As you may know, the race is VERY close. Any of the 12 have an equal shot at being in the top 6. So before the TST, each of the TST takers randomly predicts who will make the top 6. What is the probability that one of them has their prediction exactly right? Assuming that no one is rigging the contest.

This seems too easy, but is it
[hide]1/84, using combinations
Or do you mean exactly one gets it, or one or more gets it?
EDIT: 1/7?

[/hide]

Uhh..I get there are 12C6=924 total

ONly 1 works so 1/924 (tahts regarless of how top 6 is in order, just as a group)

so atleast 1 of the 12 eppole get it right is 1-(923/924)^12

[hide]
I'm assuming none of the TST takers is pessimistic and gives himself dead last every time, and that all of the TST takers have a different prediction.
There are $\binom{12}{6}\cdot6!$ ways of choosing the top 6.  Out of those, only 12 are the predictions.  So the answer is $\frac{6!}{11!}$ which I'm not bothering to expand.
[/hide]

Hope I understood that right, to me it was kinda vague.

Eh, Im guessing
[hide]
$\frac{1}{77}$
[/hide]
Edit: Ok, I'll post how I got that answer.
[hide]
Since there are 12 contestants, there are $\binom{12}{6}$ ways of guessing the correct members.
Then, since there are 12 people who guess, the chances of having at least one right would be $\frac{12}{924}$ which simplifies to $\frac{1}{77}$
It's problably wrong though
[/hide]

Guys, please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=814211#814211]this[/url] post which you should've read when starting posting in this forum anyway. Your posts are not helpful at all, and 'guessing' at an answer is utterly pointless. However, giving an answer and [i]some[/i] reasoning would be nice. Thanks.

[hide="is this a solution or an answer?"]
There are 12 choices for 1st place, 11 for 2nd, and so on, for $12\times11\times10\times9\times8\times7$.  Possibilities.  If everyone guesses differently then it might be $\frac{6}{665280}=\frac{1}{110880}$.

yeah, that has a pretty good chance of being wrong.
[/hide]

question 1: does top 6 order matter?  Or is it just the top 6 as a group? (it looks like group from what im reading)
question 2: what do you mean one gets it right?  do you want exactly one to get it right? (it looks like only one from what im reading)

There are $\binom{12}{6}$ total ways you can pick 6.  Only 1 group will contain all top 6, so $\frac{1}{924}$.  If you want only 1 to be right, we have

${\frac{1}{924}\cdot{(\frac{923}{924}})^{11}}\cdot \binom{12}{1}$ as it can be any one of the 12 that gets it right.  Multiplying that out, the answer is (obviously with a caculator) $0.0128$ when rounded to nearest ten thousandth, if you want exact value go with the answer with all the fractions in it.

If you want [i]alteast one[/i] to be right, I think that answer will be $1-({\frac{923}{924}})^{12}$

@Art of Owna, you assumed correctly for those questions. Good Solutions! Please hide them now though (you can leave your questions unhidden)


integration
limit
calculus
calculus computations

Recently I came across this nice monotonic sequence of real numbers defined by an integral.

Let $a_{n}=\frac{1}{n!}\int_{0}^{n}e^{-x}x^{n}dx$.

Prove that:

a) $a_{n+1}-a_{n}\geq 0$, for all $n\in N$.

b) Find $\lim_{n\rightarrow \infty}a_{n}$.

The sequence increases, therefore it cannot converge to 0.

b) Continuing integration by parts we get
\[a_{n}=1-\sum_{k=0}^{n}\frac{n^{k}}{k!}e^{-n}\]
Well-known that the last sum tends to $\frac12$ when $n\to\infty$, hence $\lim_{n\to\infty}a_{n}=\frac12$


linear algebra
matrix
inequalities
function
linear algebra unsolved

Let $A,B\in\mathcal M_{n}(R)$ with the propriety $AB=BA$ and $^{t}A=A,\ ^{t}B=-B$. Show that $\det(A^2+a\cdot B^2)\geq 0$ for any real $a$.

$^{t}A$ is te transpose matrix

If $^{t}A=A$ then all the eigenvalues of $A$ are real numbers. If $^{t}B=-B$ all the eigenvalues of $B$ are of the form $iy$ where $i^2=-1$. (I can prove those statements). So $\det(A^2+\alpha B^2)=(\lambda_1^2-\alpha \beta_1^2)(\lambda_2^2-\alpha \beta_2^2)....(\lambda_n^2-\alpha \beta_n^2)$ where $\lambda_k,i\beta_k$ are the eigenvalues of $A$ and $B$.
If $n=2p$: knowing that the eigenvalues of $B$ are complex then they are conjugated so if $iy$ is an eigenvalue of $B$ then $-iy$ is also an eigenvalue so in the formula of $\det(A^2+\alpha B^2)$ a term of the form $(\lambda_k^2-\alpha \beta_k^2)$ has multiplicity of $2$ so the determinant is positive ($\det(A^2+\alpha B^2)=\prod (\lambda_k^2-\alpha \beta_k^2)^2\geq 0$).
if $n=2p+1$ then one of the eigenvalues of $B$ is $0$ so $\det(A^2+\alpha B^2)=\lambda_m^2 \prod (\lambda_k^2-\alpha \beta_k^2)^2\geq 0$ where $m$ is chosen such that the $m$-th eigenvalue of $B$ is the null one. Hope it`s ok.Nice problem.

for $a$ real  $det(A+iaB) det(A-iaB)\geq 0$ and $det(A+aB) det((A+aB)^t)\geq 0$

I know it`s true but how does it help?We have $a$ not $a^2$. :roll:

[quote="MoHaMeD"]Yeah i know ,the $a$ in my answer isn't the one of the question,the first inequality is for $a\geq 0$ and the second for $a$ negative ;)[/quote]

I saw a solution that analised the given inequality function of $a$ (poziteve, equal to $0$ or negative)...

(I) If $a=0$ then $\det(A^{2}+a\cdot B^{2})=\det(A^{2})=(\det A)^2\geq 0$.

(II) If $a<0$ then let $b\in\mathbb{R}^{*}$ such that $a=-b^2$.

Thus $\det(A^2+a\cdot B^2)=\det(A^2-b^2\cdot B^2)=\det(A+b\cdot B)\det(A-b\cdot B)$.

We know the statments $\boxed{\det(A)=\det(^{t}A)}$ and $\boxed{^{t}(A+b\cdot B)=^{t}A+b\cdot ^{t}B}$ so we use them... like this

$\det(A+b\cdot B)=\det{^{t}(A+b\cdot B)}=\det(^{t}A+b\cdot ^{t}B)$ but from the problem conditions we obtain $\det(^{t}A+b\cdot ^{t} B)=\det(A-b\cdot B)$.

Hence $\det(A+b\cdot B)=\det(A-b\cdot B)$ and $\det(A^2+b^2\cdot B^2)=[\det(A+b\cdot B)]^{2}\geq 0$.

(III) If $a>0$ we take $a=c^2$ with $c\in\mathbb{R}^{*}$ and ... i'm thinking  :blush:

[quote="Slizzel"]
$\det(A^2+b^2\cdot B^2)=\det(A+b\cdot B)\det(A-b\cdot B)$.
[/quote]

hmmm.... $\det(A+b\cdot B)\det(A-b\cdot B)=\det(A^2-b^2\cdot B^2)$ why is $\det(A^2-b^2\cdot B^2)=\det(A^2+b^2\cdot B^2)$?

I don't get it... that's true I guess but what's its connection with the quoted text  :huh: ?

EDIT: I modified... for the last part I'm still thinking  :roll:

[quote="Slizzel"](III) If $a>0$ we take $a=c^2$ with $c\in\mathbb{R}^{*}$ [/quote]

A,B commute then
$det(A^2+c^2B^2)=det(A+icB)det(A-icB)=|det(A+icB)|^2 \geq 0$


I GREET ALL THE WORLD
I AM NEW HERE
I WANT IF  YOU CAN SAY TO ME
THE ANSWER IN THE FOLLOWING EXERCISE
APOLOGY IF I HAVE MADE SOME ERROR
IN MY ENGLISH!

A. In a test of qualitative control, a router of network (router),
undertakes not distinguished network "parcels" to four different channels.
With how many ways it can become a test routing of 300 parcels in one
second if the first channel has "inexperienced" capacity,
second and third 50 parcels/second while fourth 40 parcels/second
should be dispatched ten at least parcels in each channel?
(write the generator interelation and solved it so that is found
the factor of term that gives the right answer).

B. In a half test, the router blames moreover different (distinguished)
heading in each one from the parcels front him it undertakes. ¶With how
many ways it can become the routing of 60 such parcels in one second if
we doesn't interest the line with which they will pass from the each
channel? (write the generator interelation and determine
the term of the factor that gives the right answer.(Is not required
the calculation of factor).With how much ways it can become the routing
of 60 such parcels in one second if we are interest for the
line with which they will pass from the each channel?
(write the generator interelation and determine the term
of the factor tha gives the right answer.(Is not required the calculation of factor).

thank you!


The standard two-column proof taught in many geometry classes in schools is, in my opinion, far less clear than the paragraph proofs found in, for instance, AoPS books. Just want to see other people's opinions.

Possibly the only good thing about two-column proofs(that I can think of, in any case) is that, especially for non-gifted students just starting proofs, it can help keep your information organized.

The bad thing about two-column proofs is that sometimes something you write doesn't really have a postulate or theorem to back it up that you can think of, yet it's definitely true, so you end up wasting 20 minutes or more digging through your geometry textbook looking for the name of that postulate or theorem, and not always with success. Another thing is that it really hinders your creativity. As an experiment, I took a proof question I found on the forums somewhere(this was a while ago - I don't remember the problem), and tried to do it first like a two-column proof, and later like a paragraph proof. I had some ideas I wanted to explore to see if it would help with the proof, but I ended up spending a lot more time trying to find the name of the postulate/theorem of some step, and in the end forgot the ideas I had. A few days later I returned to it, and instead used the more natural paragraph form of a proof, and I was able to prove it(hopefully rigorously.)

I'm pretty interested in what a teacher has to say to say on this subject - both an "ordinary" school teacher and one that teaches like AoPS books do.

I know in my Honors Geometry Class, our teacher requires us to do two-column proofs.  I'd much rather do a paragraph proof though.

Two-column proofs give students the ability to organize the proof, as well as back up their assumptions. It's very nice to introduce the concept of proving mathematical ideas to students who may not be as indepth in math as they could be.

Paragraph proofs are easier to do, but harder to understand. Complicated proves that could wind up pages and pages long are often done in paragraph proofs, but these lack the organizations that two-columns provide.

A version of the paragraph proof is the proof-without-words format. Usually, these are algebraic calculations that prove that one thing leads to another (most common are verifying trigonometric identities) without having to use any words. These are easier to format, easier to read, and even more organized that two-column proofs. Unfortunately, errors are harder spot in the proof-without-words format because no reasoning is given.

[quote="isabella2296"]The bad thing about two-column proofs is that sometimes something you write doesn't really have a postulate or theorem to back it up that you can think of, yet it's definitely true, so you end up wasting 20 minutes or more digging through your geometry textbook looking for the name of that postulate or theorem, and not always with success.[/quote]

That's in the nature of the beast, though; the reason geometry has been taught from the ancients until now is to teach critical thinking, not necessarily to memorize a body of facts about various figures. To that end, if you cannot tie a given statement to the axiom set, you cannot prove it, period.

Seeming to be true is not sufficient.

No, what I meant was that you can explain it in a paragraph proof, but not so much in a two-column proof.

I wonder if you can include "lemmas" to prove theorems without names (or, in some books, numbers). :maybe:

I agree that paragraph proofs are easier to do, but here's my two cents:

Paragraph proofs seem to be more about logic and reasoning than two-column proofs, while two-column proofs have the property that any incorrect assumption is quickly revealed. This is, perhaps, one reason that the two-column proof is so emphasized in schools; they want to ensure that budding geometers do not develop a habit of making leaps in reasoning that result from unprovable assumptions.

Ultimately, though, paragraph proofs seem more suitable for hard problems that involve a lot of reasoning. Two-column proofs are sometimes too restrictive for this.

How much does the added rigor in two-column proofs hamper problem solving? In addition to hunting and pecking for theorem names, are there any problems that are simply unprovable via a two-column approach? (for instance, a geometry problem calling on a lot of logic and reasoning?)

Roughly speaking, Mathematics [i][u][b][size=150][color=red]is[/color][/size][/b][/u][/i] two-column. If any of you have a chance of studying the Foundations of Math, two-column and paragraph forms will be clear to you. The "logic" you talk about is different from the mathematical logic employed in proofs, which is purely logical deduction, in a philosophical sense. So, rather than talking about the rigor and logic behind the forms of proof as some of you mentioned about 'logic and reasoning', our argument here should be directed towards proof preference and convenience.

[b]Advantages:[/b]
- Simple, consistent structure
- Forces the writer to justify every step [i](rule, application, conclusion is the structure of any argument)[/i]
- Forces the writer to work towards the conclusion, instead of working backwards
- Forces the writer to work in a linear fashion

[b]Disadvantages:[/b]
- Clutter (explicit justification for trivial steps)
- Little flexibility to place emphasis on/add discussion to important steps
- Difficult to make an aside to prove a lemma

In proving theorems in basic geometry, you probably won't need to add too much discussion to any step or make an aside to prove a lemma, so there are no disadvantages to the two-column format in that context. On the other hand, the advantages are particularly significant in the context of introductory geometry, since this is usually the class in which students are introduced to proof-writing.

What is a bit silly is that Geometry textbooks will all have their own naming scheme/numbering scheme for theorems, and teachers will sometimes require students to cite theorems by these names rather than by stating the theorem. This is not a shortcoming of the two-column proof format. In any case, this is not necessarily a bad thing from a teaching point of view, because it emphasizes the rigor behind the concept of justification -- students cannot just write down a fact that they intuitively think is true.


summer program
PROMYS
geometry
symmetry
quadratics
Harvard
college

My experience at Promys for the past three weeks has not been good. I really don't like to flame anything, so I will try to be as objective as possible.

Pros:
(*) you can go wherever you want in Boston until 11 PM
(*) you can go to sleep really late, if you want to
(*) most of the first years are quite nice
(*) the counselors are helpful with respect to number theory
(*) the teachers are really smart
(*) we have advanced seminars (this year Galois Theory, Geometry & Symmetry, Combinatorics) that are challenging and interesting
(*) most of the first years and returning students like this place

Cons:
(*) no air conditioning
(*) most of the counselors and returning students are quite arrogant
(*) the teachers strongly dissuade first years from going to the advanced seminars
(*) the number theory homework takes a long time, is much easier than Olympiad problems, and is tedious
(*) occasionally there is a serious lack of honor code
(*) the counselors do not answer any of my non-NT math questions
(*) the returning students and counselors have a lot of math secrets (like the law of quadratic reciprocity!) that they do not allow anyone to discuss in public
(*) there is a strong class structure against first years

hm, from the sound of it, I'm not that mad now that I couldn't go to PROMYS :maybe:

If you want to apply to PROMYS in the future (and you have not been there before), I strongly advise you to think [u]carefully[/u] about it because a lot of things happen there that they do not advertise. If you know little number theory and you really want to learn it, then I encourage you to apply. Otherwise, I would definitely advise against it; the attitudes of many of the counselors might shock you, and it is already too late for me.  :wallbash_red:

Okay Phil, just because you happen to know "everything" about number theory and you got a 221/366 on the midterm, doesn't mean that it's a bad program for others. If you already knew so much number theory, why did you even come to Promys. Didn't you read the description of the program "before" you decided to come?

On the other hand, I think the number theory is quite tough, but i'm pretty stupid anyways... so, i guess it is okay though.

Here's a run through:

All the counselors are really smart, and most are math majors, and go to schools like Harvard, Yale, MIT, Caltech, etc. plus, there is a ratio of 4 students per counselor.

You get to experience the "joys" of mathematical rigor, and write completely rigorous proofs.

Mostly though, you get problem sets and its supposed to be "discovery" learning, which is that you are supposed to solve the problems before learning about it in class, which can be a pain, but is quite interesting.

I disagree that the program is against first years, because the biggest part of the program ARE the first years.

Finally though, you get to hang out with some really smart people who actually enjoy math. I didn't even know that aops existed before promys.

Oh and also, first years are only allowed to suck at mafia, 'nuff said.

To Phil:

Shut up first year!  Jk lol.

Your story of the talent show...priceless.  Surely you weren't up there Mr. Leszczynski?  :D

A lot of things come up that we may not've realized by the time we arrive at our respective summer programs.  Personally, I have been less than satisfied the past couple of weeks at my program, and it's definitely refreshing to get some new perspectives - no matter how outrageous they may seem!  I'm tired of biased opinions that all follow the same pattern:  "Wow this is awesome!"...

Thanks for posting this Phil.  While I don't know much about PROMYS, hopefully this'll help others make more informed decisions.

[quote="Philip_Leszczynski"]
Cons:
(*) no air conditioning[/quote]
Annoying, but not a fatal flaw.
[quote="Philip_Leszczynski"]
(*) most of the counselors and returning students are quite arrogant[/quote]
How?
[quote="Philip_Leszczynski"]
(*) the teachers strongly dissuade first years from going to the advanced seminars[/quote]
Ok, this is just a 'What the h*ll?' bullet. You go to math camp, presumably to learn math, and yet you dissuaded from doing so?
[quote="Philip_Leszczynski"]
(*) the number theory homework takes a long time, is much easier than Olympiad problems, and is tedious[/quote]
Tedious in what way? How much easier?
[quote="Philip_Leszczynski"]
(*) occasionally there is a serious lack of honor code[/quote]
How?
[quote="Philip_Leszczynski"]
(*) the counselors do not answer any of my non-NT math questions[/quote]
Wow. Are you asking too many non-NT questions, or are they lazy/mean?
[quote="Philip_Leszczynski"]
(*) the returning students and counselors have a lot of math secrets (like the law of quadratic reciprocity!) that they do not allow anyone to discuss in public[/quote]
Math Gestapo!
[quote="Philip_Leszczynski"]
(*) there is a strong class structure against first years[/quote]
How so?

The counselors and teachers want first-year students to focus on Number Theory, so they strongly discouraged us from taking any of the advanced seminars since those would distract from our Number Theory problem sets.

I'm enjoying Galois Theory right now though :D

Phil, I wouldn't call the counselors and returning students arrogant.  There's just a very specific social structure here.  If you take it all in good fun then it's enjoyable, but I suppose it's not for everybody.  That part of your criticism is valid.  But it's not so much a "class structure" against first-years here in particular as it is just a very general veterans vs. newcomers thing that happens all the time everywhere.  You might not be used to being treated this way (at PROMYS, seniority and age are correlated less well than in other similar social structures), but that's the way it is.

I will also agree that the Number Theory sets are tedious.  :|

The imposed self-discovery here I find more a good idea than bad.  It should be a lot of fun when we get to figure out that really nice proof for Quadratic Reciprocity all by ourselves, right?  ;)

Edit:  And what exactly do you mean by the serious lack of honor code?

Phil, I should tell you that a lot of people are angry about your posts (several counselors...and they DO know who posted them, by your screen name.  I recommend you watch out for Lucas...)

The reason the second years keep secrets is that the entire focus is to let you prove it on your own.  That's why they reveal as little as possible.  You won't have someone to give you a solution or even hints when you're exploring a new branch of math as a research professor.

The reason some counselors (mine didn't, really) try to prevent advanced seminars is that they want you to focus on your number theory.  If you try to juggle three classes, you won't have enough time to simply sit and think, conjecture, discuss with other people, and prove things.  Not to mention you won't have time to try numerical examples.  :lol:
I mean, imagine trying to do 15 Olympiad problems in one day.  You can't just give up on an olympiad problem after a half-hearted effort of 15 minutes...
If you found the problem sets too easy, you should have entered a lab group.  That would be good and solve your problem.  Also try to generalize and prove your results.  I mean, have you rigorously proven every proof thus far, and solved every problem?

For other people?  just know that at PROMYS, most of what you learn is outside the classroom.  You have one hour of number theory lecture in the morning (which isn't exactly a lecture, but the first years figure it out themselves), and through the day, you have freedom, but you learn mostly from the problem sets and from experimentation.  The idea is to think deply about a single topic, and not only figure out things yourself, butdevelop a new way of thinking.  You see what thinking went into your proofs, and it teaches you a lot about rigor.  Unlike some other camps, where you're taught the things, you're basically given a prod into the unknown, and you have to navigate yourself.  I strongly recommend PROMYS.
You also do a lot of other things...frisbee etc.

PS: If you don't know who t0rajir0u, Grandiloquence, and I are, try to figure it out  :D

Phil, I strongly disagree with your  depiction of PROMYS. (Full disclosure, I have been involved with PROMYS for many years). I entered PROMYS with a number theory background which I suspect was about your level or higher and found PROMYS to be one of my best highschool experiences.

To respond to your individual points:

[no air conditioning]

Even this minor detail is innacurate. While the BU dorms are not air conditioned, almost all the classrooms are air conditioned as are the major study and social spaces. I'm sorry, but if you can't get this sort of minor detail correct, I have trouble seeing why we should take your other comments seriously. Nevertheless, I will endeavor to respond to them.

[most of the counselors and returning students are quite arrogant]

This is not true in my experience. In fact, in my experience the only generally arrogant people were first years who thought they knew much more than they did. I don't mean to make an ad hominem attack, but is it possible that you are confusing arrogance with actually knowing something you don't know?

[the teachers strongly dissuade first years from going to the advanced seminars]
Other people have already responded to this point, but I'll note that in my experience they only discourage first years from attending when in the opinion of the counselors a) the first year has shown a lack of attention or success at the number theory and/or b) the teacher believes that the student will not is not gaining significantly from the advanced seminar. My first year I sat in on a seminar on modular forms. In any event, even if they discourage a student from attending, like all events at PROMYS, the ultimate decision is left to the student.

[the number theory homework takes a long time, is much easier than Olympiad problems, and is tedious]

I would disagree with this statement, and suggest that you wait until the end of the program at least before making this complaint.

[occasionally there is a serious lack of honor code]
This is possibly the most serious complaint (I would have ranked it well before a complaint about air conditioning, but that's just me). There seem to be every year a small segment of students who either share answers indiscriminately and/or use the internet and books to find solutions to problems. This makes the experience useless for other students, and can be demoralizing. In general, such practices are strongly discouraged by the counselors. However, it is by nature difficult to distinguish collaboration and and simply sharing answers, and this is a genuine problem.

[the counselors do not answer any of my non-NT math
questions ]

I have to say that this claim is surprising to me. In my experience, counselors were often willing to discuss almost any sort of mathematics, as long as one's number theory sets were done well and turned in for the day. Furthermore, I did some of my first actual research (that is, research which lead to published papers) as a first year in my free time. In that regard, counselors were incredibly helpful and informative, full of suggesttions and ideas, books to read and sometimes even some very obscure sources which they knew about at the tips of their fingers.

[the returning students and counselors have a lot of math secrets (like the law of quadratic reciprocity!) that they do not allow anyone to discuss in public ]

Again, this point has already been addressed by others, but it bears repeating: PROMYS is supposed to be about self-discovery and learning how to discover. They want students to discover major results for themselves, and discussing those results around people who haven't discovered them yet is a sure way to ruin that. As long as no first years are around, I'm sure that counselors would be more than willing to discuss some of the more secret topics.

[there is a strong class structure against first years]

I have trouble understanding this claim. Is this a claim of class structure socially or what? The majority of the students are first years, in fact many of the returning students spend a lot of time helping first years, and all the minicourses (the small, generally 1 to 3 lecture sets given by counselors) are geared to be understandable by first years, even though much more could be covered and accomplished if some were done of a more advanced nature. So learning focuses on the first years. There is also no social class structure, even very little between counselors and students. At meals first years, second years and students all sit together. This is not be design or by force, people happen to sit together. Similarly, at more structured social events all the students mingle freely. When students play Mafia or Illuminati or other games on weekends, they play together without regard for what years people are (with the exception of midterm writeup weekend, where first years need do their midterm writeups, but other than this single case the statement holds).

In summary, I have to find your comments at best innaccurate or misleading. The inability to even give accurate information about even such minor details as the state of the air conditioning forces me to wonder if your complaints reflect more on you than on PROMYS.

- Vincentius

Really, if you've been struggling with a problem for a few hours and
you have no knew ideas, this is the correct time to get the answer. Discovery learning is great, but we all aren't Gauss. Enforcing some kind of math gestapo is BS.

However, the original poster needs to expand on some of his points. It is quite possible he is having a different experience from most you. On the other hand, it might just be a frusteration fueled rant. Could the OP clarify his points?

Vicentius, no insulting please.

[quote="PenguinIntegral"]Really, if you've been struggling with a problem for a few hours and
you have no knew ideas, this is the correct time to get the answer. Discovery learning is great, but we all aren't Gauss. Enforcing some kind of math gestapo is BS.
[/quote]

Professional mathematicians who work on the real difficult problems in mathematics of today don't give up after a few hours.  They don't even give up for many years.  Now is the time to [b]learn[/b] how to work at a problem for a while and get nowhere with it, the time to learn how to solve problems which take a lot of time and thought.  You don't have to be Gauss to do that: you have to be persistent and diligent, willing to experiment and conjecture, share ideas, try all sorts of approaches, and then find yourself discovering many more questions in the same general area.  I'll give you an example.  For my Exploration lab, we got the Sums of Squares problem.  The first general thing to think about was "When can a number be expressed as a sum of two perfect squares?"

Our group thought a lot about this.  I experimented, and found patterns when one multiplied such numbers by certain primes.  We first conjectured that any product of two such numbers also had the property, and that was easy to prove.  After some experimentation, we conjectured that [hide]each prime 1mod4 can be expressed as a sum of squares in exactly one way[/hide]  But as we delved deeper into this question, we found many other things which we proved, and found these things linked very deeply to our number theory.  We also found other, more complicated problems, which could be reduced to this problem.  I managed to prove our conjectures, finally.

The goal isn't just to figure out the answers to these questions and problems we encounter, but [b]how[/b] to look at such problems.  It changes the way you look at such problems, and you find lots of connections at a fundamental level when you actually understand these things fully (and the thinking which goes into proving such things).

You have to be not allowed a solution booklet.  Then only will you think deeply about these problems, and not give up after a couple hours.  You claim we aren't all Gauss, but did Gauss give up so easily?

I'm sorry if my posts sounded a bit too dramatic; perhaps I was wrong about many aspects of the program. The course definitely challenges you with many many problems, there is a large amount of rigor involved, and the amount of number theory covered is quite extensive. And first years are free to go to whatever advanced courses they choose. So I apologize for my initial criticisms.

> My experience at Promys for the past three weeks has not
> been good. I really don't like to flame anything, so I will try
> to be as objective as possible.

Your incendiary remarks reek of "flaming"

Pros:
(*) you can go wherever you want in Boston until 11 PM
> The fact that you know this means you've been gallivanting Boston rather than doing the psets.  good job.

(*) you can go to sleep really late, if you want to
> The fact that you are sleeping really late is evidence that the psets are difficult.

(*) most of the first years are quite nice
> suggesting that the first years are quite nice insinuates there is some "teamwork" at play (referenced later on)

(*) the counselors are helpful with respect to number theory
> The only way you could know this is if you turned to counselors for help on number theory, meaning you needed help with number theory (this point is revisited later)

(*) the teachers are really smart
> You wouldn't think that if they didn't know much more about math than you.

(*) we have advanced seminars (this year Galois Theory, Geometry & Symmetry, Combinatorics) that are challenging and interesting
> Again, this means you don't know everything (point revisited later)

(*) most of the first years and returning students like this place
> again, your dissention is indicative of an exception to the rule

Cons:
(*) no air conditioning
> they tell you upfront that there is no air conditioning.  But of course, you have to read the papers they send you.

(*) most of the counselors and returning students are quite arrogant
> If you think everyone is arrogant, more likely than not you are the arrogant one

(*) the teachers strongly dissuade first years from going to the advanced seminars
> teachers wouldn't dissuade if you knew your number theory well enough to justify going to the advanced seminars

(*) the number theory homework takes a long time, is much easier than Olympiad problems, and is tedious
> olympiad problems are "harder" because they don't give any contextual clues.  For example, one olympiad program boiled down to the observation that sin x <= 1 (very hard, i understand)  On the other hand, in promys, the ideas are meant to build up, so there are 20+ intermediate questions which help you build up to the salient point of the problem set.

In simple terms, its easier to add a roof to a house than to build the house from scratch.  of course, if you paid attention when doing the problem set, youd notice this

(*) occasionally there is a serious lack of honor code
> the only way you could know of any lack of honor code is if you were part of the debauchery

(*) the counselors do not answer any of my non-NT math questions
> the counselors want you to grok the number theory before dealing with other math

(*) the returning students and counselors have a lot of math secrets (like the law of quadratic reciprocity!) that they do not allow anyone to discuss in public
> The reason why they don't tell you is because they want you to conjecture it.  this is tantamount to you pissing and moaning because the kid next to you in the theater won't tell you how the movie ends before the movie ends

(*) there is a strong class structure against first years
> conspiracy theory?  most first years probably wouldn't like promys, and there probably wouldn't be many returning students.

The salient point of this discussion is that Philip Leszczynski is the exception to the general rule: PROMYS is an amazing place and you will learn a lot if you pull your head out of your arse.

This calls for my expertise. I think God's answers to Phil's points raise some interesting points about God himself.

[quote]> My experience at Promys for the past three weeks has not
> been good. I really don't like to flame anything, so I will try
> to be as objective as possible.

Your incendiary remarks reek of "flaming"

Pros:
(*) you can go wherever you want in Boston until 11 PM
> The fact that you know this means you've been gallivanting Boston rather than doing the psets.  good job.[/quote]

This clearly means you are merely a false god.

[quote](*) you can go to sleep really late, if you want to
> The fact that you are sleeping really late is evidence that the psets are difficult.  [/quote]

You obviously have difficulty accepting the fact that some people have more money than you.

[quote](*) most of the first years are quite nice
> suggesting that the first years are quite nice insinuates there is some "teamwork" at play (referenced later on)[/quote]

This is evidence of rampant promiscuity at this program.

[quote](*) the counselors are helpful with respect to number theory
> The only way you could know this is if you turned to counselors for help on number theory, meaning you needed help with number theory (this point is revisited later)[/quote]

You eat too much fresh fruit.

[quote](*) the teachers are really smart
> You wouldn't think that if they didn't know much more about math than you.[/quote]

You got drunk last night and woke up with your head in a toilet.

[quote](*) we have advanced seminars (this year Galois Theory, Geometry & Symmetry, Combinatorics) that are challenging and interesting
> Again, this means you don't know everything (point revisited later)[/quote]

You stole a rocket pack and accidentally blasted yourself to the moon. The point you just made emphasizes that you have very limited access to oxygen there.

[quote](*) most of the first years and returning students like this place
> again, your dissention is indicative of an exception to the rule[/quote]

I sense a possible gang structure here. Perhaps PROMYS is to blame for the rampant gang terrorism across that part of the country?

[quote]Cons:
(*) no air conditioning
> they tell you upfront that there is no air conditioning.  But of course, you have to read the papers they send you.[/quote]

You obviously lost your reading glasses, possibly while your head was in the toilet.

[quote](*) most of the counselors and returning students are quite arrogant
> If you think everyone is arrogant, more likely than not you are the arrogant one[/quote]

Yes

[quote](*) the teachers strongly dissuade first years from going to the advanced seminars
> teachers wouldn't dissuade if you knew your number theory well enough to justify going to the advanced seminars[/quote]

Well, let's just say that we have witnesses who saw you carrying a baby into a forest a few weeks ago.

[quote](*) the number theory homework takes a long time, is much easier than Olympiad problems, and is tedious
> olympiad problems are "harder" because they don't give any contextual clues.  For example, one olympiad program boiled down to the observation that sin x <= 1 (very hard, i understand)  On the other hand, in promys, the ideas are meant to build up, so there are 20+ intermediate questions which help you build up to the salient point of the problem set.

In simple terms, its easier to add a roof to a house than to build the house from scratch.  of course, if you paid attention when doing the problem set, youd notice this[/quote]

Olympiad problems often give contextual clues; in fact, experienced solvers can usually look at easy (even some medium) problems and immediately tell you the right method to use. I must also confess that I fail to understand how sin x is always less than or equal to one. Perhaps you are not familiar with complex numbers (very hard, i understand). Finally, in addition to eating too much fresh fruit, you also probably have yearnings to be a lumberjack.

[quote](*) occasionally there is a serious lack of honor code
> the only way you could know of any lack of honor code is if you were part of the debauchery[/quote]

The only way you would know a word like "debauchery" is if you were part of the debauchery.

[quote](*) the counselors do not answer any of my non-NT math questions
> the counselors want you to grok the number theory before dealing with other math[/quote]

"Grok" is not a word.

[quote](*) the returning students and counselors have a lot of math secrets (like the law of quadratic reciprocity!) that they do not allow anyone to discuss in public
> The reason why they don't tell you is because they want you to conjecture it.  this is tantamount to you pissing and moaning because the kid next to you in the theater won't tell you how the movie ends before the movie ends[/quote]

The law of quadratic reciprocity is pretty cool.

[quote](*) there is a strong class structure against first years
> conspiracy theory?  most first years probably wouldn't like promys, and there probably wouldn't be many returning students.  [/quote]

In fact, it is strongly believed that the returning students at PROMYS were directly responsible for the Kennedy assassinations.

[quote="PenguinIntegral"]Training teenagers to become researchers is absurd. Higher mathematics and number theory can be learned quickly, creativity and unusual thinking methods are much more important.[/quote]

This is a little mixed up.  From what I've gathered while here, the ultimate goal of PROMYS is precisely to inspire creativity and unusual thinking in the face of problems where the solutions aren't given.  Number theory is simply a very, very good medium for allowing that creative thinking to work in a rigorous setting.  It trains good habits of mind.

The difference between preparing for an Olympiad and the experience at PROMYS is this:  when answering an Olympiad question, you can rely on all of your knowledge of well-known theorems to help you answer your question.  When writing a PROMYS proof, you can rely only on axioms, and on theorems proven from the axioms.  It forces you to develop a deeper understanding of why all those theorems are true in the first place!

For example, take the Totient theorem.  Before I came to PROMYS, I used it to aid me in my proofs on AoPS all the time, but I didn't have the slightest idea how to prove it.  Halfway into the program, the proof became completely obvious and practically a necessary consequence of group axioms.

[quote="K81o7"][quote="PenguinIntegral"]

Olympian. IMO problems are way cooler than research problems. While they aren't mutally exclusive, having a firm foundation is very important. Training teenagers to become researchers is absurd. Higher mathematics and number theory can be learned quickly, creativity and unusual thinking methods are much more important.
[/quote]

It isn't training to become researchers.  What we do IS thinking creatively in unusual ways.  Not like the thinking we do can't help with olympiad problems.

But I ask just out of curiousity: what's the point of just training for Olympiads?

Perhaps you misunderstood what I was saying.  T0rajir0u clears it up well.[/quote]
I'm sorry, I misunderstood. Training is not a bad thing, but from what the OP says, PROMYs doesn't do a good job of it. I'd just like to see if the accusations have some merit or not...

On the topic of training for olympiads, I'd say mainly fun and betterment of myself are my motivations. Why is that relevant?

The philosophy here, from what I've gathered, is that Olympiads are only one of the many diverse aspects of enjoying math at a high level, and that the things we do here allow us to better understand and enjoy math beyond high school Olympiads, through to college, and even after that.  :)

To say that PROMYS doesn't do as good a job of training for Olympiad problems as one would like is just to say that there are things more important than Olympiads to consider in the world of mathematics.

Wow, after reading this thread it's hard to believe that everyone is talking about the same PROMYS that I went to two years ago... I went there with very little background in Number Theory; the program made me work and think day after day to come up with new ideas to prove these problems they gave me. I learned a lot about, well, learning. PROMYS is a great way to get a sense of "real" mathematics; olympiad problems are almost like manufactured results. The math at PROMYS is built up step-by-step.

And the biggest pro I see missing about PROMYS is that it's FUN. Whether it's staying up all night playing mafia or any of the other billion games that people bring, you can have lots of fun with people that are smart. If you're trying to get hardcore olympiad training, PROMYS is not for you. But if you love math, there's no doubt that you'll get at least a good experience out of PROMYS.

[quote="t0rajir0u"]:o

PROMYS mafia wasn't fun at all.  Everyone takes it too seriously  :P[/quote]

That's what makes it fun :wink:. That way people actually think instead of just randomly accusing; it makes mafia a psychological game.

It DID take too long...even with fewer people, it's a psychological game, and although you can't blend in as well with too few people, nobody agrees with too many, and we end up lynching a random person anyways to keep the game moving.  The first night of mafia was about 3 hours to a game.  That's why most of us resorted to other games such as grape, illuminati, set, card games, etc...

Btw, this is off topic, but  :o  happy birthday Qiaochu!
heh yeah this makes me look like a stalker but heck they gave that information on the contact sheet so what can you expect with 3 weeks left until school

man you people would not like hardcore mop mafia

dude good job on increasing the floor function of your age

[quote="mathfanatic"][quote="corinao"]Yeah, but that way it takes too long.  isn't it easier to shout out a random name and lynch that person  :wink:  ?  hehe[/quote]

Only at RSI, Corina.  (Hey, you post on AoPS?!)[/quote]

Yeah. Gosh, you don't know all the AoPS 2006 Rickoids yet?

[quote="K81o7"]I  The first night of mafia was about 3 hours to a game.  That's why most of us resorted to other games such as grape, illuminati, set, card games, etc...
[/quote]

The Illuminati games took all night on 2 nights. The mafia games never did.

The Illuminati games never took two nights, at least in my experience.  And if the people actually were playing and not just walking around, doing random stuff, the games of illuminati went pretty quickly.  At least in the Illuminati games, we were doing something, and actually playing, whereas in mafia, we were mostly sitting there.  When, in games of mafia, things actually happened, it WAS fun...there were so many people we didn't know where to turn...


abstract algebra
calculus
integration
group theory
category theory
Ring Theory
superior algebra

Let $ k$ be a field and $ G$ a finite group. Let $ U$, $ V$ and $ W$ be three finite-dimensional representations of $ G$ over $ k$ such that $ U\oplus W\cong V\oplus W$ as representations. Does this yield $ U\cong V$ as representations?

If not, does one still say that $ U\equal{}V$ in the representation ring of $ G$ over $ k$ ? Does one speak of a "representation ring" at all, in this case?

This is well-known in the case when $ \left|G\right|$ is not divisible by $ \mathrm{char}k$, but what if it is? I don't really believe it holds then, but I can't find a counterexample either. Thanks a lot (I need this for an introduction into lambda-rings).

  darij

Everything is fine over a field.  U,V,W all have composition series so the Krull-Schmidt-Azumaya theorem holds, and the indecomposable direct summands of U+W, U, V, and W are uniquely determined up to isomorphism.

This is the main reason for looking at the representation ring for finite dimensional algebras over fields and for artinian rings.

It even holds when k is a complete discrete valuation ring, A is a finitely generated k-module, A is a k-algebra, and then every finitely generated A-module is expressible as a direct sum of indecomposable A-modules, and the isomorphism types and multiplicities are uniquely determined.  In particular cancellation holds.

This is the normal setup for modular representation theory: k is the ring of integers in a finite extension field of the p-adic numbers containing the |G|th roots of unity, and A=kG is the integral group ring.  - (x)_k Frac(k) and - (x)_k k/P convert between modular and ordinary representations, where Frac(k) is the field of fractions of k and P is the maximal ideal of k.

Nagao-Tsushima describes this setup on page 230 of their Representations of Finite Groups.  It is also described in Curtis-Reiner, Chapter XII, though they boldy do not require a splitting field.

Thank you a lot! So proving unique decomposition using Schur's lemma was a stupid idea after all.

  darij

No problem.  You are right that dealing with simple modules makes things much, much easier.

I think for you lambda ring thing, probably any "Hopf algebra" is fine (finite dimensional over a field).  It should have a representation ring where it makes sense to want to decompose exterior products.  The representation ring itself is studied for artinian algebras.  Benson's book "Modular Representation Theory" has some nice bridging material from the group theory world to the Artinian algebra world.  Gordon James (I think) has a book on the representation theory of Sn or GL (with the idea that they are the same), and he has some fairly fancy things to say about representation rings.

The unique decomposition for a semisimple ring is outstandingly nice:  The additive module of the representation ring is finite rank free, with basis the simple modules.  Expressing a module in that basis is just finding a composition series, which is often quite possible to do computationally.

The unique decomposition for a general artinian ring might be ridiculously uselessly complicated.  For instance, for the group ring of a not-cyclic group of order 9 over a field of characteristic 3, the additive module of the representation ring is infinitely generated free, and expressing a module in that basis can be quite challenging (writing down the basis is even quite hard).

For the group ring of a group with a cyclic Sylow p-subgroup over a field of characteristic p, the representation ring again has an additive module that is finite rank free, and one can even fairly easily write down the basis using what is called the "Brauer tree" (forest) of the group.  Expressing a module in the basis is not trivial, but it is much less daunting than in general.

Curtis-Reiner has a nice chapter on ZG modules, and the appendix of that chapter is on the indecomposable ones.

I am only needing representation theory as an easy example for $ \lambda$-rings, and I wanted to avoid the $ \mathrm{char}k\nmid\left|G\right|$ condition. Thanks anyway, I am going to study modular representation theory one day.

  darij

[quote="-oo-"]since this also has to to with the structure of the rep. ring:

does every short exact of representations split?

again, the only crucial case is when the characteristic divides the group order.[/quote]

If the characteristic of the field k divides the order of the finite group G, and I is the submodule of kG generated by { 1-g : g in G }, then 0 → I → kG → k → 0 is short exact, but not split.

A reasonably easy proof that kG is not semi-simple is on page 12 of Alperin's Local Representation Theory:

Basically in a semi-simple ring the regular module contains any particular isomorphism class of one-dimensional module only once as a composition factor.  However, kG contains it at least twice, since kG/I contains the trivial module k once, and I contains x = Sum(g,g in G) = Sum(1+g,g in G), and kx is again a one-dimensional trivial (central) module.

The particular sequence above does not split, because if it did then k would be projective by page 29, Theorem 2, and k is not projective by Corollary 7, page 33.

thanks. but why are your arguments so complicated? kG -> k splits <=> there is a in kG with ga=a for all g and mapping to 1 in k <=> there is b in k such that b|G|=1 <=> char(K) does not divide |G|.

Ok. Another question, this time probably harder.

Often, one proves isomorphism of representations by showing that their characters are equal. This is based on the fact that for $ k \equal{} \mathbb{C}$, the map

$ \text{representation ring of }G\text{ over }k \to \left(kG\right)^{\ast}$,
$ \text{representation} \mapsto \text{its character}$

is injective. Of course, this cannot hold in the modular case, or else the $ p$-th direct power of any representation over $ \mathbb{F}_p$ would be the zero representation.

The question is, what to do then. For instance, suppose we want to prove that $ \mathrm{Sym}^p\left(\mathrm{Sym}^q V\right)\cong\mathrm{Sym}^q\left(\mathrm{Sym}^p V\right)$ for any $ 2$-dimensional representation $ V$, regardless of $ k$. Or we want to prove that the representation ring is a special $ \lambda$-ring (what means certain isomorphisms involving $ \wedge^k\left(V\otimes W\right)$ and $ \wedge^k\left(\wedge^{\ell}\left(V\right)\right)$). We can prove equality of characters, but what use is it?

The best answer (and the logical thing to do, according to the strategy "if it doesn't work, functorialize it") would be a theorem along the following lines: Let $ \mathrm{Rep}^i G$ be the category of pairs $ \left(A,V\right)$, where $ A$ is a ring (commutative, with $ 1$), and $ V$ is a finitely-generated free $ A$-module of rank $ i$ with a $ G$-action on it. A functor $ \Phi$ from a product of finitely many such categories with the same $ G$ but possibly different $ i$'s into another such category is called an [i]operation[/i] if the way it transforms the $ G$-actions is compatible with ring homomorphisms $ A\to A^{\prime}$. Then, two operations are isomorphic (i. e., there exists a functorial isomorphism between them) if and only if their characters are equal, which means that whatever representations you feed into the operation, you obtain two representations with same characters. Is anything known of this sort?

  darij


[hide="hint"] Construct a polygon of constraints [/hide]

[quote="aspiringchemlete"]Let $ x$ and $ y$ be positive integers such that $ x+y\leq 13$ and $ 3x+y\leq 24$. Find the max value of

a) $ x+y$
b) $ 4x+y$
c) $ 2x+y$.[/quote]

Not sure if this method is correct.

[hide][asy]draw((0,0)--(17,0)); draw((0,0)--(0,27)); draw((8,0)--(0,24)); draw((0,13)--(13,0)); fill((0,0)--(0,13)--(5.5,7.5)--(8,0)--cycle,red);[/asy]

The area in red is where we are restricted to. Looking at the "corners", we see that the max/min occur at
$ (0,0)$
$ (0,13)$
$ (5.5,7.5)$ (The coordinates aren't integers, we'll deal with that soon.)
$ (8,0)$

Now, just plug in and see which works best. We get answers of:
(i) $ (0,13)\implies \boxed{13}$ (duh)
(ii) $ (8,0)\implies\boxed{32}$
(iii) $ (5.5,7.5)\implies 18.5$,but the coordinates have to be integers. So let's try to get $ 18$. We see that $ (6,6)$ produces $ \boxed{18}.$[/hide]

:) Nice solution vishalarul though for part C, are not $ x \ , \ y \in \mathbb{Z}^{+}$ ?
[hide="Making it"]
$ 2x+y = 16$ Where $ ( 8 , 0 )$ gives the maximal value for the expression.
[/hide]


[quote="stergiu"]If for some year the 1st day is Sunday , how many times in the year(not leap) we will have Frinday and 13 nth?

 I saw this in a competition in USA. Do you have any simple answer(no modules)?

 Thank you[/quote]

sorry but mods is just easiest
[hide]
January 13 is a friday
add 31 days = +3 mod 7
add 28 days = +0 mod 7
add 31 days = +3 mod 7
add 30 days = +2 mod 7
add 31 days = +3 mod 7
add 30 days = +2 mod 7
add 31 days = +3 mod 7
add 31 days = +3 mod 7 (it's a multiple of 7 again here, so once again it is Friday the 13th)
add 30 days = +2 mod 7
add 31 days = +3 mod 7
add 30 days = +2 mod 7 (and again)
add 31 days = +3 mod 7

so it's 3 times? :huh:

[/hide]

edit: AFSDGDFASFDGSDFF $ 5\not\equal{}7$
ignore this

Thanks to both of you :)  !

 I wil read the solution carefully.  Is there perhaps a general way to find the number of such days for every not leap (or leap) year , if we know the day at first (using all intermidiate  level theory  of numbers )?

 Babis

CircleSquared, you are wrong, the answer is 2.  To confirm, just look at a calendar of the year 2006, in which such a scenario occurred.

[quote="JRav"]You mean Friday the 13th, yes?

So we're assuming that January 1st is a Sunday.  It's not hard to see that all of the Sundays of January then are the 1st, 8th, 15th, 22nd, and 29th.  Since the 15th lies on a Sunday, the 13th lies on a Friday.  $ \boxed{1}$.  Take note of this pattern.  Now, all we have to do is find all the months in which the Sundays follow a similar pattern to January.  For February, we have that the 5th is a Sunday, no good.  Since February has 28 days (4 weeks), we know March will follow the same pattern so we can immediately skip to April.  April 2nd is a Sunday so we are no good.  We now see that April 30th is a Sunday, so May 7th is a Sunday.  Again, no good.  May 28th is a Sunday, so June 4th is the next Sunday.  No good.  Now June 25th is a Sunday so the next one is July 2nd.  Still no good but we know that July 30th is a Sunday so August 6th is a Sunday.  Nope.  August 27th is a Sunday, so the next one is September 3rd.  Still no good.  But now we can easily see that October 1st is a Sunday, so October 13th is a Friday.  $ \boxed{2}$.  You can quickly check for yourself that there will be no more that year.

So there is your solution, no modulos.[/quote]

 Really simple solution and nice explanation !
 Last night I was lloking for past mathcount problems and I saw a simmilar for year 2001 I think.This happens every year and there is a proof.
 Thanks a again and good luck an your math competitions!

 Babis - Greece


For a,b,c,d real positive numbers such that

   abcd(1/a+1/b+1/c+1/d)=a+b+c+d

prove that

(a+1) 2 /(a+b) + (b+1) 2/(b+c) + (c+1) 2 /(c+d) + (d+1)^2/(d+a)  \geq 2(a+b+c+d)

I think the direction of ineq should be reversed. This inequality can be proven by using nice identity:

(a^2+1)/(a+b) + (b^2+1)/(b+c) + (c^2+1)/(c+d) + (d^2+1)/(d+a) = a + b + c + d. (Showed by Harazi)

Now, using (a+1)^2 <= 2(a^2+1) we have done.

Namdung

A little clarification: this problem was published in the Arhimede magazine by me and was not given in any mathematical olympiad. And yes, it is another application of the identity seen by Namdung.


counting
distinguishability
derangement
function
combinatorics unsolved
combinatorics

How many undirected graphs are there in which every vertex has valency 2 (2-regular), where $ m$ the the number of vertices.

Thanks in Advance.

Any such graph decomposes into a disjoint union of cycles. There are two possible interpretations here: either the vertices are distinguishable, or they're not. The graphs can be encoded by a permutation (mod reversing each cycle) in the former case, and a partition in the latter. I'll focus on the first version, with the vertices labeled $ 1,2,\cdots,m$.

Cycles in the graph translate to cycles in the permutation, or the inverses of those cycles. We can't have any fixed points in the permutation, as those would give degree-zero vertices in the graph. We can't have any 2-cycles in the permutation, as those would give degree-1 vertices in the graph. Cycles of length 3 and up are fine.

Excluding fixed points leads to the derangement problem. By analogy, we shouldn't expect a clean closed form, unless we get lucky by rounding something irrational.

OK, let's build a recursion. Call the number of such permutations $ G(n)$. Consider the cycle that the newest element $ n$ belongs to. If it has length $ k$, there are $ \binom{n \minus{} 1}{k \minus{} 1}$ ways to choose its elements, $ (k \minus{} 1)!$ ways to cyclically order those elements, and $ G(n \minus{} k)$ ways to handle the other $ n \minus{} k$ elements, but identifying the cycle with its inverse cuts this in half. That gives $ G(n) \equal{} \sum_{k \equal{} 3}^n \frac12\cdot\binom{n \minus{} 1}{k \minus{} 1}\cdot (k \minus{} 1)!\cdot G(n \minus{} k) \equal{} \sum_{k \equal{} 3}^n \frac {(n \minus{} 1)!}{2(n \minus{} k)!}G(k)$.
Now, bring in a generating function. Let $ H(n) \equal{} \frac {G(n)}{n!}$, and consider the exponential generating function $ P(x) \equal{} \sum_n H(n)x^n$. The recursion for $ G$ can be rewritten as $ nH(n) \equal{} \frac12\sum_{k \equal{} 3}^n H(n \minus{} k)$, which becomes the functional relation $ P'(x) \equal{} \frac12\cdot(x^2 \plus{} x^3 \plus{} x^4 \plus{} \cdots)\cdot P(x) \equal{} \frac {x^2}{2(1 \minus{} x)}P(x)$. Solve this (separable) differential equation to get $ P(x) \equal{} \frac {\exp\left( \minus{} \frac{x}{2} \minus{} \frac {x^2}{4}\right)}{\sqrt{1 \minus{} x}}$.

Closed form for the coefficients? Don't make me laugh. On the other hand, we can find an asymptotic estimate. $ P$ has no singularities on the circle except at $ 1$, where it looks like $ \frac{e^{\minus{}\frac34}}{\sqrt{1\minus{}x}}$. The $ n$th coefficient $ \frac12\cdot\frac34\cdot\frac56\cdots \frac{2n\minus{}1}{2n}$ in the power series for $ \frac1{\sqrt{1\minus{}x}}$ is approximately $ \frac1{\sqrt{\pi n}}$, so the coefficient of $ x^n$ in the series for $ P$ should be approximately $ \frac{e^{\minus{}\frac34}}{\sqrt{\pi n}}$. I've confirmed this numerically, but it's pretty slow convergence.

[Edited- corrected overcounting]


Nancy spent 3 hours painting her dormitory room. Her roommate, Marcia, didn't like the color. So Marcia did the job over again in 5 hours. Working together, how long would it have taken to paint the room?
The answer is 9/8, but can someone show me how you get it?

Nancy can do $\frac{1}{3}$ of the work in 1 hour. Marcia can do $\frac{1}{5}$ of the work in 1 hour. Thus, together they can do $\frac{8}{15}$ of the work in 1 hour, so they take $\frac{15}{8}$ hours to do all the work.

It is very well known that the amount of work done (or the distance traveled, etc.) is equal to the product of the rate (work/time) and the time.

So you can say $rt=w$, where $r$ is the number of hours required to paint the room, $t$ is however much time is spend, and $w$ is the number of rooms painted during that time.

For this problem you have $r_1=\frac{1}{3}$ (1 room per 3 hours is the rate) and $r_2=\frac{1}{5}$ and you want to find $t_0$ such that $r_1t_0+r_2t_0=1$, so you have $t_0=\frac{1}{r_1+r_2}$, which in this case is $\frac{1}{\frac{1}{3}+\frac{1}{5}}=\frac{15}{8}$.

In general, say you have $n$ people and they need to do $y$ jobs. Then you want $t_0$ such that $r_1t_0+r_2t_0+\ldots+r_nt_0=y$, so $t_0=\frac{y}{\sum_{i}{r_i}}$.

[quote="3X.lich"]I don't get the idea of 2 same exact posts. The other one is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25166]here[/url]

Please post [b]ONE[/b] topic regarding same one problem.[/quote]

Sorry about that! I posted it twice because I wanted a fast reply. I was assuming that I could delete my post after I got an answer in one of them, but I realised that only moderators can do that...


How many different isosceles triangle have integer side lengths and a perimeter of 81 square units?

[quote="Ars"][hide]20[/hide][/quote]
Next time post a solution, not just an answer. I would do this problem but I don't feel like it rite now, and I have to go find a bunch of stuff for school.


algebra
function
domain
polynomial
complex analysis
complex analysis unsolved

If there exists $ p$ integer such that $ a_n \equal{} a_{n \plus{} p}$, what can we say about the complex power power series $ \sum_{n \equal{} 0}^\infty \frac{a_n}{n!} z^n$?.
I don't think its periodic in the complex domain but I do not manage to find a counter-example... There are some well-known properties for this kind of series?

You can write a fairly explicit expression describing all such functions within the open unit disc:
\[ \sum_{n \equal{} 0}^\infty a_n z^n \equal{} \frac {p(z)}{1 \minus{} z^p}
\]
where $ p(z) \equal{} \sum_{n \equal{} 0}^{p\minus{}1} a_n z^n$ is just a polynomial.  So all such series are just rational functions of degree at most $ p$ within the unit disc, and do not converge anywhere else.

[quote="JoeBlow"]You can write a fairly explicit expression describing all such functions within the open unit disc:
\[ \sum_{n \equal{} 0}^\infty a_n z^n \equal{} \frac {p(z)}{1 \minus{} z^p}
\]
where $ p(z) \equal{} \sum_{n \equal{} 0}^{p} a_n z^n$ is just a polynomial.  So all such series are just rational functions of degree at most $ p$ within the unit disc, and do not converge anywhere else.[/quote]

Thanks very much for the answer but I made a mistake when I wrote the question, I edited it. I forgot to divide by  n!.

About your answer, I knew in fact that if there is a linear recurence relation between the coefficients, we still get a rational function.

My question is a little annoying because we find all the classical functions $ e^z$, $ cos z$, $ sh z$, ...

In that case all such series are exactly the solutions of $ \frac {d^p}{dz^p}f(z) \equal{} f(z)$, which again is a family of functions you can write down fairly explicitly for any given $ p$:
\[ f(z) \equal{} \sum_{n \equal{} 0}^{p \minus{} 1} c_n e^{r_n z}
\]
where $ c_n\in \mathbb{C}$ are arbitrary and $ r_n$ are the $ p$ roots of the polynomial $ r(r^{p \minus{} 1} \minus{} 1)\equal{}0$.

[quote="JoeBlow"]In that case all such series are exactly the solutions of $ \frac {d^p}{dz^p}f(z) \equal{} f(z)$, which again is a family of functions you can write down fairly explicitly for any given $ p$:
\[ f(z) \equal{} \sum_{n \equal{} 0}^{p \minus{} 1} c_n e^{r_n z}
\]
where $ c_n\in \mathbb{C}$ are arbitrary and $ r_n$ are the $ p$ roots of the polynomial $ r(r^{p \minus{} 1} \minus{} 1) \equal{} 0$.[/quote]

Very nice! I should have thought about a differential equation.


Euler
geometry
circumcircle
trigonometry
angle bisector
geometry proposed

Show that if $\angle BAC=60^{\circ}$ then the Euler line cuts off an equilateral triangle with vertex $A$ from the original triangle $ABC$.

I found it here: [url]http://www.cut-the-knot.org/Curriculum/Geometry/Euler60.shtml[/url].

[Edit: there was a mistake pointed out by Darij; it's been corrected]

[color=red][[b]Moderator edit:[/b] LevonNurbekian solves the problem in the following equivalent form: Show that if < BCA = 60°, then the Euler line cuts off an equilateral triangle with vertex C from the original triangle ABC. Actually, the problem was originally posted in this form, only with a typo, and later Grobber replaced C by A.][/color]

The solution is pretty simple. Define by $H$ and $O$ the orthocenter and the circumcenter of the triangle $ABC$. It is  easy notice that if $\angle{BCA}=\frac{\pi}{3}$ then $CH=CO$ (as $CH=2R\cos{\frac{\pi}{3}}=R=CO$). We will also need the fact that $\angle{HCA}=\angle{OCB}$ (which is simple as well). Using the last one we can say that the angle bisectors of angles $\angle{DCE}$ and $\angle{HCE}$ are the same, where $D,E$ are the points of intersection the line of Euler with the sides $CA$ and $CB$, respectively. So , if $CH=CO$, then $CD=CE$ , as the common angle bisector   of the angles $\angle{DCE}$ and $\angle{HCE}$ is perpendicular to the Euler's line. $CD=CE$ and $\angle{C}=\frac{\pi}{3}$ means that triangle $CDE$ is equilateral.
Using the reverse of the proof above we can also solve the following problem:
[i]Given triangle $ABC$. Let Euler's line intersect the sides $CA$ an $CB$ at the points $D$ and $E$, respectively. It is given that $CD=CE$. What is the value of the $\angle{C}$?[/i]
I think you guess the answer :) .

This problem is similar to:
http://www.mathlinks.ro/Forum/viewtopic.php?t=13502
I have posted a simple proof.


quadratics
function
Vieta
algebra
quadratic formula

The sum of the zeros, the product of the zeros, and the sum of the coefficients of the function $ f(x) \equal{} ax^{2} \plus{} bx \plus{} c$ are equal. Their common value must also be which of the following?

$ \textbf{(A)}\ \text{the coefficient of }x^{2}\qquad \textbf{(B)}\ \text{the coefficient of }x$
$ \textbf{(C)}\ \text{the y \minus{} intercept of the graph of }y \equal{} f(x)$
$ \textbf{(D)}\ \text{one of the x \minus{} intercepts of the graph of }y \equal{} f(x)$
$ \textbf{(E)}\ \text{the mean of the x \minus{} intercepts of the graph of }y \equal{} f(x)$

[hide]
By Vieta's Formulas, the sum of the zeroes is $-\frac{b}{a}$ and the product of the zeroes is $\frac{c}{a}$. If these are equal, multiplying by $a$ gives $-b=c$.

$a+b+c$ is also equal to the common value.  Since $-b=c$, $a+b+c=a+b-b=a$.  So the common value is equal to $a$, the coefficient of  $x^{2}$.  Thus the answer is A.
[/hide]

If one had never heard of Vieta's formula, could one solve this question? Cuz i never heard of it, and thats not fair  :P .

I feel the pain lol.

I haven't learned about Vieta's either.

You don't have to learn about Vieta's if you understand that if $r$ is a root then $x-r$ is a factor (the Factor Theorem).

[hide="Because then..."] $f(x) = ax^{2}+bx+c = a(x-r_{1})(x-r_{2}) = ax^{2}-a(r_{1}+r_{2}) x+ar_{1}r_{2}$

Which implies Vieta's formulas for quadratics (in fact, this is how you prove Vieta's formulas generally).  Then, from our givens,

$r_{1}+r_{2}= r_{1}r_{2}= a+b+c = a-a(r_{1}+r_{2})+ar_{1}r_{2}= a$. [/hide]

If you have the equation $ax^{2}+bx+c=0$, then by the quadratic formula (which I'm sure you know),

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$, or

$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$.

These are the only two solutions (by the fundamental theorem of algebra).

When you add them (the sum of the zeros), you get $-\frac{b}{a}$.

When you multiply them (the product of the zeros), you get $\frac{c}{a}$.

Am I getting smarter or is this an unusually easy #21? From the looks of it, I would have ownt this test.....but then it snows for the first time all year  :cursing:

You would've pwned the test, because I did. :P

Vieta's formula tells you that

In

$P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$

The sum of the roots is

$-\frac{a_{n-1}}{a_{n}}$

I'm beginning to think that the problem makers were targeting people who didn't know Viete's?? Because if you did, it was incredibly easy.


hm

i'm trying to learn foodtongue.
i read the wiki, but it doesnt help much.

for instance, what is tofu? and what is eggs?

and how do you say "i lost" (possibly the most important phrase i need to learn!)

This sounds like a pretty tasty language.

Hmm I would also like to actually learn FoodTongue. I feel like a horrible Mathcamp alumna not knowing it.

[quote="mathfreak77"]apple=tofu
[/quote]
Not really?

Tofu:
Apple clam tofu cheeseburger-food?

Eggs:
Mint eggs, mint wine: Eggs plantain wine grass wine plantain eggs.  Eggs grass eggs.  Meat eggs grass wine, wine grass eggs.

[code]E      E
  G G
  G G
S      S[/code]

pi cheetoes potato monkey liver horse brain.

Mango tongue? @E_A Pear sauce mango tongue?

I'd say apple spinach tofu eggs, Eggs grass mango pear wonton.


AMC
AIME
USA(J)MO
USAMO
AMC 10
AoPS Books

How are the hardest problems in the AMC 10, 12, and AIME, respectively to the hardest questions in the AOPS books Volume 1 and 2? (the ones with a eye of needle?)

Those are pretty hard problems...if they are even harder, then I'm totally screwed.  :rotfl:

AMC 10 (I just took one) is about plus or minus a quite a bit (the level varies for problems  1-10 and 15-25 and stuff in the middle) the level of volume one from what my experience is.

Hardest problem on AIME is much harder than problems in Vol. 1.  Haven't worked through Vol. 2 yet so not sure.  But you don't need to get the hardest problem to score well and qualify for USAMO  :)

There are some volume 2 problems from MOP, so they are pretty hard.  On the other side, some of the other problems are just as easy as Volume 1 problems.


[quote="TripleM"][quote="galois"]given a value K find the minimum value N such that
?1?2?3?4?.....?N=K
where ?=+ or -
input K<=10^9
output:
single line giving N[/quote]

Choose the smallest $N$ with $\frac{N(N+1)}{2}\ge K$ and $\frac{N(N+1)}{2}-K$ even. Obviously $N$ must be at least this large. To prove it works, just notice that $\frac{N(N+1)}{2}-K < \frac{N(N+1)}{2} - \frac{(N-3)(N-2)}{2} = 3N+3$. So start with $1+2+\ldots +N$, easily choose up to two numbers which add to half the above difference, and turn them into negatives.[/quote]

What if the difference isn't even? Like say I give you 27, you'd say N = 7, what do I turn negative? You'd have to go up to N = 8, then you get a sum of 36, which is still an odd amount off, so you have to go to N = 9, with a sum of 45. Now your difference is 45 - 27 = 18, and you can swap the 9 to get your necesarry sum. However, I believe you can prove that you'll never have to check any more cases than 3.

[quote="JSteinhardt"][quote="TripleM"]

Choose the smallest $N$ with $\frac{N(N+1)}{2}\ge K$ and $\frac{N(N+1)}{2}-K$ even. Obviously $N$ must be at least this large. To prove it works, just notice that $\frac{N(N+1)}{2}-K < \frac{N(N+1)}{2} - \frac{(N-3)(N-2)}{2} = 3N+3$. So start with $1+2+\ldots +N$, easily choose up to two numbers which add to half the above difference, and turn them into negatives.[/quote]

What if the difference isn't even? Like say I give you 27, you'd say N = 7, what do I turn negative? You'd have to go up to N = 8, then you get a sum of 36, which is still an odd amount off, so you have to go to N = 9, with a sum of 45. Now your difference is 45 - 27 = 18, and you can swap the 9 to get your necesarry sum. However, I believe you can prove that you'll never have to check any more cases than 3.[/quote]

Thats exactly what I said ;) You choose the smallest N such that it *is* even. And yes, of course there are only 3 or less cases to check.


modular arithmetic
number theory unsolved
number theory

Find all prime triplets (p,q,r) such that:

p|r^q+1, q|p^r+1 and r|q^p+1.

I think it's just casework, but I don't have the time to complete it right now :(.

I guess we could start like this: assuming $p\not|r+1,q\not|p+1,r\not|q+1$ yields a contradiction, because we get $2r|p-1,2p|q-1,2q|r-1$. So, if we assume $p$ is the smallest among the three numbers, we must have $p|r+1\ (*)$.

There are now to cases: either $r\not|q+1$, so $2p|r-1$, and from $(*)$ we get $p=2$, or $r|q+1$, and $q\not|p+1$ (I [b]think[/b] we get a contradiction otherwise :)), so $2r|q-1$, and $r=2,p=2$ (because we assumed $p$ was the smallest one, getting a contradiction).

This means we need to focus on the case $p=2$, so $r|q^2+1,q|2^r+1$. Now $q$ can be $3$ and $r=5$, or, if $q\not|2+1$, we get $2r|q-1$, and together with $r|q^2+1$ we get a contradiction.

This is pretty much what we have to do, although as I said, I'm in a hurry, and I bet I missed quite a few things  ;) .

Let me rewrite something which seems clearer (plz check although i'm not in hurry but hungry  :D )

1st) if there is one for exam. $p=2$ then he said
2nd) $p,q,r$ odd we assume that $p$ min of them.
$p=3$ then $3\mid r^q+1$, $q\mid 3^r+1$, and $r\mid q^3+1$.
Note  $q\mid(3^{2r}-1,3^{q-1}-1)=3^{2r}-1$ so $2r\mid q-1$ then $r\nmid q+1$.
then $r\mid q^2-q+1$ so $r\mid 1$
3rd) $p>3$ then

$p\mid (r^{2q}-1,r^{p-1}-1)=r^2-1$ or $r-1$ but the second can't hold.

so $p\mid r+1$.

Now

$q\mid (p^{2r}-1,p^{q-1}-1)=p^2-1$ or $p^{2r}-1$
The first implies $q=p+1$ this contract cause both are odd! So the second holds. We have $2r\mid q-1$ .
Now
$p\mid r+1$, $2r\mid q-1$ and $q\mid p^{r}+1$

so from $q^p+1\equiv 2\pmod{r}$ this contracts!!!!
So we are done!

[quote="al.M.V."]hmm...I can understand where you are coming from grobber, but I really have difficulties following you. I see that, if assuming p the smallest of (p,q,r), since r^(p-1) = 1 (mod p) (FLT), and r^(2q) = 1 (mod p), then p-1 | 2q. (This is the opposite of what you had: 2q | p-1.) Then I don't see how this leads to p | r+1. (r?) Could you please explain in more details? (maybe when you are not as busy?) Thanks.  :)[/quote]

We have $r^{2q}\equiv 1\pmod p$, so the order of $r\pmod p$ is $1,q,2$ or $2q$, and the order must divide $p-1$. If $p\ne 2$ and $p\not|r+1$, then the order can't be $1$, it's not $q$, and it's not $2$ either, so it's $2q$, and $2q|p-1$.
 Something like that :). Sorry: I'm sure there are lots of things I missed,

Yeah, it seems somewhat clear now, I will go through it again (first time it worked,  :D )...thanks grobber, and pluricomplex.. :lol:

btw, this problem was proposed by Reid Barton :)


Is there a positive integer $ n$ for which $ n^7 \minus{} 77$ is a Fibonacci number?

@PhilG -nice :lol:.......just out of curiosity ,how did you think of applying $ \mod 29$ :D ?

I looked for primes that are $ 7k\plus{}1$ to minimise the number of residues of $ n^7$
I only tried this one so there may be smaller numbers that work.


Let $f$ and $g$ be polynomials defined for all real numbers. For $-2 \le x \le 2$, $|f(x)|+g(x)=2x+9$ and for $x>2$, $f(x)+g(x)=2x^{2}+2x+1$. Find the minimum value of $|f(x)+2g(x)|$.


Let $ a,b,c$ be non-negative real numbers. Prove that

[b]1/[/b]  $ \frac1{a\plus{}b}\plus{}\frac1{b\plus{}c}\plus{}\frac1{c\plus{}a}\ge\frac 1{\sqrt{2(a^2\plus{}bc)}}\plus{}\frac 1{\sqrt{2(b^2\plus{}ca)}}\plus{}\frac 1{\sqrt{2(c^2\plus{}ab)}}$

[b]2/[/b]  $ \frac a{b\plus{}c}\plus{}\frac b{c\plus{}a}\plus{}\frac c{a\plus{}b}\ge\frac a{\sqrt{a^2\plus{}3bc}}\plus{}\frac b{\sqrt{b^2\plus{}3ca}}\plus{}\frac c{\sqrt{c^2\plus{}3ab}}$

1/ See here: http://www.mathlinks.ro/viewtopic.php?t=116853&start=20
2/ After using Cauchy Schwarz, we can see that the inequality follows from
$ \sum \frac{a}{b\plus{}c} \ge \sum \frac{a(b\plus{}c)}{a^2\plus{}3bc},$
that is
$ \sum \frac{a(a^2\plus{}bc\minus{}b^2\minus{}c^2)}{(b\plus{}c)(a^2\plus{}3bc)} \ge 0,$
or
$ \sum \frac{a^3(b\plus{}c)\minus{}a(b^3\plus{}c^3)}{(b\plus{}c)^2(a^2\plus{}3bc)} \ge 0.$
Without loss of generality, we can assume that $ a \ge b \ge c$, then
$ a^3(b\plus{}c) \minus{}a(b^3\plus{}c^3) \ge 0 \ge c^3(a\plus{}b)\minus{}c(a^3\plus{}b^3),$
and
$ \frac{1}{(b\plus{}c)^2(a^2\plus{}3bc)} \ge \frac{1}{(c\plus{}a)^2(b^2\plus{}3ca)} \ge \frac{1}{(a\plus{}b)^2(c^2\plus{}3ab)}.$
It follows that
$ \frac{a^3(b\plus{}c)\minus{}a(b^3\plus{}c^3)}{(b\plus{}c)^2(a^2\plus{}3bc)} \ge \frac{a^3(b\plus{}c)\minus{}a(b^3\plus{}c^3)}{(c\plus{}a)^2(b^2\plus{}3ca)},$
and
$ \frac{c^3(a\plus{}b)\minus{}c(a^3\plus{}b^3)}{(a\plus{}b)^2(c^2\plus{}3ab)} \ge \frac{c^3(a\plus{}b)\minus{}c(a^3\plus{}b^3)}{(c\plus{}a)^2(b^2\plus{}3ca)}.$
Therefore
$ \sum \frac{a^3(b\plus{}c)\minus{}a(b^3\plus{}c^3)}{(b\plus{}c)^2(a^2\plus{}3bc)} \ge \frac{\sum (a^3(b\plus{}c)\minus{}a(b^3\plus{}c^3))}{(c\plus{}a)^2(b^2\plus{}3ca)} \equal{}0.$
Our proof is completed. :)

[quote="can_hang2007"]1/ See here: http://www.mathlinks.ro/viewtopic.php?t=116853&start=20
2/ After using Cauchy Schwarz, we can see that the inequality follows from
$ \sum \frac {a}{b \plus{} c} \ge \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc},$
that is
$ \sum \frac {a(a^2 \plus{} bc \minus{} b^2 \minus{} c^2)}{(b \plus{} c)(a^2 \plus{} 3bc)} \ge 0,$
or
$ \sum \frac {a^3(b \plus{} c) \minus{} a(b^3 \plus{} c^3)}{(b \plus{} c)^2(a^2 \plus{} 3bc)} \ge 0.$
Without loss of generality, we can assume that $ a \ge b \ge c$, then
$ a^3(b \plus{} c) \minus{} a(b^3 \plus{} c^3) \ge 0 \ge c^3(a \plus{} b) \minus{} c(a^3 \plus{} b^3),$
and
$ \frac {1}{(b \plus{} c)^2(a^2 \plus{} 3bc)} \ge \frac {1}{(c \plus{} a)^2(b^2 \plus{} 3ca)} \ge \frac {1}{(a \plus{} b)^2(c^2 \plus{} 3ab)}.$
It follows that
$ \frac {a^3(b \plus{} c) \minus{} a(b^3 \plus{} c^3)}{(b \plus{} c)^2(a^2 \plus{} 3bc)} \ge \frac {a^3(b \plus{} c) \minus{} a(b^3 \plus{} c^3)}{(c \plus{} a)^2(b^2 \plus{} 3ca)},$
and
$ \frac {c^3(a \plus{} b) \minus{} c(a^3 \plus{} b^3)}{(a \plus{} b)^2(c^2 \plus{} 3ab)} \ge \frac {c^3(a \plus{} b) \minus{} c(a^3 \plus{} b^3)}{(c \plus{} a)^2(b^2 \plus{} 3ca)}.$
Therefore
$ \sum \frac {a^3(b \plus{} c) \minus{} a(b^3 \plus{} c^3)}{(b \plus{} c)^2(a^2 \plus{} 3bc)} \ge \frac {\sum (a^3(b \plus{} c) \minus{} a(b^3 \plus{} c^3))}{(c \plus{} a)^2(b^2 \plus{} 3ca)} \equal{} 0.$
Our proof is completed. :)[/quote]

Very nice this proof, my dear friend. :)

I like a lot this method. :wink:

Thank you very much. :)

Yes, excellent solution. But I cannot follow the first step :blush: :
[quote]
[b]2/[/b]  $ \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b}\ge\frac a{\sqrt {a^2 \plus{} 3bc}} \plus{} \frac b{\sqrt {b^2 \plus{} 3ca}} \plus{} \frac c{\sqrt {c^2 \plus{} 3ab}}$[/quote]
[quote="can_hang2007"]
2/ After using Cauchy Schwarz, we can see that the inequality follows from
$ \sum \frac {a}{b \plus{} c} \ge \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc},$:)[/quote]

Please can you elaborate how do you use C-S? Can you use it for showing $ \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc}\ge\frac a{\sqrt{a^2\plus{}3bc}}\plus{}\frac b{\sqrt{b^2\plus{}3ca}}\plus{}\frac c{\sqrt{c^2\plus{}3ab}}$ ?

Is is so  :wink:

$ \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc} \sum \frac{a}{b\plus{}c} \geq (\sum \frac a{\sqrt {a^2 \plus{} 3bc}})^2$

Thank you very much, Manlio. I see  :blush:  :blush:

So we can say, $ \sum \frac {a}{b \plus{} c} \ge \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc}$ implies by C-S the original $ \sum \frac a{b \plus{} c} \ge \sum \frac a{\sqrt {a^2 \plus{} 3bc}}$ .

The question remains: is it stronger than the original one, i.e. does

$ \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc}\ge \sum \frac a{\sqrt {a^2 \plus{} 3bc}}$

hold?

[quote="spanferkel"]

The question remains: is it stronger than the original one, i.e. does

$ \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc}\ge \sum \frac a{\sqrt {a^2 \plus{} 3bc}}$

hold?[/quote]

It's not true, try $ a\equal{}2; b\equal{}c\equal{}1$.  :wink:
But I think the following is true for all $ a,b,c\ge0$:
$ \sum \frac {a(b \plus{} c)}{a^2 \plus{} 3bc}\le \sum \frac a{\sqrt {a^2 \plus{} 3bc}}$

Nice proof, Can. :)

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers. Prove that

[b]1/[/b]  $ \frac1{a \plus{} b} \plus{} \frac1{b \plus{} c} \plus{} \frac1{c \plus{} a}\ge\frac 1{\sqrt {2(a^2 \plus{} bc)}} \plus{} \frac 1{\sqrt {2(b^2 \plus{} ca)}} \plus{} \frac 1{\sqrt {2(c^2 \plus{} ab)}}$

[b]2/[/b]  $ \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b}\ge\frac a{\sqrt {a^2 \plus{} 3bc}} \plus{} \frac b{\sqrt {b^2 \plus{} 3ca}} \plus{} \frac c{\sqrt {c^2 \plus{} 3ab}}$[/quote]

Same form with the first, I conjecture the following Inequality
$ \frac1{a \plus{} b} \plus{} \frac1{b \plus{} c} \plus{} \frac1{c \plus{} a}\ge\frac{\sqrt{1\plus{}k}}2\left(\frac 1{\sqrt {a^2 \plus{} kbc}} \plus{} \frac 1{\sqrt {b^2 \plus{} kca}} \plus{} \frac 1{\sqrt {c^2 \plus{} kab}}\right)$
for all $ k\in[\frac{2}5,\frac{5}2].$

A nother one : http://www.mathlinks.ro/viewtopic.php?p=1449531#1449531

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers. Prove that

[b]2/[/b]  $ \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b}\ge\frac a{\sqrt {a^2 \plus{} 3bc}} \plus{} \frac b{\sqrt {b^2 \plus{} 3ca}} \plus{} \frac c{\sqrt {c^2 \plus{} 3ab}}$[/quote]

Infact, we have the following Inequality for all $ a,b,c\ge0$

\[ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}\ge\frac a{\sqrt {a^2 \plus{} 3bc}} \plus{} \frac b{\sqrt {b^2 \plus{} 3ca}} \plus{} \frac c{\sqrt {c^2 \plus{} 3ab}}
\]


For a square matrix $ A$, is it necessary that $ (A^{\minus{}1})^n\equal{}(A^n)^{\minus{}1}$, for any integer $ n$?

This is true for any invertible matrix, or for any element in a group; if we have enough structure for an inverse to make sense, inverses will commute with powers.
$ B\equal{}A^{\minus{}1}$ when $ BA\equal{}AB\equal{}I$ (the identity element); $ B^nA^n\equal{}B^{n\minus{}1}B\cdot A\cdot A^{n\minus{}1}\equal{}B^{n\minus{}1}A^{n\minus{}1}\equal{}\cdots\equal{}I$, and similarly $ A^nB^n\equal{}I$.

Nothing here is specific to matrices.

The statement is equal to $ \left(\left(A^{ \minus{} 1}\right)^n\right)^{ \minus{} 1} \equal{} A^n$, so just prove $ \left(A^{ \minus{} 1}\right)^nA^n\equal{}I$ and it's very easy to see that this is true.

Now, for which $ n\in Z$ is it true that $ A^nB^n\equal{}(BA)^n$? What about $ A^nB^n\equal{}(AB)^n$?  :D

(Suppose $ A,B$ are square and invertible)


If $ a,b,c\geq1$, then

$ \frac {a}{2b \plus{} ca} \plus{} \frac {b}{2c \plus{} ab} \plus{} \frac {c}{2a \plus{} bc}\leq 1,$

with equality if and only if $ a\equal{}b\equal{}c\equal{}1$.

[quote="Ji Chen"]If $ a,b,c\geq1$, then

$ \frac {a}{2b \plus{} ca} \plus{} \frac {b}{2c \plus{} ab} \plus{} \frac {c}{2a \plus{} bc}\leq 1,$[/quote]

Let $ f(a,b,c)\equal{}\frac {a}{2b \plus{} ca} \plus{} \frac {b}{2c \plus{} ab} \plus{} \frac {c}{2a \plus{} bc}$

Suppose the maximum value of $ f$ occurs at $ a_0,b_0,c_0$

$ \frac{a_0}{\min(a_0,b_0,c_0)},\frac{b_0}{\min(a_0,b_0,c_0)},\frac{c_0}{\min(a_0,b_0,c_0)}\geq 1$
It is easy to see that $ f(a_0,b_0,c_0)\leq f(\frac{a_0}{\min(a_0,b_0,c_0)},\frac{b_0}{\min(a_0,b_0,c_0)},\frac{c_0}{\min(a_0,b_0,c_0)})$

But since $ f(a_0,b_0,c_0)$ is minimum, equality must occur in the above inequality.

Equality occurs iff $ \min(a_0,b_0,c_0)\equal{}1$

WLOG $ a_0\equal{}1$

It is now sufficient to maximize $ \frac{1}{2b\plus{}c}\plus{}\frac{b}{2c\plus{}b}\plus{}\frac{c}{2\plus{}bc}$

Case I: $ b\geq c$

Since $ (b\minus{}2)(c\minus{}1)\geq 0\implies bc\plus{}2\geq 2c\plus{}b$

Thus $ \frac{1}{2b\plus{}c}\plus{}\frac{b}{2c\plus{}b}\plus{}\frac{c}{2\plus{}bc}\leq \frac{1}{2b\plus{}c}\plus{}\frac{b\plus{}c}{2c\plus{}b}$

It is sufficient to show that $ g(b,c)\equal{}\frac{1}{2b\plus{}c}\plus{}\frac{b\plus{}c}{2c\plus{}b}\leq 1$

Suppose maximum of $ g(b,c)$ occurs at $ b'_0,c'_0$

Again $ g(b'_0,c'_0)\leq g(\frac{b'_0}{c'_0},\frac{c'_0}{c'_0})$

Equality occurs iff $ c'_0\equal{}1$

Thus we need to prove that $ \frac{1}{2b\plus{}1}\plus{}\frac{b\plus{}1}{2\plus{}b}\leq 1$
Which is obvious

Case 2: $ b\leq c$

Since $ (b\minus{}1)(c\minus{}2)\geq 0\implies bc\plus{}2\geq c\plus{}2b$

Thus $ \frac{1}{2b\plus{}c}\plus{}\frac{b}{2c\plus{}b}\plus{}\frac{c}{2\plus{}bc}\leq \frac{c\plus{}1}{2b\plus{}c}\plus{}\frac{b}{2c\plus{}b}$

It is sufficient to show that $ g(b,c)\equal{}\frac{c\plus{}1}{2b\plus{}c}\plus{}\frac{b}{2c\plus{}b}\leq 1$

Suppose maximum of $ g(b,c)$ occurs at $ b'_0,c'_0$

Again $ g(b'_0,c'_0)\leq g(\frac{b'_0}{b'_0},\frac{c'_0}{b'_0})$

Equality occurs iff $ b'_0\equal{}1$

Thus we need to prove that $ \frac{c\plus{}1}{2\plus{}c}\plus{}\frac{1}{2c\plus{}1}\leq 1$
Which is obvious

We can do it by simple expanding as well.
$ \Rightarrow 4(ab^2\plus{}bc^2\plus{}ca^2)\plus{}2(a^3b\plus{}b^3c\plus{}c^3a)\plus{}abc(2a\plus{}2b\plus{}2c\plus{}ab\plus{}bc\plus{}ca) \leq abc(abc\plus{}8)\plus{}4(b^2c^2\plus{}c^2a^2\plus{}a^2b^2)\plus{}2(a^3bc\plus{}b^3ca\plus{}c^3ab)$
Which holds because the following ineq.s hold good for $ a,b,c \geq 1$, with equality occuring in each case iff $ a\equal{}b\equal{}c\equal{}1$
$ a^3bc\plus{}b^3ca\plus{}c^3ab \geq a^3b\plus{}b^3c\plus{}c^3a\\
a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \geq a^2c\plus{}b^2a\plus{}c^2b\\
abc\plus{}8 \geq 2(a\plus{}b\plus{}c)\plus{}ab\plus{}bc\plus{}ca$
 :)

Hope there is no typing error....

[quote="Sunkern_sunflora"]We can do it by simple expanding as well.
$ \Rightarrow 4(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} 2(a^3b \plus{} b^3c \plus{} c^3a) \plus{} abc(2a \plus{} 2b \plus{} 2c \plus{} ab \plus{} bc \plus{} ca) \leq abc(abc \plus{} 8) \plus{} 4(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2) \plus{} 2(a^3bc \plus{} b^3ca \plus{} c^3ab)$
Which holds because the following ineq.s hold good for $ a,b,c \geq 1$, with equality occuring in each case iff $ a \equal{} b \equal{} c \equal{} 1$
$ a^3bc \plus{} b^3ca \plus{} c^3ab \geq a^3b \plus{} b^3c \plus{} c^3a \\
a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \geq a^2c \plus{} b^2a \plus{} c^2b \\
abc \plus{} 8 \geq 2(a \plus{} b \plus{} c) \plus{} ab \plus{} bc \plus{} ca$
 :)

Hope there is no typing error....[/quote]
Are you sure,for the last,try $ a\equal{}2,b\equal{}c\equal{}1$ :)

nice proof
we prove that:

$ \frac {a}{ac \plus{} 2b}\leq\frac {a}{a \plus{} b \plus{} c}$
so if you some these inequalities you reach:
$ \sum \frac {a}{ac \plus{} 2b}\leq \sum \frac {a}{a \plus{} b \plus{} c} \equal{} 1$

$ \frac {a}{ac \plus{} 2b} \leq \frac {a}{a \plus{} b \plus{} c}\longleftrightarrow (ac \plus{} 2b)\geq(a \plus{} b \plus{} c) \longleftrightarrow (ac \plus{} b)\geq(a \plus{} c)$
but $ (ac \plus{} b \minus{} a \minus{} c)\geq(ac \plus{} 1 \minus{} a \minus{} c) \equal{} (a \minus{} 1)(c \minus{} 1)\geq0$
so we have equation when ($ a \equal{} b \equal{} c \equal{} 1$)

@akashnil: you have written $ b\ge c \Rightarrow (b\minus{}2)(c\minus{}1) \ge 0$ How does that come about?

My proof is similar to sumita's
But how about
/ Let $ a,b,c\ge0$ such that $ abc\equal{}1$. Prove that
$ \frac a{2a \plus{} 3bc} \plus{} \frac b{2b \plus{} 3ca} \plus{} \frac c{2c \plus{} 3ab}\le \frac {a \plus{} b \plus{} c}5$
It is from $ dduclam$,a nice ineq (but i cant find the link which was posted it,sorry authour

[quote="hsbhatt"]@akashnil: you have written $ b\ge c \Rightarrow (b \minus{} 2)(c \minus{} 1) \ge 0$ How does that come about?[/quote]
Sorry. That was wrong. :blush:

Yes Akashnil, I have seen my mistake, sorry! :(
But why doesn't it follow from expansion? It ought to follow. Perhaps, ther is some thing wrong in my arranging of terms... :maybe:

[quote="sumita"]$ \sum_{cyc}\frac {a}{2b\plus{}ca}\leq \sum_{cyc}\frac {a}{a \plus{} b \plus{} c} \equal{} 1.$[/quote]Nice proof, sumita. This is also my method, see: http://www.mathlinks.ro/Forum/viewtopic.php?t=214281


induction
number theory open
number theory

I thought of the problem below but I don't know if it is correct:
Given the sequence: $ {a_{n}}$
$ a_{1}\equal{}2 ; a_{n\plus{}1}\equal{} a^{a_{n}}\minus{}1$
where a is an odd natural number.
We call the set S of primes such that there exists n such that p|n.
Prove that the set S has infinitely many elements.

$ \mathbb{P}\equal{}\{primes\}$

we have $ 2|a_{1}$
if $ 2^{n}|a_{n}$ then $ a_{n}\equal{} 2^{n}b$ and $ a_{n\plus{}1}\equal{} a^{2^{n}b}\minus{}1$
so $ 2^{n\plus{}1}|(a^{b})^{2^{n}}\minus{}1 \equal{} a_{n\plus{}1}$
then $ \forall n\in\mathbb{N}:\ 2^{n}|a_{n}$
NOW we take $ S \equal{}\{p\in\mathbb{P}|\ \exists n\in\mathbb{P},\ p\minus{}1 \equal{} 2^{n}\}$ and $ A \equal{}\{p\in\mathbb{P}|\ \exists n\in\mathbb{N},\ p|a_{n}\}$
$ \forall p\in S,\exists n\in\mathbb{P}:\ p\minus{}1 \equal{} 2^{n}|a_{n}$ so, $ p|a^{a_{n}}\minus{}1 \equal{} a_{n\plus{}1}$
then $ S\subset A$
NOW if we prove that $ |S| \equal{}\plus{}\infty$ (i don't know if it's true or false) the probleme is done.

We use lemma :
Let (m,n)=1 then
$ gcd(a^{m}\minus{}1,a^{n}\minus{}1)\equal{}a\minus{}1$
Then Induction :)

I think that if $ 2^{n}\plus{}1\in P$ then $ n \equal{} 2^{k}$
Then proving that there are infinitely many primes which have the form of $ 2^{2^{k}}\plus{}1$ is not simple.

I 've thought of using the lemma $ gcd(a^{m}\minus{}1,a^{n}\minus{}1)\equal{}a^{gcd(m,n)}\minus{}1$ but it does not work. Please post your full sollution, TTsphn

[quote="mathangel"]I think that if $ 2^{n}+1\in P$ then $ n = 2^{k}$
Then proving that there are infinitely many primes which have the form of $ 2^{2^{k}}+1$ is not simple.

I 've thought of using the lemma $ gcd(a^{m}-1,a^{n}-1) = a^{gcd(m,n)}-1$ but it does not work. Please post your full sollution, TTsphn[/quote]
Ferma deside, that all ${ F_{n}=2^{2^{n}}}+1$ are primes.
New conjecture, that only finite of numbers $ F_{n}$ are primes.

[quote="mathangel"]I think that Fermat numbers are $ 2^{2^{n}}\minus{}1$ and the conjecture is there are finite primes which have the form of $ 2^{2^{n}}\minus{}1$.
Anyway, this does not help solving my previous problem.[/quote]
Obviosly $ 2^{2^{n}}\minus{}1\equal{}F_{n\minus{}1}F_{n\minus{}2}...F_{0}$ is not prime.

I 'm sorry for the mistake. We may imagine how the sequence behaves but we don't know how to prove that it 's true or false.

we take $ f(x)\equal{}a^{x}\minus{}1,f^{k}(x)\equal{}fofof..k\ times..of(x),f^{0}(x)\equal{}x$ and $ V_{n}\equal{}gcd(a_{n\plus{}1},a_{n})$
$ {V_{n}\equal{}gcd(a_{n\plus{}1},a_{n})\equal{}gcd(a^{a_{n}}\minus{}1,a^{a_{n\minus{}1}}\minus{}1)\equal{}a^{V_{n\minus{}1}}\minus{}1\equal{}f(V_{n\minus{}1})}$
so $ V_{n\plus{}1}\equal{}f(V_{n})$ gives $ {V_{n}\equal{}f^{n\minus{}1}(V_{1})\equal{}f^{n\minus{}1}(gcd(a_{2},a_{1}))\equal{}f^{n\minus{}1}(2)\equal{}a_{n}}$
then $ a_{n}|a_{n\plus{}1}$ (i don't know if it help)
so $ \forall n\in\mathbb{N}:\ a^{a_{n}}\equiv 1\ (mod\ a_{n})$

I 've already known that $ a_{n}| a_{n\plus{}1}$ because it 's the purpose of the conjecture. It makes the conjecture difficult. First I thought that I could prove something like $ a_{n\plus{}1}\equal{} a_{n}. k$ where $ gcd(a_{n},k) \equal{} 1$ but I couldn't find such k.
It 's a hard problem, isn't it ?

The problem reminded me of the Euclide 's proof for the existence of infinitely many primes.
I also thought of the following problem when I saw the proof:

      Given a sequence:
                  $ p_{1}\equal{} 2$
                  $ p_{n\plus{}1}\equal{}\prod_{i \equal{} 1}^{n}p_{i}\plus{}1$
Is it true that the sequence $ p_{n}$ has infinitely many primes ?

let ${ \mathbb{P}=\{\prime s}$,and $ f(n)=\prod_{p\in\mathbb{P},p|n}p$
$ a_{n}|a_{n+1}$
if we suppose that $ S=\{p\in\mathbb{P}|\exists n\in\mathbb{N},\ p|a_{n}\}=\{p_{1},p_{2},...,p_{h}\}$ is finite.
then $ \exists N\in\mathbb{N}:\ \forall n\ge N,\ f(a_{n})=\prod_{i=1}^{h}p_{i}$
so $ n\ge N$: $ f(a_{n})=f(a^{a_{n}}-1)$ (it can hlep)


calculus
integration
inequalities
calculus computations

I spend all day tring to solve this integrals but... no luck...

$\int {\frac{dx}{1+x^{2n}}}$, $\int {\frac{dx}{1-x^{2n}}}$  and $\int {\frac{x^{n}}{1+x^{2k}}dx}$

where k and n are integers.
SOS :(
Besides, does somebody have a proof for Chebyshev integral: $\int x^m(b + ax^n)^pdx$.

first integral is here, second and third are similar...
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=447646#p447646]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=447646#p447646[/url]