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How long do you think Wikipedia will continue to be the "go-to site for information", to put it in Larry Sanger's words (not sure if I remember that quote correctly...)?

Until someone comes up with something better or someone comes up with some good reason not to.  It's convenient, it's readable and it's pretty accurate (not perfect, but pretty good).  Many teachers and university professors aren't too fond of their students using wikipedia but others, particularly math and science professors, frequently have a link to wikipedia on their own websites because it nicely explains many difficult concepts.

Wikipedia is a wonderful site. As far as I'm concerned, it's accurate since every article (should) have citations, making it easy to verify. Plus, it's free.

Non-free online encyclopedias might as well not even exist, because even if you were willing to pay for them, you can't link them to anyone else. It goes against the basic principle of the Internet.

Until we don't use the internet anymore.

I'd hope it's like everyone said. But keep in mind, not many things last very long, especially on the internet. Take all those social networking sites, for example. Wikipedia used to be my life, like, literally, but I got tired of it. Now I contribute when I'm bored, or procrastinating. I hope those hard-core Wikipedians don't get tired of it too.

I'm not convinced that Wikipedia won't just die out like fashions do. I'm especially concerned with the philosophy of Wikipedia - completely and honestly open and free? How long can the culture keep up? How long will it be before some businessman takes over and decides to add advertisements to the site? How long before Wikipedia becomes a thing of the past?

BTW, I encourage people here to go help out on the mathematics articles. There's a lot to do if you look around. Activity in the Mathematics WikiProject is rather slow I'd say, and not only can the content be improved, conventions are not very consistent throughout the articles.

[quote="leoxnlin"]I'd hope it's like everyone said. But keep in mind, not many things last very long, especially on the internet. Take all those social networking sites, for example. Wikipedia used to be my life, like, literally, but I got tired of it. Now I contribute when I'm bored, or procrastinating. I hope those hard-core Wikipedians don't get tired of it too.[/quote]The problem with this is that Wikipedia is and remains the easiest form of information and the biggest too. I'll rephrase what I said previously. Wikipedia will die only if there becomes something bigger and better than wikipedia (which I don't see happening in the near future,) or the information on it becomes untrustworthy. (I don't see this happening either considering the safeguards they put up.)

[quote]I'm not convinced that Wikipedia won't just die out like fashions do. I'm especially concerned with the philosophy of Wikipedia - completely and honestly open and free? How long can the culture keep up? How long will it be before some businessman takes over and decides to add advertisements to the site? How long before Wikipedia becomes a thing of the past?[/quote]I don't really see what's so bad with advertisements. Yeah, it might look a bit uglier, but not much has changed. And ads don't kill websites. Look at google, look at any other search engine. Look at youtube, facebook, any website that uses ads. It hasn't hurt them in the slightest. The only ways I can see wikipedia going down are how I said above. And both seem [b]very[/b] unlikely.

[quote="12markkram34"]The problem with the math articles is that they are often over-technical. Math professors editing Wikipedia can take a simple subject and add calculus and complex ideas to a simple explanation. This makes it hard to effectively "learn" from Wikipedia. A good example are the statistics articles. Wikipedia's articles on stats make no sense to me, but stats textbooks do.[/quote]
I don't think anyone should (want to) learn math from Wikipedia. That's what textbooks are for. You don't go to Wikipedia to learn group theory, for example. Wikipedia is really useful as a quick reference, where all kind of relevant facts and related subjects are grouped together.
For example, if you know statistics, but don't remember some particular property of a particular distribution, Wikipedia is your friend.

Wikipedia is great for quick, on-the-go information. For example, the other day, I forgot the name of the "law of large numbers" (thought it was the "law of large proportions") and Wikipedia corrected me. It's also where I get general info about stuff I'm not sure about; although for specifics I usually depend on textbooks.

One disadvantage about the Wiki is that it can be editted by everyone, and I do understand why some teachers refuse to see it on bibliographies. I believe that Wikipedia will be replaced by more definite and accurate articles. Perhaps one that only professional users may edit (similar to WebMD). That way, it could provide more accurate, reliable information.

[quote="BlueVelvet"]I don't think anyone should (want to) learn math from Wikipedia. That's what textbooks are for. You don't go to Wikipedia to learn group theory, for example. [/quote]
That's pretty much what I did.  :P

Okay, okay, so it's not true that I didn't have any other sources, but technical Wikipedia articles tend to only be written by people who really know what they're talking about and some of them are quite good.  Wikipedia can be a good resource on some topics - for example, I've been picking up a little category theory.  I don't pretend to be able to do any proofs in the subject, but having learned some of the basic terminology and concepts has put a lot of things that I am learning into perspective.

And I value perspective.  For example, there is a certain class of polynomials that all satisfy identities formally similar to things like the binomial theorem, and you can spend a long time working with specific examples of these polynomials without ever realizing [url=http://en.wikipedia.org/wiki/Umbral_calculus#The_modern_umbral_calculus]that they're all related[/url], but I bumped into this article on Wikipedia one day and it clarified several interesting things for me.

[quote="12markkram34"]The problem with the math articles is that they are often over-technical.[/quote]

that's pretty subjective.  you would expect that certain topics would be more technical than others (e.g. an article on pseudodifferential operators will be written at a higher level of sophistication than an article on differential calculus).  in a lot of cases it is natural to talk about a topic in a certain level of sophistication, and it can hardly be called a problem if the reader may happen to lack the requisite mathematical background.

[quote="12markkram34"]Math professors editing Wikipedia can take a simple subject and add calculus and complex ideas to a simple explanation. This makes it hard to effectively "learn" from Wikipedia. A good example are the statistics articles. Wikipedia's articles on stats make no sense to me, but stats textbooks do.[/quote]

wikipedia is intended to be an encyclopedia first, and in particular one wouldn't expect the exposition to be the same as that of a typical textbook.  on the other hand, it does make a very handy reference, which is what i usually use it for.

I would agree that wikipedia can sometimes be over technical.  I browse to pages on things that I understand, and see confusing extensions, mountains of LaTeX.

Also, someone at my school demonstrated the shortfalls of wiki, he edited the page on the Republican Party with something along the lines of The Republican party is a christian fundamentalist party that can no longer relate to people and will die out in the next 50 yrs, and, it stayed up for 2 weeks.

Even though it has its cons, I love wikipedia as a site for information
(I love it when I get scholar's bowl questions just because I randomly read the wiki page one day  :D )  For me, it's especially helpful in deciphering abbreviations, it feels good when an abv is mentioned and it's possible to find it out yourself before the other person can explain  :lol:

[quote="Quickster94"]
Also, someone at my school demonstrated the shortfalls of wiki, he edited the page on the Republican Party with something along the lines of The Republican party is a christian fundamentalist party that can no longer relate to people and will die out in the next 50 yrs, and, it stayed up for 2 weeks.
[/quote]

I don't buy that this would happen with such a high-profile topic (American politics). You'll have to show me the edit logs for proof.

Usually, when something like that happens, it is corrected almost immediately.  For example, a few years ago, when Al Sharpton called for the resignation of Don Imus, someone edited Sharpton's page to say something to the effect of "Sharpton claims to be a man of God so he should act like it more often."  It was corrected within a few hours.

In addition, heavily edited articles are closed down very fast.

I agree with JRav. Errors are fixed correctly. My friend posted on the article about MarbleBlast, "MarbleBlast is where you blow Marbles up". It was corrected in 2 hours.

Anyways, I don't think Wikipedia will close very soon. This is a hard core Wikipedian saying this and I don't think I will get tired soon. Also, it is quite accurate. And in no other encyclopedia will you find specific stuff about things such as companies, universities etc. It has a lot of information in it. So, I can totally say it won't die down.

Two unexpected things for me:

-So much faith in the... longevity, shall we say?... of Wikipedia
-So much discussion as to the accuracy and speed of correction of vandalism (I unconsciously made the assumption that everyone had as much experience as I do with Wikipedia and knew all about that stuff)

Yes, for the most visited articles, Wikipedia appears incredibly flawless. I guess I'm the type that would look up obscure topics in which the articles are... not so perfect. Despite quick vandalism reverting, Wikipedia remains rather unpolished. A perfect illustration - take a look at the [url=http://en.wikipedia.org/wiki/Portal:Mathematics]Mathematics Portal[/url]. Scroll down to the "Topics in Mathematics" box at the bottom. As of time of post, there is a BIG MISSING TABLE CELL! There are problems everywhere in Wikipedia, and you only have to look a little harder to find them. So help Wikipedia!

I believe that most have not looked into Wikipedia very deeply to really take a look at what is going on. People seem to just look at Wikipedia, and most will think its awesome, and then others will reject it no matter what you say. I encourage everyone to examine Wikipedia and its community (don't just read articles; see special pages, project pages, and talk pages) more carefully. It gives one a sense of wonder and awe at the amount of... goodwill. But one also realizes just how ad-hoc and messy the whole business is. JBL's post is another example (although I wish the author would have posted a link to history).

[quote="Ihatepie"]I don't really see what's so bad with advertisements. Yeah, it might look a bit uglier, but not much has changed. And ads don't kill websites. Look at google, look at any other search engine. Look at youtube, facebook, any website that uses ads. It hasn't hurt them in the slightest.[/quote]

For me, ads would be one of the worst things that could happen to Wikipedia. Anything with ads tends to mark it as too connected with the world of money, impure, and perhaps even dishonest. Wikipedia is supposed to be a noble collaborative endeavor to compile knowledge... it doesn't seem to quite fit. Wikipedia has a distinctive feel because of its lack of advertisements. Even the occasional banner at the top asking for donations really annoys me.

Interestingly enough, my and my colleagues' attempts to correct the errors in [i]mathematical[/i] articles have never met any resistance. Something like Buddhism is a different matter: the only really qualified person to edit such a page with confidence is Dalai-Lama (and even he may be incorrect as far as the historical "facts" are concerned :P). So, if you want to make a major edit of such pages, you should always add a log entry to the discussion page explaining why it was an error, what sources of information you used, etc. and not be surprised if you'll be met with the doubts about your corrections: after all, the information put on Wikipedia is, mostly, not the result of somebody's wild imagination, those who put it online got it from what they believed to be reliable sources and they have no a priori reasons to assume you to know the things better. I had some positive experience with just putting my doubts on the discussion page without an attempt to edit the article itself and see it corrected in a few days or weeks by those "old guards" themselves, but you need to be very clear in your requests and objections to achieve this effect.

[quote="JBL"]Worth reading re: errors in Wikipedia: http://science-professor.blogspot.com/2008/03/wikid-stupid.html[/quote]

I think about all that can be concluded is that the quality of articles on wikipedia can vary wildly.  I can't think of the last time I saw an obvious (non-vandalism) error in a math-related article, and it seems that more technical topics don't usually suffer 'negative' edits by dilettantes with an axe to grind.

As an anecdote, I looked at the page for [url=http://en.wikipedia.org/w/index.php?title=Loop-erased_random_walk&action=history]loop-erased random walk[/url] the other day and was pleasantly surprised to see that the majority of the article had been written by Kozma and Schramm (separately), who have proved some of the most important results for LERW.  Maybe the state of affairs is different in the physical science that female science professor refers to, but we have it pretty good in math (thanks to the efforts of fedja and people like him), it seems.

[quote="blahblahblah"]I think about all that can be concluded is that the quality of articles on wikipedia can vary wildly.[/quote]  Yes, I agree.  And the problem is compounded because it's not always clear to nonspecialists which articles are the good ones and which are the bad ones.  I mean, it's an encyclopaedia -- they're useful primarily for getting a quick overview of something and for pointers to other resources.

http://science-professor.blogspot.com/2008/03/wikid-stupid.html

Are you guys just saying that false information is not corrected, or do you have any proof? I haven't seen one piece of information on this article that shows where the wiki is wrong.

[quote="leoxnlin"]Yes, that article is rather suspicious; it makes only vague references to exactly what happened.[/quote]

I'm pretty sure that the vague references are intended to preserve the anonymity of the author of the blog.

Wiki is a great quick resource...but you can't always be sure of everything.  I know an entire group of people who make it a game to see how much false information they can add into wikipedia, and how long it lasts.  It's pretty amusing actually.

[quote="funtwo"]they don't always ix vandalism super fast
a couple of months ago i edited the simple english wikipedia article about waterfall to say that angel falls was in colombia. it's still there.[/quote]We're talking about the regular wikipedia, not the simple one.

I think it's important to realize what Wikipedia is and can be, and what it isn't and can never be.

What it *is* is the world's largest encyclopedia.  Most of it is accurate, but much of it is poorly written and/or oddly organized.  As discussed, it's inconsistent in terms of quality, depth, breadth, and standards.  What it *can be* is a higher quality and even bigger version of what it already is.  What it *isn't* and never can be is a summary of the world's knowledge, organized so that anyone can learn anything from it with enough time and effort.

I also want to comment that an encyclopedia is supposed to be a quick and handy reference--for any given topic, it will tell you a little bit about it.  Regarding students who whine that they should be allowed to use wikipedia as a research reference, I'll tell you that when I was growing up in the stone ages, there was no wikipedia, and you know what?  Any good teacher with decent standards STILL frowned upon using an encyclopedia as a reference.  The reason is that encyclopedia research is shallow and lazy research.  It's a starting point, nothing more.

But if you just want to "look something up" and nothing more, wikipedia really is the greatest thing to happen since google.

One big thing I dislike about wikipedia is the lack of citations.  You could spend the rest of your natural life surfing through wikipedia, flagging items for "citation needed."  I'd wager that at least 80% of the raw information on wikipedia is not cited at all.  On the other hand, laziness with citations is part of what allows wikipedia to grow so quickly, so at least there's a silver lining here.

[quote="leoxnlin"]Amusing!?  :o  :mad: We're talking the most widely used source of encyclopedic information! Amusing to deceive thousands worldwide? Ok, maybe it's just because I've put my heart, mind, and time on several articles, and I know how much work it is. Not only adding material, but uploading images, checking books out from the library, and spending hours just figuring out conventions and systems, and the best way to cite sources. And then to have people come and... ugh.

If you contribute in a beneficial way, even in just a small way, you will probably delight whoever created or made most of the article, or whoever has that article on their watchlist.[/quote]
there are also college professors that purposely edit wikipedia articles and tell their students to find the mistakes. this is a very good way of testing true understanding.

That brings back a good memory. My teacher did this once. A kid came in late, and by then, the article had been edited back so he got 100%

[quote="funtwo"]there are also college professors that purposely edit wikipedia articles and tell their students to find the mistakes. this is a very good way of testing true understanding.[/quote]
I think that's pretty clever but there are problems with that: first, there are no guarantees that the mistakes won't be corrected before the students get the chance to fix it; and if someone else is using the site and they come across this incorrect information, they may be inclined to take it as fact which might not end well.

On the other hand, one should not believe things without proof or reliable reference.

I can't express how annoying it is when people purposely write false information on Wikipedia.

[quote="funtwo"]
there are also college professors that purposely edit wikipedia articles and tell their students to find the mistakes. this is a very good way of testing true understanding.[/quote]

That's completely insane. Disseminating false information in a source that many people trust and use regularly is a horrible abuse of the internet.

[quote="funtwo"]they don't always ix vandalism super fast
a couple of months ago i edited the simple english wikipedia article about waterfall to say that angel falls was in colombia. it's still there.[/quote]

It is corrected now. Can you please NOT do that again? Wikipedia is an amazing resource that we should feel privileged to have access to. Putting errors into it for no reason other than because you can is incredibly immature. Many article-writers and vandal catchers spend hours and hours trying to undo bogus changes like yours. Give them a rest.

By the way, I have seen Wikipedia referenced both in math and physics classes and in formal seminars.

I would like to strongly second the opinion that intentionally vandalizing Wikipedia is in most conditions a completely immoral violation of trust, especially with regard to college professors vandalizing articles that are presumably of academic importance.

[quote="funtwo"]there are also college professors that purposely edit wikipedia articles and tell their students to find the mistakes. this is a very good way of testing true understanding.[/quote]

why not copy it into Word, fix, print, and have the students do it in lecture/discussion sections?

[quote="MysticTerminator"]I would like to strongly second the opinion that intentionally vandalizing Wikipedia is in most conditions a completely immoral violation of trust, especially with regard to college professors vandalizing articles that are presumably of academic importance.[/quote]

I agree with this view.

OK, since no one really seems to want to talk about what I originally intended this topic to be about (will Wikipedia last?), (whether it is because everyone thinks the question of the encyclopedia's information's validity is much more interesting/important, or because everyone thinks that Wikipedia will certainly last for a very long time), and since the discussion of Wikipedia's content's accuracy is getting nowhere (everyone is agreeing with each other), I would like to pose these questions:

1. Often, some people on Wikipedia push for changes in Wikipedia's policies that would make Wikipedia less dynamic (intended to increase the proportion of time in which any given article is displaying correct information), such as
-having a reviewed version (main) and an editable draft version of each article (periodically an authorized person will update the reviewed version)
-implementing the above on only articles that are biographies of living persons
-allowing only registered members to edit
-protecting featured articles (that is, limiting which users may edit the highest quality articles)
At the moment, Wikipedia is among the... purest, shall we, of wikis; it is very free, open, there are no ads, content of obscene nature is not censored, and it is clear that there is a very large amount of trust in the community when one reads the policies. Wikipedia adheres quite strictly to its original philosophy. It would disappoint many if this were changed.
What do you think about this?

The german version of Wikipedia already has some of that stuff. Additionally, a lot of the administrative "staff" of it has a rather stupid point of view (e.g. deleting the article about Christian Reiher, because they, most of them never having done any mathematics, think it's not important enough; same for lots of other themes).
In total, their administration together with the already mentioned faults have made it much worse. Comparing the german articles with the english ones tells me, that they are a worse and often contain useless or stupid information, whereas important things are missing (no wonder, after making editing more difficult, which annoys "good" editors but doesn't really stop trolls). A long time ago, sometimes the German one was even better than the English version!
As it is now, I do not really contribute anymore to them, and practically never use the German articles, either.

[quote="leoxnlin"]
What do you think about this?[/quote]

This is like the classic cathedral versus the bazaar question. I think it would take a lot of research to answer your question fairly, but here are my gut reactions.

Wikipedia works fairly well the way it is now. Sure, you come across some flagrant errors here and there, and some articles are poorly written, and articles are sometimes entirely deleted and replaced with a swear word, but in general, when you go to an article that you know something about, it usually corresponds quite well to what you know, and when you go to an article to learn something, you usually learn what you want. So, it seems like the "good guys" are beating the "vandals" and the encyclopedia is maintaining a pretty good quality standard.

As you said, it is a really pure model of knowledge-collection to allow anyone to contribute from any computer connected to the internet. I think its worthwhile to try to preserve this, because it empowers lots of people and teaches them responsibility. Putting more restrictions on who can edit which articles may or may not improve the quality of the encyclopedia, but it would definitely make the internet feel less flat and open and more exclusive and controlled.

[quote="jain_harshi"]I agree with JRav. Errors are fixed correctly. My friend posted on the article about MarbleBlast, "MarbleBlast is where you blow Marbles up". It was corrected in 2 hours.[/quote]

I saw that on the page [url=http://en.wikipedia.org/wiki/deoxyribose]deoxyribose[/url], a claim that ribose destroys DNA remained for almost a month. Also see http://en.wikipedia.org/wiki/User:Shii/Hoaxes.

An interesting article that was on the front page of yahoo today:
[url]http://finance.yahoo.com/news/Irish-student-hoaxes-worlds-apf-15201451.html;_ylt=Aj3kM2ooL3OndSx0CKhgNnEazJV4?.v=1[/url]

[quote="aadil"]it is approximately 90% accurate[/quote]
[citation needed]


A bullet of mass m1 strikes a pendulum of mass m2 suspended from a pivot by a string of length L with a horizontal velocity v.  The collision is perfectly inelastic and the bullet sticks to the bob.  Find the minimum velocity v0 such that the bob(with the bullet inside) completes a circular vertical loop.

I got
[hide]
$ \sqrt\frac{4gL({m_1\plus{}m_2})^2}{m_1^2}$
[/hide]

If I interpreted it correctly.

Because they are under de sqrt.

[hide="Actually"] Minimum velocity of the bullet is $ {{m_1\plus{}m_2} \over {m_1}} \sqrt{5gL} .$[/hide]


algebra
polynomial
quadratics
trigonometry
Rational Root Theorem
system of equations

Hi, I have recently been working on exercises of factoring completely, including finding the rational, irrational, and complex roots of the polynomials. I have learned about different methods, theorems, and tricks used, and understand everything in the book. my question regards a curiosity not in the book.

In the text, problems given to solve with an odd degree are in the form-
f(x)=x^3+8     where you can use synthetic to make it an even degree polynomial
or
f(x)= x^3-2x^2+2x      where you can factor out an x and make the rest of even degree

I thoroughly understand how to factor higher degree evens, ex. substituting a "y" for say, x^4 to solve for roots. However, I don't understand how to factor an odd degree when it is not in the above two forms.

For example, how do I factor completely, say f(x)= x^3-8x+3 ???
Thanks

If one of the roots is rational, use the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]rational root theorem[/url] to find it.  If it exists, divide out; the quotient is a quadratic.

Unfortunately, there exist cubic polynomials that are irreducible, i.e. they can't be further factored with rational coefficients, and you can't do any better than the [url=http://mathworld.wolfram.com/CubicFormula.html]cubic formula[/url].  (Don't use the cubic formula.)

Not necessarily. Sometimes you are given quartics (or higher even powers) that can be factored into smaller even powers. In a case of a quartics, quartics w/o real solutions can be factored into two irreducible quadratics. There can be many problems that you might be given that involve polynomials with no real roots and those tend to be the hardest.

So, if we know that there are no linear factors over the rationals, how do we find the quadratics?

EDIT: What about something like $ x^4 \minus{}4x^2 \plus{} 4 \equal{} (x^2 \minus{} 2)^2 \equal{} (x \plus{} \sqrt{2})^2(x \minus{} \sqrt{2})^2$? There aren't any linear factors with rational coefficients, but there are quadratics with rational coefficients.  :maybe:

[quote="ThinkFlow"]So, if we know that there are no linear factors over the rationals, how do we find the quadratics?[/quote]
That's the point.  If a cubic equation doesn't have a linear factor over the rationals, it can't have a quadratic factor either.

[quote="pinkmuskrat"]Thanks. So, if none of the roots I try with synthetic division work, is it irreducable? Does this mean it will not come up in textbooks, competitions, etc.?[/quote]
Sometimes it's possible to write the roots of an irreducible cubic in terms of sines and cosines of certain angles; occasionally contest problems may ask you to do this.  The key is to be very familiar with the sine and cosine triple-angle formulas.

@ThinkFlow: We always want to factor polynomials into other polynomials with integer coefficients. Once we know that a polynomial has no rational roots, then we know that it can't be factored into a product of a few polynomials and a binomial. For example, if we knew that $ x^4\minus{}x^2\minus{}2x\minus{}1$ had no rational roots, then we know that it must be the product of two quadratics. Since the coefficient of the term that has the highest degree ($ x^4$) is 1, then we know that the two quadratics are monic (the term with the highest degree has coefficient 1). We now have $ x^4\minus{}x^2\minus{}2x\minus{}1\equal{}(x^2\plus{}ax\plus{}b)(x^2\plus{}cx\plus{}d)$, where $ a$, $ b$, $ c$, and $ d$ are integers. We also have a system of equations:

$ x^4\minus{}x^2\minus{}2x\minus{}1\equal{}x^4\plus{}(a\plus{}b)x^3\plus{}(b\plus{}ac\plus{}d)x^2\plus{}(bc\plus{}ad)x\plus{}bd$

$ a\plus{}b\equal{}0$
$ b\plus{}ac\plus{}d\equal{}\minus{}1$
$ bc\plus{}ad\equal{}\minus{}2$
$ bd\equal{}\minus{}1$

We can solve this if we put our minds to it. [hide="The solution happens to be"]$ (a,b,c,d)\equal{}(\minus{}1,\minus{}1,1,1)$.[/hide]

[hide="A practiced eye would notice that"]the polynomial was $ x^4\minus{}(x^2\plus{}2x\plus{}1)\equal{}(x^2)^2\minus{}(x\plus{}1)^2\equal{}(x^2\minus{}x\minus{}1)(x^2\plus{}x\plus{}1)$.[/hide]

That is how you should solve higher degree polynomials. But it won't work with 6th degree or higher polynomials; it could be a product of a quartic and a quadratic, or two cubics, or even three quartics. Stick to this method only for quartic and quintic polynomials or else it will get ugly.

Sorry if I have made any mistakes, I'm typing this in a hurry.

[quote="1=2"]That is how you should solve higher degree polynomials. [/quote]
For polynomials with small coefficients and of moderate degree, it's often more efficient to substitute various values for $ x$ and compute the prime factorizations of the corresponding integers.  Then you can guess and interpolate the irreducible factors from there.  (For example, if the polynomial is prime for more than a few values of $ x$ then you can be pretty sure that the polynomial is irreducible.)

For even larger polynomials, it's often much more efficient to factor them [url=http://en.wikipedia.org/wiki/Polynomial_factorization]over finite fields[/url] instead; this restricts the possible irreducible factors by CRT.

there are some tools to use like grouping. But for x^3-8x+3, i would try the factor theorem or just put it in a TI-84 and look on the table for zeroes.

Ok thanks everyone.
Some of those things you talked about are beyond me though.
I get the basic idea of these odd un-orthodox factoring problems now.


Is there an official site to one of these two where I can find official solutions?

Thanks

Please type next time in this way Belarus.
There is no official math olimpiad site. As I understand it is impossible or very, very hard to find it in the internet.

Try mateforum.ro, but it is in Romanian and I think that it isn't official.

[url]http://www.rms.unibuc.ro/[/url]

Romanian mathematical society. Could help.

[quote]http://www.rms.unibuc.ro/ [/quote]
Useless. If you want subjects from the past, try http://www.olimpiade.ro .But it's not only for math. And http://www.mateforum.ro has a section dedicated to math contests and olympiads ,but it's not official.

[quote="gordon_c"][quote]http://www.rms.unibuc.ro/ [/quote]
Useless. If you want past subjects, try http://www.olimpiade.ro .But it's not only for math. And http://www.mateforum.ro has a section dedicated to math contests and olympiad ,but it's not official.[/quote]

I see now, after browsing the pages... :D

Can someone translate some romanian olympiads (1990+) ?  :blush:


function MATHCOUNTS LaTeX calculus derivative

ok this is the part where i get confused..

Given F(X)=3x^2+2   find [f(x+h)]-f(x)]/h   [6 points]

so i start off like this..
[3(x+h)^2+2]-[3x^2+2]/h

see the part where it says 3(x+h)^2? im not quite sure how to simplify that. In my previous exam i did 3x^2 + 3h^2, which i got deducted -2.5 points.
so yeah if someone can explain that to me, would be grateful~!

I think this could be moved to mathcounts :D
in latex:  given $f(x)=3x^{2}+2$, find $\frac{f(x+h)-f(x)}{h}$
[hide="solution"]$\frac{f(x+h)-f(x)}{h}=\frac{(3x^{2}+3h^{2}+6hx+2)-(3x^{2}+2)}{h}=3h+6x$
[/hide]
this problem is basically finding the derivative of $3x^{2}+2$ except skipping the important part about limits:D


Express the product $ 12\frac{1}{2}\times10\frac{1}{6}$ as a mixed number.

Ok.
So we first express them as common fractions:
$ \frac{25}{2}*\frac{61}{6}$. So then we multiply and we get:
$ \frac{1525}{12}$.
So if we express this as a mixed number, it would be $ 127\frac{1}{12}$.


quadratics
function
conics
ellipse
parabola
special factorizations

How do you convert this equation:

f(x)=x^2+6x+9

into the h and k form:

f(x)=a(x-h)^2+k

 :?:

This is a bit too advanced to be in Classroom math, and should probably be moved to a higher section..
to do that though, you need to use the technique "Completing the square"
This is the technique itself:
$ f(x) \equal{} ax^2 \plus{} bx \plus{} c$
$ f(x) \equal{} a(x^2 \plus{} \frac {b}ax) \plus{} c$
Then you add and subtract he square of half the coefficient of x inside the brackets.
$ f(x) \equal{} a[x^2 \plus{} \frac {b}ax \plus{} (\frac {b}{2a})^2 \minus{} (\frac {b}{2a})^2] \plus{} c$

$ f(x) \equal{} a[x^2 \plus{} \frac {b}ax \plus{} (\frac {b}{2a})^2] \plus{} c$
And finally get it to the final form:
$ f(x) \equal{} a(x \plus{} \frac {b}{2a})^2 \minus{} \frac {b^2}{4a} \plus{} c$

Now, all we need to do is plug in your equation ($ x^2 \plus{} 6x \plus{} 9$)to that one.

So we get

$ f(x) \equal{} 1(x \plus{} \frac {6}2)^2 \minus{} \frac {6^2}4 \plus{} 9$
$ f(x) \equal{} 1(x \plus{} 3)^2 \minus{} 0$
or just
$ \boxed{f(x) \equal{} (x \plus{} 3)^2}$
which is the completed form.

 :D

I think this is OK for Classroom Math, BlackSwordsman. It's a bit too easy for high school..

Actually Blackswordsman,

You can easily see that $ f(x)$ is a factorable function because $ \left(\frac{6}{2}\right)^2\equal{}9$, so you don't really need to complete the square, even though it is the way to go for this style of problem.

Usually completing the square is the best way to go whenever you have a function with an $ ax^{2}$ term in there, such as circles, ellipses, parabolas,...

For conics (excluding parabolas), you wind up with pesky $ y^2$ terms that can be in the way. But yes, for simplifying quadratics in to $ (x\minus{}h)^2\plus{}k$ then the completing the square is the best method.


function
algebra
domain
inequalities proposed
inequalities

Let be $ a,b,c>0$ . Prove that :

$ \sum_{cyc}\frac {\sqrt {a^2 \plus{} 2b^2 \plus{} c^2}}{a^2 \plus{} 2bc} \geq \frac {6}{a \plus{} b \plus{} c}$

[quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ \sum_{cyc}\frac {\sqrt {a^2 \plus{} 2b^2 \plus{} c^2}}{a^2 \plus{} 2bc} \geq \frac {6}{a \plus{} b \plus{} c}$[/quote]

First, by Cauchy, we have:
$ \sum_{cyc}\frac {\sqrt {a^2 \plus{} 2b^2 \plus{} c^2}}{a^2 \plus{} 2bc} \geq \frac{\left( \sum \sqrt[4]{2a^2\plus{}b^2\plus{}c^2}\right)^2 }{(a\plus{}b\plus{}c)^2}$
So we only need to show that:
$ \sum \sqrt[4]{2a^2\plus{}b^2\plus{}c^2} \geq \sqrt{6(a\plus{}b\plus{}c)}$

Because this is homogeneous, let $ a^2\plus{}b^2\plus{}c^2 \equal{} 3$, and consider the function $ f(t) \equal{} \sqrt[4]{t^2 \plus{} 3}$ defined on $ 0 < t < \sqrt{3}$
Turns out that $ f''(t) \equal{} \frac{(6\minus{}t^2)(\sqrt[3]{t^2\plus{}3})}{4t^4\plus{}24t^2\plus{}36}$ which is positive in the domain of $ f$.

So
$ LHS \equal{} f(a) \plus{} f(b) \plus{} f(c) \geq 3 f(1) \equal{} 3 \sqrt{2}$
But $ \sqrt{a\plus{}b\plus{}c} \leq \sqrt{3}$ from AM-RMS, so $ RHS \leq 3 \sqrt{2}$

[quote="alex2008"]Let be $ a,b,c > 0$ . Prove that :

$ \sum_{cyc}\frac {\sqrt {a^2 \plus{} 2b^2 \plus{} c^2}}{a^2 \plus{} 2bc} \geq \frac {6}{a \plus{} b \plus{} c}$[/quote]
It's obviously trues because with $ a\plus{}b\plus{}c\equal{}3$, easily to prove: $ (a^2\plus{}3)(b^2\plus{}3)(c^2\plus{}3) \geq\ 64$
 :)


Hi,

Can't solve this problem - need help. This problem is in the Mathcounts 2008 Prep, Warm Up #17, Problem #2.

Question:
Two different circles are each tangent to both the X-axis and the Y-axis, as shown in the figure. One of the intersection points of the two circles is (1,5). What is the product of the lengths of the radii?

http://www.stutzfamily.com/mrstutz/mathcounts/Resources0708/WK06HWKProbs25-27.pdf

Thanks very much.

Since it's a request for help, here you go, but hidden incase others want a crack at it.

This is from this year's hand book. There are a few tricky geometry problems near the end of volume II of the handbook.

Frost

[hide]
The equation for a circle is  : $ (x\minus{}x_1)^2 \plus{} (y\minus{}y_1)^2 \equal{} r^2$  with $ (x_1, y_1)$ being the center and r the radius.
Since the two circles are tangent to both axis, the line x=y passes through their centers and the x=y=r.  Also since they both pass through (1,5), I can write this equation
$ (1\minus{}r)^2 \plus{} (5\minus{}r)^2 \equal{} r^2$
$ 1\minus{}2r \plus{}r^2 \plus{} 25\minus{}10r \plus{} r^2 \equal{} r^2$
$ r^2\minus{}12r \plus{} 26 \equal{} 0$
This isn’t factorable, but the product of the two possible r’s is 26.  Since the equation doesn’t specify which circle that equation is for, it really is for both circles.  So the product of the two radii is 26
[/hide]

Thanks very much for your help, frost13!


linear algebra
matrix
probability
expected value

How the game works:

This is a game for two. For Player 1 can only choose from the columns (A,B,C) and the row player can only choose from Row 1 and Row 2.

In each round, the players choose their columns and rows simultaneously. If the resulting number is postive, Player 1 wins. For example, player 1 chose B and Player 2 chose 1, so the matrix cell is B1 which contains 17. In that case, player 1 wins 17 points and player 2 looses 17 points.

Score:

Player 1: 17
Player 2: - 17

However, if the number chosen is negative player 2 wins.
For example, the cell chosen is C1, then, player 2 gains 61 points and player 1 looses 61 points.

Score:

Player 1: -44
Player 2: 44

1.) Is the matrix fair?
2.) Assuming that the game is played 10,000 times (rounds), and the players are equally intelligent and equally discerning? Who do you think will win?

[quote="diablos3"]It seems fair since the expected value is 0. After 10,000 trials it's impossible to say who would win because theoretically they would be tied.[/quote]

[hide="but..."]
What if player 1's strategy is to never choose column C?[/hide]

How would you design a board such that the probability of winning of the Row player is $ 2/3$?


If you got $n$ ones and $m$ zeros, how many permutations are there in which we have $x$ consecutive ones?

 :cool:

let me announce "the return of peter VDD"

Thanks  :lol:

Yup, I'm back here. I couldn't go here during exams :( anyway those are over now :D

(but it's only for some days too bad, I'm leaving on holidays July 5th early in the morning :P)


function
topology
real analysis
advanced fields
advanced fields unsolved

Hello there,

I'm desperately searching for an (elementary!) proof of the following problem: Define the Hopf fibration by $ h: S^3\equal{}\{(x,y)\in\mathbb{C}^2: |x|^2\plus{}|y|^2\equal{}1\}\to\mathbb{CP}^1,(x,y)\mapsto [x: y]$. There doesn't exist a smooth function $ j: \mathbb{CP}^1\to S^3$ with $ h(j[x: y])\equal{}[x: y]$ for every $ [x: y]\in\mathbb{CP}^1$.

laxton

Hi. A continuous section would yield a homeomorphism $ {\mathbb S}^3 \cong {\mathbb S}^1\times {\mathbb S}^2$.


i think that $\lim(u_{n})=-2 $ but i can't prove it.

Draw $f(x)=1/2.x^2+x -2$ and $y=x$
See what happen for the sequence $u_n$ monotony ? bounded ?
Its limit verify $L=1/2L^2 + L -2$

Can you explain anymore about your proof?
This one is not monotonous, so it's not easy at all.

nice problem. set $x_n=u_{2n+1}-2$. we get $x_{n+1}=x_n(\frac 18 x_n^3-\frac 14 x_n^2+1)$. we will prove inductively that $-1<x_n<0$. indeed, this is true for $x_1$. it follows that $0<\frac 18 x_n^3-\frac 14 x_n^2+1<1$ by a little algebra. thus, we multiply $x_n$ by a factor between 0 and 1 to get $x_{n+1}$, so we get the inequality. it follows that $x_n$ is strictly increasing since $x_n$ is negative, but it is bounded by 0, thus it is convergent. thus $x_n$ converges to 0 by a little algebra. thus $u_{2n+1}$ converges to -2. finally, by the recursion for $u_n$, we also have $u_{2n}$ converging to -2, so we get the result. i hope i didn't miscalculate.

Peter


a) Let $ p,q,r$ be positive real numbers such that $ p^2\plus{}q^2\plus{}r^2\plus{}2pqr\equal{}1$ Show that there exists an acute triangle $ ABC$ such that $ p\equal{}\cos{A}$, $ q\equal{}\cos{B}$, $ r\equal{}\cos{C}$
b) Let $ p,q,r>\equal{}0$ with $ p^2\plus{}q^2\plus{}r^2\plus{}2pqr\equal{}1$. Show that there are $ A,B,C$ $ \in$ $ [0,\frac{\pi}{2}]$ with $ p\equal{}\cos{A}$ $ q\equal{}\cos{B}, r\equal{}\cos{C}$, and $ A\plus{}B\plus{}C\equal{}\pi$

I don't really see the difference between [b]a)[/b] and [b]b)[/b] now. Maybe it's too late? :maybe:

Anyway, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=151902#151902]http://www.mathlinks.ro/Forum/viewtopic.php?t=23888 post #8[/url] for a proof (maybe not the best).

  darij

I didn't see it also, that's why I posted it. By the way, my solution was like yours.
 :)

In fact the following identity is true for angles $ (A,B,C)$ of any triangle:

$ \cos^2{A}\plus{}\cos^2{B}\plus{}\cos^2{C}\plus{}2\cos{A}\cos{B}\cos{C}\equal{}1$

others interesting sobstitutions:

c)Let x,y,z real number such that $ x^2\plus{}y^2\plus{}z^2 \plus{} 2xyz \equal{} 1$ then there is a triangle ABC such that $ x \equal{} \sin{\frac{A}{2}}$, $ y \equal{} \sin{\frac{B}{2}}$ and $ z \equal{} \sin{\frac{C}{2}}$.

d)Let x, y, z real numbers with $ xy \plus{}yz\plus{}zx \equal{} 1$, then there exists a triangle ABC such that $ \cot A \equal{} x$, $ \cot B \equal{} y$, and $ \cot C \equal{} z$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

What is an example of a simple group $ G$ such that $ K\triangleright H\leq G$ such that $ H\diagup K$ is not isomorphic to a subgroup of $ G$ (if it exists).

I feel like $ G$ should be large. Small examples don't seem to work, like any of the $ A_n$ that are simple. If you look at say $ A_4\leq A_5$, then quotienting by the only normal subgroup gets you the group of order $ 3$, which is in $ A_5$. This becomes a problem for any small group. You end up getting such small things that they are forced to be in the big group. Possibly infinite would work?

(Also, I've simplified the search. I am actually looking for a group that you can write as a quotient of a subgroup but can't write as a subgroup of a quotient).


[quote="Persia"]Hi and thank you for providing such a opportunity.

Question:

[img]http://i177.photobucket.com/albums/w229/Persian_Empire_2007/series2.jpg[/img]

I know this question makes the use of e^j0 = cos0 + jsin0 (0= theater) but I cant quite undrestand how.[/quote]

Look here for some ideas:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=120999[/url]


Let  ABC  be a triangle. Let  X  be a point on  AB  and  Y  be a point on  AC  such that  AX=XY=YB=BC  . If the angle  ABC  measures  128  degrees, find the number of degress in the measure of the angle  BAC .

Must be easy, i even saw the solution from the mathcounts drills site. But i dont understand it, can someone explain it to me (and perhaps, draw a diagram - i just cant 'see it')?

Since I can't get an image, you'll have to draw it. I think this is the solution.

[hide]So we have a triangle ABC, with AX=XY=YB=BC. There are a bunch of isosceles triangles. Label the one at angle at A a, at BYC b, and at XBY c.

a+b+(180-2c)=180
a+b-2c=0

a+b+128=180
a+b=52

2a+(180-c)=180
2a-c=0

Solve and get a=13, b=39, and c=26.

We want BAC, which is A, so it is 13.

Is that right?[/hide]

yes that is right-thanks, and yay i finally got it :)


geometry
3D geometry
linear algebra
matrix
algebra
polynomial
invariant

I think this has been posted before, but since the interest was not too big and there wasn't a complete solution and since it is really nice problem, here it goes:
  For which $3x3$ matrices $A$ there exists a real matrix $B$ such that $B^3=A$?

First of all, the invertible matrices are all cubes, I think. In order to see this, it suffices to check Jordan cells of dimension $1,\ 2$ and $3$ with non-zero diagonal entries. It's clear for those of dimension $1$, which are simply scalars, those of dimension $2$ are similar to a matrix of the form $\begin{pmatrix}\alpha^3&0\\3\alpha^2&\alpha^3\end{pmatrix}$ with non-zero $\alpha$, which is the cube of $\begin{pmatrix}\alpha&0\\1&\alpha\end{pmatrix}$, and those of dimension $3$ are similar to a matrix of the form $\begin{pmatrix}\alpha^3&0&0\\3\alpha^2&\alpha^3&0\\3\alpha&3\alpha^2&\alpha^3\end{pmatrix}$, which is the cube of $\begin{pmatrix}\alpha&0&0\\1&\alpha&0\\0&1&\alpha\end{pmatrix}$.

If the matrix has all eigenvalues equal to $0$, then it must be the null matrix.

If it has exactly two null eigenvalues, then it's a cube iff its minimal polynomial has degree $2$, i.e. iff it's diagonalizable. Otherwise, a third root of such a matrix would be similar to some $\begin{pmatrix}\alpha&0&0\\0&0&0\\0&1&0\end{pmatrix}$, whose cube is diagonalizable and thus not similar to our matrix.

Finally, if it has precisely one eigenvalue equal to zero, then it's a cube and this can be shown just like in the first paragraph, since its null eigenvalue does not belong to any Jordan cells, and the argument used there works.

In conclusion, it looks like the only matrices which are not cubes are the non-diagonalizable ones which have at least two null eigenvalues. Is it correct? :? It sounds rather strange (not a very nice answer :)).

Is every matrix in $GL_4(\mathbb{R})$ a cube.

i guess not
if $\lambda$ is an eigenvalue of $B$ and $B^3=A$ then $\lambda^3$ is an eigenvalue of $A$
but the invertible matrix $A$ has no real eigenvalues, with $A$ :

$\begin{array}{cccc}0&1&0&0\\-1&0&0&0\\0&0&0&1\\0&0&-1&0\end{array}$

It's pretty clear that fredbel's matrix satisfies $A=(-A)^3$, so.. :)

I'm almost sure that by using something similar to what I did in post #2, one can prove that every invertible real matrix, no matter what its order is, is a cube, but I'll have to recheck the proof, and I don't have the time to do that now. I'll post it if I manage to convince myself :).

I'm not sure a similar method works as well since you have to work in the real case with invariant factor, and thus prove that any invertible companion matrix is a cube, wich is not as easy than Jordan blocks.

I think it does work, it just needs a bit more work, and it's harder to explain. It suffices to prove that every block in the rational canonical form over $\mathbb R$ of our matrix is a cube. Such blocks are of the form $(*)\ \begin{pmatrix}A&0&0&\cdots&0&0\\T&A&0&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0\cdots&T&A\end{pmatrix}$, where $A$ is a real $2\times 2$ invertible matrix with non-real eigenvalues, and $T=\begin{pmatrix}0&0\\1&0\end{pmatrix}$.

Lift a matrix $B$ of this form and of order $2k$ (it's clear that the order is even) to the $3$'rd power and you'll get a matrix whose Dunford decomposition is $D+N$ is such that the smallest exponent $t$ such that $N^t=0$is $t=k$. If the  diagonal blocks of $D$ are chosen to be cube roots of $A$, the matrix $B^3$ will be similar to $(*)$.

It thus suffices to prove that every invertible $2\times 2$ matrix is a cube. I haven't even checked this, but I think it's true :).

Sorry if it's wrong and/or unclear, but I really can't explain it better :).

I might be wrong, but I really do not see the mistake in the following proof of the fact that we can choose the root to be a polynomial in $A$ if $A$ is real invertible:
   The characteristic polynomial of $A$ has of course the form $\prod_{i}{(X-a_i)^{m_i}}\cdot\prod_{i}{(X^2+b_i X+c_i)^{n_i}}$ for some $a_i,b_i,c_i$ reals such that $a_i,c_i$ are not 0 and $b_i^2<4c_i$. If we manage to prove that for any $a$ non-zero and any $k$ there is a real polynomial $P(a,k)$ such that $(X-a)^k| P^3(a,k)(X)-X$ and for any $b^2<4c$ there is a polynomial $P(b,c,k)$ for which $(X^2+bX+c)^k| P^3(b,c,k)(X)-X$ I think we are done by the chinese remainder theorem and Hamilton-Cayley. The existence of $P(a,k)$ is immediate by Hensel's lifting method. Let's prove the existence of $P(b,c,k)$ by induction. We will speak about the case $k=1$ later.  Suppose now that we prove the existence of a polynomial $P$ in $R[X]$ such that $P^3(X)-X=Q(X)( X^2+aX+b)^k$. We may assume that $Q$ is not a multiple of $X^2+aX+b$. The expression $-\frac{ Q(z)}{3P^2(z)}$ is a polynomial in $z$ since $P$ is relatively prime with the irreducible polynomial $X^2+aX+b$. Call $H$ this polynomial. It is then immediate that $P(x)+H(x)(X^2+aX+b)^k$ will do the job for $k+1$.
   Now, the case $k=1$. This comes down to: for any real numbers $a,b$ with non-zero $b$  we can find $P$ with real coefficients such that $X^2+b^2| P^3(X)-X+a$, which can also be interpreted as: any $z$ has a third root which is a real polynomial in $i\cdot Im(z)$. But this is very simple: just write $z$ in trigonometric form.
   What is wrong here?

Am i getting old or what?
Oh my grobber you are right for some reason I was still half working with 3*3 matrices which always have a real eigenvalue :o


Show that each number in the sequence $49$, $4489$, $444889$, $44448889$,$...$ is a perfect square.

write the general term, and make square. it's easy.

Try to use formulas like $\underbrace{111\cdots 111}_{n \textrm{ times}} = \frac{10^n - 1}{9}.$


Problem 2.7

2.7	An optically active alkyne D has 89.52% carbon. Compound D can be catalytically hydrogenated to n-butylcyclohexane. Treatment of D with C2H5MgBr liberates no gas. Catalytic hydrogenation of D over Pd/C in the presence of quinoline (a catalyst poison) and treatment of the product with ozone and then H2O2 gives an optically active tricarboxylic acid E (C8H12O6). Compound E on heating undergoes dehydration by loss of a molecule of water to give F. Draw the possible structures of D, E and F.								(3 marks)

Problem 2.8

2.8	When equimolar amounts of 1,3-butadiene and bromine are reacted at high temperature, two compounds G (major) and H (minor) with the formula C4H6Br2 are formed. Compound G reacts with more Br2 to form compound I (C4H6Br4) which proved to be a meso compound. Compound H reacts with more Br2 to form both I and a diastereomer J. Draw the structures of G and H. Draw the Fischer projection for I and J indicating the stereochemistry (with R/S designation) of the chiral carbons.
											(4 marks)

Problem 2.9

Diels-Alder reaction of 2,5-dimethylfuran and maleic anhydride gives compound K which can exist in two stereoisomeric forms. Draw the structures of the two isomers.   (1 mark)

Problem 2.10

2.10	Compound K undergoes acid catalysed dehydration to give L (C10H8O3). Draw the structure of L.									(1 mark)

[hide="Problem 2.7"]From the percentage of carbon and the fact that D is hydrogenated to butylcyclohexane, we can conclude that the molecular formula of D is $ \ce{C_{10}H_{14}}$, from which we get IDH = 4: this this means that, besides the triple bond and the ring, there is an additional double bond. The triple bond cannot be on the ring, and it cannot also be on the end of the side chain, since there is no liberation of gas by reaction with the Grignard. Also, the double bond cannot be on carbon 1 of the ring, otherwise there would be no place for the stereocenter. On the other hand, the stereocenter must be on that carbon 1.
Now, from the molecular formula of E, and from the fact that E is also optically active, we can conclude that E is $ \ce{HO2C\minus{}CH2CH2CH2CH(CO2H)\minus{}CH2CO2H}$, from which it follows that D is 3-(2-butynyl)-cyclohexene.[/hide]

Problem 2.9.

The two stereoisomers of K are shown bellow:

Problem 2.10.

The structure of L is shown bellow:

2.8

G is trans 1,4-dibromo but-2-ene, H is 3,4-dibromo butene, I is S,R-1,2,3,4-tetrabromo butane, J is R,R-1,2,3,4-tetrabromo butane

I remember solving this in the exam! :D  :D


If you can draw any number of circles in the plane, how many colors do you need at least to color the entire plane in such a way that no 2 adjacent regions have the same color? [we call 2 regions adjacent if their shared border is more than some points, thus is really a curved line]

Although this has not yet been proven, it is assumed that 4 colors are enough to color in any 2-dimensional pattern.

I think.

4 are indeed enough. But 4 are enough on any map. :) (the famous 4-color problem)

But for this case, you can do it with less... how many and how to do?

Two colors are enough. Use one color where an odd number of circles intersect. Use the other color where an even number of circles intersect.

If that is wrong, provide a counterexample.

Eum... 2 is correct, but I don't understand your construction. The circles divide the plane in some regions, you have to color the regions in those colors... not the points (which is what I assume you did?)


[quote="mdk"]Let x, y > 0. Show that x^y + y^x >1[/quote]

[url=http://www.qbyte.org/puzzles/p044s.html]Here[/url]

[quote="http://www.qbyte.org/puzzles/p044s.html"]
It can be shown that the minimum value of $ x^{y}+y^{x}$ for positive $ x$ and $ y$ is $ 2e^{-\frac{1}{e}}$ [/quote]
I think, it's wrong. It cann't be shown. :wink:


abstract algebra
function
superior algebra
superior algebra unsolved

if $\pi : G \to G/N,$ and $H,N \leq G$
and $Ker (\pi)=N$
Is it true that $Ker(\pi |_{H})=H\cap N$

$N$ must be normal but $H$ does not need to be normal, and the answer is yes.


inequalities
trigonometry
inequalities unsolved

[quote="darkmaster"]A,B,C are three angles of triangle, prove that:
     $\sum_{sym} \frac {3-sin(A/2)}{3sin(A/2)} <= \frac 5{8sin(A/2)sin(B/2)sin(C/2)}$[/quote]

Note that $ cos(\frac{B+C}{2}) = sin(\frac{A}{2}) $
We now have to prove this inequality for acute triangle
  $\sum_{sym} \frac {3-cosA}{3cosA} \leq \frac 5{8cosAcosBcosC}$ (*)
 Use 2 indentities :
$\sum cosAcosB = \frac{p^2+r^2-4R^2}{4R^2} $
$  cosAcosBcosC = \frac{p^2-(2R+r)^2}{4R^2} $

(*) can be written as
$ 4r^2 + 8Rr \leq \ 5R^2 $
$ (R-2r)(5R+2r) \geq \ 0$

So the inequality is proved. The equality holds iff ABC is an equilateral triangle.

[quote="tranminhhoang"]Note that $ cos(\frac{B+C}{2}) = sin(\frac{A}{2}) $
We now have to prove this inequality for acute triangle
  $\sum_{sym} \frac {3-cosA}{3cosA} \leq \frac 5{8cosAcosBcosC}$ (*)
[/quote]

Hoang, very cute substitution!!!

Another way to look at this after your beautiful substitution:

Using $\cos^2A + \cos^2B + \cos^2C + 2\cos A\cos B\cos C =1$,

(*) is equivalent to $(\cos A+\cos B+\cos C)^2 \leq \frac{9}{4}$ which is clear.

Thanks fuzzylogic. But darij grinberg had used that substitution before in the topic "Acute triangle". And your solution is very simple and nice.
However, I think we use this kind of substitution because expressions with $cosA $ is rather easy to work with than $ sin(\frac{A}{2}) $.


Does anbody know where can I find full results of IOI 2004?

On official page, there`s only "Coming soon"..

Hope this is a right place for this question..

I don't think it's over yet.  It says September 11-18.  They would probably have the results after the awards ceromony.

i believe that the official competition is over as of this moment.  the official results will be (should be) announced this weekend

Thanks.

I found just a little bit about Croatian team results..

If there are sites like usaco, acm.uva.es and many others wiht lots of informatics problem for practise, how come there`s no forum like mathlinks` one?

[quote="idioteque"]If there are sites like usaco, acm.uva.es and many others wiht lots of informatics problem for practise, how come there`s no forum like mathlinks` one?[/quote]

I dunno...perhaps b/c there wouldn't be as many people interested?...i'm not sure how many do USACO...

The US team's results are posted at [url]http://www.usaco.org/[/url].  Looks like Anders Kaseorg (who was the US IMO team alternate) and Brian Jacokes won gold. Alex Schwender and Eric Price (both MOPpers) won silver.  Seems like a decent result for the US team.

If anyone is interested, you can see full results at http://www.ioi2004.org/html/competition_score.html .


I live in Florida and i will be in sixth grade. Does Florida have to be participating in PMWC for me to take the entrance test?

This forum is not the best place for this question. This is a [b]mathcounts[/b] forum.

EDIT: Try asking a moderator where this would go best.

there isn't really a [b]PMWC[/b] forum either

Just contact Texas Mathworks at mathworks@txstate.edu; http://www.txstate.edu/mathworks/ for more info.  They can send a test to your math teacher/coach to give to you.  I know that people from out of state are allowed into the summer Mathworks program, and, if you qualify, might be accepted onto the team.

Looking on the internet I also see that California (with MathEdge) also has a program. I am not acquainted with that program.  You can contact them at http://www.mathedge.org, MathEdge@gmail.com.

Or ask in the International Contests and Programs forum.


I encounter a difficult limit, and I need someone help me calculate  it. Waiting for your reply.

\[\text{Calculate the limit:} \displaystyle\lim_{x\to 0} \&\&\&\&\&\& \frac{\sqrt[p]{1+px}\sqrt[q]{1+qx}-\sqrt[r]{1+rx}}{\sqrt[s]{1+sx}-1} \&\&\&\&\&\text{where p,q,r and s} \in \mathbb{N}^4.\]

just use $\lim\frac{(1+x)^a-1}{ ax}=1,x \to 0$
my answer is 1.

Shouldn't $p,q,r,s \in \mathbb{N}$, not $\mathbb{N}^4$?  p,q,r, and s aren't quadruples of integers, are they? :D

i cannot understand zhaobin resolution. Could you please post the complete resolution?

let me try to explain my idea.
we can $(1+x)^a=1+ax+o(x),x \to 0$
then $ \lim_{x\to 0} \&\&\&\&\&\& \frac{\sqrt[p]{1+px}\sqrt[q]{1+qx}-\sqrt[r]{1+rx}}{\sqrt[s]{1+sx}-1}=\lim_{x \to 0} \frac{(1+x+o(x))(1+x+o(x))-1+x+o(x)}{x+o(x)}$
the answer sure =1
where  $lim_{x \to 0}\frac{o(x)}{x}=0 $

There is a slight typo in the numerator of the last term there -- it should be $-1 - x + o(x)$, not $-1 + x + o(x)$.


Hi, I wanted to know if I needed to submit any forms for Round 2. I already Submitted the Permission form, Under 13 form, Entry Form, whatever else in Round 1. Do I need to submit anything else besides my solutions?


um...why am i the charachter in your post?  Can't it be you?...oh well anyways
[hide]
i did
561+344 and that gives me 905
but i think that's wrong for some reason.[/hide]

maybe this is an easier example: you have 3 balls: 2 has red colors in it, 2 has blue colors in it  :what?:  :what?:  :what?:
yep so this means that....(please reply my post and yes you can make a short reply with less than 6 words)

[quote="10000th User"]math92 posted 561 math posts and 344 science posts. If 289 of his posts are both math and science posts, what is the total number of posts by math92?[/quote]

[hide]  We have to add posts that are only-math, posts that are only-sceince, and posts that are both math and science related.  Thus we see:

$ (561-289)    +   (344-289)   +  289=  616$  Thus the total number of posts by math92 is $\boxed{616}$
[/hide]

[hide="answer"]

[quote="math92"]i did
561+344 and that gives me 905

but i think that's wrong for some reason.[/quote]
your answer - $289*2$ is right.
[/hide]


function topology geometry calculus integration trigonometry

Talk about Song Contest..

Who thought Lehigh Valley's song rocked? ME!!!!! :D  :rotfl:

Kurt was a natural performer.. Haha..

Haha, we didn't qualify, but we went to the Preliminaries and did a rap.  Or actually, I did a rap while my teammates wached.  We did pretty well, considering I wrote the lyrics 30 seconds before we had to get up and start.

I thought "Hey Snake" was better on Friday night than it was on Saturday, but they had a piano and much better audience participation then too (the entire room was singing the "Na Na Na" part, which was a very cool experience.)  I think the Friday night atmosphere overall is much more conducive to good performances.

Once again, TJ's shot at Song Contest glory was lost for want of a microphone.  We were the only one with a guitar solo, though; that must count for something!

The song contest was a lot of fun.  I couldn't hear the lyrics very well, though, especially for the prelims on Friday night.  What I heard was clever, so I was wondering if people would mind posting the songs here.  It would be fun to see them in all their glory. :-)

Best,
Catherine

Song contest did not occur at Las Vegas since the award ceremony place was moved to the cafeteria where there was no room (due to Mr. Rusczyk's food poisoning + heat exhaustion). Some ppl did perform outside, just not for an award, including our team.

[quote="mathgeek2006"]I only watched the A-R-M-L song (went to the tune of YMCA...) at the Las Vegas site.  Then we had to go start packing.  That song had pretty good lyrics, though.[/quote]
That was us: lyrics due to Jacob Levine of the Southern California team.

Hmm, I guess most ARMLers don't listen to rap.  The SC rap "Workin' These Problems" was a parody of "Poppin' My Collar."  I think it would have went down better if people had heard the original song.  As it was, I saw only one girl in the audience who got the reference.  Oh well...

I don't remember the first and second verse completely, but the refrain was something like this (dance choreography in parentheses):

Cause when we do those proofs, I get this feeling (flaired legs and bye-bye-bye move except as if writing)
Solving step by step, it's the peak of my life (footsteps, then reach up, pull down to left leg, and wave)
Don't cha know my world's all a blur, when I'm tangent to your curves (jazz hands over eyes and then hip roll with right hand)
I need you to survive (point out and then draw back to chest)

When your curves revolve around my axis, (hand jive/spin in clockwise motion)
I can swear the volume multiplies (count out fingers)
I think about you all through class, (hands on temple, nod head back and forth from right to left)
I wanna feel your ___ (right foot step out and slap on hip)
I need you in my life (point out, draw back to chest, and look down)

It was a parody of Everytime We Touch by Cascada, which is an awesome song by the way.  We had the karoke music on an ipod so we used that...although, it still had some background vocals occasionally which threw us off.

Maybe I'll post the whole song later.

No, you guys just sucked. :D It was too short. lol funny lyrics though

Lehigh Valley Fire's song was amazing though. And Hey Snake was pretty good too.

Our team has never come before, so we didn't realize that the 30 second time limit was meaningless, that we could include recorded music and dancing, or that we would have to be very loud to be heard in the preliminaries.  Also we (actually, I) wrote the lyrics two days before we left.  They were:

(Tune: "Modern Major General" by Gilbert and Sullivan)

 I am the very model of a modern math'matician
 I know all of the generating functions for partitions
 I understand the details of the non-Hausdorff topologies
 I'll quickly list all of the groups with singular homologies

 In fields of math from algebra to spherical geometry
 I'll tolerate abstractions and definition pedantry
 But should a number enter in my usefulness evaporates
 Subtraction isn't kind to me, I cannot find the square of eight

 I know the trig'nometric integrals  but sine of pi eludes me well
 I work with Grassman algebras but can't solve three x equals twelve
 In short, on matters relevant to non-academicians
 I am the very model of a modern math'matician

yeah, mysmartmouth, we would've won if you used my lyrics instead...or did you? i don't remember...

and yeah, the only really good thing about our song was Arnav's piano skills.

Yep.  Our song was amazingly short, because we were the only team that abidedby the 30 second time limit.  Also, we thought we weren't allowed to have music, besides what we played ourself.

Also, the judging at Penn State was flawed for the final vote.  The judge stood closer to one side, and one side clapped for Lehigh Valley, the other side clapped for Massey Ontario.  I liked Lehigh Valley's song better anyway, but to be fair, Massey Ontario probably should have won, based on applause.  Oh well...

Just so everybody who watched the Penn State preliminaries knows, the rest of us have disowned Sean.

3 don't underst3nd it... SC should h3ve g3tten past the prelims.. And th3n, in the finals....Ontari3 shoulda w3n to3.

[quote="dakyru"]
It was a parody of Everytime We Touch by Cascada, which is an awesome song by the way.  We had the karoke music on an ipod so we used that...although, it still had some background vocals occasionally which threw us off.

Maybe I'll post the whole song later.
[/quote]

The beat of "Everytime We Touch" is cool, just the lyrics stink. You should have used a midi file if you had one.

[quote="Xantos C. Guin"]Lehigh Valley should have used a midi file instead of the format they used. That way, the words of the original song wouldn't be heard.

Personally, I didn't like any of the three songs at all. Someone on our team said that Hillary Duff and Jamie Spears could do better.  :lol:

Georgia couldn't think of a line to rhyme with $dx$ besides "the obvious".
BTW: Boy Soprano II had nothing to do with that line.[/quote]

Personally, "lots of math, no vertex" would have worked just fine in my opinion. And I am in no way associated with that song of GA's: I neither encouraged it, nor helped in writing it, nor sang it.

I thought it was kinda lame how a bunch of people on the GA team thought that the song was really awesome. Personally I didn't think it was that great. Lehigh Valley's song was better—they chose a better song to parody, their lyrics were better, and they actually had choreographed moves. The only problem was that when they performed it they were kinda desynchronized with the recording.

The Georgia team did overreact a little.  I thought that it was a mistake to judge by the applause when the performing teams were sitting in the audience, because that adds too much of Team Enthusiasm Contest.  To be blunt, I found Lehigh's song a little creepy.  I liked "Hey, Snake", although performance at the actual event was rather hurt by lack of piano.  Also, Mr. Koppelman [a GA coach] pointed out after the event that when you change \ks\ to \kst\, you get a whole new series of rhymes :  text, context, etc.  To be honest, I think that line was just there for a cheap laugh—which it got.  It would have been fairly easy to change the previous line to match a new one.  Anyhow, it was originally $x$ this, $x$ that, not $x$ this, $\sim x$ that.

Well, you guys didn't hear the "Lobatchevsky" section (replaced "Galileo") :D   Plus, it was probably funnier for Georgia people, because there were certain references (later omitted) to Georgia coaches, jokes, counties, etc.  (Honestly, what other state has a county called Lumpkin?)


trigonometry
complex numbers
complex analysis

Evaluate the following expression using complex numbers
$ \cos\frac{2\pi}{15}.\cos\frac{4\pi}{15}.\cos\frac{8\pi}{15}.\cos\frac{16\pi}{15}$

[quote="RAGHAV GROVER"]Evaluate the following expression using complex numbers
$ \cos\frac {2\pi}{15}.\cos\frac {4\pi}{15}.\cos\frac {8\pi}{15}.\cos\frac {16\pi}{15}$[/quote]

$ 16\sin\frac {2\pi}{15} .\cos\frac {2\pi}{15}.\cos\frac {4\pi}{15}.\cos\frac {8\pi}{15}.\cos\frac {16\pi}{15}$
$ \equal{}8\sin\frac{4 \pi}{15} .\cos\frac {4\pi}{15}.\cos\frac {8\pi}{15}.\cos\frac {16\pi}{15}$
$ \equal{}4\sin\frac{8\pi}{15} .\cos\frac {8\pi}{15}.\cos\frac {16\pi}{15}$
$ \equal{}2\sin\frac {16\pi}{15}.\cos\frac {16\pi}{15}$
$ \equal{}sin \frac{2 \pi}{15}$

So the expression $ \equal{}\frac{1}{16}$

It is more complicated to use complex numbers in this case. Isn't it?

yes, of course..

your solution is more easier..  :P

Plz post the solution involving complex system..

I wnt to know how it is done???

Haha! This reminds me of a combinatorics problem ! :P
The solution is almost the same :lol:
Let $ x \equal{} e^{i\frac {2\pi}{15}}$. Then $ x^{15} \minus{} 1 \equal{} 0$ and $ |x| \equal{} 1$.
$ A \equal{} 16\cos\frac {2\pi}{15} \cdot \cos\frac {4\pi}{15} \cdot cos\frac {8\pi}{15} \cdot \cos\frac {16\pi}{15} \equal{} (x \plus{} \overline{x})(x^2 \plus{} \overline{x^2})(x^4 \plus{} \overline{x^4})(x^8 \plus{} \overline{x^8})$
Since it's real we see:
$ A \equal{} |x \plus{} \overline{x}| |x^2 \plus{} \overline{x^2}| |x^4 \plus{} \overline{x^4}| |x^8 \plus{} \overline{x^8}| \equal{}$ $ |1 \plus{} x^2| |1 \plus{} x^4| |1 \plus{} x^8| |1 \plus{} x^{16}| \equal{} |1 \plus{} x| |1 \plus{} x^2| |1 \plus{} x^4| |1 \plus{} x^8|$
(Since $ |x| \equal{} 1|$ and $ x^{15} \equal{} 1$)
Remember the identity $ 1 \plus{} a \equal{} \frac {1 \minus{} a^2}{1 \minus{} a}$.
So $ A \equal{} \frac {|1 \minus{} x^2|}{|1 \minus{} x|} \frac {|1 \minus{} x^4|}{|1 \minus{} x^2|} \frac {|1 \minus{} x^8|}{|1 \minus{} x^4|} \frac {|1 \minus{} x^{16}|}{|1 \minus{} x^8|} \equal{} \frac {|1 \minus{} x^{16}|}{|1 \minus{} x|} \equal{} 1$.
Hence $ \cos\frac {2\pi}{15} \cdot \cos\frac {4\pi}{15} \cdot cos\frac {8\pi}{15} \cdot \cos\frac {16\pi}{15} \equal{} \frac {1}{16}$

Stephencheng: Not much more, is it ? :)

[quote="RAGHAV GROVER"]Mine is much more simpler than yours :P
let $ z = \cos\frac {\pi}{15} + i\sin\frac {\pi}{15}$

note that,$ \cos\frac {2\pi}{15} = \frac {1}{2}(z^{2} + \frac {1}{z^{2}})$

$ \prod_{i = 1}^{4}\cos\frac {2^{k}\pi}{15} = \frac {1}{16}\frac {(z^{4} + 1)(z^{8} + 1)(z^{16} + 1)(z^{32} + 1)}{z^{30}}$
but $ z^{15} = 1$, so   $ z^{30} = 1$

now multiply and divide by $ (z - 1)$

to get $ \frac {1}{16}\frac {(z^{16} - 1)}{(z - 1)}$
but since $ z^{16} = z.z^{15}$   so  $ z^{16} = z$

done :P

do you liked it????[/quote]
It is not any better in my opinion. You just skip some of the steps and did them mentally instead. (This too lead to an error)
I could have written: $ (x + \overline{x})(x^{2} + \overline{x^{2}})(x^{4} + \overline{x^{4}})(x^{8} + \overline{x^{8}}) = 1 + x + x^2 + ... + x^{15} = \frac {1 - x^{16}}{1 - x} = 1$, but I wouldn't have shown you any ideas which you could use to solve other problems.

In fact: $ z^{15} = - 1$ and therefore $ (z^{4} + 1)(z^{8} + 1)(z^{16} + 1)(z^{32} + 1) = (z^4 + 1)(z^8 + 1)(1 - z)(1 + z^2)$ so you probably want to multiply and divide by $ 1 + z$,  cause then you get $ \frac {1 - z^{16}}{1 + z} = \frac {1 + z}{1 + z} = 1$.


Find the natural numbers $ m$ and $ n$ such that $ A\equal{}\frac{m}{n} \plus{} \frac{n\plus{}1}{m^{2}}$ is an integer.

In order for this expression to be an integer, we must have $ n|m$ or $ n\equal{}m^2$. However, if $ n\equal{}m^2$, then we have $ m|1$, which means that $ m\equal{}1$. Thus, $ n\equal{}1$ also. Hence, $ (1,1)$ is an ordered pair which satisfies the equation. Next we consider $ n|m$. Clearly, $ n^2|m^2$ and since $ m^2|n\plus{}1$ (in order for the reslut to be an integer). This implies that $ n^2|n\plus{}1$, which means that $ n\equal{}1$ is the only solution in natural numbers. In this case $ m\equal{}1$. Since we are asked for the one ordered pair satisfying the equation, and $ (1,1)$ satisfies it, the only solution in natural numbers is $ \boxed{(1,1)}$.

your solution is wrong since $ n\equal{}8,m\equal{}6$ is another solution.
[hide="in fact..."]
 we can prove that if there exists such $ m,n$ with $ A \in N$ then there exists $ (l,k) \in N^2$ such that $ \frac{l^3\plus{}k^3\plus{}1}{l^2k^2} \in N$ for example $ (l,k)\equal{}(1,1),(2,3)$ are some of the solutions.
[/hide]

$ m \equal{} d.m_1 \; , \; n \equal{} d.n_1 \; ,\; (m_1,n_1) \equal{} 1$

Since $ \frac {m^3 \plus{} n^2 \plus{} n}{m^2n}$ is an integer so

$ m^2 | (n^2 \plus{} n) \; , \; n | m^3$

$ \Longrightarrow \quad d \;|\; n_1 , \; n_1 \;|\; d^2$

$ \Longrightarrow \quad n_1 \equal{} ku^2, \; d \equal{} ku, \quad k,u \in \mathbb{N}$

$ \Longrightarrow \quad n \equal{} k^2u^3, \; m \equal{} kum_1$

$ \Longrightarrow \quad (k^4u^5m_1^2) \;| \; (u^3k^3m_1^3 \plus{} k^4u^6 \plus{} k^2u^3)$

$ \Longrightarrow \quad k | km_1^3 \plus{} k^2u^3 \plus{} 1 \; \Longrightarrow \; k \equal{} 1$

$ \Longrightarrow \quad u^2m_1^2 \; | \; (u^3 \plus{} m_1^3 \plus{} 1)$

Let $ a \equal{} m_1 , \; b \equal{} u$

We must find that the $ a, b \in \mathbb{N}$ such that

$ a^2b^2 \; | \; (a^3 \plus{} b^3 \plus{} 1)$

$ b^2 | (a^3 \plus{} 1) , \; a^2 | (b^3 \plus{} 1) \; \Longrightarrow \; a^2 \leq b^3 \plus{} 1 , \; b^2 \leq a^3 \plus{} 1$

$ a^3 \plus{} b^3 \plus{} 1 \geq a^2b^2 \; \Longrightarrow \; \frac {a}{b^2} \plus{} \frac {b}{a^2} \plus{} \frac {1}{a^2b^2} \geq 1$

$ a \equal{} b \; \Longrightarrow \; a \equal{} b \equal{} 1$

$ max\{a,b\} \equal{} a \; , \; a > b > 4 \; \Longrightarrow \; \frac {1}{a^2b^2} < \frac {1}{256}$

$ t \equal{} \frac {1 \minus{} \frac {1}{256}}{2}$

$ \frac {a}{b^2} \geq t \quad \Longrightarrow \; b^3 \plus{} 1 \geq a^2 \geq b^4t^2$

$ \Longrightarrow \; b^3(bt^2 \minus{} 1) \leq 1 \; \Longrightarrow \; 125(5t^2 \minus{} 1) \leq 1$ since $ b \geq 5 \; , \; 5t^2 \minus{} 1 > 0$

However, if we calculate that;

$ 125(5t^2 \minus{} 1) > 1 \; \Longrightarrow \; \frac {b}{a^2} < \frac {a}{b^2} < t \; \Longrightarrow \; \frac {a}{b^2} \plus{} \frac {b}{a^2} \plus{} \frac {1}{a^2b^2} < 1$ is a contradiction.

$ \Longrightarrow \; b \leq 4$

The rest is easy and all solutions are:

$ (a,b) \equal{} \{(1,1),(2,3),(3,2)\}$

$ (m,n) \equal{} \{(1,1),(6,27),(6,8)\}$

Thank you very much and nice solution :lol:

as i said in my last post we have $ A \in N \implies \frac {l^3 \plus{} k^3 \plus{} 1}{l^2k^2} \in N$ so we have
$ l^3 \plus{} k^3 \plus{} 1^3 \equal{} nl^2k^21^2$
but this equation is just a special case of [url=http://www.mathlinks.ro/viewtopic.php?t=150629]H19[/url] PEN...what a bad memory i have! i've forgotten it at that time! :(


For $ a,b,c$ positive reals

$ \sum a^2(b\plus{}c)(c\plus{}a)(c^2\minus{}b^2) \geq 0$

Does it exist a one line or two line solution?

[quote="manlio"]For $ a,b,c$ positive reals

$ \sum a^2(b \plus{} c)(c \plus{} a)(c^2 \minus{} b^2) \geq 0$

Does it exist a one line or two line solution?[/quote]
\[ \sum_{cyc} a^2(b \plus{} c)(c \plus{} a)(c^2 \minus{} b^2) \equal{} \frac {1}{6}\sum_{cyc}(a \minus{} b)^2(a \plus{} b)^2\left(3c^2 \plus{}a^2\minus{}b^2\right)\]

I proved it this way

$ \sum a^2(b \plus{} c)(c \plus{} a)(c^2 \minus{} b^2) \equal{} \sum ab\sum a^2(b^2 \minus{} c^2)$
$ \plus{} \sum a^2c^2(c^2 \minus{} b^2) \equal{} \sum a^2c^4 \minus{} 3a^2b^2c^2 \geq 0$

true by AM-GM.

This inequality is useful to prove:

$ \sum \frac {a^2}{(a \plus{} b)^2} \plus{} 2\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq 1$

an inequality of my friend can_hang2007. :)

for manlio
\[ \sum a^2(b \plus{} c)(c \plus{} a)(c^2 \minus{} b^2) \equal{} \frac {1}{2}\sum b^2(a^2 \minus{} bc)^2\]


geometry
circumcircle
vector
geometry proposed

Let $ H$ and $ G$ be an orthocenter and a centroid of triangle $ ABC$ respectively and $ R$ be the circumradius of this triangle. $ F$ is the midpoint of $ GH$.
Prove that $ AF^{2}\plus{}BF^{2}\plus{}CF^{2}\equal{} 3R^{2}$.

I know solution using vector one. Does someone know other solutions?

[quote="Leonhard Euler"][quote="mathpk"]Let $ H$ and $ G$ be an orthocenter and a centroid of triangle $ ABC$ respectively and $ R$ be the circumradius of this triangle. Prove that $ AF^{2}\plus{}BF^{2}\plus{}CF^{2}\equal{} 3R^{2}$.
I know solution using vector one. Does someone know other solutions?[/quote]
what is $ F$?[/quote]

Sorry I've edited it :blush: .

Let $ O$ be a circumcenter. We can easily see $ GO\equal{}GF$
$ FA^{2}\plus{}FB^{2}\plus{}FC^{2}$
$ \equal{} GA^{2}\plus{}GB^{2}\plus{}GC^{2}\plus{}3GF^{2}$
$ \equal{} GA^{2}\plus{}GB^{2}\plus{}GC^{2}\plus{}3GO^{2}$
$ \equal{} OA^{2}\plus{}OB^{2}\plus{}OC^{2}$
$ \equal{} 3R^{2}$

[quote="mathpk"][color=darkred]Let $ H$ and $ G$ be an orthocenter and a centroid of triangle $ ABC$ respectively and $ R$ be the circumradius of this triangle.
$ F$ is the midpoint of $ GH$. Prove that $ AF^{2}+BF^{2}+CF^{2}= 3R^{2}$.[/color][/quote]
[color=darkblue][b]Proof.[/b] I"ll use the [b]Lagrange's identity[/b] (you can prove easily) : for any point $ X$ from the $ ABC$'s plane, $ \sum XA^{2}= 3\cdot XG^{2}+\frac{1}{3}\cdot\sum a^{2}$.
Particularly, $ \{\begin{array}{ccc}X: = G &\implies & GA^{2}+GB^{2}+GC^{2}=\frac{1}{3}\cdot\sum a^{2}\\ \\ X: = O &\implies & OG^{2}= R^{2}-\frac{1}{9}\cdot\sum a^{2}\end{array}$. Observe that $ \overline{HO}= 3\cdot\overline{GO}\implies FG = GO$.
Apply the relation of $ A$- median in the triangle $ FAO$ a.s.o. : $ 4\cdot AG^{2}= 2(AF^{2}+AO^{2})-FO^{2}$, i.e. $ 2\cdot GA^{2}= FA^{2}+R^{2}-2\cdot GO^{2}$ a.s.o.

Therefore, $ 2\cdot\sum GA^{2}=\sum FA^{2}+3R^{2}-6\cdot GO^{2}$ $ \Longleftrightarrow$ $ \frac{2}{3}\cdot\sum a^{2}=\sum FA^{2}+3R^{2}-6\cdot(R^{2}-\frac{1}{9}\cdot\sum a^{2})$ $ \Longleftrightarrow$ $ \sum FA^{2}= 3R^{2}$.

[b]Remark.[/b] Generally, if $ F\in GO$ so that $ \overline{FO}= m\cdot\overline{FG}$, then $ \boxed{(m-1)^{2}\cdot\sum FA^{2}= 3R^{2}+\frac{m(m-2)}{3}\cdot\sum a^{2}}\ \ (*)$.

The relation $ (*)$ is interesting because for any $ F\in\overline{HGO}$ exist two real numbers $ \alpha_{f}$, $ \beta_{f}$ so that $ \alpha_{f}\cdot\sum a^{2}+\beta_{f}\cdot\sum FA^{2}= k$ (constant), where $ k = 3R^{2}$.

Particular cases : $ \{\begin{array}{cccc}1\blacktriangleright & m: =0 &\implies &\sum OA^{2}=3R^{2}\mathrm{\ (\ obviously\ )}\ .\\\\ 2\blacktriangleright & m: = 2 &\implies &\sum FA^{2}=3R^{2}\mathrm{\ (\ initial\ problem\ )}\ .\\ \\ 3\blacktriangleright & m =\frac{3}{2}&\implies &\sum HA^{2}= 12R^{2}-\sum a^{2}\ .\\ \\ 4\blacktriangleright & m\rightarrow\infty &\implies &\sum GA^{2}=\frac{1}{3}\cdot\sum a^{2}\ .\\ \\ 5\blacktriangleright & m = 3 &\implies & 4\cdot\sum EA^{2}= 3R^{2}+3\cdot\sum a^{2}\mathrm{\ (\ E\ -\ Euler's\ point\ )\ .}\end{array}$[/color]


arithmetic sequence
number theory open
number theory

Is it true that if $a_1,...,a_n$ are non-zero integers and $ x_1,...,x_n$ are different rationals such that $ a_1 x_1^k+...+a_n x_n^k$ is integer for infinitely many k, then $x_i$ are integers?

Maybe you think it's true,but then you must add more conditions for $a_1,a_2,...,a_n$

 and$x_1,x_2,...,x_n$.The reason in the following case:

  $n$is even , $a_1=...=a_n$,and,$x_1=-x_2=\frac 1 2,...,x_{n-1}=-x_n=\frac 1 n$

 The expression is $0$ for all $k$ odd.

 I think we need to add that $\sum^{n}_{i=1} a_ix_i^k\neq 0,k\in\{1,2\} $

  Indeed,I haven't tried my best to solve this problem,but i'm expecting to know a
solution for this nice problem.

Your new condition is unsufficient either. Just add an integer $x_1$ to the set.

BTW, is this problem inspired by http://www.mathlinks.ro/viewtopic.php?p=177316#p177316 ?

Indeed, it is inspired by that problem. And indeed, the hypotheses must be stronger. Sorry.

[quote="fedja"]What if we ask that $a_j$ and $x_j$ be positive? This seems to eliminate all trivial counterexamples  :) .[/quote]

This may or may not help, but you may assume that the $k_i$ converge p-adically, the $a_j$ are integers, and the denominators of the $x_j$ are all some fixed power of $p$.
The divisibility by very many powers of $p$ seems unlikely, so yes, the result seems like it should be true.

I think that I can show the following (but it's 3a.m. here, so ...  :) ):

[b]Claim. [/b] Let $\alpha_1,\ldots,\alpha_n$ be different algebraic numbers and let $c_1,\ldots,c_n \in \mathbb{C}^*$ such that
\[
c_1\alpha_1^k + \cdots + c_n\alpha_n^k
\]
is an algebraic integer for infinitely many $2^n$-tuples of consecutive integers $k$, then each $\alpha_i$ is an algebraic integer.

Conjecturally, I think that it is sufficient to have that $c_1\alpha_1^k + \cdots + c_n\alpha_n^k$ is an algebraic integer for $k$ on a set which intersects infinitely many times every arithmetic progression with difference at most $n$; this eliminates the trivial examples.  :)


inequalities
number theory
algebra theorems
algebra

Can any1 give me link to some book containing problems fom olympiads??

something like hojoo-lee number theory or inequalities blabla
does there exist such a book (for freeeee)?

thanks in advance ;)

ebook   mathematics   links  http://www.mathcenter.net/forum/showthread.php?goto=newpost&t=5252


geometry
integration
analytic geometry
calculus
inequalities
calculus computations

Find the area of the figure G, bounded by the curve: $ \left (\frac {x}{ a} \plus{} \frac {y}{b} \right)^4 \equal{} 4xy (a> 0, b> 0)$

Substituing $ \left\{x\equal{}\frac{1}{2} a^{3/2} \sqrt{b} (v\minus{}u),y\equal{}\frac{1}{2} \sqrt{a} b^{3/2} (u\plus{}v)\right\}$
$ u^2\plus{}v^4\equal{}v^2$
$ \int_0^1 \sqrt{w^2\minus{}w^4} \, dw\equal{}\frac{1}{3}$

Scaling the integral. $ A \equal{} a^2 b^2 \frac{2}{3}$

[quote="EulerIntegral"]Substituing $ \left\{x \equal{} \frac {1}{2} a^{3/2} \sqrt {b} (v \minus{} u),y \equal{} \frac {1}{2} \sqrt {a} b^{3/2} (u \plus{} v)\right\}$
$ u^2 \plus{} v^4 \equal{} v^2$
$ \int_0^1 \sqrt {w^2 \minus{} w^4} \, dw \equal{} \frac {1}{3}$

Scaling the integral. $ A \equal{} a^2 b^2 \frac {2}{3}$[/quote]
 I don't  understand.
Where the Jacobian determinant of the coordinate conversion formula has to be considered: $ J\equal{}\frac{2}{3} a^2b^2$
Why do u know integral from  0 to 1?

The jacobian gives how much scaling the area is.

For example $ \left(\frac{x}{a}\right)^2\plus{}\left(\frac{y}{b}\right)^2\leq 1$.
Following substituition.
$ \int _{\minus{}a}^a\int _{\minus{}\sqrt{b^2\minus{}\frac{b^2 x^2}{a^2}}}^{\sqrt{b^2\minus{}\frac{b^2 x^2}{a^2}}}1dydx\equal{}\int _{\minus{}a}^a\int   _{\minus{}\sqrt{1^2\minus{}u^2}}^{\sqrt{1^2\minus{}u^2}}a bdvdu$

When I calculated the integral, I left the jacobian out of the integral, so It is need to plug it in to get the integral.

Where did I get the values?
The same way I got above that the value of $ x$ is between $ \minus{}a$ and $ a$.
Using the Inequality


geometry
3D geometry
tetrahedron
geometry unsolved

Let $ T$ be an isosceles (i.e. equifacetal) tetrahedron which is not regular. Find the group of the isometries $ f$ s.t. $ f(T)\equal{}T$. (There exists 2 cases).

Isn't it $ \mathbb{Z}/2\times \mathbb{Z}/2$

Yes it is in the general case; give your proof.
And in the special case ?

Well for the general case the group is generated by $ (AB)(CD)$ , $ (AC)(BD)$ and $ (AD)(BC)$.

For the special case its $ S_4$ right?

No; $ S_4$ is the result only for a regular tetrahedron.


geometry
inequalities
trigonometry
inequalities proposed

Given two triangles $ ABC,A'B'C'$ with area $ S,S'$ prove that
\[ aa'+bb'+cc'\ge 4\sqrt{3SS'}+a(\sqrt{b'}-\sqrt{c'})^{2}+b(\sqrt{c'}-\sqrt{a'})^{2}+c(\sqrt{a'}-\sqrt{b'})^{2}\]

[color=purple]Quite hard ? I don't think so gemath  :D ! This problem is easy and the method to prove it is familiar as the method to prove the following problem :
  $ a[x-(\sqrt{y}-\sqrt{z})^{2}]+b[y-(\sqrt{z}-\sqrt{x})^{2}]+c[z-(\sqrt{x}-\sqrt{y})^{2}]\ge 4\sqrt{3S.S'}$
 Dear gemath do you need me post all the solution  :P ! I think it's not nescessary and I hope you can give me the generalization of two problems above.  :) [/color]

[quote="evarist"]Quite hard ? I don't think so gemath
 Dear gemath do you need me post all the solution  :P ! [/quote]
yes of course I want to see your solution post it if you can!
[quote="evarist"]
:D ! This problem is easy and the method to prove it is familiar as the method to prove the following problem :
  $ a[x-(\sqrt{y}-\sqrt{z})^{2}]+b[y-(\sqrt{z}-\sqrt{x})^{2}]+c[z-(\sqrt{x}-\sqrt{y})^{2}]\ge 4\sqrt{3S.S'}$
[/quote]
what do you mean? $ x,y,z$ and $ S'$ :rotfl: or you want to rewrite my problem?

[quote="gemath"][quote="evarist"]Quite hard ? I don't think so gemath
 Dear gemath do you need me post all the solution  :P ! [/quote]
yes of course I want to see your solution post it if you can!
[/quote]
 [color=purple]   Ok I'll post my  solution tomorrow  :D
About my problem x,y,z are three side length of different triangle which has are S'  :) [/color]

[color=purple]To make your problem simple we prove the equivalent inequality :
       $ a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}\ge 16SS'+a(x-y)^{2}+b(y-z)^{2}+c(z-x)^{2}$
$ \leftrightarrow 2a^{2}yz+2b^{2}zx+2c^{2}xy\ge a^{2}(y^{2}+z^{2}-x^{2})+b^{2}(z^{2}+x^{2}-y^{2})+c^{2}(x^{2}+y^{2}-z^{2})+16S.S'$
$ \leftrightarrow 2a^{2}yz+2b^{2}zx+2c^{2}xy \ge a^{2}.2yz.cosX+b^{2}.2zx.cos Y+c^{2}.2xy.cosZ$
$ \leftrightarrow 2yza^{2}(1-cosX)+2zxb^{2}(1-cosY)+2xyc^{2}(1-cosZ)\ge 16S.S'$
$ \leftrightarrow 2yza^{2}.2{\sin\frac{X}{2}}^{2}+2zxb^{2}.2{\sin\frac{Y}{2}}^{2}+2xyc^{2}.2{\sin\frac{Z}{2}}^{2}\ge 16S.S'$
$ \leftrightarrow 4(s'-x)(s'-y).c^{2}+4(s'-y)(s'-z)a^{2}+4(s'-z)(s'-x)b^{2}\ge 16S.S'$
 ( with s' is the semiperimeter of triangle XYZ )
$ (s'-x)(s'-y).c^{2}+(s'-y)(s'-z)a^{2}+(s'-z)(s'-x)b^{2}\ge 4S.S'$
  Now according to the well- known inequality :$ ma^{2}+nb^{2}+pc^{2}\ge 4\sqrt{mn+np+pm}S$ we get
  $ (s'-x)(s'-y).c^{2}+(s'-y)(s'-z)a^{2}+(s'-z)(s'-x)b^{2}\ge 4\sqrt{(s'-x)(s'-y)(s'-z)(s'-x+s'-y+s'-z)}S$
 $ \leftrightarrow (s'-x)(s'-y).c^{2}+(s'-y)(s'-z)a^{2}+(s'-z)(s'-x)b^{2}\ge 4\sqrt{S'^{2}}.S=4S.S'$
  We finish the proff.  :) I think you know me gemath, because I am busy so I can't show the solution for your problem it's very bulky and need some indentities. So I post this problem and the solution  :D [/color]

[quote="evarist"]To make your problem simple we prove the equivalent inequality :
       $ a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}\ge 16SS'+a(x-y)^{2}+b(y-z)^{2}+c(z-x)^{2}$
[/quote]
The first your inequality is false, maybe you mean $ a^{2}(x-y)^{2}+b^{2}(y-z)^{2}+c^{2}(z-x)^{2}$  :D
I don't understand why my problems is equivalent to yours  :D . Actually there is a problem say
If $ ABC$ is an acute triangle then $ a^{2},b^{2},c^{2}$ is three sides of a triangle whose circumradii and inradii are $ R',r'$ then $ 4S_{ABC}^{2}=4R'r'+r'^{2}$, thus if you want to change to triangle of $ \triangle (a^{2},b^{2},c^{2})$ you must prove the following for two acute triangles:
\[ \sum a^{2}a'^{2}\ge \sqrt{3r(4R+r)r'(4R'+r')}+\sum a'^{2}(b-c)^{2}\]
or you thought triangle of $ a^{2},b^{2},c^{2}$ has area $ S^{2}$  :rotfl:

[color=purple]Oh It's my mistake  :D ! Ok may be I am very lazy  :rotfl: ! But the method is equivalent I think you agree with me  :)
[hide]
   Gemath , em nghĩ anh biết em bận nhưng anh cứ bắt em giải thích thế rõ ràng anh hiểu mà :D Hi vọng đc chat Vietnamese cùng anh ! [/hide] [/color]


Can I mod the next game? Or do I have to ask randomdragoon?
If I can, signups for the next game.

I'll start if at least 7 people want to play.

sign up me and anirudh
yay
playerlist
i_like_pie
BanishedTraitor
worthawholebean
ckck
now a ranger
ccy
junggi
anirudh

I'm gonna wait for a few more minutes just in case anyone else wants to play.


Hi everyone,
I just bought a new Macbook Pro and I was wondering if there was a Mac OS X compatible LaTex or something similar for the Mac.

Thanks

Yes, there is. [url=http://www.tug.org/mactex/]Here[/url] is a complete Mac distribution. I use TexShop for the editing, which is included in that package.


$ 6\minus{}iodo\minus{}6\minus{}methylhept\minus{}1\minus{}yne \plus{} \ce{Bu3SnSnBu3}/hv \longrightarrow \,\,?$

we have formation of tributyl tin radicals, which then abstracts the iodine from the compound to form the tertiary radical. The radical should then attack the 2-carbon in a 5-exo ring formation. I'm not sure what happens now -- usually I would expect this new radical to abstract a hydrogen from tributyl tin hydride but that's not what we're using here. Do we just get a whole SnBu_3 group attached? That would yield ultimately, I think, 2,2-dimethylcyclopentane with a double bond at the one position to a carbon and then the tributyltin group.

I want the name of the final product (which is a hydrocarbon).


There are 3 natural numbers that give the same result when they are added and when they are multipled. 1 + 2 + 3 = 1 x 2 x 3. There is only one set of four natural numbers that has this property. 1 + 1 + 2 + 4 = 1 x 1 x 2 x 4. There are 3 sets of 5 natrural numbers all less than 6 that have this property. What are they?

These are the 3 sets i was able to find with 5 natural numbers that are less than 6

1. 1 + 1 + 2 + 2 + 2 = 1 x 1 x 2 x 2 x 2
2. 1 + 1 + 1 + 2 + 5 = 1 x 1 x 1 x 2 x 5
3. 1 + 1 + 1 + 3 + 3 = 1 x 1 x 1 x 3 x 3

I was able to find all 3 sets, but only through trial and error. Is there a rule or forumla to solve for this sets of natural numbers without going through trial and error?

this seems like a very specialized problem without big theorems that would really help. Just look at what you know, for instance there is probably a 1, stuff like that.


$ x;y$ are positives numbers and $ 1\leq x^2\minus{}xy\plus{}y^2 \leq 2$
1).--> prove that:  $ \frac{2}{9} \leq x^4\plus{}y^4 \leq 8$
1).--> prove that:  $ x^{2n}\plus{}y^{2n}\geq \frac{2}{3^{2n}}$ for all $ n\geq3 \in N$

hello, the inequality $ 1 \le x^2 \minus{} xy \plus{} y^2\le 2$ is equivalent with $ 1 \plus{} xy \le x^2 \plus{} y^2\le2 \plus{} xy$ and this with $ (1 \plus{} xy)^2\le x^4 \plus{} y^4 \plus{} 2x^2y^2\le 4 \plus{} x^2y^2 \plus{} 4xy$. This is equivalent with $ 1 \plus{} 2xy \minus{} x^2y^2\le x^4 \plus{} y^4\le4 \plus{} 4xy \minus{} x^2y^2$. Now we have to prove that $ 4 \plus{} 4xy \minus{} x^2y^2\le 8$ This is true because the inequality above is equivalent with $ 0\le (xy \minus{} 2)^2$. Now we want to prove that $ \frac {2}{9}\le1 \plus{} 2xy \minus{} x^2y^2$. This is equivalent with
$ |xy \minus{} 1| \le \frac {4}{3}$. If $ xy\geq1$ our inequality is equivalent with $ xy \le \frac {7}{3}$.This is true because from$ 2xy \le x^2 \plus{} y^2\le2 \plus{} xy$ we get $ xy\le2$. In the other case $ xy < 1$ we have to prove that $ xy\geq \minus{} \frac {1}{3}$. This is also true because we have $ xy > 0 > \minus{} \frac {1}{3}$ and so the given inequality is proven.
Sonnhard.

for $ x^4\plus{}y^4\geq\frac{2}{9}$ here is a simple resolution
we have $ \frac{3}{2}(x^2\plus{}y^2) \geq x^2\minus{}xy\plus{}y^2 \geq 1$  <=> $ x^2\plus{}y^2\geq \frac{2}{3}$
and we have with cauchy $ x^4\plus{}y^4\geq\frac{1}{2}(a^2\plus{}b^2)^2\geq\frac{1}{2}(\frac{4}{9}\equal{}\frac{2}{9}$


$ 1. (fog)(x)\equal{}6x\plus{}5$  ;  $ (hog)(x)\equal{}2x\minus{}1 \Rightarrow (foh^{\minus{}1})(2)\equal{}?$

$ 2. f(x\times y)\equal{}f(x)\plus{}f(y)$  ;  $ f(2)\equal{}a \Rightarrow$ define $ f(256)$ in terms of $ a$.

$ 1. (fog)(x) \equal{} 6x \plus{} 5$  ;  $ (hog)(x) \equal{} 2x \minus{} 1 \Rightarrow (foh^{ \minus{} 1})(2) \equal{} ?$

I would calculate
$ (hog)^{\minus{}1}(x) \equal{} (g^{\minus{}1}oh^{\minus{}1})(x)$
and then compound it:
$ [(fog)o(g^{\minus{}1}oh^{\minus{}1})](x) \equal{} (foh^{ \minus{} 1})(x)$

good! Heres another kind of easy:
$ f(2x)\equal{}f^2(x)\minus{}2f(x)\plus{}2$  and given  $ f(1)\equal{}3$
Find $ f(5)$


A survey of 300 high school students regarding their music preferences determined that 140 liked rap music, 115 liked contry music, and 125 liked rock music, while only 50 liked all three.  It was also determined that 40 liked only rap music, 25 liked only country music, 45 liked only rock music.  How many liked none of the three

I kind of semi-brute forced it and it took me like 5 mins to realize what I was doing.  Is there a easier way to do this?

that's not really brute force. a Venn diagram is the way it's supposed to be used. you can do the problem in one minute (max)

what we know:
140 like rap
115 like country
125 like rock
50 like all
40 like only rap
25 like only country
45 like only rock

therefore, 140 - (40+50) = 50 like rap/country or rap/rock
and similarly, 115 - (25+50) = 40 like country/rap or country/rock
and finally, 125 - (45+50) = 30 like rock/rap or rock/country

that means (50 + 40 + 30)/2 = 60 students like only TWO music genres
which means there are 40 + 25 + 45 + 50 + 60 = 220 students who listen to music.

therefore, 300 - 220 = [b]80 students[/b] do not listen to music.

Brute force is like guess and check kind of stuff, using a venn diagram is a real way to do the problem. Btw, that's the method shown in the AoPS Counting and Probability book (they also show undefined's way since it really just originates in a venn diagram if you think about it).

yeah, a venn diagram helps to visualize the solution i posted above..


[quote="bubala"]The generating functions mathjam  was replaced by the WOOT mathjam.  Huh?[/quote]Yes. WOOT is a very important project for AoPS, so it takes priority.

[quote="bubala"]what is WOOT?  Oh and will it be to hard for me?  I noticed it was rated 7.[/quote]:w00t: is the WOOT emoticon. Obviously it has nothing to do with the WOOT project :)

It means Worldwide Online Olympiad Training. Sort of like an online all-year round MOP. More details in the upcoming Math Jam, announces, etc.  ;)

[quote="Valentin Vornicu"][quote="bubala"]yeah, but it's tonight--so please give me more details so that I know whether to come or not.[/quote]All details will be there. There's no math (I think) involved, it's an informational Math Jam.[/quote]

In the other thread Mr. Rusczyk said that there will be one Olympiad problem discussed.

That's right - there will be one math problem.


Solve in $\mathbb{N}$ the following equation:

     \[ 2^{x}-3^{y}=7. \]


calculus
integration
logarithms
calculus computations

Can somebody help me express the integral of x^x, using the [b]taylor[/b] series? I know that you can not express x^x using elementary functions.

The Taylor series of $ x^x$? Not practical. The series expansion you might use comes from writing $ x^x\equal{}e^{x\ln x}$, and using the series for the exponential: $ x^x\equal{}1\plus{}x\ln x\plus{}\frac{x^2\ln^2 x}{2}\plus{}\frac{x^3\ln^3 x}{6}\plus{}\cdots$

Next comes integrating those $ x^k\ln^k x$ terms, which is repeated integration by parts. $ \frac1{k!}\int x^k\ln^k x\,dx\equal{}\frac1{k!}\frac{x^{k\plus{}1}}{k\plus{}1}\ln^k x\minus{}\frac1{k!}\int \frac{x^{k\plus{}1}}{k\plus{}1}\cdot\frac{k\ln^{k\minus{}1}x}{x}\,dx$
$ \cdots\equal{}x^{k\plus{}1}\left(\frac{\ln^k x}{k!\cdot (k\plus{}1)}\minus{}\frac{\ln^{k\minus{}1} x}{(k\minus{}1)!\cdot (k\plus{}1)^2}\plus{}\cdots \plus{}(\minus{}1)^k\frac1{(k\plus{}1)^{k\plus{}1}}\right)$

It's a horrible mess. In the one special case $ \int_0^1 x^x\,dx$, all of the terms with logarithms go away and you get the nice series expression $ \int_0^1 x^x\,dx\equal{}1\minus{}2^{\minus{}2}\plus{}3^{\minus{}3}\minus{}4^{\minus{}4}\plus{}\cdots$.


calculus
integration
limit
logarithms
calculus computations

Express the following the limit values in terms of a definite integral and find them.

(1) $I=\lim_{n\to\infty}\frac{1}{n}\ln \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\cdots \left(1+\frac{n}{n}\right).$

(2) $J=\lim_{n\to\infty}\frac{1}{n^{2}}(\sqrt{n^{2}-1}+\sqrt{n^{2}-2^{2}}+\cdots\cdots+\sqrt{n^{2}-n^{2}}).$

(3) $K=\lim_{n\to\infty}\frac{1}{n^{3}}(\sqrt{n^{2}+1}+2\sqrt{n^{2}+2^{2}}+\cdots+n\sqrt{n^{2}+n^{2}}).$

[quote="kunny"]Express the following the limit values and find them.

(1) $I=\lim_{n\to\infty}\frac{1}{n}\ln \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right).$[/quote]

(1) $I=\lim_{n\to\infty}\frac{1}{n}\ln\left[ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right)\right] = \lim_{n\to\infty}\frac{1}{n}\ln\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)$
$= \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\ln\left(1+\frac{k}{n}\right) = \int_{1}^{2}\ln x\,dx = \left[x(\ln x-1)\right]_{x=1}^{2}= 2\ln 2-1$

[quote="kunny"]Express the following the limit values and find them.
(2) $J=\lim_{n\to\infty}\frac{1}{n^{2}}(\sqrt{n^{2}-1}+\sqrt{n^{2}-2^{2}}+\cdots+\sqrt{n^{2}-n^{2}}).$[/quote]

(2) $J = \lim_{n\to\infty}\frac{1}{n^{2}}\left(\sqrt{n^{2}-1}+\sqrt{n^{2}-2^{2}}+\cdots+\sqrt{n^{2}-n^{2}}\right) = \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=1}^{n}\sqrt{n^{2}-k^{2}}$
$= \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{1-\left(\frac{k}{n}\right)^{2}}= \int_{0}^{1}\sqrt{1-x^{2}}\,dx = \frac{\pi}{4}$


Let $a, b, c$ be non-negative real numbers with $a+b+c=1$. Show the inequality

$\frac{1+a^{2}}{1+b^{2}}+\frac{1+b^{2}}{1+c^{2}}+\frac{1+c^{2}}{1+a^{2}}\le \frac{7}{2}$.


What branch of math is this exactly? I understand what the problems are asking; it's just that I've never covered any math like this, nor even heard of this. Could someone fill me in on combinatorics and the other material that is tested on the olympiad. I'd like to learn and compete on it, but I've never seen math like this before.

Wait...do you mean contest math in general? That's what a lot of the problems come from. There's very little teaching of just fundamentals from normal school math classes. You can get some good books, like the Art of Problem Solving Vol. 2 is a good book.

Yeah, I'm asking about olympiad/state math competition level math. Could you recommend a good list of books where I can begin to learn all of the math required for those types of competitions?

Well, if you're more interested in just improving contests skills on contests like AMC and AIME, the AOPS Vol. 2 and the AOPS intermediate or introduction books would be a pretty good choice. If you're more interested in Olympiad math, like USAMO level, you could try The Art and Craft of Problem Solving by Paul Zeitz or maybe Engel, since I think these are more advanced and proof-based.