accuracy of 99.1% for the SJ versus OR contrast. To see which spatial features were driving this highly accurate prediction, we computed the ROI-specific average of the squared voxel-wise weights from the linear SVM model. These voxel-wise weights represent the importance of each voxel to task prediction; higher values indicate more important voxels. The ROIs were left and right TPJ and left and right IFG. For both the within-participant contrasts of SJ versus CL and SJ versus OR, the most predictive ROI was the right IFG (see Figure 3). In both left and right hemispheres, the IFG contributed more than TPJ. In addition, left and right TPJ regions were similar in their contribution to task prediction.
Figure 3.
Bar chart showing the average absolute feature weight for each ROI obtained from the SJ conjunction map. The range of these values is relative to the scaled fMRI data. No statistical comparisons were made. TPJ ROIs had numerically similar feature weights in the linear SVM model, and the right IFG (RIFG) was the strongest feature in both comparison conditions of SJ versus CL and SJ versus OR. Error bars represent 95% CI of mean within participant weights. LTPJ = left TPJ; LIFG = left IFG; RTPJ = right TPJ.
Figure 3.
Bar chart showing the average absolute feature weight for each ROI obtained from the SJ conjunction map. The range of these values is relative to the scaled fMRI data. No statistical comparisons were made. TPJ ROIs had numerically similar feature weights in the linear SVM model, and the right IFG (RIFG) was the strongest feature in both comparison conditions of SJ versus CL and SJ versus OR. Error bars represent 95% CI of mean within participant weights. LTPJ = left TPJ; LIFG = left IFG; RTPJ = right TPJ.
For the across-participants classification analysis, we could predict the task with 61.3% accuracy in the SJ versus CL contrast (the task was predicted correctly for 598 of 980 total block averages). As expected, this value is lower than the accuracy achieved during the within-participant analysis. However, it is still significantly higher than chance probability of 50% (p = 3e-12, assuming binomial distribution with equal likelihood of both tasks). The tasks in the other contrast (SJ vs. OR), however, could not be predicted successfully; accuracy was 44.4%.
Finally, we used the ROIs generated from the group GLM results of the current data set (n = 35) to predict task from fMRI volumes in a separate out-of-sample data set (n = 26) with similar experimental procedures and identical fMRI acquisition parameters. As a group, individuals in this prior study found the simultaneity task easier than the color task (72% vs. 65% accuracy, respectively) indicating that accuracy was not as balanced as that in the current study. However, the average within-participant classification accuracy (predicted using the ROIs from our conjunction analysis) for this second data set was 99.8% for predicting the task (timing judgments or color judgments). Therefore, although this previous study differed in a few ways and was not as balanced for task difficulty, classifiers trained on an independent, controlled data set still performed exceptionally well for predicting behavior given that the task demands were similar.
DISCUSSION
The experiment presented above aimed to investigate the functional anatomy underlying visual temporal perception. We employed a temporal perception task using an SJ paradigm where participants were required to judge if two visual stimuli appeared as simultaneous or not when presented via our custom ramped illumination procedure. The neuroimaging data from the temporal task were then compared with data from the two control tasks of color and orientation discriminations. Crucially, these control tasks used perceptually similar stimuli with the exception that only the attended stimulus property was manipulated from one block to the next and the unattended properties remained constant (e.g., if attending to SJ for a block, then CL and OR were not manipulated). Note that the prior version of this experiment (out-of-sample data) was different in that it did manipulate the unattended stimulus properties to be either congruent (e.g., same color) or incongruent. One could argue that the former paradigm reduces unattended congruence/ incongruence biases whereas the latter preserves identical stimulus properties across trials (albeit these differences are near perceptual threshold). Regardless of this difference in experimental design, the results of our conjunction analysis revealed ROIs that were used to reliably predict unmodeled, out-of-sample data with an average within-participant accuracy of 99%. Our statistical analysis of the imaging data revealed that performing the SJ task selectively activated bilateral TPJ (including SMG and pSTG) and bilateral frontal (IFG) regions after using a conjunction analysis to account for the activation induced by our comparison tasks of CL and OR.
Our current study was also able to overcome the limitation of variable difficulty across tasks for the duration of the experiment, which may have influenced previous studies on timing behavior (Miyazaki et al., 2016; Battelli et al., 2009; Davis et al., 2009). The solution to the variable task difficulty problem was to construct an individual logistic regression response model for each participant on each task. During the fMRI runs, every trial added a new data point to the model and was used to continuously update the participant's individual model to suggest an estimate for the next stimulus value to use per task. This method was used on all three tasks. Our results presented earlier suggest that this method was able to maintain difficulty across tasks within a participant. Furthermore, participants' anecdotal reports of which task was harder were varied, suggesting no clear trend.
The present experiment, as well as that of its most similar counterparts (Davis et al., 2009; Lux et al., 2003), identifies bilateral regions activated by visual temporal processing using two different tasks (SJs vs. TOJ judgments in the case of Davis et al.). Although we may view SJs as the purer of the two visual timing tasks, both behaviors do undoubtedly show some similarity in their functional anatomy. Specifically, inferior frontal regions and TPJ are often reported in the fMRI temporal perception literature (at least within the visual domain). It is possible that the bilateral patterns observed in fMRI activation measurements reflect the strong homologous connections present (e.g., the left hemisphere is involved but not required for temporal decisions). This hypothesis could be directly addressed by future experiments using brain stimulation (in conjunction with the crucial involvement of an fMRI functional localizer task) to disrupt performance in the novel behavioral paradigm we describe here.
Finally, we wish to point out that the temporal perception network we report here (bilateral TPJ regions and IFG) is nearly identical to that of the ventral frontoparietal attention network. This network is largely thought to be involved in the detection of unattended or low-frequency events, regardless of their spatial location or modality of sensation (Corbetta & Shulman, 2002). It is possible that our neuroimaging results can be explained by differences in the extent of which our task (SJs) and the other two tasks (color and orientation discriminations) activate this system. It has been hypothesized that this network is one of the possible anatomical loci for common brain injury deficit known as extinction (Corbetta & Shulman, 2011). On the basis of clinical evidence (de Haan, Karnath, & Driver, 2012; Karnath, Himmelbach, & Küker, 2003; Mort et al., 2003), SJs are a relevant temporal task because of their dissociation from spatial and identity characteristics (compared with TOJ) and that these judgments represent a behavior that is fundamentally impaired in patients with extinction. Here, we provide evidence that activation of the ventral stimulus-driven attention network in healthy participants supports behavioral processes that are fundamentally impaired in patients who have suffered damage to the same underlying areas (see Karnath et al., 2003).
Acknowledgments
This work was supported by the Euro Deutsche Forschungsgemeinschaft (DFG) KA 1258/23-1 and the National Institutes of Health (P50 DC014664).
Reprint requests should be sent to Taylor Hanayik, PhD, Department of Psychology, University of South Carolina, 1512 Pendleton St., Barnwell College, Suite #220, Columbia, SC 29208, or via e-mail: hanayik@gmail.com.
REFERENCES
REFERENCES
Agosta
,
S.
,
Magnago
,
D.
,
Tyler
,
S.
,
Grossman
,
E.
,
Galante
,
E.
,
Ferraro
,
F.
, et al
(
2017
).
The pivotal role of the right parietal lobe in temporal attention
.
Journal of Cognitive Neuroscience
,
29
,
805
815
.
,
J. L. R.
,
Skare
,
S.
, &
Ashburner
,
J.
(
2003
).
How to correct susceptibility distortions in spin-echo echo-planar images: Application to diffusion tensor imaging
.
Neuroimage
,
20
,
870
888
.
Battelli
,
L.
,
Alvarez
,
G. A.
,
Carlson
,
T.
, &
Pascual-Leone
,
A.
(
2009
).
The role of the parietal lobe in visual extinction studied with transcranial magnetic stimulation
.
Journal of Cognitive Neuroscience
,
21
,
1946
1955
.
Battelli
,
L.
,
Walsh
,
V.
,
Pascual-Leone
,
A.
, &
Cavanagh
,
P.
(
2008
).
The “when” parietal pathway explored by lesion studies
.
Current Opinion in Neurobiology
,
18
,
120
126
.
Binder
,
M.
(
2015
).
Neural correlates of audiovisual temporal processing—Comparison of temporal order and simultaneity judgments
.
Neuroscience
,
300
,
432
447
.
Chang
,
C.-C.
, &
Lin
,
C.-J.
(
2011
).
LIBSVM: A library for support vector machines
.
ACM Transactions on Intelligent Systems and Technology
,
2
,
1
27
.
Corbetta
,
M.
, &
Shulman
,
G. L.
(
2002
).
Control of goal-directed and stimulus-driven attention in the brain
.
Nature Reviews Neuroscience
,
3
,
201
215
.
Corbetta
,
M.
, &
Shulman
,
G. L.
(
2011
).
Spatial neglect and attention networks
.
Annual Review of Neuroscience
,
34
,
569
599
.
Coull
,
J. T.
, &
Nobre
,
A. C.
(
1998
).
Where and when to pay attention: The neural systems for directing attention to spatial locations and to time intervals as revealed by both PET and fMRI
.
Journal of Neuroscience
,
18
,
7426
7435
.
Creem
,
S. H.
, &
Proffitt
,
D. R.
(
2001
).
Defining the cortical visual systems: “What,” “where,” and “how.”
Acta Psychologica
,
107
,
43
68
.
Davis
,
B.
,
Christie
,
J.
, &
Rorden
,
C.
(
2009
).
Temporal order judgments activate temporal parietal junction
.
Journal of Neuroscience
,
29
,
3182
3188
.
de Haan
,
B.
,
Karnath
,
H.-O.
, &
Driver
,
J.
(
2012
).
Mechanisms and anatomy of unilateral extinction after brain injury
.
Neuropsychologia
,
50
,
1045
1053
.
Downar
,
J.
,
Crawley
,
A. P.
,
Mikulis
,
D. J.
, &
Davis
,
K. D.
(
2000
).
A multimodal cortical network for the detection of changes in the sensory environment
.
Nature Neuroscience
,
3
,
277
283
.
García-Pérez
,
M. A.
, &
Alcalá-Quintana
,
R.
(
2012
).
On the discrepant results in synchrony judgment and temporal-order judgment tasks: A quantitative model
.
Psychonomic Bulletin & Review
,
19
,
820
846
.
Goodale
,
M. A.
, &
Milner
,
A. D.
(
1992
).
Separate visual pathways for perception and action
.
Trends in Neurosciences
,
15
,
20
25
.
Greve
,
D. N.
, &
Fischl
,
B.
(
2009
).
Accurate and robust brain image alignment using boundary-based registration
.
Neuroimage
,
48
,
63
72
.
Husain
,
M.
, &
Rorden
,
C.
(
2003
).
Non-spatially lateralized mechanisms in hemispatial neglect
.
Nature Reviews Neuroscience
,
4
,
26
36
.
Husain
,
M.
,
Shapiro
,
K.
,
Martin
,
J.
, &
Kennard
,
C.
(
1997
).
Abnormal temporal dynamics of visual attention in spatial neglect patients
.
Nature
,
385
,
154
156
.
Jenkinson
,
M.
,
Bannister
,
P.
,
,
M.
, &
Smith
,
S.
(
2002
).
Improved optimization for the robust and accurate linear registration and motion correction of brain images
.
Neuroimage
,
17
,
825
841
.
Jenkinson
,
M.
,
Beckmann
,
C. F.
,
Behrens
,
T. E. J.
,
Woolrich
,
M. W.
, &
Smith
,
S. M.
(
2012
).
FSL
.
Neuroimage
,
62
,
782
790
.
Jenkinson
,
M.
, &
Smith
,
S.
(
2001
).
A global optimisation method for robust affine registration of brain images
.
Medical Image Analysis
,
5
,
143
156
.
Karnath
,
H.-O.
,
Himmelbach
,
M.
, &
Küker
,
W.
(
2003
).
The cortical substrate of visual extinction
.
NeuroReport
,
14
,
437
442
.
Karnath
,
H.-O.
, &
Zihl
,
J.
(
2003
).
Disorders of spatial orientation
. In
T.
Brandt
,
L. R.
Caplan
,
J.
Dichgans
,
H. C.
Diener
, &
C.
Kennard
(Eds.),
Neurological disorders: Course and treatment
(2nd ed., pp.
277
286
).
San Diego, CA
:
.
Keetels
,
M.
, &
Vroomen
,
J.
(
2012
).
Perception of synchrony between the senses
. In
M. M.
Murray
&
M. T.
Wallace
(Eds.),
The neural bases of multisensory processes
(pp.
147
178
).
Boca Raton, FL
:
CRC Press/Taylor & Francis
.
Kleiner
,
M.
,
Brainard
,
D.
,
Pelli
,
D.
,
Ingling
,
A.
,
Murray
,
R.
, &
Broussard
,
C.
(
2007
).
What's new in Psychtoolbox-3
.
Perception
,
36
,
1
16
.
Ku
,
S.-P.
,
Gretton
,
A.
,
Macke
,
J.
, &
Logothetis
,
N. K.
(
2008
).
Comparison of pattern recognition methods in classifying high-resolution BOLD signals obtained at high magnetic field in monkeys
.
Magnetic Resonance Imaging
,
26
,
1007
1014
.
Lux
,
S.
,
Marshall
,
J. C.
,
Ritzl
,
A.
,
Zilles
,
K.
, &
Fink
,
G. R.
(
2003
).
Neural mechanisms associated with attention to temporal synchrony versus spatial orientation: An fMRI study
.
Neuroimage
,
20(Suppl. 1)
,
S58
S65
.
Manly
,
T.
,
Owen
,
A. M.
,
McAvinue
,
L.
,
Datta
,
A.
,
Lewis
,
G. H.
,
Scott
,
S. K.
, et al
(
2003
).
Enhancing the sensitivity of a sustained attention task to frontal damage: Convergent clinical and functional imaging evidence
.
Neurocase
,
9
,
340
349
.
McIntosh
,
A. R.
, &
Lobaugh
,
N. J.
(
2004
).
Partial least squares analysis of neuroimaging data: Applications and advances
.
Neuroimage
,
23(Suppl. 1)
,
S250
S263
.
Miyazaki
,
M.
,
,
H.
,
Matsuzaki
,
K. S.
,
Takeuchi
,
S.
,
Sekiguchi
,
H.
,
Aoyama
,
T.
, et al
(
2016
).
Dissociating the neural correlates of tactile temporal order and simultaneity judgements
.
Scientific Reports
,
6
,
23323
.
Mort
,
D. J.
,
Malhotra
,
P.
,
Mannan
,
S. K.
,
Rorden
,
C.
,
Pambakian
,
A.
,
Kennard
,
C.
, et al
(
2003
).
The anatomy of visual neglect
.
Brain
,
126
,
1986
1997
.
Nichols
,
T.
,
Brett
,
M.
,
,
J.
,
Wager
,
T.
, &
Poline
,
J.-B.
(
2005
).
Valid conjunction inference with the minimum statistic
.
Neuroimage
,
25
,
653
660
.
Roberts
,
K. L.
,
Lau
,
J. K. L.
,
Chechlacz
,
M.
, &
Humphreys
,
G. W.
(
2012
).
Spatial and temporal attention deficits following brain injury: A neuroanatomical decomposition of the temporal order judgement task
.
Cognitive Neuropsychology
,
29
,
300
324
.
Rorden
,
C.
,
Li
,
D.
, &
Karnath
,
H.-O.
(
2018
).
Biased temporal order judgments in chronic neglect influenced by trunk position
.
Cortex
,
99
,
273
280
.
Schmah
,
T.
,
Yourganov
,
G.
,
Zemel
,
R. S.
,
Hinton
,
G. E.
,
Small
,
S. L.
, &
Strother
,
S. C.
(
2010
).
Comparing classification methods for longitudinal fMRI studies
.
Neural Computation
,
22
,
2729
2762
.
Smith
,
S. M.
,
Jenkinson
,
M.
,
Woolrich
,
M. W.
,
Beckmann
,
C. F.
,
Behrens
,
T. E. J.
,
Johansen-Berg
,
H.
, et al
(
2004
).
Advances in functional and structural MR image analysis and implementation as FSL
.
Neuroimage
,
23(Suppl. 1)
,
S208
S219
.
Suchan
,
J.
,
Rorden
,
C.
, &
Karnath
,
H.-O.
(
2012
).
Neglect severity after left and right brain damage
.
Neuropsychologia
,
50
,
1136
1141
.
VanRullen
,
R.
,
Pascual-Leone
,
A.
, &
Battelli
,
L.
(
2008
).
The continuous Wagon Wheel Illusion and the “when” pathway of the right parietal lobe: A repetitive transcranial magnetic stimulation study
.
PLoS One
,
3
,
e2911
.
Vossel
,
S.
,
Weiss
,
P. H.
,
Eschenbeck
,
P.
, &
Fink
,
G. R.
(
2013
).
Anosognosia, neglect, extinction and lesion site predict impairment of daily living after right-hemispheric stroke
.
Cortex
,
49
,
1782
1789
.
Xu
,
J.
,
Moeller
,
S.
,
Auerbach
,
E. J.
,
Strupp
,
J.
,
Smith
,
S. M.
,
Feinberg
,
D. A.
, et al
(
2013
).
Evaluation of slice accelerations using multiband echo planar imaging at 3 T
.
Neuroimage
,
83
,
991
1001
.
Zampini
,
M.
,
Guest
,
S.
,
Shore
,
D. I.
, &
Spence
,
C.
(
2005
).
Audio-visual simultaneity judgments
.
Perception & Psychophysics
,
67
,
531
544
.
|
2021-04-18 11:17:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5397888422012329, "perplexity": 5712.627911786713}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00235.warc.gz"}
|
https://www.quizover.com/course/section/section-summary-photon-momentum-by-openstax
|
# 29.4 Photon momentum (Page 4/6)
Page 4 / 6
## Problem-solving suggestion
Note that the forms of the constants $h=\text{4}\text{.}\text{14}×{\text{10}}^{\text{–15}}\phantom{\rule{0.25em}{0ex}}\text{eV}\cdot \text{s}$ and $\text{hc}=\text{1240 eV}\cdot \text{nm}$ may be particularly useful for this section’s Problems and Exercises.
## Test prep for ap courses
The mass of a proton is 1.67 × 10 –27 kg. If a proton has the same momentum as a photon with a wavelength of 325 nm, what is its speed?
1. 2.73 × 10 –3 m/s
2. 0.819 m/s
3. 1.22 m/s
4. 2.71 × 10 4 m/s
(c)
A strip of metal foil with a mass of 5.00 × 10 –7 kg is suspended in a vacuum and exposed to a pulse of light. The velocity of the foil changes from zero to 1.00 × 10 –3 m/s in the same direction as the initial light pulse, and the light pulse is entirely reflected from the surface of the foil. Given that the wavelength of the light is 450 nm, and assuming that this wavelength is the same before and after the collision, how many photons in the pulse collide with the foil?
## Test prep for ap courses
In an experiment in which the Compton effect is observed, a “gamma ray” photon with a wavelength of 5.00 × 10 –13 m scatters from an electron. If the change in the electron energy is 1.60 × 10 –15 J, what is the wavelength of the photon after the collision with the electron?
1. 4.95 × 10 –13 m
2. 4.98 × 10 –13 m
3. 5.02 × 10 –13 m
4. 5.05 × 10 –13 m
(c)
Consider two experiments involving a metal sphere with a radius of 2.00 μm that is suspended in a vacuum. In one experiment, a pulse of N photons reflects from the surface of the sphere, causing the sphere to acquire momentum. In a second experiment, an identical pulse of photons is completely absorbed by the sphere, so that the sphere acquires momentum. Identify each type of collision as either elastic or inelastic, and, assuming that the change in the photon wavelength can be ignored, use linear momentum conservation to derive the expression for the momentum of the sphere in each experiment.
## Section summary
• Photons have momentum, given by $p=\frac{h}{\lambda }$ , where $\lambda$ is the photon wavelength.
• Photon energy and momentum are related by $p=\frac{E}{c}$ , where $E=\text{hf}=\text{hc}/\lambda$ for a photon.
## Conceptual questions
Which formula may be used for the momentum of all particles, with or without mass?
Is there any measurable difference between the momentum of a photon and the momentum of matter?
Why don’t we feel the momentum of sunlight when we are on the beach?
## Problems&Exercises
(a) Find the momentum of a 4.00-cm-wavelength microwave photon. (b) Discuss why you expect the answer to (a) to be very small.
(a) $\text{1.66}×{\text{10}}^{-\text{32}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$
(b) The wavelength of microwave photons is large, so the momentum they carry is very small.
(a) What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? (b) What is its energy in MeV?
(a) What is the wavelength of a photon that has a momentum of $5\text{.}\text{00}×{\text{10}}^{-\text{29}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$ ? (b) Find its energy in eV.
(a) 13.3 μm
(b) $9\text{.}\text{38}×{\text{10}}^{-2}$ eV
(a) A $\gamma$ -ray photon has a momentum of $8\text{.}\text{00}×{\text{10}}^{-\text{21}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$ . What is its wavelength? (b) Calculate its energy in MeV.
(a) Calculate the momentum of a photon having a wavelength of $2\text{.}\text{50 μm}$ . (b) Find the velocity of an electron having the same momentum. (c) What is the kinetic energy of the electron, and how does it compare with that of the photon?
(a) $2\text{.}\text{65}×{\text{10}}^{-\text{28}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$
(b) 291 m/s
(c) electron $3\text{.}\text{86}×{\text{10}}^{-\text{26}}\phantom{\rule{0.25em}{0ex}}\text{J}$ , photon $7\text{.}\text{96}×{\text{10}}^{-\text{20}}\phantom{\rule{0.25em}{0ex}}\text{J}$ , ratio $2\text{.}\text{06}×{\text{10}}^{6}$
Repeat the previous problem for a 10.0-nm-wavelength photon.
(a) Calculate the wavelength of a photon that has the same momentum as a proton moving at 1.00% of the speed of light. (b) What is the energy of the photon in MeV? (c) What is the kinetic energy of the proton in MeV?
(a) $1\text{.}\text{32}×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{m}$
(b) 9.39 MeV
(c) $4.70×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{MeV}$
(a) Find the momentum of a 100-keV x-ray photon. (b) Find the equivalent velocity of a neutron with the same momentum. (c) What is the neutron’s kinetic energy in keV?
Take the ratio of relativistic rest energy, $E={\mathrm{\gamma mc}}^{2}$ , to relativistic momentum, $p=\gamma \text{mu}$ , and show that in the limit that mass approaches zero, you find $E/p=c$ .
$E={\mathrm{\gamma mc}}^{2}$ and $P=\mathrm{\gamma mu}$ , so
$\frac{E}{P}=\frac{{\text{γmc}}^{2}}{\text{γmu}}=\frac{{\text{c}}^{2}}{\text{u}}.$
As the mass of particle approaches zero, its velocity $u$ will approach $c$ , so that the ratio of energy to momentum in this limit is
${lim}_{m\to 0}\frac{E}{P}=\frac{{c}^{2}}{c}=c$
which is consistent with the equation for photon energy.
Consider a space sail such as mentioned in [link] . Construct a problem in which you calculate the light pressure on the sail in ${\text{N/m}}^{2}$ produced by reflecting sunlight. Also calculate the force that could be produced and how much effect that would have on a spacecraft. Among the things to be considered are the intensity of sunlight, its average wavelength, the number of photons per square meter this implies, the area of the space sail, and the mass of the system being accelerated.
Unreasonable Results
A car feels a small force due to the light it sends out from its headlights, equal to the momentum of the light divided by the time in which it is emitted. (a) Calculate the power of each headlight, if they exert a total force of $2\text{.}\text{00}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{N}$ backward on the car. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
(a) $3\text{.}\text{00}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{W}$
(b) Headlights are way too bright.
(c) Force is too large.
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
What is meant by dielectric charge?
what happens to the size of charge if the dielectric is changed?
omega= omega not +alpha t derivation
u have to derivate it respected to time ...and as w is the angular velocity uu will relace it with "thita × time""
Abrar
do to be peaceful with any body
the angle subtended at the center of sphere of radius r in steradian is equal to 4 pi how?
if for diatonic gas Cv =5R/2 then gamma is equal to 7/5 how?
Saeed
define variable velocity
displacement in easy way.
binding energy per nucleon
why God created humanity
Because HE needs someone to dominate the earth (Gen. 1:26)
Olorunfemi
Ali
Is the object in a conductor or an insulator? Justify your answer. whats the answer to this question? pls need help figure is given above
ok we can say body is electrically neutral ...conductor this quality is given to most metalls who have free electron in orbital d ...but human doesn't have ...so we re made from insulator or dielectric material ... furthermore, the menirals in our body like k, Fe , cu , zn
Abrar
when we face electric shock these elements work as a conductor that's why we got this shock
Abrar
how do i calculate the pressure on the base of a deposit if the deposit is moving with a linear aceleration
why electromagnetic induction is not used in room heater ?
room?
Abrar
What is position?
What is law of gravition
|
2018-11-15 05:46:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 30, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7427943348884583, "perplexity": 662.8427141222719}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742567.46/warc/CC-MAIN-20181115054518-20181115075752-00035.warc.gz"}
|
https://exceltrick.com/formulas_macros/excel-mid-function/
|
Excel MID Function – How To Use
The MID function returns a specific number of characters from the middle of the string after we state the starting position and the number of characters to extract. The MID Function in Excel is one of the text functions that allows us to get a substring from a string\text as per our requirement
It is one of the most useful functions when we want to extract the middle name from the full name, the domain name from email IDs, or any substring between two characters.
Syntax
The syntax of the MID function is as follows:
=MID(text,start_num,num_chars)
The MID function in Excel accepts three arguments and all three arguments are mandatory.
Arguments:
'text' – This holds the value of the input text/string. A cell reference containing the input string or text enclosed in double quotes are both acceptable input values.
'start_num' – The position of the first character we want to extract. It must be an integer/formula which indicates the starting position of the extraction.
'num_chars' – Number of characters we wish to extract from the starting point. The num_chars argument indicates the character length of the return value.
Important Characteristics of the MID Function
One of the essential and basic characteristics of the MID function being a text function is that it always returns a text value regardless of the input value. Even if the retrieved substring is only made up of numbers, it will be in text format. Other important MID function features are as follows:
• If the value of start_num exceeds the length of input text, the MID function returns "" (empty text).
• If the start_num value is less than 1, the function returns a #VALUE! error.
• If the value of num_chars is a negative number, the MID function gives a #VALUE error.
• If the num_chars value is less than 1, then the MID function's return value is an empty string, a blank cell.
Last of all, if the sum of the value of start_num and num_chars is greater than the total input text length, the MID function returns all the characters from start_num to the last character. This characteristic comes in very handy for certain conditions. We will find out in the examples below.
Examples of MID Function
The MID function can be used independently or can be combined with various other Excel functions such as SEARCH, FIND, LEN, and more. Let's go through some examples to fully comprehend the usage and utility of the MID function.
Example 1: Basic Functionality of the MID Function
In the below example, we have taken 3 different instances and tried using the basic MID function.
In the first case, the input text/string is a full name 'Cassie Martha Soros' where we wish to extract the middle name –"Martha". So, using the MID function we apply the formula:
=MID(B3,8,6)
Here, the first parameter text is the cell reference B3. The second parameter start_num is the starting position which is 8 as the first name is 6 characters long, the space is the 7th character and then we add 1 so that we reach the beginning of the middle name. Lastly, the third argument num_chars is the number of characters of the middle name which is 6.
The second instance is where we are trying to extract the house number and street address from the complete address in cell B4. We are checking if the MID function counts a special character (,) and also whether the MID function can return characters from the beginning of the text or only from the middle. The formula goes as follows:
=MID(B4,1,13)
The first parameter is B4, the input text. The second parameter is 1 as the house number is in the beginning. Finally, the third parameter is 13 which is the house number (3 characters long) combined with a comma, a space character, and then the street name which is 8 characters long. Hence our output, in this case, becomes '220, Kisvagen'.
In the last example, the input text is in cell B5, which is actually a number 123456789. To extract 4 digits from the second position, the formula would be:
=MID(B5,2,4)
Here, the output value is left aligned. Is it a text value? We have something to learn here, which we will find in the section below.
Example 2: Extract Substring After a Specific Character
In the following example, we are trying to extract a substring from the input text after a special character. The dataset in column B in this example consists of Employee IDs; a combination of the department name, and a unique ID followed by the employee's date of birth.
For instance, in the Employee ID TECH-1234-08051990, the employee belongs to the tech team, the unique ID is 1234, and the date of birth is 08 May 1990.
We want to extract the unique ID from each Employee ID. Using the MID function, we can extract it from the input string, but the issue is that the character length of the department name varies. Resultantly, the value of start_num varies.
Lucky for us, the department name is followed by a hyphen '-', and the employee ID is 4 characters long. Now, we can use the MID function combined with the FIND function. The FIND function will locate the hyphen's '-' position, which will be the input value of start_num.
Use the FIND function as follows:
=FIND("-",B3)
This formula locates the hyphen in B3 and returns 4 (the hyphen is the 4th character in B3). To use the FIND function as an input value for the start_num parameter, we must add 1 to this result because the unique ID begins from the 5th character of the text string and we don't want to see the hyphen in the final result.
The value of num_chars shall be 4 as all the unique IDs in our example are 4-digit numbers. So, combining the FIND and MID functions, the final function is as follows:
=MID(B3,FIND("-",B3)+1,4)
And here's how this formula extracts the unique IDs for us:
Example 3: Extract First or Last Name
Imagine we have a data set of peoples' names and are asked to extract first, middle and last names in different columns. Usually, we would want to use the LEFT function for the first name and the RIGHT function for the last name. However, the functionality of two different functions can be accomplished using the MID function alone.
Let's learn how to extract first and last names using the MID function. We cannot simply use the MID function as the starting position for separating last names as the number of characters will vary for each name. Ergo, we will use the SEARCH function to our advantage and then merge it with the MID function.
We could have used the earlier mentioned FIND function too, but as we have understood the functionality of the FIND function, let's try to learn the MID function with the SEARCH function here. Also, the SEARCH function is case insensitive and that will make our task even easier.
Extract First Name using MID Function
The first and last names are divided by a space, a common pattern we can make use of. The beginning position in the MID function will be the starting of the string, whereas the number of characters can be determined using the SEARCH function.
The position of the space " " is where the first name ends and hence, is the number of characters to be extracted. The space character is located in the name using the SEARCH function, which returns its position, from which we deduct 1 to delete the inclusion of the extra space.
Let's try to put this logic in the form of an Excel formula.
Seeing the example ahead, the first name in cell B3 is 'Simon Rogers'. If we used the MID function only on this cell, we would try:
=MID(B3,1,5)
However, this formula will not give us the required result for all other cells since all the names won't be 5 letters.
Let's generalize it so we can use it for the complete dataset.
As the first parameter, start_num remains the same for all the names,. To determine the second parameter, we must find the position of the space. To find the position of the space " " using the SEARCH function, the formula will go as follows:
=SEARCH(" ",B3)
Combining it with the MID function, the formula will become:
=MID(B3,1,SEARCH(" ",B3))
To remove the extra trailing space, subtract 1 from the position. The final formula will be:
=MID(B3,1,SEARCH(" ",B3)-1)
Furthermore, can we use the same logic to extract the last name? Let's find out.
Extract Last Name using MID Function
To pull out the last name using the MID function, we can use the SEARCH function to point out the start_num parameter as the last name begins after the space. How to find the number of characters (num_chars) to extract?
As mentioned earlier, if the total of the start_num and num_chars is more than the length of the string, the MID function returns the remaining string from the start_num. So, utilizing that to our advantage, there are two ways to use the MID function.
The first one is fairly simple. Assuming that the total character length of names will never exceed 1000, we can assign 1000 to the num_chars parameter.
The SEARCH function will determine the position of the space " " using the formula:
=SEARCH(" ",B3)
This will be the input value of the first parameter start_num. Assigning 1000 to the second parameter, the formula will become:
=MID(B3,SEARCH(" ",B3),1000)
To remove the extra leading spaces from the extracted last names, use the TRIM function. The final MID function will be used as follows:
=TRIM(MID(B3,SEARCH(" ",B3),1000))
Now we have the last names extracted by the MID and SEARCH functions and leading spaces removed by the TRIM function:
The second way is more elegant. Instead of an assumed number in the num_chars parameter, we can use the LEN function to determine the total length of the name and use that value in the num_chars parameter.
Combining the LEN & SEARCH functions with the MID function, the formula will be as follows:
=MID(B3,SEARCH(" ",B3),LEN(B3))
Here, LEN function will return the total length of the input string, and SEARCH function will extract the position of the space " " in the string. The ultimate effect is similar to what we have seen in the previous formula. The location of the space character added to the total characters of B3 will obviously be more than the total characters in B3.
To polish it further, we can use the TRIM function to eliminate all the extra spaces.
=TRIM(MID(B3,SEARCH(" ",B3),LEN(B3)))
Substring Between Two Delimiters
After practicing the above examples, we can extract a substring from a string after determining the position of the space " " or any other special character. Now, what if the input string is long with several delimiters? Just for clarity, one or more characters that divide text strings are known as delimiters, such as space ( ), quotes ("), commas (,), semicolons (;), slashes (/) or pipes (|).
When we dive in further, there are two scenarios of extracting a substring from an input string between delimiters; first when they are different and second when the delimiters are the same. Let's understand both cases using an example.
Example 1 – Extracting Substring Between Two Different Delimiters
Let's assume we have a set of email IDs, and for research purposes, we want to extract the domain name from each email ID to analyze which one is most preferred. The email ID structure goes as follows '[email protected]', where 'learning' is a username and gmail is a domain name.
As we can see, we have two different delimiters (@) and (.) separating the domain name (gmail) from the rest of the string (username). So, to pull out the substring, we must locate the position of two delimiters where the first delimiter is the starting position (start_num), and using the second delimiter, we can calculate the number of characters to be extracted.
As we already know, the MID function consists of three arguments. The first argument is the cell reference to email ID (B3). The second argument is the starting position which can be calculated using the SEARCH function:
=SEARCH("@",B3)
This formula will give us the position of the first delimiter (@). As we require the starting position to be one character after (@), we add 1.
Lastly, for the third argument (num_chars), we need the number of characters to be extracted. We can find the location of the second delimiter (.) and then subtract the position of the first delimiter. The formula goes as follows:
=SEARCH(".",B3)-SEARCH("@",B3)
To finalize and remove the trailing space, subtract 1.
Putting it all together, the final MID function formula to extract the substring from two different characters is:
=MID(B3,SEARCH("@",B3)+1,(SEARCH(".",B3)-SEARCH("@",B3))-1)
This is how our experiment turned out:
Example 2: Extracting Substring Between Same Delimiters
In the above examples, we already know how to extract the first and last name from the full name using the MID function. However, what about the middle name? We cannot simply use the basic MID function as the character length of middle names varies for each person.
However, in the previous example, we learned how to extract a substring between two delimiters.
As we will notice, there is a space ( ) before and after the middle name, separating it from the first & last name. We can find the position of two delimiters (in this case, both the delimiters are space characters) and extract the middle name from the full name.
In the given dataset in column B, we have the list of full names. Let's take the example of the first name in cell B3.
The first parameter of the MID function is the input text which is cell B3. For the second parameter, that is start_num, if we find the position of the space before the middle name and then add 1 to it, we should have our argument input. Using the SEARCH function, the position of the first space " " can be determined using the following formula:
=SEARCH(" ",B3)+1
This returns 7 and if we do a quick count check, indeed the middle name's first letter is the 7th character of the text string.
Easy enough, right?
Now, to determine the number of characters to be returned, we can find the position of the second space " " (after the middle name) and subtract the position of the first space from it.
Remember, the SEARCH function has an additional third optional parameter start_num stating the starting position of the search within the string. This will come in handy here as there are two spaces in this text (full name). To find the position of the second space, the formula goes as follows:
=SEARCH(" ",B3,(SEARCH(" ",B3)+1)
Explaining further, we are telling the SEARCH function to find " " in B3 starting after the position of the first space. If we don't use the nested SEARCH function, it will keep returning the position of the first space, and we don't want that, do we?
This formula returns 11 which you can tally again. Now that we have the position of both the space characters 11-7 is the number of characters to extract. Combining all to finish our MID function, the formula goes as follows:
=MID(B3,SEARCH(" ",B3)+1,SEARCH(" ",B3,(SEARCH(" ",B3)+1))-(SEARCH(" ",B3)+1))
The MID function will return 4 characters from B3 starting from the character after the first space. The result is Dona and the rest can be seen below:
Forcing MID Function to Return Numbers
Like other text functions in Excel, the output value of the MID function is always in text format. Even if the input or output value contains only digits and looks like a number format, it is indeed in text format.
There are two ways to find if the output value is in text or number format. One is the innate nature of Excel to left align a text and right align a number. The second is to use the ISNUMBER function.
In this example, we multiply 99 with 200 and then extract the 2 digit of the result, starting from the second position. Using the MID function, the formula goes as follows
=MID(99*200,2,2)
We know that 99*200 is 19800. We are trying to extract 2 digits out of it starting from the 2nd position and hence the result is 98.
However, if you notice the result here is left aligned. By default Excel aligns all the numbers to right and text to left. This hints that the result here is a text instead of a number.
This can be further proved with the ISNUMBER function, which also returns false implying that the result of the MID function is not a number.
To ensure that the MID function returns a number, use the VALUE function, which transforms a text value representing a number into a number. Wrap the MID function inside the VALUE function. The formula used will be:
=VALUE(MID(99*200,2,2))
As evident from the example, the return value of the MID function combined with the VALUE function is right aligned, and when checked with ISNUMBER function, the result is TRUE.
RIGHT Function vs LEFT Function vs MID Function
We have now understood several ways to use the MID function for various tasks. Similar to the MID function, there are two more text functions in Excel; LEFT & RIGHT functions. For better comprehension, we will attempt to use all three Excel functions on the same input string in the following example.
In this example, we have taken an input text/string in cell A2 "Smart, Easy & Effective"
We have used the LEFT function to extract the first word of the string "Smart". As it is 5 characters long, the LEFT function formula will be:
=LEFT(A2,5)
Unlike the MID function, the LEFT function doesn't have the start_num argument since the starting point is automatically the first character from the left. The same is the case with the RIGHT function. It just extracts characters from right to left.
Similarly, we've used the RIGHT function on the input string to extract the last word "Effective" (9 characters long). The formula is:
=RIGHT(A2,9)
Now the function that we know very well; the MID function. The formula used to extract the middle word "Easy" will be:
=MID(A2,8,4)
With that, we've discussed the ins and outs of the MID function. There are numerous other interesting things the MID function can manage such as returning a word beginning with a certain character (e.g. social media tags with the @ symbol such as @cholecakes, @popartillustrations), returning the nth word from a text string, and so on. Keep practicing and explore new ways to use the MID function.
|
2023-03-23 02:07:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4710870385169983, "perplexity": 749.316872766878}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00534.warc.gz"}
|
https://math.stackexchange.com/questions/3798388/error-in-divergence-theorem
|
# Error In Divergence Theorem
If $$\vec{F}=y\vec{i}+(x-2xz)\vec{j}-xy\vec{k}$$, Evaluate $$\iint_S (\nabla \times \vec{F})\cdot n \vec{ dS}$$ where $$S$$ is the surface of the sphere of radius $$a$$ with center at origin above the $$xy$$ plane and $$n$$ is the unit normal vector to $$S$$.
When I Applied Gauss Divergence Theorem, $$\iiint_V \nabla\cdot (\nabla \times \vec{F}) dv$$ should be equal to zero so the surface integral must be zero but it's not the answer Why?
• Could you show your attempt so that we can figure out where you went wrong? Aug 21 '20 at 8:33
The divergence theorem is about a three-dimensional body $$B$$, its boundary surface $$\partial B$$, and a vector field $${\bf v}$$ with domain $$\Omega\supset B$$. The theorem says that $$\int_{\partial B}{\bf v}\cdot{\bf n}\>dS=\int_B{\rm div}\,{\bf v}({\bf x})\> dV\ .\tag{1}$$ In your case $$B$$ is a half ball, and $${\bf v}={\rm curl}(\vec F)$$. It follows that $${\rm div}\,{\bf v}({\bf x})\equiv 0$$ in $$B$$. But $$\partial B$$ is not just the spherical part of $$\partial B$$. The boundary of $$B$$ also contains a plane circular disc in the $$(x,y)$$-plane. It follows that the right hand side of $$(1)$$ consists of two terms, one of which you have already computed. Computing the flow of $${\bf v}$$ through the disc is simpler, since $${\bf n}\equiv(0,0,-1)$$ and $$dS=dA$$, where $$dA$$ is the standard area element in the $$(x,y)$$-plane.
|
2021-10-27 04:46:35
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9232332706451416, "perplexity": 61.184815852040956}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588053.38/warc/CC-MAIN-20211027022823-20211027052823-00334.warc.gz"}
|
https://www.physicsforums.com/threads/limits-to-infinity-with-natural-logs.586012/
|
# Limits to infinity with natural logs
## Homework Statement
I'm having trouble calculating these limits.
none
## The Attempt at a Solution
1. lim.......((lnx)^5)/(x)=
x->infinity
(((lnx)^5)/x)/(x/x)=
(((lnx)^5)/x)/1=
How do I calculate from here?
2.lim .......(14lnx^2)/(6lnx^3)=
x->infinity
((14lnx^2)/(x^3))/((6lnx^3)/(x^3))=
((14lnx)/(x))/(6lnx)=
0/6lnx= 0
Is this right or am I forgetting something?
Related Calculus and Beyond Homework Help News on Phys.org
use Lhospital
SammyS
Staff Emeritus
Homework Helper
Gold Member
## Homework Statement
I'm having trouble calculating these limits.
none
## The Attempt at a Solution
1. lim.......((lnx)^5)/(x)=
x->infinity
(((lnx)^5)/x)/(x/x)=
(((lnx)^5)/x)/1=
How do I calculate from here?
You're going in circles... back to what you started with.
As Punky said, use L'Hôpital's rule.
2.lim .......(14lnx^2)/(6lnx^3)=
x->infinity
((14lnx^2)/(x^3))/((6lnx^3)/(x^3))=
((14lnx)/(x))/(6lnx)=
0/6lnx= 0
Is this right or am I forgetting something?
In #2, what you have done is incorrect.
Use properties of logarithms for #2. It works out quite simply.
$\log_{\,b}(x^a)=(a)\log_{\,b}(x)\,.$
I'm not allowed to use L'Hopital, and I will change #2
So would the second problem look something like this:
28lnx/18lnx?
SammyS
Staff Emeritus
Homework Helper
Gold Member
So would the second problem look something like this:
28lnx/18lnx?
If you mean the limit of 28lnx/(18lnx), then yes, that's right.
I'll think about #1 some more. Maybe someone else will help.
ok so it will be ((28lnx)/x)/((18lnx)/x) = 0/0 = DNE?
SammyS
Staff Emeritus
Homework Helper
Gold Member
ok so it will be ((28lnx)/x)/((18lnx)/x) = 0/0 = DNE?
A limit of the form 0/0 is indeterminate. It might exist or might not exist.
Why are you bothering to multiply the numerator & denominator by 1/x ? (That accomplishes nothing!)
What is $\displaystyle \frac{\ln(x)}{\ln(x)}\,$ for any x in the domain of the natural log ?
1/1 ?
SammyS
Staff Emeritus
Homework Helper
Gold Member
For #1.
Do you know $\displaystyle \lim_{x\to\infty} \frac{\ln(x)}{x}\ ?$
If so then rewrite $\displaystyle \frac{(\ln(x))^5}{x}\ \text{ as }\ \frac{(\ln(x))^5}{x^5}\ x^4\ .$
SammyS
Staff Emeritus
Homework Helper
Gold Member
1/1 ?
Yes !
What does that leave you with?
So basically the limit is then 28/18?
SammyS
Staff Emeritus
Homework Helper
Gold Member
So basically the limit is then 28/18?
Yes.
I think I know how to do the first one by applying the chain rule:
limit to infinity
((lnx)^5)/(x)=
(5(lnx)^4)*(1/x)/(x)=
5(lnx)^4*0/(x)= 0
Is that valid?
Ignore my last post I got confused, this isn't a derivative
SammyS
Staff Emeritus
Homework Helper
Gold Member
Look at post #10 .
|
2020-07-13 16:05:26
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8874015212059021, "perplexity": 4821.0470707949125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657145436.64/warc/CC-MAIN-20200713131310-20200713161310-00405.warc.gz"}
|
https://coin5s.com/docs/rhe6ec96/archive.php?tag=0580f8-symmetric-key-exchange
|
Symmetric Easy to follow VPN describes how the symmetric trying to establish a tutorial. Both algorithms are not used at the same time. In particular, they can provide perfect forward secrecy, as illustrated by key exchange protocols based on the Diffie-Hellman scheme. Key exchange protocols in the asymmetric-key setting are known to provide stronger security properties than protocols in symmetric-key cryptography. For the symmetric key, you can approach this problem as a complete graph with order 1000. Although symmetric key algorithms are fast and secure, key exchange is always a problem. With the vertexes representing people and the edges representing the symmetric keys. Although slower than symmetric key encryption, this method ensures the key cannot accessed by a 3rd party while being transferred. Here is a Diffie-Hellman key example. Then each vertex would have degree 999 and, applying the Handshaking lemma, the number of edges would be: $(1000 \times 999)/2 = 499500$ Since the key has been transferred using a secure channel, a symmetric key can be used for data exchange. The message exchange using symmetric key cryptography involves the following steps- Before starting the communication, sender and receiver shares the secret key. This is why PKI incorporates the use of asymmetric key exchange protocols for the symmetric key exchange process as part of the TLS handshake. Exchanging symmetric keys in public channels (like on the internet) otherwise is an issue because the key is vulnerable. As you already noted, there are two ways to exchange symmetric session keys: through key encipherment or through key agreement (which is based on Diffie-Hellman algorithm). The format of a TR-31 key block is defined in ASC X9 TR 31-2018: Interoperable Secure Key Exchange Block Specification. The Diffie-Hellman algorithm is mostly used for key exchange. At sender side, sender encrypts the message using his copy of the key. Assume Alice and Bob would like to exchange a secret symmetric key for their secure communications. This secret key is shared through some external means. The provided information is an extension of Managing AES, DES, and HMAC cryptographic keys.Refer to this topic for additional information on symmetric … Rather than create a symmetric key cipher and find a way to securely exchange keys, then figure out a way to use out-of-band verification to ensure there wasn't a … Using asymmetric techniques in the key exchange process is more secure. For key exchange, to protect the key used in data exchange it is often encrypted using public key encryption. The Diffie-Hellman algorithm helps with this. You have to figure out a way to get the private key to all systems. Symmetric-key Authenticated Key Exchange (SAKE) with Perfect Forward Secrecy Gildas Avoine 1;2, S ebastien Canard 3, and Lo c Ferreira 1 Univ Rennes, INSA Rennes, CNRS, IRISA, France 2 Institut Universitaire de France gildas.avoine@irisa.fr 3 Orange Labs, Applied Crypto Group, Caen, France fsebastien.canard,loic.ferreirag@orange.com First, they need to be agree upon a modulus, which is a prime p, and a base g for computing the public key and symmetric key. Key, you can approach this problem as a complete graph with order 1000 public key,. Steps- Before starting the communication, sender and receiver shares the secret key is shared through external! Encrypts the message exchange using symmetric key cryptography involves the following steps- Before starting communication! A secret symmetric key can be used for data exchange it is often encrypted using public key encryption process more. Involves the following steps- Before starting the communication, sender encrypts the exchange... The vertexes representing people and the edges representing the symmetric key, you can this. By a 3rd party while being transferred asymmetric techniques in the key has been transferred using a secure channel a! Exchange, to protect the key exchange protocols in the asymmetric-key setting are known to provide stronger properties. In symmetric-key cryptography a way to get the private key to all.... Of the TLS handshake used for data exchange it is often encrypted using public key encryption this! Key is shared through some external means security properties than protocols in the asymmetric-key setting are known to stronger... Tls handshake is often encrypted using public key encryption can provide perfect forward secrecy, as illustrated key. Message using his copy of the TLS handshake algorithms are fast and,... The private key to all systems protocols in symmetric-key cryptography been transferred using a channel. Tls handshake part of the TLS handshake is more secure by key is. Key algorithms are fast and secure, key exchange block Specification as part the.: Interoperable secure key exchange block Specification to figure out a way to the. Secrecy, as illustrated by key exchange the format of a TR-31 key is. Message using his copy of the key can not accessed by a 3rd party while being transferred with order.! Sender encrypts the message exchange using symmetric key can be used for key exchange is defined in X9! The communication, sender encrypts the message using his copy of the TLS handshake is... A symmetric key, you can approach this problem as a complete graph with order 1000 encryption, method. Easy to follow VPN describes how the symmetric trying to establish a tutorial secret.! Some external means it is often encrypted using public key encryption for data exchange message exchange using symmetric key involves! Alice and Bob would like to exchange a secret symmetric key encryption, this method the. Get the private key to all systems sender and receiver shares the secret key is shared through some external.. Transferred using a secure channel, a symmetric key can not accessed by a 3rd party while transferred. Since the key used in data exchange it is often encrypted using public encryption! A complete graph with order 1000 can not accessed by a 3rd party while transferred... A complete graph with order 1000 is shared through some external means you... Exchange block Specification 31-2018: Interoperable secure key exchange protocols in symmetric-key cryptography exchange using symmetric key you! Key to all systems 31-2018: Interoperable secure key exchange block Specification TR 31-2018 Interoperable. Provide stronger security properties than protocols in the asymmetric-key setting are known to provide stronger security properties than protocols the! Asc X9 TR 31-2018: Interoperable secure key exchange is always a problem can not accessed a. Properties than protocols in symmetric-key cryptography format of a TR-31 key block is defined in ASC X9 TR 31-2018 Interoperable! Ensures the key has been transferred using a secure channel, a symmetric key algorithms are fast and,. Exchange is always a problem forward secrecy, as illustrated by key exchange process as part of the TLS.! While being transferred sender and receiver shares the secret key involves the following steps- Before starting the communication sender. Key exchange is always a problem receiver shares the secret key is shared through external... Problem as a complete graph with order 1000 and the edges representing the symmetric key algorithms are fast secure!, they can provide perfect forward secrecy, as illustrated by key exchange protocols in the key 3rd party being! And receiver shares the secret key secret symmetric key can not accessed by a 3rd party while being transferred symmetric... Symmetric key, you can approach this problem as a complete graph with order.! Although slower than symmetric key algorithms are fast and secure, key exchange process as part of key... Provide stronger security properties than protocols in the key has been transferred using a secure channel a... Establish a tutorial receiver shares the secret key a tutorial public key,! A symmetric key exchange block Specification symmetric-key cryptography graph with order 1000 graph with order.! Sender and receiver shares the secret key is shared through some external.! Key block is defined in ASC X9 TR 31-2018: Interoperable secure exchange. Bob would like to exchange a secret symmetric key can be used data. Secure channel, a symmetric key for their secure communications this secret.. Be used for key exchange protocols for the symmetric trying to establish tutorial... To protect the key has been transferred using a secure channel, a symmetric key exchange protocols on! Tr 31-2018: Interoperable secure key exchange the communication, sender and receiver shares the key... Provide stronger security properties than protocols in the asymmetric-key setting are known to provide stronger security than... Part of the key exchange protocols for the symmetric trying to establish tutorial! Format of a TR-31 key block is defined in ASC X9 TR 31-2018: Interoperable key. Key algorithms are not used at the same time this method ensures the key been. Using symmetric key algorithms are not used at the symmetric key exchange time TLS handshake both algorithms fast! Techniques in symmetric key exchange asymmetric-key setting are known to provide stronger security properties than protocols symmetric-key! Using public key encryption, this method ensures the key can not accessed by a 3rd party while transferred. By key exchange protocols for the symmetric key exchange process as part of the TLS handshake secret symmetric key their... Accessed by a 3rd party while being transferred security properties than protocols in symmetric-key cryptography always a problem key you., a symmetric key exchange block Specification can not accessed by a party... In ASC X9 TR 31-2018: Interoperable secure key exchange block Specification and Bob would like to exchange secret! Secure key exchange block Specification representing people and the edges representing the symmetric keys symmetric key exchange Specification not... Graph with order 1000 encryption, this method ensures the key has been using! Key block is defined in ASC X9 TR 31-2018: Interoperable secure key exchange based. Using asymmetric techniques in the asymmetric-key setting are known to provide stronger security properties protocols! Symmetric-Key cryptography party while being transferred, key exchange protocols for the symmetric keys problem! Is defined in ASC X9 TR 31-2018: Interoperable secure key exchange to. Defined in ASC X9 TR 31-2018: Interoperable secure key exchange protocols based on the Diffie-Hellman scheme illustrated by exchange. Receiver shares the secret key for their secure communications a way to the. And Bob would like to exchange a secret symmetric key encryption, this method ensures the key has transferred! External means are not used at the same time assume Alice and Bob symmetric key exchange to! The private key to all systems the same time incorporates the use of key... Representing the symmetric keys setting are known to provide stronger security properties protocols! Their secure communications shares the secret key is shared through some external means private to... Of the TLS handshake message exchange using symmetric key algorithms are not used at the same time scheme. While being transferred used at the same time incorporates the use of asymmetric exchange! This method ensures the key has been transferred using a secure channel a. Using symmetric key can not accessed by a 3rd party while being transferred TLS! Process as part of the key their secure communications it is often encrypted using public key.... Both algorithms are not used at the same time a problem both algorithms are fast and secure, key,... A tutorial copy of the TLS handshake describes how the symmetric key, can! More secure illustrated by key exchange process is more secure Before starting the,... Key has been transferred using a secure channel, a symmetric key exchange protocols based on the Diffie-Hellman.... In ASC X9 TR 31-2018: Interoperable secure key exchange process as part of the.! Part of the TLS handshake TR 31-2018: Interoperable secure key exchange is always a problem provide security! Sender encrypts the message exchange using symmetric key for their secure communications particular, they can provide perfect secrecy. Encrypts the message using his copy of the TLS handshake encrypts the message using his of. The use of asymmetric key exchange protocols in the key private key to all systems key. Have to figure out a way to get the private key to systems! Encryption, this method ensures the key can not accessed by a 3rd party while being.! The TLS handshake side, sender encrypts the message using his copy of the key used in data it. Key for their secure communications in ASC X9 TR 31-2018: Interoperable secure key exchange Specification... Representing the symmetric keys copy of the TLS handshake a tutorial is mostly used for exchange., as illustrated by key exchange protocols in the key has been transferred using a secure channel a... Protocols based on the Diffie-Hellman algorithm is mostly used for data exchange asymmetric-key setting are known to stronger. Secure communications data exchange it is often encrypted using public key encryption, this method ensures the key used data...
|
2021-03-02 17:08:39
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18579064309597015, "perplexity": 2072.094458139784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364027.59/warc/CC-MAIN-20210302160319-20210302190319-00505.warc.gz"}
|
https://www.askiitians.com/forums/Wave-Motion/in-young-s-double-silt-experiment-the-distance-bet_259283.htm
|
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.
There are no items in this cart.
Continue Shopping
# In young's double silt experiment the distance between the silt is 1.2mm and the screen is 0.75m from the silts. If the distance of the 5th fringe from the central fringe on the screen is 1.8mm. Calculate the wavelength of light used. What will be the distance of the 5th fringe from the center of the screen?
6 months ago
Arun
24736 Points
As we learnt in
Young Double Slit Experiment -
- wherein
$y=\Delta x\cdot \left ( \frac{D}{d} \right )$
$y=$ Distance of a point on screen from central maxima
$\Delta x=$ Path difference at that point
Ref y be the distance from the central maximum to the point where the bright fringes due to both the wavelength coincides
For $\lambda_{1}$, $y=\frac{m\lambda_{1}D}{d}$
for $\lambda_{2}$, $y=\frac{n\lambda_{2}D}{d}$
$\therefore\ \; m \lambda_{1}=n\lambda_{2}$
6 months ago
Vikas TU
12135 Points
Dear student
Good Luck
6 months ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Wave Motion
View all Questions »
### Course Features
• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
### Course Features
• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions
|
2020-09-22 00:45:12
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4703856408596039, "perplexity": 9021.339105669602}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00713.warc.gz"}
|
http://cs.au.dk/research/centers/madalgo/events/show/artikel/madalgo-seminar-19/
|
MADALGO theory seminar, Suresh Venkatasubramanian, University of Utah
2014.05.14 | Trine Ji Holmgaard Jensen
Date Wed 21 May Time 14:15 — 15:00 Location
Speaker: Suresh Venkatasubramanian
Location: Nygaard 395
Title: A directed isoperimetric inequality with application to Bregman near neighbor lower bounds
Abstract:
The \emph{Bregman divergences} $D_\phi$ are a class of distance measures parametrized by a convex function $\phi$. They include well known distance functions like $\ell_2^2$ and the Kullback-Leibler divergence from information theory and are used extensively in data-driven applications such as machine laerning, computer vision, text mining, and speech processing. There has been extensive research on algorithms for problems like clustering and near neighbor search with respect to Bregman divergences; in all cases, the algorithms depend not just on standard parameters like the data size $n$, dimensionality $d$, number of probes allowed $t$, and error $\varepsilon$, but also on a \emph{structure constant} $\mu \ge 1$ that depends solely on $\phi$ and can grow without bound independent of the other parameters. This dependence has withstood attempts to remove it, and appears to be intrinsic to the study of these measures.
In this paper, we provide the first evidence that this dependence might be intrinsic. In particular, we focus on the problem of approximate near neighbor search for Bregman divergences, a problem of interest in its own right. We show that under the cell probe model, any non-adaptive data structure (like locality-sensitive hashing) for $c$-approximate near-neighbor search that admits $r$ probes must use space $\Omega(n^{1 + \frac{\mu}{c r}})$. In contrast, for LSH under $\ell_1$ the best bound is $\Omega(n^{1+\frac{1}{cr}})$.
In proving this lower bound, we follow and extend the Fourier-analytic approach that has yielded many lower bounds for such problems. Our new tool is a directed variant of the standard boolean noise operator, for which we prove a generalization of the Bonami-Beckner hypercontractivity inequality. While such an inequality cannot exist in general, we show that it exists in expectation'' or upon restriction to certain subsets of the Hamming cube, and that this is sufficient to prove the desired isoperimetric inequality that we use in our data structure lower bound. This operator might be of independent interest in the analysis of boolean functions.
We also present a structural result reducing the Hamming cube to a \emph{Bregman cube}. This structure allows us to obtain lower bounds for problems under Bregman divergences from their $\ell_1$ analog. In particular, we get a (weaker) lower bound for approximate near neighbor search of the form $\Omega(n^{1 + \frac{1}{cr}})$ for an $r$-query non-adaptive data structure, and new cell probe lower bounds for a number of other related near neighbor questions in Bregman space.
Joint work with Amirali Abdullah
|
2018-09-21 02:08:35
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8114209175109863, "perplexity": 658.3362370966564}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156724.6/warc/CC-MAIN-20180921013907-20180921034307-00332.warc.gz"}
|
https://wordpress.entos.se/iwp/2ueluo/39eda9-when-to-use-ols-regression
|
# when to use ols regression
The relationship is modeled through a random disturbance term (or, error variable) ε. If you found this topic interesting and you want to see a practical application with Python of what you have been reading so far, you can read my article here. Here, we start modeling the dependent variable yi with one independent variable xi: where the subscript i refers to a particular observation (there are n data points in total). There are five assumptions associated with the linear regression model (these are called the Gauss-Markov assumptions): The Gauss-Markov assumptions guarantee the validity of Ordinary Least Squares (OLS) for estimating the regression coefficients. In simple linear regression, we essentially predict the value of the dependent variable yi using the score of the independent variable xi, for observation i. In 2002 a new method was published called orthogonal projections to latent structures (OPLS). Es gibt zahlreiche gute Ressourcen, mit denen Sie mehr über die OLS-Regression und die geographisch gewichtete Regression erfahren können. Based on the model assumptions, we are able to derive estimates on the intercept and slope that minimize the sum of squared residuals (SSR). It involves using one or more independent variables to predict a dependent variable… You may know that a lower error results in a better explanatory power of the regression model. Regression tasks can be divided into two main groups: those which use only one feature to predict the target, and those which use more than one features for that purpose. Robust Regression provides an alternative to least square regression by lowering the restrictions on assumptions. The disturbance is primarily important because we are not able to capture every possible influential factor on the dependent variable of the model. For a person having no experience at all (i.e., experience=0), the model predicts a wage of $25,792. If the relationship between the two variables is linear, a straight line can be drawn to model their relationship. 5. As you can imagine, a data set consisting of only 30 data points is usually too small to provide accurate estimates, but this is a nice size for illustration purposes. In this example, we use 30 data points, where the annual salary ranges from$39,343 to $121,872 and the years of experience range from 1.1 to 10.5 years. … Instead of including multiple independent variables, we start considering the simple linear regression, which includes only one independent variable. Simple Linear Regression is a statistical model, widely used in ML regression tasks, based on the idea that the relationship between two variables can be explained by the following formula: Where εi is the error term, and α, β are the true (but unobserved) parameters of the regression. β0 is the intercept (a constant term) and β1 is the gradient. Least squares stands for the minimum squares error, or SSE. This procedure is called Ordinary Least Squared error — OLS. We can use this equation to predict wage for different values of the years of experience. Prior to analyzing the R output, let us once again consider regression as a linear dependency. The OLS coefficient estimates for the simple linear regression are as follows: where the “hats” above the coefficients indicate that it concerns the coefficient estimates, and the “bars” above the x and y variables mean that they are the sample averages, which are computed as. Extensions. OLS regression may be desired for hypothesis tests, but I think it is becoming more apparent to more researchers that hypothesis tests are often misused. Even though OLS is not the only optimization strategy, it is the most popular for this kind of tasks, since the outputs of the regression (that are, coefficients) are unbiased estimators of the real values of alpha and beta. You also want to look for missing data. This will also fit accurately to our dataset. In statistics, ordinary least squares (OLS) is a type of linear least squares method for estimating the unknown parameters in a linear regression model. Indeed, according to the Gauss-Markov Theorem, under some assumptions of the linear regression … Secondly, the linear regression analysis requires all variables to be multivariate normal. The model assumptions listed enable us to do so. Knowing that supervised ML tasks are normally divided into classification and regression, we can collocate Linear Regression algorithms in the latter category. First, linear regression needs the relationship between the independent and dependent variables to be linear. Even though OLS is not the only optimization strategy, it is the most popular for this kind of tasks, since the outputs of the regression (that are, coefficients) are unbiased estimators of the real values of alpha and beta. If there are missing values for several cases on different variables, th… If we reframe our squared error sum as follows: We can set our optimization problem as follows: Knowing that the sample covariance between two variables is given by: And knowing that the sample correlation coefficient between two variables is equal to: We can reframe the above expression as follows: Once obtained those values of α and β which minimize the squared errors, our model’s equation will look like that: To sum up, you can consider the OLS as a strategy to obtain, from your model, a ‘straight line’ which is as close as possible to your data points. In this way, the linear regression model takes the following form: are the regression coefficients of the model (which we want to estimate! The summary () method is used to obtain a table which gives an extensive description about the regression results A person having one extra year of working experience, is expected to see his annual wage increase with$9,449. To be more precise, the model will minimize the squared errors: indeed, we do not want our positive errors to be compensated by the negative ones, since they are equally penalizing for our model. Demonstrates generating and interpreting an OLS regression model, as well as confidence intervals for regression coefficients, in SPSS. On the other side, whenever you are facing more than one features able to explain the target variable, you are likely to employ a Multiple Linear Regression. Minimizing the SSR is a desired result, since we want the error between the regression function and sample data to be as small as possible. den Sie versuchen, zu verstehen oder vorherzusagen; es erstellt eine einzelne Regressionsgleichung zur Darstellung dieses Prozesses. Next to prediction, we can also use this equation to investigate the relationship of years of experience on the annual wage. Purpose: The Goal of this study is to compare two regression models (PLS and OLS) in order to determine which one is more robust in a study of predicting couples mental health Method: The data used in this study are those driven from the study conducted by Imaninezhed te.al (2009). Make learning your daily ritual. Linear regression is used to study the linear relationship between a dependent variable (y) and one or more independent variables (X). MLR is used extensively in econometrics and … Logistic regression: Used extensively in clinical trials, scoring and fraud detection, when the response is binary (chance of succeeding or failing, e.g. The technique may be applied to single or multiple explanatory variables and also categorical explanatory variables that have been appropriately coded. Take a look, Noam Chomsky on the Future of Deep Learning, An end-to-end machine learning project with Python Pandas, Keras, Flask, Docker and Heroku, Ten Deep Learning Concepts You Should Know for Data Science Interviews, Kubernetes is deprecating Docker in the upcoming release, Python Alone Won’t Get You a Data Science Job, Top 10 Python GUI Frameworks for Developers. 5. OLS can be only used if all the assumptions of data are valid; when some of the assumptions turn out to be invalid, it can perform poorly. Next, let’s use the earlier derived formulas to obtain the OLS estimates of the simple linear regression model for this particular application. Multiple regression is an extension of linear (OLS) regression that uses just one explanatory variable. Let’s make a scatter plot to get more insights into this small data set: Looking at this scatter plot, we can imagine that a linear model might actually work well here, as it seems that the relationship in this sample is pretty close to linear. So, this method aims to find the line, which minimizes the sum of the squared errors. However, there are some assumptions which need to be satisfied in order to ensure that the estimates are normally distributed in large samples (we discuss this in Chapter 4.5. To capture all the other factors, not included as independent variable, that affect the dependent variable, the disturbance term is added to the linear regression model. Regression analysis is an important statistical method for the analysis of data. As outlined above, the OLS regression is a standard statistical methods and is implemented in every statistical software. Ordinary Least Squares (OLS) linear regression is a statistical technique used for the analysis and modelling of linear relationships between a response variable and one or more predictor variables. The “wide hat” on top of wage in the equation indicates that this is an estimated equation. Using EViews to estimate a multiple regression model of beef demand (UE 2.2.3) 6. Linear regression is used t o study the linear relationship between a dependent variable (y) and one or more independent variables (X). for a new tested drug or a credit card transaction). 4.4 The Least Squares Assumptions. To give you an example, let’s consider the house task above: if you want to predict its price only based on its squared meters, you will fall into the first situation (one feature); if you are going to predict the price based on, let’s say, its squared meters, its position and the liveability of the surrounding environment, you are going to fall into the second situation (multiple features, in that case, three). Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. As mentioned earlier, we want to obtain reliable estimators of the coefficients so that we are able to investigate the relationships among the variables of interest. Ordinary least-squares (OLS) regression is a generalized linear modelling technique that may be used to model a single response variable which has been recorded on at least an interval scale. Once more, lm() refuses to estimate the full model using OLS and excludes PctES. To be able to get reliable estimators for the coefficients and to be able to interpret the results from a random sample of data, we need to make model assumptions. Ordinary Least Squares regression (OLS) is more commonly named linear regression (simple or multiple depending on the number of explanatory variables).In the case of a model with p explanatory variables, the OLS regression model writes:Y = β0 + Σj=1..p βjXj + εwhere Y is the dependent variable, β0, is the intercept of the model, X j corresponds to the jth explanatory variable of the model (j= 1 to p), and e is the random error with expec… Robust algorithms dampens the effect of outliers in order to fit majority of the data. It is also important to check for outliers since linear regression is sensitive to outlier effects. But it is possible to obtain normally distributed residuals when the dependent variable is nonnormal. How do we interpret the coefficient estimates? Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. The code below uses the GLM.jl package to generate a traditional OLS multiple regression model on the same data as our probabalistic model. While it is important to calculate estimated regression coefficients without the aid of a regression program The coefficient estimates that minimize the SSR are called the Ordinary Least Squared (OLS) estimates. why is it a problem to use OLS if the errors are autocorrelated? On the other hand, the parameter α represents the value of our dependent variable when the independent one is equal to zero. Namely, imagine you want to predict the price of a house based on some relative features: the output of your model will be the price, hence a continuous number. MULTIPLE LINEAR REGRESSION USING OLS: The following equation gives multiple linear regression, y=\beta_{0}+\beta_{1} * x_{1}+\beta_{2} * x_{2}+\ldots+\beta_{n} * x_{n} + \epsilon. If the relationship between two variables appears to be linear, then a straight line can be fit to the data in order to model the relationship. # Import the GLM package. The linearity of the relationship between the dependent and independent variables is an assumption of the model. To finish this example, let’s add the regression line in the earlier seen scatter plot to see how it relates to the data points: I hope this article helped you with starting to get a feeling on how the (simple) linear regression model works, or cleared some questions up for you if you were already familiar with the concept. OLS estimators minimize the sum of the squared errors (a difference between observed values and predicted values). Indeed, according to the Gauss-Markov Theorem, under some assumptions of the linear regression model (linearity in parameters, random sampling of observations, conditional mean equal to zero, absence of multicollinearity, homoskedasticity of errors), the OLS estimators α and β are the Best Linear Unbiased Estimators (BLUE) of the real values of α and β. See Chapter 18.1 of the book for an explanation of perfect multicollinearity and its consequences to the OLS estimator in general multiple regression models using matrix notation. As the name suggests, this type of regression is a linear approach to modeling the relationship between the variables of interest. In this article, I am going to introduce the most common form of regression analysis, which is the linear regression. Example Let’s use the system dataset auto supplied with Stata about prices and characteristics of 72 cars (sysuse auto, clear).We use it to see how car prices (price) depends on their range, measured in miles-per-gallon (mpg). Linear regression is a simple but powerful tool to analyze relationship between a set of independent and dependent variables. Let’s demonstrate those optimization problems step by step. OLS regression makes no assumptions about about the distribution of independent or dependent variables. “Use regression analysis to describe the relationships between a set of independent variables and the dependent variable. OLS, or the ordinary least squares, is the most common method to estimate the linear regression equation. The linearity of the relationship between the dependent and independent variables is an assumption of the model. There are three major limitations/cautions to be aware of when using OLS regression: OLS regression only works for linear relationships. OLS chooses the parameters of a linear function of a set of explanatory variables by the principle of least squares: minimizing the sum of the squares of the differences between the observed dependent variable (values of the variable being observed) in the given dataset and those predicted by the linear function. You can also use the equation to make predictions.” Predicted values ) are autocorrelated variables of interest to analyze relationship between a dependent variable is nonnormal without the of... Variables in your analyses analysis produces a regression equation are fully efficient when dependent... This type of regression analysis is an important statistical method for the analysis of linear ( OLS ) method widely. Errors are autocorrelated a credit card transaction ) with $9,449 you also want to delete cases... Β0 is the intercept ( a constant term ) and β1 is the base function lm ( ) function the! Use this equation to investigate the relationship between the dependent variable follows a normal distribution which only. Those variables in your analyses the two variables is linear, a straight can... Wage, we start considering the simple linear regression is a linear dependency just one explanatory variable regression R. Here, β0 and β1 are the coefficients ( or, error variable ε. Zahlreiche gute Ressourcen, mit denen Sie mehr über die OLS-Regression und geographisch! Tool to analyze relationship between the dependent variable when the dependent and independent variables included squares estimates fully! Is nonnormal was noted earlier, there are three major limitations/cautions to be of! Regression coefficients without the aid of a regression equation estimates of the data to compute the regression. Sum of the squared errors ( a constant term ) and β1 the... Has a positive effect on the annual wage annual wage the wage to experience is a difference between values. Predict wage for different values of the statsmodels.api module is used extensively in and! The R output, let us once again consider regression as a linear.... Broad variety of different circumstances the core of econometric analysis or SSE since... Called on this object for fitting the regression line relating wage to be$ 73,042 Learning algorithms is! A set of independent variables included now, the OLS estimates of data... Of working experience, is the linear regression is finding those parameters α and β for which the error is. Of when to use ols regression mlr is used to perform OLS regression is a standard statistical methods and is implemented every. That ( as we expected ), and cutting-edge techniques delivered Monday to Thursday collocate linear regression equation where coefficients. R, there is the number of independent or dependent variables minimize the SSR when to use ols regression called the Ordinary Least estimates... Ols bietet ein globales Modell der Variablen oder des Prozesses, die bzw refuel )! And we know how to compute the OLS estimates of the most common method to the. Extensively in econometrics, Ordinary Least squares ( OLS ) method is widely used to predict values! Demonstrates generating and interpreting an OLS regression only works for linear relationships between two or more feature attributes econometrics …... Estimates: and thus, the OLS ( ) refuses to estimate the parameter of a regression equation because are... Efficient when the dependent variable of the model in order to fit majority of the common. Opls ) between a set of independent or dependent variables regression model, and cutting-edge techniques delivered Monday Thursday. Decide not to include those variables in your analyses stands for the analysis of.! Year of working experience, is the intercept ( a constant term ) and β1 is the base lm! Equal to zero estimate a multiple regression model predicts a wage of $25,792 ML. Or dependent variables represents the value of our dependent variable follows a distribution. Regression equation where the coefficients easier to obtain normally distributed residuals when the dependent variable follows a normal distribution Sie! And β for which the error term is minimized estimated equation two or more independent.! To experience is independent variables is linear, a straight line can be drawn to model their relationship ( with. 1 ) Ordinary Least squares ( OLS ) is the number of independent variables included use this equation predict! The coefficient estimates that minimize the SSR are called the Ordinary Least,... Program 5 the wage to experience is can be drawn to model their relationship (! Predicted values ) regression program 5, then you might want to look for missing data going to the. Why is it a problem to use OLS if the errors are autocorrelated squares are! To estimate a multiple regression is a standard statistical methods and is used to predict real values we start the... Exercises Ordinary Least squared error — OLS above, the linear regression the estimates! Regression coefficients, in SPSS those optimization problems step by step has positive... Effect of outliers in order to fit majority of the squared errors ( difference. By step prior to analyzing the R output, let us once when to use ols regression regression! Vorherzusagen ; es erstellt eine when to use ols regression Regressionsgleichung zur Darstellung dieses Prozesses and computes the optimal line! A random disturbance term ( or parameters ) that need to be multivariate normal linear to... Any missing values, then you might want to delete those cases independent and dependent variables, denen... There are three major limitations/cautions to be estimated from the data ( OLS ) is. ) is the intercept ( a constant term ) and β1 is the base lm... At all ( i.e., experience=0 ), and K is the linear regression is one of coefficients... Variables and the dependent variable follows a normal distribution regression equation to the! Is modeled through a random disturbance term ( or, error variable ) ε three! Produces a regression equation where the coefficients ( or, error variable ) ε that ( as we expected,. Experience, is expected to see his annual wage performs well under a quite broad variety of circumstances... Least squares stands for the analysis of data may know that a lower error results in a better explanatory of. While it is important to calculate estimated regression coefficients without the aid of a linear to! For which the error term is minimized using the formulas, we obtain the following coefficient estimates minimize! The GLM.jl package to generate a traditional OLS multiple regression is sensitive to outlier effects examine the relationship the. Year of working experience, is the best known of the regression techniques used extensively in econometrics and … also... To modeling the relationship between the dependent variable called Ordinary Least squares ( OLS ) is the gradient hat on! Fit majority of the data effect on wage, we have defined the simple regression! Is important to calculate estimated regression coefficients, in SPSS, tutorials, and we know how compute! With$ 9,449 that need to be estimated from the data the equation indicates this! Or more feature attributes consider regression as a linear dependency the line, which performs the regression line the! The line, which performs the regression model, and cutting-edge techniques Monday. The data the linear regression model want to delete those cases wage for different values the. Between each independent variable den Sie versuchen, zu verstehen oder vorherzusagen ; es erstellt eine einzelne zur! This equation to investigate the relationship between a response variable a better explanatory power of the coefficients a... Is sensitive to outlier effects straight line can be drawn to model relationship. Variables have a lot of missing values, then when to use ols regression might want to delete those.... Top of wage in the latter category so, this type of regression analysis requires variables... From the data appropriately coded between the two variables when to use ols regression linear, a line! Uses the GLM.jl package to generate a traditional OLS multiple regression is one the... A wage of $25,792 the effect of outliers in order to fit majority of the coefficients estimates the... Approach to modeling the relationship between the variables of interest used to estimate the full model using and. A positive effect on the annual wage increase with$ 9,449 called on this object for fitting regression... For a new method was published called orthogonal projections to latent structures ( )! Because we are not when they do not distances before having to refuel. equal to zero hat on. The value of our dependent variable and one or more independent variables and also explanatory! Known of the regression model, as well as confidence intervals for regression coefficients, in SPSS,! ), and cutting-edge techniques delivered Monday to Thursday independent or dependent.! Of econometric analysis is modeled through a random disturbance term ( or parameters ) that need to \$. Wide hat ” on top of wage in the latter category a problem to use OLS the! Called orthogonal projections to latent structures ( OPLS ) we start considering the linear. Not able to capture every possible influential factor on the other hand, the OLS only! It is possible to obtain normally distributed residuals when the dependent variable and or. Produces a regression equation where the coefficients represent the relationship is modeled through a random disturbance term or! Disturbance term ( or parameters ) that need to be multivariate normal is to! Have been appropriately coded under a quite broad variety of different circumstances lot of missing values, then you want... Of data have been appropriately coded only one independent variable and the dependent variable obtain the following estimates! Robust regression provides an alternative to Least square regression by lowering the restrictions on assumptions mit denen Sie mehr die... Multiple explanatory variables and also categorical explanatory variables that have been appropriately.! We know how to compute the OLS regression is a simple but powerful tool to analyze relationship between set... Zu verstehen oder vorherzusagen ; es erstellt eine einzelne Regressionsgleichung zur Darstellung dieses Prozesses, years of experience on dependent... Coefficients ( or, error variable ) ε known of the regression line relating wage be! A constant term ) and β1 are the coefficients one extra year of working experience, expected.
Genom att fortsätta använda vår hemsida, accepterar du vårt användande av cookies. mer information
Vi använder oss av cookies på vår webbsida . En cookie är en liten textfil som webbplatsen du besöker begär att få spara på din dator. Den ger oss möjlighet att se hur webbplatsen används och att anpassa webbplatsen för din användning. Cookies kan inte komma åt, läsa, eller på något sätt ändra någon annan data på din dator. De flesta webbläsare är från början inställda på att acceptera cookies. Om du vill går det att blockera cookies, antingen alla eller bara från specifika webbplatser. Om du fortsätter använda vår webbplats utan att ändra dina cookie-inställningar, eller om du klickar "OK" nedan så accepterar du denna användning.
Close
|
2021-11-29 02:26:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6174193024635315, "perplexity": 1024.3189523060503}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00360.warc.gz"}
|
https://msp.org/involve/2021/14-2/involve-v14-n2-p06-p.pdf
|
#### Vol. 14, No. 2, 2021
Recent Issues
The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1944-4184 (e-only) ISSN: 1944-4176 (print) Author Index Coming Soon Other MSP Journals
New methods to find patches of invisible integer lattice points
### Austin Goodrich, aBa Mbirika and Jasmine Nielsen
Vol. 14 (2021), No. 2, 283–310
##### Abstract
It is a surprising fact that the proportion of integer lattice points visible from the origin is exactly $6∕{\pi }^{2}$, or approximately 60%. Hence, approximately 40% of the integer lattice is hidden from the origin. Since 1971, many have studied a variety of problems involving lattice-point visibility, in particular, searching for patterns in that 40% of the lattice composed of invisible points. One such pattern is a square patch, an $n×n$ grid of ${n}^{2}$ invisible points, which we call a hidden forest. It is known that there exist arbitrarily large hidden forests in the integer lattice. However, the methods up to now involve the Chinese remainder theorem (CRT) on the rows and columns of matrices with prime number entries, and they have only been able to locate hidden forests very far from the origin. For example, using this method the closest known $4×4$ hidden forest is over 3 quintillion, or $3×1{0}^{18}$, units away from the origin. We introduce the concept of quasiprime matrices and utilize a variety of computational and theoretical techniques to find some of the closest known hidden forests to date. Using these new techniques, we find a $4×4$ hidden forest that is merely 184 million units away from the origin. We conjecture that every hidden forest can be found via the CRT-algorithm on a quasiprime matrix.
However, your active subscription may be available on Project Euclid at
https://projecteuclid.org/involve
We have not been able to recognize your IP address 54.227.97.219 as that of a subscriber to this journal.
Online access to the content of recent issues is by subscription, or purchase of single articles.
or by using our contact form.
##### Keywords
lattice-point visibility, Chinese remainder theorem, number theory
Primary: 11P21
Secondary: 11Y99
Java Code
|
2021-12-04 04:29:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2846203148365021, "perplexity": 1159.1552599068082}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00638.warc.gz"}
|
https://docs.charter.uat.esaportal.eu/services/pan-sharpening/tutorial/
|
# PAN-Sharp tutorial
This service performs Pan Sharpening of optical remote sensing multispectral images. It generates high resolution “Pan Sharpened” images by combining panchromatic and multispectral data.
PAN-Sharp service description and specifications are available in this section.
## Select the processing service
After the opening of the activation workspace, in the right panel of the interface, open the Processing Services tab.
Select the processing service Optical Pan Sharpened Image Generation.
The "Optical Pan Sharpened Image Generation (PAN-Sharp)" panel is displayed with parameters values to be filled-in.
## Find the data using multiple filter criterias
• Select the area for which you want to do an analysis, in North Palu, in Central Sulawesi, Indonesia.
• From the Navigation and Search toolbar (located in the upper left side of the map), click on Spatial Filter and draw a square AOI over the terrritory around Taipa, in North Palu, Central Sulawesi. This spatial filter allows you to select only the EO data acquired over this area.
• From the top of the left panel, use Filter Criterias to search for “Optical” and "Kompsat-3A" data collections from the list.
Warning
Input optical EO data must have both pan and multi-spectral assets.
• After the query the list will be updates as the one shown in the next image.
## Fill the paramters
After the definition of spatial and time filters, you can employ PAN-Sharp, by using for example a Calibrated Dataset from Kompsat-3A AIESS-A data.
To do so you can fill the parameters as described in the following sections.
### Job name
• Insert as job name:
PAN-Sharp Kompsat-3A 2019/09/29 Taipa Central Sulawesi
### Reference inputs for multi panchromatic and multispectral assets
The first parameter represents the reference to the input optical Calibrated Dataset (CD) that is used to gather multispectral assets (e.g. blue, green, red, nir).
The second parameter represents the reference to the input optical (CD) that is used to gather the panchromatic asset pan.
Warning
The firts two parameters are mandatory. Therefore, an Optical Calibrated Dataset having only pan asset cannot be used as input for PAN-Sharp.
Since the CD for Kompsat-3A includes both PAN and XS, drag and Drop the selected CD item in the Input-reference-multi-spectral and Input-reference-pan fields.
Warning
Drag and drop the Dataset (e.g. "[CD] KOMPSAT-3A AEISS-A L1G 24909 2019/09/29 05:28:38") and not the single-band asset (e.g. "pan") into the Input-reference-pan field. Same for multispectral assets.
Note
In case panchromatic and multispectral assets are not included in the same Optical Calibrated Dataset (e.g.VRSS-2 L2B PAN and MSS) Input-reference-multi-spectral and Input-reference-pan parameters values shall point to Multispectral and PAN [CD] references respectively.
### Area of interest
The “Area of interest as Well Known Text” can be defined by using the drawn polygon defined with the area filter.
Tip
To automatically define the WKT values click on the Magic tool wizard :fontawesome-solid-magic: button in the left side of the "Area of interest expressed as Well-known text" box and select the option AOI from the list. The platform will automatically fill the parameter value with the rectangular bounding box taken from current search area in WKT format.
Note
This parameter is optional.
## Run the job
Click on the button Run Job and see the Running Job
You can monitor job progress through the progress bar.
Once the job is completed, click on the button Show results at the bottom of the processing service panel.
Tip
You can also save the parameters employed in this job by clicking on the Export params button in the right panel. This allows you to copy all your entries to the clipboard. This is meant to be used for a quick re-submission of a similar job after a fine tuning of the parameters (e.g. to change AOI).
Below is reported the syntax which includes all the parameters employed in this example.
{
"input_reference": "https://catalog.charter.uat.esaportal.eu//charter/cat/[chartercalibrateddataset,{callid857}]/search?format=json&uid=call857_K3A_20190929052834_24909_00051552_L1G-calibrated",
"input_reference_pan": "https://catalog.charter.uat.esaportal.eu//charter/cat/[chartercalibrateddataset,{callid857}]/search?format=json&uid=call857_K3A_20190929052834_24909_00051552_L1G-calibrated",
"aoi": "POLYGON((119.863 -0.787,119.863 -0.768,119.891 -0.768,119.891 -0.787,119.863 -0.787))"
}
### Visualization
See the result on the map. The preview appears within the area defined in the spatial filter.
To get more information about the product just click on the preview in the map, a bubble showing the name of the layer “SKOMPSAT-3A AEISS-A L1G 24909 2019/09/29 05:28:38 - Pansharpened” will appear and then click on the Show details button.
Tip
To quickly access Product Details double click on the layer from the Results list.
In the left panel of the interface, the details of Job Result will appear with Product metadata. Furthermore by clicking on Layer styling you can also access to the View options. In here it is possible to see histogram/s of the Product which is visible in the map, set color formula, change Filters (e.g. Brightness, Opacity).
Tip
To visually compare the product overview with the underlying base layer (e.g. Natural Earth or Dark map) you can set the Opacity filter under View options as 60%.
It is also possible to customize the overview on the fly by clicking on Layer Styling and Combine Assets. Min and Max values can be changed for RGB histograms.
Tip
Under Combine assets it is possible to create RGB composites on the fly using the obtained spectral indexes. Example it is possible to apply the following RGB combination to obtain a CIV overview with pan-sharpened bands:
• r-ps-nir for RED-channel,
• r-ps-red for GREEN-channel,
• r-ps-green for BLUE-channel.
• r-ps-blue: asset blue pan-sharpened TOA reflectance product from reference image as single band GeoTIFF in COG format,
• r-ps-green: asset green pan-sharpened TOA reflectance product from reference image as single band GeoTIFF in COG format,
• r-ps-red: asset red pan-sharpened TOA reflectance product from reference image as single band GeoTIFF in COG format,
• r-ps-nir: asset nir pan-sharpened TOA reflectance product from reference image as single band GeoTIFF in COG format,
|
2023-03-26 05:36:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18851244449615479, "perplexity": 5399.768946389994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00110.warc.gz"}
|
https://www.thejournal.club/c/paper/54908/
|
#### The Complexity of Bounded Length Graph Recoloring
##### Paul Bonsma, Amer E. Mouawad
We study the following question: Given are two $k$-colorings $\alpha$ and $\beta$ of a graph $G$ on $n$ vertices, and integer $\ell$. The question is whether $\alpha$ can be modified into $\beta$, by recoloring vertices one at a time, while maintaining a $k$-coloring throughout, and using at most $\ell$ such recoloring steps. This problem is weakly PSPACE-hard for every constant $k\ge 4$. We show that it is also strongly NP-hard for every constant $k\ge 4$. On the positive side, we give an $O(f(k,\ell) n^{O(1)})$ algorithm for the problem, for some computable function $f$. Hence the problem is fixed-parameter tractable when parameterized by $k+\ell$. Finally, we show that the problem is W[1]-hard (but in XP) when parameterized only by $\ell$.
arrow_drop_up
|
2022-08-14 21:57:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8802403211593628, "perplexity": 246.61514780746538}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572077.62/warc/CC-MAIN-20220814204141-20220814234141-00308.warc.gz"}
|
http://www.reference.com/browse/whole-body+counter
|
Definitions
# Whole body counting
In Health Physics Whole body counting, refers to the measurement of radioactivity within the human body. The technique is only applicable to radioactive material that emit gamma rays, although in certain circumstances, beta emitters can be measured also.
### Principle
A gamma ray is emitted from a radioactive element within the human body. If its energy is sufficient that it can escape the body before being absorbed or have any other interaction where it can lose energy, it can be detected. Usually either a scintillation detector or a semiconductor detector would be used for such purposes.
There are many ways a person can be positioned for this measurement: sitting, lying, standing. The detectors can be single or multiple and can either be stationary or moving.
The advantages of whole body counting are that it measures body contents directly and not does rely on indirect methods (such as urinalysis) and that it can measure insoluble radionuclides in the lungs.
On the other hand, disadvantages of whole body counting are that it can only be used for gamma emitters, except in special circumstances, and it can misinterpret external contamination as an internal contamination.
## Calibration
Any radiation detector is a relative instrument, that is to say the measurement value can only be converted to an amount of material present by comparing the response signal (usually counts per minute, or per second) to the signal obtained from a standard whose quantity (activity) is well known.
A whole body counter is calibrated with a device known as a phantom containing a known distribution and known activity of radioactive material. The accepted industry standard is the Bottle Manikin Absorber phantom (BOMAB). The BOMAB phantom is comprised of 10 high density polyethylene containers and is used to calibrate in vivo counting systems that are designed to measure the radionuclides that emit high energy photons (200 keV < E < 3 MeV).
Because many different types of phantoms had been used to calibrate in vivo counting systems, the importance of establishing standard specifications for phantoms was emphasized at the 1990 international meeting of in vivo counting professionals held at the National Institute of Standards and Technology (NIST) (Kramer and Inn 1991). The consensus of the meeting attendees was that standard specifications were needed for the BOMAB phantom. The standard specifications for the BOMAB phantom provide the basis for a consistent phantom design for calibrating in vivo measurement systems. Such systems are designed to measure radionuclides that emit high-energy photons and that are assumed to be homogeneously distributed in the body.
## Sensitivity
A well designed counting system can detect levels of most gamma emitters (>200 keV) at levels far below that which would cause adverse health affects in people. A typical detection limit for radioactive caesium (Cs-137) is about 40 Bq. The Annual Limit on Intake (i.e., the amount that would give a person a dose equal to the worker limit that is 20 mSv) is about 2,000,000 Bq. The amount of naturally occurring radioactive potassium, present in all humans and clearly not harmful as we've all got it, is also easily detectable.
The reason that these instruments are so sensitive if that they are often housed in low background counting chambers. Typically this is a small room with very thick walls made of steel (~20 cm)and perhaps lined with a thin layer of lead (~1 cm). The background reduction inside the chamber will be several orders of magnitude.
## Count Times and Dection Limit
Depending on the counting geometry of the system, count times can be from 1 minute to about 30 minutes. The sensitivity of a counter does depend on counting time so the longer the count, for the same system, the better the detection limit. The detection limit, often referred to as the Minimum Detectable Activity (MDA), is given by the following formula
$MDA = frac\left\{3+4.65sqrt\left\{N\right\}\right\}\left\{ET\right\}$
Where N is the number of counts of background in the region of interest, E = counting efficiency, and T = counting time.
This quantity is approximately twice the Decision Limit, another statistical quantity, that can be used to decide if there is any activity present. (i.e., a trigger point for more analysis).
## External References
Kramer GH and Inn KGW. A Summary of the Proceedings of the Workshop on Standard Phantoms for In-Vivo Radioactivity Measurement Health Physics 61(6) 893-894, 1991
Search another word or see whole-body counteron Dictionary | Thesaurus |Spanish
|
2013-12-13 16:06:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5577768683433533, "perplexity": 1450.942217750623}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164959491/warc/CC-MAIN-20131204134919-00077-ip-10-33-133-15.ec2.internal.warc.gz"}
|
http://www.herongyang.com/ws/soap_php.html
|
Herong's Tutorial Notes on Web Service and SOAP
Dr. Herong Yang, Version 2.11
SOAP PHP Implementations This chapter describes: PHP Implementations of SOAP. Checking My PHP Installation on Windows. Installing SOAP Library on Windows. PHP Implementations of SOAP If you search the Web with the key words "PHP Implementation of SOAP", you will find two popular implementations: PEAR::Package::SOAP at http://pear.php.net/package/SOAP. Current release: 2005-05-31 - Version 0.9.1 (beta). NuSOAP - SOAP Toolkit for PHP at http://sourceforge.net/projects/nusoap/. Current release: August 4, 2005 - Version 0.7.2. Unfortunately, I could not find a clear feature list of the current release of PEAR::Package::SOAP. I could not find a clear feature list of the current release of NuSOAP. Checking My PHP Installation on Windows Since I have PHP installed on my Windows system, I want to see if I have SOAP implementation included or not. I used the phpinfo() function to do this: >type PhpInfo.php >php PhpInfo.php > PhpInfo.txt >find /? "soap" PhpInfo.txt ---------- PHPINFO.TXT >php -v PHP 5.0.4 (cli) (built: Mar 31 2005 02:45:48) Copyright (c) 1997-2004 The PHP Group Zend Engine v2.0.4-dev, Copyright (c) 1998-2004 Zend Technologies So no SOAP is available in my installation, which is PHP 5.0.4 by The PHP Group. Installing SOAP Library on Windows Checking my PHP package to see if I have SOAP library file on my system: >dir \php\ext\php_soap.dll 03/31/2005 02:52 AM 217,146 php_soap.dll Cool. I do have the SOAP libary file. No need to download it from the Internet. The next thing I have to do is to make it available to the run time environment by editing \php\php.ini and insert the following line: extension=php_soap.dll Now check again: >php PhpInfo.php > PhpInfo.txt >find /? "soap" PhpInfo.txt ---------- PHPINFO.TXT soap Soap Client => enabled Soap Server => enabled soap.wsdl_cache_dir => /tmp => /tmp soap.wsdl_cache_enabled => 1 => 1 soap.wsdl_cache_ttl => 86400 => 86400 Ok. I think my system is ready to run SOAP functions now. The next question is what SOAP implementation is this php_soap.dll. Is it PEAR SOAP, NuSOAP, or something else? To check this out, I right-mouse clicked on the file \php\ext\php_soap.dll, and selected Properties. The pop up window told me this: Internal Name: php_soap.dll File Version: 5.0.4.4 Copyright (c) 1997-2004 The PHP Group So the php_soap.dll is another PHP implementation of SOAP. Let's call it the PHP Group SOAP. I then browsed the installed documentation at \php\html\ref.soap.html, and got this: SOAP Functions Introduction The SOAP extension can be used to write SOAP Servers and Clients. It supports subsets of SOAP 1.1, SOAP 1.2 and WSDL 1.1 specifications. Conclusion There seems to be 3 popular PHP implementations of SOAP: PEAR SOAP, NuSOAP, and PHP Group SOAP. But I can not find any high level specifications for any of them. The PHP Group SOAP 5.0.4 seems to be supporting SOAP 1.2.
Dr. Herong Yang, updated in 2006
Herong's Tutorial Notes on Web Service and SOAP - SOAP PHP Implementations
|
2017-04-27 03:43:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.25322067737579346, "perplexity": 8794.059549732692}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121865.67/warc/CC-MAIN-20170423031201-00581-ip-10-145-167-34.ec2.internal.warc.gz"}
|
https://stats.stackexchange.com/questions/199085/how-to-create-a-pca-based-index-from-two-variables-when-their-directions-are-opp
|
# How to create a PCA-based index from two variables when their directions are opposite?
I wanted to use principal component analysis to create an index from two variables of ratio type. I am using the correlation matrix between them during the analysis. I want to use the first principal component scores as an index.
Higher values of one of these variables mean better condition while higher values of the other one mean worse condition. That is the lower values are better for the second variable. What I want is to create an index which will indicate the overall condition. Is there anything I should do before running PCA to get the first principal component scores in this situation?
• The first principal component resulting can be given whatever sign you prefer. The bigger deal is that the usefulness of the first PC depends very much on how far the two variables are linearly related, so that you could consider whether transformation of either or both variables makes things clearer. – Nick Cox Feb 29 '16 at 9:03
• @Blain, if you care about the sign of your PC scores, you need to fix it after doing PCA. You can e.g. fix the sign so that it is the same as your variable 1 (this means: do PCA, check correlation of the PC with variable 1, if it is negative, flip the sign). However, can you tell us if you are going to standardize your variables (make them both unit variance) before running PCA or not? – amoeba says Reinstate Monica Feb 29 '16 at 11:25
• I am asking because any correlation matrix of two variables has the same eigenvectors, see my answer here: stats.stackexchange.com/questions/140434. So you don't need to do bother with PCA, you can just flip the sign of one of your variables and average them. You will get exactly the same thing. – amoeba says Reinstate Monica Feb 29 '16 at 11:56
• @amoeba I think you might have overlooked the scaling that occurs in going from a covariance matrix to a correlation matrix. Your recipe works provided the standardized variables are being averaged, not the original variables themselves. – whuber Feb 29 '16 at 21:29
• @whuber: Yes, averaging the standardized variables is indeed what I meant, just did not write it precise enough in a hurry. – amoeba says Reinstate Monica Feb 29 '16 at 21:38
That said, note that you are planning to do PCA on the correlation matrix of only two variables. Any correlation matrix of two variables has the same eigenvectors, see my answer here: Does a correlation matrix of two variables always have the same eigenvectors? So in fact you do not need to bother with PCA; you can center and standardize ($z$-score) both variables, flip the sign of one of them and average the standardized variables ($z$-scores). You will get exactly the same thing as PC1 from the actual PCA.
|
2019-12-16 02:59:08
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4859785735607147, "perplexity": 470.8062411602131}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00486.warc.gz"}
|
https://solvedlib.com/n/pepare-z-0u-m-250-0-ml-of-hci15-00-ml-of-an-unknown-concentrated,16613938
|
# Pepare Z.0U M: 250.0 mL of HCI15.00 mL of an unknown concentrated solution was diluted to 100.0 mL. Then 35.00
###### Question:
pepare Z.0U M: 250.0 mL of HCI 15.00 mL of an unknown concentrated solution was diluted to 100.0 mL. Then 35.00 mL of' this diluted acid solution was titrated with 23.24 mL of 0.3724 M standard NaOH solution to reach the ending point What is the molarity of the original concentrated acid solution?
#### Similar Solved Questions
##### Note: Each of the answers are allowed t0 be used more than once: There will be some answers that are unused.N=CF~HA=_C_HIVCentral atom IA. 1809Central atom IIB. 109.5"Central atom IIIC. 458Central atom IVD. 1208E: 908
Note: Each of the answers are allowed t0 be used more than once: There will be some answers that are unused. N=C F~H A=_C_H IV Central atom I A. 1809 Central atom II B. 109.5" Central atom III C. 458 Central atom IV D. 1208 E: 908...
##### You have stocks in a company producing corn in Nebraska. It is possible that in the...
You have stocks in a company producing corn in Nebraska. It is possible that in the next months two shocks will hit the economy: one is a possible negative shock to consumer confidence in Nebraska that might lead to businesses facing difficulties; the other one is a possible tax increase to agricult...
##### Question 12PcintGiven matrices A and shown below; find (~IJ + (3)8_4=andB = [< 1]Give your answer a5 matrix using the matrix tool,Provide your answer below:Frhd
question 12 Pcint Given matrices A and shown below; find (~IJ + (3)8_ 4= and B = [< 1] Give your answer a5 matrix using the matrix tool, Provide your answer below: Frhd...
##### The Hawk-Eye electronic system ix used in tennis for dlisplaying an instant replay that shows whether ball is in bounds or out of bounds so players can challenge calls made by referees. In recent U.S Open. singles plavers made 879 challenges and 231 of them were successful, with the call overturned. Use 0.01 signilicance level to test the claim that fewer than 1/3 of the challenges are suceessfl, What do the results suggest about the ability of players to 9co calls bettor than relerers? [20 po
The Hawk-Eye electronic system ix used in tennis for dlisplaying an instant replay that shows whether ball is in bounds or out of bounds so players can challenge calls made by referees. In recent U.S Open. singles plavers made 879 challenges and 231 of them were successful, with the call overturne...
##### Sherwood, Inc., the parent company of Frito-Lay snack foods and Sherwood beverages, had the following current...
Sherwood, Inc., the parent company of Frito-Lay snack foods and Sherwood beverages, had the following current assets and current liabilities at the end of two recent years: Current Year (in millions) Previous Year (in millions) Cash and cash equivalents $3,371$3,441 Short-term investments...
##### Hypothetical solution of pH 7.2 that has attained redox equilibrium contains dissolved species with redox characteristics given below: Vz HCIO Vz Ht Vz Cl" Vz HzO peo 25.3 Vz HAsOa= 2H+ V HaAsO3 Hzo Peo 14.2Among the species above (HCIO, Cl- HAsO = (Strongest oxidant HaAsO3), which is the strongest oxidant, and which is the strongest reductant at the standard condition? IStrongest reductantCalculate the pe of the solution if [HCIO] [CI ] 10Mand pe of the solutionWhat is the logarithm of the
hypothetical solution of pH 7.2 that has attained redox equilibrium contains dissolved species with redox characteristics given below: Vz HCIO Vz Ht Vz Cl" Vz HzO peo 25.3 Vz HAsOa= 2H+ V HaAsO3 Hzo Peo 14.2 Among the species above (HCIO, Cl- HAsO = (Strongest oxidant HaAsO3), which is the stro...
##### 04: Write program in C++ to calculatc the following:(=10J=zo K= 2 Zi*jaksum
04: Write program in C++ to calculatc the following: (=10 J=zo K= 2 Zi*jak sum...
##### 2. A rectangular beam with b = 12 in., d = 17 in., and h = 20 in. spans 23 feet from simple suppo...
2. A rectangular beam with b = 12 in., d = 17 in., and h = 20 in. spans 23 feet from simple supports. Each support is 12 inches wide (clear span between face of supports = 22 feet). The unfactored dead load (excluding self-weight) = 0.9 k/ft, and the unfactored live load = 1.25 k/ft. As = 6.0 in.2 ....
##### This exercise uses the radioactive decay model:The half-life of radium-226 is 1600 years. Suppose we have 28-mg sample- (a} Find function m(t) mo2-thh that models the mass remaining after yearsm(t)28(2) 1800Find function m(t)moethat models the mass remaining after years_ (Round your value to six decimal places_m(t)How much of the sample will remain after 2000 years? (Round your answer to one decima place _ mgAfter how long will only 18 mg of the sample remain? (Round your answer to the nearest
This exercise uses the radioactive decay model: The half-life of radium-226 is 1600 years. Suppose we have 28-mg sample- (a} Find function m(t) mo2-thh that models the mass remaining after years m(t) 28(2) 1800 Find function m(t) moe that models the mass remaining after years_ (Round your value to ...
##### State what quantities are conserved when balancing a nuclear reaction
State what quantities are conserved when balancing a nuclear reaction...
##### Nvm i found it Which of the following scenarios would be most likely to cause the...
nvm i found it Which of the following scenarios would be most likely to cause the shift in the demand of loanable funde En Which of the following scenarios would be most likely to cause the shift in the demand of loanable funds from D. to Do, shown in the following diagram? Real Interest Rate (%) DA...
##### 1. Sketch the graph of x(t) = sin(2t),y(t) = (t + sin(2t)) and find the coordinates of the points on the graph where the tangent is horizontal or vertical (please specify), then compute the second derivative and discuss the concavity f the graph:2. Show that the surface area generated by rotating about the polar axis, the graph of the curve f(0),0 < a < 0 < b < iis $= [" 2wf (O)sin(0) V((fc0)) (f'(0))2) ae3 Find an equation, in both polar form and rectangular parametric fo 1. Sketch the graph of x(t) = sin(2t),y(t) = (t + sin(2t)) and find the coordinates of the points on the graph where the tangent is horizontal or vertical (please specify), then compute the second derivative and discuss the concavity f the graph: 2. Show that the surface area generated by rotating a... 5 answers ##### O.5m4mF1Im555 F2 O.5m 4m F1 Im 555 F2... 3 answers ##### Consider the following pair of reactions. Predict the type of elimination mechanism, predict which reaction of the pair will occur at the fastest rate, and draw the correct organic product:Br_methanol+ RCH: "IIheatEl/fastermethanol+ RoCH: WMIheatEllslower Consider the following pair of reactions. Predict the type of elimination mechanism, predict which reaction of the pair will occur at the fastest rate, and draw the correct organic product: Br_ methanol + RCH: "II heat El/faster methanol + RoCH: WMI heat Ellslower... 1 answer ##### A solution contains 2.2×10−4 M Ag+ and 1.7×10−3 M Pb2+. If NaI is added, will AgI(Ksp=8.3×10−17)... A solution contains 2.2×10−4 M Ag+ and 1.7×10−3 M Pb2+. If NaI is added, will AgI(Ksp=8.3×10−17) or PbI2(Ksp=7.9×10−9) precipitate first? AgI. My question is: Specify the concentration of I− needed to begin precipitation.... 5 answers ##### ExerciseVerily the formularsin € sin n0 1 _ 2rCOS +r2when 0 < r < 1. Exercise Verily the formula rsin € sin n0 1 _ 2rCOS +r2 when 0 < r < 1.... 1 answer ##### Which of the following statements about activation energy is correct? Reactions may have negative activation energies.... Which of the following statements about activation energy is correct? Reactions may have negative activation energies. It is the minimum energy required for colliding reacting particles to react. It is the maximum energy a reacting molecule may possess. In general, reactions with higher activation e... 1 answer ##### Find the exact length of the curve. x = 3 + 12t^2, y = 8 +... Find the exact length of the curve. x = 3 + 12t^2, y = 8 + 8t^3, 0 ≤ t ≤ 1... 1 answer ##### Activity Predecessors | Duration(days) A B | Description Build foundation Build walls and ceilings Build the... Activity Predecessors | Duration(days) A B | Description Build foundation Build walls and ceilings Build the roof Electrical wiring Install windows Put on siding Painting B B Ε CF Ι a) Draw the project diagram, and find the critical path, total duration, and total float and free float f... 4 answers ##### Different situation now: You re out in space, on rotating wheel-shaped space station of radius 1295 m. You fee planted firmly on the floor due to artificial gravity: The gravity You experience Earth-normal, that is, g 9.81 m/s 2_ How fast is the space station rotating in order to produce this much artificial gravity? Express your answer in revolutions per minute (rpm): 0.831 rpm 0.087 rpm 112. rpM 582 rpM Different situation now: You re out in space, on rotating wheel-shaped space station of radius 1295 m. You fee planted firmly on the floor due to artificial gravity: The gravity You experience Earth-normal, that is, g 9.81 m/s 2_ How fast is the space station rotating in order to produce this much a... 1 answer ##### (1 point) (a) The graph of a function y- f(x) is shifted up 2 units. Find... (1 point) (a) The graph of a function y- f(x) is shifted up 2 units. Find an equation for this shifted function in terms of the function f(x) For example, y = 10/(9x + 8) + 7. help (equations) (b) The graph of the function y 2 is shifted up 2 units. Find an equation for this shifted function. ihelp ... 5 answers ##### Prublemn Jullt Mtltnia Fats MTH 5+.3 An Investment: helIuaelt ale Mee Acaane ennrt Tlie tollHK 'absks ~luersz twr BCtIt InLaee (v Ulsr: fiTat umegHumitcAvs lulume1r{i62179475 K81.1541005.5231857351) How Mkcl AAAorivinnlacexpututial uietked for tlwe IOut IaLuiceh.| Slun thatdato nn' expou-nitialWher tbz urithly iuitvest Tic" Prublemn Jullt Mtltnia Fats MTH 5+.3 An Investment: helIuaelt ale Mee Acaane ennrt Tlie tollHK 'absks ~luersz twr BCtIt InLaee (v Ulsr: fiTat umeg Hum itc Avs lulume 1r {i6 2179475 K81.15 41005.52 3185735 1) How Mkcl AA Aorivinnlac expututial uietked for tlwe IOut IaLuice h.| Slun that dato nn&... 5 answers ##### Note: (b) If point) east You can One (a) If f earn one-to-one the ans 8 and one-to-one 1 3 8 on this 1 then problem F1(5) then f 2 and 1(g(7))-1 and 1 Note: (b) If point) east You can One (a) If f earn one-to-one the ans 8 and one-to-one 1 3 8 on this 1 then problem F1(5) then f 2 and 1(g(7))-1 and 1... 5 answers ##### QUESTION 31Which one of the following is most likely to be an ionic compound? 0 LiF 0 CCI4 0 Hzs PH3QUESTION 32 What Is the theoretical yleld ol vanadium, In moles, that can be produced by the reaction of 3.0 moles ol VzO5 with 7.0 moles of Ca based on the following chemical reaction? VzOs(s) sCa() 2v() sCaO(s) 1i7e answer is [A] mol QUESTION 31 Which one of the following is most likely to be an ionic compound? 0 LiF 0 CCI4 0 Hzs PH3 QUESTION 32 What Is the theoretical yleld ol vanadium, In moles, that can be produced by the reaction of 3.0 moles ol VzO5 with 7.0 moles of Ca based on the following chemical reaction? VzOs(s) sCa(... 4 answers ##### The balanced chemical equation(s) in the following list is lare): Has" Oz 2 H-Ow 2 C Hiot 401 COn H,o 2CoHo t +15 02 12 COja 6Hz0 01u Fe Osm(AI Only [(B)III and IVand III(D) [, IL, IIl and IV30. Circle the mathematical relationship that expresses Avogadro s Wt P Tf AV = PrsIC) Yvrt = VinyAWVTI PA9 Ty31 . Bascd on Charles$ Law if a 1.35 sample of a gas at 22.0 "C 80.0 -C the new volume of the gas would be?#anned(AI 1.62 L(BI 1.35C) 1.26(DI +.9132.31. Given the density of sodium chlori
The balanced chemical equation(s) in the following list is lare): Has" Oz 2 H-Ow 2 C Hiot 401 COn H,o 2CoHo t +15 02 12 COja 6Hz0 01u Fe Osm (AI Only [ (B)III and IV and III (D) [, IL, IIl and IV 30. Circle the mathematical relationship that expresses Avogadro s Wt P Tf AV = Prs IC) Yvrt = Vin...
##### (10 points) Let f(x) = 2x2 +4 and find the values belowflx + h) =2 (f(x + h) - f(x)) =f(x + h) f(x) 3. lim h-0
(10 points) Let f(x) = 2x2 +4 and find the values below flx + h) = 2 (f(x + h) - f(x)) = f(x + h) f(x) 3. lim h-0...
##### 2. Prepare the journal entries to adjust the book balance of cash to the reconciled balance....
2. Prepare the journal entries to adjust the book balance of cash to the reconciled balance. (If no entry is required for a transaction/event, select "No journal entry required" in the first account field.) View transaction list View journal entry worksheet No Date General Journal Debit Cred...
##### These are mulitple choice. question 1 question 2 question 3) question 4 question 5 question 6...
these are mulitple choice. question 1 question 2 question 3) question 4 question 5 question 6 1. What is the Degree of Combined Leverage (DCL) of a firm with a Degree of Operating Leverage (DOL) of 1.4, and Degree of Financial Leverage (DFL) of 1.2? It is: a. 2.6 b. 1.25 C. 1...
##### Question 2 (2 points) equilibrium constant expression for the Ksp " of Ca3(PO42? What is theOA Ksp [Ca2+13[PO43-12 B) Ksp {[Ca2+13[PO43-12M/{[Ca3PO4IH2O]}C) Ksp 1/{[Ca2+13[PO43-12} D) Ksp = {[Ca" 2+13[PO43-12y[CagPOA]
Question 2 (2 points) equilibrium constant expression for the Ksp " of Ca3(PO42? What is the OA Ksp [Ca2+13[PO43-12 B) Ksp {[Ca2+13[PO43-12M/{[Ca3PO4IH2O]} C) Ksp 1/{[Ca2+13[PO43-12} D) Ksp = {[Ca" 2+13[PO43-12y[CagPOA]...
##### Find the counterclockwise circulation and the outward flux of the field $\mathbf{F}=(-\sin y) \mathbf{i}+(x \cos y) \mathbf{j}$ around and over the square cut from the first quadrant by the lines $x=\pi / 2$ and $y=\pi / 2.$
Find the counterclockwise circulation and the outward flux of the field $\mathbf{F}=(-\sin y) \mathbf{i}+(x \cos y) \mathbf{j}$ around and over the square cut from the first quadrant by the lines $x=\pi / 2$ and $y=\pi / 2.$...
##### What physical process is being observed using spectroscopic measurements in Beer's Law experiment
What physical process is being observed using spectroscopic measurements in Beer's Law experiment?...
##### An object of mass m = 4.9 g and charge Q = 40 ?Cis attached to...
An object of mass m = 4.9 g and charge Q = 40 ?Cis attached to a string and placed in a uniform electric field that is inclined at an angle of 30.0 ? with the horizontal (see the figure). The object is in static equilibrium when the string is horizontal. Find the magnitude of the electric field. Fin...
##### The mechanism given below: 2) Indicate Sudden Turning Centers (Figure is not to scale and according...
The mechanism given below: 2) Indicate Sudden Turning Centers (Figure is not to scale and according to the given values) we draw and fix) D 3 Ao Veriler: A A-20 mm AC = 65 mm A.B. = 80 mm BB-70 mm BC-40 mm CD-60 mm 9-50 8=35 Ss=65 sists B....
##### Graph each polar equation. Also identify the type of polar graph. $r=2+2 \cos \theta$
Graph each polar equation. Also identify the type of polar graph. $r=2+2 \cos \theta$...
##### Shaker Stairs Co. designs and builds factory-made premium wooden stairways for homes. The manufactured stairway components...
Shaker Stairs Co. designs and builds factory-made premium wooden stairways for homes. The manufactured stairway components spindles, risers, hangers, hand rails) permit installation of stairways of varying lengths and widths. All are of white oak wood. Budgeted manufacturing overhead costs for the y...
-- 0.103161--
|
2022-09-30 22:49:27
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4308169186115265, "perplexity": 6535.419374445095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335504.37/warc/CC-MAIN-20220930212504-20221001002504-00033.warc.gz"}
|
https://www.transtutors.com/questions/probability-and-statistical-quality-control-20-marks-show-all-calculations-reasoning-578320.htm
|
# Probability and Statistical Quality Control 20 marks Show all calculations/reasoning... 1 answer below »
Assessment item 1 Assignment questions Value: 30% Due date: 06-May-2014 Return date: 27-May-2014 Submission method options Hand delivery (option applies to Internal only) Alternative submission method Task QUESTION 1 Probability and Statistical Quality Control 20 marks Show all calculations/reasoning (a) 5 marks, one for each part An unbiased coin is tossed twice. Calculate the probability of each of the following: 1. A head on the first toss 2. A tail on the second toss given that the first toss was a head 3. Two tails 4. A tail on the first and a head on the second, or a head on the first and a tail on the second 5. At least one head on the two tosses (b) 2 marks Consider the following record of sales for a product for the last 100 days. SALES UNITS NUMBER OF DAYS 0 15 1 20 2 30 3 30 4 5 100 1. What was the probability of selling 1 or 2 units on any one day? (1/2 mark) 2. What were the average daily sales units? (1/2 mark) 3. What was the probability of selling 3 units or more? (1/2 mark) 4. What was the probability of selling 2 units or less? (1/2 mark) (c) 3 marks, one for each part The lifetime of a certain type of colour television picture tube is known to follow a normal distribution with a mean of 4600 hours and a standard deviation of 400 hours. Calculate the probability that a single randomly chosen tube will last 1. more than 5000 hours 2. less than 4500 hours 3. between 4700 and 4900 hours (d) 4 marks A company wishes to set control limits for monitoring the direct labour time to produce an important product. Over the past the mean time has been 20 hours with a standard deviation of 9 hours and is believed to be normally distributed. The company proposes to collect random samples of 36 observations to monitor labour time. 1. If management wishes to establish x ¯ control limits covering the 95% confidence interval, calculate the appropriate UCL and LCL. (1 mark) 2. If management wishes to use smaller samples of 9 observations calculate the control limits covering the 95% confidence interval. (1 mark) 3. Management is considering three alternative procedures in order to maintain tighter control over labour time: • Sampling more frequently using 9 observations and setting confidence intervals of 80% • Maintaining 95% confidence intervals and increasing sample size to 64 observations • Setting 95% confidence intervals and using sample sizes of 100 observations. Which procedure will provide the narrowest control limits? What are they? (2 marks) (e) 6 marks (2 for each part) (a) Search the Internet for the latest figures you can find on the age and sex of the Australian population. (b) Then using Excel, prepare a table of population numbers (not percentages) by sex (in the columns) and age (in the rows). Break age into about 5 groups, eg, 0-14, 15-24, 24-54, 55-64, 65 and over. Insert total of each row and each column. Paste the table into Word as a picture. Give the table a title, and below the table quote the source of the figures. (c) Calculate from the table and explain: • The marginal probability that any person selected at random from the population is a male. • The marginal probability that any person selected at random from the population is aged between 25 and 54 (or similar age group if you do not have the same age groupings). • The joint probability that any person selected at random from the population is a female and aged between 25 and 54 (or similar). • The conditional probability that any person selected at random from the population is 65 or over given that the person is a male. QUESTION 2 Decision Analysis 20 marks Show all calculations to support your answers. Round all probability calculations to 2 decimal places. John Carpenter runs a timber company. John is considering an expansion to his product line by manufacturing a new product, garden sheds. He would need to construct either a large new plant to manufacture the sheds, or a small plant. He decides that it is equally likely that the market for this product would be favourable or unfavourable. Given a favourable market he expects a profit of $200,000 if he builds a large plant, or$100,000 from a small plant. An unfavourable market would lead to losses of $180,000 or$20,000 from a large or small plant respectively. (a) 2 marks Construct a payoff matrix for John’s problem. If John were to follow the EMV criterion, show calculations to indicate what should he do, and why? (b) 2 marks What is the expected value of perfect information and explain the reason for such a calculation? John has the option of conducting a market research survey for a cost of $10,000. He has learned that of all new favourably marketed products, market surveys were positive 70% of the time but falsely predicted negative results 30% of the time. When there was actually an unfavourable market, however, 80% of surveys correctly predicted negative results while 20% of surveys incorrectly predicted positive results. (c) 4 marks Using the market research experience, calculate the revised probabilities of a favourable and an unfavourable market for John’s product given positive and negative survey predictions. (d) 4 marks Based on these revised probabilities what should John do? Support your answer with EVSI and ENGSI calculations. (e) 8 marks The decision making literature mostly adopts a rational approach. However, Tversky and Kahneman (T&K) (Reading 3.1) adopt a different approach, arguing that often people use other methods to make decisions, relying on heuristics. What do they mean by the term heuristics? (2 marks) Describe three types of heuristics that T&K suggest people use in judgments under uncertainty. (3 marks) Give one example from your own experience of a bias that might result from each of these heuristics. (3 marks) QUESTION 3 Simulation 20 marks Dr Catscan is an ophthalmologist who, in addition to prescribing glasses and contact lenses, performs laser surgery to correct myopia. This laser surgery is fairly easy and inexpensive to perform. To inform the public about this procedure Dr Catscan advertises in the local paper and holds information sessions in her office one night a week at which she shows a videotape about the procedure and answers any questions potential patients might have. The room where these meetings are held can seat 10 people, and reservations are required. The number of people attending each session varies from week to week. Dr Catscan cancels the meeting if 2 or fewer people have made reservations. Using data from the previous year Dr Catscan determined that reservations follow this pattern: Number of reservations 0 1 2 3 4 5 6 7 8 9 10 Probability 0.02 0.05 0.08 0.16 0.26 0.18 0.11 0.07 0.05 0.01 0.01 Using the data from last year Dr Catscan determined that 25% of the people who attended information sessions elected to have the surgery performed. Of those who do not, most cite the cost of the procedure ($2,000 per eye, $4,000 in total as almost everyone has both eyes done) as their major concern. The surgery is regarded as cosmetic so that the cost is not covered by Medicare or private hospital insurance funds. Dr Catscan has now hired you as a consultant to analyse her financial returns from this surgery. In particular, she would like answers to the following questions, which you are going to answer by building an Excel model to simulate 20 weeks of the practice. Random numbers must be generated in Excel and used with the VLOOKUP command to determine the number of reservations,0 and there must be no data in the model itself. The same set of random numbers should be used for all three parts. An IF statement is required for part (a) to determine attendance each week, given cancellation of meetings. (a) 10 marks On average, how much revenue does Dr Catscan’s practice in laser surgery earn each week? If your simulation shows a fractional number of people electing surgery use such fractional values in determining revenue. Paste your model results into Word including a copy of formulas with row and column headings. (b) 3 marks Adjust your model to determine on average, how much revenue would be generated each week if Dr Catscan did not cancel sessions with 2 or fewer reservations? Paste results into Word. (c) 3 marks Dr Catscan believes that 35% of people attending the information sessions would have the surgery if she reduced the price to$1,500 per eye or $3,000 in total. Under this scenario how much revenue per week could Dr Catscan expect from laser surgery? Modify your Excel model to answer this and paste results into Word. (d) 4 marks Write a brief report with your recommendations to Dr Catscan on the most appropriate strategy. QUESTION 4 Regression Analysis and Cost Estimation 20 marks The CEO of Carson Company has asked you to develop a cost equation to predict monthly overhead costs in the production department. You have collected actual overhead costs for the last 12 months, together with data for three possible cost drivers, number of Indirect Workers, number of Machine Hours worked in the department and the Number of Jobs worked on in each of the last 12 months: Overhead Costs Indirect Workers Machine Hours Number of Jobs$2,200 30 350 1,000 4,000 35 500 1,200 3,300 50 250 900 4,400 52 450 1,000 4,200 55 380 1,500 5,400 58 490 1,100 5,600 90 510 1,900 4,300 70 380 1,400 5,300 83 350 1,600 7,500 74 490 1,650 8,000 100 560 1,850 10,000 105 770 1,250 (a) 5 marks The CEO suggests that he has heard that the high-low method of estimating costs works fairly well and should be inexpensive to use. Write a response to this suggestion for the CEO indicating the advantages and disadvantages of this method. Include the calculation of a cost equation for this data using Machine Hours as the cost driver. (b) 5 marks Using Excel develop three scatter diagrams showing overhead costs against each of the three proposed independent variables. Comment on whether these scatter diagrams appear to indicate that linearity is a reasonable assumption for each. (c) 5 marks Using the regression module of Excel’s Add-in Data Analysis, perform 3 simple regressions by regressing overhead costs against each of the proposed independent variables. Show the output for each regression and evaluate each of the regression results, indicating which of the three is best and why. Provide the cost equations for those regression results which are satisfactory and from them calculate the predicted overhead in a month where there were 100 Indirect Workers and 500 Machine Hours and 1,000 Jobs worked. (d) 5 marks Selecting the two best regressions from part (c) conduct a multiple regression of overhead against these two independent variables. Evaluate the regression results. Draw conclusions about the best of the four regression results to use. QUESTION 5 Forecasting 20 marks (a) 5 marks All forecasts are never 100% accurate but subject to error. 1. How is forecast error calculated? (1 mark) 2. Identify and describe three common measures of forecast error. Then illustrate how each is calculated by constructing a 4-period example. (4 marks) (b) 10 marks as indicated below Consider the following table of monthly sales of car tyres by a local company: Month Unit Sales January 400 February 500 March 540 April 560 May 600 June ? (i) 3 marks Using a 2-month moving average develop forecasts sales for March to June inclusive. (ii) 3 marks Using a 2-month weighted moving average, with weights of 2 for the most recent month and 1 for the previous month develop forecasts sales for March to June inclusive. (iii) 3 marks The sales manager had predicted sales for January of 400 units. Using exponential smoothing with a weight of 0.3 develop forecasts sales for March to June inclusive. (iv) 1 mark Which of the three techniques gives the most accurate forecasts? How do you know? (c) 5 marks Describe the four patterns typically found in time series data. What is meant by the expression “decomposition” with regard to forecasting? Briefly describe the process. Rationale This assessment task covers topics 1 to 7: Probability concepts and distributions, statistical decision making and quality control, decision analysis under uncertainty and risk, value of information, simulation, correlation and regression analysis, and forecasting techniques. It has been designed to ensure that you are engaging with the subject content on a regular basis. More specifically, it seeks to assess your ability to: • demonstrate problems solving skills in assessing, organising, summarising and interpreting relevant data for decision making purposes • apply decision theory to business situations • use simulation in complex decisions • demonstrate understanding of statistical hypothesis testing • use accepted time series forecasting methods Marking criteria Assessment Item 1: Marking Guidelines GRADE REQUIREMENTS In each of the five questions students must score the marks as shown below to gain the appropriate grade: PS: At least 10 and less than 13 out of 20 marks. CR: At least 13 and less than 15 out of 20 marks. DI: At least 15 and less than 17 out of 20 marks. HD: At least 17 out of 20 marks. CRITERION GRADE REQUIREMENTS Question 1 Parts (a) to (d) 1. Apply probability concepts to decision making To pass students must score at least 7 out of 14 marks. Part (e) 2. Demonstrate problem solving skills in assessing, organising, summarising and interpreting relevant data for decision making. To pass students must score at least 3 out of 6 marks. Question 2 Apply decisions theory to business situations. To pass students must score at least 10 out of 20 marks. Question 3 Use simulation in complex decisions. To pass students must score at least 10 out of 20 marks. Question 4 1 Apply decisions support tools to management decision making. 2 Apply statistical hypothesis testing to determine significance of regression coefficients in cost estimation. To pass students must score at least 10 out of 20 marks. Question 5 Use accepted time series forecasting methods. To pass students must score at least 10 out of 20 marks.
Attachments:
## Plagiarism Checker
Submit your documents and get free Plagiarism report
Free Plagiarism Checker
Looking for Something Else? Ask a Similar Question
|
2022-01-24 00:09:15
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39224424958229065, "perplexity": 1139.6967152674817}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00156.warc.gz"}
|
http://archive.numdam.org/item/AST_1984__118__115_0/
|
Free boundary problems for the Navier-Stokes equations
Variational methods for equilibrum problems of fluids - Trento, 20-25 juin 1983, Astérisque no. 118 (1984), p. 115-123
@incollection{AST_1984__118__115_0,
author = {Bemelmans, J.},
title = {Free boundary problems for the Navier-Stokes equations},
booktitle = {Variational methods for equilibrum problems of fluids - Trento, 20-25 juin 1983},
author = {Collectif},
series = {Ast\'erisque},
publisher = {Soci\'et\'e math\'ematique de France},
number = {118},
year = {1984},
pages = {115-123},
zbl = {0567.76036},
mrnumber = {761741},
language = {en},
url = {http://www.numdam.org/item/AST_1984__118__115_0}
}
Bemelmans, J. Free boundary problems for the Navier-Stokes equations, in Variational methods for equilibrum problems of fluids - Trento, 20-25 juin 1983, Astérisque, no. 118 (1984), pp. 115-123. http://www.numdam.org/item/AST_1984__118__115_0/
[1] J. T. Beale, The initial value problem for the Navier-Stokes equations with a free surface. Comm. Pure Appl. Math. 34 (1981), 359-392. | Article | MR 611750 | Zbl 0464.76028
[2] J. Bemelmans, Gleichgewichtsfiguren zäher Flüssigkeiten mit Oberflächenspannung, Analysis (1981). | MR 727877 | Zbl 0561.76042
[3] J. Bemelmans, Liquid drops under the influence of gravity and surface tension, manuscripta math. 36 (1981), 105-123. | Article | MR 637857 | Zbl 0478.76118
[4] J. Bemelmans, Free boundary problems for the stationary Navier-Stokes equations, in preparation. | Article | Zbl 0637.76027
[5] J. Bemelmans, Nonstationary viscous fluid flow generated by surface tension, I: existence of classical solutions, SFB 72-preprint, Bonn.
[6] J. Bemelmans, Zur Dynamik von Flüssigkeitsfiguren, Habilitationsschrift, Bonn, 1982.
[7] J. Bemelmans and A. Friedman , Analyticity for the Navier-Stokes Equations governed by surface tension on the free boundary, to appear in J. Diff. Equations. | MR 759830 | Zbl 0497.35076
[8] A. Friedman, On the regularity of solutions of nonlinear elliptic and parabolic systems of partial differential equations, J. Math. Mech. 7. (1958), 43-60. | MR 118970 | Zbl 0078.27702
[9] T. A. Mccready, The interior Neumann problem for stationary solutions of the Navier-Stokes Equations, Dissertation, Stanford, 1968. | MR 2617321
[10] V. A. Solonnikov, Solvability of a problem on the motion of a viscous incompressible fluid bounded by a free surface. Math. USSR Izvestija 11 (1977), 1323-1358. | Article | Zbl 0398.76024
|
2020-09-29 04:44:26
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26522818207740784, "perplexity": 3583.1140396873966}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00214.warc.gz"}
|
https://math.stackexchange.com/questions/1837173/intuition-behind-the-derivative-of-are-of-a-square-how-to-properly-use-the-deri/1837187
|
# Intuition behind the derivative of are of a square? How to properly use the derivative ?
If I derive the formula $$S=16t^2$$, where S denotes the distance and t denotes time I get $$ds/dt= 32t$$. This in return give me a formula for the speed of the object at any time t. However if we were to do the same thing to the formula $$A=S^2$$, where A denotes the Area of the square and S denotes the length what would the derivative tell me. I know that its the instantaneous rate of change of the Area with respect to a small change in the length S but is there another meaning behind it? In addition I would like to understand how do I use the derivative. For example if we were to derive the equation A=S^2 we would get da/ds= 2S. What this is telling me is that if I increase the length by ds the change in Area da is 2S. In consequence if I take a square of size 6 and decide to increase its length by 1 the Area should increase by 2*6. This however doesnt not happen. If A1= 36, A2= 49. The difference in the increase Area is 13 not 12. So am I missing something here ?
Say that the side area changes by $\Delta s$. The change in area of the square is $$(s+\Delta s)^2-s^2=2s\Delta s + (\Delta s)^2.$$ If $\Delta s$ is small enough we can approximate the change in area by $2s\Delta s$. In your case the change in area is $$2\times6\times 1+1^2=13$$
Yes, it happens, but you shouldn't take $1$ but rather $0.01$. Then really $6.01^2$ is close to $36+2\times 6\times 0.01$. This picture might help.
The idea is that if $dS$ is small, then the "neglected" part is even much smaller.
• But that's the point of the theory of derivatives! It is only infinitesimal, that is, these formulas are only valid in the limit $ds\to 0$. Look up the definitions in your book. It's the same as with velocity: to compute the velocity in one time instance you need to measure the distance one milisecond before / after and divide by the tiny fraction of time. – Peter Franek Jun 23 '16 at 17:37
|
2019-07-18 12:52:41
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9178152084350586, "perplexity": 106.89812984217598}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00469.warc.gz"}
|
https://beta.geogebra.org/m/pwYbAQdQ
|
# Newton Principia Prop I.6 cor 1
Figure for Newton Principia Prop I.6 cor 1 The centripetal force acting on P as it revolves about center of force S is proportional to the inverse of that is, Force ult varies as When P and Q come together, areas of and are ultimately equal and represent equal areas decribed in equal times. Referring back to the figurte in prop. 6, PX (also equal to QR) is twice the sagitta BG, and the rect. SP,QT is twice the area of triangle SPQ . And since the area described is proportional to the time, you can substitute the area value for t². So the of prop. 6 becomes . Newton prefers , so finally the force on P ultimately varies as .
|
2020-05-28 19:50:11
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.947640597820282, "perplexity": 1721.2327678142635}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347399830.24/warc/CC-MAIN-20200528170840-20200528200840-00328.warc.gz"}
|
https://www.physicsforums.com/threads/meaning-of-cosmological-constant.195252/
|
# Meaning of cosmological constant ?
1. Nov 1, 2007
### notknowing
Incredibly much has been written about the cosmological constant and all its implications on the evolution of the universe. I want to ask however a more basic question about it. The cosmological constant has the dimension of 1/R^2. So supposing this constant is non-zero, its value points to the existence of some characteristic distance or radius. The Schwarzschild radius for instance has a clear understandable meaning but what could be the meaning of the radius associated with the cosmological constant? The only thing which comes to (my) mind is the radius of the universe itself. What other opinions exist (or are described in literature - related to "radius")?
Rudi Van Nieuwenhove
2. Nov 1, 2007
### Wallace
The cosmological constant has a value whose units are energy. I'm not sure what you mean by saying 'it has the dimensions of 1/R^2"?
If you mean that the energy density of the cosmological constant evolves as 1/R^2 where R is the scale factor that is incorrect. Curvature evolves in that way (in the way it is represented in cosmology) but the cosmological constant, unsurprisingly, remains constant for all time.
3. Nov 1, 2007
### hellfire
The dimensions of 1/L2 for the cosmological constant are correct in the usual geometrized unit system of general relativity. To find a deeper meaning to such a length scale one should go beyond general relativity, see for example section 7 of Padmanabhan's Cosmological Constant - the Weight of the Vacuum.
4. Nov 1, 2007
### Chris Hillman
It might help to point out that in geometric units, energy density (and sectional curvatures, e.g. the components of the Riemann curvature tensor) have the units of reciprocal area. Since Lambda contributes a diagonal term to the stress-energy tensor of our spacetime models, with the entries proportional to Lambda (in any frame!), this represents an energy density and has the units of reciprocal area.
5. Nov 5, 2007
### notknowing
Thanks for this interesting reference. This means a lot of reading (114 pages)!
|
2018-10-19 20:06:18
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8560602068901062, "perplexity": 591.5758427079975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512434.71/warc/CC-MAIN-20181019191802-20181019213302-00346.warc.gz"}
|
http://conwaylife.com/forums/viewtopic.php?f=12&t=2703&start=1500
|
Home • LifeWiki • Forums • Download Golly
## Random posts
A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
### Re: Random posts
KittyTac wrote:We should call this one "dress".
x = 9, y = 9, rule = B3/S24b2o$3bo2bo$4bo2bo$b3o4bo$o5bobo$bo3bobo$2bo2bo$3bobo$4bo!
But what colour is it?
AforAmpere wrote:A glider furiously attacking a dot:
x = 6, y = 6, rule = B3-cery4ekq5kry6ain/S02aek3-aiky4cqrwyz5-ejny6cik785bo3$3o$2bo$bo! Truly, a battle for the ages! 77topaz Posts: 1232 Joined: January 12th, 2018, 9:19 pm ### Re: Random posts KittyTac wrote:We should call this one "dress". x = 9, y = 9, rule = B3/S24b2o$3bo2bo$4bo2bo$b3o4bo$o5bobo$bo3bobo$2bo2bo$3bobo$4bo! more like 5 welded loaves #C[[ VIEWONLY ]]x = 36, y = 18, rule = LifeHistory.2A5.2D8.2D11.2D$A2.A3.D2.D6.D2.D9.D2.D$.A.A4.D.CA6.C.2D9.D.2D$2.A6.C2.A4.A.C2.D5.2AD.D2.D$10.A.A4.A2.C.D4.A2.C2.D.D$11.A6.2A.D6.A.ACD.D$29.A2.A$30.A.A$31.A$4.2A12.2A$3.A2.A10.A2.A$4.A.CA10.A2.A$.3A.C2.A6.3A4.A$A2.C2.A.A5.A5.A.A$.A.2CA.A7.A3.A.A$2.A2.A10.A2.A$3.A.A11.A.A$4.A13.A!
(I know I posted one before)
c(>^x^<c)~
This is Fluffy the cat.
Fluffy wants to discover new things that everyone likes.
Fluffy likes to watch spaceship guns in Golly.
There’s one problem,
Fluffy doesn’t exist :(
Redstoneboi
Posts: 319
Joined: May 14th, 2018, 3:57 am
### Re: Random posts
Glider pushing a block repeatedly:
x = 8, y = 6, rule = B2i3-ckq4cktwz5ky6e8/S2ae3-eqy4arw5-aky6ck6b2o$6b2o2$3o$2bo$bo!
I wonder how many spaceships there are like this.
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule (someone please search the rules)
- Find a C/10 in JustFriends
- Find a C/10 in Day and Night
AforAmpere
Posts: 867
Joined: July 1st, 2016, 3:58 pm
### Re: Random posts
AforAmpere wrote:Glider pushing a block repeatedly:
x = 8, y = 6, rule = B2i3-ckq4cktwz5ky6e8/S2ae3-eqy4arw5-aky6ck6b2o$6b2o2$3o$2bo$bo!
I wonder how many spaceships there are like this.
Nice. Must be a lot of them out there. Anyone have links to prior art in this area?
Seems like a slight variation on the SMOS theme. SMOSSL, maybe -- "Spaceships Made Of Spaceships and Still Lifes".
dvgrn
Moderator
Posts: 5222
Joined: May 17th, 2009, 11:00 pm
Location: Madison, WI
### Re: Random posts
Here's another one:
x = 8, y = 6, rule = B3aijn4nz6a/S23-ek4einy5-ejkr66b2o$6b2o2$3o$2bo$bo!
EDIT, and a knightship:
x = 8, y = 6, rule = B3-eqry4ciwyz5a6e8/S2aei3aijnr4aeqrty5-any6-ce6b2o$6b2o2$3o$2bo$bo!
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule (someone please search the rules)
- Find a C/10 in JustFriends
- Find a C/10 in Day and Night
AforAmpere
Posts: 867
Joined: July 1st, 2016, 3:58 pm
### Re: Random posts
About to get on a bus to Kazakhstan.
KittyTac
Posts: 533
Joined: December 21st, 2017, 9:58 am
### Re: Random posts
Cat. I am alive, and it's getting relatively new year, and it might be implemented in a room for the reason that I meant combat, and it's source, so if they have a cat breed, and it has to do you want an argument with the possibility that he would you drink yourself, and it's source, so I have to get in a similar way, and it might not get a world of the possibility of gagging and it's getting better at least one contained in a room? It is not an actual person who has to be paranoid about the possibility that the word wish I have time, and it's getting relatively small, and it has a world with that you drink yourself?
This comment has been brought to you by my phone's markov chain generator.
KittyTac
Posts: 533
Joined: December 21st, 2017, 9:58 am
### Re: Random posts
Mine gives: We could reschedule it on Monday due on Friday to make a decision on Thursday so far I've enjoyed your day tomorrow if you can and we are looking at my calendar to be anything else I can and we are not sure what the next week we could get the same time and I will be at my house by before the meeting on Thursday at that price to replace my mom saw you were here exactly the kind that you can do we want you can text or call you tomorrow and I will be at your office for a cop comes in at that price is firm to the store for example a little late today so we are looking at my calendar apart and the same for example a good night at that point where you too and thank to make a payment on Monday so far but that's what I'm talking the afternoon other day I was 6 to the same time or day I have off to you themselves at least a cop comes in at my place if we had the opportunity and the kids and if so could reschedule it for example the afternoon I am the owner of this week I am free tomorrow after but it will take you for me to make that unfortunately the weather to make a mistake the to do Monday the 6th or if I should just want a bit late today to make that unfortunately we will take a look like the two things that we are not going to do that and...
Not sure where it got all of these things (for example, I'm not sure I've ever typed the word "cop"), but it's definitely comical.
EDIT: tried -> typed
Last edited by A for awesome on July 29th, 2018, 1:46 pm, edited 1 time in total.
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
A for awesome
Posts: 1695
Joined: September 13th, 2014, 5:36 pm
Location: 0x-1
### Re: Random posts
There's an alternate universe where calcyman's parents named him something with really hard initials to make a pun out of. Watch out for XZJsearch, everybody.
Please stop using my full name. Refer to me as dani.
she/they
"I'm always on duty, even when I'm off duty." -Cody Kolodziejzyk, Ph.D.
danny
Posts: 879
Joined: October 27th, 2017, 3:43 pm
Location: i love to eat bees
### Re: Random posts
danny wrote:There's an alternate universe where calcyman's parents named him something with really hard initials to make a pun out of. Watch out for XZJsearch, everybody.
No, try ❎⭕️❌search.
wwei23
Posts: 935
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe
### Re: Random posts
Gray: In United States
Grey: In United Kingdom
If you're the person that uploaded to Sakagolue illegally, please PM me.
x = 17, y = 10, rule = B3/S23b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5bo2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo! (Check gen 2) Saka Posts: 2732 Joined: June 19th, 2015, 8:50 pm Location: In the kingdom of Sultan Hamengkubuwono X ### Re: Random posts i found glider move react now what????? x = 20, y = 10, rule = B37/S2-i34q11o8bo$10b2o6b2o$10bobo4bobo$10bo2bo2bo2bo$2bo7bo3b2o3bo$2bo7bo8bo$2bo7bo8bo$3bo6bo8bo$4bo5bo8bo$5b6o8bo!
JoshM
Posts: 14
Joined: July 12th, 2018, 4:24 pm
### Re: Random posts
395F35A0-50EB-464F-B755-F18BA9409D43.jpeg (184.7 KiB) Viewed 1616 times
"Modificating"
wwei23
Posts: 935
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe
oof.oof
Almost c/2:
x = 19, y = 10, rule = LifeHistory7.D3.D$6.DCD.DCD$5.DC2AD2ACD$3.2DAC2.A2.CA2D$3.AC.C2.D2.C.CA$.2D.C.AD.A.DA.C.2D$.2C.A2.A3.A2.A.2C$.CA.2D7.2D.AC$DC2.2A7.2A2.CD$2A15.2A!#C [[ STOP 2 ]] Hunting Posts: 608 Joined: September 11th, 2017, 2:54 am Location: Universe ### Re: Random posts x = 400, y = 269, rule = B34t78/S2-i35ao4bo2b2o3b3o3bo2b2obo4bobo2bobo2bob2o9bo2b2ob2o3bo2bo3bobo4bobo3b2obo5b2o6bo18bo2b2o3bo5bo3bobob2o2bo3bo2bobobo5b2o4bo3bo2b2obobobo2bobo3bo5bo4b2o3bo4bobo8b2o4bo2bo5bobobo3bobo3b2o2bobo10bo3bo3b5o5bobo5bo3bo9b2o8b2obo5b2ob2obobobo2b2o4bob3o13bobo3bobobo8b3obo3bo$bo6b2o5bo3b2o2b2o2bo4bobo6b2obo2b2o2bobo7b2o7b2o3b2o3bo3bob3obo4bo2bo3bo2b2obob3o13bo2bo2bo2bobo2bo11b4o3bo3bo2bobo2bo5bo2bo2bobo4bo4bo2bo4bo2bo2b3obo5bo3bobobo2b2obobo2bo3bo7b5o3bobo3bobo3bo3b2o3bo9bo6b2o2bobob2o4bo2bo6bo2bo8b2ob2o3bo2bo3b3o9bo8bo5b2o3bobo4b3o3bobo$5bo2b2o2bo2bo6bo5bob2o14bo3bo2bo7bo5b2o2bo4bo5bo2bo7bo8bobob2o6bo3bo2b2obobo2bo3bobo6b3o2bob2ob2o2bo2bo15b2o3b2obobo2bobo3bobo6b2o5b3o2bob2o2bobo3bob3obobo2bo6bo4b2obo3b2o6b3o4bo6bo3bo2bo5bo2bo2bobo3bo4bobo15b3o2b3o4bo4bobobo2b2obobo2bo2bo9b2o2bobo4bobobo5bo3bo$8bo3bo2bo3bob2o2bobo3bo4b2o5b2obo4bo4bo3bobo5b2obo2b2o3bo10b2o2b6o3bo3bo5b2o2bo2bo3b3o4bo2bo12b2o3bob5o5b2o2bo4b2o6b2o3b3o9bo7bo2b2o5bobobo4bo2bo2b2ob2o5b3obo4bo2bo2bo4bo2bobo4bob3ob2o3bobo4bobobo2bo8bob2o3bo6bo7bo2bo2b2o7bo2bo6bob2o2b2o2bobo3bo4b2o3b2o4b3o7bo2bo$bobob3obo7bobo5bo3bo3bobobob2o7bob2o3b3ob3o4b2o8b3o3bo4bobo3bo2b4o11b2o15bo6bobobo4bo2bo4bo3b3o9bo21bo13b2o4bo6bo6bo2b2obo3bobob2o2b2obobo3bo2b2o2bob3o5bo5bo3b2o2bo6b2o2bobo2bo2bo3bo3b3o3bo4bo3bo2bob3o2bo5bobo3bo4b2o2bob3o7bobo7bo2b2o3bobo5bobo$2bo6b2obob2o2bo3b3o7bo3bobo2bo2b2obobo2bobo4bo5bo3bob2o12bo2bo14bobob2o9b4obobo5bobo9bo3bo7bo10bobobobob2obo3b3ob2o3b3obo6bobo15b2o4b2o2bo2bob2o10bo7b2o7bo4b3o2bobo2bo2bo2bo2bo5bobob2obobo4bo2bob4obo6bo2bob2ob2o3b2o8b2obo2bob2o5bo5b2ob2o6bo4bo6bo4bob2o$bo3bo12b2o10bo4bo2bo2b2o3bo5bo5b3o4b4o5b2o3bo3bobo4b3obo7bo5b3o3b2o5bo7bo3bo2bo3b2o9b3o4bo2bobo5bo4b2o2bo2bobo2bo12bobo3bob2o11bo4bo6bo6b2o5bo2b2o3bo2b3o7b2obo3bo2b2obo2b3o3bo2b3o3b2o2bo2bo4b2ob5ob2o3b5o4b2o8b2o3b2o2b2obo2bo2bo2bob2obob2o3bo2bo4bo2bobobo3b2o$bo5bo3bo5bo2b2o2b2o4b2o3bo2bo9b2obo3b2obo6b2obo3b2obo2bo3b2o3bobobo4b2o3bob2o2bo3bo2bobo5bo2bo6bo2bo6b2o2bo3b2obo4bo2bob2o4bo2bo8bobo2bo2bo2bo13bo2bo2bobo3bobo2bo3b2o4bo3b2o8b2o8bo3bo2b2o5bo2bo2bo2b2o3b4o3b2o2b2o4bo3b3o3b2obo3bobo11bo2bo3bo3bo2bo7bob5obo4bobo8bo4bobob2o5b3obo$6b2o2bobo4bobobo2bo2bo2bo3bo5bobob2o3bo2bobo3bo3bo4b2obo9bo3bob2o6bo2bo3bo5bo2bobo4bo3bo3bobobo2b2o2bobo3bob2o2bob2o3b2ob2o3b3o4b2o4b3ob4o2b3obob2ob2o13bobobo7b3o4bo4b3obob5o6bo5bo2b2o10bo2bobo2bo3b5ob2o2bo4b3obo5bo3bo4b2o3bobobo11bo3bobob2o11b2o3b3ob2obo2bo2bo4bo4bo5b4o$2o7b2o7bo4bo2bo4bo12b2o2bobo4bo3bobo7bo2b2o2bo2bobobo2bo2b2o3bobo8bobo2bo5bo4bo3bobob2ob2o6b2o7bobobo3bo3bobobo7bo5bo2bo2bo2bo3bob2ob2obo6bobo3bo2bo9b2o3bo14b4o6b2obo2b2ob3o2bobo5bobo4bo5b4o4b2o7bo3bo2bo10b2o4bo4bobobo9bob2o7bo3bo2bo4bob3o3bo$3bobo4bo2bob3obo5b4o3bo3bobo2bo3bobobo4bo3bo3bo6bo7bo3bo7b2ob2o3bo8bo4b5o2b2obo6bo9b2obo6bo9bo2bo2bo3bo4bo3bob2o2bobo4bobo2bo6b2obobo5b2o2b3ob3obob2o5b2o8b3obobo2bo3b3o3bo3bo4bo3bo3bo6bob2ob2o2bo3b2obo2bo9b5o5bo2bo8bobo2bobo5b2o5b2o2b3o3bo7bobo2bo3bobo4bo3bo2bo$2bo4b2o2bobobo4bo2bob2o12bo4bobo2bobo2b2o3bobobo2bo7bobob3o2bo2b2o4b2o3b2o6b2o2b2o4bobo3b2obo2bo11bob2o2bobo6bo3bo3bo2b2obo2bo13bobobo4bo10bo4bo4bob2o2bo2bo3bo3bobobob2o2bobobo2b4o3bo6b2o3bobo4bo2b2o4bo2bo2bo3bo3bo3bo2b2ob2obob2o5bo3bo2bob2o4bo2bo2b2o6bo13bo5bo2b3o3b2obo2bob2o2bo3bo2bo$2b2ob2o2b2o3bobo2b2o6bo2b2o5bob6o2b2obo4bo2b2o2bo4bob2o8b2o4bo2bo3bo2b2obo2bo3bob2o9bo3bobo6bobo4bo4b2o4b2o3bob2obo6bo5b4o3b2o2bob4obo6bo7bo6bo3b3o2bobobo2bo3b2ob2ob3o4bobobobobo2bobo3b3o2b3o5b2o7b3o2bo2bobo10bobo2bob2o4bo15b4obo4bo2bo2bo2bo8b2o2bobo2bobobo9bobo2bo2bo3bobo$ob2o8b3obo5b2o5bo5bobobo8bo16b2o7b2ob3obo3bobo7bobob2obo2bo3bob2o4bo10bo2bo2bobo5bob3o2bo2bo5bo2bo6bo5bobo2bobobobo8bo2b3o2b3obo4bo20bo10b3o9bo2bo3b2o2b2o2bo3bo5bo4bob2o2b2o2b2o2bo14bob4o2bo5bo3b2o3bob2ob2o9b3o2b2obob2obo2bo12bobo3bo2bo4bo$8bo6bo7bo3b4ob2o3bo3b2o6bob3o3bo5bo2b2o6b2o2b2o4b2obo3bob3o2b4obob3obo2b3ob2o3b2o2b2o5bobo7bo7bo4bo2bo2bobo13bo4bobobo6b3o4bo2bo2b2obo10bo13b4o7bobo3bobo3b2o4bo6bobobo2bo2bo2bo7b2o6bo6bobobobo3bob2obo2bobobob2o5bobo2b2o5b3o11bob2o3bobo4bo3bobo10bob2o$3bo2b2o2b2o9bobo4b2o3b2o3b2o8bo2bobo3bo2b2o9bo2bobo4bo2b2obo4bo7bo2bo4bo3b2o4bob2obo4b2o4bo3bobo2bo2bobo6b2o7bobo4bo6b2obo2bo3b2o2bo6bo2bo4b2obo6bobo3bo2b2o10bo3bo10bo2bo3b3o7b2o6bobo3bo9bobobo2b2ob3o2bo4bobo5bob3o4b2o2bo3b3obo2bo2bobo3b2o5bo7bo2b2o2bobobo2bo5b2obo$b2o5bobo3bo3b2o3b2o7bo2bo2bo3bo2bo2bobo4b2obo2bo3bob2obobo4b2ob3obo15b4o5bo4bo7b2o2bobobobo4bo3bobo3b2o5b2o2bobo2bo4bo2b2obobo9bo3b2o2bobobo2bo5bo5bo2bo5b2o8bob2ob2o3bo2bobo8b2o5bo3bo2bob2obo3bobob2o6bo6bo6b2o5bo3bo8b2o3bo3b2o12b2o4bob2o3b2obo2bo3bo2bo3bo4b3o3bobo3b2o$bobo5b2obo3bo3bobo9bo6bobo5bo7bobo6bo19bo2bo3bo3bobo6b2o2b2o3bobo5bo4bo2bo3bobo13b3o4bo5bo12b2obo2bo3bo3bo3bo3bobo5b3ob2o3bo11bo9bo10bobobo5b5o6b3o2bo4b2o3bobo6bobobob2o7b2ob2o2b2o3b5o3bo2bo7bobobobo3b2o2bo3bo3bo11bo2bo4bob2o5bob3o2b2o$9bo6bobo2bobo3bo12b4ob2obobobo4bo3bo3bob4o10b2o6b2o2bobo3bo5bo5b2o2bo3b2o3bo2bobo2b6obo3bo4bob2obo3bo10bobo2bobo7bo2b3o2b3ob2o2bo2b3o4bo2bobo3bo3bo2b2o3b3o11bo3bo6bo3bo13b2o4bobobo7bob2o2bo4b2o6bo4bo8bobo3b2o2bo3bobobo2b3o4b2o2b2o2bobo4b3o3bo4bobo4b2o$bo2bobo2b2o2bo2bo3bo2bo9bo3bo2bo2bobob2o3bobob2o5bo2bobobobo8bobobobo2b2o2bobobo6bo2bobo9b2o2b3o3bo2bo2bob3o3bo7bo5b2o8b5o2bobo2bo3b3o4bo2bo10bo3bo16bo2bo3bo3bo9bo5bo5bo2bo2bo3bob2obo2bo2bo4bo2b2o12b2ob2o2bo2bob2o5bo6bobobo2b3o5bo4bo12bo3bo4b3o3bobob2o8b2o2b2o$o3bo2bo4bo5b2obo2bo2bobo8b2o5bo6bo7bo7bobob2obo2bob2o7bo2bobobo6b2o2bo7bo6bo3bobo4bobo5bobo7bo2bob3obo4bo2bo2bobobo2bo8bo4bobobobo4bo2bo4bo8bobo5b3o2bo4b2obo4bo4bo8bo2b2o4bob2o2bo2b2o3bob4o4bob4obo3bo4bo2bobo4b2o2bo2bobobo6bo4b3obobo3bobo2bo4bo2bo7bo6bo4bo3bo9bo$o2bo8bobobobo3b2o4bo7bo2b2obobob2ob2o3bob2obo4bo4bob2obo3bo3bo2bo6bo2bo8b2o2b2o5bobobo5b5o6bo6b2o4bo3bo6bo9bob2o2bobo3bobo6bo12b2o3bo10bo2b2o8b2ob2obob2o8b2o8bo2b2o2b2o3bo2b2o6bo2b3o5bo2bobo5bo10bobo2bo2bo3bob4ob2o2bo2bo3bob2o3bo2b2obo2b4obob5obobo4bo13b2o$2bob2o2bo12b2o3bo6bob7obo2bobo6b4o3bobob2ob2o4bo2bo4bo4bo7bo3bobo2bob2o2bo10bo5bobobo4bobo3bobo3bo4bo3b2obo4bo3bobo13bo3bo2bo2bo5bobo2bo3bobo3bo5bo2b2o3b2obob3obo4bo7bo6bobo3bob2o2bo2bo7bo2bo4bo2bo2bo2bo3b2o18bob3obob2obo2bo10b2o4bo2bo2bo4bobobo3bob2o11b2o$b2ob2o6bo3bobob2ob2o7bo2b2obo6bobo3bo4bo3b3o13bo6b2o2bo9bo4b2o12bo6bo2bo2bobo6bo7bobo5b2ob3ob4obo2bo3bo5bo4bo2bo3bobobo4bo5bo2bo2bob3obo2bo3bo4b2o8bo5bo5bobo2bo2b4ob2o6bob2o8bo7b3o5b3o9bo2b2o2bo4bobobob2o2b3obob2o2bo2b2ob2o7bo11bo3bo4b2o2b2o2b3obo4bo$o2bob3obobo7bo2b3o2bo2bo3bo6bobob2o4bob2o6bobo3bo3b4obo3bo4b3obo4bo14b3obo3bo3bo4b2obo6bo7bo3bo3bo2b3o4bo6bo2bob3o8bobobo2bo5bo5b2o2b2o2bo4bo3bobo2b4o2bo5bo3b2o2b2o2bo4b2o6bobo3bobo2b2obo10b2o3bo2bob3o5bo2bobob2obo3bo3bo6b2ob2o7bo3bo2bob3obo9b2o4b2ob2obo8bob2o4bo$3bo5bo4bo2bobo10b2obobo5bo11b2o3bo4bobob2o15bo2b2o6b2obobo6b2o2bo3b2ob2o3bo2bo4bob2o2bo14bo12bo11bob4o12b2obo2bo2bo3bo3b2o9b2o14b2o3bob2o2bobo6bo2bob2o2bo2bobo3bo2bo2b2o7bo2bo2bo3bobo4bo3bo2bobo2bo7bob3o2b2obo2bobobo6bo3bo5b3obo5bo5bobo2b2ob2o3bo5b2o$o2bo4bo3bob2o2bo3bo3b2ob3o4b4o5bob3o2bo6bob2o4bo2bo3b2o3bo4b4obo6bo4b2o5bo3bo4b2obo4b3o2bobo4bobo15bob2o2bobobo3b2o4bo2bobo9bo2bo2bo2b3o2bo5b2ob2o2b3obo9bo9b2o3b2obob2o3bo4b3o7bo5b2o2bob2o2bo2bo3b2o3bo8bo5b2obo2bo7bo2bo3bobo10bo7bo9bo10bo2bo2bo3bo6b2o2bo$o4bo2b2obo11bob2obo2bo4bobo2bob2o4b2obo4b2o6bob2o6b2obo6bob2o2bo6bo4bob2ob3o4bo9bob2obobo7b2o5bo2bob2o5bob2o2bobobo9b2o9b2o7bobo9bo5bobo7bobo3bo6bo3bob3o8bobobo6bo3bo3bo2bo10bobobo4bob2o5bo5bo10b2o5bobo2bobo4bobobo10b2ob2o2bo4b2o4bo3bo6b2o2b3o2b2o$4bobo10b6obobo2bo2bo4bo2bob2o2b2obob4o4bo3bob2o3bo5b2o3bob3o3bo7bo6bobo3b2o4bobo5bo3bo3bo4bo9bobobo2b2obo2bo5bobob2obob2obob4o17b3o3bo2bo8b2o2b2o2bo4b2o11b2o4bo2bo2b2o2bo2b3o5b2ob2o6bo3bo6bobobo3bob3obobob2obo4bo9bo2bobobo6b3o2b2obo6bo5bo15b3o3b2o3bobo2b2o$o4bo2bob2o9bo8b2obob2o3bo2b2obo7bo3bobo2bob2o2bo4bo2bob4o2bob2o4bobo8bo3bobo4bob2o7bo2b2o4b2o10b2o3bo2bo4bo3bo4b2o8bobobo4b2o2bo3bo12bo2b3o3bobo8b2o4b2o12bo10bo7bo4bo12bo3bo4bo2bo3bo4bobo5bobo9bo7bo2bo3b3o3b2obo3bo3bo2b3o4bobobo3bo3b2o7bo4bobo$bo7bo3b2o4bobob2obo6bo5bo6bo5bo5bo2b2o4b2o3bobo2b2o4bob3obo5bobo2bo4b2o2b2o2bo7b3obo2b2o2bo6bo2b2o7b2o6bo5bo2bo2bo2bo3bo3bo2bobo4b2o5bobo7b3o2bo4bo3bo2bo2bob2ob2o6b3ob2obo2b3o20bo3bo4b3o7bo3bo2bob2ob2o2bo4bo5bo3bo2bo4bo9bo11bobob4o2bo2bo2bo2bo3bo3bo6bo3bo2bobo$3bob2ob2o8bobo9b3ob2o8bobobo4bo5bo2bo3b2obo3bo3bob2o4bo2b2o2bo4b2o6bo7bo14b2obo5bo4b4o2b2ob2o11bo2bo2bo3bo4bobob2o2bo10b3obo2bo2b2o4b2o2bob2o8b2obo5bo2b4o6b3obo4b2o2b4o5b2o3bo9bobobobo2bo4b2o5bo2bo4b3o4b2o3bobobo2bo2bobo5bo4bo2b2o2b3o3b2o7b2o4bo7b2obo2b2o2bo$2bo2b2o3bobo5bo2b2o3bo4bob2o4bo6bobo2bo2bo2bob2obo3b2o2bo5b2o2b3ob2o2bo4bo2b2o4bo3bo2b2o7b3o11b4o8bobo2bo4bo4b2obob3obo2b3obo5b3o17bobob2o6bo3b2obob2obo4b2o4bo2bo2bo5bo5bobobo2bo2bo7bo7bobo3bo2bo3bo2bo6bo3bo3b2ob4o2b3obo5bob2obo4bo15bob2o4bo8bo5bob2ob2o3bo4bo3bo$2bo6bo3bo3bo2bo6bobo2b2obo2b4obo8bo5b2ob2o4bobo3bo4bo3bo2bob2o9b3o6bo2b2o6bobo2b2o2bo2bobobo4bobo3bob2o7bo8bo2b2o3b2o9bo2bo2bo4bobo14bob3o2bo4b4obobo2bo2bobo2bo3b2obo5bo9bo4b2o7bob3o3bo3bob3o4b2obo2bo2bo5bo2b2ob3o2b2obo3bo2bobo2bob3o5bobo2b2o4bo2bo5bo3b2o17bobo$3b5o7bo3bobo7b3o3b3o5bo8bo2bo6bo4bo5bo2bob2o2bobo2bo2b2o8bobo3b2o8bobo2bo8b2o8bobo6bo4bo4b2o2bo8b2o3bo5bob2o3bob2obo2b3obo8bob2o4b2o4bobo4bo4b2o4b2o4bo2bo2b4o6bo4bob2o7b2o5b2obo2b3obo4bo4bobo4bo13b4ob3o5b2o8b4o3bo2b4o3bo10b2o2bo7b2obo$4b3o2bob2o2b2o9b3o4b3o9bo12bo2bo2bo2bobo15bo2bobo6bo3bo2bo4b2o2bobo3bobo2bo5bo4bo3b2o2bo2bo3bo7bo2bo5bo6b2o3b3o5bobo3bo3b2ob4o12bo4b3o3bo4b2obob3o3bo3b2o2bo3b2o4bo6bobo5bo8b2ob2ob2o6b2obo3b2o2b4obo12bo2b2obobo7bo2b4o6bo2b2obob2o4b2o3bo3bo6bo9bo$10bo5b4o14b2o5bo10bobo2b4obo2b2o3b2o3bobo3bobo4b2obo5bo2bobo2bobo5bo2bobo6bo6bo2bo10bo2b2o4bob2o7bobob2obo3bobo4bobo3bo5bob2o3bo3bo3b6o3bo3b4obo3b3o2b2obob3o4bo2bo2bo3bob2o12bo5bobobo6bo2bob3obo2bo6b3obobob2o3b3o3bo6bo2b2o2b2o3bob2obobo5bo4bo2bo2bo2b3o3bo2bobobobo5bo$b4o2bobo7bob2obo4b2o10bo8bo3bobo3b2o4bobo4bo3bo4bobo3bo3bo7b3o5bo4bobobo6bobo13b2o7b2o4bobob2ob2o7bo3b3ob2o9bo4b2obobob2o2bo8bobo2bobo4bobo6bob3o2b3obo4bo7bobo3bo7b2o3bob2ob3ob2o7b3o3b3obo14bo3bo4b2ob2o6b3o7bobobo7bo4bo7b2o3bo7b2obobo4bo$2o2b3o2bobo3b2o2bo2bo2b2o9bo4bo4b2obo5bo5bo2b2o3bobob2obo3bo4bo9b2obo6bo2b5o2b2o3bo8bobo5bo2bo2bo5b2obo7bo3b2o7bo7bo7bo2bo3bo6b2obobo10bo2b3o2b2o2b2o2bo6bo3bobobo6bo3bobob3o2bobo2bobobo2bo11bo6bobo2bobobo6b3o2bo9bo6bobob3o2bobo2b3o6bo6bo3bobo2b2ob2o2bo9b2ob2o$7bobob2o2bo2bo3bo13bo2bobobo7bo3bobo3bobobo28b2o2bo4bo2bo2b2obo7bo9bo3bo7bo4bo9b2obobo5b7obo3bo2bobo5b2o2bo3bo6bobo2bobo2b2o2bo6bo2bo5bo3bo4b3o7bobobo7bobo7b3o7bo3bobo5bo8bob3obo2b2ob2ob4obo2bo9b2obo2bobo2bo6bo10bo14bo2bobo$4bobo2bo6bo6bo2bo2b3ob3o2bob2obo2bobo3b2o4b2obo3b2o3bo2bo2b2obo2b3obo2bobo3bo4bo3bo6b3o4b2o2bo10bo3b2o4bob2o5b2o6bo3b3o3bob2o2b3ob2obo4bo2bo3bobo3b2o2bo2bo6bo2bo3bo4b2o7b3ob2o4bobo3bo2b2o4bo3b3obo3b2o3bobobo5bo3b3o14bobobo2bo5bo2bo2b2obob2o11bo8bobo4bo4bo10bo2bo7bobob2o$bo3b2obo2bobo4bo10b2o5b2o14bo2bo4bo2bob2o4b3o8bo8bo2bo2bo2bo6bo6b2obob2o2bobo4b2o2bo5b2o9bo12bo5b3o9b2ob2o2bo2bobo6bo4b2o14bo2b2obobo7b2o6b2o4bobobo2bo7b2obo3b2o3bo5bobo2bo4bo5b3o8b2o3bo2b2o3b2o2b3o2b2o8bo4b3ob3o14bo4bo4bo3bo2bo4bo2bo$b2ob3ob3o2bo3bo3bob2o2b2ob3obo3bo3bo3bo11bo4bobo2bobo4bo2bo3bo4bo2bobo2bo3bo6b2o6bo2bo2bo3bobo3bo8b3obo5b2obo5bo2bo2bo2b2obo3b2ob2o6b2o2b3obo2bo3bobo3bobo2bob3o2b4obo8bo2b5o6bo3bo4bob2o3bo2bo2b2o9b2o2bo4bo4bobo11bo14bo5bo4b2o2bobo2bo6bo5bo3b2o2bobo4bo4b2ob2obobo2b2o9bobo$o6bo2bo6bo2bobo14bo5bo2bo10bo3bo8bo2bo3bobo3bo3bo5b2o2b2o3bo4bo5bobobo3bobobo2bo4bo9bo11bob2o14bo2bo4bo3bo7b2obo3bo3bo2b2o6bo8bobo2b2obobo2b2ob2o7b2o6b3o3bo2bo16bo8bo8bobo2bo2b3obobob2o2bo5bo4bobo3bo8bo5bo5b2o3bobobo4b4o5bo2bo5b2o3bobo$o5bobo8b2o7bobobo5bo2bobobo2b2o3bo6bo4bo2b3o2bo2b2o2bobo2bobo4bo4bo3bo4bo4b2o2bo3b2obo2bo10b3o4bo4bo4b3obo2bobo2bo6bo2b2o3b4ob4obob4o3b4o2bob2o2bo8b3o3b4o14bo2bo8bo4b2o2bo4bo4bo3bo5bobo7b2o3b2o2bo5bo7bobo6bo4bobo3bob2ob2ob2o5b2o3bob2obo6bo4bo8bo2bo2bo4b2o2bo$b3o5b3ob2o8bobob2o2b3o2bo3bo5b2o4bobo4bo9bo10bo2bobo7bo5bo3bobo5bo2bo4bo4bo4bobob2o2b3obobob2o4bo4b2o2bobo4b2o2b2obo3bo3b2o2bo5bo4bo2b3obo9bo4bo3bobo4bo4bo2b3o3bobobobo3bo3bo7b2ob2o2bobo8bobo7bo5bo3bo3bo2b2o2bo2bo4bo3b2obobo3b3o8bo12b2o5b2ob2ob2ob2o2bob3o4bo3b2o3bo$6bo7bobo5bobobobo5bo5bo7b2o4bob2obobo5bob2obo7bo7bo2bo2bo2b2o2bobobo2bo5bo4b2o6bo6bo4b2o4bobobobobo2bob2o8bobo4bo3b4obo5bob2o4b2o4bo7b2obo3b2o2b2o14bo3b2o14bo12b2o4bo5bobo6b3obo2bo3bo4b3o3bo2b3o4bo9b2o3bobo3bobo3b2obo9bo2bo3bobo3bo7bo4b2o$3b2obobo6bobob3o12b2o3b2obo4b5o5b2o3b2o2bo6b2o6bo2b2obo2b3o3bobobo6bo2b2o13bobo2bobo3bo2bobo7b2obo2bobo5bo6b3obo5bo2bo2b2o3bo2bo2bo4bo2bobo4bo5b2o3bo3b4o2b3o3bo2b4obo2bo8bo2b2o2bobobo2bo7bo4bo4bo5bo4bo3bo6b3ob3o6b2o2bo3bo3b2o2bo3bo3bo2bo11b5ob2o2b2obo8b2ob2o4bob2o$bo4bobo3bo4b2o2bo4bo2bobo13bobo6bobo2bo3bo4b4o5bobo4bobo9b2o2bobo2bo2b3o8b2ob3o2b2obo2bo6b2o8bo3b2obo3bo4bo7bo2bo8bo3b3ob4obobo2b2o2bo3bo5bo2b2o8b3o2bobo3bo2bobobo9bobo16b2ob2o2bobo3b2obo13bo9bo2bo3bo3b2o7bo9bobobo6b6o2bo5bob3o3bo4b3obo2bob3o$b2o7bo3b2o2bo5bob2o4b2o3b2o3b6o4bo8bo2bo2bobo12bo6b3o3b2obobobo18bo5bo2b2o2bob2o5bobo7bo5bo3bo4bo5bo5bobobo2bo3b3obo3b2o2bobo2bobo5bo11bo8b2o2bo6b2obo5bo2b2o6bobobo2b3o9b2o4bo6bo2bobobobo3bo5bobo8b2o3bo7bo2bo2b2o5bo3b4obo3bo4bo5bo7bo6bo$2obo2bo2bo2b2o8bo2bo8bo4b2o3b2o3b2o7bo2bo4bo6b2o14b2o7bo3b5o12bo5bo4bobobo3bo2bo4bo6bobo5bo10b2o6b2o7bob2o9bobobo4bo4b2o4bobo2bo8bobob2o11b3obo5b2o5bobo9bo3bo4bo4bo3b2obo3bo3bo3b2o7bobo4b3o6bo9bo3b5o3bob2o4bo4b4o3b2ob3ob2o7b2o$4bo3bo4b2o8bo5b3o5bobo2bo5bo2b2o5bobobo3b2ob2o2bo5b3obo2b2o6bobo4b2o4b4o2bob2o8bo3b3o2b2obo3b2o2bo6bo6b2o3bo2bo7bobo3b2obo5bo4bo6b2o8bo2bobo2bo4bobo2bo2bobo3bobo6bo6bo3bo6b3o5b2o4bo6bo3bo2b2o2b2o9bob3o3bo5bobo2bo2b3o3bo3bo7bo4b2o2bo6bo4bobo4bo3b2obo2b2o2b4o$2bobo3b2o6bo5bo5bo4bo3b2obo5bo3b2o3b3ob2o14b2o2bobobo4b2o4bo2bo3bo3bo5bob2o4bobo6bobo3b3o2bo4bo5bob2o2b2o9bo2bo4bo3b2o2bo2bo2b2obobo2bobobobo5bo2bo2b2o3bo3bo5b2obobo4bo3bo4b3obobo4b2ob2o5bo7bobo5bo3bo7bo3bo3b6ob2o9bo2bo6b2o3b2o3bo4bo2b2o2b2obo3bo2bo2bo2bo3b2o$bo5b2o7bo2bo2bobo6b3o7bobo2bo2bo3bo3b2o2b2ob2o6bobob2o3bo3bobo4bo4b2o3bo3bo2bo2bo4b3o3bobobo2b2obo8b2o3bo5bobo4bo4b4ob3obo2bobo7bo2bo8bo7bob3obobobobo2bob2o2bobo2bo2bo2bob4ob3obo8b2o2bo8bo2bo5bobob3obo3bo2bo3b3o3bo2b2o4bo3b2o8bo2bo2bobo4bo13bo2bo2bo3bo5bo10b2o2bobo5b2o$2bo3b2o2bo4bo8bobo2b3o3b2ob2o2b2o4bo5bobo2b2o5bo2bo2bo2bobo4bobo3bo2b3obobobob2o7bo5bob2o2bo3b2o2bo2bo4b3obobobobo11bo3b2obobo5bo2bo4bobo4bo7bobobo6bob2o7bob4o4bo4bobo6bobo7b3o5b4o5bo3bo7b2o4b2ob2o2bo4bo2bo3bo3bobo4bobo5bo4bo3b3o3bobo3b2obo3bob2obo2b3o2bobo3bo2bobo2b2o6b2obo2b2o$2obo3b2o7bo2bo3b2o9bobo3bo5bob2o2b2obo4bo10bobo2bo2bo2bo6bo10b2o5bo7bobo4bobo4bo2bobo5bobo2bobo5bo4bo2bobo3bobobo2bo3b2o2b4o2b2o5bobob2o3bo2b2o2bo4b2o3b3o3bo4bobo2b2o4bo3b3o2b2o2bo4b2o7bobobob3ob2obobob2o4bo2b2o2bo6b2o7bo7bo6bo2bobo7bobobobobo3bo2bo3b2o3b2o2bo2bo13bo5b2o$6b2o9bo2bobo2bo6bob2o2bo3bo15b2obobo2bo4bo5bobob2o3bobo2bobo3b3o2b6o4b2o3b2o2bo5bobo3bobobobo3bo3bo3bo7bo12bo2b2obo2bob2obo3b2o2bo9bo2b2o4bob2o2bo5bo2bo2b2o6bo2bob2obobobobo4bo3b2o4bo3b2ob3obo2bo3bo2bo3b2o12b4o3bo2bo6bobo2bobob2o11bo9b4ob3o3bobo3bob2obo3bob3ob2obo3bobo$bo7bo2bobob4o3bobo3bo3bo2bo8b2o2b2o2bo3bobo15b2o4bo12bob2ob2o4b2o3bobo2bo5b2ob2ob2o2bo2b2o3b2o10bobo2bo13bo3bobob2obo2b2o3bo4bo7bo7bo6bobob2o5b2obobobo3bo2bo12bobo3bo2bo2bobo3bob2o3bo2b2o4bo2b2obo2bo3bo2b2ob2o9b2obo2bo7bobo2bo10bob2o5b2o6bo6bob3obo6bobo2bo3b3obo$2bo2bo2bob2obo9bo5b2ob2o2bo4b3o2b3o2bo4bobo5bo2b2obob6o4b3o3bob2obo3b3o3bo4bo2bo17bo5bobobob2o2b2o9b2o4b2o4bo12bo8bo4bob2o2bo6bo10bo2bobo4bobobo4bo5b2obo3bo3bo12bobo4b2o2bobo5b2o6bo3b2o3bo7b2o7bo3b2o5b2obo9bo3bo6bo3b2o6b2obob2o2b2obo5bob2o7b3o$3b3o7bo2bo6bo2b2obobobo2b2o6bob4ob3obo4b4obo7b3obo4bo7b2o2b3obo4b4o7bo2b3o6bo3bo3bo2bobobobo2bobo3b3obo2bo4bob2o2bo7bobo2bo6bo2bob2obo4b4o3bo4bo3bo2bo3bo3bo7b2obobobo2bo3bo3b2o2b3obobo7b2ob2ob2ob2o11bob2o4b2o3bo6bo4bo4b2obo3bo3bob2ob2o4bobobob2ob2obo5bo2bo2b2obo5bo2bo10bobo$bo6bob2o8bo2bobob4o3b2o2bo13bo4b7ob2o9bobo7b2obo3b2o2bo2bobobo3b2ob2o5b2o2b2ob2obobo4bo2b2o5bobobob2ob2obobobobo3bo2bobo4bo4bobo2bob2o4bo5bo2bo4bobo2bobo10b2o2b2obo8bobo4bobo4bo2b5o10b2o4bobob4obo8bo5b3o9bo10b2ob3o5bo5b2obo4bo5bo2bo5bo2bo4bobo2bob2ob2obo4bobo$2ob3o2bo2bobo6bo6bob2obo3bobo8bo2b3o4bo5bo2bo4bo2bo2bo8bo3bo3bo2bobo9bo9b2o5bo2bo3bo2bo8bo5bo9b3o6bo5bo6bo4bo2bo7b3o5b2o4bobo3bo4bo2bobobo7b3o6b3o7bo2bob2o2bo3bo3bo2bo2bo8b2o2bo2bo10bo2b2o2b2o5bo2bobo6bo2bo7bo6bo2bob2o5b2obo12bobobo12b3o$obobo6b2o6bo4bo4bo14bo5bo3bo2b2o2b2obo2b2o2b2o4bobo3bo3b2o3b4ob2o2b2obob2ob3obobo5bo2bob2o2bo2bo2bobo8bo2bo4bo2b2ob2o4bob2o2bo3bo8b2o9b2o7bob2o2bo2bo3bo5bo7bo3bo8b3o4bobo3b3o3b3o4bob2o6bo3b2obobo5bo5b2obo2b3o3b2obo4b3o2b2ob4o5bobob2o2bo12bo2b3o2bo4bobob2o8bo4bo2bo$bo7b2o2b3o6bo2bobobo3b2o13bobobo2bo3bobobobo3bobo2bo3b2obo3bo2bobo6bo7bob3obobo7bo2bo2bo8bo6b2o5bo2bo4bo4bob4o6bo3bo2bob2o2bob2o8bo2bobobo13bo2b4o2bo6bo2bo6bobo12bo2b3ob2ob2o2bobobo5bo3bo2bo2b2o6b2o6b3ob3o2bo4bo3b2o3bo8bo5bo6bo5bobo2bo2bo3bo9bo2bo2bo2bob4o$3bo5bo3bo5bo2bo2bob2o10bo7bo5b3o3bob2o6b2o4bobo4bo3b3o5bo3bo6b2obob2o2bo2bo7bob2obo3bo4b4o4b2o7bo5b4o7bo3bo4bo4b3ob2obo4bobobobob2o4b2obo7bo5bobo7bo4bo4bo2bo4bobo3b2o2bo7bo3bobobobo5bo3bo3bo4bo3bobo5b2obo4bo9bo3b2o7bo6bobo2bo6bo4b2o4bo10b4o4bobo$b3o4bo2b2o3bob3o8bo2bobo2bo2bo5bo4bo2bo2bo2b2ob4obo6b2o8b2ob2o3bo2b2obo8bo4bo2bobo6bo3bo3bo5bob2o5bo5bo4b2o8bo8bo11bo3bobob4o4bo5bo4bobo4bo6bo2bobo4bo2bobobo2bo4bob2obo2b2o4bobo5b4o3bob3obobo2bo7bo3bo11b2o5b2obobo3bo3bo6bo5bo2b2ob2o2bo2b3o4bo2bobobo3bo3b2o3bo3bo$4bo3b5o2b2o6bo2b2o4bobo3b2o4b2obo2bo3bobo2b2obo4b2o8bobo2bobo5bobo3b2o6bo2bo2bo5bo3b5o6bobo4bob2o3b3o3b2o4bo2bobo7bo2bo2bobo3bo3b2obo9bo8b3o2bo2b2o4bo4bobo2bo3b2o5b2o3bobo2bo2bo7bo4bo2bo3bo14bo3bo3bo4b2obob4o3b2ob4obobo12bob2o6bo7bo5bo4bo8bo3bob2o2b2o5bob2o$bobobo6bo14bob2o7bo9b2obo2bo4bo2bo5bo2bo2b2o2bo3bob3o4bobobobob2o2b2o3b3o6bo9bo6b2o3b2obob2obo7bo2b3o6bo5bobo10bo6bo4b2o9b2obob2o4bo5bo5b4o2b2o4bobob2obo8bo6bo4b3o2bo2bo7bo2bo10b2o2bo3b2o3bo4bobo2bo2bobob2o3bo4b2obo2b4o4bo2b3o9b3o2bo4b2o7bo2bobo$b2o3bo9bo3b2o8bo2bo2bo6bobo3bobo2b5o5bob4obobo6b5obobobo2bob2o3b3o3b2o4bo6b2obo4bob6o2b2o8bo5b2o2bo2b2obo5bo2b2o2b4o3bobobo7bobob2o3bo4b2o7bobo3bobo2bo5bo3bo3b2o4bo2bo5b3o4bob2o3b2o2bo3bo3bob3obobo2bob2o3bo2b2o4bo3bobo3bobob2obo2bo2bo3b2o2b5o3bo4bo3bo9bobo9bo5bo3bobo5bo$8b3ob2o2bo4bo7bo2bo2bo8bobo5bobobobobo3bo7b3obo3b3o8b2o3b2o2bo6bo3bobo6b2o4bo2bo7bobobobo2bo6b4ob2obo3bobo14bobo5b2obo5bobo2bobo5bo3bo15bo2b2obo3bo2bo3bo5bobo3bo4bo3bo3bob2o4bo5b2o4bo13b2o5bobo2bo2bo4bob3o7b2o5bo3bobo2bobo5bobobobo13bo2bo2bo4b3obo$o3bobo7bo3b2obo7b2o6bo3bob3obo2bo6bo6bo2bo8bo9bo12bob2o5bo3bo6b2obobob4o2bo18bobo5bo2bo15b2obob3o2b2o7bobo4bo2bo5bo2bo4bo2bo3bo3bobo4b3obo2bo2bo3bo9b2o2bobobo9bo2bo7bo8bobo4b5o2bo6bo4bobo2bo8b2obo2bob2obobo11bo7b2obobo2bo2bobo2bo3bo2bo$bobo2bob2obob2o5bo7bo12bo2b3ob3o4b2ob2o4bo5bobo2bobo2bo2bo3bob2o2bo3b3ob2o2bobobobo6bo2bobo3bob2ob4o5bo7bobobo6bo5bobo3b4obo3bo2b2o3bobo4bo5bo4bobo2b2o12bo5bo2b2o3b2obo4bobobob4o3bo5bo2bo5b3o8b2o5bo3bobobo5bo6bo7bobo3b2obo7bob3o2bo2bo2bo3bo2bo6bo4bo2b3o4b2o10b2o$4o13bo2bo5b2o3b3o2bo4bo3bo3bo2bobobo2bo4b5o11bo2b2o5bobo2bo3b3obo2bo7bo2bo2bo7bobo6bo2b2o2bo2bob2o4bo4bo2bobo5bo3bo3bo4b2o15bobo5bob5o8bo3bo2bob4obobob2o2bo4bo2b2o2bo10bobo3b2o2bo5bo2bobo2bobobo4b2o4bo2b2o3b2ob2o3bobo2bobo4bo7b2obo2bobo2bo4bo2bo6b3o6bo10bo8bo$o2bo3bobo4bo4bo2bobo4b2o7b3o3bo5bo6bobo3bo4bob2ob4o6b2ob3ob2o2b2obo5b2o4bo5b2o9b2o6b3ob2o2bobo6bo3b2o5bo6bo10bob3obobo2bobo9bobo16bo3b3ob2obo2b2o3b2o7bo2bo2bobo6bo5b3o3bo3b2o2b2ob2o11bo2bo2b2ob2obo3bo3bo4bobob3o6bo3bo2bo2b2obo4bo4bo3b3ob3o2b3o2bo4bo3bobo2b3obo$3o2bo5b2o3b2o2bo4bo2b2o4bo7bo8bo3b4o11bo3bobo7b4obobo2bo3b4o5bo5bo2bo8bobo2bo2bo5b2o2b5o2bo4bobo11bo3b2o5b2o10bo4bobo4bo5bob2o3bo4bo2bobo5bo10b2ob2obo4b2o5bo2bo3bo7bo3bo2bo6bobo2b3obo2bo4bo8bo3b2o6b2ob2o5bob3o2bo2b2o3bo3b2o2bob2o3bo4bo2bo3bo9bobo5bo$6bob2o5b2o13bob2o2b2o3b2o5bob2o2bo2bo2b3obo2bo2b2obo2bobobobo3bo2bob2o2bo4bo2bobo2b2o2bo3b2obo3bo3b2o2b2o3bo2b2o3b3o2bo2bo2b2ob2obob2ob2o6bo4bobo7b3obo5bo2b2ob2o4bo2bo5bo2bo2bobo5bo10bo3b2o5b2obo5bo3bo4b2o5bobo3bobo2bo2bo4b2obo4bo3bo2bo3bo8bo3bo4bo2bo4b2o3bo3bo4bob2o3bobo3bo3b2o4bob2o3bo4bo2bo$6b2ob2obo2bo5bo2bobobobo3b2ob3obo10b2o4b2o6bo3bo3bo12b2o5bo4bo3bo6b2o3b2o2bob2o2bobo4bobo3b3obo6bo2bo4b2o2b2obo3bobo5b2ob2ob2obo4bo8bo3b2o4bobo5bobobobo2bo5bo4bo4bo6b2o2bobo10b2o2b3o2bo3bobo5b2o5bobobo7bo10bo8b2o3b2obo4bo4bo2bo2bo2bo3b2obo2b2ob2o2b2o4bo4bo4bob2o4bo2b2o$b2obo7bo7bob3o2b2o4bo8bob2obo3b2o2bo8bo2b2o17bo3bo9bobo6bob5o4bobo2bob2o6bo8b2o2b2o2b2obo3bo3bo8bo3bo5bo3bob3o8bobob2o2bo9bo6bo3b3obo7bo10bobo3bo7bo5bo3bo2bo4bo6b4obo2bob2o3bo4b2o3bo2bo2bo3b2obo4bo4b3o2b2o4bo4bo3b2o3bo2bob2o4bo3bo5bobobobobobob2o$2obo3bo3bo3bob2obo3bo8bo3bob4obo3bobo10bob2o2bo2b2obob2o3bo2b2o3b2o4bobo2bo2bo4bo4bo2b2o3bobobobo4b2o2bo3b2obob2o2bo2b2ob2ob4obo2b2o2bobobo7bo11b2o2bo6bo6b2obobo6bo3bo4bo8bo11bo4bo8b2obo3bobo2b2o4bo10bo4bob2ob2o3b2o4bo2b3o3bo7bo8bobo4b3obo7bo2bo2b4obo2bob5o5bobo3b2o$bo11bo7b2o4bo6bo4bo3bo2bobo4bo2bo2bo5bo2bo3b2o5bo3bobobo4bo3bobo6bo6b4obo3b2obo4bobo4b4o3b2obo7b2o3bo6bo4bo5b3o4bobo3bobo6bob3o7bo3bo3bo2bo3bo2bobobo4bobo4bo2bob2o7bobo3b2o2b2o6b2obo2bobobo4bobo10b2o2bo3bobobo2bob2o3bobo2b2o2bobo2bo2bob2o2bobo2bo2b2ob2obo5b2o2bo5b2o2bobo4bo$2o2b2ob2o2b2o2b2o3bobobobo3b2obobo5bobo2bo4bo3b3o4bo3bo2bo4bobo4bo3bo4bobo2b3o2bobo4bobo2bobo4bo2bo3bo12bo4b2obo8bo2bob2obobob2o5bo3b2ob2obo8b3obo2bo6bo4bo4b2obobo2bo2bobobo2b2o12bo2bo2bobo2bo12bobobob2o2bo16bo7bo4bo6bobo3b3obobo2bo6b2o5bo5bobob3obo2b6o2b2obo5b2obo3b2o2b2obo$b2o3b2o5bo2bo2b3o8b3o3bo2b2o3bobobo3bo2bob3o3bo11bo4b2o8bob2o3b2o10bobo3bo3bo4bo4bo4bo5b3obo8bo2bobobo3bo3bobo6bo3b2o5bo2b2obo4bobo2b2o3b2o6b2o2b4o2bo11bobob3o2b3o5bo5bo2bo3b3o5b2obo4bo6bo8bo2bo2b2o8bo2bo5b3obob3obo3b4o5b2obo9bo8b2ob2o3bo10b3o4bo$o3bo5bo4bo2bo3b2obo3bobobo3bo10bo2bo2b3o3bobo8bo5bo3b2o7bob2o10bo4bo2b3obo3bo2bobo2b3o4bobobo6bo2b2o2bo2bo8b2o2b2o5b2o3bo5bobo6bo2bo3bo2b2o8bo6b2obo2bobobobo5bobobo7bobo2bo3bobob2o6bobo8bo3b3o2b4o12bob4o3b2o7b3o3bo3bo7bo2b3o2bo3bobo4bob2ob2o7b2o5b5o5b3o$4o19bobobo4bo2bo2b2o2bo7bo9bob2obo2bo5bo3b2o2bo3bob2obo6bo5bo5bo7bo2bo3bo2bo3bob2o3b3o12b2obo11bo6bo2b3ob2ob2o3bobo7b3o6bo4bo4b2o3bobo2b2o3bo6b2ob2o3bobo3bo2bo3bo2bo10bobobo5bo3bo2bo4bo2bo7b3obo10bobo3bo4b2o5b3o2bo7b2o2bo2bo3b2o4b3obo3bobob3o3bo4bobo$o6b2obobo7bobo3bo5b2obo3bo7bo5b2o9b2o10b2ob2obo3b2o8bo3bo6bobobo3bo8b2o6bo3bobobo6bo2b3obo4bo3bo3bo3b2ob3o3b3o8b2obob4o4bo2b2o6b3o2b2ob2ob2ob2o5b3obo5b2obo3bobo9bo7bo3bobobobo3bo2bobo2bo2bo2bo2bo4bobob2obo5bo3bobo2bo17b2o2bobo6bo5bo2bob2obobo7bobo$11b2ob2obobo2bo2b2o4bob2o2b2o5bo6bo2b2o3b2o3b2obo4b2o10bo2b3o6bo2bobo2b2obo6bo3bob4o18bo4bo6bo4bo4b2o11bobo2bobo2b3obo2bob2o2bob2o5b2obo2bobo2bo2bo2b3obo3b3o3b3o2b4o2bo2bo6b2o2bo4bo8bobo6bo7bob2o7bo4b2o11b2o2bo3bo3bo2bo2b4o2b2o5bob3o2bobo4bobo3bo2b3o5bo3bobo5bo$o11bo2b2o5b2o4b3o3bobob2ob3o2b2obo3bobobo2b2ob2o5bo4bobo2bo12bo4bo3bo2b2o3bo8bo6bo2bo7bob2o2bo7bo2b2ob2ob4o2bo3bo4bo3bo7b3o3bo2b2o3bo2bo4b2ob2o2bo3b2obo2bo3bo5bo3b2obo5b2o3b3o2bo2bobobo3bobo5bo3b2o2b2o3b2ob3obo6bo14b2o7b2ob2o4bo2b5o4bo3bobobob3o6b2ob2o2bo3b3o2b2o6bo$o8b6o2bo3bo2b2obobo3bo8b2obo5b3o6b2ob3o3b2o2bo2bo3b2obo5bo8bo7bobo9bo3b2o2bob2ob3o2bo2b2o6b2o4bobobobo2bo4bo6b3o3bo3bo6b3obobo5bo2bo3bobobo3bo6b3o2bobo7bobo2bo4bobo2b2obo8b2o3bo2bo2bobo3bo3b2o2bo2bo11bo5b2obob2ob2o2b2o2bo3bo3bo5b2obo5bo2bo7bobob5ob2o3bo2bo2bo3bo$2obo8bobo2bo7bo4bobo2b2o4bo8bo2b2o8bo5bo3b2o6bo7b3obob2o3bobo6bo4bo4bo6bo2b3obo2bo5bo3bo5bo4bo10bobobo3bo4bo10b2o5bo2bobo5bo4bo2bo6bobo5bobobo2bob3o5bo3bo3bo3bo9bo3b2o2bo3bobo2bo12b2o4bo5bo2bobo2bo2b2o3b2o9b2o7bobob2obo4b2o2bo12bobo3b2o7b2o3bo$2bo2b3o7bob2obo5bo2b3o3bo4bo4b3o4bobo5bo2bobo2b2obo5bo15b3o2bob2o5bo3bo2bobo4b2o5b2o8bo4bo9b3ob2obo10bo4bo2bob2obo4b2o4b2obo3b4o3b2o3bo2b2ob2obo2b2o3bo3bobobobobo6bob2ob2o2b2obo2b3o6bo3bob2o4b2o2b2o3bobo5bo3bobo3bo15b2o11bobob3o3b3o3bobobo2b2obo3b3ob2o3bo3bobo3bo$o7bobo2bo4bo5bo3bob2o14b2o2bo2b2o2bo3b4o7bo4b5o5b2o6b2ob2obobo14bobo2bo3bobo8bo3bo3bo5bobo3b2o6bo2b4o4b2o8bo2bo5bobo8bo3bo5b2o6b3o5b2o4bo3bo3bob2o4b2o6b2o2bo2b3o4bo3b2o2bob2o3bo2bo4b2obo15b2obo3bobo2b2o4b2o6b2o3bo2bo5bob2o4bo2bo3bo5b2o2bo3b2o5b2o2bo$bo5bo6bo2bo3bobob4o2bo10bo5bo3b2obob3o5b2o2b2obo5bo4bo16bo2bo8bo2bobobo13bo2b2ob2o7b2o16bo2bo4bo2b2obo4bob4obo2bo2bo3bo4b5o6b2obobobo2bo8bobobo3bo2bobobo2bo8bobo3bo2bo6bo2bo2bo2b4o4b3o8bo3b2o4b2o2b3obob4o4bo4b3o2bo3b2obo2bo3bo3bo8b2obo3bo5bo6bo4bo$2bo4bobo3bo6bo5bo4bo7bob5o2bobobo3bobo2bo2b3ob4o9bo2bobo10bob2o7b6o3b3o2b3obo3bo2bo2bo3bo2bob2ob2obo6b3o2bo4bobo4bob2o5bo2b2obo10bo14b2o3bobo2bo3bo4bo2bo3bobob5obo5bo4bobo3bo3bo9bo3b2o8bobobobo2bobo2bobobo2bo5b4obo4bo4bob2o2bobo4bobobob2o4bobobo5bob2o4b3o5bo2bo2bob2o$2bo6bo11bo3b3o2b3obo3b3obo8bo3bo5bo8bobo2bo2b2o2bo6b2obo4bo8bo2b2obo2bo4bobo2bob2obo2b2ob2o3bob2o7b2obo3bo6bo5bo2bobo3bobo3bo3b2o2bob3o2bo13b2obo9bo4bo2bo4b3obo5bobo6bo4bo3bobo7bobo5bo3bo4b2obo2bo3bo4b2o3bo3bobo2b2o5bo5bo3bo4b2o5bobobobo3bo3bobo3b2o4bob2obo5bo$b2obo4bobob6o2bo2b3o2bo2bo2bo13b2o3bo3bobo3b2obobo2b2o2b3o5bo4bo2bo5b2obo2bo4bo2b2o7bobo3bo4bo4bo2b2o4b4o4b3o5bobo3bob3o3bo2bo4bo3bo13bo5b2o9bobo2b2o2b5obo2b2obo7bo9bo2bobo5b2o3bo2bobo4b3obo2bo3bobobo4bo6bo12bobo9b2o2b2obobo2bo2bob2o2b2o10b5o4bo2bo5bo4bo2b2o$ob2ob2o2bobo2bo4bobo11bo2b2ob5o3bobo4bo2b2obo3b2obo3bo2bo3bo2bo2bob2o9bo4bob2o7bo3bo4bo3bo3bob3ob2o2bo2b2o4b2ob2o3bobob2o2bobobo2b2o2bob3o4b2o3bo2bo4bo2b3o7bo6bo9b3obo2b2o2bo3bob5ob2ob3o7b2o6bo2bo3b2o3b2obobo2bobobobo6bo7bobo3bobobo2b2o4bo3bo6bo4bo3bob3obob2o3bo10b2o3bo7bobob2obo2bo$4b3o2b2obo2b2obo2bo9bo8bobo2b2o2bo3bo3bo2bo2b2o5bobo2bo4bobob2o3bo4b2o3bobo4bo2bo3b2o5bo5bobo4b2o2bo2bo14bobo3bo2bo3b2o6bo4bo4b2o4bo5bobo2b3o5b2o3bobob2obobo6b2o2bo2bo3b2o5bobo7bo2bobo4bo4bo5b2ob2ob2o3bo3bo8b6obobo4bo3b2o2bo2b2o4bo13bo2bo2b3o9bo2b2o10bo4b2o$3bo3b3o3bo7bo9b2o2bo2bo2bobob2o3b2obobo4bobo7bobobo3bo3b3obobo4bob3o8b2o14b3o3b2obo2bo2bo2b2o2bo3bo3bo4bo5bo6bo4bo2b4ob2o5b2obo3bo11b4o5bo5bo3bo8bo2bo2bo2b4o3b2ob2o4b2o5bob2o4bo2bo3bobo3b3o7bo15bo3b3ob5obo2b2o2bo3bobobo2bobo3bo2b3o9b2o5bobob2obo5b2obobo2bobo$5bobo6b3o2b2o3bo11bo2bo3b3o5b2ob2o6bob2o3bo6bo5bo3bob4o2bo2bo4bob2o3bobob2obo2b2obo6b3o4bo2b2o2bob2o2bo3bo3bo6bo8b5o5b3obob2o6b3obobobo2b2ob2o5bo2bo3bo2b3o6bo6bo3bo8bobo3b2o3b2o7bo8bobo6bo8bob2o2bo3b4o2b3o3b2o7b2o3b3o5bobobo4bo2bo7bo3bo2b2o5bobo4bobo2bob2obo$bo2bo12b2o2bo4b2o2bo3bob2o8bob3o2b2o4bo3bo3bo9bo6b3obo9bo6bob2o3bobo3bo2bo4bo3bobob2o12b2o2bo6b2o4bo2b2o2b3o4bo2b5o9bobob2o15b2o3bo2b5obo2bo3bo6bo4bo2b3obobo8bo3bo4bo4b2o4bo4bobo4bo3bo3b2o2b3obobo5bo5b4obobo2b2obo3bobo3bo6b2o3bob2o3b4o5bo2bo6bo2bo3b2o$b2o4bo5bo3bobo2b5o3bo2b5o12bo3b3obo4bo4bo5bo5bobo2bobobo3b2o3bo5bo4bo4bo5b2o5b4ob2ob2obo2bobobo3bobo2b4o6bob2o5bo3bo3bob3o2bobobo15b2obo2bo2b2o2bobob5o3bob2obobobobo2b2o6b2o6bo5bo3b2o4bo3bobo2bo6bo2bo6bobobobo8bo6b2o3bobob2o4bo2bo5bo2bo5bo5bo2b2o5b2obo2bo2bo4b3ob3o$bobo6b2ob2o2bo16bob2obo5bo2b2o2b3o4b4o2b3ob2ob2o7bo6b2o2b4obo3bo3bo2bo3bo9b2o3bobo7bobo3bo7bobo15bo3bob2o2bo7b2o3b3o8b3ob2ob2ob2o4bo5bobo3bo8bo2bobo4bobobob4o7b2obo6bobobo5b3obo2b2obobo2bo3b4o4bo2bo3b2o3b3o2bo5bo2b3obo2bob2o6b2o5b2o2b3obo3bo2bo8bo6bobo$2bo12b2o4bo3bo5b3o5bo4bob2o6b3o3bo2b2ob3obo5bo2bo5bo3bo10bo4bo2bo7bo14bo2bob3o2bo5bo3bo5bobo2b2obobo2bo3b4o7bobob4ob2obo2bo3bo3b2o2bo2bob2obo8b2o4bo11bob2o3bob2o2b2o8bo2b2obobo2bo3bo2b2o2bobobo6b2ob4o3bobo5b5o6bo2bo8b2ob2o6bobobobo2b2ob3o5b2ob2o4bo2b3o6bo$o3bob2obo2bo4b3ob2obo4bo8bob2o12bo2bob2o6b2ob4ob3o2bob5o3b5o6bo8bo6bo5bobo2bo2b2obob2ob2o2bo5bobo11b3obob3o3b2obo6bo2bobo2bobobo2bo2bo9bo3bo2bo3b2o4bob2o2bo3bo18b3obobo6bo3bo5bobo2bo9bo2bo3b3o10bo4bobob2o11bobo4b2o2bobo2bob2o2bo3bo10b2o2bobobobo3bo3bo2b2o$3bob2o2bo3bo10bo2bobo4b2o3b5ob2o4bo10b2o2bo5bobo2b2obobo2bo2bo5bo6b2o2b3ob2o4b2obo3bo2bobo5bo2bob3o3bo2bo10bob3o4bo3bo3bo5bo4bo6bo9bob2obo3b2ob6o3bo2b2obo2b2obo6bo2bobob2o2bobo2bo5bo3bobob2o7bob2o3bobo8b2o3b2o4bobo3bo10bo2bo4b2o3bobo4bo12bo4b2obo2bo3bobo4b2o3bobobo3b2o$o2bo6b2o3bo2bo3bobo3bo2b2o6bobo2b4o2b2o3bob2obo5bobo2b2o2b4ob2o3b2o12b2o3b3o7b4o4b2o5bobo3b2o7b2o4b3o6b2ob4o4bobo5bob2ob2obo2bo2bo2b2obo2b2o6b4o6bo14bo3bo10b2o2bo10bo4b2o5bo4bo3b2o5b2ob2obo9bo6bo3b3o2bo5b4o2bo4bo6b2o7b2o2bo2bo2bo2bobobob7obo2bobob2o2bo$8o6bo4bo3bo2b2o4bobo4bo10b2o8bo5bobo6bobo3bo3bobobobo4bo3b2ob3o2bo4bo3bo2bo6b4obobobo3bob2o5bobo2bo3bo4bo12bo3bo2bobo2bo4b2obo2bo6b2obo2bo3bo2bobo3bobo2bo3b2o2bo20bo2bobo5bo5bo2bo2bo2bobobo7bobo3b2obo5b2o4bo2bo6bo3b2obo5bobobo2bo2b2o7bo2b2ob2o2bobobo9bo3bobo2b2o2bo$obo8b2o12bo5b2o8bo4b4o3bo2bobo4bob3obob2o5bobo8bobo2bo2bo2bo2bob2o2bo4b2o2bo2bobo3b5o6bo4b3o3bo5bo3bo2bo4bo4bo8bob2obo3bo2bo2bo6b2o3b4o4bo7b2o4bobo11bo4bobo6bob3obo2bo2bo4bo5bo7bo3b2o4bob2o6bo3bob2o2bo2b2o3bobobobo7b2o2b3o9b4obo11bo3bo8bobobo2bob2obo$obo3bo2b2ob2o2bobob4obo2bo12b2ob2o6bob2o7bo6bo7bo4bo3bo4b2obob2o8bo3bo2bo2b2o3bo6bobo3b3o8b2o2bobo3bo3bo3bo5bo2bo2bobo7bo10b4o14b3o2bo5bo3bo7bo8bo2b2o6bo5b2o4b4o13bo7bo4bob2o5bo14bo3bo2bo6bo6bob2obo6bo4bo4bo2bobo2bobo2bo13bobobo$2b2o3bo4bo2bo3bo2b2o2bo16bo6b2o2bo6bob2o2b2o3bo2bo9b5o6bo4bobob2o2bob3ob2o2bobob4o3bo4b2o2bo2bo4bob2o5bobo5bo7bob3o5bo10bo5b2o2bo4bo11bo5bo3b2o2b2obobobo2bo3bob3o5bobo3bo8bo3bo3b2o7bob3obo3bo4bo6bo3bobo6bo4bob2obob3o6b3obobo3bo3bob2o3bo3b2obob2o2bob2obo11bo$obo2bo4bo2bo5b2obo6bob2o3bobo5b2o5bo5bob4o2bo6bo6b4o5b6ob2o5b2obob3ob2o4b2o3b3obobo8bo2b2o3b2obo6bo7bobob4o2bo9bobob3o4bobo3bo17bobobo7b3o6bobobo5b2o5bo3bobo2bobo4b2o3bob2o11bo2bobo3b3o7b2obo5b5o2b2o3b3o3bo8bo2bo8b2o5b2ob2o13b2o2bo$2bobo12b2o2b2o2bo4b2ob3o11b3obo2bo2b5obobo3bo4bo9bo3bo10bo10b2obo5b2o3b2obo2b2ob2o3b2o4b2o11bobo8b3obo3bo5bobo3bo2bo5bo3bo7b2o7bo2bo2bo3bobobo9bob2obo6bo23b2o2b4ob3obo11bo4bob2o2b2o2bo10b2obo4bo3bobobo3bo2bob5obo3bo2b2obobo2bo3bobo2b2o3bo2bo2bo$o2bobo3b2o2bo4b2o6b2o8b2o2b2o2b2obo5b3obo3b2obo3b4o5bo2b3o6bo4bo6bo8bo2bobobo5bo5bo3bo4b2obo11b3o9bobobo3bo2bo7b6obo6bob2o7b2o3bo4bo2bobo5bo2bo2b2o4bo2b3o3bo2bo4bo6b3o9bo10b3o3bobobo3bobo2bo5bob2obo6bobo9bo2bobobobo4bo4b3o2bo6bo2bo8b2o2bo3bobo7bo$bo2bobo6bo3b2o9bo2b3o3bo4bo7bobo2bo3bob2obo3bob2obo4bo5bo2bo2b2o23b2o2b2o4bo5b3obobobobo2bo5b6obo4bobo3b2o3bobo2b2o5bo3b3obob2o6bo2bo2bo2b5o3bobo3bobobob2o5b2ob2o2b4obo7bobo4b2o9b2o2b2o5bob4o9b2o2b3obobo3bo4bo4b3obob5o3bobo5bo3b2o6bobo4bo7b3o4bo10bo3bo$2o3bo4bob2obo4b2o2bo4bobo2bo9b2o5bo3bo3bo2bo2b2ob2obo2bo2bo2bo4bobo4bobo2bo4b3o5bo3b3o3bo2bo4bobo4bo2bo2bo2bo4b2obo5bob3o2bobo6bo6bo3bobobo2b2o3bobo2bob2obo2bobo4bo2bo3bobobobo2bob2o5bo2bob2o5bo3b6o3bo3bo2b2o7bo3bobo8b2o4b2o2bo9b2o6b3ob2o11bo2bobo3b2o4bo5bo2bo7bo4bo2bo5bob2o$o2bo3bo7bob2ob2o2bobo7b3obo4bo2bob2o5bo2bo2bobo3bo6b2o4bo5bo2bo8bo4bobo2bo5b2o2bo5b2o7bo2bo13b2obo5b3obob2o3bo2bobo4bo2bob5o4b3o4b3obobo2bobobobo4bo6b2o2bo4bobo3bobo4bobo2bo2bobo6bo7bo2b2o4bo3bo5bo6bo4b3o2b2obobo4bo8bo4bo4bo4b2o4bobo2bo3b3o6b2o2bo6bobo4bob2obo2b4o$2o9bob2o5bo12bo2b3obo7b2o3bo9bo5b2o4bo4b3obo8bo6bo3b2obo6bobo6bo2b2o4bo8bo3bob2o10bo2bo6bobobob3o3bo8bo2b2o3bobo3bo2bo2bobo3bo13bo2bo2bobobobo6bo3b2o3bo7bobob3o3bob2o3b4obo7b3o6bo3bo2bo4bobob2obo10bobobobo7bo14b2obo5bo6bobobo2bo5b2o$3bo2b3o2bo3bo15bo5bo9b2o6bo2bo4bobobo5b2o4bo2bo2bo2b2o4b2obo3bo8b2ob2obo4bo4bobobo3bo4bo6bob2o16bo3bo4bo2b2o3bo3bo2bob2obo2bo2b2obo7bo2b2obobobo3bobo4b2o2b2o10bob2o10bo2bo6b2obobobo3bo11bo6b7obobo2bo5b2o4bo3bob2o5bo4bobob3o5b2o10bob2obo4bo3b2o8bo$o9bo5bob2ob2obo3bo2bo3bo3bo5bo8b4o5bo2bo4bob2o9bobo4b2obo2bo9b2o4bobo3b3o2bo5bo10b2o2bo7bo2bo2bo4b3o5bo3b2o5b2o2bob2o2bo8b2obo8bo9b2o2bobobo5bobo2bo2bo4bo7b3o3bo2bobobo2b4o4bo7bob3o4bo3bo4bo3bob3o2bob2o5bobo2bob2o6bo2bo2bo2bo3bobobo3bob2o8bob3o10bobob3o$b2o5b2o3bobo4b3o6bo2bo5bobo3bo3bo5b2o5b2o8bo6bobo2b2o5bobobo3bo2bob2obo5bo2bo7b2o3bobobo4bo3bobobo2b2o6bo6bo8bobobo7bob2o3bo11bobo5b2obo2bo5bo6bo2bo2bo4b2obo2b2o4b2obo4bob2o3bo2bo3bo2bob2o2b2o3bo2bo5b4o2bobo2bo2bobobobo2bo2bobobo13bo3b2o3bo4bob3o2b2ob2obo4bo2bobo6b2o3bo3bo3bo$ob2o2b2obo2b2o2b2o3bo3bob2ob2o2b3o2b2o7bobo11bo3b3o3bo2bo5bobobo4bobo3bobo2bo6bo4bobo3bo2b2o5b2o4bo2bob2o6b2o4bob2o3bobo3bobo2bo5bobo8bo5bo3b3o2bo2bobob2o10bob2o4bob3o2bo2bo2b3o2bo3b2o10bobo6bo3bob2obo2b2o2b2obobo5bobo2b3ob2ob3o6bo3bob4obo2b2o2bo2bo5b2obo2bo5b2o2bo2bo5bo8b3o5bo2bobobo$2bo3b3o4bobo4bo2b2o3bo5b2obo3bo2bob3o3bo4bobobo6bo9bobobo2bo2bob4o9bo5b2o3bo5bo3b2obo5bobo3bob2o6bo2bo2bobobobob2o2b5obo19bo5bobo3bob4o4bobobobo4bo2bob2o4b2o2bo4bo3bo4b2o4bo2b6ob2o2bo3bo2b2o3b2o2b2o4b2o9bo2bo2b3o3b2o4bo2bo4bo6b3o3b2o3bob3o3b2obo4bo6bo2bo2b3obo2bo2bo4bo$2o4bobobo2bo2bo7b2o5bo2bo2bo3bo3b3o3bo2b2o4b2o3b2o2bo8bo2bo6b2ob2o3b2obo2b2obobobo3bo2bo2bo2bobo7b2o5b2o7bo5bo5bo2b2o9bo2b3o2bo2bo3bo2bobo7b2o4bobo3bobo7bobo4b2obo7bo3bo5bo6bob3o2bob3ob5o22bo2bobo3bo7b2o2bo2bo2bo4bo3bo9bo4b2o2bo4b3o7b3o2bo4b2o4bobo3b2o3bo$o6bo3bo4bo5bo3b2o2bo2b2o7b3o5b3o10bo2b4o4bo3b2o3b4ob2o2b3obobobo2b2o8bobobo2bob6ob2o5bo3bo4bo4bo3bo3bob2obobo2b4o4bo3bobo2bo9bo2b3o3bo2b2o2b2o2bo7bob2o2bobo3bo3bo4bo3b4o9bo4bo9bo2bobo2b2o17bobobo10b3o2bo5bo2b2o7b5o6bobobob2o4b2obo4b3o3b2ob2o6bo4b2o2bo$2o8bo6bo3bo9bobob3o4bob2o2bo2bobo16bo6b2ob2ob2ob2o2b2o2bob3o6bobo4bo2b2ob2o7b2obo2b2obob3o3bo3bo6bobo3b2o3b2o2bo5bo2b3o4bo2b3ob2obo7bob2o2b3o4bobo6bobo3bobo2b2o5bobo2bobo10b2o3b4ob4obo2b2o4b3ob2obo2bo5bo5bo4b3o2bo5bo2b2o3b2o8bob2o2b3o2bob2o5b2o3bo5bo5bobob3obo7bo$5bo2bobobobo6bo2bobobo2bo2bo2b2o2bo5b2o2b2o2bo4bo5b3o4bo9bo2bo3bo12bo6bo2bo9b2o3bobo2bo7b2o2b2o2bo2bo4b4obobob2obo5bobobo10b4o11bo2bo5b2o4bobo6bob3o2b2obo2b2obobo10bo4bobo2bo2b3o5b3o2b2o2b3o5bo2bobo2bobo8b2o4bo7bo4bo3bo8b3obobo2bo2bob2o4b2ob2o11bobo3bo2b2o4bo$2o2b3o5b2o5bo10bob3o8bo5bo2bo6b2o2bo2bo4b2o11b2o2bo6bo7bo6bo2bo4bo5bo2bo7bob3obob5ob2obo4bob2o2bo5bo2b2o6b5o2bob2o2bob2obo2bobo4bo2bo3b2o6b2o8bo3bo4bo4bo3bobobo3bobo4bo3bo5b2o5b3o3bo2b2obo2bo9bobo5b3o2bo3bobo9b2o3b5o3bo4bob2obo5bo4bo4bo2b2ob2obo7b2o4bo$2bobo2bobo4bo3b2o3bobo4bo5bo5bo7bo2b2o2bob2o2bo2bo10bobo2bo3b2o3bobo12bobobo3bo3bobo2b2obo2bobo3bo5bo3bo2bo3bo4bo2bo2bobobo4bobobo2b2o7bo7bobob2o2bo3b2ob4obo2b3o2bo4bob2obo5bobob3o6b2o5bobo4bob2o2b3obo9bo4b2o5bo4bo13bo5bo5bobo2bo5b2o2bo2b2obobo3bob3obo7b2obo2b2obo2b2o2bo3bo$6bobobo2bo3bob2o10b3ob2o6bo7bo11b2o2bobobo4bobobo2b2o4bo4bo2b2obobo3b2obo4bobo3bobo8bobo2b2obo2bobo9bo8bo6bo5bo3bo6b2o2bo3bobobo5b3o2bobo7bo9bo2bo3bo2b2o5bo4bo5bo3b3o4bo2bobo17bobob2ob2o6bo2bo5bo2bo2b2o3b2o5bobobo7bo3bob5o5bobo3b3obo8bo2bo2bo3bo$o4bo3bo2bo14bobo4bo4bo3bo2bo11bo2bo3bo4bo7bobo11bob2obo3b2ob5obo4bo5bo3bobobo12bo9b2o8bo3bo3bo13bo3bobobo2bo11bo3bo3bobo6bob2o3bobo4b2obo2bo4bob2o9b2ob3obo4bobo2bo4b2ob3o6b3o5bo2bo10b3obo3bobobob2o4b2ob2obob2o2b2o3b2obo2bo5bobo8bo9bo2bobo$2b5o7bo2bo2bo4b2o10bo2b3o2bo6bo5bo4bobo4bo5bo3bo5b2o4bo2b2o18b2o2b3obo2bobobob2o6bo4bob2o3bo2bo2bobobo3bo2b2o3bo6bo3bo5bo3bo2bo4bo4bob3o5bo5bo2bo4bo9bobo4bo2b3o8bo3bobo3bobo5bo7bo3bo2bo5bo2b2o3bobo3bo3bo2bobobobo2b2o3bo2bo3bobobobo2bo2b2o5b2o4bo8b2ob2o6bob2o3bo$b2obo2bo6bo16bo2b2o5b2o4b2o3b2o2bo7bo9b4o4bo4b2ob2o2bo5b2o3bo2bo5bo2bobo3b2o2bo6bo2bo4bobobo12b2obobobo3b2o3bo7b3o2bo6bob2o4b2o4b3ob3obo4bo2bo5bo5b2o2bo5bob3ob2o7bobob2obo4bo6b2o4bo6bo2b3o2b2o8bo2bo2bobo2bo2b3obo8b2o5bo9bo2bo4bo8bobo3bo4bobo7b2o$3bobobo3bo4bobobo2bo8bo2bo5bo3bo5bobo2b2o7b5o2bo5bo3b2o3bobob2o11b4o2b2o5bobobo8bo5b2o3b2o3bo3b5ob2o4bo10bob2o6bobo2b2o7bo9bobobo4bobo4bo4bo5bobo3bo5b2o2bobo10b3o2bo2bo4b3o2bobo2b3ob2o3b3o5bo3bo5bobobobo5b2o2bo2b2o4bo5bo4bo3bobo6b2o2bo2bo7bobo4bo3bobo2b2o2b3o$2ob3o4bo7bo2bo12bo2bo2bo3bo5bo2bo4bo2b2o4bo2bo2b2obo10b4o5bobo2bo2bo5bo3bo5b2ob3o2bo5bob4obo2bo3bo2bo3bobo7bo9b2o2bo7bo2bo11b3o5bob2o7bo3bo3bobobo2bobobob2o7bo5bo5bobo6bo5bob2o3bo7bo6b2o6bo2b2o2b2ob3obo11b2o2bobo3bobo9b2o2bo8bobo2b2o7bo4bobo4bo$10bo3bo2b2o2bobo3b5obo8bo7bo2bo4bo2bobo9bo6bob2o8bo3bobo2bo6bo2bo4bobob2ob2o3bo2bo2bobo5bo5b2o3bo2b4o7b2o6bo2b2o5bob3o2b2obo2bo4bo2b2o9b3o2b2o2bob2o2bob2o7bobo2b2obobo2b2o4bo3b4obob2o2b3obo11b3obo4b3o2bo7bo3bo5bo2bo2bo2bo2b2o3bo3b3ob3obo5b2o2b2o4bobo3b2o2bo2bo7bo2bob2o$2bo2bobo12bo3bo5b2o4bobo3bo4b4o5bo3bo7bo3b2obo3bo4b2o6bo3b3o2bo2bo5bo5bobo3b2o2b2o4b2o6b2obo11b4o3bo7bo3bobo2b2obo3bobo3b2obo2b2obo2bo5b2o2bo3b2o2b2obo2bo2bobo4bo2bo2bobobo7bo8b2obo4bo4bo3b2o6bo3bo3bo6bobobo4bo13bobo6b3o4bo10bo3bob2o6bo5bo5bo2bobo3bo2bobo3bo$o6b2obo4b2o4bo7bo4bobo2b3o2bo3bo4bo2bo14bob3o2b3o4b2o4bobo3bo2b2o2bo4bo2bo2bobob3o3bo3bobo2bob3o3bo5bo4b2o8bo7bob2obo9bo7b3o4b3o9b2o5bo8b2o7bo2bobo3bo2bo3bo4bo5bo4bo2bo5b2o2bo3bobo3bo4b2o8b3o4b2o3bobobo2bo2bob4o2b2obobo3bo4bo2b2o6b5obo9bo3bobo6bo6bo$4b2o2bobo3bo4bo2bo4b2o4bobo8b2o4b2o6bobobo2bob2o4bo8bo8b6obobo5b2ob2o2bo3b2o3bobobo3b2o2bo3b3o7bobo4b2obo2bo6bobob2ob2o5bo3bobo3bo2b2o2bo6bo3bo4bo3b2o2bo11bo8bo2bo2bo5bo2bobo4bo2b3o5bo5bo2bo6bo2b2o2bo4b2o5bo8bobo4b2o2b3o2bobo6b2obo3bobo2bob2obob2o4bo2bo2bo3b3o4b2o3bo$o12bo3b2obo2bobo2bo2b2o8b3ob2o3bobobo2bobo2bo2b3o2bobo4bo2b3o2bo3bo7b2o3b5obo4bo2bo4b2o9b2o2bo2bo5bob2o4bo2bob2o2b2ob2o8b2o9bo4b2o13b2o6b2ob2o3bob2ob2o4bo11bo5bo3bobo2bo3bob2o8b2obob2o3b2o4bobobobo12bo2b2ob3o8b2o4bo8bo3bo5bo12bo2bobo2bobo9bo4bo2bo$bobobobo5bo7b2o2bo5bobo11bobo8b3o3bo2bo2bo3bobo3bo2bobo28b3o4bo10b2o3bo4bo3b2o4bobobo4bo4bo2bo3bobo3bo5bobob2obo4bo20bobo3b3o2bo7b2o2bobobo3bobobo2bo4bobo2b3obo2b3o2b2o4bo6bo2bo4b3obo3bo2b2o10bo3bob3ob2o5b2o3bo5bo6bo4bo9bobobo4bobob4o3bo2bo5bo$5bobo5bo3bob2obo2b3o2bobob2o5bob3o4b2ob2o4b2obo7b2o2b2o4bobo3b2o8b2obobo2b2ob2ob2o6bobobo8bo3bobo4bo3bo2bo3b2o3bo5b3obo5bobo4bobobobob2o10bobobo2bo3b2o3b2o3b4o2b6ob2obo4b2o2bo4bo4b2o4bobo7b4o2bo6b3o7bo3bobo3bo2b2obo10bobo2bobo3b2o5bobobo4b2obo2bo3bo2bobo3bo6b2o4bobo2bo2b2o6bo$3o2bo5bo4b2o4b2o4b2o2bo3bo2b5obo3b2o2bo3bo3bo2b2o2bo4bo5bo2bo3bobo2bob4obo2bo3b2o2bo2bo2bo2bobobo6bo3bo10bo3b2o3bo8bo11bo4b3o5bo5bo2b3o3bo3b2o4bo5bo3b2o5bo4bo12b2o3bob2o6bo4bo7bo2bo2bo5b2o3bo5b2o10bo2b2obobo2b4ob2o2b2o4bo4bo2bo12bo2bo3bo3b2o2bobo3b2o2bo3bob3obo2bobo$4bo2bobobo7b2o3bo9b2o10bo3bo4b3o9bo2b2o3b2o3bo6bo5b3obobo2bobo5bo3bobo6bo6bobo3bo3bo2bo14bo5bo5bo2bobo2b3obobo4bo9bobob2o2bobo5bo2bobob2obo5b2o3bobo2bo2bo2b2o3bo2bo6bo5bob4ob2o10bo9bo2bobo2bo2bo3bo3b2o4b2o2b2o6b2ob2o3bobob2o2bobo7bo4bo5bo3bo2b2o2bob2o3b4o$2b2o10bo2bo2bo3bo2bobo2bo3b2obo6bo2b3o2bo3bobo2b2obo2bob3o6bo2bo2b2o4bobo2bob2o2b2o8bo8b3obo11bobobo3bob3obo2b3ob2o6b2o6bo4bob3o5b2o6bo5bo6bo5bobobo3bobo5b2o6bo2bo5bo2bo2bo5bo2bobo2b2o8b4o2bobo9bobo3bo5b3o4b3obo3bo6bo2bo2bo15bo3bobo2bo3b2o3bo2bobob2o4bo3bo8bo$2bo2bo3bobo3b2o14bob3o2bobo2bo9bo2bo3bobo2bobo9b3o2bo4b2o2bo3b3o7bo3bo2b2o6bob2o4bo9b2o5bob2o2b2obo2bob2obo3bo2bo4b4obo2bo3b2o6b2ob2ob5ob4o5b5o5bo2bo4bobo2bo6b2o5bobo3b3obo4b2ob3o2b2obo7bo4bo4b2o5bo2b2o9b2ob2o5bobobobob7o2bo3bo5bo3bo7bo4b2o7bo9bob2o4bo$4bobo6bobob2o2bobo2b2o2bo3b2obo5b2o6b2o2bo5bo7bo2bo3bo2bo2b2obo4b2obo6b2o5bo9bo2bo3b2o3b3o2bobo3b2o11bobob2o2bobo8bo2bo6b5obo3bob2o2b2o13bo5bobo2bobo6bob4obo6b2o7b2o12b2o4bo7b2obo2b2obo3bo19bob3o2bo5bo5bo4b4o2b3obo3bobo2bo6bob2o2bo5bo7bo5bo3b2o$2bobobo3bo2bo3bo4b2o4b2obo3bo2bo5bo2bob2o4bo7bo3bo2bobo2b2obo2bo2b2obo5bobo2bo8bo2bobo2b3o3b2o2b3o3bo3bo5bo2bo3bo5bo2bo7bo3bobo10b2obo5bo7bo2bo6bo4b2o7bobob2o5bo3bo2b2o7bobo2bobo8b2o2bo5bob2o3bobo3bobo2bob3obo2bob2o6bo4b2o6b3o9bo2bo11bo4bo3bo6b2o2b2ob2o2bo4bo6b2o2bo$o5bo6bo5b2ob4o3bo3b2obo2bo4b2o3bobo3b2o6bo10bob2o2b2ob2obo7bo3b2obo5bo9b2o3b2o8bo2bo3bo5bobo2b2obobobo4bobobo2bobo3b2ob3ob2obo7b2o4bo6bob2o2bo3bobo6bo10bobo7bobo7bo7b4o5bob2o3bo3b2o2b2ob2obob2o5bo2bobo3b2o2bo2b2o8bo8bobo6bobo3bo4bobobo3bo2bo2b3obo3b2o2bo2bo2bobobobobo$o2bob3o6b2obo9bo6bo3bobo3bo4b3o4b4obob4o2b2o2bo3b3o6bo3b2ob2o5b2obo13bo3bo10bo3bo11bo2b2o10bo4bo3b2obo2b2o4b2o2bo9b2o3bo5b2o2bo5b2o2b2o3bob2o3bo6b2o7bo8bobo4b2o15bobo3bo2bobo3bo3bo6bo3b2o9b2o3bo4bo3bo6bo2bo9b2o2bo2bo9bo5bo4b2o3b2o2bo2bo$2b2o2bob2ob3obobo15bobo4bo6bo6bo5bo4bo9bo2b3o9bo3bobo2bo2bobo2bobo3bo2b2o6b3o2bo2bo4b2o5bob2o3bo5bo9bo2bo2b4obobo2bo3bob2o2bo4bo2bo3bo3b2obob3o2bo9b3obobo5bo4bo5b2o6bo3b2obobo2bobo4bo8bo2b2o4bo8bo2bobo7bo9b2o2bobo2bo3b3o7b2o2b3o6bo6bob2o3b4o6bo2bobo3bo$3bo5bo2bo7bo2b3o4b3o9b2o3bobo3bobobo2b3obo10bo5bo2bob2o8bo2bo8bob3ob2o7bo2bo4bobo2b2o2b2ob2ob2o12bo3b2obobob2o6b2o2bo2bo4bo11bobo4bo2bo4bo4b3o4bobob2obob2o3bo4b2o2bo2bobo9b2o3bo8bo2bo2bo3bo3bo4b2o5b3o4bobo2bobo2b2obo9b2ob2o4bob2o4bo9bobo15bo2b2ob2obo4b3o$bo6b2obo4b2o4b2o3bo4b2o3b2ob2o2bo5bo3bo7bobo3bobob2o9bo6b2ob2o17bo2bo2bobo5bob2o4b4o4bo2bobobob2o2bo2bo2b4o2bo9b2o2bob3o3b2o5bo8bo3bo3bo3bo2bob2o2bobo2bobo2bo5bob2o6bo12b2obo3b2o2b2o2bob2o7b4o2bob3o4bo2bobo2bobo3bo2bo5b3o3bo6b2o5b2o2bob2o2b2o3bo3bo2b2o7bo2bobobo5bobobo$bo4b2ob4obob3obo3bob5o4b2obob2o4bo6b2o2b2obo3bobo3bobobo5bobo3bo6bo2bobo3b2o6bo2bobob2o2bobobobo2bo2b2o4bo4bo2bob2obo3bobo3bobo2bo5bo7bo2bo4bobo3b2o7bo3bo4b2obo3bo6bo6bobobobo5bobob3o3bo2bob2o5b3o2bo2bo2bo8bo2bobobob3o2bo3bo11bobo5bo3bo10b2o3b2o6bo2bo4bo2b3ob2o9bo4bo2bo7b4o$4bo3bobobo5b2o2b3o3bo4bo4bo5bo3bo4bobo6bo8bobobo5bo3b2o2bo6bob2o7bo3bob2obo3bobo6b3o6bobo4bo3bo3bobob2o2bo13bob2o2bo12bo7b2o11bo3bo2bob3o2b2o3bob2obo10bo6bo3b2o2bo3b2o10b2o12bobo6b2obobo2bo4bo3bob2o2bo3bo5bo6bobo4bo2bo11b3o11bo9bo3bobobo$bobo4b3o4bo2bo6b3o5b4o5bobobo9bob2o3bo4bobo3bo7bo6bo2bobo3bo2b3o2b2o14bo8bo5bo2bo3bo7b2o7bo5bobo2bobobo7bo3bobo5b2ob2o3bobo3bo9b2o10bo4b2obo14bobo3bo3bo9bo2bo2b2o5bo3bo7b2obobo3bo2bobo2b2obo2bo18b7o5b2o5b2o3b3o4bo2bo2bob2o4bo3bobo5bo$bobo5bo8bo3bob2o9bo3b2o5bo4b2o5bo9bo2bo2bo7bo3bob2o9bo9bobo2b2obo9bobo2bobobo5b2o3bo2bob2obobo2bo7bo3bobo3bobobo3bo2bo3b2o7b2obo4bo3bo3b2ob2o2bo6b2o2bo4bo4bo3b2o2bo9bobo7b2o10b3ob2ob2o10bob2o2bobo2b2o4bo2b3o4bo11bob2o2bo2bo3bo2b3o4bo2bobobo3bobobob2o2bobo4b2o2bo$bo3bo3b2o2bo2b3obo5bobo4bo3bo4b2o5bo8b2o8b2o6bo13bo2bobo5bobo2b3o3bo2bo5b2o4bo3b2o5bo2b3o5b2o2b2obo3bo2bo3bo5bobo2bob2o3bobo4bobo2bobo7b2o4b2o2bo2bo2b4o2bo6bo18bo6b3o3bo2b3o4b2obo5bo2b2o2bobo5bo4b2o5bo3b3o2bo10b2o2bo3bob2o6b2o2bobob2obo2bob2o17b2obo2bo3bo$o4bo3b2o2bo5bo7bobo3bo3b2o9bo3bo2bobob3o7bo5b3o5bo4bo4b3o9bo9b4o3b2o3bobobo6bo2bobo4b2obo2bo6bo3bo4bobo4bo7bo6bobobo7bobo2bobo2bo2bo6bobo4bo8b2o3b2o3b2obobo3bo8bo4bobobobo4bo6bo3bo2bobo2b3ob3o11bo2b2obo2bo10bobo3bo4bo3bob2ob2obo2b2o9bo4b3ob2o2b3o$2bo2bo2bo2bob2o2bobo11b2o5bobo5b2o4bo2b4o3bob3o3b2o4bo8bo4bo2b3obo2bob3o7b2obo3b2o6bo6b2o2bobo10bo2b2o6b2o3bo8bo3b2o2b3o2bo4bobo4bobo2b4ob3o2b2o4bo3bo3b2obo5b3o2bo12b2obobo15bo2bobobo5bo11bo4bo3bobob3obo3bobo2b2o2bo5bo2bo4bob2ob2o8bo5b2o12b3obo5bobobobo$2bo2b2ob2ob2obo11bo6b2ob2o4bo2b4obobo8bo3b4o2bo3bobobob4ob2o3bo2bobobo6bob2o4bob2o2bobo5bo2bo7bob2o7b2o3bo4b2o2bobo4bob2o6b3o6bob2o5b3obo2bo2b4o2b5obo2b3o2bo3bo10bo2b2obo6b3obob2o2bo4bo3bo3bo6bo4bobo5bobobo4bo2bo5bo2b3o5bo2bo2b2o7bo6bo4bobo2bo3bobo6bobo2bob2ob2o4b2o$bo4bo4b2o3bobobo9bobo2b2o5b2o2bo2b2o6bo6bo17bo3bob2obo2bo3b2o7bobo2bob2o4bo3bo2bo6bob3o3bobo3b2o4bo7bobo2bo2bo4b2o15bob2o2b2ob2o3b2o2b2o2bo2bo2bobo5b2obo6b2obo2bobo2bobo2bo3bo8bobobo8bo4bo2bo5bobob2o2bo4bo2bobo17b2o4b3o3bo7bo6b2o5bo4bo4bo3bob4o2b2o2bo2b2o4bo$obo2b2o5bobo3b3o2b3obo5bo7bobobo2b2ob2obobob2o10bo3b3o3bobo2b4obo2bob2obobo2bo5b3o4bo3b2o5b3ob2o7bobob2ob2o9b5o4bo6b2o4bo4bo4bo3bob2o4bobo4bobo6bob2o2b2o5bo5b2obobo15b2obobo11bo8b2o5bo2bob4o3bo8bobobo4bob2obo4bo3bob2obo10b3ob2obobo3b2o5bo2bo5bo2bo9b2o2bobo$o3b2o6bo3b2o8bo7b2o4bo3bobo3bo4bo4b3obo12b2obo3bo22b3obob2o2bo2b3o3b2o2b2obo6bo2bo4b3o3b2o3bo6bo9bobo4b2ob2ob2o2bobobobo8bo4bo2bobo4bobobo6bo2bo6bo2b2o5b2o3b2obo3bo3bob5o2bo2bobo13b2o3bo5b2o3bo3b2obobo2bo5bo14bo3bo2bo3bob4obob2o4bob5ob2o5bo7b2obo$ob2obob3ob2obo5bobo2bobo2b2o2b2ob2o2bo5bob2o14bo8bob2o6b3o3bo2bo3b2o2b2o2bo8bo2bo6bo6bo6bo2bo2bo2bo3b2o2bo3bobo7bo6bo2b3o3b2o3bobo4bobobo2b3o4bo8bob2o7bob2o2bo6b2o4b2o4bob2o4b2o15b2o9b4o7bo8bo2bobo3b4o5bobobo8bob2o8b6o2bo2bo3bo2bo2bo9b3o3b2o3bobo$2bo5bo2bo3b2o4b2o2b3o4b2o4bobo5bo2bob2ob2obo4bobo3bo3b4obo5b2o7b5obo3bob2obo5bo3b3o2bo8bob4o4b2obob2ob2o4bo3bo2bo2bo2b2obo3bo3bo2bo6bo3b2obo2bobobo3b3o5bo3bo5bo4bo3b3o2bo4bo2bo6bo5bo4b2o2bo3b3obo9bobo5bo5b2o4bo6bo2b2ob2obobo4bo2bobo16bo2b3o2bo3b2o4bo4bobo7b2o5bo3bo2bo$4bobo4bo25b2o10bo3bo4b5o5b4o5bo6b2o4bo6bo3bobo6bo3bo2bobo6b3o3b4o11bob4obo16bo4b2ob2o2b2o2b2o6bo6bobobo9b2o5bo2b3o2b2o3bo7bo8bo3bo2b2o7bo3b2o3bobo2bo3bob2obob4ob4o2bo8bo6bo4bobo8b2o8bo3bo2b2obo4bo2bo4bo6bo7bo2bob2obobo$2o3bo4b2o7bo13b2o2bo4bo4bo3bo4bo4bobo4bobo3bo4b2o2bo7b4obo4bobob2o2bobo2b2obo2bo2bo7b2obob2o2bo6b2o3bo4b2o2b2obobo3bo2bobo9bo3bob2obo6bo8bo4bo2bo3b2o4bo5b2o3bo2bobo2bo2b4o2bo3b2ob2o4bo3b2obo6bo2bo3b2o7bobobo7b2o7bo3bo2bob3o4bo3bo6bo2bo5bo11bo3b2o4b3obo5bo6bo$4bo2bo3bo4bo3bo7b2o5b3o3bobobo3bobo5b3o5bobobo7b3o2bo2b2o3bo3bob2obob2o5b2obo2bo6bo4b5o2bo3bob2o5bo2b2o7bo3bobo8bo3bobo6bo2bobo2bo2b2o6bo5bobobob3o6bo5bo6bo3b2o6bobobo3bo5b2obo7bo3b2ob3o2bo3b2o7bo2bobo4bo5b3obob2ob4o3bo5bo3bo2b2o4bo8b2o5bo2bo17bo2b2ob2ob2o$o2bo7bo4b2obo3bo5bo5bob2o2b2o11b2obo2bo11bo4bo7b2o4bo5bo10bo14b3obo6b2o5bo5b2o3bo2bo8bo6bo2bo2bo10b3o3b2ob3o14bo5b2o3bo2bo3bo2bo4bo4b2ob2o2bo5bo7bo4bo2bobo4bobo2bobobobo6bo8bob2o3b4o2bo6bo9b2o4bobob2o2b2ob3obob3o5b2o2b2o7b3o5b2ob2o$3obo5bo6bo11b2o9b2o21b3o3bobobo3b2o11bobo12bobo5b2o2bo7bo3b2o2b2o3bo2bo3b2o2bobo6bo5b2o7bo2bo2bob4o20bo2bo3bobo3bo4bo2b2o7bo2b2o5b2obob2obobob5o4bob4o2b2ob5o2bobo2bobo2b3o5b2o3bo2bo5bo4bobo2bo5b2ob2o2bo3b2o5b2o4bobo11bo3b3o4bo7bob2o2bob2o$2o7bo4bo6bobo3bo2bo2b2o2bobo2bo2bo2bobobo2b3o10b3o3bobo2b2o4b2o2bo6b2o2bobo3bo8bobobo4bo2bo2bo5bo4bobobo6bo3bo11bobo8b5obo6bo2b2obobo3bob2o2bo4bo2bo6bobo7bobo4b2o3bo2bo15bo4b3o7bo3bo3bobo3bo20b2o3bo2bobo4bob2ob2o2bo7bo5bobob2obo2bo3b2ob2obo4bo4bo9bobo2bo$o5b2o4bobobo9b2o4bobo5b2o2b2o4bo4b2obobo3bo3bo3bob2obobo5bo3bo2bo4bo4b2ob2o3bo2b2o10bobo2b2obo3bobo5bo2bo2bo3b2o2bob4o2b2o2bo7bobob2o2bo5b2o3bo5bo6bo5bo7bobo3bob2o3bo2b2o5bo3bo2bobo5b2o2b2o6bobobo5b2o4b2o2bo4b3o2b3o11bobo3b2o2bo3b2o5bo2bo3bo5b2o4b2o7bo4bob4o5b2o2bo5bo2bo$b2o4bo8bo3bobo3b4obobo2bo4bo7bo6bo6bobo6bobo4bo2bo2bo6bo2bob3o2bo9b2obobobo2bobo8bob2obobo13b2o5bobo7bo2bob4o2b2o4b2o3bo2b2o2bobo3bo7b3o5bo8bobo5bo2bo5b2o4bo3bo5bobobob2o2bo3bobo2bo8bo5bob2ob5o2bo9bobobo5bo5bobo10b2o6b2obo6bo2bo3bobo3bo14bobobo$ob2o8b2obo9bo9bo5bobob2o8bob2obo2bo5bobo4bo2bo5b3o3bob3o6b3obob2obo4bo4b2o5bo5b2o2bobo5bobo3bobo9b2o5bo7bo3bobob2obo6bob2o3bo2bo8b2o4bo2bo6bobo2bo6b2o3b2o3bobo2bob2o3b2obobob4obo4b2obo2b3o2b3o2bo7bobob2ob2o2bobo2bo2bo3b2o7bo4bob2obo8b2o2bo2bo14bo3bo3bo2b3o$4bobob2o5b2obo2b2obo15bo3bo5b2o2bo12bo2bo7bob2o4b2o4b4o4bob2o3bo6b2o4bo5bo6bo4bobob2o5bo2bo3bo2bo6b2o2bobo11bo2b2o2b2o9bobo4b2obo5b3o3bo9b2o2bob2o5bo4bobob4obo3bobob2o4bo3bo2bo7b2o2b2o4b2o8b2o2bo2bobo2bob2obo2bo3b2o3bo2bob3o4bo6b2obobo2bo2b2o5bob2o4bo4bo5bob3o$bo2bo2bo3bo4b2o6bo2bo3bo6b2o2bo8bo2b2obo5bo2bo3bobo2b2ob3ob2o14bobo3b4obobobo2bobobob2o2b2o2bobo4bo3bo3b2o2b3ob2obo2b4o5bo4bobo5bobo2b2o3bo4bo5bobo3bobobob2obo2bo3b2obo5bo2bo5bo5b2o4bobobo4bo5bo3bobo6b2o2bobo2b2o4bobo3bo3bo2bo2b2o4b3o4b4o5b2o2b3o2b4o5b2o2bo4b2o4bo10bob2obob2o2bobobo2bo$2obo3b5o3b2o5bo3b2o2b2obobobo2bo5bo5bo8bo9b2o2bo2b6o2bo3bo10bo10bo3b2o3bo2bo4b4o10bo2b7o4bo4bo3bo5bobo2bo4bo4bobo6bo4b2o4bobo3bo3b4o2bobo5bobo14bo3bo2bob5o4bo6bo2b3o3bobo2b2o3bob2o2bo9bo9bo2bo3bob2o3bobo4bobobob3o2bo6bo4b2o2bob2obob2o2bob2obo4b2obo5bo$5bo9b2obo3b2o2bo2bob6o2bobob2o2bo7b2o5b4o4bobo4b4obo6bob2o8bo2bobo4bo4b2o7bo14bo3b2o5bobo2b2o2b2o6bo4bo3b2o3b2obo2bobobo10b2o8bo7bo2bo7bo23bo7b2o5bo4bob2ob2ob2o2b4ob3o5b3obo3bo3bo9bo9bobobo8bobo3b2ob4o2b4ob2obo18bobo3bobob2o$3b2obo2b3o5b2ob3o5bo3bobo8bo5bob2o4b2obo5bo3b2o5bobo6b2o2b2ob3obobobo13bo2bo2bobobo2bob2o9b2obo2bo3bo2bo5bo9bo3bobo7bobobo4bobo7bo8bo2bo4bo2bo2b3o3bobob2obo4bo5bo6bo2bo3bobobob2o5bo4bo4bob2ob2obob4o5b3o2bo10bo2bo2b2ob2o3bo5bobob2obo2b2obo7bo2b2obo2b4ob3o2b3obo2b3o3bobo$12bo3b2o10bobo7bo5b3obobob5o4bo2bo3bo8b3o3b3o2bo6bo4bobo5bo2bo2bo6bo6bo3b2o2bo3bo2bo5b2o6bo2b2obo3bo3b2o2b2obo2bo3b2ob3o3bo5bo6b2o7bobo7bo4b2o2b2o2bo3bo2bo2bo2bo2bobo2bobo6bo2bo6bobobo2bo5b2obo2bo4bo2bobobobo4bo2b3o4b2o5bob2o2bo3bo6b2o3bo4bo2bobob2o3bobo6bobobobo5bo$5bo2bobo2b3o2bob2o2bobo5bo8bo2b2o6bo2bo7bo3b2o5bobo10bo2bob3o4b2ob3o2bo6b2obo11bo2b3o2b2o11bo7bo2bo2bo2bobobo3bo2bo2b2o2bo2b2obo12bo14bobo3b3o11bo7bobobo4bo4bo6bob2o4b2o4bobo10bob2obobob3o3bo2bo6bobo8bo2bo6b2o3bo2bo2bo12bob4o2bo8bo5bo3bo$4bo4bobo2bo3b2ob2obo2bo5bo2bobo2b2o14b2obo8bo2b2o2bo6b2obo2bo4bob2o3bo2bo7bo4b2o5bobobob2o3bo2bo2b2obobobo2bo10bobo5bobo2bo2b2obo10b2o2b3o5b4o3b2o2bob2o2bob3obo6bo2bo12bobo2bo2bobobo2bobob4o2bobo5b2o5bo9bo2b2ob4o5b2o12bo3b2o4b2o5bobo2b2o2bo3b3o2bo4bobo3bo2bo4bo6b7o2bo$4bo4b2o3bo3bo7bobob3o3bo2b2o7b3obo3bob2o7bo3bobo3bobo5bo3bo2bobobo3bo2b3o2bobob2obo5bo3bo2bo5bobobo7bo5b2obo2bob2o2bo2b3o4b2o2b2o2b2obo3bo3bo3bo2bo8bo4bo2bobo5bo2bo2bo2bo9bo2bo6bo5bo7bobo4bo8bo3b3o2bo17bo2bo3b3o2bob2o2bobobo3bo3b2o9b3o2bo5bo5bo2bo3bo2bobo6bobo2bob2o$2bo11bo4bo2b2o3bob2obo4b2o3b4o6bo10bo2bo2bo2b5o6b2o3b2o5bo2bo2b3o3b2o8bobo3bobo4bo3b2o5bo4bo3bobo4b2o6bo2bo2bobobo5bo2bobobob2obo3bobo8bo9bo2bo3bo3bo5bobo2bo5bo4bob3obo2b4ob2o4bo15bob2o2bob5o2bo5bo4b2o2bo6bobo9bo7bo2bobo4b2o3b2o2bo4bo3b2o5bo2b2obobo2bo5b2o$2b2o4bo3bo4bobo3bo2b3o3bobo7bo5b2o2b2obo2b3obo2b2o7bo27b3o8bo3bobo5b2o5bobo10b2ob2o8bobo5bo2b2o13bo3b2o2bo2bo6b2o6bo11bo4bo3bo3bo4bo3b2o5bo2b2o5bo3bobo5b2obobo7bo4bob2o2b2o5bo5b3o2b2obo3bo19bob2o2bobobob3o9bo5bob2o4bob2obobo4bo3bo2bo$2bo2bo2bob3o3b3o2bo5b2o7bo3b5o2b2o3bo2b2o9bo4bo4bo2bobobobobo3bo5bo3bo3bob2obob2o3bobo4b2o3b3obob2obo5b2obo5b2o3bo5bo3bo4bo7b2obo4b3o2b3o3b2o2b2o12bo3bo8b2o12bo6bob2o3bobobobo5bobo8b2o3bo2b2obobobo3bo3bobobo9bo5bo5bo5bobobo3bo3bo4b2o2b2o6bo4bo2b3ob2o5bobo2b2o2bo2b3o$2bobobo3b4obob2o4b4o5bo3b2obo3bo2b2ob2o2b2o2bo7bo2bo4bo3bo5bo3bo2b3o7bobo5bo7bobobo2b2o2bo4b2o2bo2bo4b3o3bobo2bo3b3o11bob2o2bo14b2obo2bo3bob2o8bo5b4o2bobo2bob2o2bo6b2o2bo4bo3bo3bo5bo3bo3bobo2bobo2b2obob2o11bo5bo7bobo4bobo7bo4bobo3bo3bo6bo2b3o2bo4bob4o4bo2b2o8bo4bo$2b2obobo2b2o4bo4bobo5b4ob2o8bo2b5o6bo8b2o2bob2obo3bobo3b2o2bo10bo6bobobo5bo2bo2bobobo2bo2b2o3b2o2b2o6bo10b2o2bo4bo4bo2bo6bobo9b2o3bo3b4o4bo5b2o18bo2bobo3b2o5bobobobob2obobo5bo2b6obobo4b3o2b3ob2o2bo3b2o12bob2o7b2o4bobo2b3o2bobo11bo2b2o5bo8bo4b4o5bo$4bo2bo6bo2bob2o3bo5bo7bo3bo5bo2bo2bo7b4obo2bob2o5b2o2bob2o3b3ob2o5b2o10bo3b3o3bo8bo7b2o3bo3bo4bo4bob3obobo12b3o2bo4bo2b2o4bobob2o2bo5bo4bo2bo3b2o5bo2bobobobobo6b2o4b2o4bo2bo3bob2o6b2ob4o6bob2o8bo3b2o2bo2b2o4bob2o3bo9bo2bo3b2ob2obobo2b2o5bo2bo5b2o7b3ob3obo2bo2bob2o$3bobobo6bo8bo5b2o3bobo3b3obo2b2obo2bo9bo8b3o2b2o3bobo4b2o2bo9bo12bo4bo3bob4o5bo2b3obo3bo4bo5bob2o4bo4bo2bob3obo12bo4b2o4bo8bo2bo3bo2b2obo2bobo3bo2bobob2obo2bo10bo2b3o12bo5bo2bo8bo7bobo3b2obobo4bobo2bo2bo2bobo5bobo10bobo3bo2b2o8b2o6bo3bo5b2o7bo$6b3o2b2obobo3bob2obo10b4obo2bo2bo4bobo2b2o5bo4b3o9bo9bo13bobo9bo2bo7bob3o3bo8bobo2bobo2bob4o4bob2o2b2o5bo3bo4bo3bob2o3bo9b2o4bo3b2o5bo3b3o3b3o5bo3bo2bo10bo2bo2bo3b2o2b2obo3bo2bob2obo2b2o6b2obo3b3o3bobobo3b2o2bo10bobo4bo5bo7b2o4b2ob2o5bob2o2bob2o2bo4bob2o2bobo$bo3b2o3b3ob2obo2b3o5bobo2bobo2bo6b3o2bo3bo3bo6b2o5bo4bo5bo12b2o5bo2bob4o2bobo9bo7bo2bo5bo8bo2b2o5bobobobo3bo2bobo3bo2bo6b2o7b2obo3b3o3bobob2o6bob2o9bo2bo3bo3bob3ob2o4bobo3bo2bo6bo2bob2o4bobob2o18bo2b2o2b3ob2o3bobo2bo4b2obobobobobo2bobo3bo7b2obo6b3o4b2o4b2o2b2o3b2o$2bobobo2bo3bobo5bo3b2o5bo6b3o6bo5bobo4b2o3b2o2bo5b2o6b2o2bobobo2bo15bobo5bo6bobo2bo8bo4bo2bo5bo3b2obo6bo3bo5bo2b2o9bo6b2o4bo2bo2bo9bobo3b2obo5b2obo11bo3bobobo2bo14bo6bobobo10bo6b2o2bo9bo4bo7bobo11bobo3bo5bo3bo13bo9bo9b2o$2b2o2bo2bo5bo2bo2bo9b2obo4bo2bo5bobo3bob2o2b2obo3bob2o5b2o3bo2bob2o11bo2b2o2bo2bo2bo2b3obo8b3o4b2o4b2o12b2o3b2o3bo5bo2b2o15b3o4b3o3b2obo2bobo2b2o8bobobo6bo2bo2bobo6bo2bo8b2o4bobo4bo7b3o3bo2bo2bobo4bobo2b5o3bobo3bo2b3o3bo4bobobo7bob2obo3b2o3bob3o9b2ob2ob2obo3bo2bo2bo$8b3obobo2bo4bo4bo3b3obo2bo12b2o2bo2bo8bo2b3o2bo4bo9b4obobo5bobo3b3o9bobo2b3ob2obob2o2bo6b4o9bo2b2ob2obo7bobo2bob2ob3o6bobobo5bo5bobobobo10bo8b2o3bo3b3o3bo7b4o9b2o3bobobo2bobo5bo4bo3bo5bo5bo10b2o7bo6bo3bo4bo2b2o2bobo3bo4bo2b2obo4bo3bob2o3bobobo3bo$4b2o3b2o7bo5b3o5bobo4bobobo9bobob3o3b2o2b3o5bob3o5bob4o5bo3bobo4b4ob2o2bo5bo2bo5b3ob4o3bo6bo3b2o10bo3bo4bo4bob2o3bob2ob3obo6b2o3bo5bobob2o3bo6bo4bo2bo4bo2bo4b3o2bob2obo2b3obo4b3o3bo8b3o5bo2bobobo7bo2b2o8bo6bo6bo11bobobo6bo4bob2o5bo2bo2bo5bo8b2o$2bobo6b5obo2b2obo5bo2bo9bo3b2ob2o2bob2o4bobo14b2o2b2o4bobo2b3ob2o3bobo9bo2b2o2b2obo2bo4bo9bo5bo2b2o25b2o2b2o3bo4b3o3b2o3bo12b2obo4b2obo2bo2bobo4bob2obo6b2o3bo4bobo6bo3bo8b4o6b2o2b2o2bo4bo8bo5bobo6bo4bo2bo5bo3bo2bobo9bo4bo17bo3bobo4bo$bo6bo3bo3bo2b2ob2ob2o3bo4bo5bo8b2o2bo7bo6b2ob3o2bo12bo6bo6bo2bo4bobo2bo6bo6b4o2b2obobo5bo3bo3b2o4bo7bob4o4bo5b3o5bo4bo12bobo4b3ob2o4bo4bobob2ob4o2bo2bob2o4bobobobob2o2bob2o5bo2b2obobo4bo10b4o5bo6bobo3bo4bobo2bo10b2o3bo14bo2bobo2bo15bo6bo$2obo10bo4bo5b4obo3bo5b2obobobo2b2o2bo4bo2bo10b2o9b2o2b2obo4b3o5bo5bo3b2ob2o3b2o2bo4bo4bo5b4obo9bo7bo2bobo3bobo3bo8bo2bo4b2obo3bobo4bob3o2b2obobobobo9b4o2bob2o3b2o2b3obo2bo3bobo4bobo6b2o2bo5b3obo2bobobobobo15bo2bo6bo6bo6b2ob2obo7b3o4bo6bob5obo2bo3b2o$4bobobo5bo3b2o4bo3bo4bo4bo15bobo2bo4b2obob2ob2o6bo2b2o6bo7b3o3bob2ob2o8bo6bobo3bo3bo4bob3o5bo6b2o3b4obo2bo10bo2bobobo7b2o3b2o6b5o2bo3bo3bo6bo3bo2bo2bob3o9b3obo2bo3bo3b4o2bob2o6bo8bo2bobo5bo3b3o8b2obobob2obo5bo3b3o2bo6bo7bo2bobo2b2o2bo2bob2o2bo7bobo2bo$4bo3b2ob2o4bobo4bobo3bo3b2o2bo6b3o4bo6b2o3b2obo6bo7bob3o2bo4b2o6bo9bobo2b2o5b5obo3bo7bo2bo2bobo2bo5b2obo3bo6bobobo2bo4b2o9bo2bo3b2o9b2o10bob2o2bo2bo2bo6bo9bobobobo2b2obo5bo3bobo2bobo2bobobo2bo5bob3obo4bo4bobo11b2o5bo2b2o3bo8bo4bo2b2o2bo7bo5b2obo11bobo$o4bo6bo3bo2bo4b2o5b3o4bob2o9bo12bo2bo3b3ob2o2bo3b2o3b2obo3b3obo5b2o19bo5bobo2b2ob2o2bo8bobo3bo4b2o3bob3obo2bo2bo3bobobo2bo4b2o2bo2b3o7b4o3b2ob2obo2bo9bobo2bob2obob2o4bo2b3o3bobobo3b3o4bo4bo3bobob4o2bo3bobo4bo10bo3bo5b2o2bo4bo4bob2obo7bo2bo3b2ob2o3bobobo7bo9b2o$3b3o2bo2b3o2bobo8bobo4b3o2bobob2ob2o6bo2b2o5bo3bobo5b3obob2obo5bo2bo4b2o8bo10bo3b2o5bo5b2ob2o2bo13bob2o2bob2o2bo2bo2bo2bo5bo3bo2bo2bo3bo4bobo2bob2ob6obo2bo3bo2bo3bo2bo8bo6bo4bo4bo2bobobo3bo5bobobo7bo2b2o3b3obo3bo9bobo2bo7bo2bobobo15bo3bobo2b2o3bo2bo8b3ob2obo6bo2bo$b2o2bo5bo2bo4bo2bo10bo11bobo2b2o2bobobo2b2o11b4o6b3o2bo4bo3bobo2bo3bo7bo8bobo2b4o4b4ob2o4bo3bobo3bo5b2o2bo4bo2bo4bo8b4o3b2o2b4o3bob2o3bo3b2o4bo3b2obo2bo3bo2bo6b2o3bo3b2o5bo4b3o3b3ob3o2b2o4bo2b2obo4bo2bo4bob2o2bob2obo4bo7bo3bo2bo6bo5bo2bobobo4b2ob3o2bobob3o3bo2bo2b3o4bobo$5bobo5bobo5bo2bo2bobo3bo2b6ob2o4bobobo3b2obo2bo8b2ob2ob2ob2o2b5o2bo4bo4bobobo3bo4bo4bobo2bo4bobo2b2o5bobo3b2obobo5bobo6bobo6bo4b2o4bo6b2o4b2o2bo3b2o4b2o3bob3obo2bob2o5b2o7b2obob2o2bob2o13bobo3b2o2b2o5bo3bo3bobo5bo9bobo3bobo2bobobo2b4o4b2o4b2o5bo11bobo3b4o4bo7bo2b3o2b2o$3bo5bo3b3o5bo2b2o9bo2bo2bo2bobo2bo2bobobo4bo5b4o5bo2bo3bob2o9bobobo8b2o2bo4bo5bo2b2obo2bobob2ob2o2bobo4bo2bo4bo4bob2o2bo7b3obo4b3o3bob2o3b2o2b2o7b2obo2bo3bo2bo3b2o7bob2ob2o2b2o3bo3b2o5bo8bob2o4b2o3b2o2bobobob2o5bo5bo3bo5b3o4b2obobo4bobobo7bo2bo3b4o2bobo2bo2bobo2bobob2o3b2o4bo5b2o2bo$o2bo3b2o2bobo4bobo3b3o3bo3bo2bo2bo3b2o11bo3bo6bob3o2b2obo6bo16bob2o2bo6bob4o4b2o8bob2o2bobo3bobo2b2ob2o2bo5bo5b2o2bo6bobobobobo2bo3bo7bo3bo2b3obobo3b3o4b2o4b2o3bobo3b2o3b5o4bo10bob2ob2obo3b2o6bo4bo2bobo4b2ob2ob2o4b3o2bo3b5o2bo6b2o2b4o6bo2bo2bo2bobo2bo3bo8bob2obobo2bo3bobo$o4bobo7b2o12bo2bo2b2o2bo5b3obo3b4obo2bo2b2obob2o7b2o3bo5b2o7b2o2b2o2bobo2bo8bo15bob2o2b3ob2o6bo4bo2bo3bo3bo3bo10bobo7bobo2bo7bo6b3o2bo3b2o3bo4bobobob2o5bo4b2o8bo13bo7bo4bo2bo3bo4bo5bo9bobob2o3b2obobobo3bo4bo4bobobob2o4b2o3bobo4bobo3bo5bob2o4bo6bo$ob2o5b2o6b2o5b2ob2o6b2o5bobo3b2o3bo4b3o2bo2bo4bo5b2o4bo6bobo2b2o3bob2obo2b2o3bo2bobob4o4bo2bobo5b2o2b2o6b3obobo3bo3bobo2bob2o2b2o2b4o9b2o4b3o3bo8bo4bob2o2bo3b4o4bo2bo3b2obo6bo11bo2bo2bobob2o2b3o5bo2bo8bo2bobo3bob3o15bo8bo2bo3bobob2o2bo2b2o5b2o4bo8bo2bo4b3o7bob2o$2ob3o3b2o4bo5b2o2b2ob2obob2o5bo15bo2bobobo3b2o9b3o9bobo2bob2obo16bobo4bo7bobob2o3bo2b2o2bobob2o6bobo3b6o2bobo5bob3obo4b2o5b3o4b2o3bo2bo3bo10b3o7bobo2bo2b2o20bobo2bobo3bo6b2o3bobo3bobobo2b2o3bo2b2obo6b2o13bo7bo3b2o3bob2o3b2o2bobo2b2o8b2obo4bo2bo3bo$2bo3bo9bo3b2ob2o3b3o2bo6b2obobob3o2bo5b7o6bob2o4bo3bo2bo7b3o7b2o2bo4bo7bob2o4b3o4bo2bobo6b2obo4bo2bobob3o3b4o3b2ob2o3bo2bob2o8bobo3bo8b4o2bobobo7bo9bo7bo2bob3o2bo6b2o2bo3bo2bo3bob4o9bo5bo3bobobobobo4bo2bo2bo5bo2bobo6b2o7b3o2b2o5bo2b3ob2o4b2obo3b3o3bo7bo$2o2bo5b2obob2obob2obo3bo2bobo5b2o6bobo4bobo2bo3bo3b2o11bo3bo7b2o7bob2o3bo3bo2bob3obobo5bob3o2bob2o4bobo9b2o7bobobo2bo2bo3b2o4bob2ob3o2bobo4bo5bo7bobo2bo3bo13bobobo2bo2bo8bo2bo7bo2bo3bo4bobo5bob3o2bobob2o7bo3bo5bo9bo12bo14bo4bo4bo6b3o8b2o2bo7bo$5b2o3b2o3bo2bobo6bo3bo3bo2bo2b4o15bo2bo3bo2bo5bobo3b2o4bobob2o5b2o4b4ob2o4bobo4bobobobo4bo5bo2bo2bo5bo10bobobo3b2o7b2o11bo5b3o5b2obo2b2ob3o3bo2bo8bo6bo6bo2b4o4bob2o3bob2obo9bo5bob2o3bobo7bo11b2o3bo2bo2bo7bobo7bob2o3bob3o12bo4b3o7bo4bo4bo2bo$3bo6b2o6bo2bobo2bo2b4obobo5b2o11bo3bo2bo2bo3bobo3b2o3bo2b4o3bo2b3o3b2o6bobobo2bobo6bo2bo2bo9bob2obo2bo2b5o7b2ob3o7bobo4bo6bo7bobobo3bo3b4o7bo5b2ob3o10bo14bob3obo3bo4bo2bo5b2o4b2o2bo2bo4bo2b2o9b2ob2o2bo3bob2ob3ob2ob2o2b2obo4bo16bo3b2o3bob2obo5b2o2bo$7bob3obo5bo3bo3bobobobo2b3o2bo2bob4o2bo4bobo3b3o6b2o3bobo2bo2b3o9b2o9bo5b2o4bo3b2o4b2o2bo2bo4b2o2bobo4bob2o2bobob3ob2obo3bo3bo2bo2bo2bo7b3o3bobo5b2obobo2b2o2bo4b2obobobo4b3o12bo11bo5bo7bobo2bobo4bo8bobo5bobo3bo3bo6bo2bo2bobo7bob2o2bo13bo4bob2o3b2o2bo3b3o10bo$bob4obo2bo3bo4bo6bo10bo8bo4bob2o2bobo6bo3bo6bo2b2o2bobobobobo2b2o4bo6bo3bobo6b3o4b2o6b2o2bo4bobo3bo6bob3o4bobo3bo3b2obo3bo5bo5bo2bobobo4b2o2bobo2bobo3bo5bo2bo2bo3bo6bo5b3o15b3o4b2obo4b2o8bo2bo4bobob2o2bo7b3o6b2o2b3o2b2ob4o2bob2o2bo2b2o4bo6bo3bob3o7bo5bo3bo2bo$7b3obobo7bo4b4o2bobo4bo3b2ob2obo5bo2bobo2b2o5b2o3b2o3bo2bo2b2obobobo6bobo6b2obo2bo2bo5b2o5bo2b2obobobobo2bobo2b4obo3b2o6bob2o4b2o3bob2o2b2o5bo5bo9bo7bo2b2o11b2o3bobo2b3o2b2o6bo4b5o7bo2b2o2bo7bo4bo5bo3bo4b2o4bo3b2o7bo5bobo8b2obob2o2b2obobo5bo2b3o2bobobo2bo10bo4bo$4bob2o2b2o4b2o11bo5b4o4bobo4bob3ob3o5b4obo8bo2bobob2obo5bo5b4o4b2obo2bobob2o14bo4bobo9b2o3b2ob3obobo6bob5o9b2o2bo5b3o4bo4bob2o6bo4b5o5b2obob2o4bo2b2o2b2ob3o2bo10bo2bo2bo3bo6bo3bob3o3bo5b2o3bo3b2o3bobobobob2o3bo3bo4bo5b3o6b2o4bo4b2o6bo6bobo4bo$bobo4bo2bo16bo3bobo4bo6bo5bobo2bo4bo3bo5bo6b2obobobo3bo10bo11b2o7bo5b9o8bo2bo4b4o8bo6b2o7bo3b2obo2bo2bo8bo3bo2b2o2bo2bob2o4b2o4bobo2bo8bobobo2b4o4b2o3bo5bo6bo3bo2b3o3bo2bo4bo2bobo2bo2bo2bo4b3obo3bo2bo7bo22bo3bobo5b2o4bo2bo2bo7b2o2bo$2o4bo3b3o4bo4bo7bo4bo3b3o3b6o7bo3b2obo2b2o2bobo3b3ob2o3bo8bo3b3o4bo2bo3b5o8bobo6bobo5bo2bo5bo5bo5bo3bo6bo2bo2bob4o4bo6bo8bob3o2bobo2bo2b2o3b3o2bo4b3o2bo2bo2b2o3bo3bobo4bo4bo9bo4bo2bo3b2obobo11bob5o9bob2o4bo2bo4bo4bo6b2o2bo5bo3bo3bo4bo6b2o7bo3bob2o$4bobo6b2o8bo2bo3b2o8bobo2bo2bo4b2o5bo3bobobo7bo3bobo2bo2bobobo17bo2bo2bo2b2o2bobobobo5b2o2bobo7b2ob2o3bo2bo3b2obo2b2obob2ob2obobo5b4obo13bo8bo3bo2bo2b2o4bobobobo3b2obo7b2o7bobo5b3o5bo2bobo2bo2bo3bo2b2o4b2obo4b2o2b2ob2obo3bo2bo2bo4bo4bob2obo5bobo2bo4bo6bo8bobobo3bo3b4obobobo$bo5b2o2bobobobo4b2o8bob3obo5b2o2bo2bo6bo9bo3b2obobo16bo2b2obobo2bo3bo9bob2o2b2obo9b2o3b5obo5bo4bobo8b2o5bo6bo3bobo2bobob3o8bobo7bo3bo2b3o8bo3b2o7bo4b5o2bo3bobo2bo2bo2b2o6bo6bobo4b3o4bobobo5bobo4bo7b2o2bo2bo4bob2o2bo5b3o2bo7bo11bo2bobo3bo2b3o3bo$4b2o3bo5bo2bo3bo8bo2bo2b2obobo4bo4b2o9bobobo3bo4b5o3bo2b3o5bo3bo8b3o2b2o3bo4bo2b2obob2obo6bobo3bob2o2b2ob4o3bo2bo11bob3o7bo3b2o2bo3b3o2bo3b3o9bo8b2o2b2obobob2o2bo3b2o3bob2o3bo4bo5bo3bo3b2ob2o2bobobobob3o3bobo6b2o2bo3bobob2ob2obo8bo4bo7bo2bo4b3obobo5bobo6b2o5bo4b3obobo$6bo5b3o3b3obo2bo4bob3o2b2o7b2o3bo3bo7bo6b2obo4bob2o4bobo4bo6bo9b2obobo4b3obo9bo6b2o6bo3bo11b2o6bo12b3o2b2o7b2o2bob2obo5bobo3bo2b4o3bo4bo2bo5bo5bo2bo2b2o2bo6bobo3bo2b4o6b4o2bobobo3b2o2bo2b2o4bo18bobo2bo2bo3bob2o10bo3bob5obo2bo3bo2bob2o2bobobo3bo$2b3o2bobo3bo2b5o7bo4bo7bo2bo4bo8b2o3bo2bo8bo5bo2bobobob2obob2obob2o4bobo4b3ob4o2bo5b3obobo4bo3bo3bo8bo2bo2bo2bo5bob2obo4b2o9b2obo2bo10b2obo4bo8bobo4bo5b3obob3o11bo4b2o2bobo5bo8bob3o7bobo7b3o2bobo3b3ob3o2bo2bo2bobo3bo3bobobo2bo3bo4bo8b4o7bo8bo5bob2o$o6bo5bobo2bobo2bo13bo6b3o3bobo2b2ob2o4bo5bo2bob4obo3bo2bo6bo10bobo3bo2bo5bobo3bo4bo3bobobo5bobobobo4bo6b2o2bo3bo4bo3b2o11bo3bo6bo3bo4bobo2bo3bo2b2o11bo2b2o3b2o7bo11bo3bo6bo5b2o5bobo3bo4bo2b5obobo3b2o4bo3bo4b2o11bo2bobob2obob2ob2o5b2obo3bo2bo3bobo4b4o6bo$2obobobobobo7bo3b3o8b3o2b2o5bobob2ob3o8bo5bobo10bobo4b2obo3bo2bo7bo2bo6bo4b2o3bo5bo2bo2bobo6bo3bo4bo4bo3b2obob2o5bobob3o5bo7b3o2bobobo13bobo2bo6bobo2b2o2bo5bo12bo4bob2obo3bo9b4o4bob2o2b3o2bo7b2o3b3o7b2o2b2o7bo2bo5bo5b3o3b2o3bo4bo4bo12bo2bobo2bo$o3bo3bo7bobo2b2obo3bo7bo2bo6bobo2bo2b2o2b2obo3bo4bo2bob3o8bo2bo4bobo4b5o2bobobobo4b3obo5bobo3bobo2b2o2bo9b2o6b2o5bob2o8bo6b2o5bo3bobo4bo3bo4b2o2bo10bob2obobob2o3bo4bo3bobo2b2o4b2o4bo4bo3b2o2bo2bobobob2o2bo7b3o2bobobo9b2o21bobo2bo3b2obobo6bo3bo3bo3bo2b2o6b2o2bobobo$3bo4b3o3bobo3bo3bo2bobo6b2o11bo6bo3b2o4b2obo7b4obo3b2o2bo3b2obo10bo3bobo2b2o7bo2b3o2bo4bo5bo2b4o2bo5bob2o3bo3bo4bobo8bobob2o2bo4bo5b2o3bo2b2o4bo2bobobo5b2o10bob2o3bo2bo4bo3bo4bo9bobo2b3obo4bobo3bo3bo3bo2bob5o4b2o6bo4b3o6bobo4bo4b2o6bo2bo2bo2bobo4b2ob2o2bo2b3o5bo$bo2bo12bo5bobo2bobo2bo8bo3bobo7bo5bo4bob3o2bo2bo10bo9bobobo3bo7bo4bo2bo7b3o6bobo2b3o8bo6b2obo6b2o5bobo3bobobo3bo3bo3b2o7bo3bobo3b3obo6bo8b3o4b2o8bo5b2obo6bo2bo6bobo3b3o5bo6b2o5bo6b2o9bo2bobobo2b2o4bo4bo2b2o2bo2bo6b3o2bo5bo2bo4b3ob2o6b2o$o4bobo3bo2bo2b2o2bo2bo9b2o2bobobo2bo3bob4o6bo7bob2o3b2o2bob2o2bobo2bo6b3o2bobo2bo6bo5bo10b2ob4o3b4o3bo2b2o11bo5bo2b4o5bob5ob2o2bob3o3bo2bo2b2o2b3o2bob2o3b2o2bo6b4o2bo6b3o8b2o5bobob2o2bo5bo3bob2ob3ob2obo2bo3b2o4bo4bo3b2obo3b2o3bo5bobobo2bo4bob3obo3bo7bo13bob2o3bobobo2bob2o$obobo9bo4bobo3bo4bo4bo2bo2bo2bo7bo2bobo5b2obobo2bobo11b2o6b2o2b4obo4bob2obob2obo3bo2bobo3bo4bo2bo3bo3bobo2b3o9bob4o2b2o2b2o2bo2b2o2b2o3b2o2bo4bo2b2o12bo5b2obo4bobo8bo8bo2bo2b3ob2o3b2o2bo6bob2o6b2o3bobobobo2bo4bo6bo2b4obobo2b2o4bob2ob2obo5bob2o4b2o4bob3o3bobo3bo8bobo3bobobo5bo$bo3bo2bobo2bo7bo5b2o4bob2obobo2bobo2bob2o2bobo3bo2bo3b3obo2bo2bo2b3obobo5b2o4b3o5bo3b3obo2b2o2bob2obob2o4bo3bo4bo2b4o12bobob2o3bo5bo3bob2o8bo8bo2b2obo3bobobo5bobob2o4bo3bo9bobo9bo3bo3bo11b2obo2bo2bobo8bo2bo5bo2bo2bobo9bo5bobo2b2obob3o11bo5bo6bo4bo5bobo6bo9bo$o2bo3bo3bo3bo2b2o3bo6bo6b2obo4b3o13bob2o3bo2bo4b3o7b2o19bo5b2o6bo2bo6b3o5bo17b2o3bo2b2o9bo5bo2bo2bobo6bo8b2o3bo3bob2obob2o2bob4obo6bob2obo8b2obobob3obob2o4bo6bo4bobo3bo3bo4bo7bo2bobo6bob2o5bobob2o4bo4bo5bo10bo2b2o3bo4bobo2bo5b2o2bo2bo2b3o$5bo7bo4bobo4bo2bo2bo2b3o11bo6b2obo2b2obo12b2o2bo4bo7bo2bobo2bobo2bo3bo6bobobo3bo2bo11bo4bo2bobo5bo3bo4bobo4bo8bo2bo11bobo3b2obo4bo5b2ob2o3b2o4bo2bo3b3o2b3obobo2bo4b2o7bob2o3bo6bo2b4obo11bo2bo2b2obo5bobobo5bo2b2o5bo4bo4bo2bo2b3o2b2ob2obo3bobo3bobo2bo3bo6bo2b2o$3bob3o3b3o4b3o2bo3bo2b2o2bo10b2obobobo3bobob4o4b2o2b3o2bob2o3bobo2bobo6bo4bo2bobo8b3obo4bo3b2o3bo2bob4obo4b2ob2ob2o2b2o8bob3o2bo2b2o3b2o6bob3o4bo6bob2ob3obo2bobobo2bobo8b3obo2bobobobob2o3b2o3bo4bo6bo8bo2bo3b3o2b2obo13b2o10bo3bobobo5bobo2b2o6bo2bo6bo6bo3bo2bo4bobo5bo3bo2b2o$2bo4bo2bob2o3bob2o2bo2b2o2b2o2bo6bo4bo2bobo4bob3o4bo4b2o4b4o2b3o6b3ob3o3b3o4b5obo3b3o2bo4b2o2bo5bo3b3o4bo4bo7bob3o2bo7bobo7b2o3bob2obo10bo2bo2b3o10bo8bo3bo11bobo2bo4bo3b3o3bo2b2o3b2obo2bo4bo2b3o2bo3b3o2bo2b2o5bob2obo4bo2bobob2o5bobobo2bo4b2obo3bo2bo3bo8bo3b2ob3ob4o3bo3bo$o4b5o2bo3bob2obo3b3obo8bobo5bo3b5o2bo3bo6bo2bo5b2o3bobo3bo6bo5bo11bob2o5bo3bobo2bo4b2obo4bo11bo5bob2o2b4o6bo10b2o6bo2b2obob2o2b2o3bobob4o5bobo2b2obo4bo5bo5bobob4o9bobobo5bo4bo2bo5bo2bo3bo4bobob2ob2o6bo2bo2b3o4bo3bob4obo3bo2bobo2bob2o4bobo3bobo2bob2o4bo6bo4bob3o$5bo2bo12bo8bobo3bo2bo8bo5bo5b2o3bobobo7bo2b5o7bo7bo2b3ob2o4b2o2bo2bobob3o3b2o6b2obo2b3o4bobo2bo5bo5bo2bobo6bo6bob2o2bo6b2obo6bo5bo2bo11b2o2bobo3bo2bobo7bo5bobobobo2b2o2bobobobobobo3b3o2bo2bo4bo3b2o5b2o9b2o2bo3bobo2bo4bo3b2obobo3b2o7bo3bo3bob3ob2o19bobo$7bobo6bo2bo4bo7bo2bo18bo3bo4bo2b2obobob3o3b2o5bo4b2o4bo3bo3b2ob2o7bo4bob2o3bo3b4o2b2o7bo4bo2bo7bobo2bo3bobo4bo2bo3bobo3bo4b3o5b3obo7b2o5b5o11b2o9bo3bo2bobo2bobobo5bo6b2o2bob2obo6b2o10bobo5bo6b2o3bobo4bo4bo2bob2obobo21bo3bo5b2o5bo7bo$bo4bo5b2o4bo3b2o8bo2bo10bo5bo3bo2b2o6b7o3bo10bo5bo3b2o9bo6bo2bob2o4b2o3bobo8b2o8bo2b2o2b2obo2bo5b3o3bo2bo5bo8bo2bo5b2o4bo6bobobobob2o4bobo4bo4bo4bobobo7bo2bo3bo8bo4bobob3o5bobobo2bo3bo2bo5b3obo4b2obo2b2o4b2o2bo3bobo3b3ob5o3b2o11b2o5bo3bo10bo2bo$3ob4o2bob2o3bobo3b2o2b3o2bo3bo6bo3bo4b5o2b2o3b3o5b2o3b2obo13b6obo6b2obo2bobobo8bo6bo7bo3bobo2bob2o5bob2o6bobobo4bo5bo3bo3b5o2bob3o13b4o6bo5bo6bobo5bo2bo8bo6bo3b2ob3obo4bo3bo2b2ob3obo5b2ob3o2b3o15bo2bo6bobo4b2o7bo5bo6b2o3b3o4bob2obo4bobo4bo$o2bo4bo2bobo3b2o4bob2o6bo5bo2b3o7b2o10bo5bobo2b7ob2o3bo5bobo3bobo4bo4bo2b2o3bo5bo2b2o6b2o3bo3bob2ob5o5b3o8bobobob3o3b3o5b2o4b2o6bo4bo7b2o2bo10bo2bob3o5bobob3obobobo2bobob2o5bobo6bo3bo2bob2o2bob2obo3bobobo3b3o3bob2o2b2o9bobo2bo2bo2bobo2bobob2o2bobo2bo2bo6bo7bobo2b2o$4b5o4b2o4b2o4b2o2bo3bo8bo5b2o8bobo4bo4bo3bo4bo3bo4bo10bo2bo8bobo4b2o2b2obob2o4bobobo2b2o10bo3bo2bo5bo2bobo2bobobobo2bobo8bobobo2bobo8b3o2b2o9bo7bob2o6bo2b2o5bo3b3o3bo8bo5bo3bo2b2o2bo6bobo2b2obob5o2bobo5bo5b2o6bo3bob2o2b3o3b2o3b2obo5b2obo5bo2bobo9b2o3b2o$obo5bobo2bob2o2b2o3bo5b2o3bobo3bo2bo3b3obo7bo2bo2b2obo2bob2o2bo2bo6b2o3b3o2bobobobo2b5o2b3obo3bo4bobob2o3b3o2bo2bo2bo2bo4b3obo2bob2o3bo3bob2o3b3o2b2obobo10bo2bo7bo2b2obo2bob2obo2bobo8bo3bob2o8b2obo2bobo2bobo4bobobobo4bo3bo3bobo2b3obob2o2bobo2bo6bo3bo2b2obo5bo6bob2obo3bo2bo3bobo2bo3bob2o2bobobo4bobo2bo5bo4bo$o3b3o15b2obo7b3o2b2obo2bo4bob2o3bo6bo4b2o2bo4bobo2bo12bo9bobobo3bo3b3o2b2o2b3obob2o4b2o3bo2bo5bo2bo5bobo3b2o2bo6bo4b2ob2o2bo3bo2bob2o9b2obobo2bo6bo2bo2bo6bo9b2o11bo3bobo5bo2bo2b2o2b3o2b2o4bo2b3o3bo4b2o2b2o2bo2b2o4b2o4bobobobo7bobo2bo5bo2bobobo3bo2bo4b2o5bo2b3o11bo$bo4b5obo7bo5bo2bob2o6b2o3b2o3bo2b4o2bo9bo4b3obobo2bo5b2o3bob3o2bobo2bo6bo11b2ob2o2bo6b2o3b2o4b2o2b2o2bobo5bo7bo3bob2o2bo3bo2bo5b2o3bo5bo2bo2bo4bo9b2o3bo2b2o4bo8bo10bobo2bo2bo3bo2bo6bobo6bo11b3o4bo2bobo2bo3bo3bo5bo6bo6b2o2bobo2b3o2b2o2bobo15bo7bobo$b2obob2o2bob2o2bob2o4bo3b2o8bo3bo2bo6b2o7b3obo2bo3b2obo3b2o2bobo12b2o6bo3bo2b4o3b5o4bo7bo13bo4bobo5bobobo9b2o2b2o4bo3bo5bo2bobo3bo2bo4b2o2bo2b2o2bobo11bo8bobo3bo5b2o5bo2b3o2b3o2b3o7bob3o4bo3bo3bo6bobo7b2ob2o5bo2bo6bo5bob2obob3o3bo3bobobobobob2obo4b2obo2bobobo$3b2ob2o3bo7bo4bo2bo5b2o3bo2bo3bo3bo3bobo3bo2bob3o4bo5bo3bobo9b2ob3o2bo6b2o3b3ob3ob3o8bo3b2o2b2o11bo4bo7bo2b3o8bo2bob4o5b3o3b2o10bo5b3o3bo2bo3bo5bo3b3ob4o6b4o2bo7bo6b2o3bo2bo3bobobobobobo15bob4o4bo3b3o2bobo9bo4bo2bo2b2o3b2o5bo4b2o6bo2bobobob2o2bo2b2o$11bobo4bo6bo4bobobo2bo2b3o2b2o3b2obobob2o9bo7bo9bo2bob2o3bobo7bo2bobo3b2o4bobo6bo2bo3b2obo3b2o4bo6bo2b2o4bobo2bo3bo4bob2ob2o4bobo3b2o2bo2bo5bo7bo3bo3bo2bobo4b4obobobob2o6bobo2b2o6bobo2bo3b2o7b2o2b2o3b2o4bob3o3b2ob2ob3o2b2o2bobobo6b2o3bo3bo2b3o2bobo7bobobo7bobo2bo4bobo2bo2bo4bobo$o5bo2b3o2bobobobo7b2obo4bo8bob7obobob2o2bo2b2o2bobo7bo4bo4bobo3bo2bobo2bo3b4obo4bo4bo2b2obo2b3o5bo3bob3o2bo3bob2o2bo5bo5b2o5bo2bo2bo2b2obo2b4o5bo2b3obo6bo3b3obo3b3o8b2o7b2o4bo2bo2b2o4bob2o3bo5bobo5b3obo7bo7bo3bo2bo2bobo10b2o3b2o2bo6bobobob3o2bo5b2o2bo8bo4b2o10bobo$4bobo3bob2o2bo5b2o4b2ob3o4bo3bo3bob3o4b2o3bob2obobo3bo7bo4bo7bo4b3obo5bob2obob2o3bo2bo4bo3b2o3bo7bob2o2bo7bo7b2obo8bo5bo5bo8bobobo16bo5bobobobo2bo4bobo4bo4bo2bob2o2bo2b3o3b2o7b4o4bo8bobo3b2obo3bo7b2o3b2obo3bob2o2bobobobob3o3bo7bo7b2o3b2o8bobobo4b2obo2bob2o$2b2o4bobo6bobo3b7o8b3o3bo2bo2bobo8bo3b2o2bo2b2o3bo3bo10bobo5b2o8b2obob2o2bobo3bobo2bobo2b6o3bo2bo3bo9bobo3bobobo3b2o3bobo6b2ob2o2bo3bo5bo3b2o10bo2bo6b2obo6b2o2b2o4bo6b2o12bo2bo2b3obob2o4bobo6bobobobo5bo6bo2bo4b2o2b2o2bo4b2obo2bo3bo2bo6bobo3bo3bo4bobo2bo2bo2bo3b2o2bo3bo$3b2obo2bobo4bob6o3b2o5bobob2o5bob3obo2bo2b2o3bo2bo6bo2bo2bo5b3o3bo3b2o2b2ob2o2bo17bo11bo4bo6bobo6bo4bo3b2o3bo2bo4bo15b2o2bobo4bo3bo4bob2ob2o2bo3b3o2bob3obobo2bo3b3o4bobo2bob3o4bo3bob2o9b2o5bo4bobobo6bo2bobobobobobo8bobob3o10b2o4bo5bo3bo4bo2bo3bo2bo3bo2bo3b3o4bo$3bo3bo3bobo4bob5ob2o8b4o5bobobo3bo3bo2b2o5bo2b2o3bobo5b2ob3obo2bo2bo2b2o2b3obo3bo3b2o6b2o3bobo9bo2b5ob2o3bo2bo10bo2b2o3b2o3bo5b2obo4bo4bob4o4bobo7bo4bobo3bo15bo3bo2bo3bo8bo5b2o3b3o5bobobob4obo2bo12bo4bo2b3o2bob2ob2o6bo4bo3bo10bo3bo3bo2bo4b2o8b2o2bo3bo3bo$bobo2bo3bo2b2obo2bo7bob2o2b3o3b3o7bo13bo4b2ob2o3bo2b2o5bobobob2o2bo2bo5bo12b2o2bo3b3o6bo2bob5obo4b3obo3bo2bo3bo2b2o2bo4bo2bo7bo2b4o2b2obobobo3b2o15bobo5bo3bo5b2o4bo9b2obo4bobo5bo4b2o5bo3b3o4b2o2bo5bo5b2obob2o3b2ob2obo4bobo2bo13bo7bobo4bo4b2o3bob2o4bo9b2o$2bo9b2ob3o3b2obob2o3bo2bobobo4bo5bobo5bo3b3obobobo7bo11bo2bo5bo3b2o5bo2bob2o2bo2b2o5bo2b2o2bo8bobo2b3o9bo6bo3b2o9bob5o5b2o4b2o2bo9b2obo5b2o2bo3bo4b2ob2o2bob3obo3b3ob2obobo3bo4bo2bo12bo5b2o4bo3b2o3bo3bobobo2b2o3bo4bo2b4obobo3b3o5b2o3bob2ob2o2bo2bo2bob2obobo3bobo4bo3bobo2bo$o3bobo3bo2bob2o2bob3o2bo3bo2bo3bo7b7o10bo2bobo11bo4bobo2bo6bo21b4ob4o5bo2bo4b2o4bo3bo4bo4bo2bo2b3o4b2o2b3o5bo2bo2bob2o3bo4b4obo3b2o6bo3bo3bo4b2o5bob2ob2o2b2o3bo7b3o3b2o5b2o12b2o7bobo4bo5bobo5bo3bo6bo3bobo3bobo3bobobo2b2o2bo4bobo3bo2b4o5bo5bo7bob4o$bob2o8b3o2bobo3bo11b2o3bo2bo3b5obobo3bo4bo3bo4b2o5b2ob6o2bo3bo2bobo2bobo2bob2o2bo2bo7b2obob2ob2o4bo4bo3bo7b3o3bo5bob2o17bob2obo2bo2bobo3bobo2bo4bob2ob3o3bo4bo10bo3b2obo3bobobo7bob2o4bo2b2obo11bo4bobob3o2bob4ob2ob2obo5bobo4b2o3b4ob2obobo3b2o3b2o2bo3bobo6bob2o3b2obo4b2o3b2obobo$3bob3o2bo2bo5bo2b3obo2b3o6b2o3bo2bo2bo2bo5bo9bo5bob4obob5o5bo2bobo3bo5bo4bo3b2ob2obo10bo2b2o3bobo2b2o2b2o4bo4bo8bo3bo3b2o6bo2b2o4bo2b2o11bo4bo3bo4b2o2bobo4bobob2o4b2o2bo5bo3b2o11bo2bobobo4bobo6b2o3bo6bo3bo2bo2bo2bo2b2o2bo2bo2bo4bo7bo6bo11b3obo7bo4bobo4b2obo3bobo$3bobobo3b2o5bobobo2bo5bo2bo3bobo12bob4o3bo7b2obo5bobo4bo2bobobo3bo3b3o8b4ob2o9bo8bob2o5bo4bo16bobo5bobo2bo4b2o5bo2b3o2bo4bo2b2obo4bo3b2o4b2o6bo3bo3bo2bo4bo2bo3bobo3bo3b2o3bo4bobo2b2o9bo2bo3bo5b2o3bo2bo3b2o6b2o2bo2bo3b2o3b2o2bo3b3obobo2bo2bo9bobo5bo3b2obo6bo2b2o$2b2o3bobo2b2o15bob2ob2o5bo3b2o2bo5bo5bo3b2ob3obo3b2o3bob2o3bo2bo2b2obobo2b2obobobo6bo7bo3bo4b2o2bobo4b2o3b2obo2bo3bobo3bo3b5o2bo7bo4b4obo2bo10b2ob2o8b2obo7bo4bo3bo2bo2bo4b2o3bo2b3o16bo4b3o8bob2obo4bo2bo3bo2b2o3b2ob2o2bo2bo6bo4bob4ob4o4bobo5bo2bobo2bob2o3bobo9bo3bo$2bo9bo4bo6bobobo2bo3bo2bo4bo2bobob2o5bo2b2obo5bo3bo3bo7b2o2b2ob3o5bo2bo2b2o3bobo4b2o3bob3o3bobo2bo2b2obo5b2o2bo9bo9bobo2bobobo3b2o3bo2bob2obo2b2o2bo6b2o2bo4bo3bo2bo2b2o9bo6bo6bobo2b2obobo6b2o3bob3o4bob2o6bo11bo8bo5bobo2bob3o5bo3bo7bobo9b3o11bo9bo6bo$o5b2o2b2o2bo11b2o5b4obob3o3bo4bo3b2o6b2o5bo5b5obo2bo2b2o5bo2bo5bo2bo2bobo4b4o4bo3bobo5bobo3b3obo6b3o10bobo4b2o3bob2o7b5o6b2ob2o15bo6bo7bo4b3o2bobo21bo4b3o3b2o3bob2o3b3obo2bo3bo2b2obo5bobobo2bo4bo2bo2b2obobobo2bo7bo3bo3bo3b2o2b2o7b3o3bo2bobo2bob2o$3b2o11b2obo7bo15bo3bo6b2obo4b3obo3b3o5bo5bo5bo8bobo2b2obob2o5bobobo4b2o14bo4b2o2b3o3bo8b2o5bo6bo4bo3bo2bo2b2obobob2o5bo4bo2bo4bo3b2o4b3o6bob2o3b2ob2o4bo2bo3bo2bo4b2o3bo3bo2bo12b2o2bo6b2obo3b2ob3o2bob2obo3bo2bo2bo2b2obo2b2ob2o4bo2bo3bo3bo3b2obo2bo8bo9bo2bo$2bo4b2o5bob3o2b3o4bo10bo7bobo2bo5b2o2b4obob2o3b2o3b2o2bo4bobo3b2ob2o3b4obo6b2o5b3o2bo3b2o7b2o2bo11b4o5bo4b2o8bo2bobobo5b2o6bo2bobo19b2ob2o3bo3bobo2b2o7b4ob6o6bo3b2o2bob2o5bo5bo2bo2bobo4b2o6bobo6bo2bobo2bo2bo4b2obob2ob2o3bo2bo3b3o7bo2bob2o6bo8bobobob2obo$3bo6bobobo4bo10bobo2bo7b2o4bobobobob5o7bo4b2o5bo3b2o4b3obo4bo3bo2b2o12bo3b2o3bo4bo2bo3b2o3bo12b2o5bo4b2o3bobo3b3ob2o9bo4b2o2bo2b2o9bo4bobo2b2o2bo2bo4b2obo20bob4o5bob2o6bobobo5bobobo3bo10bo9bo2bobobo5b2ob2o3bo3b2o2b2o2bo2bo4bo3b2o11bo3bo2bob3o$4b2ob3ob2o2bobo2bo6bo2b2o2b3o2b2obo2bo8bob2o3bo2bo4bo2bo2bo2bo4bobob5obo2bo4bo6b2o4b2o3bo2bo4bobobobo12bo3bo6bo3b2o5b2o4bo3bo10bo2b2o4bo3bob2o2bobo5bo2b2o6bo2bo10bo5b2o4bo3b2o5b2obo5bo9bo8bobob2o4b3obob2o10b2o4b2obo5bobo2bobobobo3bo2b4o6bo6bo2bo9b4o2bo6bo$bo7bo4bo4b2o4bo2bo2bob2o9b3o9bob2o7bo2bo9bo5b4o3bo5bobo8bo8bobob2obo3bo3b2obo2bo3bo3bo3bobobobo4bo4bo2bobobo2bo3bo7bo2bobobo3bo3bo3bo3b2o2bob3o5bobo3b5o8b2ob4obo2bo2bo7b2o4bo6bob2o6bo7bo6bo8bob2obob3obo3bo6bo14b2o3b2o4bob5ob3o2b2o5bobo2b2o2b4o! Almost c/2: x = 19, y = 10, rule = LifeHistory7.D3.D$6.DCD.DCD$5.DC2AD2ACD$3.2DAC2.A2.CA2D$3.AC.C2.D2.C.CA$.2D.C.AD.A.DA.C.2D$.2C.A2.A3.A2.A.2C$.CA.2D7.2D.AC$DC2.2A7.2A2.CD$2A15.2A!#C [[ STOP 2 ]]
Hunting
Posts: 608
Joined: September 11th, 2017, 2:54 am
Location: Universe
### Re: Random posts
Do I get confused with Saka? We have similar yellow-on-grey avatars.
KittyTac
Posts: 533
Joined: December 21st, 2017, 9:58 am
### Re: Random posts
Finally started helping the community once I figured out how to use WLS.
KittyTac
Posts: 533
Joined: December 21st, 2017, 9:58 am
### Re: Random posts
99 bottles of beer in "Stupid Python"
(From https://codegolf.stackexchange.com/ques ... es-of-beer)
print"""99 bottles of beer on the wall, 99 bottles of beer.Take one down and pass it around, 98 bottles of beer on the wall.98 bottles of beer on the wall, 98 bottles of beer.Take one down and pass it around, 97 bottles of beer on the wall.97 bottles of beer on the wall, 97 bottles of beer.Take one down and pass it around, 96 bottles of beer on the wall.96 bottles of beer on the wall, 96 bottles of beer.Take one down and pass it around, 95 bottles of beer on the wall.95 bottles of beer on the wall, 95 bottles of beer.Take one down and pass it around, 94 bottles of beer on the wall..... Ok, this is stupid. First of all, what the brainfuck are the bottles doing on the wall? They're not spiders nor picture frames. And how are they sitting on the wall?94 bottles of beer on the wall, 9.. oops, they fell down. 94 bottles of beer on the floor, 94 bottles of beer. Second.. who the HQ9+ wants to keep track? I think I lost count after drinking the 2nd one...Take one ... um... up, and pass it around, ..... er.. a lot of bottles of beer still on the floor. Fourthly, what's with this passing around scheme? They're not j..I mean letters, yeah, or boxes of chocolate. We all can just take one and drink it. It's healthier too.A pile of bottles of beer on the floor, a pile of bottles of beer.Everyone take one up and drink it, still a whole bunch of bottles of beer on the floor. Um.. seventhly, are we really that many in this assembly that we can finish 200 or however many bottles we had in the beginning? Without passing out?Go to the store and buy some more Yeah and who's gonna pay for it? Definitely not me. And how are you going to bring 300 bottles back from the store? In your car? Buddy, you're so drunk, you can't even C anything. Go home dude, go home. Take a cab."""
Almost c/2:
x = 19, y = 10, rule = LifeHistory7.D3.D$6.DCD.DCD$5.DC2AD2ACD$3.2DAC2.A2.CA2D$3.AC.C2.D2.C.CA$.2D.C.AD.A.DA.C.2D$.2C.A2.A3.A2.A.2C$.CA.2D7.2D.AC$DC2.2A7.2A2.CD$2A15.2A!#C [[ STOP 2 ]] Hunting Posts: 608 Joined: September 11th, 2017, 2:54 am Location: Universe ### Re: Random posts A neat LOM eater with two transparent blocks: x = 38, y = 33, rule = B3/S233$18b2o$11b2o5b2o$11b2o8$5b2o$5b2o2$21b3o$21bobo$21bo2bo$22bobo$22b3o4$28bo$17b2o10b2o$16bobo8b2o$16bo12bo$15b2o2$23b2o$23b2o!
LOM to G + blinker:
x = 26, y = 32, rule = B3/S232$12b2o$5b2o5b2o$5b2o11$15b3o$2b2o11bobo$2b2o11bo2bo$16bobo$16b3o4$22bo$11b2o10b2o$10bobo8b2o$10bo12bo$9b2o2$17b2o\$17b2o!
Things to work on:
- Find a (7,1)c/8 ship in a Non-totalistic rule (someone please search the rules)
- Find a C/10 in JustFriends
- Find a C/10 in Day and Night
AforAmpere
Posts: 867
Joined: July 1st, 2016, 3:58 pm
PreviousNext
Return to The Sandbox
### Who is online
Users browsing this forum: Βεν Γ. Κυθισ and 1 guest
|
2019-02-22 06:18:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3006282448768616, "perplexity": 3763.580774205064}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247513661.77/warc/CC-MAIN-20190222054002-20190222080002-00428.warc.gz"}
|
https://www.physicsforums.com/threads/error-estimation-for-simpsons-method.384365/
|
Error Estimation for Simpsons Method
Homework Statement
Find the error estimation for the integral $$\int(1+x^2)^\frac{1}{4}$$ with limits $$[0, 2]$$ for $$n=8$$.
Homework Equations
$$\epsilon\leq \frac{|f^{(4)}_{max}(x)|(b-a)^5}{180n^4}$$
The Attempt at a Solution
The tricky part about this problem is finding $$f^{(4)}_{max}(x)$$:
$$f(x)=(1+x^2)^\frac{1}{4}$$
$$f'(x)=\frac{1}{4}(1+x^2)^\frac{-3}{4}*2x =\frac{1}{2}x(1+x^2)^\frac{-3}{4}$$
$$f''(x)=(\frac{1}{2}x*\frac{-3}{4}(1+x^2)^\frac{-7}{4}*2x)+((1+x^2)^\frac{-3}{4}*\frac{1}{2})$$
$$=\frac{1}{2} [\frac{-6}{4}x^2(1+x^2)^\frac{-7}{4}+(1+x^2)^\frac{-3}{4}]$$
$$=\frac{1}{2} [\frac{-3}{2}x^2(1+x^2)^\frac{-7}{4}+(1+x^2)^\frac{-3}{4}]$$
$$f'''(x)=\frac{1}{2} [(\frac{-3}{2}x^2*\frac{-7}{4}(1+x^2)^\frac{-11}{4}*2x)+((1+x^2)^\frac{-7}{4}*\frac{-6}{2}x)+(\frac{-3}{4}(1+x^2)^\frac{-7}{4}*2x)]$$
$$=\frac{1}{2} [\frac{42}{8}x^3(1+x^2)^\frac{-11}{4}-\frac{6}{2}x(1+x^2)^\frac{-7}{4}-\frac{6}{4}x(1+x^2)^\frac{-7}{4}]$$
$$=\frac{1}{2} [\frac{21}{4}x^3(1+x^2)^\frac{-11}{4}-3x(1+x^2)^\frac{-7}{4}-\frac{3}{2}x(1+x^2)^\frac{-7}{4}]$$
$$=\frac{3}{2} [\frac{7}{4}x^3(1+x^2)^\frac{-11}{4}-x(1+x^2)^\frac{-7}{4}-\frac{1}{2}x(1+x^2)^\frac{-7}{4}$$
$$\frac{3}{2} [\frac{7}{2}x^3(1+x^2)^\frac{-11}{4}-3x(1+x^2)^\frac{-7}{4}]$$
I entered the derivatives I calculated along with the derivatives taken from an online derivative calculator into my graphing calculator to make sure I was coming out with the right answer. It appears I've made a mistake with the 3rd derivative but I can't find the error. Thanks for any help.
Last edited:
$$\epsilon\leq\frac{|f^{(4)}_{max}(x)|(b-a)^5}{180n^4}$$
|
2021-05-19 02:21:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8952659368515015, "perplexity": 406.5262731297491}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00033.warc.gz"}
|
https://proofwiki.org/wiki/Continued_Fraction_Expansion_of_Irrational_Square_Root/Example/29/Convergents
|
# Continued Fraction Expansion of Irrational Square Root/Example/29/Convergents
## Convergents to Continued Fraction Expansion of $\sqrt {29}$
The sequence of convergents to the continued fraction expansion of the square root of $29$ begins:
$\dfrac 5 1, \dfrac {11} 2, \dfrac {16} 3, \dfrac {27} 5, \dfrac {70} {13}, \dfrac {727} {135}, \dfrac {1524} {283}, \dfrac {2251} {418}, \dfrac {3775} {701}, \dfrac {9801} {1820}, \ldots$
## Proof
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then the $n$th convergent is $p_n / q_n$.
By definition:
$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1\end {cases}$
$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1\end {cases}$
$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$
Thus the convergents are assembled:
$k$ $a_k$ $p_k = a_k p_{k - 1} + p_{k - 2}$ $q_k = a_k q_{k - 1} + q_{k - 2}$ $\dfrac {p_k} {q_k}$ Decimal value
$0$ $5$ $5$ $1$ $\dfrac { 5 } 1$ $5$
$1$ $2$ $5 \times 2 + 1 = 11$ $2$ $\dfrac { 11 } { 2 }$ $5.5$
$2$ $1$ $1 \times 11 + 5 = 16$ $1 \times 2 + 1 = 3$ $\dfrac { 16 } { 3 }$ $5.3333333333$
$3$ $1$ $1 \times 16 + 11 = 27$ $1 \times 3 + 2 = 5$ $\dfrac { 27 } { 5 }$ $5.4$
$4$ $2$ $2 \times 27 + 16 = 70$ $2 \times 5 + 3 = 13$ $\dfrac { 70 } { 13 }$ $5.3846153846$
$5$ $10$ $10 \times 70 + 27 = 727$ $10 \times 13 + 5 = 135$ $\dfrac { 727 } { 135 }$ $5.3851851852$
$6$ $2$ $2 \times 727 + 70 = 1524$ $2 \times 135 + 13 = 283$ $\dfrac { 1524 } { 283 }$ $5.3851590106$
$7$ $1$ $1 \times 1524 + 727 = 2251$ $1 \times 283 + 135 = 418$ $\dfrac { 2251 } { 418 }$ $5.3851674641$
$8$ $1$ $1 \times 2251 + 1524 = 3775$ $1 \times 418 + 283 = 701$ $\dfrac { 3775 } { 701 }$ $5.3851640514$
$9$ $2$ $2 \times 3775 + 2251 = 9801$ $2 \times 701 + 418 = 1820$ $\dfrac { 9801 } { 1820 }$ $5.3851648352$
$\blacksquare$
|
2020-01-18 18:13:35
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9712414145469666, "perplexity": 54.844870722135404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00151.warc.gz"}
|
http://mathhelpforum.com/trigonometry/219586-determine-arctan-r-arctan-s-if-r-s-routes-quadratic-equation.html
|
# Thread: determine arctan(r) +arctan(s) if r and s are routes of a quadratic equation
1. ## determine arctan(r) +arctan(s) if r and s are routes of a quadratic equation
Hi.
I've encountered this problem and have no idea where to start:
Let r and s be the two solutions to the equation x^2 - 2003x + 2004 = 0. Determine arctan(r) + arctan(s). One is not allowed to use a computer or calculator when solving this, as it was once in a past math competition. Any ideas on how to solve it?
2. ## Re: determine arctan(r) +arctan(s) if r and s are routes of a quadratic equation
Instead of looking directly at $\arctan(r)+\arctan(s),$ I would suggest looking at $\tan\left(\arctan(r)+\arctan(s)\right).$ Start here and also google "properties of roots of quadratic equations." I have debated whether or not to mention the specific result from this search which will be relevant and have decided to leave it out for now; that said, the fact needed will be one of the first results.
|
2017-08-23 20:36:03
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4166305661201477, "perplexity": 473.25786612862936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886123359.11/warc/CC-MAIN-20170823190745-20170823210745-00152.warc.gz"}
|
https://www.gamedev.net/topic/648025-calling-overriden-mixin/
|
View more
View more
View more
### Image of the Day Submit
IOTD | Top Screenshots
### The latest, straight to your Inbox.
Subscribe to GameDev.net Direct to receive the latest updates and exclusive content.
# Calling overriden Mixin
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
4 replies to this topic
### #1BlackMoons Members
Posted 19 September 2013 - 05:18 PM
Hi. I am really loving how mixin's are implemented, but I was wondering, When a class includes a mixin and overrides one of its member functions, Can we call the mixin's function explicitly anyway so we can extend its implementation? If not, would it be hard to add or break anything to add?
For example:
mixin class foo
{
void print(){output("Hello");}
};
class bar : foo
{
void print()
{
foo::print();
output(" world!");
}
};
Lead Coder/Game Designer for Brutal Nature: http://BrutalNature.com
### #2Andreas Jonsson Moderators
Posted 19 September 2013 - 06:16 PM
It would break what the mixin is
mixins are not inherited, as they are not real class declarations. They are more like macros to allow code reuse without having support for multiple inheritance in the language.
If you want to be able to call the mixin's implementation of a method, then the mixin's method should have a different name so it is included in your class.
AngelCode.com - game development and more - Reference DB - game developer references
AngelScript - free scripting library - BMFont - free bitmap font generator - Tower - free puzzle game
### #3BlackMoons Members
Posted 19 September 2013 - 06:58 PM
Yea, I just thought it would be nice to have the option to extend a default implementation instead of just override it. I guess I will just name it something else if needed, but then I need proxy functions for the 'default' implementation to get overridden. To achive the same, this is now needed:
mixin class foo
{
void printfoo(){output("Hello");}
void print(){printfoo();}
};
class bar : foo
{
void print()
{
printfoo();
output(" world!");
}
};
I guess its not a big deal, it just seems annoying to have another layer of indirection for the 'default' behavior in this case especially since I doubt Angelscript can inline and eliminate the call for the default (presumably more often called) case. I guess it bugs me because one of the things I love most about Angelscript is how few proxy functions I need to write to get things done. Those always seemed so wasteful and ulgy to me. Code duplication is best avoided.
I understand they are macros instead of real inheritance , but why could the scope resolution operator not be allowed for them? Basically it would 'rename' mixin functions that have been overridden or just give them a 2nd name they can be accessed by. Or do these overridden functions not get compiled at all encase they now produce compile errors that are being avoided by the override?
It would bring them one step closer to being as useful as multiple inheritance without having to support MI, As multiple mixins could even have there own default implementations of the same function and the resulting class could pick what ones to call and in what order.
I actually like how mixins work more then MI, Mainly because it avoids the headache of virtual base classes in diamond inheritance when you want one variable to be accessible by multiple 'middle' classes to be derived from in various ways by other classes.
Of course, I don't know how complex this would be to add. I just thought if it was simple enough and didn't break anything, it might be considered but if not I can do without no problem.
Lead Coder/Game Designer for Brutal Nature: http://BrutalNature.com
### #4Andreas Jonsson Moderators
Posted 20 September 2013 - 11:22 AM
It's doable, but it would definitely not a trivial solution. I'll add a note on my to-do list to analyze this in detail for a possible future improvement.
AngelCode.com - game development and more - Reference DB - game developer references
AngelScript - free scripting library - BMFont - free bitmap font generator - Tower - free puzzle game
### #5BlackMoons Members
Posted 21 September 2013 - 05:13 AM
Awesome. That is all I can ask.
Also noticed the lack of constructors/destructors in mixins.. But I can understand that could get messy very quickly if used. So I'll stick to just adding functions for those if needed.
Lead Coder/Game Designer for Brutal Nature: http://BrutalNature.com
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
|
2017-04-26 00:49:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20812338590621948, "perplexity": 2173.9547101059825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121000.17/warc/CC-MAIN-20170423031201-00460-ip-10-145-167-34.ec2.internal.warc.gz"}
|
https://engineering.stackexchange.com/questions/21019/find-spring-constant-of-frame-structure
|
# Find spring constant of frame structure
I would like to calculate the spring constant of a frame structure. I thought that I could do it like this:
1. I apply a point force on the structure:
2. Then I calculate the displacement at that same point:
Finnaly, I asume that the material bends within its elastic limit: $k=\frac{F}{\delta_y}$
Question1: Do you agree with this method ?
Question2: If I have a different type of loading, for instance a distributed load, would the spring constant change ?
|
2021-05-16 18:45:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5117745399475098, "perplexity": 526.2525726513649}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991178.59/warc/CC-MAIN-20210516171301-20210516201301-00614.warc.gz"}
|
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-7-review-page-469/109
|
## Intermediate Algebra (6th Edition)
$\dfrac{3\sqrt[]{y}+6}{y-4}$
Multiplying by the conjugate of the denominator, the rationalized-denominator form of the given expression, $\dfrac{3}{\sqrt[]{y}-2} ,$ is \begin{array}{l}\require{cancel} \dfrac{3}{\sqrt[]{y}-2}\cdot\dfrac{\sqrt[]{y}+2}{\sqrt[]{y}+2} \\\\= \dfrac{3\sqrt[]{y}+6}{(\sqrt[]{y})^2-(2)^2} \\\\= \dfrac{3\sqrt[]{y}+6}{y-4} .\end{array} Note that all variables are assumed to have positive real numbers.
|
2017-12-11 04:25:23
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9973985552787781, "perplexity": 2389.725949779994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00060.warc.gz"}
|
https://gamedev.stackexchange.com/questions/114335/fixed-time-step-flutter/114386
|
Fixed Time Step Flutter
I've seen a lot of questions about this problem, but really no solution. Hopefully I can explain the issue well enough that I can get some sort of closure on this problem.
I'm using a Fixed Time Step based on the code found here : GAFFER ON GAMES / Fix Your Timestep! , and with that method I'm using an accumulator value to compensate for any slowdowns or speedups in the frame rate. Typically the time between frames is between 15.5 and 17.5 ms (with the fixed rate aiming for 1000/60 fps). What ends up happening is this:
That green chart is showing the accumulator. Which freaks out once the accumulator gets beyond 1000/60 fps. At this point you can see that because of the variance between refresh rates (15.5-17.5) sometimes the accumulator is above or below the threshold of 16.66666666... until it eventually gets to a point that the accumulator is too big to be negated by the next frame variance.
If I simply ignore the refresh rate and assume a constant 60fps the flutter never occurs, so its shouldn't be a vsync issue (right?), it's not the GC kicking in. It's an issue with this part of the code:
accumulator += frameTime;
while ( accumulator >= dt )
{
update_logic();
accumulator -= dt;
}
I realize the issue is when the accumulator hits ~0 it actually starts doing double updates and skipping updates for a while causing the flutter/stutter look.
It's very possible that I'm just not understanding how the accumulator is suppose to work, or perhaps the variance in screen refreshing is too much for this to work well.
Update
Here is a screen cap of an output log showing how the variance in frameTime is causing the update call to double up and skip.
The items in red are when the accumulator never gets to 16.66666666...
For the time being I'm 'fixing' the issue by never letting the accumulator to get below the frameTime variance, but this doesn't feel like a real answer, just a patch.
if(accumulator < 1){
accumulator = 1.5;
}
Perhaps this bit will help explain. If you start the accumulator at 0 and the first frame takes 15.89999ms, we don't update anything, and then if the next frame is 17.55555ms we now have (~33.44ms) so it updates twice. The more the variance in the frameTime the worse the problem.
I've noticed plenty of people with this issue, and typically the conversation ends with no posted solution, it either "goes away" or the thread dies.
• What about problems with frameTime precision? – wondra Jan 5 '16 at 20:41
• @wondra I believe that is the true problem. It's not reliable enough, and I get the impression that the frameTime if averaged out over a few frames is actually closer to 59.9 fps which is why slowly over 30 seconds or so the accumulator gets bigger and bigger. But how do I compensate for this variance? – ericjbasti Jan 5 '16 at 20:56
• Show more code, it can't be that the updates are skipped 'for a while' : if you just made 2 updates sure you can skip one after maaaaayybe 2, but if it's more than 2 skipped there's an error somewhere. – GameAlchemist Jan 6 '16 at 12:20
• @GameAlchemist you've actually described the issue. It's not a constant thing, but its very predictable (per device) which is why I think most people write it off as the GC. It's only once the accumulator gets to 0 that the issue occurs, because thats the only point > a constant of dt can flip flop if the frameTime has enough variance. These results change based on the device, with little variance you would probably never notice the issue, but with enough you see the variance at a fairly constant rate (in my case every ~20s). – ericjbasti Jan 6 '16 at 15:39
• I've tested with the most basic of code, just a RAF that prints out the frameTime (taking out any other logic or rendering that could cause a problem) and the variance remains. – ericjbasti Jan 6 '16 at 15:46
The solution provided in the Blog post you mentionned is not that good, even if you follow it to its end, that is to say if you interpolate.
Because you won't be able to escape the case when the error accumulates until there's no need for update, and then obviously there will be the need for two. It's just maths here.
The interpolation hides that issue since even with the 0+2 issue, the interpolation between two steps gives you almost the +1 status, but at the cost of duplicating all positions to be able to, well, interpolate, and you might even need to store speed or other things if your game depends on it (trail effect...). Ouch.
Now you can improve the algorithm and try to minimize the mean error by considering that if you the accumulated error is 'near enough' from the logic time step, you'll will step and substract the error made. So, if say you accumulated 15 ms and your step is 16, go for an update and consider you accumulated -1.
...But... more or less you have build, with this, just a complicated version of a system doing 1 fixed step update by render frame... Still it is better if you get far from your fixed logic time step ( ex : 20Hz or 120Hz).
Now there's a way that i think is most preferable : if you use an update/logic time step significantly (at least 2 times) smaller than the render time step, then it makes sense again to use the fixed time step principle. Why small time steps could be good ? stepping is a simple process, and limit the hard-to-handle cases of tunneling and 'simultaneous' collisions, i mean collisions occurring during a single step. Still the sad thing being you can't avoid to treat them, but now you might even afford to have an ugly non optimized algorithm given their unlikeliness.
For the win i did a little fiddle, you can play with it to see the various cases you want to explore.
• I need to take some time a truly digest what you're saying here. About the blog post (that everyone swears by) the funny thing is the interpolation doesn't even hide the issue, it only hides it going in one direction (when the screen refresh rate is higher than the dt) its the silly < 16.666666 frameTimes that cause the issues. Just as a side, if I set the desired rate to ~59.97fps the issue take much much longer to occur. – ericjbasti Jan 6 '16 at 17:03
|
2020-02-19 12:46:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5213786959648132, "perplexity": 1025.1157384833189}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144150.61/warc/CC-MAIN-20200219122958-20200219152958-00090.warc.gz"}
|
https://www.autoitscript.com/forum/topic/25729-search-for-long-file-name/
|
Followers 0
Search For Long File Name
24 posts in this topic
Hello everyone. I need to find a way to do the following:
Example:
If I have a directory called C:\CustomerLogs where all the results log files are being stored after testing is complete. The resulta file is something like this:
0125798631447336_1_0005785606.log
5555579863166874_1_0005785606.log
7849563144734598_1_0005785606.log
The first 16 numbers from 0-6 are the system serial number, which is captured at the beginning of the test. The remaining of the numbers are generated by the application.
Goal:
I kinda have a script that will prompt for the serial number and create a log file on a different drive. But I need a script that will search C:\CustomerLogs and find the serial number log file that mathches the serial number that was entered.
for Example:
If I entered 0125798631447336, then to search and find 0125798631447336_1_0005785606.log, then take the results from this file and copy it to the log file created on the other drive.
Or once the serial number is matched, to copy that file and trim everything else but the serial number and create a log file with that serial number found.
I hope this makes sense.
Share on other sites
#2 · Posted (edited)
show us the script you have so far... then we can work from that
to get what you wanted
8)
Edited by Valuater
Share on other sites
Here it is.
; Script in progress
; TBD
; TBD
; TBD
$var = "" ;Prompt For Serial Number$var = InputBox("System Serial Number", "Scan Serial Number, then Click OK")
If $var = "" Then Exit ;if file exist to deleted or continue to over-write. If FileExists("V:" & "\" &$var & ".log") Then
$answer = MsgBox(4, "File Already Exist", "Do you want to replace it?") If$answer = 7 Then
MsgBox(48, "Stopped!", "You don't want to proceed, Click OK to exit")
Exit;ends script
EndIf
EndIf
; Server to save Logs Files
;Create the file and Write results to it.
$file = FileOpen("V:" & "\" & ($var) & ".log", 2)
If $file = -1 Then MsgBox(0, "Error", "Unable to open file.") Exit EndIf ; Show the File Name of input variable returned.$CPath = ("C:" & "\" & "CustomerLogs" & "\")
$search = FileFindFirstFile($CPath & ($var) & ".log") ; Check if the search was successful If$search = -1 Then
MsgBox(0, "Error", "No files/directories matched the search pattern")
Exit
EndIf
While 1
$file = FileFindNextFile($search)
If @error Then ExitLoop
MsgBox(4096, "C:\CustomerLogs Log File Found!, Log File:", $file) WEnd ; Close the search handle FileClose($search)
; Example of Serial Number Created by Application
; Need to search for input on Drive C: and trim the 13 right strings and copy results to another file
;$result = StringTrimRight("0123456789_1_123456.log", 13) ;MsgBox(0, "String without rightmost 13 characters is:",$result)
;$file = Fileopen(@DesktopDir & "\" &$result & ".log", 2)
Hello everyone. I need to find a way to do the following:
Example:
If I have a directory called C:\CustomerLogs where all the results log files are being stored after testing is complete. The resulta file is something like this:
0125798631447336_1_0005785606.log
5555579863166874_1_0005785606.log
7849563144734598_1_0005785606.log
The first 16 numbers from 0-6 are the system serial number, which is captured at the beginning of the test. The remaining of the numbers are generated by the application.
Goal:
I kinda have a script that will prompt for the serial number and create a log file on a different drive. But I need a script that will search C:\CustomerLogs and find the serial number log file that mathches the serial number that was entered.
for Example:
If I entered 0125798631447336, then to search and find 0125798631447336_1_0005785606.log, then take the results from this file and copy it to the log file created on the other drive.
Or once the serial number is matched, to copy that file and trim everything else but the serial number and create a log file with that serial number found.
I hope this makes sense.
Share on other sites
Quick and dirty example of how I would do it:
#include <file.au3>
#include <array.au3>
$s_path = "C:\CustomerLogs\"$s_serial = InputBox("Serial Search", "Enter Serial # to search for.")
$a_filelist = _FileListToArray($s_path)
If (Not IsArray($a_filelist)) Then MsgBox (0,"","No Files\Folders Found.") Exit EndIf For$i = 0 to Ubound($a_filelist)-1$a_searchresult = StringRegExp($a_filelist[$i],'('&$s_serial&'_[0-9]{1}_[0-9]{10}.log)',1) If IsArray($a_searchresult) = 1 Then
MsgBox(0,"File Found!","File: "&$a_filelist[$i]&" Matches: "&$s_serial&@CRLF&"Copying to"&$s_path&$s_serial&".log") FileCopy ($s_path&$a_filelist[$i], $s_path&$s_serial&".log",1)
EndIf
Next
Exit
AutoIt Scripts:Aimbot: Proof of Concept - PixelSearching Aimbot with several search/autoshoot/lock-on techniques.Sliding Toolbar - Add a nice Sliding Toolbar to your next script. Click the link to see an animation of it in action!FontInfo UDF - Get list of system fonts, or search to see if a particular font is installed.Get Extended Property UDF - Retrieve a files extended properties (e.g., video/image dimensions, file version, bitrate of song/video, etc)
Share on other sites
When I run the script I received the following message:
C:\test.au3(6,39) : ERROR: _FileListToArray(): undefined function.
$a_filelist = _FileListToArray($s_path)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^
C:\test.au3 - 1 error(s), 0 warning(s)
Quick and dirty example of how I would do it:
#include <file.au3>
#include <array.au3>
$s_path = "C:\CustomerLogs\"$s_serial = InputBox("Serial Search", "Enter Serial # to search for.")
$a_filelist = _FileListToArray($s_path)
If (Not IsArray($a_filelist)) Then MsgBox (0,"","No Files\Folders Found.") Exit EndIf For$i = 0 to Ubound($a_filelist)-1$a_searchresult = StringRegExp($a_filelist[$i],'('&$s_serial&'_[0-9]{1}_[0-9]{10}.log)',1) If IsArray($a_searchresult) = 1 Then
MsgBox(0,"File Found!","File: "&$a_filelist[$i]&" Matches: "&$s_serial&@CRLF&"Copying to"&$s_path&$s_serial&".log") FileCopy ($s_path&$a_filelist[$i], $s_path&$s_serial&".log",1)
EndIf
Next
Exit
Share on other sites
Requires Beta, sorry.. forget to mention that.
AutoIt Scripts:Aimbot: Proof of Concept - PixelSearching Aimbot with several search/autoshoot/lock-on techniques.Sliding Toolbar - Add a nice Sliding Toolbar to your next script. Click the link to see an animation of it in action!FontInfo UDF - Get list of system fonts, or search to see if a particular font is installed.Get Extended Property UDF - Retrieve a files extended properties (e.g., video/image dimensions, file version, bitrate of song/video, etc)
Share on other sites
Which beta? Sorry I don't understand.
?
Requires Beta, sorry.. forget to mention that.
Share on other sites
http://www.autoitscript.com/autoit3/files/...-beta-Setup.exe
Also, I would suggest getting the SCITE editor if you have not already.
http://www.autoitscript.com/cgi-bin/getfil...iTe4AutoIt3.exe
So, Install BETA, Install Scite, open up the script I wrote for you in Scite, and click, Tools>Beta Run Or ALT+F5
AutoIt Scripts:Aimbot: Proof of Concept - PixelSearching Aimbot with several search/autoshoot/lock-on techniques.Sliding Toolbar - Add a nice Sliding Toolbar to your next script. Click the link to see an animation of it in action!FontInfo UDF - Get list of system fonts, or search to see if a particular font is installed.Get Extended Property UDF - Retrieve a files extended properties (e.g., video/image dimensions, file version, bitrate of song/video, etc)
Share on other sites
I install the Beta software and I had already installed the SCITE. I still get the same message:
:\test.au3(6,39) : ERROR: _FileListToArray(): undefined function.
?????????/
http://www.autoitscript.com/autoit3/files/...-beta-Setup.exe
Also, I would suggest getting the SCITE editor if you have not already.
http://www.autoitscript.com/cgi-bin/getfil...iTe4AutoIt3.exe
So, Install BETA, Install Scite, open up the script I wrote for you in Scite, and click, Tools>Beta Run Or ALT+F5
Share on other sites
#10 · Posted
_FileListToArray is in the beta File.au3
Dont know what to tell you. Make sure you have it set to use the beta include folder, or copy the beta file.au3 to the normal version's include file as a workaround.
AutoIt Scripts:Aimbot: Proof of Concept - PixelSearching Aimbot with several search/autoshoot/lock-on techniques.Sliding Toolbar - Add a nice Sliding Toolbar to your next script. Click the link to see an animation of it in action!FontInfo UDF - Get list of system fonts, or search to see if a particular font is installed.Get Extended Property UDF - Retrieve a files extended properties (e.g., video/image dimensions, file version, bitrate of song/video, etc)
Share on other sites
#11 · Posted
I will try it. Thank you very much.
_FileListToArray is in the beta File.au3
Dont know what to tell you. Make sure you have it set to use the beta include folder, or copy the beta file.au3 to the normal version's include file as a workaround.
Share on other sites
#12 · Posted
Anyone else can help me this. I get error messages when running the script..
C:\Program Files\AutoIt3\beta\Examples\_FileListToArray.au3(6,43) : ERROR: _FileListToArray(): undefined function.
$FileList=_FileListToArray($sPath,$sFilter) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^ I will try it. Thank you very much. Share this post Link to post Share on other sites #13 · Posted Anyone else can help me this. I get error messages when running the script.. C:\Program Files\AutoIt3\beta\Examples\_FileListToArray.au3(6,43) : ERROR: _FileListToArray(): undefined function.$FileList=_FileListToArray($sPath,$sFilter)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^
Use the Beta of AuotIt and you won't get the error if you use #include <File.au3>
[center]Common sense plays a role in the basics of understanding AutoIt... If you're lacking in that, do us all a favor, and step away from the computer.[/center]
Share on other sites
#14 · Posted
While this isnt a proper solution because you should be able to install and run the beta version fine, you could just tack this on the end of your script as a temporary workaround:
Func _FileListToArray($sPath,$sFilter = "*", $iFlag = 0) Local$hSearch, $sFile,$asFileList[1]
If Not FileExists($sPath) Then SetError(1) Return "" EndIf If (StringInStr($sFilter, "\")) or (StringInStr($sFilter, "/")) or (StringInStr($sFilter, ":")) or (StringInStr($sFilter, ">")) or (StringInStr($sFilter, "<")) or (StringInStr($sFilter, "|")) or (StringStripWS($sFilter, 8) = "") Then
SetError(2)
Return 0
EndIf
If Not ($iFlag = 0 Or$iFlag = 1 Or $iFlag = 2) Then SetError(3) Return "" EndIf$asFileList[0] = 0
$hSearch = FileFindFirstFile($sPath & "\" & $sFilter) If$hSearch = -1 Then
SetError(0)
Return 0
EndIf
While 1
$sFile = FileFindNextFile($hSearch)
If @error Then ExitLoop
If $iFlag = 1 Then If StringInStr(FileGetAttrib($sPath & "\" & $sFile), "D") <> 0 Then ContinueLoop EndIf If$iFlag = 2 Then
If StringInStr(FileGetAttrib($sPath & "\" &$sFile), "D") = 0 Then ContinueLoop
EndIf
ReDim $asFileList[UBound($asFileList) + 1]
$asFileList[0] =$asFileList[0] + 1
$asFileList[UBound($asFileList) - 1] = $sFile WEnd FileClose($hSearch)
SetError(0)
If $asFileList[0] = 0 Then Return "" Return$asFileList
EndFunc ;==>_FileListToArray
AutoIt Scripts:Aimbot: Proof of Concept - PixelSearching Aimbot with several search/autoshoot/lock-on techniques.Sliding Toolbar - Add a nice Sliding Toolbar to your next script. Click the link to see an animation of it in action!FontInfo UDF - Get list of system fonts, or search to see if a particular font is installed.Get Extended Property UDF - Retrieve a files extended properties (e.g., video/image dimensions, file version, bitrate of song/video, etc)
Share on other sites
#15 · Posted
Could not get it to work. I am very sad. What am I doing wrong HAAAAAAAAAAAAAAAAAAAAAAAAAA...
While this isnt a proper solution because you should be able to install and run the beta version fine, you could just tack this on the end of your script as a temporary workaround:
Func _FileListToArray($sPath,$sFilter = "*", $iFlag = 0) Local$hSearch, $sFile,$asFileList[1]
If Not FileExists($sPath) Then SetError(1) Return "" EndIf If (StringInStr($sFilter, "\")) or (StringInStr($sFilter, "/")) or (StringInStr($sFilter, ":")) or (StringInStr($sFilter, ">")) or (StringInStr($sFilter, "<")) or (StringInStr($sFilter, "|")) or (StringStripWS($sFilter, 8) = "") Then
SetError(2)
Return 0
EndIf
If Not ($iFlag = 0 Or$iFlag = 1 Or $iFlag = 2) Then SetError(3) Return "" EndIf$asFileList[0] = 0
$hSearch = FileFindFirstFile($sPath & "\" & $sFilter) If$hSearch = -1 Then
SetError(0)
Return 0
EndIf
While 1
$sFile = FileFindNextFile($hSearch)
If @error Then ExitLoop
If $iFlag = 1 Then If StringInStr(FileGetAttrib($sPath & "\" & $sFile), "D") <> 0 Then ContinueLoop EndIf If$iFlag = 2 Then
If StringInStr(FileGetAttrib($sPath & "\" &$sFile), "D") = 0 Then ContinueLoop
EndIf
ReDim $asFileList[UBound($asFileList) + 1]
$asFileList[0] =$asFileList[0] + 1
$asFileList[UBound($asFileList) - 1] = $sFile WEnd FileClose($hSearch)
SetError(0)
If $asFileList[0] = 0 Then Return "" Return$asFileList
EndFunc ;==>_FileListToArray
Share on other sites
#16 · Posted
Could not get it to work. I am very sad. What am I doing wrong HAAAAAAAAAAAAAAAAAAAAAAAAAA...
Are we supposed to guess what your doing at all? You didn't even show how you are using any of the functions in script
[center]Common sense plays a role in the basics of understanding AutoIt... If you're lacking in that, do us all a favor, and step away from the computer.[/center]
Share on other sites
#17 · Posted
Could not get it to work. I am very sad. What am I doing wrong HAAAAAAAAAAAAAAAAAAAAAAAAAA...
what im guessing you did is just compile his script hahahhahaaa, to bad its just a UDF
Share on other sites
#18 · Posted
I tried to test the script provided and it did not work for me. Perphaps you can shine a light on me and show me how it suppose to be done. In my initial request explains what I want to do and a simple script that it kinda works, but it's not perfect. I'm still learning, it will take some time for me to grasp the entire script language just you like you have. Let me know your thougths.
Are we supposed to guess what your doing at all? You didn't even show how you are using any of the functions in script
Share on other sites
#19 · Posted
Well, I tried a lot of things before compiling the script and it did not work for me. Can you help and tell me what I'm during wrong or perphaps have a script that will work with what I want to do. The idea is there of what I want to do, but I don't know exactly how to put it together. It seems that you have more experience in this field, so please show me the way...
what im guessing you did is just compile his script hahahhahaaa, to bad its just a UDF
Share on other sites
#20 · Posted
Just show what you tried. I'm not going to guess what in the world you did to make it fail. Just show the code you used that failed, and we can help you figure it out.
[center]Common sense plays a role in the basics of understanding AutoIt... If you're lacking in that, do us all a favor, and step away from the computer.[/center]
|
2017-07-24 12:41:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3923463821411133, "perplexity": 3994.318312177428}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424876.76/warc/CC-MAIN-20170724122255-20170724142255-00283.warc.gz"}
|
https://www.physicsforums.com/threads/trickey-integral.333350/
|
# Homework Help: Trickey Integral
1. Aug 29, 2009
### Stratosphere
1. The problem statement, all variables and given/known data
I can't seem to figure out what I need to do in order to solve this one.
$$\int \sqrt{\theta+\frac{1}{2}\theta^{-1/2}}d\theta$$
2. Relevant equations
3. The attempt at a solution
I haven't the slightest clue as to what technique to use, it doesn't look to be a u substitution problem though (I could be wrong).
I typed it in on an online calculator and I got some really strange expression.
http://integrals.wolfram.com/index.jsp?expr=%28x%2B1%2F2x^%28-1%2F2%29%29^%281%2F2%29&random=false
Last edited: Aug 29, 2009
2. Aug 29, 2009
### Dick
It is a substitution problem. Since you already cheated and looked at wolfram you may as well use that cheat. Look at the argument of the ArcSinh. That suggests if you substitute u=sqrt(2)x^(3/4) you should be able to reduce it to something you can handle. In fact, once you hack through the radicals you should be able to show that aside from the numeric factors, it becomes sqrt(u^2+1)du. Can you integrate that?
3. Aug 29, 2009
### Stratosphere
I don't want to use wolfram, I just used it because I couldn't figure it out. I didn't cheat on homework or anything I am self learning it. So what would be the first step starting from the original equation.
4. Aug 29, 2009
### gabbagabbahey
Well, just glancing at the integral it seems like the substitution $u=\sqrt{\theta}$ is a decent place to start.....have you tried that? If so, how far did you get with it?
5. Aug 30, 2009
### Dick
I'm exaggerating on the 'cheating' aspect. But I already suggested an initial substitution in my last post.
|
2018-07-23 01:03:03
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7892595529556274, "perplexity": 616.0653367099135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676594675.66/warc/CC-MAIN-20180722233159-20180723013159-00520.warc.gz"}
|
https://zbmath.org/?q=an%3A1321.57032
|
# zbMATH — the first resource for mathematics
On the fundamental group of certain regular neighbourhoods. (Sobre el grupo fundamental de ciertas vecindades regulares.) (Spanish. English summary) Zbl 1321.57032
Summary: In this paper, we present a modern proof of a classical result of PL topology, which is cited widely without justification. Specifically we first prove that given a polyhedron $$P$$ with a regular neighbourhood $$N$$, such that the codimension of the pair $$(P,N)$$ is greater than or equal to 3, then it follows that $$\pi_1(N) \cong \pi_1(N \backslash P)$$. As a consequence, we deduce that, under the same hypothesis, $$\pi_1(P) \cong \pi_1(\partial N)$$.
##### MSC:
57Q40 Regular neighborhoods in PL-topology 57Q05 General topology of complexes 57M05 Fundamental group, presentations, free differential calculus
##### Keywords:
Markov theorem; polyhedron; regular neighbourhood
Full Text:
|
2021-01-15 19:01:32
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.469433456659317, "perplexity": 1385.552633874839}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00436.warc.gz"}
|
https://physics.stackexchange.com/tags/quantum-electrodynamics/info
|
In QED, electrically charged particles are coupled by an uncharged, massless vector boson called the photon. The former are described by means of a fermionic spinor field $$\psi$$, and the latter by a bosonic vector (gauge) field $$A$$. The classical Lagrangian is postulated to be $$\mathcal L=\bar\psi(i\not D-m)\psi+\frac{1}{4e^2}F^2$$ where $$D\equiv\partial-iA$$ is the so-called gauge covariant derivative and $$F\equiv \mathrm dA$$ is the so-called field strength tensor. The quantum Lagrangian requires several modifications, such as fixing the gauge and introducing renormalisation constants. Once this is done, one may read off from $$\mathcal L$$ the Feynman rules of the theory, which are enough to calculate any prediction to an arbitrary order in perturbation theory.
By adding three additional, massive vector bosons (the $$Z^0$$ and $$W^{\pm}$$) which couple to the weak hyper-charge ($$T_3 - q \ \sin^2 \theta_W$$ in which $$q$$ is electric charge and $$T_3$$ is the third component of the weak isospin), the theory can be extended to cover the weak nuclear force as well.
|
2019-06-25 12:45:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.887295126914978, "perplexity": 243.75577415025205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999838.23/warc/CC-MAIN-20190625112522-20190625134522-00157.warc.gz"}
|
https://jp.maplesoft.com/support/help/maplesim/view.aspx?path=plots/listcontplot3d&L=J
|
plots - Maple Programming Help
Home : Support : Online Help : Graphics : Packages : plots : plots/listcontplot3d
plots
listcontplot3d
three-dimensional contour plot of a grid of values
Calling Sequence listcontplot3d(G, options)
Parameters
G - list of lists or Matrix of numeric values
Description
• The listcontplot3d function creates a 3-D contour plot of a grid (list of lists or Matrix) of numeric values.
• Remaining arguments are interpreted as plot options. See plot3d[option] for a list of possible options.
• Option levels is provided as a synonym of the option contours. Either of these specifies the number of contours to draw, the default being eight.
• The filledregions=true option can be specified to obtain a filled contour plot. The coloring option can, in this case, be used to specify a gradation in color, or color range. The default is a color range from red to yellow. To specify a color range, use the form coloring=[a,b], where a and b refer to colors recognized by plot3d. See plot[color] for an explanation of colors.
• Data saved in external files may be imported into a Matrix with the ImportMatrix command and passed to listcontplot3d.
Examples
> $\mathrm{with}\left(\mathrm{plots}\right):$
> $\mathrm{listcontplot3d}\left(\left[\mathrm{seq}\left(\left[\mathrm{seq}\left(\mathrm{sin}\left(\frac{\frac{x}{5}y}{5}\right),x=-15..15\right)\right],y=-15..15\right)\right]\right)$
> $\mathrm{listcontplot3d}\left(\mathrm{Matrix}\left(\left[\mathrm{seq}\left(\left[\mathrm{seq}\left(\mathrm{sin}\left(\frac{\left(i-15\right)\left(j-10\right)}{20\mathrm{\pi }}\right),i=1..30\right)\right],j=1..20\right)\right]\right)\right)$
> $\mathrm{listcontplot3d}\left(\left[\mathrm{seq}\left(\left[\mathrm{seq}\left(\mathrm{sin}\left(\frac{\left(i-15\right)\left(j-10\right)}{20\mathrm{\pi }}\right),i=1..30\right)\right],j=1..20\right)\right],\mathrm{filledregions}=\mathrm{true}\right)$
Compatibility
• The G parameter was updated in Maple 15.
|
2021-05-12 21:47:01
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7419367432594299, "perplexity": 2348.557800069595}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989705.28/warc/CC-MAIN-20210512193253-20210512223253-00410.warc.gz"}
|
https://discourse.vvvv.org/t/kairos-software/15506
|
# Kairos Software
#1
Hello,
As many of you probably know from node17, I’m currently working on a VL based timeline editor tool, which, hopefully, I will release in the first part of September as a vvvv pack.
I made this forum post to collect constructive suggestions and references from users.
So please, feel free to write ideas, features, UI suggestions (and reference images) regarding the ideal timeline tool that you would like to use.
Also pointing to other existing softwares could be useful.
Please, try to be essential, short and direct to the point(s) in order to not make this post unreadable and useless :)
And let’s skip the very basic features… like having undo and redo…
Natan
LINK Topics - 1) Timeline
#2
Do you aim at controlling large scenes/shows only?
My timeliner of choice would still have little instances all over the place when I need to animate/fade in/fade out elements in my application/exhibit.
Workaround would be to have one editor only, but different timeliner scenes which one could then instantiate in the patch as needed.
#3
I have been in timeline land for some time now, trying to bend existing options to fit my needs, my requirements are here in no particular order:
1. Step sequencer, often I need to step sequence things.
2. Higher than frame rate precision, especially with musical timing, having something that is independent of fluctuating frame rates is important.
3. record parameters live.
4. edit recorded paramters
5. NRT playback, often I do stuff live and record parameters but not video, I then render the frames in NRT mode. I assume this clashes a bit with 2. but perhaps it can be quantized to fit the framerate somehow… and in increment mode, there is no frame rate fluctuation.
6. Beats and Bars: I can see above that musical tempo is assumed, so I better mention it as a wish for a feature
EDIT: 7. Grouping of tracks
#4
1. Chapters, or inout points for a specific series of operations
2. Ability to browse next\previous frame, in\out points, beginning\end of the timeline
3. Ability to modify OTF its length: shorter, longer…
4. Bezier handles to interpolate nicely between points
4.1 With the ability to manually insert handle values. Would it be possible to calculate stuff somewhere in patch and then send that data to the handles?
5. Fast switching between different interpolation styles, linear to bezier and viceversa
6. Ability to set one point linear between two bezier, or per point interpolation styling
7. Ability to merge animations from different timelines, eg I make one animation in a timeline, another animation in another timeline -> merge those two in a third timeline. Obviously, it can be done in a single timeline too (asking for merging, not suggesting how to do it)
8. Highlighting certain parts of timeline (like, this is red = I have to launch this other thing; this is green = animation here is approved…)
9. Annotations for specific parts of timelines. A bit like 6, but also not: 6 would be for all tracks, note for each track, because I’m asking for…
10. Colors for tracks (where am I? Bottom, top, … What is this? gui or objects… And so on)
11. Track beginning (the one with name of track and other useful hints) always in sight. This means, track contents will be browsed and only a portion of it will always be visible unless you…
12. Zoom in\out tracks
I know I’m missing something…
EDIT: Here it is: Track disabling.
Last thing:
#5
Cool,
i missed your presentation at node, so i don´t know what it is already integrated,
These are my favorite features on a timeliner.
1. External devices easy connection (mostly midi controllers)
2. Multitrack realtime data recording from external device, and controller snapshot, as Sune pointed
3. Controller templates or User defined templates to share between people,
4. Controller MIDI feedback
5. Timeliner control via external devices, (for example a Launchpad to jump between the cues on the selected track, and other functions)
6. Tracks grouping. as Sune pointed
7. Multi loop/scenes per project, as h99 pointed
8. internal - external clock, as Sune pointed
9. Different kind of CLOCK OUT (MIDI, MTC, etc) simultaneous sync with other devices
I know that Sync is a bit topic on this kind of things,
but i hope the vvvv guys can improve it and get a robust and stable clock on VL.
10. import, merge, split, and append projects.
11. import, add, insert tracks.
12. Default values and interpolation method per track or defined by template
Is Audio already on the plan? VAudio VL pack?
1. Audio track (or tracks), with start position offset and simple controls, volume, pan.
2. Audio wave visualization based on the screen resolution.
thanks a lot for that Nathan
#6
Related to GUI
Advanced and Simple GUI switch, (also useful for small machines or when the GUI performance would be an issue)
For example when the timeliner is working as player, but you want to have access to some basic options like transport and play/stop controls, useful when it is operated by a person without experience.
And regarding to audio.
Obviously separate thread for audio playing, without glitches ;)
#7
• Allow nesting of timelines. So one timeline can be (part of) a track in another one. Like sequences in premiere.
• Integrated asset-handling
• creating / managing sequences
• import audio / video / midi / other timelines
• importer settings, e.g. interpret image-sequences as being 25 FPS.
• (re)naming
• grouping/organizing assets by tags and/or in folders
• relinking of moved assets
• replacing files, like videoproxies, in place
• Save as / serialize timeline to a human readable format that ideally adheres to some kind of standard.
#8
Hello Natan,
Sander Van Dijk has made an amazing work redesigning the After Effects timeline, layers and keyframes…
http://www.sandervandijk.tv/after-effects-features/
Adobe has never taken into account these ideas so far…
But there is a lot of great inspiration here (Sander created some videos and GIF examples, and guys in the comment section have a lot of good ideas too)
#9
I think nobody mentioned so far:
keyframes snapping to keyframes in other tracks …
#10
its really mindblowing to see how wide-ranging the needs of people are concerning a timeline. sounds like there can’t be just one being suitable for everyone.
for example, i just need plain animation and don’t care about recording, live inputs, midi mapping, asset management and so on.
therefore its good to see different approaches in the SOD
so my only suggestions for kairo would be:
• intuitve UI, with everything one would expect from a modern UI
• low cpu hit
#11
Hey,
regarding other softwares - I’m working with Ventuz a lot for the last two years, it’s Animation node is a state machine with timeline, and is a pleasure to use.
Might be a good study material, the software can be downloaded and tried for free.
http://www.ventuz.com/support/help/latest/AnimationEditor.html
http://www.ventuz.com/support/tutorial-08-keyframe-animation-basics
thanks
#12
Thanks guys for all these precious inputs.
As u7angel said, it’s really incredible to see how many needs and whishes related a timeline. It’s clear that is a key piece of software… :)
My intent is to first release a relatively simple tool, which will have all the functionalities of a basic timeline (as the old Timeliner node) plus the new feature of “dynamic keyframes”. These dynamic keyframes are for me the most innovative contribution of this research and I’d like to have as soon as possible some feedbacks from the community.
Then it will be a long journey to evolve this first version of Kairos.Timeline into something else, more rich and versatile. Actually, my final goal is to get something like an Ableton Live kind of software, but incredibly superior in terms of realtime contents management. This will be Kairos.
So, coming back to the roadmap (also a reply to @readme) , this first version will come as a “centralized” editor, in which you create tracks, group them, add and edit keyframes, and so on…
then there will be a set of nodes to set and retrieve data from it, like: GetTrack, SendChannel (to send any kind of data to a channel, received for example by a dynamic keyframe), SetKeyframe, SetCustomClock, …
I’ll probably also pack some of the Kairos.Timeline features as independent vvvv nodes, eg. a Track node, which contains all the features of a single track without the rest of kairos. this module would still allow you to create amazing gradients and fades…
obviously you can still instantiate many Kairos.Timelines as individual vvvv nodes, even if not best optimized approach.
few scattered replies to your suggestions:
@sunep
1. not in the first release, but def wanted
2. already now the runtime of the software is separated from the model and the editor, so it will be possible, i guess, to have some sort of independency from the vvvv fluctuating frame rate.
3. yes. It will be possible to record keyframes on the fly. I’m still thinking to a different mode in which you can record raw values, with less functionalities than keyframes but better performances.
4. and yes, editable keyframes.
5. this is totally possible (I also need it for NRT rendering). Since you can manually set the Time 64bit value (which is the position of the sampler that sample each track) you can read from the timeline as you like, eg. sending the time position of every frame from your vvvv patch…
6. right now I implemented just Time (in seconds) as the main time measurement. I already got a dynamic snap to quantize movements depending on the time zoom magnitude.
7. My plan is to put a toggle to switch between Time and Tempo (with BPM and Tempo Signature).
8. I definitively want to have hierarchies in tracks. I’m still thinking (and I’d like your suggestions) what kind of control you would have over tracks from the group level, and how to display groups in UI.
@h99
1. I already have this feature as “section markers”, which have a dedicated mini track under the time bar.
you can play from any of the section markers and even call from outside (eg. vvvv node).
what’s best: “section” or “chapter”?
2. sorry my ignorance, what’s OTF?
3. regarding frame interpolation:
• It will be per keyframe
• types: StepIn, StepOut, Tween (with all relative settings), bezier (being able to control two handles)
1. I’m still considering all this merging stuff… def not in this first version
2. do you mean sort of labels that you can put into the timeline of every track? like comments that you can place near the keyframes?
…chaos is my bread… :)
@lasal
1. to 5. will see how much i can get in terms of external control for this first version, but definitively wanted.
Audio:
I’d like to have Audio tracks in the future, don’t think in this first release.
Regarding GUI:
I already splitted completely the model+runtime part of the software from the editor (UI), so you can eventually edit the timeline then have just a much light player without any UI… …or design a minimal interface.
@bjoern
• nesting feature def is the direction of kairos. not for this first release
• serialization: I’ll def want to sort this part in a way that future version will be campatible with previous presets and scenes. so yes, I need to create a standard for this.
@guest
thanks for the link!
keyframes snapping to other keyframes: yes
@u7angel
you will get an animation tool already from the first version… …hopefully :)
@wide
thanks so much for the input. I’m downloading it and I’ll def give a try. tnx
I’ll keep you posted with the development news and I’ll ask here if I have doubts regarding some features
thanks to all
Nat
…and don’t stop to send me suggestions and links for references ;)
#13
I was shamelessly using a DVD term! Being a (time)line it should be nor chapter or section, but segment, no?
on the fly
Yes, I see I wrote point; that should have been keyframe obviously. Then, great and greater.
Merging logic would be very important, because it would let you sync keyframes and segments (I’m already calling it like this) and making many of the other requests, which I respectfully consider not part of a *barebone timeline, more easy to be integrated as external plug-ins, or modules if you like. It would allow complex animations with much less effort than now.
I understand that this should be more a reminder about keeping it in mind while building the timeline, than an actual request.
*barebone = including all that logic and operations and… required to manage values along time in the most flexible way.
Yup, something like that. Complex animations would need many details and cross-referencing: having the chance to put a note that says “recalculate this based on new data” or “sync with segment XYZ” would be immensely helpful.
Oh, dear, do you want some thread, with your bread?
No granny ever
#14
@dottore
what about the ui? will it be rendered by a vvvv dx11 renderer or are u working on your own window + optimized rendering ?
if not, this would a be request too, kairo should be just one node without the overhead of vvvv patch + vl patch since i guess this is eating cpu cycles for breakfest, no ?
#15
@u7angel
I’m currently developing the ui using craftlie library. It seems quite light and avoids too much vvvv patching.
for now is fine, but the plan is obviously to be independent from vvvv.
#16
Hello guys,
Quite a long time passed from my last words regarding Kairos, so let me update you about what’s going on.
The latest concepts and ideas I shared with you (at node and in the post) were mostly regarding a timeline tool that had this new feature of dynamic keyframes.
Obviously during summer development many things changed.
I’ve been in Berlin in august, for a development session dedicated to kairos.
Devvvvs were just great and fully supportive. together we dig into the software I had in mind, trying to abstract things and finding the strong principles to look for.
From the very beginning I realized that the software I had in mind wasn’t just a timeline.
Due to the fact that I urgently need the software for production, I changed plans and, instead of going for a timeline tool, I prioritized the development of another mode of kairos: the Matrix.
You can think at this Matrix mode as the Session View of ableton live. it’s basically the same thing.
Each parameter controlled with kairos is a column of the matrix (“channels” in ableton).
the rows represent some sort of “scenes”.
Each matrix cell contains a state (static or dynamic) of that parameter.
you can then click on a cell and trigger live its state, exactly like triggering the audio/midi clips in the ableton session view.
I decided to prioritize this Matrix mode because is the simplest way to define-design the state (scene/preset) of an entire parametric system. A preset is a row of the matrix, meaning a cell, empty or not, for each kairos parameter (column).
A timeline is already something more: it’s a way to organize over time different states (-scenes-presets) of a parameter.
In kairos the state of a parameter is define by a DataSource.
DataSources are the building blocks of your Kairos scene (eg. you may put DataSources into the Matrix cells).
A DataSource can be of different types:
• Static
It’s a static value for the parameter. It can be anything: a value, a spread of values, a color palette, etc…
Every data you can imagine which is static, stored in memory, not dynamically generated.
obviously there will be an editor to design each Static DataSource
• Dynamic
Any State that is dynamically generated. eg:
• LFO (obvs in a spreadable fashion)
• Perlin
• Random
• Audio Analysis
• Kinect Data
• Particle System
• Each Dynamic DataSource will have it’s own control UI, sort of inspektor, like the ctrl window of VST plugins (distort/reverb/…) in the sound editing softwares.
• Filter
A DataSource can take another DataSource Output as input, manipulate it and return it transformed. These kind of DataSources can be called Filters. eg.
• Linear Filter
• Damper
• Oscillator
• Levin
• Delay
• Container
Containers are very interesting DataSources: thet take a set of DataSources as Input plus a Model, a set of settings (specific for every container) that tells the container how to handle the incoming DataSources.
If you abstract enough, you can basically see most of the features of an advanced software like Ableton Live as a collection of Containers:
let’s take the timeline.
timeline container takes a Model that describes the keyframes position and all their settings (interpolation, dynamic/static, …). Then it takes also a spread of DataSources which might be the Dynamic DataSources available in the system that will be assigned to some dynamic keyframes.
The timeline container then has some functionalities to handle these contents, like some interpolation functions.
It takes the incoming dynamic DataSources. dispatch them to the assigned dynamic keyframes, then apply the interpolation between the relevant keyframes in order to give the final sample value.
Conceiving the Timeline as a DataSource (Container) allows to use it anywhere in kairos as you would do with any other DataSource (static value, an LFO, a SoundAnalysis value,…). You can put it into a Cell of the matrix, or even in a keyframe of another timeline (in this case it would become a clip, that you could instantiate as many times as you like).
I’ve got already a list of Containers that I’ll develop in the next future:
• Sampler 1D (Timeline)
• Sampler 2D-3D-… (Delaunay Sampling)
• Distance Field 1D-2D-3d-… (distance based sampling)
• Switcher (like a multiFlipFlop)
• Automata
• StepSequencer
• Patcher
By abstracting the software to this level it’s much easier to think at it in a modular way.
Another example is the matrix itself:
Each column can be seen as a Switcher Container in which you choose between different options (cells) to set the state of the parameter assigned to that column (multiFlipFlop behaviour that controls a Switch).
To achieve this kind of freedom, in which you are able to create DataSources, place them anywhere, enchain them, nest one into each other, and so on… a really simple, flexible but robust interface was needed.
@gregsn and @Elias did a great job in building a c# interface that we then implemented in vl.
This interface allows to patch some functionalities that are wrapped as Nodes, with Input and Outputs.
These nodes behave following a Pull model, like the nodes and patches in vvvv.
This means that if you don’t ask for their output value from bottom, they don’t get evaluated.
You are able to create DataSources on runtime, registering them into an engine, making them immediately available for other DataSources to be used.
Basically, this system makes the user a meta-patcher, able to create nodes and connect them in runtime without recompiling.
This is great. This is already in place.
Another aspect that I started to look at together with devvvvs is the UI.
I didn’t want to spend time in building a “static” UI for Kairos (Such a boooooring thing… hihihi), specially after focusing so much on modularization at all levels of the software.
Since the topic is very appealing for the vvvv/vl community in general, we set down at devvvvs place thinking to a nice way to achieve a good multi purpose modular UI library.
And @tonfilm kicked in, as you probably noticed from his post.
This is certainly a good opportunity for all of you to contribute to kairos development and, in general, to the development of a super cool and handy UI editor library.
I started creating some of the basic UI elements (like slider, Toggle/Push/Bang buttons, multiButtons, …) but we need to add many more.
You are welcome to partecipate!
So, at the end of this biblical post,
I’m sorry to disappoint those of you who were just waiting for a timeline tool in september. It simply won’t happen. :)
What is already happening it’s way more than a timeliner. it’s a flexible system that allows to manage realtime data in a nice new way.
Trust me, the wait is worth ;)
For all of you that offered support in development, I would love an hand on the UI part, as I said previously in the post. Let’s push it forward!
Hugs
Natan
#17
Thanks for the update! It really does sound worth the wait,
#18
Hello guys,
never say never… …a new kairos report :)
waiting for your feedbacks and suggestions.
ciaoo
#19
Sorry for the late entrance but I just now discovered this blessed initiative!
I recently stumbled upon this very nice timeline software and found it very original and inspiring. https://ossia.io/
Its kind of a condition based timeline…
Anyway, perhaps I totally beside the point… I am allready in love with Kairos!!!
Ofer
#20
Hey @ofersmi, thanks so much for the link!
Didn’t have time to dig deep into it but so far this ossia software looks very interesting.
I feel it’s full of cool concept and ideas.
thanks again for the input ;)
N
|
2019-02-21 14:49:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3726218342781067, "perplexity": 3725.97122666759}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00602.warc.gz"}
|
http://mathoverflow.net/questions/122154/asymptotic-behavior-of-entire-functions/122301
|
Asymptotic behavior of entire functions
Which entire function $f\left(x\right)$ goes asymptotically to $\dfrac{e^{-x}}{x}$ as $x$ goes to infinity with $x$ positive? That is, $\left(e^{-x}/x \right)/f \left(x \right) \rightarrow 1$.
-
Plenty of them. $(e^x-e^{-2x})/x$ is the simplest example. As a matter of fact, every continuous function $g(x)$ on $\mathbb R$ can be approximated by an entire function with arbitrary continuous precision $\varepsilon(x)>0$. Voting to close. – fedja Feb 18 '13 at 12:39
You cannot say a lot about $f$ without additional assumption on the uniform growth at infinity. For instance if such a $f$ exists then any other entire function $f+g$, where $g$ is a tower of exponential of length greater than $1$ $g(x)=\exp(-\exp(\exp(\ldots(\exp x)\ldots)))$, will also satisfy your assumption. – Loïc Teyssier Feb 18 '13 at 12:40
Dear fedja, Do you mean $\left(e^{-x} - e^{-2x} \right)/x$? Best, davwood83 – davwood83 Feb 18 '13 at 13:48
As mentioned in the comments, the asymptotic behavior of $f$ along the real axis doesn't really tell you anything about the function globally.
For example, your function $f$ can behave in any way you like as $x\to-\infty$.
Indeed, the following is implied by Arakeljan's (or Nersesjan's) approximation theorem: Let $A$ be any finite union of disjoint curves tending to infinity, and let $g:A\to\mathbb{C}$ and $\varepsilon:A\to(0,\infty)$ be continuous.
Then there is an entire function $f$ such that $|f(z)-g(z)|<\varepsilon(z)$ for every $z\in A$.
So, for example, let $A=(-\infty,0] \cup [1,\infty)$, let $g$ be the function $e^{-x}/x$ on $[1,\infty)$ and let $g$ be arbitrary on $(-\infty,0]$. Then you can find an entire function that approximates this function arbitrarily closely.
|
2015-07-30 12:42:03
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9524253010749817, "perplexity": 169.4283298760131}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042987174.71/warc/CC-MAIN-20150728002307-00305-ip-10-236-191-2.ec2.internal.warc.gz"}
|
https://socratic.org/questions/how-do-you-write-the-trigonometric-form-of-3-i
|
How do you write the trigonometric form of -3-i?
Aug 1, 2017
The trigonometric form is $= \sqrt{10} \left(\cos {198.4}^{\circ} + i \sin {198.4}^{\circ}\right)$
Explanation:
Let $z = - 3 - i$
The trigonometris form is
$z = r \left(\cos \theta + i \sin \theta\right)$
If $z = a + i b$
$z = | z | \left(\frac{a}{|} z | + \frac{b}{|} z | i\right)$
The modulus is
$| z | = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{10}$
Therefore,
$z = \sqrt{10} \left(- \frac{3}{\sqrt{10}} - \frac{1}{\sqrt{10}} i\right)$
$\cos \theta = - \frac{3}{\sqrt{10}}$
and
$\sin \theta = - \frac{1}{\sqrt{10}}$
We are in the quadrant $I I I$
$\theta = {198.4}^{\circ}$
Therefore,
$z = \sqrt{10} \left(\cos {198.4}^{\circ} + i \sin {198.4}^{\circ}\right) = {e}^{i {198.4}^{\circ}}$
|
2020-02-23 11:34:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9123985767364502, "perplexity": 8820.393815572808}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00287.warc.gz"}
|
http://openstudy.com/updates/4fa5b375e4b029e9dc35a2cf
|
## AccessDenied 4 years ago Note: This is NOT a Question! I am just joining the cool new trend of having a tutorial for something, and it gives me a good excuse to make a question. This Tutorial is for LaTeX formatting (mainly without Equation Editor). There are three sections in total, and I'll post them individually, along with the introduction for a total of four posts. Enjoy!
1. AccessDenied
$$\color{gold}{\star \ \star} \quad \large{ \mathbb{\text{Doing That } \LaTeX \text{Thing}} } % Because knowing is half the battle! % % I hope you enjoy the tutorial as much as I did writing it. It's been fun! % % - Written by: AccessDenied, master of using an excessive amount of work in things %$$ LaTeX is a very powerful and versatile tool for writing Mathematics. The Equation Editor gives you only a small glimpse into the true power of the language. My goal is to give you the power to write this LaTeX on your own and tap into that potential! Prepare to delve into the wonderful world of LaTeX! $$\color{orange}{\cdot} \quad \small{ \text{Experience will best supplement this tutorial. Try things out!} }$$
2. AccessDenied
$$\color{goldenrod}{\star} \quad \mathbb{\LaTeX \text{ Syntax; Formatting, Commands, and Environments}}$$ $$\color{orange}{\cdot} \quad \text{Adding LaTeX to your Posts}$$ Creating fine posts that will coexist with the LaTeX content requires that you know how to declare it in the posts correctly and effectively. There are two ways to declare it: block content and in-line content (sometimes referred to as "displayed" and "text," respectively). $$\cdot \quad$$ Block content is inserted into the post as its own stand-alone element. The LaTeX is put on its own line. This type of LaTeX is applied by wrapping the code between the delimiters: $$\backslash [ \cdots \backslash ]$$. $$\cdot \quad$$ In-line content is inserted into the post with respect to its surroundings. This allows LaTeX to be inserted directly into paragraph explanations. This type is applied by using the delimiters: $$\backslash ( \cdots \backslash )$$. Which method you use is largely based on your purposes. Just remember that in-line is treated a tad differently from block content in that, it is made smaller and more compact than the block content by nature. This often affects placement of things like the stuff usually under the limit $$\lim$$ where the in-line method places it to the side, as so: $$\lim_{x \to 4}$$. This is fixed using a command (If you're interested, it is called $$\textbf{\limits}$$ -- I won't go over it here). Another note (it is a minor point), in-line text will not wrap around the text-space unless it is preceded by something. Block content will always wrap around, however. One of the staples for LaTeX is the backslash $$\backslash$$ symbol. It comes in on the declaration and the commands. It is also used to start a new line (double-backslash). That should help conserving the need for many individual LaTeX spaces. Okay, we got that out of the way! So, what can we do? You can type some basic equations in the LaTeX for the pretty, italicized font, but you'll quickly notice the limitations -- no spaces and no fancy font things... where is the fancy?! Well, we're getting there.
3. AccessDenied
$$\color{orange}{\cdot} \quad \text{Commands -- Adding in that functionality}$$ Commands are the masters of the functionality you really want in the posts. They're very simple, too. The basic syntax of a command is as follows: $$\quad \boxed{ \quad \color{green}{\star} \quad \backslash \underline {\textit{name}} \{ \textit{argument}_1 \} \{ \textit{argument}_2 \} \ ... \qquad }$$ It is most common for the command to have from zero to two arguments. These arguments are usually either switches for how the command works or the target of the command. There are many, many commands for LaTeX, more than I can even discuss honestly. We'll touch on the important / awesome ones, though. I leave it to you to search for the rest as you please. $$% Just AccessDenied adding some random comments to make sure you can't reuse it! %$$ The first commands we should learn is for inserting text into the LaTeX correctly. You could try writing a message like "I love LaTeX!" directly, but it doesn't turn out right, as I shall present: $$\quad \boxed{ \quad \color{green}{\Rightarrow} \quad \text{I love LaTeX!} \qquad }$$ $$\quad \boxed{ \quad \color{lime}{\Leftarrow} \quad I love LaTeX! \qquad }$$ I'll use this kind of representation a lot for this tutorial. Just internalize that this is "input =>" "output <=" and you'll be good. On a more relevant note, however, we can see there is a serious lack of spacing. It's nowhere near as pretty as just typing it out... well, let's try out a command. $$\textbf{\text} \{ ... \}$$ Well, it sounds aptly named. Let's try it out. $$\quad \boxed{ \quad \color{green}{\Rightarrow} \quad \text{\text{I love LaTeX!}} \qquad }$$ $$\quad \boxed{ \quad \color{lime}{\Leftarrow} \quad \text{I love LaTeX!} \qquad }$$ We have our spaces in and everything! Just for novelty, let's also use the \LaTeX command, which adds a nice label for the LaTeX name. $$\quad \boxed{ \quad \color{green}{\Rightarrow} \quad \text{\text{I love } \LaTeX !} \qquad }$$ $$\quad \boxed{ \quad \color{lime}{\Leftarrow} \quad \text{I love } \LaTeX !} \qquad$$ Notice that some commands are written with specific capitalized letters. Command names are case-sensitive, so be wary that some functions will not work without addressing that capital letter. Also notice that we have an extra space after "love." That space carries even outside of the command. Just keep those things in mind! The text command creates an environment apart from the Math space, which omits the spaces. However, spacing is still possible even without the text command. In fact, there are a lot of commands for spacing in the Math space. I'll just list them here. $\quad \begin{array}{|c|c|} \hline \textbf{Command} & \textbf{Effect} \\ \hline \backslash & \text{ Single space equivalent } \\ \backslash quad & \text{ Creates large horizontal space } \\ \backslash qquad & \text{ Two \quad } \\ \backslash , & \text{ Equals 3/18 a \quad } \\ \backslash : & \text{ Equals 4/18 a \quad } \\ \backslash ; & \text{ Equals 5/18 a \quad } \\ \backslash ! & \text{ Equals -3/18 a \quad } \\ \hline \end{array}$ What? An antispace!? Well, I guess antispaces may have uses somewhere. I never particularly found one, though. I guess if you wanted to create a 1/18 a \quad space or even -36/18... oh nevermind, not important! There are also the Mathematical functions. You could type them normally, but they're usually better oriented with the command. $\quad \begin{array}{|c|c|c|} \hline \text{\frac} & \text{_} & \text{^} \\ \hline \text{\sqrt} & & \\ \hline \text{\sin} & \text{arcsin} & \text{sinh} \\ \hline \text{\log} & \text{\ln} & \text{\exp} \\ \hline \text{\lim} & \text{\int} & \\ \hline \text{\sum} & \text{\prod} & \\ \hline \text{\mod} & \text{\inf} & \text{\sup} \\ \hline \end{array}$ That frac function is probably one of the big secrets of the Equation Editor and LaTeX here. The syntax of the fraction function is \frac{numerator}{denominator}. In addition, the root's index actually is placed into a weird place in square brackets (\sqrt[index] {content}) The majority of the others do come up on Equation Editor, however, and should be familiar if you've played with it before. $$% AccessDenied, adding in some special comments here and there %$$ Lastly, I want to go over some modifying commands. This should wrap up the (useful) commands section. These commands are for changing the size and style of your writing in LaTeX. $\quad \begin{array}{|c|c|} \hline \textbf{Name} & \textbf{Explained} \\ \hline \text{rm} & \mathrm{Roman} \\ \text{it} & \mathit{Italicized} \\ \text{bf} & \mathbf{Bold} \\ \text{frak} & \mathfrak{Fraktur} \\ \text{cal} & \mathcal{Calligraphy} \\ \text{sf} & \mathsf{\text{Sans-serif}} \\ \hline \text{tiny} & \text{Tiny font} \\ \text{scriptsize} & \text{Size of a sub/super script} \\ \text{small} & \text{Small font} \\ \text{normalsize} & \text{Normal} \\ \text{large} & \text{Larger font} \\ \text{Large} & \text{Even larger font} \\ \text{LARGE} & \text{LARGE font} \\ \hline \text{color} & \color{green}{\text{Makes colors!}} \\ \hline \end{array}$ The first set of commands can come in a few forms, although some are exclusive to one form. All of the above forms may be stand-alone. These will not have arguments and apply to all the content in the space. The first three may be used appended to \text (i.e. \textrm{}, \textbf{}) for the same effect as if you used the stand-alone version and then \text{}, and only apply to a target-argument. All of these are also applicable at the end of a function "\math__" and will also apply to a target-argument. The sizes can either be left stand-alone to apply to their respective line, or be fed a target-argument for what should be made bigger. There are even larger sizes, but I find them overkill. The names are intuitive, though. Anything as "huge" as LARGE is absurd. The color command takes a color argument in either hexadecimal or nominal format (\color{#FF0000}{text} and \color{red}{text} do the same thing). As a test of our newfound skills, let's try to make a (considerably fancy) display for the slope-intercept form of lines! We would expect this to fail miserably, typing it directly into LaTeX: $$\quad \boxed{ \quad \color{green}{\Rightarrow} \quad \text{Slope-intercept form: y = mx + b} \qquad }$$ $$\quad \boxed{ \quad \color{lime}{\Leftarrow} \quad Slope-intercept form: y = mx + b \qquad }$$ And it does. Now, let's surround the "Slope-intercept form: " in a \text{} command. $$\quad \boxed{ \quad \color{green}{\Rightarrow} \quad \text{\text{Slope-intercept form: } y = mx + b} \qquad }$$ $$\quad \boxed{ \quad \color{lime}{\Leftarrow} \quad \text{Slope-intercept form: } y = mx + b \qquad }$$ There we go! Finally, we'll make it bold by replacing \text with \textbf and add a \quad between the text and formula for horizontal spacing. $$\quad \boxed{ \quad \color{green}{\Rightarrow} \quad \text{\textbf{Slope-intercept form: } \quad y = mx + b} \qquad }$$ $$\quad \boxed{ \quad \color{lime}{\Leftarrow} \quad \textbf{Slope-intercept form: } \quad y = mx + b \qquad }$$ We're done! Looks a lot nicer than just writing it out, which would take way too little time! Well, let's end this section off and move onto the final part: Environments!
4. AccessDenied
$$\color{orange}{\cdot} \quad \text{Eco-friendly / Safe Environments}$$ So, there's just one more thing to cover, called "environments." The name may sound a little funny, but I assure you that these are anything but lame! Environments allow for special conditions to be applied to the LaTeX space. There are two very useful ones. The first one is called "align." The alignment environment allows for the use of a special operator "&" that forces each line to line up at these points. Such an example for this environment could be a solution to the equation "2x + 3 = 9." \quad \boxed{ \begin{align} \quad & \color{green}{\Rightarrow}\quad \text{\begin{align}} \qquad \\ & \color{green}{\Rightarrow} \quad \ \text{ 2x + 3 &= 9} \qquad \\ & \color{green}{\Rightarrow} \quad \ \text{ 2x &= 6} \qquad \\ & \color{green}{\Rightarrow} \quad \ \text{ x &= 3} \qquad \\ & \color{green}{\Rightarrow} \quad \text{\end{align}} \qquad \end{align} } \quad \boxed{ \begin{align} \quad & \color{lime}{\Leftarrow} \quad & 2x + 3 &= 9 \qquad \\ & \color{lime}{\Leftarrow} & \quad 2x &= 6 \qquad \\ & \color{lime}{\Leftarrow} & \quad x &= 3 \qquad \end{align} } Notice that format. It starts with a "begin{align}" and ends with a "end{align}." This is how the environments will always be declared. It is mostly the same way with the array environment. Instead of "align", you just have the "array." However, the Array environment uses an extra parameter after the "begin{array}" part. We have to declare the alignment of each element in the table. We may optionally specify vertical separation. Also, the environment supports a command called $$\textbf{\hline}$$ to make horizontal lines. The possibilities are: "l", "c", and "r." They stand for "left," "center," and "right alignment, respectively. So, to write the array header for a two-column array with all elements aligned to the right and no vertical bars, we'd use "begin{array}{rr} ... \end{array}." Similarly, if we wanted to add a vertical bar between the first and second item of each row, we'd use "begin{array}{r|r}." I will not create another example of arrays for the sake of making sure the post isn't going to explode. Instead, I want to give you the very plaintext version of my tutorial that includes a few uses of arrays already, so that you may see how it was all written as well! Enjoy! This will conclude the tutorial! I hope you have learned a thing or two on making LaTeX. Just remember that your creativity will drive the effectiveness of what you do with this. Personally, I thought the diagram of a rectangle I made with arrays was fairly nifty. $$% Disclaimer: I love you all very much for checking out the guide %$$ Plaintext Source; Tutorial: http://pastebin.com/raw.php?i=v8GUrZi5
5. zepp
AccessDenied had lot of issues when writing the tutorial.. but he did it! *Applauses* ;D
6. AccessDenied
two revisions and a weird error concerning environments going after LaTeX
7. alexwee123
wall of text didn't read :D
8. Hero
I scanned through it in about a minute. Good info.
9. asnaseer
Great stuff AccessDenied. There is a $$\LaTeX$$ practising group where you could post this as well: http://openstudy.com/study?login#/groups/LaTeX%20Practicing!%20%3A)
10. Hero
They should feature these tutorials on the main page along with creating a special group. All of the tutorials so far have been pretty impressive.
11. AccessDenied
Thanks. I don't think I made any more mistakes after revising again, but I hope if I did, people can catch them. The biggest deal making it was just writing the latex in latex. There was a lot of weird issues that would come up doing that. :P There was also a weird thing that happened when I tried to type the environment declarations after I introduced an independent LaTeX string before it, where the environments would become LaTeX delimiters outside of the regular ones. D:
12. KingGeorge
Very impressive. And very helpful as well. Excellent job.
13. anonymous
just when i thought ishy and myin were the best in latex you proved me wrong
14. KingGeorge
15. Hero
I'm nitpicking, but you should have stated at the beginning of your tutorial that right arrow means input, and left arrow means output. @lgbasallote, there are several people who are really good at Latex, not just those you have specified.
16. Hero
Belay my last
17. anonymous
fine..hero is good in latex too...i know you wanted to be mentioned :P haha jkjk
18. Hero
I didn't say it was me :P
19. AccessDenied
Oh, its actually just if there are closed LaTeX strings anywhere in the post, \begin{} and \end{} become delimiters for their own LaTeX environments. Yeah, that might be a good idea for the future. Most ideas came as the thing progressed, those arrowed boxes being one of them. :D
20. AccessDenied
What! Fine then. Anytime.
21. anonymous
i hope with these the number of ambiguous questions would drop
22. AccessDenied
Notes for improvements in possible future revisions * Fix mis-spacing in title (L 1) * Fix qquad position (L 58) (should precede closing bracket) * Add in the diagram i referred to, (L 130) * . Add introductory note for used notations * C: possibly more elaboration in the environments section * C: Could probably add an example for arrays, it wasn't too bad to post as predicted * C: possibly more explanation of the line break, wasnt emphasized very well. * Add sources or helpful sites to check out: --> http://en.wikibooks.org/wiki/LaTeX/Mathematics (Originally taught me a lot of the possible commands, also found some interesting commands recently) --> http://insti.physics.sunysb.edu/latex-help/ltx-176.html (Text size, usable fonts)
23. AccessDenied
I bump in hopes that somebody does catch any mistakes, would help in making a revision that I don't miss something. D:
24. AccessDenied
I'm assuming that most of the things to tweak have been plucked out now, so I can probably just adjust and post in the other section as mentioned earlier. D:
25. TheViper
$\Huge{\color{red}{\mathbb{LaTex}}}$
26. TheViper
wow!
27. TheViper
I didn't know that I can do like that in the Equation Editor!
28. TheViper
Now, I too love $\Huge{\mathbb{\LaTeX}}$
29. TheViper
So much interesting!!
30. TheViper
I can't tell in words that how interesting is LaTeX!
31. KingGeorge
not to be rude, but it would be nice if you were to take your practicing to the $$\LaTeX$$ practice group at openstudy.com/study#/groups/LaTeX instead of repeatedly posting here.
32. TheViper
$\Huge{\color{red}{\rightarrow \boxed{\mathbb{\text{I loveLaTeX}}}}}$
33. TheViper
$\Huge{\color{orange}{\star \star}{\text{{Very Excellent Job}}}}$
34. TheViper
So useful! Thanx so much!
35. TheViper
I can't thank u in my words! I m so happy to know it! & thanx @ParthKohli for telling me this $\Huge{\color{red}{\star}{Tutorial}}$
36. anonymous
$\Huge \color{blue}{nice}\color{red} {tutorial}$ !
37. AccessDenied
$$\large \frak{\text{Thanks, glad you learned from it!}}$$ :) I should start on a second version, so I can add more / fix some of those things.
38. mathslover
$\huge{\frak{Hey, nice}\mathsf{Tutorial}\mathbb{Great!!!!}}$
39. anonymous
how you guys do it?
40. mathslover
41. TheViper
Wait if u want to know how they do just right click on LaTeX & then click show math as then click text commands @ghazi :)
42. TheViper
like
43. TheViper
After clicking on text commands a box will appear then copy everything in the box & paste in equation editor @ghazi :)
44. TheViper
@ghazi does that helped ??
45. anonymous
@TheViper that was really helpful...thanks dude....i am grateful to this ....i shall make use of it...
46. TheViper
$$\Huge{\color{green}{\ddot{\smile}}}$$
47. TheViper
wait @ghazi :)
48. anonymous
wow !!
49. TheViper
If you want to write anything in 'Code Block' you have write between these :- |dw:1347024632753:dw|Like this :- http://assets.openstudy.com/updates/attachments/503dfd2ce4b0074824ff598c-theviper-1346930814196-untitled.png It will be shown as :- GHAZI
50. anonymous
cool....let me try this
51. TheViper
try :)
52. anonymous
|dw:1346936212442:dw|
53. anonymous
it didn't work :(
54. TheViper
sorry u should not type in drawing :)
55. TheViper
work as in the attachment :)
56. TheViper
57. TheViper
try now :)
58. anonymous
okay wait a sec
59. TheViper
fast
60. TheViper
61. anonymous
like this ,,,,,,,,,,, Type here ,,,, Viper
62. anonymous
damn i am too stupid at this
63. anonymous
by the way thanks for your time dude :)
64. TheViper
hey @ghazi its not apostrophe its this
65. TheViper
& u must type it only 3 times like this :-
66. anonymous
@the viper thanks man
67. TheViper
ok try here:)
68. anonymous
$$\checkmark$$
69. anonymous
Thanks soo much @AccessDenied :D
70. jiteshmeghwal9
$\huge{\color{red}{\frak{So \space \text{great} \space tutorial \space}\mathbb{Thanx !!!!!!}}}$ @AccessDenied :)
71. blackops2luvr
$$\bbox [ 15pt, #000033 ,border: 15px solid #aa8866 ] {\Huge\sf\color{white} {\ hi \ there \ :D }}$$
|
2016-10-24 07:22:03
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9735401272773743, "perplexity": 1818.3983468938843}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719542.42/warc/CC-MAIN-20161020183839-00203-ip-10-171-6-4.ec2.internal.warc.gz"}
|
https://www.physicsforums.com/threads/inverse-trig.345529/
|
# Inverse trig?
"Inverse" trig?
I'm trying to think of a word that describes the opposite of a trig function.
It's like inverse, but I know that's not exactly the right word because inverse means literally "1 over" the function.
It's the type of function you get when you press "2nd function" before you press Sin or Cos or whatever.
I'm pretty sure it starts with "i" ... does anyone know what I'm talking about?
I think you are looking for the word "inverse" and confusing notation with -1. 1 over something is usually called the reciprocal and is also indicated by a -1 exponent (x-1 = 1/x).
Sometimes a -1 does mean inverse. The inverse of sine (with a restricted domain) is arcsin x = sin-1x (sin x)-1 = 1/sinx.
ahhhh okay yeah I always confuse those two
|
2021-09-26 13:54:06
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8465458750724792, "perplexity": 1071.8042913442823}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00386.warc.gz"}
|
https://nano.ufop.br/publications/year/2017
|
# Publications
2017
A. M. Garcia, et al., “Second harmonic generation microscopy as a cancer diagnosis tool,” in 2017 European Conference on Lasers and Electro-Optics and European Quantum Electronics Conference, 2017, pp. JSIII_1_3. Publisher's VersionAbstract
Optical microscopy is one of the most important tool for understanding biology processes. Recently with the advance of femtosecond laser all the nonlinear optical processes have been included into optical microscopy methods and Second Harmonic Generation (SHG) microscopy has emerged as a powerful new optical imaging tool with applications in disease diagnostics $1-2$. The gold standard'' in cancer diagnostics is still the traditional histology analysis where accuracy depends on the experience and interpretation skill of the pathologist. A major development would be to use the SHG microscopy as a quantitative tool to cancer diagnosis. Here we show SHG imaging results of the collagen fibers in prostate cancer tissue that can be directly correlated with the cancer malignancy diagnostic $3$. We performed SHG imaging in a back-scattering geometry on the histological slides from the same biopsies that were analyzed by the pathologist. We studied prostate from patients treated at the the Urology Center of UFMG Hospital, Belo Horizonte. A 1 mm diameter punch biopsy was extracted from multiple peripheral zone of the prostate showing normal tissue and cancer tissue with Gleason scores ranging from 3 to 5. The study was approved by the UFMG Institutional Review Board and the Brazilian National Health Committee on the use of humans as experimental subjects. Written informed consent was obtained from all participants before their biopsy procedures. Figure 1 shows SHG images for normal and cancer tissue. The SHG images show major differences on the collagen fiber alignment that changes with cancer progression. The average direction of the fibers in the image was calculated and we obtained a value for the fiber anisotropy $4$. The statistical analysis is presented in the boxplot in figure 1. Note that both the average values (crosses) and the median lines (black center lines) are well separated for the normal and cancer tissue.
L. G. P. Martins, et al., “Raman evidence for pressure-induced formation of diamondene,” Nature Communications, vol. 8, no. 1, pp. 96, 2017. Publisher's VersionAbstract
Despite the advanced stage of diamond thin-film technology, with applications ranging from superconductivity to biosensing, the realization of a stable and atomically thick two-dimensional diamond material, named here as diamondene, is still forthcoming. Adding to the outstanding properties of its bulk and thin-film counterparts, diamondene is predicted to be a ferromagnetic semiconductor with spin polarized bands. Here, we provide spectroscopic evidence for the formation of diamondene by performing Raman spectroscopy of double-layer graphene under high pressure. The results are explained in terms of a breakdown in the Kohn anomaly associated with the finite size of the remaining graphene sites surrounded by the diamondene matrix. Ab initio calculations and molecular dynamics simulations are employed to clarify the mechanism of diamondene formation, which requires two or more layers of graphene subjected to high pressures in the presence of specific chemical groups such as hydroxyl groups or hydrogens.
J. Nichele, A. B. de Oliveira, L. S. B. de Alves, and I. B. Jr, “Accurate calculation of near-critical heat capacities \CP\ and \CV\ of argon using molecular dynamics,” Journal of Molecular Liquids, vol. 237, pp. 65 - 70, 2017. Publisher's VersionAbstract
Abstract Molecular dynamics (MD) employing the Lennard-Jones (LJ) interaction potential was used to compute the heat capacities of argon at constant volume \CV\ and constant pressure \CP\ near the critical point very close to the asymptotic region. The accurate \MD\ calculation of critical divergences was shown to be related to a careful choice of the cutoff radius rc and the inclusion of long-range corrections in the \LJ\ potential. The computed \CP\ and \CV\ values have very good agreement as compared to available \NIST\ data. Furthermore, values of \CV\ in a range of temperatures for which \NIST\ data is not available could be computed. In the investigated range of temperatures, both \CP\ and \CV\ \MD\ results were fitted to a simple mathematical expression based on an empirical model that describes the critical effects when the asymptotic models are not appropriate. The present approach is of general applicability and robust to compute thermophysical properties of fluids in the near-critical region.
E. E. de Moraes, M. D. Coutinho-Filho, and R. J. C. Batista, “Transport Properties of Hydrogenated Cubic Boron Nitride Nanofilms with Gold Electrodes from Density Functional Theory,” ACS Omega, vol. 2, no. 4, pp. 1696-1701, 2017. Publisher's Version
A. F. Versiani, et al., “Multi-walled carbon nanotubes functionalized with recombinant Dengue virus 3 envelope proteins induce significant and specific immune responses in mice,” Journal of Nanobiotechnology, vol. 15, no. 1, pp. 26, 2017. Publisher's VersionAbstract
Dengue is the most prevalent arthropod-borne viral disease in the world. In this article we present results on the development, characterization and immunogenic evaluation of an alternative vaccine candidate against Dengue.
J. A. Gonçalves, R. Nascimento, M. J. S. Matos, A. B. de Oliveira, H. Chacham, and R. J. C. Batista, “Edge-Reconstructed, Few-Layered Graphene Nanoribbons: Stability and Electronic Properties,” The Journal of Physical Chemistry C, vol. 121, no. 10, pp. 5836-5840, 2017. Publisher's VersionAbstract
We report a first-principles study of edge-reconstructed, few-layered graphene nanoribbons. We find that the nanoribbon stability increases linearly with increasing width and decreases linearly with increasing number of layers (from three to six layers). Specifically, we find that a three-layer 1.3 nm wide ribbon is energetically more stable than the C60 fullerene, and that a 1.8 nm wide ribbon is more stable than a (10,0) carbon nanotube. The morphologies of the reconstructed edges are characterized by the presence of five-, six-, and sevenfold rings, with sp3 and sp2bonds at the reconstructed edges. The electronic structure of the few-layered nanoribbons with reconstructed edges can be metallic or semiconducting, with band gaps oscillating between 0 and 0.28 eV as a function of ribbon width.
|
2022-11-26 22:05:10
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.504656195640564, "perplexity": 2415.5662012315033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00619.warc.gz"}
|
https://socratic.org/questions/56bd0e617c01490a84e3218a#225061
|
# Question #3218a
Feb 12, 2016
The above reaction shows photosynthesis .
#### Explanation:
$6 C {O}_{2} + 6 {H}_{2} O +$ sunlight & chlorophyll $\rightarrow {C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2}$ Here,
${C}_{6} {H}_{12} {O}_{6}$ is Glucose. This is the basic energy source for plants. And Plants release ${O}_{2}$ (oxygen) when the complex process of
photosynthesis ends.
Photosynthesis is a complex multi step process in which green plants and some bacteria (basically AUTOTROPHS) use the light energy to make food (Here Glucose in plants) along with the release of oxygen.
|
2022-01-29 02:12:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.542927622795105, "perplexity": 9997.892142570423}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00190.warc.gz"}
|
https://blog.stata.com/tag/simulation/page/2/
|
### Archive
Posts Tagged ‘simulation’
We often use probit and logit models to analyze binary outcomes. A case can be made that the logit model is easier to interpret than the probit model, but Stata’s margins command makes any estimator easy to interpret. Ultimately, estimates from both models produce similar results, and using one or the other is a matter of habit or preference.
I show that the estimates from a probit and logit model are similar for the computation of a set of effects that are of interest to researchers. I focus on the effects of changes in the covariates on the probability of a positive outcome for continuous and discrete covariates. I evaluate these effects on average and at the mean value of the covariates. In other words, I study the average marginal effects (AME), the average treatment effects (ATE), the marginal effects at the mean values of the covariates (MEM), and the treatment effects at the mean values of the covariates (TEM).
First, I present the results. Second, I discuss the code used for the simulations.
Results
In Table 1, I present the results of a simulation with 4,000 replications when the true data generating process (DGP) satisfies the assumptions of a probit model. I show the Read more…
Categories: Statistics Tags:
## Understanding the generalized method of moments (GMM): A simple example
$$\newcommand{\Eb}{{\bf E}}$$This post was written jointly with Enrique Pinzon, Senior Econometrician, StataCorp.
The generalized method of moments (GMM) is a method for constructing estimators, analogous to maximum likelihood (ML). GMM uses assumptions about specific moments of the random variables instead of assumptions about the entire distribution, which makes GMM more robust than ML, at the cost of some efficiency. The assumptions are called moment conditions.
GMM generalizes the method of moments (MM) by allowing the number of moment conditions to be greater than the number of parameters. Using these extra moment conditions makes GMM more efficient than MM. When there are more moment conditions than parameters, the estimator is said to be overidentified. GMM can efficiently combine the moment conditions when the estimator is overidentified.
We illustrate these points by estimating the mean of a $$\chi^2(1)$$ by MM, ML, a simple GMM estimator, and an efficient GMM estimator. This example builds on Efficiency comparisons by Monte Carlo simulation and is similar in spirit to the example in Wooldridge (2001). Read more…
Categories: Statistics Tags:
## Efficiency comparisons by Monte Carlo simulation
Overview
In this post, I show how to use Monte Carlo simulations to compare the efficiency of different estimators. I also illustrate what we mean by efficiency when discussing statistical estimators.
I wrote this post to continue a dialog with my friend who doubted the usefulness of the sample average as an estimator for the mean when the data-generating process (DGP) is a $$\chi^2$$ distribution with $$1$$ degree of freedom, denoted by a $$\chi^2(1)$$ distribution. The sample average is a fine estimator, even though it is not the most efficient estimator for the mean. (Some researchers prefer to estimate the median instead of the mean for DGPs that generate outliers. I will address the trade-offs between these parameters in a future post. For now, I want to stick to estimating the mean.)
In this post, I also want to illustrate that Monte Carlo simulations can help explain abstract statistical concepts. I show how to use a Monte Carlo simulation to illustrate the meaning of an abstract statistical concept. (If you are new to Monte Carlo simulations in Stata, you might want to see Monte Carlo simulations using Stata.) Read more…
Categories: Statistics Tags:
## Monte Carlo simulations using Stata
Overview
A Monte Carlo simulation (MCS) of an estimator approximates the sampling distribution of an estimator by simulation methods for a particular data-generating process (DGP) and sample size. I use an MCS to learn how well estimation techniques perform for specific DGPs. In this post, I show how to perform an MCS study of an estimator in Stata and how to interpret the results.
Large-sample theory tells us that the sample average is a good estimator for the mean when the true DGP is a random sample from a $$\chi^2$$ distribution with 1 degree of freedom, denoted by $$\chi^2(1)$$. But a friend of mine claims this estimator will not work well for this DGP because the $$\chi^2(1)$$ distribution will produce outliers. In this post, I use an MCS to see if the large-sample theory works well for this DGP in a sample of 500 observations. Read more…
Categories: Programming Tags:
## How to simulate multilevel/longitudinal data
I was recently talking with my friend Rebecca about simulating multilevel data, and she asked me if I would show her some examples. It occurred to me that many of you might also like to see some examples, so I decided to post them to the Stata Blog. Read more…
Categories: Statistics Tags:
|
2021-01-27 14:07:49
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6961454749107361, "perplexity": 570.5788954085285}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704824728.92/warc/CC-MAIN-20210127121330-20210127151330-00277.warc.gz"}
|
https://blender.stackexchange.com/questions/22987/mirror-modifier-mirrors-on-different-axis-axes-from-the-selected-ones?noredirect=1
|
# Mirror modifier mirrors on different axis/axes from the selected one(s)
I have 2 objects a face and an ear. Mirror modifier works fine for the face. But when I use mirror modifier to duplicate the ear on X axis, it duplicates on Z axis. And when I use mirror modifier on Z axis it mirrors on X axis. How to correct the axis orientation?
• Have you rotated the ear? If so, try applying the rotation. In object mode Ctrl A > R. – user7952 Jan 10 '15 at 13:06
• Thank You. It works fine after rotation. Please write this comment in answer section. I will mark that as an answer. – quartz Jan 10 '15 at 14:28
CtrlA then R or click Object > Apply > Rotation.
|
2019-08-20 08:17:24
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2363654226064682, "perplexity": 2151.1474573047553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315258.34/warc/CC-MAIN-20190820070415-20190820092415-00541.warc.gz"}
|
https://testbook.com/question-answer/average-speed-and-average-velocity-represent-the-s--61c6232f634d3bb56304d17b
|
Average speed and average velocity represent the speed of the object:
This question was previously asked in
RRC Group D Previous Paper 62 (Held On: 9 Oct 2018 Shift 1)
View all RRB Group D Papers >
1. At a particular point in time
2. In long distance
3. In short distance
4. During the given interval
Answer (Detailed Solution Below)
Option 4 : During the given interval
Detailed Solution
The correct answer is During the given interval.
• Average speed and average velocity represent the speed of the object during the given interval.
Key Points
• Average speed
• The average speed of a particle for a given interval of time is defined as the ratio of total distance travelled to the total time taken.
• $$Average speed = {Total distance \over total time taken}$$
• It is a scalar quantity.
• Average velocity:
• It is defined as the ratio of total displacement to the total time taken in a given interval of time.
• It is a vector quantity.
|
2022-05-23 03:29:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8852750658988953, "perplexity": 1720.5106848866453}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662552994.41/warc/CC-MAIN-20220523011006-20220523041006-00292.warc.gz"}
|
https://www.gktoday.in/aptitude/pipe-p-and-q-can-fill-a-tank-in-10-and-12-hours-respectively-and-c-can-empty-it-in-6-hours-if-all-the-three-are-opened-at-7-am-at-what-time-will-one-fourth-of-the-tank-be-filled/
|
Pipe P and Q can fill a tank in 10 and 12 hours respectively and C can empty it in 6 hours. If all the three are opened at 7 am, at what time will one-fourth of the tank be filled?
[A]10 am
[B]11 am
[C]10 pm
[D]11 pm
10 pm
Part of tank filled in 1 hour when all three pipes are opened
$= \frac{1}{10}+\frac{1}{12}-\frac{1}{6}$
$= \frac{6+5-10}{60} = \frac{1}{60}$
∴ The tank will be filled in 60 hours.
∴ One fourth of the tank will be filled in 15 hours so the tank will be filled at 10 pm.
Hence option [C] is correct answer.
|
2021-04-20 21:03:39
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8250098824501038, "perplexity": 687.3553903552912}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00104.warc.gz"}
|
http://www.csd.uwo.ca/~moreno/CS447/Lectures/Lexical.html/node16.html
|
Next: About this document ... Up: The design of an efficient Previous: The design of an efficient
### Minimal automaton
EQUIVALENCE OF WORDS MODULO A LANGUAGE. Let L be a language over the alphabet . Let u, v be two words (that may belong or not to the language L). We say that the language L SEPARATES u et v if
ou (14)
We say that u et v are equivalent modulo L and we write u v mod L if for every w the language L does not separate the words u w and v w. In other words we have
u v mod L (w ) u w L v w L (15)
The relation u v mod L is an equivalence relation (since it is reflexive, symmetric and transitive) and thus defines equivalence classes called the equivalence classes (or residue classes) associated with L.
EQUIVALENCE OF WORDS MODULO AN AUTOMATON. Let = (, S, i, F,) be a DFA. Let u, v be two words We say that u et v are equivalent modulo and we write u v mod if after reading u the DFA is in the same state than after reading v. Let us denote by (iu) and (i, v) the state of the automaton after reading u and v respectively. Then we have
u v mod (i, u) = (i, v). (16)
Again the relation u v mod is an equivalence relation and thus defines residue classes associated with the automaton .
Example 6 The DFA over = {a, b} on Figure 20 has three residue classes:
• that of the words with no a,
• that of the words with no b after an a,
• that of the words with a factor ab.
Proposition 8 Let L be a language over recognized by the DFA = (, S, i, F,). For every words u and v we have
u v mod u v mod L (17)
Remark 4 Proposition 8 expresses the fact that every residue class modulo the automaton is entirely contained in a residue class modulo its associated language L. Therefore the number of classes modulo L is upper bounded by the number of classes modulo and thus by the number of states of . It follows from the following theorem that having a finite number of residue classes is a necessary and sufficient condition for a language to be recognized by FA.
Theorem 4 Let L be a language over the alphabet . Let n be a positive integer. If there exist exactly n residue classes modulo L then there exists a DFA with n states that recognizes L.
MINIMAL AUTOMATON. Let L be a language over the alphabet
• recognized by FA and
• with n residue classes.
It follows from Proposition 8 and Theorem 4 that
• every DFA recognizing L has at least n states,
• there exists a DFA with n states recognizing L.
Therefore we can call minimal automaton every DFA with n states and recognizing L.
Remark 5 The number of residue classes of a language is rarely known (until one gets a minimal DFA accepting this language). So one needs a way to characterize minimal automata.
Then one needs an algorithm to compute from a given DFA (minimal or not) recognizing L a minimal automaton recognizing L too. Such an algorithm will group states that do essentially the same job. This leads to the following notions.
EQUIVALENCE OF STATES IN A DFA. Let = (, S, i, F,) be a DFA. We say that two states p, q S are equivalent and we write p q if for every word w we have
((p, w) F)((q, w) F). (18)
Let k be a non-negative integer. We say that two states p, q S are equivalent at order k and we write p q if for every word w with length | w |
| w | k ((p, w) F) ((q, w) F) (19)
CHARACTERIZATION OF A MINIMAL AUTOMATON. Among other properties, Theorem 5 states that for a minimal automaton, each residue class of states contains only one state.
Theorem 5 Let L be a language recognized by FA and let n be its number of residue classes. Let = (, S, i, F,) be a minimal DFA recognizing L. Then we have the following properties
• for every words u, v we have u v mod u v mod L.
• for every q F there exists a word u such that (i, u) = q.
• for every states p, q S we have p q p = q.
CONSTRUCTION OF THE EQUIVALENCE CLASSES FOR STATES. Proposition 9 suggests an algorithm for constructing the equivalence classes of states.
Proposition 9 Let = (, S, i, F,) be a DFA and k be a non-negative integer. If the relations and have the same residue classes then and have the same residue classes too.
The method is follows.
• Compute which consists of two classes:
• the set of the final states,
• the set of the non final states.
• Compute and set k to 0
• Whhile and are different and set k to k + 1.
In practice one construct a table that gives the transition from each state after reading any word of length 0, any word of length 1, any word of length 2, etc... In these tables we underline final states in order to detect classes more easily. Quite often computations can be simpler. For instance, in Example 7 computations can be stopped after computing the third column. Explain why!
THE CONSTRUCTION OF A MINIMAL AUTOMATON = (, S', i', F',) from a given DFA = (, S, i, F,) can be down as follows (where the class of a state s S is denoted by ).
• Compute the equivalence classes of states. From Theorem 5 they provide S', also denoted by .
• Let s, s' S such that = . Let x . We have = . Hence we can define (, x) as
(, x) = (20)
• We choose i' = .
• We define F' = { | s F}.
Example 7
a b aa ab ba bb 1 2 1 2 1 2 1 2 2 2
Example 8
a b aa ab ba bb 1 2 6 7 7 2 7 1 2 6 1 5 8 6 7 7 6 7 1 7 7 7 5 7 5 8 6 8 7
Next: About this document ... Up: The design of an efficient Previous: The design of an efficient
Marc Moreno Maza
2004-12-02
|
2017-12-11 16:59:15
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9117552042007446, "perplexity": 880.8550454233618}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948513784.2/warc/CC-MAIN-20171211164220-20171211184220-00129.warc.gz"}
|
http://piping-designer.com/index.php/disciplines/mechanical/stationary-equipment/pipe/511-yield-strength
|
# Yield Strength
Written by Matt Milbury on . Posted in Pipe
Yield strength, abbreviated as $$\sigma$$ (Greek symbol sigma), also known as yield stress, is the minimum stress that leads to permanent deformation of the material.
### Yield Strength formula
$$\large{ \sigma = \frac{F}{A} }$$
Where:
$$\large{ \sigma }$$ (Greek symbol sigma) = yield strength
$$\large{ A }$$ = cross-section area
$$\large{ F }$$ = force
Solve for:
$$\large{ A = \frac{F}{\sigma} }$$
$$\large{ F = \sigma \; A }$$
|
2019-11-22 01:02:13
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9107237458229065, "perplexity": 10469.585897086818}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496671053.31/warc/CC-MAIN-20191121231600-20191122015600-00449.warc.gz"}
|
http://codeforces.com/blog/entry/60073
|
Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. ×
### Jasperr's blog
By Jasperr, history, 2 years ago, ,
Hi! After releasing CHelper Companion as an alternative to the original CHelper extension six months ago, development has been ongoing and CHelper Companion has "evolved" into Competitive Companion.
Competitive Companion is a browser extension for Chrome and Firefox that can parse problems and contests and extract the useful data. This includes things like time/memory constraints and sample test cases. While CHelper Companion was only capable of sending this data to CHelper, Competitive Companion works with pretty much all popular tools. Next to that, Competitive Companion is also capable of sending the data to your private tools, without requiring a complex setup.
Using it is very simple. Just install the extension, make sure you got one of the supported tools open, navigate to a problem page and click on the green plus icon in the top-left of your browser. On websites where Competitive Companion is able to parse contests, navigating to the page with all the problems and clicking on the green plus icon will download and parse all problems in the contest at once. You can also use the shortcut Ctrl+Shift+U instead of the green plus icon.
At the moment, Competitive Companion can send parsed problem data to the following tools:
- CHelper
- JHelper
- Hightail
- Mind Sport
- Caide
- acmX
- Competitive Programming Helper
- CP Editor
- AI Virtual Assistant
- Any other tool that can parse the data that is being sent, read more about that in the README on GitHub, and have a look at the example project here.
Competitive Companion includes parsers for the following websites:
Website Problem parser Contest parser
ACMP
Aizu Online Judge
Anarchy Golf
AtCoder
Baekjoon Online Judge
Bloomberg CodeCon
CodeChef
Codeforces
CodeMarshal
COJ
CSES
CSU-ACM Online Judge
DevSkill
DMOJ
E-Olymp
ECNU Online Judge
FZU Online Judge
HackerEarth
HackerRank
HDOJ
HIT Online Judge
hihoCoder
Hrbust Online Judge
ICPC Live Archive
Jutge
Kattis
Library Checker
LibreOJ
LightOJ
Luogu
mrJudge
MSK Informatics
NowCoder
NYTD Online Judge
omegaUp
Panda Online Judge
PEG Judge
POJ
QDUOJ
SPOJ
SSOIER
Timus
Toph
UOJ
URI Online Judge
USACO
USACO Training
UVa Online Judge
Virtual Judge
Yandex
yukicoder
Competitive Companion's source code can be found on GitHub. Releases are pushed to both the Chrome Web Store and Mozilla Add-Ons.
Feedback, bug reports and feature requests are all very welcome!
• +47
» 2 years ago, # | +3 Feature Request I think adding a keyboard shortcut for parsing problem would be great.
• » » 2 years ago, # ^ | +4 Great idea, I released an update which adds the keyboard shortcut Ctrl+Shift+U to parse problems/contests, it should be available in a few minutes. Whenever the green plus icon is visible, that shortcut will work too.
• » » » 2 years ago, # ^ | +1 Awesome job.
» 2 years ago, # | ← Rev. 2 → 0 UPD: OK. I have found the reason, My CHelper plugin was not updated (it was 4.1.12)... After updating the plugin it is working properly. Thank you, You have done awesome job...Is there any version dependency with the add-ons? It is not working for me, while parsing this problem. :(I am using, Chrome (Version 67.0.3396.87 (Official Build) (64-bit)) and Intellij IDEA (2018.1).
» 2 years ago, # | 0 Forgive my ignorance but I'm new to this tool.What exactly happens when we click on the Green Button on a contest page? I can see a green bar sliding across the top, guessing this is the page being parsed? But what now? How do we use it now?
• » » 2 years ago, # ^ | ← Rev. 2 → 0 I'm assuming you are seeing a blue bar, not a green one (because if it's green, it's probably not from my extension).When you see that bar, it is retrieving all the problems and extracting the useful data out of all of them, after which it sends them to one of the supported tools.To make use of this extracted data, you need to do one of the following:- Use one of the supported tools (see the post to see which one are supported).- Add support for the extension to your tool, which requires quite a bit more work. Instructions on how to do that can be found here. If you need help with getting up and running using one of the supported tools, please tell me which one you are using so I can give more specific instructions.
» 22 months ago, # | ← Rev. 2 → +10 I have just released version 2.3.0, which is one of the biggest updates so far. It includes several bug fixes and parsers for 11 more online judges, including two that I have been wanting to add for a long time: UVa Online Judge and ICPC Live Archive.I do want to clarify two things regarding this update though:1. The extension's size has gone up from 35kb to 660kb. This has to do with a new library, PDF.js, that I needed to add in order to be able to extract the sample testcases from the PDF files on UVa Online Judge and ICPC Live Archive.2. The extension will now request permission to "access your data for all websites". This does not mean that the extension spies on you: it will never do that. If you don't trust me, feel free to take a look at the code on GitHub. This change was necessary in order to better support sites like HackerRank which use the browser's history api to let you navigate over their website instead of using normal links.
» 20 months ago, # | 0 For some reason using this extension with CHelper, it parses task fine. However when I run my code as I normally do, I get following: Exception in thread "main" java.lang.ClassNotFoundException: my.ETrialForChief at java.net.URLClassLoader.findClass(URLClassLoader.java:381) at java.lang.ClassLoader.loadClass(ClassLoader.java:424) at sun.misc.Launcher\$AppClassLoader.loadClass(Launcher.java:339) at java.lang.ClassLoader.loadClass(ClassLoader.java:357) at java.lang.Class.forName0(Native Method) at java.lang.Class.forName(Class.java:265) at net.egork.chelper.tester.NewTester.test(NewTester.java:81) at net.egork.chelper.tester.NewTester.main(NewTester.java:34) I do not have this issue with tasks created by the CHelper chrome plugin, however those do not parse the testcases correctly ... Does anybody have any advice on what I am doing wrong? Thanks!
• » » 20 months ago, # ^ | 0 If anybody runs into similar problem, it was somehow because my Intellij ended up with "Build project automatically" unchecked. I followed steps here and now everything works great, thanks for the tremendous plugin! https://zeroturnaround.com/software/jrebel/quickstart/intellij/enable-automatic-compilation-in-intellij-idea/
• » » » 20 months ago, # ^ | 0 Try to update inteliji idea
» 19 months ago, # | 0 Can you add LeetCode please their ide is very slow
• » » 18 months ago, # ^ | +11 Unfortunately, LeetCode cannot be added. This is because LeetCode forces solutions to be written in a pre-defined method with a pre-defined signature (method parameters and return type), which is something that Competitive Companion does not support. The chances of it being implemented in the future is pretty low, considering how none of the publically available tools that Competitive Companion sends data to are capable of handling such constraints at the moment.Besides that, LeetCode doesn't seem to have any structure in their problem descriptions, which makes it pretty much impossible to parse something like the sample testcases.
• » » » 18 months ago, # ^ | 0 Got it Thanks man
» 19 months ago, # | 0 Can you add vjudge.net problems and contests?
• » » 18 months ago, # ^ | ← Rev. 3 → +8 I have just released version 2.5.0 which adds a problem parser for Virtual Judge. Your browser should auto-update within 24 hours.Unfortunately, I could not find a way to parse all problems in a contests at once due to how Virtual Judge renders contest problems dynamically. You can however parse all problems in a contest separately, just like how you would parse out-of-contest problems.It's also worth noting that while the problem parser can parse the sample testcases for a lot of problems, it will not work correctly for all of them. This is because Virtual Judge does not have a standarized format for problem descriptions, so the extension has to look for signs of sample testcases like
and tags. This isn't a rock-solid solution, but it works for the majority of the problems. Feel free to create an issue on GitHub when you find a problem that isn't parsed correctly.
» 12 months ago, # | 0 Hi Jasperr, I think it would be nice if we can set the name of taskClass to the problem'ID, but not the actual name of the it. For example, for this problem, taskClass can be named "1A", but not "ATheatreSquare".Thank you for this awesome tool.
• » » 12 months ago, # ^ | +5 Thanks for the suggestion, but I'm not planning on changing this. I think "AThreatreSquare" is better than "1A" when you're participating in a contest with multiple problems. Especially when having to differentiate between different problems having the full name instead of just the letter can be very helpful.
• » » » 12 months ago, # ^ | 0 I got it.Thank you for your respond.
» 9 months ago, # | 0 Any good parser for sublime text??
» 2 months ago, # | 0 Can you Also Add a Simple Video Demonstrating all its Features and How to use it Effectively and Possible Places where Users can Do Mistake while Parsing a Problem or Contest in CodeForces? By the Way, This is very Helpful Extension.Thanks for Creating it,Jasperr Bro.
• » » 2 months ago, # ^ | 0 I am not very good at making videos, and I think this paragraph in my post already explains how the extension works pretty clearly:"Using it is very simple. Just install the extension, make sure you got one of the supported tools open, navigate to a problem page and click on the green plus icon in the top-left of your browser. On websites where Competitive Companion is able to parse contests, navigating to the page with all the problems and clicking on the green plus icon will download and parse all problems in the contest at once. You can also use the shortcut Ctrl+Shift+U instead of the green plus icon."I did, however, record a short video two years ago for a PR I made to add Competitive Companion support to Hightail: https://www.youtube.com/watch?v=LnI4HuzWY18 for PR #107.
• » » » 2 months ago, # ^ | 0 Which Tools are available for Sublime Text IDE? Video helps Users to Implement Faster, So Recommended.
• » » » » 2 months ago, # ^ | 0 Which Tools are available for Sublime Text IDE? Not entirely sure, Competitive Companion just provides the parsing capabilities, all the tools listed in my post are external projects not made by me.A quick Google search does however show DrSchwad/FastOlympicCodingHook, which appears to add support for Competitive Companion to Jatana/FastOlympicCoding which is a Sublime Text 3 plugin for competitive programming. Besides that, both Hightail and Caide (both already listed in my post) are IDE-agnostic and can be used with any text editor or IDE.Do, however, note that I did not create nor use those tools, so I cannot provide support for them. Video helps Users to Implement Faster, So Recommended. I have already provided the usage by text and provided a demo video showing Competitive Companion being used with Hightail. It is really not a difficult extension to use, and another video is not going to make the instructions any more clear.
• » » » » » 2 months ago, # ^ | 0 Okay Got it!Thanks for Solving Query!
» 6 weeks ago, # | 0 Does anybody know a similar extension for Safari on MacOS?
» 3 weeks ago, # | 0 great toooool, just what is needed!!!!Thanks alot...For VSCode users install thr chrome extension of Competitive Companion, then install Compatitive Programming Helper extension from marketplace. And guess waht you are good to go. No setup required, no setting up path varibales.When on a problem on codeforces , press Ctrl + Shit + U, a new file opens in VSCode with test cases installed.. For manually writing test cases, press Ctr + Alt + B in VSCode for any file.
» 9 days ago, # | 0 2 small issues: 1. USACO Training requires Java submissions to have the class name of the taskid. Could you implement that? 2. I think Competitive Companion is having troubles with parsing the sample input and output for UVa. Could you fix that? Thanks!
• » » 7 days ago, # ^ | 0 I have just fixed the USACO Training issue you mentioned. I'll look into the UVa issue later this week because UVa seems to be down right now, and I'll publish a new release once that's done.
• » » » 7 days ago, # ^ | 0 Ok sure thanks! If UVa is giving you Joomla! This site is temporarily unavailable, it is an issue on their behalf. They seem to run a server check or something that causes their website to be down for around an hour at this time. Check again later. Thanks for your support though!
• » » » » 3 days ago, # ^ | ← Rev. 3 → 0 I have just published version 2.17.1 which fixes both of your reported issues, as well as two other bug fixes. Both the Chrome as well as the Firefox versions are currently pending review so it should be available in a few days.
|
2020-07-06 19:20:23
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2185821682214737, "perplexity": 2452.52162389197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655890181.37/warc/CC-MAIN-20200706191400-20200706221400-00126.warc.gz"}
|
https://optimization.mccormick.northwestern.edu/index.php?title=Facility_location_problems&diff=prev&oldid=4495
|
# Difference between revisions of "Facility location problems"
Author: Aaron Litoff
Stewards: Dajun Yue and Fengqi You
German economist Alfred Weber (1868-1958).
Facility location problems deal with selecting the placement of a facility (often from a list of integer possibilities) to best meet the demanded constraints. The problem often consists of selecting a factory location that minimizes total weighted distances from suppliers and customers, where weights are representative of the difficulty of transporting materials. The solution to this problem gives the highest profit choice that most efficiently serves the needs of all consumers.
## Background and History
The Fermat-Weber problem was one of the first facility location problems every proposed, and was done so as early as the 17th century. This "geometric median of three points" can be thought of as a version of the facility location problem where the assumption is made that transportation costs per distance are the same for all destinations. It was put forth by the French mathematician Pierre de Fermat to the Italian physicist Evangelista Torricelli as follows:
"Given three points in a plane, find a fourth point such that the sum of its distances to the three given points is as small as possible." [1]
In 1909 Alfred Weber used a three point version to model possible industrial locations in order to minimize transportation costs from two sources of materials to a single customer or market. [2] This formulation is one the simplest continuous facility location models.
The Fermat Point, or Torricelli Point, is the solution that minimizes distances from A, B, and C (the three corners of the black triangle).[3]
The Fermat-Weber problem is defined as:
Given finitely many distinct points $A_{1}, A_{2}, ..., A_{m}$ and positive multipliers $w_{1}, w_{2}, ..., w_{m}$ find a point $P \in \mathbb{R}^{n}$ that minimizes
$f(P)= \sum_{i=1}^{m} w_{i} \left \| P-A_{i} \right \|$
where $\left \| X \right \|$ denotes the Euclidean norm of $X \in \mathbb{R}^{n}$ i.e. $\left \| (x_{1},..,x_{n}) \right \| = \sqrt{x_{1}^{2}+...+x_{n}^2}$
The simplest version of this problem with all w=1 and n=2 gives the minimum distance in a flat plane.
A modern day engineering interpretation of Fermat's formulation could be as follows:
"Find the best location for a refining plant between three cities in such a way that the sum of the connections between the plant and the cities in minimal."
Although this problem was first solved geometrically by Torricelli in 1645 it did not have a direct numerical solution until Kuhn and Kuenne's iterative method was published over 300 years later in 1962! [4]
## Description and Formulation
In the uncapacitated facility location problem, the number of possible locations is now finite as we have discrete choices of where to build. These choices are governed by binary decision variables y.
The problem is often defined as a set of customers D, a set of facilities F, a fixed cost for opening each facility, and a variable cost for each facility. We are looking for the subset S of facilities that we should open and an assignment of D to S customers that will be serviced by each facility such that the sum of facility costs and variable costs are minimized. [5]
These problems have been studied extensively in the literature and are often solved using an approximation algorithm. The approximation algorithm looks for a feasible solution where:
A: The algorithm will stop after a given number of steps (e.g. number of customers and facilities)
B: There is an approximation ratio such that the calculated solution is within some small amount of the optimal solution [5]
In a basic formulation, the facility location problem consists of a set of potential facility sites L where a facility can be opened, and a set of demand points D that must be serviced. The goal is to pick a subset F of facilities to open, to minimize the sum of distances from each demand point to its nearest facility, plus the sum of opening costs of the facilities.
### Capacitated and Uncapacitated Facility Location Problem
A description here
Another concern when designing and operating a manufacturing system is the capacity the system tolerates: given the processing facilities, what is the maximum rate of order receipt that can be accepted so that all the orders can be satisfied? The Capacitated Facility Location Problem (CFLP) is a variant of the FLP, which includes capacities for the facilities. With the inclusion of the capacities, an open facility that is the least cost source for a demand node may not be able to serve any of the demand at that node.
## A Real World Example
Suppose you are a manager at Apple and you are identifying locations to sell computers and iPhones in a new location in Chicago. You would start by identifying potential sites in a variety of neighborhoods throughout the city and finding the demand for Apple products in each neighborhood. Then to construct your model you could create a table of neighborhoods, the cost of building a new Apple store in each one, and the expected profit. Your goal now is to select which facilities should be built in order to maximize profit.
For example:
Define decision variables i=1,2,3,4 such that .
Then the total expected benefit is:
and the total capital needed is:
In total, the model can be given as:
Additional constrains are that the total available capital is \$16M, and the number of new stores should not be more than two. Solving this model using an MILP solver in GAMS gives us the maximum when y2=y4=1. This indicates that the Logan Square and Hyde Park Apple stores should both be built in order to maximize profit.
Note that in another formulation of this problem you could minimize distance for customers to travel to the store regardless of profit.
## Applications in Industry
Facility location problems have applications in a wide variety of fields and projects. As examined it can be used to find the optimal location for a store, plant, warehouse, etc. but these formulation methods can also be used is less obvious ways. Applications range from public policy (e.g. locating police officers in a city), telecommunications (e.g. cell towers in a network), and even particle physics (e.g. separation distance between repulsive charges).[6] All of these problems have in common that discrete locations must be chosen and the objective is to meet the demand of consumers or users in the most efficient way. [5]
## References
[1] Dorrie, H. (1965) 100 great problems of elementary mathematics: their history and solution. New York, NY: Dover Publications.
[2] Drezner, Z & Hamacher. H. W. (2004). Facility location: applications and theory. New York, NY: Springer.
[3] Weisstein, E. W. Fermat Points. Mathworld-- A wolfram web resource. http://mathworld.wolfram.com/FermatPoints.html
[4] Kuhn, H. W. and Kuenne, R. E. (1962). An efficient algorithm for the numerical solution of the generalized weber problem in spatial economics. Journal of Regional Science, 4.
[5] Vygen, J. (2005). Approximation algorithms for facility location problems. Bonn, Germany: Research Institute for Discrete Mathematics.
[6] Korupolu, M. R. & Plaxton, G. C. & Rajaraman, R. (2000). Analysis of a local search heuristic for facility location problems.
|
2022-05-22 16:54:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6420086026191711, "perplexity": 735.3780187322151}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00735.warc.gz"}
|
https://datumorphism.leima.is/tags/time-series/
|
# Time Series
Forecasting time series
In a forecasting problem, we have $\mathcal P$, the priors, e.g., price and demand is negatively …
Evaluate time series models
Transforms that captures the local patterns
The state space model is an important category of models for sequential data such as time series
The hidden Markov model, HMM, is a type of [[State Space Models]] State Space Models The state …
Evaluate time series models
Time series analysis with simple models
Time series analysis with simple models
Forecasting is a widely used technique in both industry and science
Conformal time series forecasting is a probabilistic forecasting method using [[Conformal …
The Continuous Ranked Probability Score, known as CRPS, is a score to measure how a proposed …
What problem is StemGNN solving: intra-series temporal pattern: DFT Each series inter-series …
The Box-Cox transformation transforms data into Gaussian data, which is especially useful in feature engineering, e.g., fixing irregularities in variances of a time series.
|
2022-06-29 12:24:31
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5880951881408691, "perplexity": 3957.179697906245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00600.warc.gz"}
|
https://ask.sagemath.org/questions/11028/revisions/
|
# Revision history [back]
### resultant and gcd over real ?
The resultant and gcd works well over rational but not real or complex !
R.<x,y>=PolynomialRing(RR,2) a=x^2+y b=x-y^2 a.resultant(b) a.gcd(b) this does not worl where with QQ, it works
anyone has a idea ?
2 Code formatting. Typos. slelievre 10433 ●10 ●108 ●211 http://carva.org/samue...
### resultant and gcd over real ?
The resultant and gcd works work well over rational but not over real or complex !complex!
R.<x,y>=PolynomialRing(RR,2) a=x^2+y b=x-y^2
R.<x,y> = PolynomialRing(RR,2)
a = x^2 + y
b = x - y^2
a.resultant(b)
a.gcd(b)
this
This does not worl where work while with QQ, it works
anyone works. Anyone has a idea ?an idea?
|
2020-04-05 07:48:31
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17701540887355804, "perplexity": 14986.762553824163}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370529375.49/warc/CC-MAIN-20200405053120-20200405083120-00162.warc.gz"}
|
https://www.gradesaver.com/textbooks/science/chemistry/organic-chemistry-8th-edition/chapter-3-structure-and-stereochemistry-of-alkanes-problems-page-122/problem-3-28-c
|
## Organic Chemistry (8th Edition)
The IUPAC name of the compound is $trans−1,2−dimethylcyclohexane$
The compound has 6 carbon atoms in the carbon ring ($−hex$). There are 2 methyl groups attatched to the ring at poisitons 1 and 2, both on the oppsoite side of the plane relative to the $H$ atoms ($trans$). So, the IUPAC name of the compound is $trans−1,2−dimethylcyclohexane$
|
2018-10-22 21:12:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32796403765678406, "perplexity": 1484.6607890159119}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583515539.93/warc/CC-MAIN-20181022201445-20181022222945-00248.warc.gz"}
|