algebra function domain geometry inequalities algorithm rectangle

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194275]This[/url] is where the current post stems from. I also think that it would be nice to make something like an online version of a research camp or whatever on AoPS. I PM-ed rrusczyk and asked for his opinion. If I understood him right, he would be in favor of it [b]IF[/b]

a) We can find enough interested students

b) We can provide enough qualified mentors (working, at least in the beginning, on a volunteer basis)

c) We can come up with some workable technical scheme about how exactly it should be done

So, at this stage I would like to hear from anybody who might be interested in the project. I set it as a poll to make it possible just to say "Yes, I'm in" without cluttering this thread but if you also have to say something beyond that, please share your opinion.

Honestly, neither from this post nor from the topic you have linked to I am getting an idea what exactly you are proposing. But chances are high I will collaborate, since I am myself interested in "fresh" research in halfways elementary (say, understandable for undergraduates) fields. The question is what we want to create:

1) A forum to ramble about university, mathematics and career in general terms. I wouldn't be averse to something like that, but at a long run I fear this might turn out quite a waste of time.

2) A forum for more concrete discussion and advices with regard to particular universities and particular mathematical subjects (where to go to study this and that, etc.). This may be interesting, but I cannot contribute anything to such a discussion.

3) A forum to really discuss mathematical research problems and obscure ideas looking for applications. Something like the Open Questions fores, but without the noobs posting olympiad problems and trivial/boring stuff. This is actually one of my dreams, but the realization should be harder than it seems at the first sight. We have to
- determine who gets reading access to the forum, because it is not quite good to have active research being held with all the planet staring and potentially stealing the results. (And I am not anxious about copyrights, I just fear that this might be considered a pre-publication.)
- determine how to filter out the trash, and what is to be taken seriously - most people can hardly evaluate their own ideas and contributions.
- make sure we find a critical mass of interested people who would sacrifice some of their time to thinking about others' problems; and as problems get solved, someone should write up the solutions and publish them rather than leaving them buried in a hidden forum.
I fear the third part is the hardest, but I am not too much up-to-date about who is contributing useful stuff to ML, so I could have overlooked some participants.

What can also be thought of is a kind of mix between the above, like

4) A forum trying to mimic the old-times ML, with only selected users allowing to post but everyone allowing to read, in the hope that the level gets higher again and one could step into more serious research.

Fedja, what of the above points have you meant?

  darij

Darij,

Number 3 is the closest to what I had in mind. I thought of something like a set of fora each devoted to one small (or not so small) project with 1 or 2 supervisors/mentors with moderating powers. The projects may be anything from middle school level to undergraduate research (that is what I primarily thought of when posting) to PhD level topics to collaborative research (which is what you seem to have in mind) projects depending on who wants to take part in them and who wants to mentor them. I am not sure about the details myself, that's why I made this short post instead of creating a big detailed proposal.

Number 1 is mostly a waste of time because, in my opinion, the only meaningful way to talk about research is to do research. As to number 2, we may have something like that too, but my knowledge there is also limited to 4-5 places, so no, that's not what I meant.

As to full reading/posting access to each project forum, I would suggest giving it to anybody who wants to participate in the project (having the general descriptions and whatever else the team wants to put in the public domain available to the general public). This calls for 2 parts in each project forum: the open part and the closed part. I guess it is not so difficult to implement technically. Also, at least in the beginning, it would be nice to have a few projects that are completely open. Honestly, I am not as much afraid of somebody's stealing the results as of lack of interest. If we ever get to something that is worth stealing, it'll be a great milestone and we can discuss the "security measures" then :P.

The trash can be filtered out by moderators/mentors as usual. I don't think it is a big problem.

The third one is the hardest indeed. One of the purposes of this post is exactly to figure out whether we have this critical mass on AoPS.

As a student about to go to college, I'll take whatever undergraduate research opportunities I can get.  Maybe I'll extend my RSI/Siemens/Intel paper...  :?:

Well, I could contribute 4-5 "hot" ideas (mostly conjectures of mine), but as my fields of research tend to stay not too far from olympiad mathematics, there is e. g. a possibility to turn such an idea into an olympiad problem. This becomes impossible if everyone is able to read. Though I could start with the less elementary ones, but then I am not sure whether we have enough people to discuss.

  darij

[quote="darij grinberg"]there is e. g. a possibility to turn such an idea into an olympiad problem. This becomes impossible if everyone is able to read. [/quote]
Do you really care? I mean you'll lose just 2, 3, or 4 good olympiad problems you would be able to submit to some competitions once otherwise. Is it such a big loss? I would prefer to start with something a simple as possible (as if I could estimate the difficulty of the question I don't know how to solve :P) so if you have research projects close to Olympiad level, that is absolutely great. I have a couple in convex geometry and a couple in analysis but they are a bit further from the usual Olympiad questions.

Well, I would rather start with the non-olympiad questions, because a higher algebra problem that I cannot solve may be absolutely trivial, while olympiad problems that I cannot solve are [b]usually[/b] more complicated.

What now, should we just start some topics in Open Questions fores until Valentin notices that it's not a bad idea, or should we wait for Valentin or Rrusczyk to open a new forum for this?

  darij

At this point I would just wait for a couple of weeks to see what other people like Kent, mlok, jmerry, pbornsztein, mathmanman, and other potential mentors (or potential students) may have to say. It would be nice to get a few more mentors first, if we can.

Then we'll ask rrusczyk if he considers the numbers high enough to justify opening a new forum (he's quite busy right now as far as I know but he knows about this thread so I guess he will just post here at some point; if not, we'll PM him). I wouldn't just use the open questions fora because I suspect we'll merely clutter them, so I would strongly prefer one subforum per project scheme.

Meanwhile, we can use these 2 weeks to prepare descriptions, etc. Then, we'll just post them in hope that some people will find our questions interesting enough to think of them and see it the idea will fly or crash immediately.

Also Xevarion seemed to volunteer to run a small math jam about research in that thread I mentioned in my first post. It would be interesting to hear from him as well.

So, just let us give other people a chance to think of the idea for a couple of weeks, prepare everything and be ready to start in April.

What areas of research are we speaking of? All the (possible) mentors mentioned above seem to favor analysis a lot :) Does this exclude other areas which are of the student's interest?

I don't quite see a qualitative difference between mentors and students here (at least in my understanding of 3), it's all about collaborative research), but sooner or later some people will contribute more questions and other ones will contribute more solution ideas and references, so one could call the former students and the latter mentors. Anyway, I don't favor analysis at all (sorry Fedja), except of inequalities which I am mostly interested in the combinatorial side of (*still writing a 40-pages paper about Popoviciu*) and complex analysis where I am unfortunately a complete newcomer. We don't have heard anything from Pbornsztein, Harazi (Fedja: add him to the list, please), Vess (ditto), Owk (ditto), ZetaX (ditto), Peter Scholze (I'll talk to him about this when he is back from Marseille) yet; all of these are rather algebra-sided. (Have you already noticed LydianRain and Lzw75, by the way? Both are new to the forum but seem be experts in algebra.)

  darij

I would love to post a session about the research I did at my REU last summer. It is very undergrad. accessible, but also I have gotten it to the point where it is quite non-trivial or "textbook-like."

It is basically in number theory of Gaussian integers. I have some great conjectures that could lead to some insights on the possible non-existence of odd perfect numbers.

Side note: The REU allowed you to pick your research and I've been assured that they will not be attempting it again next year, which would be a bad idea to post here stuff other people are working on unbeknownst to them. Also, I already have a publication on the prelim results, and I'm just plain interested in getting it out there for other minds. I'm curious about the results and am guessing that it will take some creative techniques I don't know yet.

[quote="Jacobson"]All the (possible) mentors mentioned above seem to favor analysis a lot :) Does this exclude other areas which are of the student's interest?[/quote]
Of course not! I just mentioned the names I'm most familiar with as examples :lol: The more diverse, the better...

This seems pretty interesting. I would want to participate. The topic doesn't really matter--I just think it's exciting to collaborate on something like this.

I would like to participate too, but it seems that the ideas above aren't quite concrete at the moment.. maybe I will come back again once (maybe) the forums are set up.

Nope, they aren't. We'll have to proceed by trial and error for quite a while until we find the most suitable mode for such an activity. But if you have a clear idea of how exactly [b]you[/b] want it to be done, please share it :).

(I'm partly answering here to discussions in the other topic, because I don't like the other topic. :D )

Do we really need a subforum for each research project? I think that at least at the beginning, a topic per project would be enough. The topic subject should also say what fields of mathematics the problem is about, as in:

[Ineqs, Analysis] Averaging method in R^n through majorization theory?
[Algebra, Galois] Does the "conjugacy rank" depend on the base field?
[Combi] Extensions of ISL 2001 problem C4 (historic sets)
[RepresTh, Ineqs, Algebra] A "rearrangement ineq." for A_n and S_n\A_n?
... (yes, the above are 4 real projects of mine ;) )

The first difficulty will be to find anyone interested in the projects at all, so I don't fear many of these topics will grow too long. In case one will, we can still split it.

  darij

I would prefer to set things in a workable way from the beginning. I mean if the idea doesn't work, it doesn't matter but if it does, we'll soon realize that one topic per project is a very clumsy mode. Here is one of the more elementary projects I want to offer (it should be accessible starting from middle school and has no upper limit on the difficulty of the related questions). The problem is stated in a very simple way: given a polyomino (i.e., a finite collection of cells), find out what minimal portion of cells on the infinite plane one needs to remove so that what is left doesn't contain the given polyomino. The possible activities are: pick up your favorite polyomino and figure out the answer for it (surprisingly, even for something as simply looking as
[asy]size(40);
path P=unitsquare;
D(shift(0,2)*P);
D(shift(0,1)*P);
D(shift(0,0)*P);
D(shift(1,0)*P);
D(shift(2,0)*P);[/asy]
the answer is not known!), give good bounds that work for all $ n$-mino, make an algorithm that outputs some sure estimates from above or below and works in reasonable time, prove something about the arithmetic nature of the answer (for instance, the ultimate conjecture is that it is always rational), find the answer for certain shapes (even rectangles are not at all trivial if you allow rotation) and so on. The ultimate reward is a publication in something like "Monthly". Clearly, I would prefer to have (at the every least!) one topic in which particular polyominoes are discussed, one separate topic for asymptotic bounds, etc., This doesn't necessarily call for multiple subfora but it calls at least for something like what I proposed in that other thread: the possibility to sort the topics according to the project name or code. Of course, we can always use some "poor man's solution" in the beginning but I'm afraid that it is not going to work on the long run.

Fedja, just use for subject description a certain code (even a number) for each project topic. In that way later, if necessary we can do something about it.

I'll do what I can to help. It's unlikely that I know enough about anything to be a mentor, but I could do something administrative (help moderate, help run a math jam, etc.), and I would definitely be interested in collaborating/participating/whatever in some of the analysis projects.

Bump, this looks really cool, for now to either get ideas or to actually attend this program, you can check this site out, http://euclidlab.org/programs/camp-euclid/program-description

it's only for high schoolers and below unfortunately, but it looks really interesting, :) and perhaps AoPS could incorporate something similar,


function
number theory unsolved
number theory

Let $T={0,1,...,1999}$. Find all function $f: N \rightarrow T$ satisfying:
(i):$f(t)=t$;$\forall t \in T$;
(ii): $f(m+n)=f(f(n)+f(m))$; $\forall m,n \in N$.

There are 2000 such functions:
[tex] f_{k}(n)=n [/tex],  [tex] \forall n \in T [/tex],
[tex] f_{k}(n)= (n-2000)([/tex][i]mod[/i] [tex] k) + 2000 - k[/tex],  [tex] \forall n \in N-T [/tex] ,
[tex] \forall k \ [/tex] between [tex]2[/tex] and [tex]2000 [/tex].
For [tex]k=1[/tex] , [tex] f_{1}(n)= 1999[/tex],  [tex] \forall n \in N-T [/tex].
(where [tex] f_{k}(2000)= 2000-k [/tex])


limit
floor function
trigonometry
irrational number
real analysis
real analysis unsolved

Let $l \in [-1,1]$.Prove that there exists $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ such that $\lim_{n \to \infty} \sin(n! \alpha 2\pi)=l$.

Can we try the $\frac{el}{2\pi}$?
Oh, after 6 minutes my lesson begin in University i'm in a grate hurru!!

$\frac{el}{2\pi}$ is not guaranteed to be irrational at least(i've only said that $l \in [-1,1]$).

[quote="Moubinool"]Use for every irrational positive then
$0<\alpha = \sum_{k\geq 1}\frac{a_k}{k!}$ with $a_k$ positive integer and
$0 \leq a_{k} < k$ for $k>1$
$a_{1} >0$ or $a_1 =0$[/quote]

Does  any irrational number can be writen this way?
Were i can find it? :?

@rustam: This was proved first by Cantor I think. Anyway I know two more general theorems from which the irrationality of the series of Moubinool does follow (I don't know tough if any irrational numbers can be written this way, but I assume it can ;)).
[b]Theorem 1[/b] Any real number $\alpha$ can be written as
  \[ c_0 + \sum^{\infty}_{j = 1} \frac{c_j}{a_1 a_2 \ldots a_j_{}}  \]
With $\left. ( a_n \right)_{n = 1}^{\infty}$ a sequence of positive integers and $\left( c_n \right)_{n = 1}^{\infty}$ a sequence of positive integer such that
  \[ \left\{\begin{array}{l} 0 \leq c_n \leq a_n - 1, \mbox{ for all } n \in \mathbb{N}\\ 0 \leq c_n < a_n - 1, \mbox{ infinitely often} \end{array}\right.  \]

[b]Theorem 2.[/b] Suppose that $c_n \neq 0$ for infinitely many $n \in \mathbb{N}$ and that for all prime $p$, $p | a_n$ for infinitely many $n \in \mathbb{N}$, then
\[ c_0 + \sum_{j = 1}^{\infty} \frac{c_j}{a_1 a_2 \ldots a_j}  \]
is irrational.

@others: Is this correct ?

[b]Lemma[/b] For all irrational $\beta$ there is a sequence $\left. ( \beta_n \right)$ such that $\beta_1 \geq 0$ and $0 \leq \beta_n < n$ for all $n = 2, 3, \ldots$ and
\[ \beta = \sum_{n = 1}^{\infty} \frac{\beta_n}{n!}  \]
[b]Proof.[/b] Let $\ell \in [ - 1 ; + 1 ]$ be a given real. Let $\alpha_n = \left \lfloor n \cdot \frac{\sin^{- 1} \left( \ell \right)}{2 \pi} \right \rfloor$, and
\[ \alpha = \sum_{n = 1}^{\infty} \frac{\alpha_n}{n!}  \]
It is straightforward to chech that $0 \leq \alpha_n < n$ for each $n$ positive integer, and hence $\alpha$ irrational.
Therefore
\begin{eqnarray*} \left\{ n! \alpha \right\} & = & \frac{\alpha_{n + 1}}{n + 1} + \frac{\alpha_{n + 2}}{( n + 1 ) \cdot ( n + 2 )} + \ldots\\ \lim_{n \rightarrow \infty} \left\{ n! \alpha \right\} & = & \lim_{n \rightarrow \infty} \frac{\alpha_{n + 1}}{n + 1} + \lim_{n \rightarrow \infty} \frac{\alpha_{n + 2}}{( n + 1 ) \cdot ( n + 2 )} + \ldots\\ \lim_{n \rightarrow \infty} \left\{ n! \alpha \right\} & = & \frac{\sin^{- 1} ( \ell )}{2 \pi} + 0 + \ldots\\ \lim_{n \rightarrow \infty} \left\{ n! \alpha \right\} & = & \frac{\sin^{- 1} \left( \ell \right)}{2 \pi} \end{eqnarray*}
Whence
\begin{eqnarray*} \sin \left( 2 \pi n! \alpha \right) & = & \left. \left. \sin ( 2 \pi \left( \lfloor n! \alpha \rfloor + \{ n! \alpha \right\} \right) \right)\\ & = & \left. \sin ( 2 \pi \left\{ n! \alpha \right\} \right)\\ \lim_{n \rightarrow \infty} \sin \left( 2 \pi n! \alpha \right) & = & \lim_{n \rightarrow \infty} \sin \left( 2 \pi \left\{ n! \alpha \right\} \right)\\ & = & \sin \left( 2 \pi \frac{\sin^{- 1} \left( \ell \right)}{2 \pi} \right)\\ & = & \ell \end{eqnarray*}

[quote="§outh§tar"]Can anyone prove the above?[/quote]
What do you mean?
{x} proved it, right?

I was referring to the theorems she used.


A chess master who has eleven weeks to prepare for a tournament decides to play at least one game every day, but in order not to tire himself he agrees to play not more than twelve games during any one week. Prove that there exists a succesion of days during which the master will have played exactly twenty games. :)

Να η λύση μου . Έστω $ a_i$ ο αριθμός των παιχνιδίών που έχει παίξει μέχρι την $ i$ ημέρα .

Τότε $ 1\leq a_1<...<a_{77}\leq 132$  άρα $ 21\leq a_1\plus{}20<a_2\plus{}20<...<a_{77}\plus{}20\leq 152$

Ανάμεσα στους 154 αριθμούς $ a_1,...,a_{77},a_1\plus{}20,...,a_{77}\plus{}20$ θα υπάρχουν δύο ίσοι άρα θα υπάρχουν $ i,j$  ώστε

$ a_i\equal{}a_j\plus{}20$ , το ζητούμενο . :)

Η άσκηση αυτή έχει ακριβώς την ίδια λογική με αυτή του προβλήματος 2 του Αρχιμήδη 2001-2002 :wink:
[url]http://www.hms.gr/eme/modules/wfsection/article.php?articleid=323[/url]


Να υπολογιστεί το παρακάτω ολοκλήρωμα: \[ \int_{o}^{1}\frac{\ln(1\plus{}x)}{1\plus{}x^2}dx\]

Νομίζω ότι καίριες αντικαταστάσεις είναι $ x\equal{}tanu , u\equal{}\frac{\pi}{4}\minus{}y$ είναι βιαστικό γιατί τώρα δεν έχω χρόνο. Αργότερα θα εξετάσω αν αυτό που βλέπω αρχικά δουλεύει

[quote="Carl Friedrich Gauss"]Να υπολογιστεί το παρακάτω ολοκλήρωμα:
\[ \int_{o}^{1}\frac {\ln(1 \plus{} x)}{1 \plus{} x^2}dx
\]
[/quote]

Pardon my French, αλλά... και Γ@ΜΩ τα ολοκληρώματα!!!

Έστω $ D \equal{} \{(x,y): 0\leq x,y \leq 1, 0\leq x \plus{} y \leq 1\}\subseteq \mathbb{R}^2$ κ' έστω $ I$ το ζητούμενο ολοκλήρωμα. Τότε:
\[ I \equal{} \int_0^1 \frac {\ln (1 \plus{} x)}{1 \plus{} x^2} dx \equal{} \int_0^1 \frac {1}{1 \plus{} x^2} \int_0^x \frac {1}{1 \plus{} y} dy \,dx \equal{} \int_D \frac {1}{(1 \plus{} y)(1 \plus{} x^2)} \,dA
\]
Με Fubini-Tonelli εναλλάσσουμε τη σειρά ολοκλήρωσης και βρίσκουμε:
\[ I \equal{} \int_0^1 \frac {1}{1 \plus{} y}\int_0^y \frac {1}{1 \plus{} x^2} dx \, dy \equal{} \int_0^1 \frac {\arctan y}{1 \plus{} y} dy
\]
και, μετά από μία παραγοντική ολοκλήρωση: $ I \equal{} \ln(1 \plus{} y) \arctan y\Big|_0^1 \minus{} \int_0^1 \frac {\ln (1 \plus{} x)}{1 \plus{} x^2} dy \equal{} \frac {\pi}{4}\ln 2 \minus{} I$.

Τελικά λοιπόν: $ \int_{0}^{1}\frac {\ln(1 \plus{} x)}{1 \plus{} x^2}dx \equal{} \frac {\pi}{8} \ln 2$

Cheerio,

Durandal 1707

Καλό αλλά "λυκειακό"
Παναγιώτη μάλλον ξέχασες τι λέγαμε τότε :) αλλά έκανες πολύ ωραία λύση :lol:
Λοιπόν
Θέτουμε $ x\equal{}tany$ το ολοκλήρωμα γράφεται
$ I\equal{} \int_0^{\pi /4} ln(1\plus{}tany)dy$
Θέτουμε $ y\equal{}\frac{\pi}{4}\minus{}u$ (κλασσική αλλαγή μεταβλητής του τύπου $ y\equal{}a\plus{}b\minus{}u$)
τότε
$ I\equal{}\minus{} \int_0^{\pi /4}ln(1\plus{}\frac{1\minus{}tanu}{1\plus{}tanu})du\equal{} \int_0^{\pi /4}ln(\frac{2}{1\plus{}tanu})du\equal{}\frac{\pi}{4}ln2\minus{}I$
αρα $ I\equal{}\frac{\pi}{8}ln2$

[quote="r_boris"]Καλό αλλά "λυκειακό"
Παναγιώτη μάλλον ξέχασες τι λέγαμε τότε :)[/quote]

Whoops, δεν το δοκίμασα καν με μετασχηματισμούς  :oops:

Cheerio,

Durandal 1707

Πολύ σωστά!!

Νομίζω πως το διάστημα $ D$ που πέρνεις είναι λάθος.. Νομίζω πως το σωστό είναι \[ D \equal{} \{(x,y): 0\leq x \leq 1, 0\leq y \leq x\}\subseteq \mathbb{R}^2\]

[quote="Carl Friedrich Gauss"]Νομίζω πως το διάστημα $ D$ που πέρνεις είναι λάθος.. Νομίζω πως το σωστό είναι
\[ D \equal{} \{(x,y): 0\leq x \leq 1, 0\leq y \leq x\}\subseteq \mathbb{R}^2
\]
[/quote]

Ωχ :( Δεν είναι το μόνο λάθος στη λύση μου, 150 typos έχω  :oops: :oops:

Thanks for spotting it, μόλις βρω λίγο χρόνο θα κάτσω να τη διορθώσω...

Cheerio,

Durandal 1707


superior algebra
superior algebra unsolved

Let $ k$ be a field and $ M_{n}(k)$ the k- algebra of $ n\times n$ matrices over k. For which $ m\in \mathbb{N}$ does there exist a k-algebra homomorphism $ f: M_{m}(k)\mapsto M_{n}(k)$ such that
$ f(I_{m})\equal{}I_{n}$ ?

$ m$ must divide $ n$. see
http://www.mathlinks.ro/Forum/viewtopic.php?t=90862


Today the IMO 2008 in Spain begins until the 22nd. Go USA!! :)

Good luck to Shaunak, Krishanu, Alex, Evan, Colin, and Paul!

Beast 'em xD

dude..ravi..shouldn't you be in class now?  :D

Today's the arrival of the team leaders. They won't get there until four days from now. The actual test is in six days.

http://www.imo-2008.es/programme.html

[quote="greentreeroad"]Among 6 Chinese contestant this year there is one who was my elementary school classmate for 5 years in China. What a coincidence it will be if I was in the team and would meet him!

Anyway, good luck U.S[/quote]
oh man
theres a dude on the korean team who was in my elementary school for 3ish years

yeah good luck US yay

ha on the chinese team I think there's 2 people who went to the local division of the school that I went to the international division of (Shanghai High School, International Division sucks cr4p though)! (just continuing the trend)

Dang, there is actually a Chinese team member whose name is the same as mine in English...no, I don't have any relationship with him.

Well, I have a friend who is in Shanghai High School (not international), so according to serialk11r, that might count as a very distant relationship...

Yeah, go Team USA!

A Greek and Saudi team member were both RSI '07, so good luck to them as well :)


What's your Avatar go like? If you got a good one , post here.

i win!!! :winner_first:

....or maybe not. :P

If this is annual, links to the first couple please?

maybe this is the first one of many to come...

[quote="noneoftheabove"]even I'm from texas and i still think texas is...bleh[/quote]
Shh. It's a conspiracy we're working for.

hey! noneoftheabove , nice avatar! .

      The chocobo's pretty neat, but it is rather a mismatch to
 a name like Dark Knight. :lol: .
       just what is that Chocoboom? i still haven't had it figured out?
 a warning against space pollution perhaps? How's my RedDevil, anyway?
madatmath

I created one animated gif but it is 24.7 kb so I cannot use it as an avatar.
Here it is

[url=http://server5.pictiger.com/img/206351/picture-hosting/friends.php][img]http://images5.pictiger.com/thumbs/a6/94695bc79be74f1180529a9c7037f2a6.th.gif[/img][/url]

My avatar is great.
But you shall never know unlesss you play FIFA 2005 with Cambridge. They are awesome! The two forwards Daniel Chillingworth and Daniel Webb ... I am lost for words ...

[quote="DarkKnight"]I think my chocobo beats all you guys because chocobos are just that awesome.[/quote]

I KNOW OMFG CHOCOBOS ROCK KWEHHHHH omg im changing my dp right now to a chocobo XD

right now mine is george mason,

about to change it...

You can't beat mine.I have USA's national treasure as my avatar.The funny man himself... :D

Anyone who plays Red Alert 2 should know my avatar; the Soviet Nuke Silo.

We should have some voting system!!  :coolspeak:

sorry fireforce,
there is  :D  no voting system.
autocracy prevails here and i decide after 25 contestants have joined.  :huh:
you are however ,allowed to suggest your choice, and you can change your avatar
and keep posting. may the best avatar (: read madatmath's ) WIN ! :lol:

Wow!  I lost the privelege to post an avatar!  Admins, please help!!!

Nothing beats my hand-drawn* avatar!!

*Photoshop's lens flare filter

[quote="Xantos C. Guin"]I liked that green guy with the arms moving better.

Here is mine.[/quote]

So did I

[quote="madatmath"]Klebian, glad you're flying the tricolor, inspite of being a Texan.[/quote]Actually that's the Texas, not U.S. flag  :P

My avatar is my name, using math symbols!

The awesomest avatar ever:(too big to but to be put as mine)

Right. we have a winner for the annual avatar showcase 2006.
 First place goes to $aznness$ and the dancing green guy//  :lol:
 Second goes to $shadysaysurspammed$ for the Jim Carey avatar.
 and finally, third goes to $Chocoboom$

 I've also decided to adjudge best anim.(although there were only two)
 it goes to  $easy as 3.14159...$

 congratulations. all of u get a free pm from me.
 madatmath

hey what about mine?
You can never judge it without playing fifa 2005 with cambridge!

yayayayayay

my avatar has my username on it

I win


probability
geometry
geometric transformation
reflection

I am trying to understand the basics of finance, I never took a finance class, when stock is first sold, how is the number of shares and price of each determined? I have like a million other ythings I dont understand, but I will ask just this so far. I know this isnt econ but there is no finance forum.

well economics and finance are interconnected as strategies in finance come down to economics optomisation problems.
I don't know much about finance myself (though I'm starting a masters in it in 3 weeks!!) but the value of shares will be based on the expected (discounted) future income of the firm. Shares represent a claim on future profits and the probability (risk) of them being realised. The greater the profits and lower associated risk, the greater the total value of shares.

Why is that so? I mean that they depend on profit. Btw, why r u getting a masters in finance instead of continuing with economics?

The value of a project depends on its (net) present value. The value of a firm should reflect  the sum of the value of its contracts / opportunities and the associated risk.
Its an M.A. in Econ & Finance

The number of shares released actually depends on the private firm's shareholders. Sometimes say only 25% of the shares are floated while other times a vast majority is. The pricing depends on the firm's holders as well. There is the IPO price which is what the ibanks offer to their clients PRIOR to the floating. After the floating the market determines the price. An example of this recently is BIDU which was floated around the 25-30 dollar a share mark. It rode to above 120 on the first day of trading and then subsequently collapsed. It now sits in the high 70s I believe.

The benefits of floating is that it results in profits for the shareholders. It also makes it easier to takeover and place bids for companies. This is a big advantage of having the company publicly traded.


What is the quickest way to test and see whether a quadrilateral is cyclic or not?

it depends on the information you are given

if the opposite angles add up to 180 degrees, then the quadrilateral is cyclic

there's a test that involves sides, but I don't remember it

Well, given a quadrilateral ABCD, if C and D are on the same side of AB and $ \angle ACB$ and $ \angle BDA$ are equal, the quadrilateral is cyclic.


Prove that $ \sqrt{1\plus{}2{\sqrt{1\plus{}3\sqrt{1\plus{}4\sqrt{1\plus{}...}}}}}\equal{}3.$

i see that no one answered this easy question... :(

we have,
$ (a \plus{} b)^2\; \equal{} \;a^2 \plus{} b(\;a \plus{} (a \plus{} b) \; )$

$ \Rightarrow a \plus{} b\; \equal{} \;\sqrt {\;a^2 \plus{} b\left(\;\underbrace{a \plus{} (a \plus{} b)}_{notice \;\; this}\; \right)\;}\;\;\;\quad\;\;\;\quad\ldots\quad(1)$

now, replace $ b$ with $ a \plus{} b$ in $ (1)$ to get

$ \Rightarrow a \plus{} \boxed{(a \plus{} b)}\; \equal{} \;\sqrt {\;a^2 \plus{} \boxed{(a \plus{} b)}(\;a \plus{} (a \plus{} \boxed{(a \plus{} b)})\; )\;}$

put this in RHS of $ (1)$ to get

$ \Rightarrow a \plus{} b\; \equal{} \;\sqrt {\;a^2 \plus{} b\left(\;\sqrt {\;a^2 \plus{} (a \plus{} b)(\;\underbrace{a \plus{} (2a \plus{} b)}_{notice \;\; this}\; )\;}\; \right)\;}\;\;\;\quad\;\;\;\quad\ldots\quad(2)$

replace $ b$ with $ 2a \plus{} b$ in $ (1)$ to get

$ \Rightarrow a \plus{} \boxed{(2a \plus{} b)}\; \equal{} \;\sqrt {\;a^2 \plus{} \boxed{(2a \plus{} b)}(\;a \plus{} (a \plus{} \boxed{(2a \plus{} b)})\; )\;}$

now, put this in RHS of $ (2)$ to get

$ \Rightarrow a \plus{} b\; \equal{} \;\sqrt {\;a^2 \plus{} b\left(\;\sqrt {\;a^2 \plus{} (a \plus{} b)\left(\;\sqrt {\;a^2 \plus{} (2a \plus{} b)(\;\underbrace{a \plus{} (3a \plus{} b)}_{notice \;\; this}\; )\;}\; \right)\;}\; \right)\;}\;\;\;\quad\;\;\;\quad\ldots\quad(3)$

now, replace $ b$ with $ 3a \plus{} b$ in $ (1)$ and put the resulting expression in RHS of $ (3)$...... repeat the process indefinitely... :D

finally, put $ a \equal{} 1,\;\;b \equal{} 2$ and the answer follows.


set theory
advanced fields
advanced fields theorems

Hello,
I hope my set theory questions are in the right section of your forum..
I have just finished my exams and now have time to take a look at a couple of problems that caught me out during semester.  These are probably really elementary questions, but proving things has never been a strong point, so here goes...

Questions 1,2,3 are most important to me as I have a faint hope that my answers to 4,5,6 are ok.

Thank you for your clues...

Q1
Note that A,B,U are sets such that A,B are contained in U
A "intersection" B = {}  <=>  B "is contained in" (U\A)

I see that this is very obviously true, but apart from assuming LHS for for proving =>, and assuming RHS for proving <=, I don't really know what to do, any clues would be great.

Q2
Consider the Von Neumann definition of the natural numbers.
w = {0,1,2,3,...} is the set of such numbers
w is an inductive set => 0={} is in w and for all k in w, k+ is in w
k = {0,1,...,k-1}
k+ = k union {k}

Sorry for the preamble there, just wanted to make sure we're all on the same page.

Question is a proof by induction...  Prove that for any n in w, we have n contained in w.

The base step is done, I'm stuck at the inductive step.

All I can think of so far is what you do at the start of every inductive step...

Assume that for some n=k in w, we have n=k is contained in w
RTP n=k+ is contained in w

Clues would be great.

Q3
Consider any cardinal numbers card(A)=a and card(B)=b and card(C)=c
Prove that (a^b)^c = a^(b*c)

I understand that to do this, I need to prove an equivalence of sets by finding a bijection

f: (A^B)^C --> A^(BxC)

I dont really have a clue as to what that f should be...

Am I correct in saying:

(A^B)^C is the set of all maps from C to the set of all maps from B to A, and
A^(BxC) is the set of all maps from B cross C to A

??

Q4

Give a partially ordered set (X,<=) such that X has two elements and (X,<=) has no greatest (maximum) element

I chose <= to be the standard set containment relation and
X = { {x},{y} } for any x ~= y

Is this ok?

Q5

Give a partially ordered set with a minimal element but no least element

I chose

(X,<=) where

X = { {x} , {y} , {x,y} } for any x ~= y

and <= is the standard set containment relation

So clearly {x} and {y} are minimal, {x,y} is both maximal and maximum and there is no minimum ???

Is this ok?

Q6

A partially ordered set with no maximal element

I chose (X,<=)

Where X = [0,1)  and <= is the standard <= on the reals.

Is this ok?

[quote]Q1
Note that A,B,U are sets such that A,B are contained in U
A "intersection" B = {} <=> B "is contained in" (U\A) [/quote]

Suppose $ A \cap B \equal{} \emptyset$.  In order to show $ B \subset U \setminus A$, we need to show that if $ x \in B$, then $ x \in U \setminus A$.  Okay, well, suppose $ x \in B$.  Then $ x \in U$, because $ B \subset U$.  Also, $ x \not\in A$, because $ A \cap B \equal{} \emptyset$.  Since $ x \in U$ and $ x \not\in A$, this means that $ x \in U \setminus A$.

You can give a similar kind of proof for the other direction.

Thanks Singular.  I all questions sorted now except Q3, can someone please help me with that one?


Putnam calculus AMC USA(J)MO USAMO algebra polynomial

Hi all!

I'm a HS senior and I absolutely love mathematics. I want to compete in the Putnam when I go to universtiy, but clearly since the test uses more advanced math than I've been introduced to, I'd clearly be at a disadvantage without furthering my mathematical background.

Now my question is, if I purchased books on the required topics (multivariate calculus, modern algebra, etc.) and spent a good while going through them and doing all of the exercises, would I be "ready" for the putnam? Or are the questions on the putnam much harder than the problem sets found in textbooks?

So I guess my question is: will exposure to problems in textbooks to higher math be in any way similar to the kind of problem solving on the putnam? Or is the putnam still leagues beyond what could be found in textbook math problems?

Thanks all!

-Arachnotron

Putnam problems are "leagues" beyond the difficulty of [i]most[/i] problems found in ordinary college textbooks. [Though some textbooks, or course, will have more challenging exercises than others]. That's why even college math majors (who've certainly studied from college level textbooks) cannot do many (if any) of the problems on the Putnam.

But while an understanding of the material found in college level textbooks is not a sufficient preparation for the Putnam, it is a necessary level of preparation needed for the Putnam. Meaning -- it is where you have to start.

Then, either after or concurrently with your study of the foundational material from textbooks, you can study specific problem solving books oriented for the Putnam -- like books of old Putnam problems published by the MAA, [u]Problem Solving Through Problems[/u] by Loren C. Larson, or [u]Putnam and Beyond[/u] by Razvan Gelca.

Keep in mind, though, that reaching that level of problem solving acumen usually takes years, not months. And also "success" on the Putnam (however you define that) is not a necessary step for being a successful mathematician or scientist. However, learning the basic material from the textbooks is a necessary step towards that goal.

The problem with studying from textbooks on your own is that you sometimes don't know how well you've [i]really[/i] learned the material without someone (like a professor) to discuss it with or to test you on it.

A link that may be of interest to you: http://www.math.hmc.edu/putnam/

Thanks!

I have another question; does math at the upper level undergraduate/graduate level ever become as "difficult" as contest math? (Putnam, USAMO)

I mean, I've heard how Harvard's Math 55 is brutal and involves lots of proof writing -- but is that stuff on the same level as the Putnam stuff? How about even more advanced courses?

I just want to know if problem sets from upper level texts ever require the ingenuity that the Putnam demands.

[quote="Arachnotron"]does math at the upper level undergraduate/graduate level ever become as "difficult" as contest math? (Putnam, USAMO)[/quote]
That depends on what school you're at, how hard your class is, what textbook you're using, and how hard the professor chooses to make your tests :)  (Some perspective:  the median score on the Putnam is zero.  What would a class with zero median test scores even [b]mean[/b]?)

More to the point, Putnam math is still problem-solving oriented whereas, as it's typically presented, undergraduate and graduate math is more focused on theory-building.  They're complementary approaches to mathematics, so it doesn't necessarily make sense to compare them like that.

[quote="Arachnotron"]I've heard how Harvard's Math 55 is brutal and involves lots of proof writing -- but is that stuff on the same level as the Putnam stuff? How about even more advanced courses? [/quote]
Here's how I'd describe Math 55 problem sets:  if you knew enough theory, they'd be trivial.  The challenge is that it's an introductory course:  how much of the theory can you rediscover?  How many well-known facts can you prove without having seen their proof in a textbook?  For example, one of the problems last semester was about elements of finite order in $ GL_4(\mathbb{Z})$.  It's trivial if you know that it suffices to check the cyclotomic polynomials of degree less than or equal to $ 4$ - but they got this problem before the professor had ever talked about cyclotomic polynomials.

In contrast, Putnam problems can't afford to be written this way, so they require extra steps that don't generalize easily.  (Except for some poorly written ones that really are trivial if you know enough theory.)

Edit:  "Graduate-level math" can describe research-level math, which is by definition harder than Putnam math (in some sense) since it's questions that no one knows the answer to.

Mostly in agreement with t0r, but with different emphasis:

The "understanding and being able to produce abstract proof" skills required to succeed on the Putnam are a subset of the same skills that one needs to pursue theoretical mathematics to the highest levels.  The problem-solving skills (i.e., the ability to solve tricky problems at high speed) required for the Putnam are definitely not a subset of the problem-solving skills necessary to become a mathematician, but it shouldn't be terribly surprising that the two are related to each other, i.e., interest in solving hard math problems in artificially short periods of time is well-correlated with interest in solving math problems in arbitrarily long periods of time.

Usually, "graduate-level math" refers to "coursework intended for graduate students," not "original math research."  t0r is absolutely right that you'll never take a class with exams and whatnot "just like" the Putnam in difficulty -- but certainly there exist classes that require you to solve problems that are harder (and require far more theory) than Putnam problems -- just you won't be expected to do this in one half-hour.  Research, of course, is an entirely different beast, since it's rarely* clear that "a right answer" exists when you begin.

* Unless you're doing enumerative combinatorics and your sequence is in the [url=http://www.research.att.com/~njas/sequences/index.html]OEIS[/url] ;)

The original poster wrote me via PM, but I thought that there's no reason not to respond in public (which I will do later today, hopefully):

[quote="Arachnotron"]I really appreciated your post in my thread. I was wondering though...have you ever taken a class that had harder problems than the putnam ones? If so, were these problems from textbooks? Or were they proposed by professors? Or both?

And when you say "harder", do you imply difficulty in terms of the "creativity" needed to solve them or more in understanding more advanced ideas and actually becoming proficient with them?

I plan on becoming a research mathematician which necessarily means that I will have to take those courses with those "harder" problems. I'm just worried that since I'm not exactly a math competition wizard that I will end up not succeeding.

Thanks![/quote]

As JBL mentioned, part of the key to doing well on the Putnam is to be able to solve problems [i]quickly[/i].  Even if you have all the technical tools you need to solve a particular Putnam problem, it may not occur to you which one will be most useful until after the test.  And being able to solve problems quickly is not really an issue in classes where you have a week to do problem sets and even less so in research.  So no, you don't need to be a Putnam whiz to be a good mathematician.  It just so happens that Putnam whizzes necessarily have other things going for them that also make them good mathematicians.

But what scares me, though, is that I doubt that, at this point, if I were given an unlimited amount of time that I would actually get one putnam (or IMO or USAMO) question correct. So if problem sets get to be as hard as putnam stuff, even without a small time limit, how am I supposed to do well?

Any thoughts?

That's probably never going to happen.  In a class you'll learn the material relevant to the problems, and classes are designed to be passable.  You should be aware that it's likely that the majority of math majors across the country probably can't solve a single Putnam problem - and they passed all their classes, didn't they?

The other important difference between the Putnam and your homework is that you can [b]ask for help.[/b]  You have TAs, your professors, the internet (that's us!), your fellow students...

Really, don't worry about it.  If you love mathematics, everything else will work out.

So, first, a generic point: there are far more successful research mathematicians than there are former Putnam fellows, so obviously being in the latter group can't be a precondition for being in the former.

Second, some comments on the Putnam: many Putnam problems aren't actually that hard.  Most of the 50% of people who get 1 or fewer points on the Putnam could probably score 8-10 points on any given Putnam by following the strategy, "Ignore questions 3-6.  Between 1 and 2, decide which looks more approachable and work on it for 45 minutes.  If, at that time, you don't have what you believe to be the key ideas of a proof, switch to the other and work on it for 45 minutes.  At that point, spend the next hour working on whichever was most promising, and the last half-hour writing up as clear a proof as possible."  Now, this strategy is not very much fun to implement, and will occasionally give a sub-optimal score, but I really do believe that anyone who's passed linear algebra, multivariable calculus, and seen a proof or three in an analysis or abstract algebra class should be able to get a problem correct this way.  (Recall that actually you get two attempts at this strategy in each administration of the exam.)

Third, some comments about myself: I certainly have taken classes that had harder problems than Putnam problems.  For example, freshman year I took Harvard's Math 25, and got on average about 2/3rds of everything right, which made me one of the top students (e.g., on the first midterm there was one score above 80, one more between 70 and 80, but then several scores between 10 and 20; I probably had the highest average in the class out of 10 people the second semester).  This was on problem sets that people worked collaboratively on for a week or two.

Generally, Putnam problems are "tricky" -- you don't need to know very many theorems in order to solve them, but the right theorems to apply may be very non-obvious.  "Real math" is extremely varied -- some results require trickiness, others require deep theory, plenty require some significant amount of brute-force computation time (human or computer), and still more aren't hard at all and just require someone to think about them for a little while.  Thousands of new theorems (or new proofs of old theorems) are produced each year, but these barely begin to scratch the surface of what is provable (either formally or practically) -- mathematics is truly enormous, and the boundary area of "things that haven't been proven but could be by an incremental improvement in what is known" is really quite large.  In fact, it's so large that mathematicians get to (have to?) choose among the problems they consider to find those that are most interesting.  Mathematics is not really a field in danger of losing all its low-hanging fruit.

Thanks! Hey, I have another question: could you point me to textbooks that have problems that are harder than Putnam's stuff? (i.e. the math book you used for Math 25)

I like solving hard problems, but it seems to me that working on putnam stuff may not be as important as learning harder material AND solving hard problems at the same time. So why not kill two birds with one stone?

Also, another question: I've seen in some math textbooks (Morris Hirsch's Differential Topology -- my dad owns a copy), that some of the "problems" are double starred (which according to the textbook are 'important results') and triple starred (unsolved problems). Are these the kinds of problems that must be solved in the classes that you've mentioned, or are those too extreme examples?

And finally: when you said the Math 25 stuff was harder, were the theorems that needed to be applied to the problems also non-obvious, or were the theorems just very hard to get a good grasp with?

In math 25 we didn't really use a text, except in the second semester people complained and he suggested Lang's Real and Functional Analysis as a reference text.  Stanley's Enumerative Combinatorics has lots of problems that take up several pages of published papers.  (He rates problems on a 5-point scale with +s and -s, and suggests that 2+ are the hardest problems that a class of graduate students should reasonably be expected to solve.  You need to do a certain number of 3-s in order to earn an A+ in his class.  5s are unsolved problems, with 5- meaning "unsolved but not too much attention has been given to it, so it might actually be easy.")  Any field of math has problems that are too hard to reasonably require students to do, and some textbook writers decide to sprinkle some of these problems among their other exercises -- I don't think this is particularly shocking :)

I think focusing on the difficulty of material (either of the math or the problems that come along with it) isn't really necessary -- you'll run across plenty of hard problems if you continue studying mathematics, no matter what.  I don't think there's any special need to seek them out :)

Thanks, JBL :)

And thanks to t0rajir0u and gauss202!

I love this forum.

[quote="Arachnotron"]
I love this forum.[/quote]

Yeah, this place is like a super-upgrade to CC.

I can't speak for Arachnotron, but we both post there and I'm pretty tired of the redundancy of AP exams. They are all standardized and require just memorization and comprehension to get a 5, and are not indicative of anything but a desire to work hard.

AoPS has certainly opened my eyes on a lot of things that go much farther than math. All my life I had approached problems as just rushing through alternatives until something instantly clicks. It's from this website that I learned how to, well, "think deeply of simple things."

You guys are all very helpful and amazing people.

Also, is there any "report-card" (i.e. coursework) indicator that could correlate in any fashion to how one could succeed in Putnam?

Report-card, or coursework indicator?

We're walking into an issue I've used in describing the students I know. You see, there are good students, and there are problem solvers. These terms are not mutually exclusive, nor exhaustive, but they are very much not the same.

One of the key words I would use to describe good students is [i]diligence.[/i] Good students come to class, turn their assignments in on time, never neglect a course and fall behind in it. They're never caught by surprise by a due date. The know what they have to work on, and they get that work accomplished. Where there are misunderstandings of concepts they go to the professor, or to whomever appropriate, in a timely fashion to work out those misunderstandings. A good student can survive a course with a bad instructor. Good students tend to get very high grades not just on the courses in their major but in every course they take. It's not a defining characteristic, but good students even tend to have neat handwriting. Good students are masters of the expected.

Where good students are masters of the expected, problem solvers are masters of the unexpected. They thrive on what's different, what's new.

The little one-unit course I teach (official title: "Honors Seminar in Problem Solving") that prepares students for the Putnam may be the one place on campus where the problem solvers have the advantage over the good students. The biggest difficulty for those who are problem solvers but not good students is that every other part of the academic world is made for the good students and not for them. I know some problem-solvers whose transcripts don't make for such pretty reading.

In the long run, in most avenues of life, the habits of good students should stand them in good stead. That diligence, that directness, that willingness to meet expectations - it's a good thing. It is true, sometimes, that students who are good students but very much not problem solvers have a rocky time with the transition from an undergraduate math program to a graduate program. But problem solvers who are not good students might also founder on that transition for different reasons.

Why would you want to correlate your performance on the Putnam with an even less meaningful statistic?  I would hope that the only point of the Putnam is as some kind of measure of a certain kind of problem-solving ability, and if you can't ascertain that without [b]another[/b] test then you have to wonder what you're measuring at all.

[quote="t0rajir0u"]Why would you want to correlate your performance on the Putnam with an even less meaningful statistic?[/quote]

I just asked if there was any measure outside the Putnam that could be used to correlate Putnam performance to it.

Again, I'm not implying that another off-Putnam indicator that could correlate to Putnam performance exists.

[quote="t0rajir0u"]Again, why would you want to do that?[/quote]

So I could scout for training mates... provided that the University of Montreal does it.

Hey JBL, you should have taken Harvard's famed Math 55 course while you were an undergrad ;) lol. I got some course pdf files for Math 55 (both a and b) off the course website (current one I think)...pretty insane/awesome stuff :D

Searching for posts by me that include the words "Math 25" yields the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=91379#91379
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125229
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24761
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107431
(among others) which I think explain a lot why I chose the path I did  :wink:

There are a number of links there to old math 25 and 55 webpages; don't know if any of them are still active, though.  User Tokenadult was a collector of that sort of information, but I think his interest in this forum waned as his son got older, and he hasn't posted here in a while.


$ ABCD$ is a quadrilateral. $ I,J$ are the midpoints of two diagonal $ AC,BD$. Let $ E\equal{}AJ\cap BI,F\equal{}CJ\cap DI$.$ H,K$ are the midpoints of two sides $ AB,CD$. Prove that $ EF\parallel HK$

dear mr.danh,do you have a solution?it seems nice!

It has already been posted before as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=133248]Three parallel lines[/url], by [b][size=100]gemath[/size][/b].

kostas Vittas.


[quote="junggi"][hide]
$x^{3}+8y^{3}+27z^{3}=18xyz$
$\iff \frac{x^{3}+8y^{3}+27z^{3}}{3}=6xyz$
but this is the equality case of AM-GM inequality, which means that
$x=2y=3z.$
Substituting this in the second equation and solving, we get
$x=12,y=6,z=4$
[/hide][/quote]
Yeah, that's probably the best way.

I thought of using $a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$, but it doesn't work so nicely... :|

[quote="junggi"][hide]
$x^{3}+8y^{3}+27z^{3}=18xyz$
$\iff \frac{x^{3}+8y^{3}+27z^{3}}{3}=6xyz$
but this is the equality case of AM-GM inequality, which means that
$x=2y=3z.$
Substituting this in the second equation and solving, we get
$x=12,y=6,z=4$
[/hide][/quote]

[i][color=red]AG-GM inequality you can use it only in case that $x$ $2y$ $3z$ are positive real numbers [/color][/i] :wink:

[i]Anyway , Euler's formula I think is the best way to solve it.[/i]

$a^{3}+b^{3}+c^{3}=3abc\Longleftrightarrow a=b=c$ [i]or [/i]$a+b+c=0$

[quote="Gyan"][quote="frt"]

I thought of using $a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$, but it doesn't work so nicely... :|[/quote]
as pointed out before:
 ... if a+b+c is not zero .. ($(a^{2}+b^{2}+c^{2}-ab-bc-ca)=(1/2)((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$ can be zero only when $a=b=c$ (for real numbers)
So it works out very well indeed.[/quote]
Oh, yeah... Darn, why didn't I notice that? :oops:


Hello....maybe  this  question is  silly  for most  of you , but im a  begginer  so i would like to know  , how  to  insert  a  geometric  figure (created  by another  programm  ,  i  use  a  greek one  to  create  them , called  dynageo)  in a  tex  file. Thanks  for your time  :)

Read the stickies Nickolas. For example: http://www.mathlinks.ro/Forum/viewtopic.php?t=53217


geometry
circumcircle
inradius
ratio
geometry solved

[[b]Moderator edit:[/b] This topic is about the following problem:

Let ABC be a triangle and M a point inside it. Let D be the point of intersection of the line AM with the circumcircle of triangle ABC (other than A). Find the point M such that the term $\frac{MB\cdot MC}{MD}$ is minimal.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=5658 for a discussion of this problem.]

Alekk, about that problem you said was so hard, which asks to inimise MB*MC/MD. Do you have any idea which point M could be? I think it's this point: draw the bisector of angle BAC, it cuts the circumcircle at D and take M on (AD) such that DM=DB=DC (we know that D is the midpoint of arc BC). I have some arguments, but I don't actually have a solution, so it's useless to post anything else unless I have a soln. What do you think? Is this the point?

I can't believe this! Th point I have tried so hard to decribe in the earlier post is... tada! none other than the incenter. I can't understand how I could miss that. Anyway, do you happen to know if this is the point we're looking for?

I think I have a solution for your geometry
problem.

Let R be the circumradius of ABC and in general
denote by R(UVW) = the circumradius of triangle
UVW. Also r is the inradius of ABC.

Draw the other 2 intersections , of BP and CP
with the circle (let them be E and F). Also ,
drop perpendiculars from P to the sides of the
triangle. Let X = projection of P onto BC , Y on
AC and Z on AB .

A quick angle-chase shows that XYZ and DEF are
similar . Now , compute the ratio of similarity ,
you get by a few simple calculations :

YZ/EF = (PA*PB*PC)/(2R*rho(P))

where rho(P) is the power of P with respect to
the circle ABC (rho(P) = |R^2 - OP^2|). Now ,
since PA * PD = rho(P) you also have that the
quantity (Q) you want to minimise is :

Q = (PA*PB*PC)/(rho(P))

Now it's enough to minimise Q/2R , or
equivalently YZ/EF.

But YZ/EF = R(XYZ)/R(DEF) = R(XYZ)/R , since the
triangles are similar.

Now we have to find the triangle with the least
circumradius and vertices on ABC's sides (the
triangle XYZ corresponding to the big DEF). But
R(XYZ)>=r , clearly and R(XYZ)=r is attainable.

FORTUNATELY (!!!!) the triangle for which R(XYZ)
is the least has the property that the
perpendiculars in X to BC , Y to AC and Z to AB
are concurrent . If it didn't have this property
we couldn't have disregarded the way XYZ it
obtained , by projecting a point on the sides and
we would have had to use specific properties; but
since of ALL the triangles with X on BC , Y on AC
and Z on AB , it has the least circumradius and
also has that property , it's ok.

Getting back , EF/YZ = R(XYZ)/R >= r/R , so
Q >= 2R * R(XYZ)/R = 2r

This is the required minimum and we are done. If
u need some more insight on the computations and
stuff just give me a sign , I didn't have time to
write it all up here.


modular arithmetic
number theory unsolved
number theory

let $ a,b,c$ be positive integer . $ a$ is an even and $ b$ is odd. Show that $ \forall n \in N^*$ then esixt $ x \in N^*$ to satisfy :
        . $ ax^2 \plus{} bx \plus{} c$ $ \vdots 2^n$

u mean $ 2^n | ax^2\plus{}bx\plus{}c$ ,no?
let's suppose that $ x>y$ are integers not congruent modulo $ 2^n$.
we suppose $ n \in N$.the cases $ n\equal{}0,1,2$ are trivial.
we obviously have $ P(x) \not \equiv P(y) \pmod {2^n}$ because:
if $ P(x) \equiv P(y) \pmod {2^n}$ it follows that :
$ ax^2\plus{}bx \equiv ay^2\plus{}by \pmod {2^n} \implies a(x\minus{}y)(x\plus{}y) \equiv b(y\minus{}x) \pmod {2^n}$.
so we get
$ 2^n | a(x\minus{}y)(x\plus{}y) \plus{} b(x\minus{}y) \implies 2^n | (x\minus{}y)(a(x\plus{}y)\plus{}b) , a(x\plus{}y)\plus{}b \equiv 1 \pmod {2} \implies 2^n |x\minus{}y$.
but it's a clear contradiction with the assumption $ x \not \equiv y \pmod {2^n}$.
so we have proven that $ P(x) \not \equiv P(y) \pmod {2^n}$.
so for any integer $ i$ there exists $ x \in N$ such that $ P(x) \equiv i \pmod {2^n}$.
so it remains to put $ i\equal{}0$ to get what is asked.


algebra
polynomial
inequalities
function
functional equation
algebra unsolved

Dear all,

a few days ago, I asked whether or not there was an elementary math-way to derive
$ \sqrt{1\plus{}2\sqrt{1\plus{}3\sqrt{1\plus{}4\sqrt{...}}}}$ (a problem Ramanujan posed).
Well, I have found a way to do it(though I am unsatisfied with it for reasons I'll state later):

We need to find a formula for $ \sqrt{1\plus{}n\sqrt{1\plus{}(n\plus{}1)\sqrt{...}}}$ because otherwise it is quite difficult.
Let $ f(n)$ be the desired formula. Plugging in, we have $ [f(n)]^2\equal{}1\plus{}nf(n\plus{}1)$. (This is the part I don't like)
We assume $ f(n)$ is a polynomial (this is obvious intuitively but I can not prove it). Letting k be the degree of $ f(n)$, since the degree on both sides of the equations must be the same, we have $ 2k \equal{} k\plus{}1$, $ k\equal{}1$. Therefore, our polynomial is of the form $ an\plus{}b$. Plugging in, we have $ a^2n^2 \plus{}2abn \plus{} b^2 \equal{} an^2 \plus{} (a\plus{}b)n\plus{}1$. Solving for this, we have a = 1 and b = 1, so $ f(n) \equal{} an \plus{} b$.

The only case where we could have gone wrong was if $ f(n)$ was undefined/infinite.

To show $ f(n) < \infty$, we simply need to show it for a particular case (for this particular case, let n = 2 simply because that was the original problem)
Let's do so by showing $ f(n) < \sqrt{3\plus{}4\sqrt{3\plus{}4\sqrt{...}}}$ (the latter expression is obviously finite, so proving this will immediately confirm the desired expression is finite).
We can expand into other nested rad.s in the $ RHS$; what I mean is that
$ \sqrt{3\plus{}4\sqrt{3\plus{}4\sqrt{...}}} \equal{} \sqrt{3\plus{}2\sqrt{12\plus{}16\sqrt{...}}} \equal{} \sqrt{3\plus{}2\sqrt{12\plus{}4\sqrt{48\plus{}64\sqrt{...}}}}$ (here what I'm doing is arranging the coefficient of each nested radical such that they lie in geometric progression). It is quite easy to see we can keep expanding like this, and since each term in the RHS is larger than each term in the LHS of the original inequality, the inequality is proven.
So $ \sqrt{1\plus{}2\sqrt{1\plus{}3\sqrt{...}}} \equal{} 3$.

As I said, I am unhappy with this proof because of my assumption that the function is a polynomial.
Since I am training for National Olympiads too, could you please tell me if this proof is rigorous enough (or do I have to prove some other things for it to be complete as well)?

[quote="bengal"]Plugging in, we have $ [f(n)]^2 \equal{} 1 \plus{} nf(n \plus{} 1)$. (This is the part I don't like)
We assume $ f(n)$ is a polynomial (this is obvious intuitively but I can not prove it)[/quote]

This is not at all obvious intuitively. For a proof to be "rigorous enough" it must be entirely rigorous. EDIT: it is true, I misread your polynomial.

Also, if you wish to do olympiads I recommend that you join the 3:30-5:45 class because the first one is based more on computation than proofs.

To me it would be the 1st thing I would try (why? because there is a *n on the RHS: the logic is a little confusing but it is justified intuitively). Maybe I'll be able to fish up a proof of this "intuition"  soon (Meanwhile, by all means people, please help while you can). Menwhile, thanks for your feedback on olympiads.

However, despite that assumption part, everything else is rigorous. It does end up being a polynomial in the end, so shouldn't it be all right (I really want to be able to prove the polynomial part, but that doesn't really destroy the correctness of the proof)?

It is quite obvious that this is a function, so n+1 should be the only formula, right?

The proof, quite frankly, is completely wrong because of that assumption. Mind you, I believe that was how Ramanujan intuitively found it, so its not a bad start. To see that it is not necessarily true, what if the function was $ f(n)\equal{}2^n$, or $ f(n)\equal{}n\sqrt{n}$, or any other function not modeled by a polynomial?

Well, I am rather ashamed of myself.
There is a more elementary solution (the functional equation is a little hard).
It's actually child's play.
$ \sqrt {9} \equal{} 3$
$ \sqrt {1 \plus{} 2\sqrt {16}} \equal{} 3$
$ \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {25}}} \equal{} 3$
It's easy to see that one can keep substituting in like this, and substituting infinitely many times will yield our result(this proof is not mine, I found it online because it turns out as I expected, that the functional equation required higher mathematics : according to the website I saw).
Meanwhile, my proof for the result being finite (and therefore 3) suffices to complete the proof. However, mathematicians also consider this to be unrigorous for some reason.

I know how everyone (including myself) will curse themselves for not being able to see this, so I won't post any more of what I saw to make ourselves feel worse. EDIT: I curse myself for being stupid enough to say this.

EDIT: Sorry, rofler, and differ, and everyone else for the stupid false lead.

This is not true either. For then, I could make it equal to 5 as well:

$ \sqrt{25}\equal{}5$
$ \sqrt{1\plus{}2\sqrt{144}}\equal{}5$
$ \sqrt{1\plus{}2\sqrt{1\plus{}3\sqrt{\text{blah}}}}\equal{}5$

[quote="Rofler"]This is not true either. For then, I could make it equal to 5 as well:

$ \sqrt {25} \equal{} 5$
$ \sqrt {1 \plus{} 2\sqrt {144}} \equal{} 5$
$ \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {\text{blah}}}} \equal{} 5$[/quote]

For some odd reason, that was one of the most entertaining posts I have read.

[quote="bengal"]I know this is kind of late, but your problem with this is that the function may assume multiple values, right? Or is there also something else?[/quote]

...?

1. Not every sequence can be modeled by a polynomial
2. Your argument can't work, because then my argument would work (identical except for the initial value of x), and we would have $ 3\equal{}5$.

Cheers,

Rofler

The stuff unrelated to functional equations isn't my work  :| . It's the work of some professional mathematician that was online (yeah, I got pretty desperate). Well, I'm kind of happy that didn't resolve the problem because I wanted to do so independently (of course, if anyone finds a solution, they're free to post it). I think this problem can be resolved w/ olympiad level math, but I'm not sure how (I do have a few ideas w/ the functional equations; I'll test them out and let you know. Sorry for that false lead though...


$ a_1\equal{}2, a_n\equal{}a_1a_2a_3...a_{n\minus{}1}\plus{}1$, $ b_i\equal{}\frac{1}{a_i}$  prove that for any integer $ k$ sum of $ k$ distinkt members of sequence  $ {b_i}$ is less than 1

[quote="dima ukraine"]$ a_1 \equal{} 2, a_n \equal{} a_1a_2a_3...a_{n \minus{} 1} \plus{} 1$, $ b_i \equal{} \frac {1}{a_i}$  prove that for any integer $ k$ sum of $ k$ distinkt members of sequence  $ {b_i}$ is less than 1[/quote]
$ \sum\limits_{i\equal{}1}^{n}{b_i}\equal{} 1 \minus{} \frac{1}{a_1*....*a_n}$ :)


calculus
integration
logarithms
calculus computations

this is a question on my Cal III final this summer and i cannot solve it at all, please help if you can:

Evaluate    [size=150]int((e^(-x)-e^(-2x))/x),x,0,infinity)[/size]

thanks
-Chris

$\int_0^\infty \frac{e^{-x}-e^{-2x}}{x}dx = \ln 2$. It is a special case of Froullani's integral.

If $f(0^+) = A$ and $f(\infty) = B$ then $\int_0^\infty \frac{f(ax)-f(bx)}{x}dx = (A-B)\ln \frac{b}{a}$, where $a,b>0$.

$\int_\delta^X \frac{f(ax)-f(bx)}{x}dx = \int_\delta^X \frac{f(ax)}{x}dx -\int_\delta^X \frac{f(bx)}{x} dx = \int_{a\delta}^{aX} \frac{f(x)}{x}dx -\int_{b\delta}^{bX} \frac{f(x)}{x} dx = \int_{a\delta}^{b\delta} \frac{f(x)}{x}dx - \int_{aX}^{bX} \frac{f(x)}{x} dx = \int_a^b \frac{f(t\delta)}{t}dt - \int_a^b \frac{f(tX)}{t} dt$.

Now let $\delta \to 0^+$ and $X\to\infty$,
$\int_a^b \frac{f(t\delta)}{t}dt \to \int_a^b \frac{f(0^+)}{t}dt = A\ln \frac{b}{a}$
and
$\int_a^b \frac{f(tX)}{t}dt \to \int_a^b \frac{f(\infty)}{t}dt = B\ln \frac{b}{a}$.

Here's a specific calculation in this case. We use the following fact:

$\frac1x=\int_0^{\infty}e^{-tx}\,dt$

We're going to put that into our integral, creating a double integral. Of course, the only reason I'd ever create a double integral that wasn't there in the first place is so I could switch the order of integration.

$\int_0^{\infty}\frac{e^{-x}-e^{-2x}}x\,dx$

$=\int_0^{\infty}\int_0^{\infty}e^{-tx}(e^{-x}-e^{-2x})\,dt\,dx$

$=\int_0^{\infty}\int_0^{\infty}e^{-(t+1)x}-e^{-(t+2)x}\,dx\,dt$

$=\int_0^{\infty}\frac1{t+1}-\frac1{t+2}\,dt$

$=\ln(\frac{t+1}{t+2})|_0^{\infty}\,=0-\ln(\frac12)=\ln2$


quadratics
MATHCOUNTS
algebra
arithmetic sequence

Compute the sum $(1-\frac{1}{16})+(\frac{1}{2}-\frac{1}{17})+ (\frac{1}{3}-\frac{1}{18})+\cdot +(\frac{1}{85}-\frac{1}{100})$. Express your answer as a decimal to the nearest tenth.

Is there a fast way to do these where the difference between the terms is greater than 1?
[hide]I suppose you could sum the plus terms and the minus terms: $\frac{85*43}{\text{the product of the numbers from 2 to 85}}-\frac{1930}{\text{the product of the numbers from 16 to 100}}$, but that is a dinosaur of a equation!
[/hide]

Billy

This would take years on a supercomputer! The first term in is 15/16, and the last term is 3/1700. I don't see any pattern. The second one is something like 15/34, and the third, 5/18...

Hmmm. Maybe I changed the question to be a great deal harder accidentally? The original had a difference of 5 between the denominators of each set of fractions in parentheses, and this one has 15.

[quote="tarquin"][hide]Whenever you subtract fractions like x/y - a/b the difference is always (b-x)/ay. This can be applied to this problem. The first difference is 15/16, then 15/34 and so on until 8500. Factor a 15 out so it becomes 15(1/16 + 1/34 + 1/54 + 1/68 ... 1/8500). If we go farther, we see 1/16, 1/34, 1/54, 1/76, 1/100, 1/126, 1/154 and so on. We can see that the denominators are related by a quadratic equation, x^2 + 15x. If we look at the first 3 numbers, we can apply this formula where x = 1. 1/(x^2 + 15x) + 1/[(x+1)^2 + 15x + 15] + 1/[(x+2)^2 + 15x + 30] is equal to 1/[x(x+15)] + 1/[(x+1)(x+16)] + 1/[(x+2)(x+17)]. This comes out to something really weird over (x)(x+1)(x+2)(x+15)(x+16)(x+17). We know that if were to continue, the bottom would come out (x)(x+1)(x+2)... all the way to (x+83)(x+84) times (x+99). With the numbers we have now: [x(x+15) + (x+1)(x+16)]/[(x)(x+1)(x+15)(x+16)] + 1/(x+2)(x+17) = [x(x+2)(x+15)(x+17) + (x+1)(x+2)(x+16)(x+17) + (x)(x+1)(x+15)(x+16)]/[(x)(x+1)(x+2)(x+15)(x+16)(x+17)]. If we continue farther: [x(x+2)(x+3)(x+15)(x+17)(x+18) + (x+1)(x+2)(x+3)(x+16)(x+17)(x+18) + x(x+1)(x+3)(x+15)(x+16)(x+18) + (x)(x+1)(x+2)(x+15)(x+16)(x+17)]/[x(x+1)(x+2)(x+3)(x+15)(x+16)(x+17)(x+18)]. If we substitute 1 in for x: [(3)(4)(16)(18)(19) + (2)(3)(4)(17)(18)(19) + (2)(4)(16)(17)(19) + (2)(3)(16)(17)(18)]/[(2)(3)(4)(16)(17)(18)(19)]. Remember, this is after the fourth term, and that there are 4 parts to the numerator, and there are 85 terms in total. If we were to continue all the way to the last term, then we'd notice in the first 4 terms that the numerators would be 85! * 100!/(15!*16), 85!/2 * 100!/(15!*17), 85!/3 * 100!/(15!*18), 85!/4 * 100!/(15!*19). If we continue this pattern, the last term's would be 85!/85 * 100!/(15!*100), and we know that the denominator would be 85!*100, so all of the 85!'s will cancel out, and the 100!'s will become 99!. This leaves the problem at 15[(99!/(15!*16) + 1/2 * 99!/(15!*17) + 1/3 * 99!/(15!*18) + ... + 1/85 * 99!/(15!*100)]. Bleh dunno what else.[/hide][/quote]

WOW!
RJ, you wrote a lot. I can't follow it.
Btw, was that a sprint or a target or a team question?

Target, though as you see, toughenend up a bit.

Hint: Think telescopes.

Billy

If it was target, it could be done on a calculator within 4 minutes.

Is there a way to mutiply all the numbers in an arithemetic sequence? Like, what is the product of this sequence:

1, 4, 7, 10, 13, 16, 19, 22, ..., 301?

I think the "toughened" problem is too time-consuming for a MATHCOUNTS Target Round.  IMHO, the original problem was fine for a State or National Target Round.

As solafidefarms suggested, "think telescopes".

Productive discussion of this problem seems to have stagnated, so I am offering a major spoiler:
[hide="spoiler"]By rearranging and combining terms, you will end up with 15 positive terms and 15 negative terms.  You can write the positive terms as $H_{15}=\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{15}$, where $H_{15}$ is the 15th Harmonic Number.

[hide="Further spoiler"]The negative terms can be written as $H_{100}-H_{85}=\frac{1}{86}+\frac{1}{87}+\cdots+\frac{1}{100}$

The problem can be rewritten as:  $H_{15}-(H_{100}-H_{85})$

Remember this problem asks for the answer [b]to the nearest tenth[/b].  There is no need to solve, exactly.  Using your calculator, carefully calculate and sum each of the positive terms ($H_{15}$).  You can do the same for each of the negative terms.  Then subtract.  (If you are very good with estimation, you might be able to find an estimate for the sum of the fifteen negative terms that is correct to the nearest tenth.)[/hide]
[Side note:  Harmonic Numbers are quite interesting, but a full discussion of Harmonic Numbers and their properties is beyond the level of this Forum.  If you are interested, you can look up the term on the Internet or you could ask on the High School "Intermediate" Forum.  Ask about "Harmonic Numbers" and Euler's constant, $\gamma$.][/hide]

The concept itself isn't too hard, so I'll say this much.

A harmonic sequence is an arithmetic sequence, however, the reciprocal is taken of each term.

For example...

1, 2, 3, 4, etc. is an arithmetic sequence.
1, 1/2, 1/3, 1/4, etc. is a harmonic sequence.

[hide]
If you look carefully, you see that it is 1/1 + 1/2 + 1/3 + ... + 1/85 - 1/16 - 1/17 - 1/18 - ... - 1/100.
Many of the terms can be crossed off, leaving 1 + 1/2 + 1/3 + 1/4 + 1/5 + ...  + 1/15 - 1/86 - 1/87 - ... - 1/100.
Using a calculator, these can be summed quickly.
3.156

[b]3.2[/b][/hide]

I don't see what is so time consuming about this if it is a target peoblem.


calculus
derivative
trigonometry
function
search
calculus computations

What is the derivative of e^x - e^-x / 2x?

I got xe^x + xe^-x - e^x + e^-x / x^2 , not sure if that is correct though.

hello, did you meant $ \frac{e^x\minus{}e^{\minus{}x}}{2x}?$
Sonnhard.

hello, using the quotient and the chain rule we get:
$ f'(x)\equal{}\frac{\left(e^x\minus{}e^{\minus{}x}(\minus{}1)\right)x\minus{}\left(e^x\minus{}e^{\minus{}x}\right)1}{2x^2}$.
Sonnhard.

$ \left(\frac{e^x \minus{} e^{\minus{}x}}{2x}\right)' \equal{} \left(\frac{\sinh x}{x}\right)' \equal{} \frac{(\sinh x)' \cdot x \minus{} \sinh x \cdot (x)'}{x^2} \equal{} \frac{x \cosh x \minus{} \sinh x}{x^2}$

[quote="Carcul"]$ \left(\frac {e^x \minus{} e^{ \minus{} x}}{2x}\right)' \equal{} \left(\frac {\sinh x}{x}\right)' \equal{} \frac {(\sinh x)' \cdot x \minus{} \sinh x \cdot (x)'}{x^2} \equal{} \frac {x \cosh x \minus{} \sinh x}{x^2}$[/quote]

i dont get what this means :S , i have not learnt this yet... is there any other way to do this? :huh:

$ \sinh x$ is another name for $ \frac{e^x\minus{}e^{\minus{}x}}{2}$; it's analogous to $ \sin$, and closely related using complex numbers. You don't need to know it, and another solution has already been posted in this thread.

The essential properties of $ \sinh$:
Define $ \cosh x\equal{}\frac{e^x\plus{}e^{\minus{}x}}{2}$. Now $ \frac{d}{dx}\sinh x\equal{}\cosh x$, $ \frac{d}{dx}\cosh x\equal{}\sinh x$, $ \sinh 0\equal{}0$, and $ \cosh 0\equal{}1$. We have an analogue of the Pythagorean identity- $ \cosh^2 x\minus{}\sinh^2 x\equal{}1$.
Most other trigonometric identities also have analogues involving these functions- can you find $ \sinh 2x$ and $ \cosh 2x$ in terms of $ \sinh x$ and $ \cosh x$?

For your own investigation, sinh x is a hyperbolic trigonometric function, search that in google or wikipedia and you will find out alot of information about them.

EDIT: Talking to PunPride, incase anyone thought I was talking to jmerry who obviously knows a lot more than I do.

i just dont understand why my prof would give me a question involving sinh , sinh' etc. when we have not learned or discussed it at all in lecture :S , that is why im clueless  :huh:

He didn't. The question is understandable and doable without that.

sinh(x)= isin(ix)
cosh(x)=cos(ix)
I think

[url=http://library.wolfram.com/webMathematica/Education/WalkD.jsp]This website[/url] will do step-by-step derivatives.


Let $ \{x,y,z\}\subset\mathbb N.$ Solve the following equation:
\[ (x^2 \plus{} 2)(y^3 \plus{} 3)(z^2 \plus{} 4) \equal{} 60xyz
\]

This is standard and easy problem.

Namely $ LHS \equal{} x^{2}y^{3}z^{2} \plus{} A$ where $ A$ is positive integer.

Now we have: $ x^{2}y^{3}z^{2} \plus{} A \equal{} 60xyz$ so $ 0 < A \equal{} 60xyz \minus{} x^{2}y^{3}z^{2} \equal{} xyz(60 \minus{} xy^{2}z)$
so obviously $ 60 \minus{} xy^{2}z > 0$ so $ xy^{2}z < 60$

From last inequality obviously all f $ x$ $ y$ and $ z$ can not be greater than $ 2$ at same time so at least one of them is $ 1$ or $ 2$. We have following cases:

(A) $ x\leq2$

(i) $ x \equal{} 1$
Then given equation becomes equivalent to:$ (y^{3} \plus{} 3)(z^{2} \plus{} 4) \equal{} 20yz$ so using same arguments as above we can conclude that $ y^{2}z < 20$ and now everything that remains is to check equation
$ (y^{3} \plus{} 3)(z^{2} \plus{} 4) \equal{} 20yz$ when $ y \equal{} 1,2,3,4$ (it becomes quadratic respectively to $ z$)

(ii) $ x \equal{} 2$
In this case the given equation becomes:
$ (y^{3} \plus{} 3)(z^{2} \plus{} 4) \equal{} 10yz$ so it follows that $ y^{2}z < 10$ and just take simple checking for $ y \equal{} 1,2,3$

Other cases, $ y\leq2$ and $ z\leq2$, are treated in analogous way and all solutions can be easily found.


calculus
derivative
function
analytic geometry
graphing lines
slope

Let $ x$ be a real number and let $ f(x) \equal{} \sum_{i \equal{} 1}^{10} |x \minus{} F_{i}|$ where $ F_{i}$ is the $ i$th Fibonacci number, with $ F_{1} \equal{} 1, F_{2} \equal{} 1$ and$ F_{n} \equal{} F_{n \minus{} 1} \plus{} F_{n \minus{} 2}$ for $ n > 2$.  Find the minimum value of $ f(x)$ with proof.

[hide="Hint 1"] This problem doesn't use any special properties of the Fibonacci numbers. [/hide]
[hide="Hint 2"] Compute the derivative of $ f$.  This is easier than it sounds.  Don't rush ahead; just think about it. [/hide]

I'm not sure about the solution via derivative (I think the sgn function is going to be included), but here's my reasoning:

The minimum value of $ f$ should occur when $ x\in[F_1,F_10]$
Divide the interval $ [F_1,F_10]$ into sub-intervals $ [F_1,F_2],[F_2,F_3],...,[F_9,F_10]$
$ f(x)$ should be either increasing or decreasing in each interval because says $ x_k\in(F_k,F_{k\plus{}1})$ for $ 1\le k\le10$
with the exception when $ x\in[F_5,F_6]$ where $ f$ is constant (since there are an equal number of Fibonnaci numbers smaller and greater than such x)

For the nonconstant case, note that for $ x_k\in(F_k,F_{k\plus{}1}$, if $ k>5$ then for $ t\in(F_k,F_{k\plus{}1})$ and $ t<x_k$, $ f(t)<f(x_k)$ since there are more Fibonnaci numbers in question smaller than $ F_k$, so for such smaller $ t$, it will decrease the absolute difference between $ x$ and the smaller Fibonacci numbers while increasing the absolute difference between $ x$ and the bigger Fibonacci numbers, each by the same amount of $ |x_k\minus{}t|$, but since there are more such smaller Fibonacci numbers than bigger Fibonacci numbers, the overall total is decreased.
Same reasoning for $ k<5$ (where $ f(t)<f(x_k)$ for $ t>x_k$. So the minimum must occur at when the function stop decreasing or increasing, or in other words, when it is constant, which happens in the interval $ [F_5,F_6]$.

So is this the derivative solution you meant?  :maybe:
Also how should I formulate the proof? I don't think my above proof is clear enough... :(

Introduce a function to help you out with clarity. For example, $ N(x) \equal{} n \iff F_n \leq x < F_{n \minus{} 1}$.

Your argument is basically that:
- No point at an interval of positive slope can produce a minimum, since the point slightly to the left will have a lower value.
- No point at an interval of negative slope can produce a minimum, since the point slightly to the right will have a lower value.

Thus, any minimum must occur somewhere where the function is not locally increasing or decreasing, i.e. where the derivative is zero.


Are we allowed to make our own usergroups now? or...like...is it for ppl from certain geographical locations only?

[quote="Peter VDD"]No one is :)

Only the admins can make groups, and I think they'll leave it at this one moderator group :P[/quote]

oh realli...that's odd... -.-

And this moderator group u speak of, is that only for "native" Mathlinks members?

That moderator group seems to be for Mathlinks moderators. There are other usergroups too. Each class has its own usergroup, for example.

yeah, probably only cuz the old list never got updated :D

or they only take the people moderating mathematics forums... :maybe:

Anyway if there's a need for a group just post it, if most people agree I'm sure they'll make one.


this topic sucks, i wished i hadnt posted this
no one replies.
can someone delete this ...
 :juggle:

You should not have jumped to your conclusion so quickly (you haven't even waited for a whole day!). :P

day = 24h, I would say.

Anyway, it is no wonder nobody has replied. In most countries, IMO team selection ends in spring (in Germany, it even ends in June), so most of us don't know yet whether we will participate in the IMO 2006 or not.

  Darij

Closer to 23 hrs 56 mins, i should think

Well... some already know, for example in countries not so good (like France, alas  :( ) those who already participated and can still come to IMO 2006 are sure they will be there.

well I don't know about 4 but there are many people who will go not for first time to IMO ( for example, this(if I qualify of course) will be my third IMO).
And I bet that there will be no more than 10 people with this their 4th IMO.
:) and maybe 10 has even smaller value ;)

Supposing Tiks go to 5 IMO's... is he the one that goes more times??? or there's other with 5, or more, too???

[quote="Davron"]Just read what Arne writes : :-)

http://www.mathlinks.ro/Forum/viewtopic.php?t=50712&start=180

davron[/quote]

What post exactly?

[quote="w13"][quote="Tiks"][quote="Albanian Eagle"]slovenia is the fourth imo i'll partecipate
i wanted to know if there are other people who will participate in slovenia for their fourth time[/quote]
It will be my fourt IMO too :),I hope there are other peopel beside us,hwo will fourth time :?  :).[/quote]

I thought you finished school and imo2005 was your last imo[/quote]
Why did you think that I finished school?
I havn't finished the school,but I think it doesn't need to be student of school to partisipate in IMO,you only mustn't have other educations,bisaid school education and you must not be older than 20 years!

[quote="alexei"][quote="pavel kozlov"][quote="Albanian Eagle"]slovenia is the fourth imo i'll partecipate
i wanted to know if there are other people who will participate in slovenia for their fourth time[/quote]
For all Russian IMO(and USSR)-history was only one man(Oleg Goldberg), who participted in 4 IMO(2000-2003)(+in IMO2004 in American team), since we have great concurention.[/quote]

Oleg Golberg was participated only in IMO2002 and 2003 (and 2004 in American team), but not in IMO 2000 and 2001. Nobody in the Russian (and USSR) history wasn't  participated in 4 IMO.[/quote]

Russia have very great students! It's bad one of Russia to go for 4 time in Imo!

Albanian Eagle - Ju jeni krenaria e kombit vella! Vazhdo!

[quote="alexei"][quote="pavel kozlov"][quote="Albanian Eagle"]slovenia is the fourth imo i'll partecipate
i wanted to know if there are other people who will participate in slovenia for their fourth time[/quote]
For all Russian IMO(and USSR)-history was only one man(Oleg Goldberg), who participted in 4 IMO(2000-2003)(+in IMO2004 in American team), since we have great concurention.[/quote]

Oleg Golberg was participated only in IMO2002 and 2003 (and 2004 in American team), but not in IMO 2000 and 2001. Nobody in the Russian (and USSR) history wasn't  participated in 4 IMO.[/quote]

yes you are right : http://imo.wolfram.com/contestants/teamlist.cgi?RUS

Davron

[quote="José"][quote="alexei"]Oleg Golberg was participated only in IMO2002 and 2003 (and 2004 in American team), but not in IMO 2000 and 2001. Nobody in the Russian (and USSR) history wasn't  participated in 4 IMO.[/quote]

Maybe that happened in strong countries. I mean, you won't see a student going to an IMO more than 2 times in China for example, they're all students of 11th or 12th grade[/quote]
Only one guy named Li Xin had chance,But his first time to IMO is in Taiwan,so......(In 1998,we also had the national team,although they didn't go to IMO)


rotation
group theory
abstract algebra
probability

I found a really fun and interesting problem - you may have heard of it. If you already know the solution, please don't say it!

You are given twelve coins, 11 of which are true, and 1 of which is a counterfeit. You are told that the counterfeit, or false, coin may either be heavier or lighter than the real coins. You are given a balance scale and are asked to determine, within three weighings, which of the twelve coins is the counterfeit and whether it is heavier or lighter than a true coin.

Have fun!  :)

Yes, this is a very well-known problem.  It doesn't really matter how many coins you have, as once you know the solution for 12 coins, you could easily find it for 20 coins or 100 coins.

If we know exactly the counterfeit is heavier or lighter than the real one, we can determine it within 3 weighing times with the total coins up to 81. 4 weighing times with 243 coins. n weighing times with 3^(n-1) coins.
I think you all know it but it's interesting, isn't it? :)

[quote="Nguyen Ngoc Linh"]we can determine it within 3 weighing times with the total coins up to 81. 4 weighing times with 243 coins.[/quote]

We don't know if false coin is lighter or heigher...

[hide]
1)split the weights into 4 groups of 3 weights each.
2)weigh and compare each group with another. (compare group one with group two, et cetera) repeat as nessecary
3)it may seem like step two will take forever, but it won't. you'll know which group will have the heavy weight when one of the groups is heavier than all three. after you have found the group with the heavy weight, take that group and divide it into 3 groups with one weight each. (sub-groups)
4)weigh and compare each group
5)as soon as one of the sides of the scale tips down, that side will contain the heavy one![/hide]

We don't know whether the coin is heavier or lighter, so that won't work.

Hint: What is the most number of groups to split 12 coins up such that we can determine which group the counterfeit is in one use of the balance?

Bigger hint:
[hide="DONT LOOK"]
What if two groups balance each other exactly?
[/hide]

Don't look unless you want the answer.

[hide]Split the group into 2 groups of 6 and mass them.  The heavier (lighter) one holds the coin.  Split that group into 3 and 3 and weigh them.  The heavier (lighter) group has the coin.  Then take two coins from the three and mass them.  The heavier (lighter) coin is the one you desire.  If they mass the same, then the coin you didn't mass is the one.[/hide]

Once again, we don't know whether the coin is heavier or lighter, so that might not work in three weighings.

Also, note that the technique where you take 2 coins out of 3 can be applied in other weighings as the 11 normal coins weigh the same.

EDIT i meant it as a general example
[hide="Is this right?"]
Let's call the three coins A, B, and C

(A and a) against (B and b).
(B and a) against (C and b).
There are two cases for the third weighing, according to the results of the first two weighings:

If both results are the same, then the false coin is in the minor group, and we already know from the first two weighings the result of a against b. Therefore our third weighing is to rotate the minor group (that is, balance b against c).

If the two results are different then the false coin is in the major group, and we know which A, B, C it is in, and whether it is heavy or light. The false subgroup contains three coins, so balance any two against each other. If they balance, the false coin is the third coin, and if they don't balance, the false coin is the heavy or light side depending on the fact if the fake coin is known to be heavy or light.[/hide]

[hide="Big hint"]
3 groups- two have 3 coins, third has 6 coins
Weigh first 2 groups and you get 6 coins in which is the counterfeit(or in the first 2 groups, or in the third).
Now clearer?
[/hide]

FishytheFish, like you said, it is possible you cannot figure out which coin is counterfeit, but only a 1/6 chance you will not be able to determine the counterfeit do to insufficient knowledge on whether the counterfeit is heavier or lighter.

[hide="First step"]
Divide the 12 coins into 4 groups, A, B, C, D. Let's say D has the counterfeit.
There are 4C3 or 6 different combos for weighing groups, A-B, A-C, A-D, B-C, B-D, C-D
Out of those 6, only 3 have the same weight, a probability of 1/2.

[hide="First step case one a.k.a. second step when first weighing is equal"]
If the two groups have the same weight, you will have to randomly choose one of the other groups.
Ex: first chose A-B, so you chose c or d to weigh against a or b, example: you chose c to weigh against a.
Like before, the chance that the [i]second[/i] weighing is equal has a probability of 1/2.
Therefore the probability that 1st [i]and[/i] second weighing are both equal is 1/2*1/2 =1/4

[hide="second step case one a.k.a. final step when both previous weighings are equal"]
The final step is to divide the group left over (definitely D) from the two previous weighings, which were both equal, so we know by process of elimination that the one group that wasn't weighed has the counterfeit, into 3 groups, 1 coin each. We'll name them a, b, and c. Let's say c is the counterfeit coin.
Choose two to weigh. There are a total of 3 choices, 2 of which leads to not knowing which one is the counterfeit (a-c, b-c), with a probability of 2/3. in total, probability of not knowing which coin is counterfeit is 1/2*1/2*2/3=1/6
[/hide]

[hide="second step case two, aka final step when all 3 weighhings are equal"]
like with second step case one, we split coins in group d (left over group) into 3 groups of 1 coin, a, b, c. However this time, you chose to balance a-b, which are equal, so you know that c is definitely the counterfeit.
[/hide]
[/hide]
[hide="first step case two, aka second step when first weighing is unequal"]
when first weighing is unequal, ex you weighed a-d, you know one of the two is the counterfeit group. So the second step is to pick another group (let's say, c) and the one that is lighter than the other one (let's say that a was lighter than d). If the two groups (in this case a-c) balanced, then you know that the counterfeit is heavier than the real one because the lighter group and a real group balanced. If d was lighter than a and you weighed d and c, you would see that d is lighter than both a and c, and therefore conclude that the counterfeit coin is lighter.
[hide="final step in the case that first measurement was unequal"]
split d into 3 groups. Because you know whether the counterfeit coin is lighter or heavier than the normal coin, you can now tell apart the counterfeit from the real in all final step cases (in this first step case).
[/hide]
[/hide]
[/hide]
whew! I hope that was understandable and helped!

yes i agree with a person. However, i think that a problem would tell you whether the counterfeit was heavier or lighter than a real.

But the problem is not to find the probability, it is to find the coin.

However, your method of using different cases is correct. Every time a weighing is done, there are three ways the scale can tip: left, right, or not at all (or 2 ways: it either tips or it doesn't). Whichever way you choose, try to come up with a next step for all the possible results.

i.e. If the first weighing is equal, then what will you do? If it tips, what will you do, etc.

@violin321: There are many problems where you are told whether the coin is heavier or lighter, but there are also many problems where it could be either. Being able to solve it all depends on how many coins and how many weighings there are.

Here is my solution

[hide="click to begin"]

1. split the coins (1, 2, 3, 4.....11, 12) into 3 groups, A, B, C.
2. Weigh two of them (we'll weigh A and B, made up of 1, 2, 3, 4 and 5, 6, 7, 8, respectively)

There are now two different scenarios that could happen, either
[hide="They balance (first step-scenario one)"]
If they balance, we know that the counterfeit is withing group C (9, 10, 11, 12).
We also know that 1, 2, 3, 4, 5, 6, 7, and 8 are all real.

Now, you choose 3 coins from group c (we'll choose 9, 10, 11) and balance with 3 coins from group a or b (we'll choose 1, 2, 3). Because we know 1, 2, 3 are real, we know that if 9, 10, 11 are heavier than 1, 2, 3, the counterfeit is heavier, or if lighter, counterfeit is lighter.
There are two scenarios here too:
[hide="1, 2, 3 and 9, 10 , 11 balance"]
Because we know that 1, 2, and 3 are real, if 9, 10, and 11 balance 1, 2, and 3, 9, 10, and 11 are real too. Therefore, by process of elimination, we know that 12 is the counterfeit.
In this scenario, we never need to know whether the counterfeit is heavier or lighter.
[/hide]

[hide="1, 2, 3 and 9, 10, 11 do NOT balance"]
If 1, 2, 3, and 9, 10, 11 do not balance, we know, since 1, 2, 3 have already been deemed real, that either 9, 10, or 11 is the counterfeit. Also, because we know 1, 2, 3 are real, we know that whatever relation 9,10, 11 have with 1, 2, 3(lighter or heavier), the counterfeit coin will have with the real coin (i.e. if 9, 10, 11 are heavier, the counterfeit is heavier).

Let's say 9, 10, 11 are heavier. Then you would take two of the coins ( we'll say 9 and 10) and put one on each end of the balance.

If 9 and 10 balance, then we know that they are both real, and by process of elimination, we know that 11 is the counterfeit.

if 9 and 10 do not balance, we can determine which is the coin because we've already determined whether the counterfeit is lighter or heavier than the real (see "1,2, 3 and 9, 10, 11 do NOT balance). Since in our situation, we've deemed 9, 10, 11 to be heavier than 1, 2,3, whichever is heavier of 9, and 10 is the counterfeit.

[/hide]
[/hide]
or
[hide="they dont balance (first step-scenario two)"]
If 1,2, 3, and 4 do not balance 5, 6, 7, 8, we know that the counterfeit coin is either 1, 2, 3, 4, 5, 6, 7 or 8, and therefore, 9, 10, 11, 12 are all real. Remember whether 1, 2, 3, 4 are heavier or 5, 6, 7, 8 are heavier.
In this step, you will take 3 coins from group A (1, 2, 3, 4, let's say we chose 2, 3, 4) and swap them with three coins from group B (5, 6, 7, 8, let's say we swapped 6, 7, and 8). Then replace 2, 3, 4, with three coins from group C (9, 10, 11, 12, let's say we swapped in 9, 10, and 11). Note the tipping of the balance.
Again, two scenarios:
[hide="1, 9, 10, 11 BALANCES 5, 2, 3, 4"]
If they balance, then we know that either 6, 7, or 8 is the counterfeit. Because you noticed which side was heavier in "first step scenario two", you know whether the counterfeit is heavier or lighter than the real.
Need an explanation? ok.
Because you now know that 6, 7, or 8 is the counterfeit, you know that whatever relationship between 5, 6, 7, 8 (one fake) and 1, 2, 3, 4 (all real), i.e. heavier or lighter, is the relationship between the real coin and fake coin.
Now you should pick two coins from 6, 7, and 8 and balance them (let's choose 6, 7)

then by process of elimination, you know that 8 is the counterfeit

then you should know which is the counterfeit because you know the relationship between the counterfeit and the real.

[/hide]
[hide="1, 9, 10, 11 do NOT balance 5, 2, 3, 4"]
you should have noted the tipping of the balance with 1, 2, 3, 4 and 5, 6, 7, 8.

then you know that the counterfeit is either 1 or 5. Take a real coin and balance it with 1 or 5.

 you know by process of elimination that the counterfeit is the one you didn't balance (if you balanced 1 with a real coin and they balanced, you know that 5 is the counterfeit0.

then you know the one that isn't the real coin is the fake coin (i.e if you balanced 1 and a real coin and they don't balance, you know 1 is the counterfeit

then you know that either 2, 3, or 4 is the counterfeit. you will know whether the counterfeit is heavier or lighter because the side with 2, 3, 4 will be heavier or lighter (if 1, 2, 3, 4, is heavier than 5, 6, 7, 8, the counterfeit is heavier). Now you choose 2 coins and place one on each side of the balance (i.e. 2 and 3)

then by process of elimination, you know that the one you didn't balance is the counterfeit.

then you should be able to tell which is the counterfeit because you've already determined whether the fake is heavier or lighter.

[/hide][/hide][/hide]
 i hope that was helpful

Oh, good job violin, that was a really helpful solution. I wonder if they have other problems harder than that one.

on the right track? im pretty sure he has the right solution!
good job

i had a slightly different method but the overall idea was the same
my second weighing for the 2nd scenario was 5 coins against 5

[quote="Wickedestjr"][quote="myyellowducky82"][quote="isabella2296"]I was looking through some interesting problems on the web, and I came across this and thought you guys might have some fun with it.[/quote]
Ooh! I just found the website so I'm not going to post the answer. =]
Violin's on the right track.[/quote]

What website?[/quote]
teh website with the problem.

[quote="myyellowducky82"][/quote][quote="Wickedestjr"][quote="myyellowducky82"][quote="isabella2296"]I was looking through some interesting problems on the web, and I came across this and thought you guys might have some fun with it.[/quote]
Ooh! I just found the website so I'm not going to post the answer. =]
Violin's on the right track.[/quote]

What website?[/quote][quote="myyellowducky82"]
teh website with the problem.[/quote]

Obviously, but seriously, what is the website?

I just realized that this problem is not solved just by using the knowledge of which coins we know to be real. We also have to keep in mind which coins could be heavy or light.

I won't say anymore just now... :wink:  :P  :)

geez fishy, my method (and stevenmeow's) deals with the problem of not knowing whether the counterfeit is lighter or heavier.

Fishy has a point though.
We sort of have to use trial and error to figure out the *right* way to solve it.

@a mod (preferably Izzy): Please delete the other post I made on this page [mon, sep. 15...2:52pm...after Izzy's post] Thanks.

Oh, I must have missed that post. But I've solved it (no, I will not post the whole thing), but you are on the right track.

And myyellowducky, it's not trial-and-error. It's using logic to figure out what weighing needs to be made in what situation.

[quote="isabella2296"]Just type "interesting math problem" in Google - it should come up.[/quote]

Thank you.


calculus
integration
probability
number theory
prime factorization

[quote="GameBot"]If one of the positive integral divisors of 720 is randomly selected, what is the probability it is divisible by 4? Express your answer as a common fraction.[/quote]

Prime factorization for $ 720$ is $ 2^43^25^1$ There are $ (4\plus{}1)(2\plus{}1)(1\plus{}1) \equal{} 30$ positive factors of 720
Divide the prime factorization by 4 to get $ 2^23^25^1$. Use the same #-of-factors trick to get $ (2\plus{}1)(2\plus{}1)(1\plus{}1) \equal{} 18$ factors that are multiples of 4.

$ \frac{18}{30} \equal{} \frac{3}{5}$