AMC USA(J)MO USAMO AIME email Duke college

well, first of all, welcome to AoPS.

I am not nervous myself, because I believe that cutoff is going to be 9... (big smile :D ). As for the others, I do not know, but dont be anxious, because I know everyone tried their best, and that is what counts.

[quote="GuitarGeek143"]Hey everyone.  I'm kinda new to this forum, but I'm just wondering if any of you are anxious to see the USAMO qualifiers yet.  I look forward to chatting with you all about random math topics in the future.  Good Luck to everyone hoping to qualify![/quote]

Ah greetings to you. This part of the USAMO selection is always hardest, especially if you sit on the borderline between qualifying and not qualifying. Good luck to you tomorrow.

dude, now i am checking the amc website three times a day.  i think it'll come out tonight, that is what happened last year.

[quote="chenwb"]i think it'll come out tonight, that is what happened last year.[/quote]

I'm pretty sure it didn't. I checked about once every 2.6 minutes starting at 9 AM the day of. :-D

Dang, I have three classes 8:00 to 11:00 tomorrow. Guess I'll be checking it while doing homework during my free periods tomorrow.  :D

I'm pretty sure I'll make it, but I'm still not saying anything. But that's just how I am. I would be nervous even if I had gotten a 14  :D

i know i will be checking the site compulsively tomorrow. even though i have a good chance of making it, i'm still nervous

I've been checking the site over 50 times a day for the past week. Today alone I've visited it over 100 times.....they need to post them before someone has a nervous breakdown  :P

Not that I would make it, I am interested in the list of qualifiers and AIME stats.

I am also interested in seeing who qualified from my state and the overall statistics! Grrr, why can't they just release them 18 hours earlier?!?!?! :(

when do the results of the usamo qualifiers come out?

[quote="inteluser"]I think they should put up a list of fake names and post an index that's 30 above the real one.

Then take it down and put up the real ones the next day.

Hm, I hope I'm never on the AMC committee.[/quote]

LOL, that would be awesome. Especially if the fake stats were released on April Fool's Day!!!!!

i kind of think the secretary might be wrong, but who knows

She said something like "hopefully within the next few days."

I don't blame them for taking this long to grade the papers, I'm just pissed off that that they gave the March 28th deadline so optimistically. If they had given April 1st as the release date, I'd be living my life normally right now, I wouldn't get anxious until about March 31st.

I'm in 9th and got a 8... I can't believe the floor might be a 9. Go figure...

All i can say is don't be negative, who knows, there may never even be a point to be.

The floor will be a nine. We eighters can regret together.

I just called.  The lady said that they're still working on it, and that it might be posted later this afternoon, or tomorrow morning.

Wow that's pretty weird that different people are giving different answers.

AHh I'm so confused

So will they put the results up today?

It's up! It's up!  All our lives are now complete! :)

Yay, it's up.  The qualifying score is XXX  :D

Beta, Whitehorseking, you made it!!!! And I did too!! So now your job is to give me all the answers on the USAMO so we can all go to MOP.  The floor was, I believe, a 9.  Total of 38 or so 9th graders.  That's not bad--if I can match my score/percentile last year I should make it.

Wow, the AMC Home Page has like 10000+ hits since last night. That's pretty insane.

haha, yes it did! I am proud to admit that I contributed to over 1000....no joke :P  :P


Hello,
I'm searching for documentation about Komascript but I only get it in german.
Can anyone send me something in english?

There's a link to the documentation at [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=koma-script]CTAN[/url]


So...exactly how strict is it? I had a solution to last year's #2 that I thought would have gotten a 7, maybe a 6, but it got a 1...
Basically, I had three tangent circles with radius five, put another circle between those three, and showed that it had to have a lattice point. Does it have to be more rigorous or something?

Or...do they take off for bad proof-writing and/or inelegant solutions?

How do you know it got a 1?  I wasn't aware that they released score breakdowns...

In any case, the problem is you have no discussion of what happens when any of the circles have radius greater than $ 5$.  Your solution is not a solution.

Aahh...ok. Thanks. Now I feel retarded. Could've made MOP if I had done that problem correctly...

I don't think they put the distributions online, but I believe they sent my own scores to me.

[quote="ZzZzZzZzZzZz"]But don't they also take off for bad proof-writing, like if your proof jumps around and doesn't flow smoothly?[/quote]

They can, if it affects the clarity of the proof.  Generally, if your proof is correct and complete, they won't give you below a 6, but you probably want to make it flow smoothly and make sense both for the sake of their easy understanding of the proof and for your ability to make sure it's rigorous.

So if your proof really sucks in clarity but it's correct, they'll give you at least a 6?  I'm not saying I'm going to purposely write a bad proof.  It's just in case I'm pressed for time and have to put something down.

A proof that's rushed and unclear is less likely to be correct anyway.  For example, if you handwave a key point your entire proof might be invalid; look no further than USAMO 2007 #4.


Perhaps I missed something.  Could you clarify something for me?

WHICH thousands of innocent people?

By the way, are you sure you should be putting this thread in this forum?

[quote="Rep123max"]Did you hear the news? After reviewing all the intelligence leading up to the Iraqi War, a bipartisan Senate commitee says that today, they would not have gone to war. [/quote]
Please note that a committee is not the same as the entire Senate.

Also, please ask yourself "how many Iraqis was Saddam killing each year through government sanctioned programs?" before blindly throwing out the number of Iraqis who have died in the war. I think doing so may perhaps cause you to reformulate a few of your thoughts, unless of course you base your views entirely on emotion, rather than fact. I'll provide you with a reputable
[url=http://www.hrw.org/reports/1993/iraqanfal/]starting point[/url] for your research. While you read that, think about what those numbers really mean.

Saddam was evil, drunk with power,  we can all agree to that right?  He had to be stopped.  The U.S. did that.  Yes people died, but many of them volunteered to go over there.  They wanted to help their country and others.  They died proud to have served their country.

Yes, Saddam had to be stopped, but we don't have to fight a war to stop him. Maybe we could stop him less violently.

Rep!  I shall go see Farenheit 9/11 but it won't change my mind I know.

[quote="RobinHood3000"]Isn't a film inherently biased?  You're seeing the war through a camera lens, not objective eyes.  No one in theaters will ever know what hit the editing room floor, and those involved in the making of the film necessarily project their own opinions on the issue, however subconsciously, on to the finished product.[/quote]
Biased? Certainly not. Biased is not nearly a strong enough word to describe the film. "Fraudulent" would perhaps be a better word; splicing together parts of several speeches to change the meaning, adding commentaries that are not factually supported to scenes with public officials talking, and various other techniques which are common to all Moore films is not the sign of a good film-maker and a good documentary - it is the sign of a liar, a coward, and a fraud.

Hm, nobody is going to respond to my post?  :|

So, Rep123max, you think that fahrenheit 9/11 was a great movie?  If you're still in the mood for another documentary movie, why don't you check out this upcoming documentary: [i]Michael Moore hates America[/i]?  I'm [i]sure[/i] you will come away from that movie feeling more well-informed.

[quote="Rep123max"]I cant believe I just read what I read. You Republicans have conservative radio talk show hosts who tell gay people to go have sex and get AIDS, but when Michael Moore comes out with a movie that is factual (and entertaining) that tries to show America how this administration is corrupted, and all of a sudden he is a coward. He is an American Patriot, a way bigger one than Hannity, Limbaugh (or I will remind you was the biggest hypocrite in the world) and many others. I dont know about you guys, but I feel that when 10000 innocent Iraqi's die and can never come back because of my country, well, I feel like a terrible person for not being able to stop this war.[/quote]
Pray tell, when did I make reference to Hannity, Limbaugh, or any others of the sort? I called Moore a coward; that does not preclude me from thinking that perhaps the tactics of Hannity, et al. are little better.

Also, I object to the classification "you Republicans;" if you care to call me a conservative, I will stand up and say "guilty as charged," but please do not call me a Republican. I certainly do not belong to the party, and my votes are not guaranteed to them; I'm strongly considering voting straight ticket Constitution Party. So call me conservative, but do not call me a Republican. They are not one in the same; they are nearly disjoint sets, unfortunately.

http://slate.msn.com/id/2102723/ Read that article, Rep.

As for the Iraqis who died in the war - I regret that lives were lost, as you do. However, as I've said before, I challenge you to find some estimates on how many people Saddam's regime was killing yearly, and the other heinous crimes it was committing, before you throw around your numbers. Here's one of my numbers: at least 100,000 died in [url=http://www.hrw.org/reports/1993/iraqanfal/]Anfal[/url] ALONE, but perhaps those people don't matter to you, because that was Saddam's doing, not Bush's. Perhaps the Kuwaitis or the Israelis matter to you... no, probably not. Please, though, do some research, and tell us all what you find.

Rep, I have seen the movie. While a lot annoyed me, I thought that the comic relief was enough to see it.

 [quote="rep123max"]you'll see the Iraq's whose lives were ruined by the war[/quote]

Rep, the only reason you hate the war is because you hate Bush. Your hatred for Bush overshadows EVERYTHING, inclulding the ability to accept plain facts, such as the fact that many more Iraqis were being killed by Saddam than were killed by Americans in the war.

And just for the record, I should say that I don't like Bush or Kerry.

[quote="Rep123max"]Well, how about this. Saddam attacks Kuwait, he is no being charged for war crimes. If more evidence is given that Saddam was of no threat to the U.S., shouldnt President Bush be charged for War crimes?[/quote]
All done with your research and that article I posted ALREADY? WOW! You must be very efficient.

I refer to my previous post.

What war crimes?

If you are looking for evidence, I suggest you read the article from MSN Slate that Matt gave a link to. http://slate.msn.com/id/2102723/ Now sit down and read the ENTIRE article.

And frankly, I was very angry when Moore showed the streets of Iraq before the war started. The idea that Iraq was such a peaceful, lovely country with a plethora of freedom is preposterous.

A point or two I'd like to make:

[quote="Rep123max"]I feel that when 10000 innocent Iraqi's die and can never come back because of my country, well, I feel like a terrible person for not being able to stop this war.[/quote]

Somehow, I don't think Saddam Hussein would feel the same way you do.

[quote="Rep123max"]But if you go see it, you'll see the Iraq's whose lives were ruined by the war, the soldiers who didnt want to be there, the parents who have had their kids killed in a war that they dont believe in, and best of all, Congressmen running away from Michael Moore.[/quote]

Was President Bush the ONLY one supporting this war?  Even if the parents don't believe in the war, perhaps the kids do, and fought for that which they believed; there is no shame in honor.  Did every soldier that fought in Iraq disagree with the war he/she was fighting?  And some of the Iraqis whose lives were ruined by the war were prepared to do and would have done harm to AMERICAN troops.

[quote="Rep123max"]I'll tell you though, I doubt they edited the footage of the dead Iraqi's which we killed, the mother whose son died in a war which they didnt believe in, or when a congressmen admits that they dont even read a lot of bills (including the Patriot Act) before voting on it, and the ties between Bush and the Saudi's that are undeniable.[/quote]

I'LL tell you though, I doubt they put in any footage of the liberated Iraqis whose lives are better as a result of this war or the son who fought for his country in a war in which he believed and came home victorious.  And ONE congressman?  Can one person with an apparent lack of concern for the content of a potential law properly represent the work ethics of all of Congress?

[quote="Rep123max"]Well, how about this. Saddam attacks Kuwait, he is now being charged for war crimes. If more evidence is given that Saddam was of no threat to the U.S., shouldn't President Bush be charged for War crimes?[/quote]

Compare the motives, if you will, of Saddam Hussein's attack on Kuwait and President Bush's war in Iraq.  President Bush had in mind the welfare of the American people.  If you'd prefer not to be counted among us, your presence in this country is entirely optional.  Hindsight.  Nothing more needs to be said.

[quote="Rep123max"]Half way done. I find it funny that you guys discredit Moore but believe this guy. If this guy can come out with a movie proving that Iraq had WMDs, the Bush administration did not lie at all, and that Bush didnt take vacation after vacation (which being from Texas, I know how often he is in Crawford), then I' might believe. But I will still finish this article.[/quote]
You do realize, of course, that the author of that article is a liberal, don't you? Also, he's not the only one that points that stuff out - a simple search on Google will reveal more than you have time to read.

[quote="JS1527"]
If you have had a full time career for thirty years, you might see why tax cuts would help you out.  If you are living on the edge of poverty you might want to be subsidized by the government in order to get by and raise your family.  But have [b]you[/b] really lived?  I mean, do you really think it's fair that some guy who sits in a mansion and drives to work in his nice new BMW has the right to say "the government should not give money to poor people"?
[/quote]

I might not have terribly firsthand experience, but I have secondhand experience through family members and close family friends.  We are not incredibly wealthy people, but we still get hit hard by high taxes and it impacts our lifestyle... and things do change when we get back a few extra hundred dollars from the government in tax cuts. And I do think the rich guy in the BMW has a right to want to keep his own money. We all have that right.

What I have experienced and learned has led me to believe that the best way to achieve success in life is to work hard to improve your situation in life, however it may be. I furthermore do not believe that it is the role of the government to make an attempt to redistribute wealth and punish those who are trying to advance themselves in a kind of negative sanction. Someone might be trying to get a couple extra thousand dollars a year so that they can live just a little more comfortably, or put their kids through college, by working overtime consistently, multiple hours each day, and as a result get moved into a higher tax bracket where now the returns from his extra work are significantly diminished. That just doesn't seem right nor fair to me.

[quote="Rep123max"]
5. We dont act arrogant and say "bring it on."
[/quote]

You mean like John Kerry?

[quote="JS1527"]I am really surprised by the opinions of the people on this forum.  I am a democrat, but part of being able to choose your own beliefs is not to set yourself in stone by believing all of the things that a certain political party believes or believing some other third part believes (your parents, or the general beliefs of your state, for example).  You have to ask yourself why you are arguing something.  Don't be ashamed to have different beliefs than the political party that you belong to.

All the posts about politics seem to be the same: Rep123max cites a specific example of the wrongs of the Bush administration, then confuted, miths, or nukular will cite some other statistic to try and prove him wrong, and then bhorsemath or aalindsey will make comments that are just (please and I mean no offense to either of you) devoid of personal thoughts, as if you are rehashing the same thing that someone lectures to you.

Most of you are less than eighteen.  Do you really know enough about the real world to formulate your own beliefs?  And I don't mean what you see on the news, because you can only formulate ideological beliefs from that ("war is wrong," etc).  If you have had a full time career for thirty years, you might see why tax cuts would help you out.  If you are living on the edge of poverty you might want to be subsidized by the government in order to get by and raise your family.  But have [b]you[/b] really lived?  I mean, do you really think it's fair that some guy who sits in a mansion and drives to work in his nice new BMW has the right to say "the government should not give money to poor people"? Politics is for the large part (in the REAL world) about what the party can do for you.  Why do you think a lot of rich businessmen support Bush?  Its cause his policies help them!  Without much real world experience, I dont see how many of you can be die-hard democrats/republicans

Think for yourself!  Dont spew propaganda/rhetoric (where does that really get you  cant you be happy with letting people believe what they feel is right?  Why must you try to proselytize others?) Don't be locked into a certain system of beliefs or associate with every aspect of a particular party.  And most of all, argue because you truly believe that, not just for the sake of arguing![/quote]

No offence taken.

I acctually have second hand experience.  As I have said before I have military parents.  I have cousins that are in that war.  That have died over there.  They believed it was right.  My family and I believe it was right.  I do like some of the things that Kerry has said but I still believe that Bush is better.  I am not set totally against Kerry and I am not totally for Bush.  If I could vote I am not sure which one I would vote for.

Let me start of by saying that Sadaam Hussain was a very evil person and definetly needed to be stopped.

I belive that trying to stop him was and is a good thing, President Bush lied to us about  Sadaam's connections to terrorism.  He said that he bought WMD's from Lybia, and that was found later to be false.  He also said that he owned many many WMD's, and so far, no proof has been found.  He went of trying to stop Sadaam Husaain, with little help from most countries.  He also went away from the war against terror in Afghanastan, not helping with the rebuilding of it.

The US needs to concentrate on capturing Osama bin Laden, not catching Sadaam Hussain with little evidence against him, and little help.

And even if Sadaam was connected to terror, the war has gone really bad.  Al-Sadr and all of his followers are trying to go against Bush.  Instead of telling him that they are there for a good cause, and trying to work together to rebuild Iraq, they are also fighting against them.  What good is America trying to  free Iraq if so many people are attacking the Americans at the same time?

That is why we need a better president.

[quote="Rep123max"]1. We go with the UN and dont make up 90% of the troops[/quote]
Frankly, Rep, most of your posts, and this last one in particular, are such nonsense that I'm going to quit responding to them. It's impossible to argue with a person who makes up his mind independent of facts that are presented, so rather than trying, I will spend my time on more productive activities. However, if you ever do the research I've suggested regarding the number of people Saddam killed each year, and if you read up a bit on Anfal, etc, get back to me, because it's always a pleasure to debate someone who is well informed.

However, as for the quoted comment, I can't let such ignorance stand. The United Nations has precisely zero troops; it does not have the ability to make or declare war, nor does it truthfully have the ability to prevent it. The former is clear from the Charter; the latter has been proven repeatedly throughout its history. Any actions "taken" by the United Nations are in actuality actions taken by its member nations; the UN has no sovereignty and no power, and was designed to be that way. What you say is preposterous for yet one more reason (taken directly from the first phrase of the [url=http://www.un.org/aboutun/charter/]Charter of the United Nations[/url]):
WE THE PEOPLES OF THE UNITED NATIONS, DETERMINED to save succeeding generations from the scourge of war, which twice in our lifetime has brought untold sorrow to mankind...

Surely it would not be in the spirit of such a Charter to declare war on any nation, especially not a member nation. It may be argued that it's also not in the spirit of the Charter to admit member nations responsible for ongoing genocide (such as the Sudan) to the body, but that is an entirely different argument.

Let me know if you ever do enough research to figure out what exactly it is that you're talking about.

Geez, you are all so [i]boring[/i] sometimes.  I mean, at least be [i]civil[/i] so I don't have to read through all these pointless insults hurled at people present at the conversation or otherwise.

My only actual comment, of course, falls almost under the category of things I'm complaining about:

Confuted: how can you simultaneously call yourself a conservative and support the principles of pre-emptive war and the wide variety of government powers contained in the patriot act and various other legislation and/or executive policies of which the patriot act is the principle example?

[quote="beta"][quote="h_s_potter2002"]Her cousins fought for what they belived for, and knew the risks of fighting in the war before they signed up.  If I wer her, I wouldn't be that mad, becuase you can't have a war where no one dies.[/quote]

That's why we should not have a war, too many people have died.[/quote]

War is a terrible thing but sometimes it can not be avoided.  Look  at the other wars.  Vietnam for instance.  Was it good?  No, but people felt it had to be done.

The Revolutionary War is much better example.  Sooooooooooo many people protested and thought the Vietnam War was pointless.  But it' would be very hard to find an American that though the Revolutionary War shouldn't have happenned.

Revolutionary War was right to us. But  British say we are rebellions. So who was right?? We will never know

Terrorist seen boarding English vessel and sabotaging vital beverage resources!  Still to be determined if they carried weapons of mass destruction.

[quote="JBL"]Confuted: how can you simultaneously call yourself a conservative and support the principles of pre-emptive war and the wide variety of government powers contained in the patriot act and various other legislation and/or executive policies of which the patriot act is the principle example?[/quote]
Whoa there, slugger, don't get ahead of yourself. I've never been an advocate of the USA PATRIOT Act, nor do I advocate No Child Left Behind, for that matter. As for pre-emptive war, I will quote from the Preamble to the Constitution:
"We the People of the United States, in Order to ... provide for the common defense ... do ordain and establish the Constition for the United States of America" (rest assured that I have not changed the meaning by omitting some stuff using '...')

We believed at the time we went into Iraq that Saddam posed a threat to us and our allies; whether he did or not is a somewhat different debate, but remember, hindsight is perfect, and at each moment we can only act on the information we have at that moment. When the government takes action to defend the American people from some external threat such as we believed Saddam was, it is entirely consistent with the roles laid down for it by the Constitution. So, JBL, I suppose that's how I can call myself conservative.

[quote="Rep123max"]I just thought I might add that I just found the ultimate flip flop in the world. [blah blah blah][/quote]
How exactly is that the ULTIMATE flip flop? Simply because Bush doesn't have as many as Kerry doesn't mean that it's the "ultimate."

[quote="confuted"]
We believed at the time we went into Iraq that Saddam posed a threat to us and our allies; whether he did or not is a somewhat different debate, but remember, hindsight is perfect, and at each moment we can only act on the information we have at that moment. When the government takes action to defend the American people from some external threat such as we believed Saddam was, it is entirely consistent with the roles laid down for it by the Constitution. So, JBL, I suppose that's how I can call myself conservative.[/quote]

I'm not sure which "we" believed.  Do you mean "we" as in the "the Bush administration"?  Do you mean the "we" as in the common citizens who are disallowed to view any real intelligence rendering us entirely incapable of making such a decision?  I'm quite sure this "we" does not include me because I have never for one moment believed that any nation the size of Iraq has or could ever threaten the U.S. any more than car accidents do.

[quote="RobinHood3000"][quote="JBL"]I wonder if they had the word "terrorist" in 1776 ...[/quote]

Didn't the Revolutionary War start in 1775?[/quote]

The Declaration of Independence was signed on July 4, 1776.  The Boston Tea Party took place on December 16, 1773.

Also, what MCrawford said.

Anyway I am against this Iraqi War(and all war in general), it's definitely unnecessary.

[quote="confuted"][quote="MCrawford"]I have never for one moment believed that any nation the size of Iraq has or could ever threaten the U.S. any more than car accidents do.[/quote]
Iraq -  437,072 sq km
Germany - 357,021 sq km
Japan - 377,835 sq km
(source: CIA World Factbook)
I suppose you're right. Something so small could [url=http://www.google.com/search?q=wwii&sourceid=mozilla-search&start=0&start=0&ie=utf-8&oe=utf-8]never[/url] threated something so big.[/quote]

Let's be contextual here.  I wasn't talking about land area, but rather power base.

I think it's also fair to note that even Germany and Japan found it enormously difficult to cross the oceans to actually step foot on U.S. soil despite being amongst the most industrially developed nations in the world.  I think they only serve to prove my point.

Japan is big in area at the time of WWII. Japan had 1/3 of China, much of South East Asia, and many Pacific Islands.

[quote="RobinHood3000"][quote="JBL"]
The Declaration of Independence was signed on July 4, 1776.  The Boston Tea Party took place on December 16, 1773.
[/quote]

Yes, I know the Declaration of Independence was signed in 1776, but the first battle of the Revolutionary War, the Battle of Lexington and Concord took place April 19, 1775 (the history class-staple phrase of "the shot heard 'round the world).  Contrary to apparently popular belief, the Declaration of Independence did not begin the Revolutionary War.[/quote]

It's probably just how you interpret it beginning. If you're judging it by when the US declared war on England, it was probably in 1776. Lexington and Concord were only small battles fought by the Massachussets Militia. At that time, the other states hadn't joined in the fight yet.


let T1 be a triangle having a,b,c as lengths of it 's sides and let T2 be another triangle having u,v,w  as lengths of its sides.if p,q are the areas of the two triangles,prove that 16PQ<=a^2(-u^2+v^2+w^2)+b^2(u^2-v^2+w^2)
+c^2(u^2+v^2-w^2)
when does equality hold?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2958]Pedoe inequality[/url] (probably posted elsewhere, too).

  darij


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $ l_1$ and $ l_2$ be parallel (but not identical) tangents of the circle $ \Gamma.$ Let $ \Gamma_1$ be a circle that touches $ l_1$ in $ A$ and $ \Gamma$ externally in $ B.$ Furthermore, let $ \Gamma_2$ be a circle that touches $ l_2$ in $ C,$ $ \Gamma$ in $ D$ and $ \Gamma_1$ in $ E.$

Prove that $ AD$ and $ BC$ intersect in the circumcenter of $ \bigtriangleup BDE.$

It is not difficult. First, we prove that $ A$ lies on the radical axis of $ \Gamma$ and $ \Gamma_{2}$ by showing that $ P_{A/(\Gamma)} \equal{} d_{\Gamma}.d_{\Gamma_1} \equal{} AE.AC$ (d_C denotes the diameter of the circle (C))

Hence $ AD$ is the radical axis of $ \Gamma$ and $ \Gamma_{2}$ which means that $ AD$ is the common tangent of $ \Gamma$ and $ \Gamma_{2}$.

Similarly, $ CB$ is the common tangent of  $ \Gamma$ and $ \Gamma_{1}$.

Therefore, the intersection of $ CB$ and $ AD$ is the radical center of $ \Gamma$, $ \Gamma_{1}$, $ \Gamma_{2}$.
Together with the fact that $ B, D, E$ are the three tangency points of these circles, we easily derive the conclusion.

(QED). :)

Sorry but how do you prove that the power of $ A$ to $ T$ and $ T_2$ is the same?
I see that $ A,E,C$ are collinear and therefore the power of $ A$ to $ T_2$ is $ AE.AC$.
But I still fail to see that this is $ d_T.d_{T_1}$ and that this is also the power of $ A$ to $ T$.


If $ x,y,z > 0,$ then

$ 3\left( \frac {yz \plus{} zx \plus{} xy}{x \plus{} y \plus{} z} \plus{} \frac {x \plus{} y \plus{} z}{yz \plus{} zx \plus{} xy} \right)^2\leq\left(x \plus{} \frac {1}{x}\right)^2 \plus{} \left(y \plus{} \frac {1}{y}\right)^2 \plus{} \left(z \plus{} \frac {1}{z}\right)^2,$

with equality if and only if $ x\equal{}y\equal{}z.$

ineq$ \Leftrightarrow 3[(\frac {\sum x}{\sum xy})^2 \plus{} (\frac {\sum xy}{\sum x})^2]\leq\sum x^2 \plus{} \sum\frac {1}{x^2}$
$ 3(\frac {\sum x}{\sum xy})^2\leq\frac {\sum x}{xyz}\leq\sum\frac {1}{x^2}\Leftrightarrow\sum x^2y^2\geq xyz\sum x,Right$
$ 3(\frac {\sum xy}{\sum x})^2\leq\frac {(\sum x)^2}{3}\leq\sum x^2$
We are done :)

Hey, hey. Do not just say by words! Show me your working right away,tin_124!


function
number theory
prime numbers
number theory unsolved

$(a)\Leftrightarrow(b)$
 $(a)$ [i]for all [/i]$n \geq 3$ : $2n$[i] is the sum of two odd prime numbers[/i]
 $(b)$[i]for all prime[/i] $p$[i] and for all [/i]$q$[i]and[/i] $r$ [i]two odd prime numbers[/i] $p(q+r)$[i] is the sum of two odd prime numbers[/i]

[i]the property [/i]$(a)$[i] is called the goldbach conjecture and there is no complete proof of this conjecture so can any one help me to find an other equivalente property simpler ? [/i]

If it was equivalent, it could be used to prove the Goldbach Conjecture, couldn't it?

Surely it could.
But there is http://www.mathlinks.ro/Forum/viewtopic.php?t=76454 , and I dislike reposting...

yes, i've tried to do it but it not simple.

 here is my proof for the equivalence :

 we suppoose that $(b)$ is true

   $\forall n \ge3$ we have $n=p_1^{a_1}......p_r ^{a_r}$
 w'll consider the function $f(n)=a_1+....+a_r$
 if $f(n)=1$ than $(a)$ is true
we suppose that $(a)$ is true forall $n$ such that $f(n)=m$
 for$n$ such that $f(n)=m+1$ we have $\exists p$ prime : $n=n'p$ since $f(n')=m$ we have $2n'=q+r$ ($q$and $r$ are odd primes) so $2n=p(q+r)$ so with $(b)$ we can compele the proof :D

[quote="othman"]i repost this problem because i need to know if there is an other equivalence  :D[/quote]
I don't understand this sentence, but I have deleted the old one now ;)

it's better

 i want help to find an other eqivatente proprity  :D

 i'm waiting ! :(


calculus
vector
integration
geometry
geometric transformation
dilation
LaTeX

Hi!

I have a few... interesting physics problems which I am eager to solve, but I seem to need help :) These problems are essentially of high school level, but seem to be quite hard (at least to me ;))

Well, here's one:

Consider an open vertical container of length $2L$ positioned on a horizontal surface. The container is divided into two parts by a thin horizontal wall of negligible mass. The horizontal wall can slide without friction along the vertical walls of the container. In the beggining, the container is divided into two parts of equal length $L$. The lower part is filled with air, while the upper part is filled with a liquid of density $\rho$. The entire system is at equilibrium at temperature $T_0$. Then we start slowly heating the air in the lower part, causing the liquid to spill out of the container. Determine the minimal temperature $T$ to which we must heat the air for all of the liquid to spill. The atmospheric pressure is $p_{\rm atm} = \rho g L$.

BTW, I am a new member of this forum, so I am not familiar with all of the recent posts. I hope that this problem wasn't posted before... I hope I'll be a regular visitor -- there are some very interesting problems (as I have seen so far) :D  Keep up the good work!

Mmm... I think it's something like To*(1+R/4/Cv)  where To is the initial temperature and Cv is... how do you call that... the molar heat of air?

Do you mean $T_0 \left(1+\frac{R}{4C_V}\right)$? I only have the final solution in terms of $T_0$ (which I will hold on to for a while  :diablo: )

If I got your formula right that would be approximately $\frac{11}{10} T_0$ (taking that the specific molar heat for air under constant volume is $\approx \frac{5}{2}R$), and that doesn't coincide with the answer I have...

More importantly, how should this problem be approached? As I see it, there are two ways: either by applying the first law of thermodynamics (that is, conservation of energy), or by applying the laws of ideal gases. I really don't have an idea what to do...  :(

because the air is heated slowly, i that the work done by the gas would equal nR(T - To). We first need to find the work it would take to pump all the liquid out, which requires.... calculus!!!

the amount required to pump an amount of water dx out of the container =

rho * pi * r^2 * xdx = dW

where r = radius of container and x ranges from L to 0

therefore, W = .5 * rho * pi * r^2 * L^2

Going back to the ideal gas equation (PV = nRT), we can find the number of moles of the air at the time before any heating was done by using the Patm given, or rho * g * L (is this density the same as the density of the liquid?) I guess it would have to be since the system isnt moving at equilibrium and only this pressure would cancel out the gravitational pressure of the liquid. Soooo...

rho * g * L * pi * r^2 = nRTo.

we find that n is rho * g * L * pi * r^2/rTo, which is good cuz it cancels a bunch of stuff out.

setting the work that needs to be done calculated in the first part of my solution to nR(T- To), we get, after plugging in the value of n from the thingy i just did, that

T = To + To/2g

I don't have much experience with thermo (mostly just through my ap chem class), but i would say this was a hard problem.

Ummm... Sorry, I think there's a mistake there...

[quote]

because the air is heated slowly, i assumed it would undergo a constant pressure expansion

[/quote]

I disagree. The fact that air is heated slowly implies that the system never leaves the state of equilibrium for a long time (that is, after a small (infinitesimal) movement of the horizontal wall, the system has enough time to "relax" into a new state of equilibrium). What this means is that the sum of the forces (as vectors) that act on the horizontal wall is zero at all times. What are these forces? Since its mass in negligible, the only force that acts on the wall is the result of the difference of pressures on both sides; since this force is zero, the pressures are -- [i] at all times [/i] -- the same. However, since the pressure from the upper part changes over time (since the height of the liquid decreseas) this means that the pressure of the air changes as well... BTW, yes, the $\rho$ is the density of the liquid in all formulae.

I doubt that this problem involves an extensive use of calculus. I don't know, but I suspect that the solution will require  only a bit of deriving to find the minimal $T$, and maybe there will pop up a simple integration like the one [b] psychicalchemist[/b] used above...

yeah, i realized after i made the post that pressure wouldnt stay constant because then the gas would never expand (having applied no force to the liquid). I was thinking of when a gas expands infintesimally slowly without any weight on top of it, in which case the pressure is constant. It doesnt matter, since i never assumed the pressure was constant in my mathematics. btw, you never said if i got it right or not...

At first glance I was a bit surprised with your final solution because it was dimensionally wrong (the units of $T$ and $T/2g$ are not the same, right?  :?

But on to the problems I see in your solution! First thing's first: are you certain about the work that needs to be done in order for the gas to pump the liquid out? Let me try to calculate it here:

Let $x$ be the distance between the horizontal wall and it's innitial position (the middle of the container). The pressure from the upper part of the system is $p_{\rm atm} + \rho g (L - x) = \rho g (2L - x)$, and we know that it's equal to the pressure of the gas below the liquid. This means that the gas exerts a force $\rho g S (2L - x)$ on the wall. The work done in moving the wall by $dx$ would be $dA = \rho g S (2L - x) dx$; integrating from 0 to $L$ we get $A = \frac{3}{2} \rho g S L^2$ which differs from your solution...

Even more importantly, how did you obtain the fact that the work done by the gas will equal $n R (T - T_0)$? Furthermore, even if the '$g$' in your final answer is a typo, the answer is wrong :) Let's just say it's closer to $T_0$ than $\frac{3}{2} T_0$ is...

Here's what I think the trick of this problem is: We only need to heat the gas to a certain temperature $T$ (which we have to find). That doesn't mean that $T$ will be the final temperature! After we stop heating the gas, it may still expand a little, just enough to pump out the rest of the liquid. The catch is to find the point (that is, the temperature) where we need to stop the heating. Any idea how to do this?

Of course, it may even happen that this problem is a lot simpler... I would appreciate any opinions about it! :)

OK, let's first solve the problem in one way, then move on to other ways.

Suppose that we first heat the air to a certain temperature, holding the wall. As soon as the temperature is reached, heating stops, and we release the wall. As far as I understand the problem, the container is isolated, thus, we're gonna have an adiabatical process. Gas will dilate and do work, and thus cool.

The work done is... A = \frac{3}{2} \rho g S L^2 as Djole calculated. Then, it should be equal to n*Cv*(T-To).
Thus, T should be To(1+3R/4Cv), or approximately 13/10*To.

Sorry for the ugly writing, I'll soon install LaTeX and become a normal person :)

[quote] AndersoN:

Thus, T should be To(1+3R/4Cv), or approximately 13/10*To [/quote]

Yup, that was the answer I got the first time I tried to solve the problem  :lol: Needless to say, it isn't correct :(

Let me just say that it is not stated whether the container is isolated or not, so we must assume that it is not...

FYI, the answer we're looking for is $T = \frac{9}{8} T_0$... I have no idea how to get this one...

Does the problem say anything about the way we should heat the container? First heat then let gas dilate, or heat parallel to dilation?

Ok, maybe the text I posted is ambiguous about that... Well, the air expands as we heat it.

It'd be nice if you found the original text of the problem and posted it here.

The text you have [i]is[/i] the original (but translated into english)...

I've just solved it!!!  :lol:

But I've posted this problem only yesterday, so I won't spoil the fun and post the solution soon enough if you people don't explicitly request it...  ;)

Yes, the result is $T = \frac{9}{8} T_0$ and as I suspected you don't have to know much calculus (you don't have to know it at all, actually, but it's quicker if you do...) This is one of the most interesting thermo problems I've ever done...

The fact that this problem is solved will not stop me from posting new problems of similar difficulty which I can't solve, so beware  :P

Do not consider this topic closed, pls! I want to see somebody get the same solution...

I got it too.  Quite a short solution I must say... and not very physical...

Hopefully, I'll install LaTeX tomorrow, and post my solution, if you agree :)

Ok, I'll see your solution and post mine if they are different.

In the meantime, I'll post a new problem which I'm having difficulties with. I'll alternate the fields of physics (i.e. now I will post a problem from, say, mechanics; the next time it'll be electromagnetism, etc.) I hope more people will join our discussions  :)

@ AndersoN:

Why do you need to install LaTeX so you can use it in your posts? If you know it, then just wrap your formulae in the usual dollar signs and they'll get TeXified... Of course, you can do it the old fashioned way [code] [tex] ... [/tex] [/code] but that's boring  ;)

[color=blue]The pressure of the air in the container
$\large p=\rho g(2L-x)$
The volume of the air
$V=S(L+x)$
$\Longrightarrow x=\frac{V}{L}-LV_0=\frac{VL}{V_0}-L$
where ${V_0}$ is the initial volume of the air ($=SL$)
$\Longrightarrow p=3p_0-\frac{p_0V}{V_0}$
where $p_0=\rho gL$
We get the temperature of the problem when $V=\frac{3}{2}V_0$ and $T=\frac{9p_0V_0}{4nR}$
At the initial state $2p_0V_0=nRT_0$
Consequently $T=\frac{9}{8}T_0$[/color]

Yup ;)

The only difference in my approach was that I calculated $T(x)$ and then took the derivative...

can you explain your solution a little further, giving the reason behind your mathematics. Sorry to ask, but I really have little experience with thermo outside just analyzing engine efficiency.

I'll post my solution here:

Let $x$ denote the distance between the wall and the middle of the container (I have used the same variable in one of my previous posts). The pressure of the gas below the liquid will be $p(x) = p_{\rm atm} + \rho g (L-x) = \rho g (2L - x)$ because the horizontal wall is always under equilibrium as we have already shown. The volume of this gas will be $V(x) = (L+x)S$ where $S$ is the area of the container's base. Using the famous $pV = nRT$ we get $T(x) = \frac{\rho gS(2L-x)(L+x)}{nR}$, and after using $p(0)V(0) = nR T(0) \Longrightarrow nR = \frac{2\rho gSL^2}{T_0}$ we get \[ T(x) = \frac{(2L-x)(L+x)}{2L^2}T_0.  \] Now, putting $\frac{dT}{dx}=0$ we get that the maximum temperature the gas will reach in the entire process is exactly $T=\frac{9}{8}T_0$, obtained for $x = \frac{L}{2}$. After the maximum temperature $T$ is reached, we no longer need to heat the gas (duh!). Done  :D

ah, i understand. Nice solutions by both you and phunc87!

I still have one question though: to me, it seems your function T(x) only tells you the temperature T needed to maintain equilibrium at a certain height x. Are you assuming that the gas will undergo an adiabatic expansion after you stop heating it, thereby pushing the rest of the liquid out of the container?

Not adiabatic, but slow! :)

You see, the fact that a process is quasistatical (slow) is sooo important that you mustn't ever forget about it! What I basically calculated is indeed the temperature needed to maintain equilibrium; assuming that the gas will constantly (and slowly)rise we get that it's maximal temperature is $T$.  This is a neat trick not uncommon in difficult thermodynamics: reversing the cause-reaction chain (the gas expands, [i]ergo[/i], its temperature increases vs. we heat the gas, [i]ergo[/i], it expands). With the constant equilibrium fact we get what we need. I hope you understand now; I've explained in one of my previous posts that the gas may cool and expand at the same time...

Isn't physics great  :lol:


Is there any book about barycentric coordinates for free download? Please help.

Well I need something more than only a definition and one expample, i.e. when are the points colinear in terms of barycentric coordinates, terms of isogonal conjugate...

[quote="PenguinIntegral"]Geometry Unbound by Kiran Keydlaya has a free download and there is a chapter on barycentric coordinates.[/quote]
Could you please tell me where can i download it?

[quote="April"]You can download it here: http://www-math.mit.edu/~kedlaya/geometryunbound/[/quote]
Thank you very much!
Do you know some other books about using barycentric coordinates.
Best regards.

On [url=http://www.math.fau.edu/yiu/geometry.html]Paul Yiu's website[/url], there are two lecture notes (An Introduction to the Geometry of the Triangle, 2001; Notes on Euclidean Geometry, 1998) that both feature the use of barycentric coordinates. So does his paper "The uses of homogeneous barycentric coordinates in plane euclidean geometry", also avaliable on his website.

  darij


geometry
geometric transformation
rotation
reflection
rectangle
advanced fields
advanced fields open

How many polyominoes on n squares are there? A polyomino is a connected interior-disjoint union of axis-aligned unit squares joined edge-to-edge, in other words, an edge-connected union of cells in the planar square lattice. The order of a polyomino is the number of unit squares forming it. The problem asks for the number of polyominoes of order n. The key constraint here is that polyominoes must be edge-connected. There are three variations on the problem, depending on whether two polyominoes are considered equivalent by factoring out just translations (fixed polyominoes), rotations and translations (chiral polyominoes), or reflections, rotations, and translations (free polyominoes).
 :rotfl:

such as u know :!: there isn't a general formula for the number of distinc $n-ominoes$ (polyominoes of order $n$),yet.
but here is some recurrence methods for enumerating this figures that counted the number of polyominoes to 28 by now.
you can see : [url=http://kevingong.com/Polyominoes/Math.html]http://kevingong.com/Polyominoes/Math.html[/url]
in which the writer approached to a method for enumerating the n-ominoes in higher speed.
bet still no formula is know for the number of n-ominoes.

what about to discuss on the number of ways that we can tile a $n * m$ rectangle with Dominoes?
have u any idea?it's obvious that it one of the $m$ or $n$ must be even :idea:


logarithms
real analysis
real analysis solved

Here's a cute one :).

If $a_n$ are positive reals s.t. $\sum a_n$ diverges, show that $\sum\frac{a_n}{s_n}$ also diverges, where $s_n$ is the $n$'th partial sum $a_1+\ldots+a_n$.

Let $n ,  m\in \mathbb{N}$, then
\[
\frac{a_{n+1}}{s_{n+1}}+\frac{a_{n+2}}{s_{n+2}}+\cdots+\frac{a_{n+m}}{s_{n+m}}\ge\displaystyle\frac{\displaystyle\sum_{k=n+1}^{n+m}a_k}{s_{n+m}}=
\frac{s_{n+m}-s_n}{s_{n+m}}
\]
whose limit is 1. The sequence of partial sums is not a Cauchy sequence and then the series given diverges.

I had in mind another problem when I sent reply for first time.

Consider a partial sum $s'_n$ for second series. We will prove that if we rearrange $a_1$, ..., $a_n$ in decrasing order that $s'_n$ will decrease.

Indeed, suppose $a_i<a_{i+1}$ then swapping values of $a_i$ and $a_{i+1}$ we change $s'_n$ by \[\frac{x_{i+1}}{s_{i-1}+x_{i+1}}+\frac{x_i}{s_{i-1}+x_i+x_{i+1}}-\frac{x_{i}}{s_{i-1}+x_{i}}-\frac{x_{i+1}}{s_{i-1}+x_i+x_{i+1}}=\] \[=-\frac{x_ix_{i+1}(x_{i+1}-x_i)}{(s_{i-1}+x_i)(s_{i-1}+x_{i+1})(s_{i-1}+x_i+x_{i+1})}<0.\]

So after rearrangement we will obtain \[\frac{a_i}{s_i}\geq\frac{a_{i+1}}{s_i}\geq \ln(1+a_{i+1}/s_i)=\ln\frac{s_{i+1}}{s_i}\] for all $i=1,2,...,n-1$. It means that $s'_{n-1}\geq \ln \frac{s_n}{s_1}\to\infty$ as $n\to\infty$.

I hope it is correct.

Here are two theorems of [b]Dini[/b].
[u]Theorem1[/u].Let the series $\sum_{n=1}^{\infty}{a_{n}}$ ($a_{n}>0$) diverge and $A_{n}=\sum_{k=1}^{n}{a_{k}}$.Then the series $\sum_{n=1}^{\infty}{\frac{a_{n}}{{A_{n}}^{\alpha}}}$ converge when $\alpha>1$ and diverge when $\alpha{\leq}1$.
[u]Theorem2[/u]
Let the series $\sum_{n=1}^{\infty}{a_{n}}$ ($a_{n}>0$) converge and $R_{n}=\sum_{k=n+1}^{\infty}{a_{k}}$.Then the series $\sum_{n=1}^{\infty}{\frac{a_{n}}{{R_{n-1}}^{\alpha}}}$ converge when $\alpha<1$ and disverge when $\alpha{\geq}1$.


integration
calculus
calculus computations

Find $ \int x \sqrt{\frac{x\minus{}1}{x\plus{}1}}\,dx$.

Let $ x \equal{} \text{cosh } 2t$.  The rest is easy, if tedious.

$ \int x \sqrt {\frac {x \minus{} 1}{x \plus{} 1}}\,dx \equal{} \int \frac{x}{x \plus{} 1} \sqrt {x^2 \minus{} 1} dx$

Let $ x \equal{} cost; dx \equal{} \minus{}sint dt$.

We have:
$ \int \frac{x}{x \plus{} 1} \sqrt {x^2 \minus{} 1} dx \equal{} \int \frac{cost}{cost \plus{} 1} \sqrt {(cost)^{2} \minus{} 1} (\minus{}sint) dt$

$ \equal{} \int [cost \minus{} cos^{2}(t)]dt \equal{} sint \minus{} \frac{t}{2} \minus{} \frac{sin2t}{4} \plus{} C$.


calculus
number theory unsolved
number theory

Let be given a sequence {$ x_n $} satisfying :
 $ i) x_1=1 ; x_2=2 $
 $ ii) x_{n+2}=5x_{n+1}-3x_n $ iff $ x_nx_{n+1} $ is even
 $ iii) x_{n+2}=x_{n+1}-x_n $ iff $ x_nx_{n+1} $ is odd
  Prove that :
 a) This sequence has unbounded possitive numbers , unbounded negative numbers
 b)  No number in this sequence equals 0
 c) If $ n=2^k-1 $ then $ x_n $ is divisible by 7

This isn't (in my opinion) what "Calculus Computations and Tutorials" is for. I'm moving it back to another forum - only maybe outside of calculus altogether, since is has some algebraic features, like divisibility contditions.

Anyway, about the other two parts, b is trivial as if there was a 0, the last two terms must both be equal (as we never have 2 evens in a row for rule ii to make a 0), and then every term before that must be equal as well by working backwards. So theres no 0.

I haven't fully done a, but after seeing that you are always doing (rule 2, rule 2, rule 3) repeatedly, if you look at the pair of terms you start with before doing that sequence, you go from (x,y) to (25x-12y, 20x-9y) or something like that. You can then probably work from there to show that the numbers grow indefinitely large in both directions.

For a) , I think we must determine the general formula for $ a_{3k} ,  a_{3k+1} , a_{3k+2} $

I guess the sequence will a period if n is large enough

If you mean it will be periodic, that would prove part a false. If you meant periodic mod 7 (or mod 14) then yes, it would, but it would probably take a long time and won't be the nicest way of proving part c.

Why are there two threads for this problem?

I mean it will be periodic for n >= a(a is a constant).and I also guess the period will be 2^b.so we only want to check n from 1 to a+2^b-1.
how can I guess:because a) is right.so there must appear x_k,x_k+1 and x_m,x_m+1 satisfy x_k=x_m,x_k+1 =x_m+1. :)

Did you misread unbounded as bounded? If the sequence ever becomes periodic, then there are finitely many values in the range and thus it is not unbounded.. as I said, which would disprove part a..

I'm sorry for my mistakes .As you said I mistook "unbounded" for "bounded" ;)


How many ds^3 would there be in 37.53 ms^3?

The pressure exerted on a substance is 11.605 g/(m*s2). What is the pressure in cg/(pm*ms2)?

What is "ds^3"? What is "ms^3"? Same for the other units.


If a,b,c are the sides of a triangle prove that

(a-b)/(a+b) + (b-c)/(b+c) + (c-a)/(c+a) < 1/16

Let  x = (a-b)/(a+b), y = (b-c)/(b+c), z = (c-a)/(c+a)
Then (1+x)(1+y)(1+z) = 8abc/(a+b)(b+c)(c+a) = (1-x)(1-y)(1-z). From that we have  x+y+z = xyz = (a-b)(b-c)(c-a)/(a+b)(b+c)(c+a) = P.

We can suppose that a = max{a,b,c}. If b >= c then  P < 0. If b <= c, we put
     c = b+x, a = b+x+y
From a<=b+c we have b>=y.
Now  P = (a-b)(b-c)(c-a)= xy(x+y)/(2b+x+y)(2b+x)(2b+2x+y)

Finally: 2b+x+y >= 3y+x >= 2 \sqrt 3yx
           2b+x >= 2y + x >= 2 \sqrt 2yx
           2b+2x+y >= 3y+2x >= 2(x+y)
=> (2b+x+y)(2b+x)(2b+2x+y) >= 8\sqrt(6).xy(x+y)
=> P < 1/8\sqrt(6)

We received a little bit stronger constant.


Let a,b,c is positive reals number.Prove that:
$P=\root3\of{\frac{a}{b+2c}}+\root3\of{\frac{b}{a+2c}}+2\root3\of{\frac{c}{a+b+c}}> 2$.

$a+b+c+d\ge 2\sqrt{a(b+c+d)}\to \sqrt{a}(a+b+c+d)\ge 2a\sqrt{b+c+d}\to \sqrt{\frac{a}{b+c+d}}\ge 2\frac{a}{a+b+c+d}$


A sequence of real numbers $ x_0, x_1, x_2, \ldots$ is defined as follows: $ x_0 \equal{} 1989$ and for each $ n \geq 1$

\[ x_n \equal{} \minus{} \frac{1989}{n} \sum^{n\minus{}1}_{k\equal{}0} x_k.\]

Calculate the value of $ \sum^{1989}_{n\equal{}0} 2^n x_n.$

Obviosly $ x_2\equal{}\minus{}1989$ and by induction $ x_n\equal{}0 \ \forall n\ge 2$.
Therefore sum is -1989.


Μπορεί κάποιος να δώσει τον ορισμό των κυκλικών και συμμετρικών παραστάσεων?
(Οσο γίνεται με πιο ακριβή και κατανοητό τρόπο γιατί εχω μπερδευτεί :wacko: )

Συμμετρικές είναι οι παραστάσεις στις οποίες αν εναλλάξεις δύο οποιεσδήποτε μεταβλητές μεταξύ τους δε μεταβάλλονται.

Για παράδειγμα η παράσταση $ ab \plus{} bc \plus{} ca$ είναι συμμετρική όμως η παράσταση $ a^2b \plus{} b^2c \plus{} c^2a$ δεν είναι (έλεγξέ το με τον παραπάνω ορισμό)

Επειδή εμφανίζει μια ομοιομορφία όμως (εδώ πχ. σε κάθε μονώνυμο εμφανίζονται δύο μεταβλητές με τις ίδιες δυνάμεις) λέμε ότι είναι κυκλική.

Κάθε συμμετρική παράσταση  είναι και κυκλική αλλά δε συμβαίνει και το αντίθετο.

Στις συμμετρικές παραστάσεις δε βλάπτεται η γενικότητα αν θεωρήσεις μια οποιαδήποτε διάταξη των μεταβλητών πχ. $ a > \equal{} b > \equal{} c$ και να δουλέψεις έτσι.

Στις κυκλικές βλάπτεται όμως. Μπορείς όμως να θεωρήσεις έναν μέγιστο όρο (πχ. τον $ a$) ή έναν ελάχιστο (πχ. τον $ b$). (Φυσικά όχι και τα δύο ταυτόχρονα.) και να δουλέψεις έτσι.

Όταν μιλάς για κυκλικό άθροισμα πχ τριών μεταβλητών για ένα συγκεκριμένο μονώνυμο πχ. [b]Σcyc[/b]$ a^2b$ εννοείς κάθε μεταβλητή εδώ δευτέρου βαθμού να συνδυαστεί με [u]μία μόνο[/u] μεταβλητή πρώτου βαθμού ήτοι $ a^2b \plus{} b^2c \plus{} c^2a$

Όταν μιλάς για συμμετρικό άθροισμα [b]Σsym[/b]$ a^2b$ εννοείς [u]όλους[/u] τους πιθανούς συνδυασμούς των μεταβλητών δηλαδή $ a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b$

[u][b]Προσοχή:[/b][/u]

Η παράσταση $ ab \plus{} bc \plus{} ca$ είναι συμμετρική αλλά δεν είναι συμμετρικό άθροισμα. Είναι κυκλικό άθροισμα.

Το συμμετρικό άθροισμα είναι το $ ab \plus{} ac \plus{} bc \plus{} ba \plus{} ca \plus{} cb$ δηλαδή το $ 2(ab \plus{} bc \plus{} ca)$.

Όμοια το $ abc$ είναι συμμετρική παράσταση αλλά το συμμετρικό άθροισμα είναι το $ 6abc$. (έλεγξέ το)

Αυτά έχουν μεγάλη εφαρμογή και χρειάζεται λίγη προσοχή στην ανισότητα Muirhead για να μην ξεφύγει κανένα συμμετρικό άθροισμα και γίνει κυκλικό και πάει χαμένη μια τεράστια απαλοιφή παρονομαστών... :D

Ευχαριστώ protonios με βοήθησες να ξεμπλέξω τη συμμετρική παράσταση από το συμμετρικό άθροισμα όμως μου έμειναν κπ απορείες:
1)Γιατί στις κυκλικές παραστάσεις δεν μπορούμε να θεωρήσουμε όποια διάταξη θέλουμε?
2)Τα σύμβολα $ \sum_{cyc}$ και $ \sum_{sym}$ για πόσες μεταβλητές τα χρησιμοποιούμε?
(πάντως ο κ.Στεργίου στο βιβλίο του τα ορίζει για 3 μεταβλητές)

Καλά κάνεις και διαβάζεις τα βιβλία του κ. Στεργίου γιατί είναι από τα καλύτερα του είδους.

Edited

2. Τα σύμβολα χρησιμοποιούνται συνήθως για οποιαδήποτε παράσταση οποιουδήποτε πλήθους μεταβλητών. Συνήθως το πλήθος τους θεωρείται ότι συνάγεται άμεσα από τα συμφραζόμενα της άσκησης. Το καλύτερο που μπορείς να κάνεις για να αποφύγεις την ασάφεια "συμφραζόμενα" είναι να βάλεις μηδενικές δυνάμεις για τις μεταβλητές που δεν εμαφανίζονται. Για παράσειγμα, αν θέλεις να γράψεις για 5 μεταβλητές:
$ \sum_{syc}a^3b^2$

θα ήταν καλύτερο να γράψεις:
$ \sum_{syc}a^3b^2c^0d^0e^0$

ώστε να καταλάβει ο άλλος ότι αυτό ακριβώς εννοείς. Βέβαια αυτό όταν προυσιάζεις μια καθαρή λύση, γιατί στο πρόχειρο θα σε εμποδίζει να σκεφτείς επαρκώς, γρήγορα και σωστά αυτό που θέλεις (εκτός από την περίπτωση που θέλεις να εφαρμόσεις ανισότητα Moirhead)

Όσον αφορά τον ορισμό των συμμετρικών παραστάσεων λίγο πιο αυστηρά είναι ο εξής:

συμμετρική ονομάζεται μια παράσταση $ f(X), X =${$ x_1,x_2,...x_n$}

όταν για κάθε αναδιάταξη $ Y =${$ y_1,y_2,...y_n$} του $ X$

ισχύει $ f(X) \equiv f(Y)$

Στις κυκλικές, ο ορισμός διαφέρει μόνο στο ότι ενδιαφερόμε μόνο για τις κυκλικές αναδιατάξεις:

που ορίζονται ως: $ Y =${$ x_{1 + [(1 + k) mod n]}, x_{1 + [(2 + k) mod n]}, ... x_{1 + [(n + k) mod n]}$}

Αλλά πιστεύω ότι ο Protonios δίνει πιο κατανοητά τους ορισμούς.  :lol:

[quote="maniopas"]
1. Συγκεκριμένη διάταξη μπορείς να θέσεις όταν έχεις μια οποιαδήποτε συμμετρική παράσταση, [u]επομένως και στην κυκλική.[/u]

[/quote]

To υπογραμμισμένο κομμάτι παραπάνω είναι λάθος!!! Η κυκλική ΔΕΝ είναι απαραίτητα συμμετρική (το αντίθετο ισχύει όπως τόνισε κι ο Κώστας :wink: ). Πάρε παράδειγμα το $ a^2b \plus{} b^2c \plus{} c^2a$. Στις κυκλικές δεν μπορούμε να θεωρήσουμε συγκεκριμένη διάταξη χωρίς βλάβη της γενικότητας.

Σωστά!  :blush:  Ο ειρμός των σκέψεών μου με πήγε λίγο παραπέρα απ' ότι ήθελα. Ήθελα να πω ότι μόνο όταν μια παράσταση κυκλική μπορεί να μετασχηματιστεί σε συμμετρική μπορεί να θεωρηθεί συγκεκριμένη διάταξη των μεταβλητών. Διαφορετικά, μπορούμε απλώς να θεωρήσουμε μόνο το ελλάχιστο ή το μέγιστο (που μόνο σε δύο μεταβλητές ορίζεται έτσι συγκεκριμένη διάταξη). Αυτό οφείλεται στην σταθερή απόσταση των όρων της κυκλικής παράστασης μεταξύ τους (αν μετά το τέλος συνεχίζουμε το μέτρημα από την αρχή).

[quote="maniopas"]
Στις κυκλικές, ο ορισμός διαφέρει μόνο στο ότι ενδιαφερόμε μόνο για τις κυκλικές αναδιατάξεις:

που ορίζονται ως: $ Y =${$ x_{1 + [(1 + k) mod n]}, x_{1 + [(2 + k) mod n]}, ... x_{1 + [(n + k) mod n]}$}

Αλλά πιστεύω ότι ο Protonios δίνει πιο κατανοητά τους ορισμούς.  :lol:[/quote]
Τέτοιους δεικτες δεν έχω ξαναδει  :P μπορείς να μου εξηγήσεις  τι σημαινουν.
Τωρα σε σχέση με την πρώτη ερώτησή μου ,γαιτί μπορούμε να θέσουμε όποια διάταξη θέλουμε μονο στις συμμετρικές παραστασεις?
Αν πχ μας δώσουν μια παράσταση f(x,y,z,...) όπου χ,y,z,... είναι πραγματικοι αριθμοι ,τί μας εμποδίζει να θέσουμε οποιαδήποτε διάταξη :?:

Αυτό που εννοούν αυτοί οι δέκτες είναι ότι αυξάνουμε όλους τους δείκτες κατά k. Το (mod n) μαζί με την πρόσθεση με το 1 δείχνει ότι όταν φτάνουμε στο n συνεχίζουμε το μέτρημα από το 1. Είναι αυτό που έγραψε ο protonios σε πιο επιστημονική μορφή.

Στην περίπτωση που έχουμε τρεις αγνώστους x,y,z, σε μια κυκλική μη συμμετρική παράσταση:
λες ότι μπορούμε να θέσουμε $ x \le y \le z$ αλλά όταν είναι (μπορεί να είναι)

$ y \le x \le z$,

αυτή από τον ορισμό της κυκλικής παράστασης γράφεται αλλιώς μόνο ως

$ z \le y \le x$ ή $ x \le z \le y$

και καμία από τις τρεις δεν είναι ίδια με αυτή που θεωρήθηκε ότι ισχύει.

Ελπίζω να έγινα κατανοητός.

Ευχαριστώ παιδια.Συγγνωμη που σας κούρασα :blush:


linear algebra
matrix
linear algebra unsolved

How many $10 \times 10$ matrices $A$ over $\mathbb{F}_{p}$ are there, up to similarity, with $A^{2}=1$.

If $p>2,$ I agree that they are diagonalizable. In that case, the eigenvalues are $1$ and $-1$ (the two square roots of $1$ in $\mathbb{F}_{p}$). We can permute the diagonal form to get $k$ ones followed by $10-k$ entries of $-1.$ We can do that in $11$ different ways.

But $p=2$ changes the argument. In particular, consider $A$ with block structure $\begin{bmatrix}B&0\\0&I\end{bmatrix}$ with $B=\begin{bmatrix}1&1\\0&1\end{bmatrix}.$ Then $A^{2}=I$ but is $A$ diagonalizable? And we can no longer count $-1$ as different from $1.$

you're right, I did not realize that it does not work when $1=-1$ ..

If p=2 and A^2=I then spectrum(A)={1,...,1} and A=I+N where
N is a nilpotent matrix s.t. N^2=0. By Jordan's decomposition
N is similar to [N1,...,Nr,M1,..,Ms] where 2*r+s=10,
Ni=[0,1,0,0] (row by row) and Mj=[0]. r is in {0,...,5}.
   Thus there exist 6 such matrices A, up to similarity.


calculus
integration
function
real analysis
real analysis unsolved

calculate the integral of x^x^x
it is inside the attachment in picture....
very difficult

That function has no elementary antiderivative.


There are lots of students around who cannot pay the tution fees of the top colleges. These tuition fees rather seem like an American fantasy. So either you are rich, or smart and have a lot of passion (or both). But as most people know excellent professors, e.g. Andrew Wiles in Princeton, attract excellent students and those attract outstanding professors again - basically the positively increasing "vicious circle". I guess from a certain point on professors do not care about the money that much but rather about the general high level of intelligence around, e.g. interesting people to talk to. Anyway back to the subject of this topic. Here is the [url=http://moneycentral.msn.com/content/CollegeandFamily/Savingforcollege/P36742.asp]article "Does an elite college really pay?" by Dan Akst.[/url] A nice chart of the relationship between [url=http://money.cnn.com/2002/09/12/pf/college/q_aid/popup.html]grants/scholarships and actually paid tution fees[/url] can be found here.

Thank you for finding that interesting chart from the CNN Web site. Today, the actual out-of-pocket cost of the most elite United States universities is probably even slightly lower than that year, while "list price" continues to rise. Very few American students find that their families have to pay full list price to attend a selective college. That's why I advise AoPSers to apply to the schools of their dreams, and to apply for financial aid. Only by comparing financial aid offers can a prospective student know which college is most affordable.

? tokenadult, i don't see how it is the case that very few american students have to pay the full list price to go to most of these colleges... I go to Yale, and tuition just goes up from year to year... I'm pretty sure that this semester alone cost me $21,000, not including books, snacks, and other supplies... this is going to add up to about$50,000 a year ; I get 0 financial aid, and I don't think that my family is even that well off. You may get lucky and get financial aid, it really depends. So you should apply to whatever schools you might want to go to no matter what.

Hi, Hashbrownie, I saw the Israeli flag on your user profile and at first thought that you were international. But you are an American domestic student?

I can't speak to your particular circumstances, because I don't know what information you filed with Yale when applying, but I know for sure that Yale, Harvard, and the other Ivy League schools give some financial aid each year to families with incomes above $\$$ 100,000. By any reasonable worldwide standard, incomes that high are high incomes, period. Most American families have much lower incomes. So if a college provides no financial aid whatever to a family that applied for financial aid, and that college is a richly endowed college like Yale, there must have been a judgment that the family can afford (barely, perhaps) to pay full list price. A lot of American families can afford to pay full list price at the Ivy League schools--and many of those families are happy to do that, knowing that general subsidies to the school from endowment income mean that the "true cost" of Ivy League college attendance is even higher than list price for tuition.

Congratulations on being admitted to Yale. Perhaps that is financially painful today, but I think that pain will lessen faster than the pleasant memories of your college experience--and the usual view is that the educational advantages you gain there are worth the money.

oh , sorry about the confusion... I have dual citizenship :o I am American and applied to school as an American. I understand that not everyone can receive financial aid. I just hear my father complaining all the time when I see him. I know that there are people a lot worse off then my family is, and they should definetly receive financial aid over me. But I look at Yale, and their endowment... can Yale really not afford to help out more people? I remember last year there was all this crap about a budget cut to help families with incomes between 45K and 60K. That seems absurd to me... how is it that they could force a family that makes 45K a year to pay for even any of it? Think about how much money that family has to live on...

I'm not in that situation, and I'm glad that I am not. But all I'm saying is that just because people can "barely" afford it doesn't mean that they wouldn't appreciate a little bit of help. Not only that, but for example, we are FORCED to purchase a meal plan. So that means you fork over an extra 3K or so. Seems like nothing but it all adds up... if the food was good then I wouldn't complain. But i'm in JE, and our dining hall is AWFUL. They served horrendous food all of last year. You're forced to pay 3000 yet I ended up having to buy food elsewhere because I couldn't stand their food... and i'm not even that picky of an eater. I tried putting up with it, but I actually got sick for about 3 weeks last fall and I think it had to do with me not eating enough.

Another thing that REALLY ticked me off was when the graduate students went on strike last year. All the poor grad students cried last year and went on strike right before finals week, having to TA without getting paid extra. I understand that there are certain cases where it may be difficult for those who lets say have a child already and are in grad school... but for the majority of the grad students that I know, though most of them are very nice, I sure wouldn't say that they're having it bad. Boohoo, they have to TA... which they do for approximaetly 1/3 of their time in grad school. Oh wait, maybe it's not that bad after all... they dont' pay a penny to come here AND they receive 17,000 a year. And the university puts up with that, meanwhile students are getting screwed over when they're paying 40,000 a year in the week before finals, which is a pretty important week.

All I'm saying is that these schools can afford to help out the students a little more... I love Yale and I wouldn't trade anything in the world for it, even if they don't help me or my family out one bit.

[quote="Hashbrownie"] I remember last year there was all this crap about a budget cut to help families with incomes between 45K and 60K. That seems absurd to me... how is it that they could force a family that makes 45K a year to pay for even any of it? Think about how much money that family has to live on...

Another thing that REALLY ticked me off was when the graduate students went on strike last year. All the poor grad students cried last year and went on strike right before finals week, having to TA without getting paid extra. I understand that there are certain cases where it may be difficult for those who lets say have a child already and are in grad school... but for the majority of the grad students that I know, though most of them are very nice, I sure wouldn't say that they're having it bad. Boohoo, they have to TA... which they do for approximaetly 1/3 of their time in grad school. Oh wait, maybe it's not that bad after all... they dont' pay a penny to come here AND they receive 17,000 a year.

All I'm saying is that these schools can afford to help out the students a little more... [/quote]

Two points here, one re: grad students and one re: rich schools.

First, you say that 45K is barely enough for a family to live on; 17K is about 1/3 of that amount, so if the family has only one child then these amounts are comparable.

But more importantly, a graduate student in the sciences could probably command 45K (or more) in business/industry with his or her bachelor's degree.  A grad student teaching her/his own class each term is teaching a half-time teaching load (for a Research I) and is compensated far less than 1/2 that which a beginning professor there makes. You may argue that tuition remission makes up for this difference, but that still doesn't mean that graduate students are well-compensated, just that they may be adequately compensated.  It also may be that graduate students have to pay large health-care premiums and such (this depends on the university), or that they are taxed on the amount of their tuition remission (also depends on the university).

Unless you know a lot of detail about the compensatory structure, and unless you've tried to live in New Haven on 17K, I find it odd for you to criticize the graduate student actions.  It's unlikely that they took these actions lightly.

Back to colleges with large endowments:  the cost to run a university is much larger than most of us think.  Even the list price for tuition doesn't nearly cover the expenses---that is, if all students paid full tuition then universities would still have to use endowment earnings (and perhaps also dip into the endowment and use external contributions) in order to break even.  Some years ago I saw a breakdown of the expenses for my alma mater (Haverford, which is by no means a poor school) and was shocked to see how much of a difference there is between the theoretical revenue from tuition and the cost of running a college for a year.

So no, I don't think that schools could really afford to help their students more.

[quote="Hashbrownie"]Another thing that REALLY ticked me off was when the graduate students went on strike last year.[/quote]
Having health care matters. Being able to pay for housing and food matters. As a graduate student and union member myself, I'd rather not have to strike- but it is essential as a threat.
It is a pretty good deal when you include a tuition exemption- better than any other job if you want to be a student at the same time. On the other hand, the expectation is that a TA is actually living on that money, and 1400/month isn't a whole lot.


LaTeX
algorithm
algebra
binomial theorem

[quote="now a ranger"]What is the sum of the first 3 digits to the right of the decimal in the decimal expansion of $sqrt\75$?[/quote]

The 3 digits are (in order) 2, 3, and 6. So the sum is 11.

edit: i quoted ur post and it said sqrt 75? is that what you meant?

$\sqrt{75}= 8.66025...$ so the sum is $\boxed{12}$.

[quote="now a ranger"]This is from a test with no calculators. It was from a tournament last saturday. I think there's another way to do this.[/quote]

oh yes. There is a way to find the square root of something but i forgot the method and name...

Okay so here's how to do it. I'm going to put it in code blocks so I might be able to line it up decently. (no latex either, since I'm not good with lining stuff up in latex)

This algorithm looks almost like long division, and you'd work it in a very similar way.

[code]
So we want the square root of

75 . 00 00 00 (notice groups of 2)

First we take the highest square less than 75, 64, so our first digit is 8.

Square root: 8
Now square it (64), and subtract from the first set of digits

75 00 00 00
64
-------
11 00

Notice that I appended the next two digits (00)
Now, we want the greatest A such that 160A + A^2 < 1100
The 160 is from what we currently have: 8. 8*20 = 160 (you always multiply by 20)

Trial and error gives an A of 6.
160(6) + (6)^2 = 996

Square root: 8.6 (ignore the decimal point in calculations)
75 00 00 00
64
--------
11 00
  9 96
-------------
   1 04 00

We then take 86*20 = 1720. The greatest A such that 1720A + A^2 < 10400 is 6 again
1720(6) + (6)^2 = 10356

  8   6   6
75 00 00 00
64
--------
11 00
  9 96
------------
   1 04 00
   1 03 56
----------------
            44 00

However, 20*866 > 4400, so our next digit is 0.

So to three decimal places we have 8.660, and our answer is 6+6+0=12
[/code]

[quote="diophantient"][quote="now a ranger"]What is the sum of the first 3 digits to the right of the decimal in the decimal expansion of $\sqrt{75}$?

EDIT: I was submitting, and it messed up, so I was still editting it when you posted it. Yeah I meant the square root of 75.  :wink:[/quote][hide="hint"]binomial theorem[/hide][/quote]

Can you elaborate?


calculus
integration
parameterization
trigonometry
calculus computations

Calculate the integral
$ I \equal{} \displaystyle \int\limits_0^{ \plus{} \infty } {\frac{{12{x^2} \plus{} 8}}{{{{\left[ {3 \plus{} {{\left( {{x^2} \plus{} 1} \right)}^2}} \right]}^2}}}} dx$

I got $ \frac{7}{3}\left[\frac{2 \pi}{\sqrt{3}} \minus{} 1 \right]$ for the answer. I will post my solution if this is the correct answer.

Write
$ I = \int_{0}^{\infty} \frac {12x^{2} + 8}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

$ = \int_{0}^{\infty} \frac {12x^{2} + 24x + 48}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx - \int_{0}^{\infty} \frac {24x + 40}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

$ = 12\int_{0}^{\infty} \frac {1}{\left[3 + \left(x^{2} + 1\right)\right]}dx - 8\int_{0}^{\infty} \frac {3x + 5}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

We have
$ J = \int_{0}^{\infty} \frac {3x + 5}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

$ = \frac {3}{2}\int_{0}^{\infty} \frac {2x + 2}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx + 2\int_{0}^{\infty} \frac {1}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

$ = \frac {3}{2}\int_{0}^{\infty}\left( - \frac {1}{\left[3 + \left(x^{2} + 1\right)^{2}\right]\right)'}dx + 2\int_{0}^{\infty} \frac {1}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

$ = - \frac {3}{2}\left|\frac {1}{\left[3 + \left(x^{2} + 1\right)^{2}\right]}\right|_{0}^{\infty} + 2\int_{0}^{\infty} \frac {1}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

We have

$ K = \int_{0}^{\infty} \frac {1}{\left[3 + \left(x^{2} + 1\right)^{2}\right]^{2}}dx$

$ = \frac {1}{6}\left| \frac {x + 1}{\left[(x + 1)^{2} + 3 \right] }\right|_{0}^{\infty}$ + $ \frac {1}{6} \int_{0}^{\infty} \frac {1}{\left[3 + \left(x^{2} + 1\right)^{2}\right]}dx$ which was obtained by applying integration by parts to K.

Putting everything back should give us the answer.

Edited: I am mistaken as to what the original integral was. If you replace $ \left(x^2 + 1\right)$ with $ \left(x + 1\right)$ then my calculation remains valid otherwise disregard this thread as irrelevant. If the original integral was intended as it stands, then I would be glad to see some hints.

akech thank you for posting your solution (i didnt check if you somewhere making mistake),.I have a similar biggggggg one and i didnt continue it, till the end, so i dont know the final result. I just posted it, for maybe a nice and short solution by a clever subtitution ,if we can do one.

$ I\equal{}\int_0^\infty \dfrac{12x^2 \plus{} 8}{[3\plus{}(x^2\plus{}1)^2]^2}\ \mathrm dx \equal{}\int_0^\infty \dfrac{12x^2 \plus{} 8}{(x^4\plus{}2x^2\plus{}4)^2}\ \mathrm dx$
Use the substitution $ x\equal{}t\sqrt{2}$ to reduce he given integral as the following sum:
$ I \equal{} \frac{3}{\sqrt{2}}\int_0^\infty \dfrac{t^2\ \mathrm dt}{(t^4\plus{}t^2\plus{}1)^2}\plus{} \frac{1}{\sqrt{2}}\int_0^\infty \dfrac{\mathrm dt}{(t^4\plus{}t^2\plus{}1)^2}$
Next, I deduced the general formula for evaluation of the class of integrals
$ N_{j,4}(a,m)\equal{}\int_0^\infty \dfrac{t^{2j}}{(t^4\plus{}2at^2\plus{}1)^{m\plus{}1}}\ \mathrm dt$
depending on the parameter $ a$. (The derivation being a lengthy one, I omit it presently.)
The result:
$ N_{j,4}(a,m)$
$ \equal{} \dfrac{\pi}{2^{3m\plus{}\frac{3}{2}}(a\plus{}1)^{m\plus{}\frac{1}{2}}}\times \sum_{k\equal{}0}^{m\minus{}j} 2^k \binom{2m\minus{}2k}{m\minus{}k}\binom{m\minus{}j\plus{}k}{2k}\binom{2k}{k}\binom{m}{k}^{\minus{}1}(a\plus{}1)^k$
Here, $ \binom{n}{r}$ are the binomial coefficients. Using this result we get
$ \int_0^\infty \dfrac{t^2\ \mathrm dt}{(t^4\plus{}t^2\plus{}1)^2} \equal{} N_{1,4}(1/2,\ 1)\equal{}\dfrac{\pi}{12\sqrt{3}}$
while
$ \int_0^\infty \dfrac{\mathrm dt}{(t^4\plus{}t^2\plus{}1)^2} \equal{} N_{0,4}(1/2,\ 1)\equal{}\dfrac{\pi}{3\sqrt{3}}$
Hence, ultimately we obtain the required integral as
$ \boxed{I\equal{}\dfrac{7\pi}{12\sqrt{6}}}$

Kaymant tanx for your solution
Mine was like
$ \begin{array}{l}
 I \equal{} \displaystyle \int\limits_0^{ \plus{} \infty } {\frac{{12{x^2} \plus{} 8}}{{{{\left[ {3 \plus{} {{\left( {{x^2} \plus{} 1} \right)}^2}} \right]}^2}}}} dx \equal{} \int\limits_0^{ \plus{} \infty } {\frac{{12{x^2} \plus{} 8}}{{{{\left[ {{{\left( {{x^2} \plus{} 2} \right)}^2} \minus{} 2{x^2}} \right]}^2}}}} dx \equal{} \int\limits_0^{ \plus{} \infty } {\frac{{12{x^2} \plus{} 8}}{{{{\left[ {\left( {{x^2} \plus{} \sqrt 2 x \plus{} 2} \right)\left( {{x^2} \minus{} \sqrt 2 x \plus{} 2} \right)} \right]}^2}}}} dx \equal{}  \\
  \equal{} \displaystyle \int\limits_0^{ \plus{} \infty } {\frac{1}{{\left( {{x^2} \plus{} \sqrt 2 x \plus{} 2} \right)\left( {{x^2} \minus{} \sqrt 2 x \plus{} 2} \right)}}\left[ {\frac{{2\sqrt 2 x \plus{} 2}}{{{x^2} \minus{} \sqrt 2 x \plus{} 2}} \minus{} \frac{{2\sqrt 2 x \minus{} 2}}{{{x^2} \plus{} \sqrt 2 x \plus{} 2}}} \right]dx}  \equal{}  \\
 \end{array}$[/tex]
[tex]$ \begin{array}{l}
  \equal{} \displaystyle \int\limits_0^{ \plus{} \infty } {\left[ {\frac{{ \minus{} \frac{{\sqrt 2 }}{8}x \plus{} \frac{1}{4}}}{{{x^2} \minus{} \sqrt 2 x \plus{} 2}} \plus{} \frac{{\frac{{\sqrt 2 }}{8}x \plus{} \frac{1}{4}}}{{{x^2} \plus{} \sqrt 2 x \plus{} 2}}} \right]\left[ {\frac{{2\sqrt 2 x \plus{} 2}}{{{x^2} \minus{} \sqrt 2 x \plus{} 2}} \minus{} \frac{{2\sqrt 2 x \minus{} 2}}{{{x^2} \plus{} \sqrt 2 x \plus{} 2}}} \right]dx}  \equal{}  ...\\ \end{array}$

akech, as i said i didnt check your solution ( i just saw the size of it-pretty big though smaller siza than mine :) ) . Problem of course is to calculate I. Made the post maybe for a nice solution with trig subtitution


The supplement of an angle is three times its complement. What is the number of degrees in the measure of the angle?


In a doll manufacturing company, there are three stations for building the dolls. The face painters can each paint 15 faces per hour. The hair sewers can each sew 24 heads of hair per hour. The arm-and-leg-installers can each install 35 sets of arms and legs per hour. How many employees are needed to assemble 6,720 dolls during an eight-hour day?

We need $ \frac{\frac{6720}{15}}{8}$ face painters, $ \frac{\frac{6720}{24}}{8}$ hair sewers, and $ \frac{\frac{6720}{35}}{8}$ arm-and-leg installers.  Adding these yields $ 56\plus{}35\plus{}24\equal{}\boxed{115}$ workers.


logarithms
limit
real analysis
real analysis solved

Compute the limit of the sequnce $(S_{n})_{n \geq 1}$ having the general form:

$\displaystyle S_{n}=\frac{1}{8}+\frac{1}{8+62}+\cdots+\frac{1}{8+62+\cdots+(27n^{2}-27n+8)}$.

enjoy! :)

With the asymptotic expansion

$\gamma_n = \gamma + 1/(2n) - 1/(12n^2) + 1/(120n^4) +o(1/n^4)$

we can get more precision


[b]1.[/b] If $ f$ is not continuous at $ x_0$, then $ f$ is not differentiable at $ x_0$. But if $ f'$ (or $ f''$) is not continuous at $ x_0$, then what happens to $ f$ at $ x_0$?

[b]2.[/b] The following is in the calculus notes of my classmate and it is about Taylor's theorem. I know there is something wrong, but I don't know how to fix it.

$ f(x)\equal{}e^x$

[color=red]Let[/color] $ x_0\equal{}0, x\in[0,1],$

$ f^{(n)}(x_0)\equal{}f^{(n)}(0)\equal{}e^0\equal{}1.$

So $ \forall N: e^x\equal{}f(0)\plus{}f'(0)x\plus{}\frac{f''(0)}{2!}x^2\plus{}\cdots\plus{}\frac{f^{(N)}(0)}{N!}x^N\plus{}\frac{f^{(N\plus{}1)}(\xi)}{(N\plus{}1)!}x^{N\plus{}1}$
$ \equal{}1\plus{}x\plus{}\frac12x^2\plus{}\cdots\plus{}\frac1{N!}x^N\plus{}\frac{e^{\xi}}{(N\plus{}1)!}x^{N\plus{}1}.$

Since $ |e|<\epsilon, 0<x<1,$ [color=red]??? maybe |e^xi|<3.[/color]

$ \left|\frac{e^{\epsilon}}{(N\plus{}1)!}x^{N\plus{}1}\right|\leq\frac3{(N\plus{}1)!}.$ [color=red]??? maybe epsilon should be xi.[/color]

[b]3.[/b] The following is also from her notes. It is in the proof of the method of finding the max and min point by the positivity or negativity of its 2nd derivative. I don't understand.

Since $ f'$ is cont. at $ x_0$ and $ f''(x_0)<0$, we have $ \delta>0$ s.t.

$ f''(x_0)<0, \forall x\in(x_0\minus{}\delta, x_0\plus{}\delta).$

...

1. If $ f'$ exists, but is not continuous at $ x_0$, then we can say that $ f$ is not continuously differentiable at $ x_0$. E.g. $ f(x)\equal{}x^2\sin(1/x)$.

2. Your interpretation is correct. We know that $ \xi\in[0,1]$ because it is between $ 0$ and $ x$. (Aside: I sometimes wish there was some notation for "$ c$ is between $ a$ and $ b$". Any suggestions?)

3. I don't understand either. We don't care about $ f''$ at any point other than $ x_0$. Having $ f''(x_0)<0$ and $ f'(x_0)\equal{}0$ is enough to determine the sign of $ f'$ to the left and to the right of $ x_0$, which by MVT implies that $ x_0$ is a point of strict maximum.

[quote="ifai"][b]3.[/b] The following is also from her notes. It is in the proof of the method of finding the max and min point by the positivity or negativity of its 2nd derivative. I don't understand.

Since $ f'$ is cont. at $ x_0$ and $ f''(x_0) < 0$, we have $ \delta > 0$ s.t.

$ f''(x_0) < 0, \forall x\in(x_0 \minus{} \delta, x_0 \plus{} \delta).$

...[/quote]
The whole "proof" is like the following:

[b]Theorem.[/b] If $ f(x)$ is differentiable at a nbd $ (x_0\minus{}\delta, x_0\plus{}\delta)$ of $ x_0$ and $ f'(x_0)\equal{}0$. Then $ f''(x_0)<0$ [color=red][[[/color]and $ f''$ is cont. at $ x_0$[color=red]]][/color] imply $ x_0$ is a local max point. ($ f''(x_0)>0$ [color=red][[[/color]and f'' is cont. at $ x_0$[color=red]]][/color] imply $ x_0$ is a local min point.) [color=red]??? Are they necessary?[/color]

[b][i]Proof.[/i][/b] Since $ f'$ is cont. at $ x_0$ and $ f''(x_0)<0$, there is $ \delta>0$ s.t.

$ f''(x)<0, \forall x\in(x_0\minus{}\delta, x_0\plus{}\delta).$[color=red] ??? Why?[/color]

By Taylor's formula, for $ x\in(x_0\minus{}\delta, x_0\plus{}\delta)$,

$ f(x)\equal{}f(x_0)\plus{}f'(x_0)(x\minus{}x_0)\plus{}\frac{f''(\xi)}{2!}(x\minus{}x_0)^2\equal{}f(x_0)\plus{}\frac{f''(\xi)}{2!}(x\minus{}x_0)^2,$

where $ \xi$ is between $ x$ and $ x_0$. Thus

$ f(x)\minus{}f(x_0)\equal{}\frac{f''(\xi)}{2!}(x\minus{}x_0)^2<0,$

i.e. $ f(x)<f(x_0).$ Hence $ x_0$ is a local max point.

[quote="mlok"]1. If $ f'$ exists, but is not continuous at $ x_0$, then we can say that $ f$ is not continuously differentiable at $ x_0$. E.g. $ f(x) \equal{} x^2\sin(1/x)$.[/quote]
And how about the case that $ f'(x_0)$ exists but $ f'$ is discont. at $ x_0$?

This is exactly the situation that I described in the post that you quoted.


integration
limit
calculus
calculus computations

solve  :D
$ \dfrac{dy}{dx}\; \equal{} \;\sqrt {x^2 \plus{} y^2}$

Well it's not linear, it's not exact, and it's nonhomogeneous so I give up.  :lol:

We can write : $ y'(x)\equal{}\sqrt{x^2\plus{}y^{2}(x)}\geq y(x)$.
So $ y(x)\geq C\cdot e^x$. For some constant $ C$. Now :
$ y'(x)\equal{}\sqrt{x^2\plus{}y^{2}(x)}\equal{}y(x)\sqrt{1\plus{}\left(\frac{x}{y(x)}\right)^2}$. But as we can see for big $ x$ : $ \frac{x}{y(x)}\to 0$ and this means that :
$ y'(x)\approx y(x)$.
Answer for big $ x$ : $ \boxed{y(x)\approx C\cdot e^x}$

If we put $ y(x) \equal{} e^{x}a(x)$ we get:

$ e^{x}a(x) \plus{} e^{x}a'(x) \equal{} \sqrt {x^2 \plus{} e^{2x}a^{2}(x)}$

$ a'(x) \equal{} \sqrt {\left(\frac {x}{e^{x}}\right)^{2} \plus{} a^{2}(x)} \minus{} a(x)$ and when $ x\to\infty$ then $ a'(x)\to 0$, so $ a(x)\to \textbf {const}$.

More exact solution: $ a'(x) \equal{} a(x)\cdot\left(\sqrt {1 \plus{} \left(\frac {x}{a(x)\cdot e^{x}}\right)^{2}} \minus{} 1\right)\approx a(x)\cdot\left(1 \plus{} \frac {1}{2}\cdot\left(\frac {x}{a(x)\cdot e^{x}}\right)^{2} \minus{} 1\right)$ or

$ a(x)\cdot a'(x) \equal{} \frac {1}{2}\cdot\left(\frac {x}{e^{x}}\right)^{2}$

this gives $ a(x) \equal{} \sqrt {\textbf{const}^2 \minus{} \frac {1}{4}\cdot\frac {1 \plus{} 2x \plus{} 2x^2}{e^{2x}}}$

misan, can you post solution?  :maybe: I think we can represent exect valuie of $ y(x)$ only as a series....

[quote="Yuriy Solovyov"]
$ a(x)\cdot a'(x) \equal{} \frac {1}{2}\cdot\left(\frac {x}{e^{x}}\right)^{2}$

this gives $ a(x) \equal{} \sqrt {\textbf{const}^2 \minus{} \frac {1}{4}\cdot\frac {1 \plus{} 2x \plus{} 2x^2}{e^{2x}}}$
[/quote]

$ a(x)\cdot a'(x) \equal{} \frac {1}{2}\cdot\left(\frac {x}{e^{x}}\right)^{2}$
$ \int a da \equal{} \frac{1}{2} \int \left(\frac {x}{e^{x}}\right)^2 dx$
$ \frac{a^2}{2} \plus{} C_1\equal{} \frac{1}{2}(\minus{} \frac{1 \plus{} 2x \plus{} 2x^2}{4 e^{2x}}) \plus{} C_2$
$ a^2 \equal{} \textbf{const} \minus{} \frac{1 \plus{} 2x \plus{} 2x^2}{4 e^{2x}}$
$ \therefore \ \ a(x) \equal{} \sqrt{\textbf{const} \minus{} \frac{1 \plus{} 2x \plus{} 2x^2}{4 e^{2x}}}$

Ok, misan, i don't see your solution  :mad:  :D .

For [b]big[/b] $ a(0)$ our $ \textbf{const}\approx a(0)$ because from the very beginning when $ x$ is small: $ \frac {x}{e^{x}} < < a(x)$. So we can write
\[ a(x)\approx\sqrt {a^{2}(0) \minus{} \frac {1}{4}\cdot\frac {1 \plus{} 2x \plus{} 2x^2}{e^{2x}}}
\]
For small $ a(0)$ i got a results:
$ \textbf{if}\; a(0) \equal{} 1\;\textbf{then}\;\textbf{const}\approx 1.11576741951555247$
$ \textbf{if}\; a(0) \equal{} 2\;\textbf{then}\;\textbf{const}\approx 2.06121359571601514$
$ \textbf{if}\; a(0) \equal{} 3\;\textbf{then}\;\textbf{const}\approx 3.04128359292678052$
$ \textbf{if}\; a(0) \equal{} 4\;\textbf{then}\;\textbf{const}\approx 4.03109197429583332$
$ \textbf{if}\; a(0) \equal{} 5\;\textbf{then}\;\textbf{const}\approx 5.02491346882494927$
$ \textbf{if}\; a(0) \equal{} 6\;\textbf{then}\;\textbf{const}\approx 6.02079471207307915$
$ \textbf{if}\; a(0) \equal{} 7\;\textbf{then}\;\textbf{const}\approx 7.01784158609870623$

and when $ a(0)$ is big:
$ \textbf{if}\; a(0) \equal{} 100\;\textbf{then}\;\textbf{const}\approx 100.001298359170320$
$ \textbf{if}\; a(0) \equal{} 1000\;\textbf{then}\;\textbf{const}\approx 1000.00014008449818$
$ \textbf{if}\; a(0) \equal{} 10000\;\textbf{then}\;\textbf{const}\approx 10000.0000157326104$
\[ \boxed{\lim_{a(0)\to\infty}{\left\{\frac {1}{a(0)}\cdot\lim_{x\to\infty}{a(x)}\right\}} \equal{} 1}
\]


inequalities geometry AMC USA(J)MO USAMO AIME probability

Sorry, but I couldn't find a forum in the Olympiad section to put this under.
I was just looking around, and took a look at the Olympiad section. I was startled and amazed at the same time. How did you people go about learning all this stuff? Such as Cauchy Schwarz, Holder, etc. Are there any books that you guys can suggest, not only for inequalities but just general or specific, quality books that cover some/all topics in the Olympiad section. I have had a recent hunger for learning new maths.

I am puzzled at how you guys learned all that :p Surely in my area, [i]noone[/i] goes about even hearing of such advanced maths :)
Thanks for your time. Admins, feel free to move this into a more appropriate forum as I am fully lost :p

The quickest way to learn stuff like that is to solve problems.  I mean, you need to understand exactly what the theorems are asserting, but after that, you need to solve problems; knowing that, say $p$ is prime $\Longrightarrow (p-1)!\equiv -1\bmod p$ doesn't help very much  :D

I cannot say that I understand the material in this forum either, but I am confident that I will make the olympiad next year, so I would like to get started in here. Anyway, I would suspect you just buy olympiad problem solving books or possibly math encyclopedias and read through them. Then you get on here and see if you can do problems and ask what identities state.

I learned a lot of stuff inadvertently from MathWorld (I think it's mathworld.wolfram.com).  The AoPS classes are helpful, as are the AoPS books.  I'm still trying to dig up this olympiad book I have whose name I forget.  500 Mathematical Challenges is also interesting, although its problems are generally very elementary.

thanks for the suggestions. Can you guys suggest any good quality books, that cover either many subjects well or a single subject very well?

A single subject very well I do not. But most good books cover all subjects "well enough for olympiads". (that is, for national olympiads of small and medium sized countries)

Again, that Arthur Engel book, problem solving strategies, is good but I think it's pretty expensive? (well most of them are)
I also have a copy of paul zeitz' book, the art and craft of problem solving, but that seems a bit unfit for "starting" to me. (although I didn't really train with this book, so I cannot guarantee it's good or bad).
Upon my first impression though, I think paul zeitz will rather give a "mathematically correct" background and some college-level extras, while arthur engel will show you more specifically olympiad problem solving. Which one is better may depend on personal taste.

I have heard of both these books and have been trying to get them. Just wondering, do you guys who do well at the USAMO actually think the IMO is harder? Or is it about the same level but the participants get higher scores?

[quote="Peter VDD"]Actually, I just had 2 days training (an IMO training weekend) and then got to mathlinks and started solving problems on pre-olympiad level. I still believe the beginner and intermediate forums on this site are booby traps you should keep out at all cost: they do not help you in training, they'll much rather give you false satisfaction.

If you do not have someone to explain you the basics, I suggest the very good book from Arthur Engel, problem solving strategies. Don't feel dumb cuz you can't solve them all. ;) but it is a good training resource, and contains problems of all levels + most of the necessary theorems.[/quote]

How are they 'booby traps'? The levels of these forums are clearly stated within them.  If you are already at the IMO level then why on earth would you think that these forums your level??  It would be pretty pointless to have all the forums at the same level, don't you think?

Personally I find USAMO (that is the highest level of it) harder than IMO. But they're a bit different: for USAMO you often need to know many theorems to find a good and short sol, while for IMO, that is not required that much (you need to know many, but not that many). So I like more IMO since it has a higher inventivity vs study rate in my opinion...

i learn all of my math from my dad, unless my math teacher teaches it to me... fat chance! :D

[quote="joml88"]
How are they 'booby traps'? The levels of these forums are clearly stated within them.  If you are already at the IMO level then why on earth would you think that these forums your level??  It would be pretty pointless to have all the forums at the same level, don't you think?[/quote]

On a certain level I can understand what Peter is saying. For students who can solve many of the problems in the Intermediate Forum and mid-level AIME problems, it is probably time for them to start looking at the Olympiad Forum. But those forums can be intimidating (both the problems and the people). There are many experience olympiad solvers there and also older (non-high school) people solving problems. So the attitude towards problem solving there tends to be non-pedagogical for beginning problem solvers.

Hopefully the Pre-Olympiad Forum could be a good place to rectify that. It should be for students who want to start working on Olympiad style problems, without Gold Medal winners saying "oh this was found in so and so olympiad in 1976!" or "this is trivial!" to every problem!

As a side note, I find many of those Romanian Olympiad problems to be good problems for Pre-Olympiad problem solvers. The forums for those can be found somewhere on the site.

Yes, I totally agree the pre-olympiad forum is the perfect solution.  But by the way, everyone is like the gold-medal winners in the sense that many of us say "this is trivial" to almost all AMC problems and many AIME problems, and that is probably very intimidating to many people, too.  We should just keep in mind that there will always be people better than us, and there will always be people worse than us.

[quote="bubala"]We should just keep in mind that there will always be people better than us, and there will always be people worse than us.[/quote]

However, by the Extremal/Well-Ordering Principle, there must exist someone who is the WORST at math and someone who is the BEST ;)

ok, so there are 6 billion people in the world, and 2 of them are at the extremes, and neither are on AoPS I'll wager.

For the 5,999,999,998 others of us there's someone better and someone worse.

Therefore, the probability that you're the best or worst is 2/6,000,000,000

i.e. 1/3,000,000,000

oh and by the way, how do you know that humans can be ordered? Maybe we're all complex numbers, in the land where i + 100000000000000000000000000000 is no bigger than i.  Who knows?


[hide]1 big rock is 49 pebbles, so the thing costs 5*49+2*7+3 = 262. He paid 6*49=294, so he gets 32 pebbles as change, or 4 small rocks and 4 pebbles. [/hide]

[quote="aznmonkey1992"] isn't there anybody else on? :mad:[/quote]

Nope  :rotfl:  :P

[quote="aznmonkey1992"]in thog's prehistoric village, the currency was consisted of only big rocks, small rocks, and pebbles.
1 big rock=7 small rocks
1 small rock = 7 pebbles
If Thog used 6 big rocks to buy something that costed 5 big rocks, 2 small rocks, and 3 pebbles, how much change did he get?[/quote]

1 big rock = 49 pebbles
49-17 (2 small rocks and 3 pebbles)=32 pebbles change
Or since 1 big rock is 7 small rocks
7-2=5small rocks
make 1 of those small rocks into pebbles and you get 4 small rocks and 4 pebbles

[quote="aznmonkey1992"]in thog's prehistoric village, the currency was consisted of only big rocks, small rocks, and pebbles.
1 big rock=7 small rocks
1 small rock = 7 pebbles
If Thog used 6 big rocks to buy something that costed 5 big rocks, 2 small rocks, and 3 pebbles, how much change did he get?[/quote]

[hide]6 big rocks = 49x6 pebbles = 294 pebbles
5 big rocks = 49 x 5 pebbles = 245 pebbles
2 small rocks = 14 pebbles.

245+14+3=262

294-262=32 pebbles = [b]4 small rocks and 4 pebbles[/b][/hide]

[quote="math92"][quote="aznmonkey1992"]y do u always answer my questions? isn't there anybody else on? :mad:[/quote]well, he's on all day long. :)[/quote]

From 8am to 3pm, with some breaks.

Want to answer the question  :P


This is a thread for the links to the links that contains the great problems for the beginners.

If you have a good site, post here!

Problem Solving Marathon
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11091&highlight=getting+started+marathon[/url]

Mathcounts Marathon
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11084&highlight=mathcounts+marathon[/url]

True Getting Started Problems (this is a title of the post)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13309

Another one
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12998

Getting Started Contests

They're very good. I just arranged them in order they posted or something.

#1
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9401
#2
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9564
#3
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9683
#4
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12594
#5
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12827
#6
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14022

AMC Practice Sets
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21368

I'll try to post more.