blogs MATHCOUNTS AMC AIME USA(J)MO USAMO ratio

Richard's latest post on his blog says that there will be no more calculators allowed on the AMC's next year. What are your thoughts?

I'm happy. I never even used calculators this year on 2007 AMC A and B, so this won't affect me. Other people with TIs and Graphing calculators will probably be sad though.

At least they will have to solve it the old fashioned way; with their heads only and no extra brains.  :D

I won't really care that much, other than question 1 will be a little more annoying. I don't rely on the calculator that much, but I know some people who really bust out programs during the competition who will now have to learn the elegant way to do things.

Hmm...this is interesting.  I don't mind it because it does force people to think, but I would be disappointed if they asked us to do brainless calculations by hand.  I like using my calculator to check my answers (and even to make educated guesses on some of the problems, especially ones related to graphs), but I guess that that's not an option.  Does anyone know how this will affect the test, if at all?

Aww.. now I cant type in like sin75 into my calculator and go frmo there, I actually have to write out (sqr6+sqr2)/4 and stuff(like on this years amc 12b #19, I actaully found the triangle, with the sides 2.0943.. and 5.62589.. and 6), so i just did 2.0943../6, which is pi/9.. but without a calculator, i will have to actually try and divide taht by hand and then divide pi/9 by hand to check if they are right).. and ya problems will be more annoying.. but if they take away caculator and add like 15 mins.. ill be okay..

Or if they take away the calculators and the qualification score drops by another 4.5 points.

Well, hopefully, the questions won't be as calculator-needy. The index may go down a few points or so as well. I must admit, that I am a calculator freak as soon as I hit problems 15 - 25... I sometimes program stuff. I also like to solve systems of equations using calculators... particularly with Gaussian elimination, which requires taking determinants of matrices (sometimes). And I don't really like taking determinants by hand, but that's okay, I guess.

._o gaussian elimination requires determinants? what kind of gaussian elimination are you doing...

i personally think it will be alright; as was stated, no problems required the use of a calculator, and that was really quite true

I'm happy that there are no calculators, because i've never found any need for them ever... except sometimes they might save me 3 seconds from multiplying or whatever. I prefer to do most things by hand. I mean like, in MathCounts Target and the AMCs, usually I forget that my calculator exists and I end up doing everything by hand... even if I put it 2 inches away from my paper or something I still forget it's there.

To be honest, my score will plummet. I am [i]very[/i] dependent on Mathematica and my 89; it's not that I can't do the computations by hand, but I'm slow, and make many mistakes if I don't quadruple-check.

Interesting.  I always thought the AMC system struck a good balance between calculator and no-calc events.  The first round, which has to be accessible to everyone, allowed the calculator, and I think they did a very good job of writing TI89-proof questions.  Then the AIME, which really makes the difference in USAMO qualification, disallowed them and gave plenty of time to solve anything without one.

My biggest worries are:
(1) Time might become a bigger issue on the AMC's.  It's not too big of a hindrance now, but if the calculator ban makes it much more time-consuming, time issues will get a lot more annoying.
(2) I really, really hope this doesn't scare students away.  The AMC is well-established enough that all serious students will do it, but for the kids who wake up that morning and think "hey, AMC's today, do I feel like taking it?", I feel like some of them would be a lot less likely to answer "yes" to that question if they know they can't use a calculator.  This is especially true if said students are not particularly confident in their abilities.

Strange that I'm saying this, given that I'm normally a huge advocate of no-calculator tests.  But the use of calculators on the AMC has never been much of a problem, IMHO.  Then again, they're ridiculously good test-writers, so it's quite probable that the lack of calculators won't be a problem either except for possibly (2) above.

I think that not allowing the use of a calculator will discourage some of the less-experienced students to take it.

I usually use calculators when I want to perform calculations that would be annoying to do by hand.  For example, on the 2007 AMC 10A Problem 6, I used a calculator to compute the necessary ratios to save time.

So what am I saying?  Basically, they probably should keep allowing calculators on the AMC.  Even with them, some of my friends find the test to be pretty hard.

Honestly, I don't want the AMCs to deteriorate into Mathcounts contests, because letting the AMC become a contest of speed would just be madness.

Hmm... this is my personal standpoint, so it seems I should give a little information about myself.  10th grade, 3rd time AIME qualifier, hopefully (if USAMO floor is at most 10ish) 2nd time USAMO qualifier.  i own a TI89, but used a regular 83 on the AMC 12B.

I found myself missing my 89 for 17 - didn't know how to do determinants on the 83, tried expansion by minors, left it blank b/c i didn't care about getting a perfect score anyway.  At any rate, I trust the AMC directors enough that I believe that all future problems without calculator usage will be more like those on the AIME - some casework may be involved, but most of the math will be the quiet, elegant stuff required to solve problems like #15 - a homothety, a few area formulas,  and cotangents of half-angles using the incircle made the problem a breeze.  I think that in some cases computers are valuable tools for a real mathematician - why else would Turing completeness, big/little O notation, and P/NP have come into existence?  But I feel that, except for tedious calculations (i.e. transcendental functions, graphing, matrix algebra) most real mathematicians tend to use non-computational methods to solve problems.

Enough about real mathematicians and back to the AMC.  I agree that the no-calculator policy will scare away some people.  I also feel that among AoPSers, most will not be impacted, as it will make computations far easier, and will mean that more critical thinking will be involved.

On the subject of the scientific/graphing schism, I have two comments: do little old librarians in Arkansas/other state (sorry, first thing that came to mind) read the AoPS forums? No.  Also, banning ONLY graphing calculators makes students more likely to try and sneak in usage of a 83/89/other.  Banning calculators altogether makes it far easier on the librarian.  If she sees one, then...SMACK SMACK SMACK.  :P

Oh well.  I support the decision.  Less heavy calculation, more elegance.  It benefits me.  And to any who would reply that a calculator is no smarter than its user, can you do logarithms to twenty digits in less than a second? NO.

Or, on another front, could you design a symbolic algebra program with as much functionality and breadth as that of the 89?  Probably not.

(And to whoever mentioned Mathematica - wtf?  don't think you can use that on the AMC 10/12.)

I am COMPLETELY against this ban.

1. Very few people own 89's AND know how to use them. In my grade of 500 only 3 people (including me) have 89 and can use them more effectively than someone with a 84. I think the effect of 89's on the cutoff AMC scores is quite minor.

2. I think that those of us who can't add and multiply still deserve to make the AIME. Probably not the USAMO, but we deserve to make it to the AIME.

3. The AMC is designed to be approachable. The first couple are always very easy because it is made knowing that everyone taking it isn't brilliant. Less kids will take the AMC if calculators are banned because they will say, "I only got 5 right last year with my calculator. This year it's hopeless."

4. The real issue is that the AMC is too easy. It should be made so that there are maybe 5 problems at the end that count for about 10 points each. These problems would be AIME style and would force people to find the elegant solution because their graphing calculator won't help them. Having 50 points that are "89 proof" will prevent people from making it to AIME completely due to their 89.

5. Many people have suggested a partial ban. This is a bad idea because it will only encourage mixed calculators. Kids will come in with 84's with symbolic algebra or 2 line displays that can still graph.

6. Many people say that allowing 89's favors the richer students who can afford them. This is absolutely correct and irrelevant. The AMC's have always favored the rich. To do well you must go to a good school with good teachers and be able to afford study materials. I think that everyone who makes it USAMO and probably most who make it to AIME could afford a TI-89 if they really determined that they needed it.

i completely agree with piac31. Just a suggestion, but the AMC's should also have a greater time limit, as they test mathematical ability, not speed.

This topic seems to have a variety of answers, but it seems to me that the only aid a calculator should provide on a test that tests [i]mathematical ability[/i] is computational aid.  In that sense, all this talk of TI-89's and whatnot is irrelevant; a TI-32 can do computation just as well as a TI-83, TI-84, or TI-89.

That being said, an elimination of calculators should only press the need for time, seeing as careless mistakes are not a big deal on this test, and brute force calculations that only calculators can do shouldn't be necessary in solving AMC problems.  This test already tests quick problem-solving, presenting 25 problems in 90 minutes,  a much greater problem to time ratio than the AIME or USAMO, so again, the ban of calculators should not throw people off that much.

Finally, this decision was just made recently... I doubt it would be revoked so quickly after who knows how much careful deliberation went into thinking it over.

I dunno, personally, I don't think it matters very much. I only used my TI-83 to multiply two-digit numbers that would've taken only a few seconds to do by hand anyway.
As for the 'I deserve' to make it into the AIME with calculators...as the AMCs are not very calculation-heavy, I hardly think it'll make a difference. I truly think it won't.

[quote="randomdragoon"][quote="undefined117"]on the issue of programs on graphing calcs.. they could reset (RAM and archive) all graphing calculators, and this only takes a few keystrokes on a TI-89..

but I still think at least four-function calculators should still be allowed..[/quote]
The RAM clearing doesn't actually work all that well ... for a regional chemistry contest that I took, the requirement was that all graphing calculators' memories were to be cleared. However, my friend (to save his games) archived and hid all of his games, and they only did a simple RAM reset (which does not touch the archive) and all of his games were still there. Which is to say, it would be easy to do that with cheating programs as well.

I think [i]many [/i]people would get completely pissed if you reset the archive as well.

In addition, I have two graphing calculators that look exactly the same. I could have easily switched one in for the other after they cleared one, but I didn't because I didn't have any cheating programs on either calculator. But I could have.[/quote]

ahh.. good point. of course, one could back up their calculator prior to their test, but sometimes TI Connect does not work so well.

hehe and I got #1 wrong on the AMC 8 last year because I mistyped a number on my calculator; I don't particularly mind banning them :lol:

[quote="jrshoch"]  This test already tests quick problem-solving, presenting 25 problems in 90 minutes,  a much greater problem to time ratio than the AIME or USAMO, so again, the ban of calculators should not throw people off that much.

[/quote]

Actually, I believe there are only 75 minutes allowed for AMC10/12.

I do hope that the CAMC will limit the number of calculation-intensive problems on the AMC10/12.  This seems only natural, and similar to how they decreased the difficulty of the first 10 questions while lowering the omit value to 1.5.

It is 75 minutes
I remember in 7th grade:
DANG!
75 MINUTES!!!
That's WAY TOO MUCH TIME!
haha
u have to remember it was my first time taking the AMC-10 and all the contests I had done so far stressed speed

I took my time and knew I could omit 5 questions (this was the year before they lowered the points for omitted questions) and still make AIME (my goal at the time)
Really...my TI-89 Titanium made no difference in my score
I did make AIME (with a 120.5)

For those wondering why I didn't take the AMC-10 this year was because my teacher decided that our math team shouldn't take it (or she just forgot. i don't remember)
Even if I took it
I would've been majorly disappointed still
I took a practice 2007 AMC10A and got like a 99 point something
I took a practice 2007 AMC10B and got like a 109 point something
both lower than what I got last year

I agree that taking out the calculator part out of the competition will lower many scores of math programmers
but
not the scores of math computators (who pride themselves on speed and accuracy)
and
definitely not those who are math problem solvers (who are usually very good with math concepts but sometimes make dumb /careless computation errors)

I think if math problem solvers take some time to work on their computation skills, there would be less complaining/begging for calculators to be allowed

I personally would like to have calculators on the AMC's
since
I like to double check and even triple check my computations
but then again
I am very inexperienced in high school AMC's

This just means I cant use alternate methods for solving a problem and it will take longer to solve a problem

For example on the amc 10a #18, I used a coordinate plane and I actually found the side lengths of the quadrilateral using distance formulas and then I split the quadrilateral into two and one was right triangle and the other one I used Heron's formula-there obviously a lot easier way with similar triangles, but that may not come to me very fast

also amc 10a #19-Im not going to simplify a^4+a^(-4) by hand...

and I also remember a problem I did on an old amc 12 that required the laws of cosines and it dealt with huge numbers-i can easily square the numbers and apply the laws of cosines but that would take like a couple minutes longer than using a calculator
and also on simply problemes like 12C3+10C4+8C5.. or something like that-i can do that-but with a caculator i wont make a caculatoring mistake and it would be a lot quicker

and for some problems, I get a long complicated equation. If I had a caculator, I would just plug in all the 5 answer choices and see which one fits the equation instead of trying to simplify the equaiton..

I dont care if theres no caculator, as I can solve most of the problems but my main concern is the time limit-im not very fast at doing problems (especially without a caculator)-so I might have a tough time finishing all the problems..even though I can do all of them..
So if you want to ban caculators, make the time limit like 90 mins for 25 questions,which is still very quick..

ya, i think they should make the time limit like 100 minutes. 4 min per problem seems fair...

my mistake on the 75 minute/90 minute thing.

maybe that's why i always feel rushed...

then again, thank goodness it's multiple choice. at least you can somewhat check your work. if it wasn't multiple choice i would totally be against this ban...

Lets just hope that they don't give a long computational problem with the most likely computational errors as the other four choices :roll: .  Seriously, I don't think they will do something like that (although judging from AIME 1... :mad: ).  This is AMC, not MathCounts.

I only use the regular functions (arithmetic + trig sometimes) on my calc anyways, both of which I can get around, so I hope I'll be fine with this new rule.

But it wouldn't hurt to still allow them.  :wink:  It would save me a lot of time.


Enjoy.

http://www.mathlinks.ro/viewtopic.php?t=15125
http://www.mathlinks.ro/viewtopic.php?t=164572

[quote="aadil"]so the andreescu lemma and titus lemma are the same?[/quote]

Titu Andreescu is one person.  :D

Acctually this is only a special case of famous cauchy-schwarz inequality.But
I think it is really useful ineqality.It says

a*a/p + b*b/q  >= (a+b)*(a+b)/(p + q)

With Equality iff a/p=b/q

The best example is

a/(b+c) + b/(a+c) + c/(a+b)=a*a/(ab + ac) + b*b/(ab+bc) + c*c/(ac + bc)

>= (a+b+c)*(a+b+c)/2(ab+bc+ac) >= 3/2


geometry
parallelogram
complex numbers
circumcircle
geometry unsolved

A set $A_1 , A_2 , A_3 , A_4$ of 4 points in the plane is said to be [i]Athenian[/i] set if there is a point $P$ of the plane satsifying

(*) $P$ does not lie on any of the lines $A_i A_j$ for $1 \leq i < j \leq 4$;
(**) the line joining $P$ to the mid-point of the line $A_i A_j$ is perpendicular to the line joining $P$ to the mid-point of $A_k A_l$, $i,j,k,l$ being distinct.

(a) Find all [i]Athenian[/i] sets in the plane.
(b) For a given [i]Athenian[/i] set, find the set of all points $P$ in the plane satisfying (*) and (**)

Denote by $M_i,i=1,..6$ midpoints of the segments with endpoints in the given four points.
It can be proved  (if it is not known) using complex numbers that the three lines joining midpoints of non-adiacent edges of the figure formed by $4$ points in the plane have common midpoint.
The point $P$ which satisfy the condition of the problem must lie on the three circles of diameters, with endpoints in $M_i$ of non-adiacent edges.
Since the three circles have the same center they must have the same radius, since $M_iM_jM_kM_l$ is a parallelogram with equal diagonals, it is a rectangle. And since it has edges parallel with $A_iA_j$ and $A_kA_l$ implies that, if no two midpoints coincide, than every two non-adicent edges of an athenian set, are pairwise perpendicular. Which is possible iff it is square.
Hence, in every case, two of the midpoints coincide.
Hence, the four points of an atthenian set must form a parallelogram, with pairwise perpendicular edges and the distance between every pair of parallel edge is the same, hence, a square.

Te set of the points which have the desired property, is the circumscribed circle of the square formed by midpoints of the edges.

Sadly, the above proof is wrong.
The main error in the proof lies in the following statement:
[quote="xirti"]
And since it has edges parallel with $A_iA_j$ and $A_kA_l$ implies that, if no two midpoints coincide, than every two non-adicent edges of an athenian set, are pairwise perpendicular. Which is possible iff it is square. [/quote]
This is not true. A square does not have non-adjacent edges perpendicular, only the diagonals are perpendicular, the other two pairs of opposite sides are parallel not perpendicular.
On the other hand, $A_2A_1 \perp A_3A_4, A_3A_1 \perp A_2A_4, A_4A_1 \perp A_2A_3$ is the same as saying that $A_1$ is the orthocentre of $\triangle A_2A_3A_4$, in other words, $A_1,A_2,A_3,A_4$ form an orthocentric system.
For the sake of simplicity, let's call the points $A,B,C,H$ with $H$ being orthocentre of $\triangle ABC$. Then the locus of $P$, as xirti has noted, will be the circle with midpoint of $AH$ and midpoint of $BC$ being diametrically opposite points. But this is the Nine-Point Circle.

So, to conclude, $A_1,A_2,A_3,A_4$ are an orthocentric system, and the locus of $P$ is the common nine point circle of any triangle formed by those points.


analytic geometry
trigonometry
trig identities
Law of Cosines

[quote="Ceva Theorem"]Show that the distance formula between two polar coordinates, $(r_{1},\theta_{1})$ and $(r_{2},\theta_{2})$ is given by $d = \sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta_{2}-\theta_{1})}$

[hide="hint"]
It looks very similar to the Law of Cosines.
[/hide]

[hide="more hints"]
$x=r\cos\theta$
$y=r\sin\theta$
[/hide][/quote]

[hide]Use law of cosines

Draw a line between the two points, call this distance $d$.
Draw a line from $r_{2}$ to the origin and $r_{1}$ to the origin.  These three lines form a triangle.  The angle opposite $d$ is given by $\theta_{2}-\theta_{1}$, so by the law of cosines
$d = \sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta_{2}-\theta_{1})}$

Not to belabour the point too much, but $r_{1}$ and $r_{2}$ correspond to a and b in the formula $c^{2}= a^{2}+b^{2}-2abcosC$.[/hide]


Let $ \triangle\ ABC$ and $ M\in (BC)$ so that $ \left\|\ \begin{array}{c} m(\angle MAB)\ge 90^{\circ} \\
 \\
2\cdot MC\ge BM\end{array}\ \right\|$ . Prove that $ CM \plus{} CA\ge\ AB$ .

I think with these conditions the inequality is not true, see figure

[quote="Virgil Nicula"]Let $ \triangle\ ABC$ and $ M\in (BC)$ so that $ \left\|\ \begin{array}{c} m(\angle MAB)\ge 90^{\circ} \\
 \\
2\cdot MC\le BM\end{array}\ \right\|$ . Prove that $ CM \plus{} CA\ge\ AB$ .[/quote]

I  presume triangle is acute angled. then draw M' such that BC=CM'.also draw A' such that BC=BA'/ notice, BA'M' is right angled. the inequality should now follow.

[b]Thank you[/b] and [u]sorry[/u], [b]Encyclopedia ![/b] I corrected above.

No problem Mr Virgil Nicula, but for me a question $ M\in (BC)$ it means $ M$ lie on segment $ BC$ or line $ BC$?

[b]YES[/b], [u]Encyclopedia[/u]  ! I wrote $ \boxed {\ M\in (BC)\ }$ , i.e. $ M$ belongs to the [u]open segment[/u] $ (BC)$ !


MATHCOUNTS algebra polynomial function geometry

This was adapted from a MathCounts warm-up:

Prove that the [u]only[/u] three perfect squares that add up to exactly 169 are 9, 16, and 144.

I tried brute force (educated, of course!!!). But I'm sure there's a better way.

[hide="hint"] mod 8 [/hint]  However, there is still an aspect of experimentation.[/hide] (if thats a word =)

  O' and you're signature is impossible, mathematicians have proved no polynomian has just prime roots, a contradiction.

[quote="G-UNIT"]O' and you're signature is impossible, mathematicians have proved no polynomian has just prime roots, a contradiction.[/quote]He's claiming to have a found a polynomial $p$ such that $p(n)$ is always prime. His polynomial was $p(x)=5$, but of course a constant function that generates primes is trivial; it was a joke. ;)

I tried mod 4, divisibility by 4, and a whole lot of brute force. I'm hoping to reduce the amount of brute force while increasing the other areas, or ideally to solve this in ten cases or less.

Yeah, it was a joke.

........ I was being sarcastic as well..........

I don't think mod 8 works well, there's still a ton of guess and check to do.

Mabye it was mod 4 or something, I remember seing a proof about this, that the numbers had to hold a certain parity, making it alot easier to list the squares...

$5^2+12^2=13^2$ is the only representation of 169 as a sum of two perfect squares

$3^2+4^2=5^2$ is the smallest primitive Pythagorean triple

So $3^2+4^2+12^2=13^2$ is the only representation of 169 as a sum of three perfect squares

Done

[quote="10000th User"]$5^2+12^2=13^2$ is the only representation of 169 as a sum of two perfect squares

$3^2+4^2=5^2$ is the smallest primitive Pythagorean triple

So $3^2+4^2+12^2=13^2$ is the only representation of 169 as a sum of three perfect squares[/quote]Well done.  :)

I don't understand the reasoning behind your first statement. Please prove.

Also, is there any way to solve this problem without guess and check? If you don't already know the answer, that is. As opposed to proving it.

Thanks!!!

By the way, for those commenting on my signature, I changed it already.

but the mistake with the 3,4,5->5,12,13 solution is that you are assuming that adding 2 of them gives you a perfect square

[quote="dondigo"]And how about this:
$0^2+0^2+13^2=169$[/quote]

I think he meant for positive integer . Or else you might have like this

$(-3)^2+(-4)^2+12^2=169$ or $12^2+5^2+0^2=169$ as well

 :P

[quote="Scrambled"]Theres a well known theorem that states that all primative pythagorean triples are in form a^2-b^2, 2ab, and a^2+b^2 where a and b are any number.
Clearly, the only way to sum two squares to get 13 is 3^2+2^2, and thus the sides are 9-4, 2*2*3, and 13[/quote]

Ummm... I was trying to get three squares to sum to 169, not for two squares to sum to 13.

Yes, I meant positive integers. :lol: Sorry for the confusion.


AMC AIME USA(J)MO USAMO geometry \/closed

Last year, as a freshmen, I got 105.5 on AMC 12 and 2 on AIME (would have been 4 without dumb mistakes). I had no idea what the test was until the day our math class took it, so I didn't study at all. I feel like I'm missing some basic ideas in AMC-level problems simply b/c I haven't learned the tricks yet (so Introductory-level courses might work, but I know already 75% of the stuff covered in that level, so I feel like I'm wasting money just to learn a few easy tricks, since I can deal with some AIME problems). In AIME, some of the material is way over my head, but I feel I can learn it with hard work. The USAMO problems are impossible for me at this point. I've already taken the sample tests, but I'm just curious, what class level do you forum members think I should I take: introductory, intermediate, or advanced? Thanks!

I'd recommend using the Introduction Post-tests as a guide.  If you can solve 80%+ of those, then go for the Intermediate level classes.

When will the other classes be on the Enroll list and have their own Do you need this and Are you ready tests?
For example Into to Geometry isn't up there yet.

We expect to have the rest of the classes listed next week.


limit
real analysis
real analysis solved

1   Let  $a_1,a_2,...,$be positive numbers such that $\sum_{n\geq1}a_n<\infty$. Prove that there are positive numbers $c_1,c_2,...$ such that
$\lim_{n\to\infty}c_n=\infty$ and $\sum_{n\geq1}c_na_n<\infty$

Put $s_m=\sum_{n\ge m}a_n$. We know that $s_m\to 0$ as $m\to\infty$, so we can find a sequence going to $\infty$, $(m_k)_{k\ge 1},\ m_1=1$, s.t. $s_{m_1}+s_{m_2}+\ldots<\infty$. Now take $c_n$ to be the number of indices $k$ for which $m_k\le n$.


Hi, i am a student.
tell me more about art of problem solving introduction to number theory
by mathew crawford.
Can anyone suggest some really good books about elementary number theory (in English) which degree of difficult is usamo~putnam and explain theorem include kummer's , Lucas's, Wolstenholme's , Thue's , Hensel's theorem.

which is better
 [u]Introduction to the Theory of Numbers (Hardcover)
by Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery  [/u]
or
[u]A Course in Number Theory (Oxford Science Publications) (Paperback)
by H. E. Rose?[/u]

This book is no where near IMO/USAMO difficulty. This book is more like AMC 10/12 difficulty, for example, Fermat's little theorem isn't even mentioned in it.
However, It is a great book.


There's only one thing that can be done - Sophie Germain's identity

$x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2)$

Now telescope.

thanks.
by the way can you tell me where i can find books/examples that will improve my factoring abilites.

I'm going to start deleting your posts if you keep saying 'meow'  :alien:  ;)

Okay, okay. :P  Won't do that anymore. Forgive me. Just testing something I learned in psychology in real life (in my posts). Something about how people tend to not notice words that are not important to what they are reading. Hmm..... you probably already knew how to solve the problem, then, joml88. ;)

Hmm, well I saw your meows in your other posts as well...it was starting to irritate me  :D

Well, I was seeing how much times I could post it without anyone noticing, to be honest. ;)

Anyways, a  distinctly related problem I found was to show that
$4^n +n^4$ is always composite. Okay, its barely related at all. But hopefully, it will help our Communist friend maokid (just kidding :P ).

well here is what i have for that
$4^n=2^{2n}$
$a=2^n$ so $a^2=2^{2n}$
$n^4+a^2=n^4+2an^2+a^2-2an^2$
$(n^2+a)^2-2an^2$
$(n^2+a-n\sqrt{2a})(n^2+a+n\sqrt{2a})$
so substituting back we have
$(n^2-n2^{\frac{n+1}{2}}+2^n)(n^2+n2^{\frac{n+1}{2}}+2^n)$

is that correct or do i have to simplify further. :?

[quote="mna851"]Perhaps you're a SuperTroopers fan 1234567890?[/quote]

Superwhat? :P

Maokid: you're basically right; finish it up ;). But nice factoring. :)
[hide="Hint"]When $n$ is even, its obviously composite.
When $n$ is odd.......
Can you do anything with this? You already have most of the problem done.
 [/hide]


ratio
function
geometry unsolved
geometry

$\frac{BE}{AE}= ?$

This is Inspired in Shahzada, :)

It's a good problem and the solution is very simple

Thank you yagaron nice question. :)
[hide][img]http://img47.imageshack.us/img47/9779/gregrqs2.png[/img][/hide]

This answer is another answer that I do not use a trigonometric function for.

Shahzada  thanks for the simple solution

Kouichi Nakagawa  your solution is Sugoi.
and your image is very beautifull too! yours circles have the same colors than mine. hehe

 what's program did you used? i like the way of it represent the angles. could you tell me?

thanks

Thank you for yagaron.
I prepared a color of your figure and tried to make it :)

The mathematics software that I use well mainly is as follows.

[url=http://www.wolfram.com/]Mathematica 5.2[/url]
[url=http://hp.vector.co.jp/authors/VA017172/]FunctionView[/url]
[url=http://www.geogebra.org/cms/]GeoGebra[/url]

I made it with GeoGebra this time.

kombawa Kouichi Nakagawa

this site is sugoi!! have a portuguese version.

thanks and oyasumi  :D

I got such an answer from Rascal when I introduced this problem in [url=http://mixi.jp/]mixi[/url].
In this answer is an elementary explanation and briefly, too.

mixi (ミクシィ Mikushi) is one of several SNS (social networking service) sites in Japan.

Let me post an other solution. For easier reading, I use $\varphi = \frac{\pi}{8}$
Let $w_{1},w_{2}$ be the circles with centers $B,D$ respectively and $r$ the radius of $w_{1}$.

Let $F$ be the symmetric point of $A$ w.r.t. $DE$. Then $\angle xDF = \angle xDA = \varphi+\varphi = 2\varphi$
Then $AE=FE$
The point $F$ is also symmetric of $C$ w.r.t. $BD$. Indeed, $DC=DF$ and $\angle BDC = \angle BDF = 3\varphi$

So, $\angle CFD = \varphi, \angle CFB = 2\varphi, \angle DBF = 2\varphi$, and from the right triangle $CBF$ we get $FC=r\sqrt{2}$

Also, $F\in w_{1}\cap w_{2}$

Now $\angle FDE = 2\pi-(4\varphi+2\varphi+\varphi+\varphi+2\varphi) = 6\varphi = \angle CDF$. So $FE=FC = r\sqrt{2}$
Since the triangle $CFE$ is inscribed in $w_{2}$ we get $\angle CFE = \frac{1}{2}\angle CDE = 2\varphi \Rightarrow  \angle DFE = 2\varphi-\varphi = \varphi$

$\angle BFE = 2\varphi+\varphi+\varphi = \frac{\pi}{2}$, so the triangle $BFE$ is right

$BE^{2}= BF^{2}+FE^{2}= r^{2}+(r\sqrt 2)^{2}= 3r^{2}\Rightarrow  BE = r \sqrt{3}$

Finally $\frac{BE}{AE}= \frac{r \sqrt{3}}{r \sqrt{2}}= \frac{\sqrt 6}{2}$


Will there be a message board for the special AIME problem seminar?


Go ahead, post what you think it takes to get in! :cool:

I'll send you my guaranteed ten step plan for just seven monthly payments of $12.99.

Seriously, although I didn't apply, I know that there's no foolproof method.  My only recommendation would be not to focus on getting into RSI so much.  Go do some research if you think it would enjoy it, not because you think it's what RSI admissions wants to see.  I don't imagine the admissions committee would look favorably upon an applicant whose only goal in life is to get into RSI.

I agree. And I do have research that I am conducting. However, RSI is so competitive that that is not enough. That is why I am asking people what they think it takes.

there's a huge discussion (hundreds of posts) about this very thing on [url]http://www.collegeconfidential.com[/url] Most of it was last fall, right before people applied. So just search the archives for "RSI".

What is RSI?

Edit: Turns out Google does work.  Sounds pretty rockin'.

When I searched for RSI, I got a whole bunch of web-sites about some computer disease.

Instead of "RSI," type "Research Science Institute."

Here's a link to a [url=http://www.boston.com/news/local/articles/2004/07/29/tinkering_with_their_minds/]news article[/url] about science programs for young people.

Hey I just got back from RSI around 72 hours ago, and I can say that NO ONE KNOWS HOW THEY GOT IN!!!  The truth is if you stress about it you probably won't.  I'm sure there where a few people that had an SAT score in 1300-1400 range there .  I also know a few people that had perfect scores on every College Board test they'd ever taken.  Most had of previous research experience, I personally didn't though.  Don't stress, do what you love, don't do things just to get you in.

There I've said it, now that I'm done I encourage all of you to apply to RSI it'll be one of the best times of your life I guarantee.

First, it is true that everyone at RSI has no idea how they got in.
But I compared notes with everyone, and I really have no idea how i got in.

 A lot of the people there have done previous research and have won this olympiad and that olympiad and that competition and this competition.  Almost all had SAT scores in the high 1500s - high 1600s  :)

But I dont have any of that.  I never did any previous research; i never competed in any olympiad;  i never did that well in any math competition either.  I had a 1440 when I applied.
My application essentially told the admissions committe that I had done high school.  That was about it.

So, I can confidently say that there is no foolproof way to get in.  I have no idea what the admissions committe focuses on.

However, even if you don't think you can get in, apply anyway.  It is the most amazing program on the planet and it will be the best summer of your life.  RSI is absolutely incredible!!

go to [url]http://www.cee.org[/url] and click on research science institute.  you can read all about the program there and download an application.

i called them and asked them to send me a brochure and application in the mail, which was nice because it was tangible and gave me more motivation to actually get the application done.  The brochure also has more information in it than online, i think.

Anyway, do the application, get some recs, send it in and wait.

Good luck,
David Stiebel

haha, kevin.

actually, i asked the Director, Matt Paschke, why I got in.  he said that he remembers the admissions committee discussing the dual curriculum that my school offers.  Something like that.

anyway, i don't think my sister had anything to do with my getting in.

btw, how's your mom doing?  tell her i say hi.  ;)

I'm interested in RSI.  I looked on the website, but the application seems to be for RSI 2004.

Well, you can always send them an e-mail at rsi@cee.org asking for a brochure and the application materials. I am also going to be applying to RSI this year, and hopefully I get in :)

Ok let me just say i put ALL of my efforts into getting into RSI... and my Sats were nothing special neither were any of my tests, i did alot of extra research many projects many extra curriculars many awards blah blah blah, and i come from a VERY small crappy jewish academy so that cant be it! its about 20 kids... and dual curriculum blah blah blah, i wake up 620 to get there to pray by 730 and we have class till 6:15, i get home at 730 and thursday nights i get home 930 because of extra learning, also sunday school! although my sister is twelve so she isnt hot, but as you can see its not just the dual curriculum that decided it, you probably were competing against me for diversity, which they want to see, and as seeing as you fit the Token small jewish school kid, you probably ot accepted over me.

I know how u feel.  I hate all the admissions bs so I've just decided that I am going to do what I want and what I feel I need to (I'm not going to force myself to start doing something I hate just because it will "look" good and on my application and I can say how much I love doing it...).  I guess I will end up wherever is right for me.

From my own experiences (so far) here at RSI, as well as my meeting other people and talking to them, here's how you get in:

1. Demonstrate you either a) have the knowledge to do well in your chosen field, or b) that you can learn fast, well, and under serious pressure.
2. Demonstrate a love and passion for your subject.
3. Demonstrate that you can bring something powerful to RSI. Not necesarily Olympiads or anything -- even a unique personality will do.

FYI: RSI is pure merit. There is no racial/ethnic/gender system governing RSI admissions.

Yes, absolutely. Sran hits the nail on the head with his post. Another point worth mentioning is that quotas are enforced too. So you need by far a better profile to get in from California (and any overrepresented state) than, say, Montana or any other underrepresented state.

Hmmmm....Everybody seems to say that trying won't get you in...
But if you truly love research/working in a lab and you try to emphasize that, then wouldn't that seem like trying too hard?  I mean, if you really want to spend your summer at RSI meeting cool people, learning cool stuff, and doing research, shouldn't you try to do everything to get in?

Trying will get you in. I've wanted to go since I found the program (7th/8th grade), and now I don't care about college as much, I already got what I was gunning for.

Question: will not competing in any science fairs (westinghouse, isef,...) hurt my chances?  I just started serious research, but I won't have enough time to enter westinghouse, and isef is way after the app deadline.  Can I do research and send them my project when I apply?

Virtually everyone enters Westinghouse [b]after[/b] going to RSI, not before.

My feeling (as both an RSI alumnus and a former RSI instructor) is that what they are really looking for is an aptitude for and interest in research, as opposed to a CV of pre-existing research results.  To demonstrate the aptitude, you submit your transcript, PSAT scores, contest results, etc.  To demonstrate the interest, you write essays.

Could you take a look at my stuff and tell me how to improve?

9th:
AP Stats = 4

10th:

AP Calculus BC(Self-study) = 5
AP Computer Science AB = 5
AP Physics B = 5

SAT Dec 2004 = 1410

I am entering Siemens Westinghouse this year with a project in Computer Science/Artificial Intelligence. I am not sure about making semifinalist status(the project was entirely self-concieved and done, so it isn't as good as if I had a uni mentor).

I'm taking the PSAT(haven't studied much, because I've been exhausted from Siemens and classes), and I'm thinking 2100+.

I live in NC, and go to NCSSM(North Carolina School of Science and Mathematics).

I did a research intern at UNC-Charlotte in Computer Graphic dept over the summer.

I'm looking into getting a research intern at DUKE.

My classes this trimester are:

AP Physics C
AP Chem
AP US Hist
AP Eng. Lang
Intro Spanish(am not strong enough to take AP French[took french III] last year)

and most likely Diff Equations from EPGY, Multi in winter and spring trimesters for sure.

I like doing research in Computer Science, but I'm not very great at math competitions, and I do USACO, but I didn't do to well last year(practiced a lot over the summer).

PS: Can we talk about independent projects, like chess engine i built, graphing calculator applet, etc?

Thanks, I know this is a lot, but I've never known that RSI people are on forums.  :P

What is usually the deadline for applying to RSI? :)

sometime in mid-early January -- check their website during maybe October or November next year.

[quote="kewi"]Thanks :)

And...to apply to RSI do you need to submit your SAT scores?I mean, I haven't sat the SAT yet so if I have to submit my scores then I'll have to do a little practice and stuff and then have the exam :D[/quote]

I think it is recommended (either SAT or ACT)... and I think PSAT may be required? Don't quote me on that though -- I took SAT and PSAT so I am not sure which is required or not.

On "Essays" 5-7 in RSI, the one's that asks you about extracurriculars (related and unrelated) and past programming experience. Do you just make a list and let your activities speak for themselves?  Or do you write it in prose form?

sry for asking this for the billionth time, but do you know if brits or romanians can go to rsi?
cuz i know british people attended rsi in the past but for this year, i got an email saying that brits couldn't attend rsi.

do you know why that could be?

I saw that for 2016 they accepted 3 homeschoolers, one from my home state actually (PA). I'm a sophomore now: Do you guys think I'd have a chance for next year?

So all the kids talking about their SAT scores, 1440, 1600, 1750 etc, is this the SAT that's out of 2400?

@above yes.

I think just showing a passion for research should be enough for RSI.

No. The old SAT scoring system had 1600 as a perfect score... I think

Well I know the NEW one does... At least 5 years ago, the old one was out of 2400, in which case those scores are low...

The old old one had 1600 then it was 2400 and now the new one is 1600. But by old as in the time of the OP (2004) then it should be 1600


arithmetic sequence
number theory unsolved
number theory

Find all integers $ a$ such that $ 2a^2\minus{}1$ and $ 3a^2\minus{}2$ are both perfect squares.

The numbers $ 1,a^2, 2a^2\minus{}1,3a^2\minus{}2$ are perfect squares in arithmetic progression, is well known that it must be the constant progression, thus we get $ a\equal{}1$

I wonder why it must be the constant progression.
Could you explain it?

The proof is hard. See here [url]http://www.mathpages.com/HOME/kmath044/kmath044.htm[/url].


algebra
polynomial
quadratics
search
number theory proposed
number theory

Prove that $ x^4\minus{}10x^2\plus{}1$ is reducible over $ R$ and $ F_p$ for any prime $ p$, but not over $ Z$.

$ x^4 \minus{} 10x^2 \plus{} 1$ is the minimal polynomial of $ \sqrt{2} \plus{} \sqrt{3}$ over $ \mathbb{Q}$ and in $ \mathbb{F}_p$ for any $ p$ at least one of $ 2, 3, 6$ is a quadratic residue (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=785743715&t=263897]this thread[/url]), so the minimal polynomial of $ \sqrt{2} \plus{} \sqrt{3}$ is quadratic in $ \mathbb{F}_p$ for every $ p$.


There are 4 blue marbles, 5 red marbles, and 3 green marbles in a hat. What is the probability of selecting 3 marbles without replacement, and having less than 2 blue marbles?

[hide]well, since there are 12 marbles the chances you will get 0 blue marbles will be 2/3*7/11*3/5=14/55

chances you will only get one blue marble would be 7/165

add them together and you'll get 49/165. so i guess that's the answer.[/hide]

Leeyuhang, are you sure about  your exactly one blue marble case?  That seems too small to me.

Why do you have the 1's there?

Also, you pointed to three different cases, all of which have the same probability p.  Are you sure that the probability of any one of the three cases happening is also p?  (You wrote p = 7/165, but that isn't in fact correct on account of you're having the 1's in there, per my first comment.)

i'm getting confused now,
i think the 1/12, or 1/11, or 1/10 should your probability that you will get a blue marble on your first, second, or third draw.

But there are four blue marbles to choose from, not 1..

By the way, I get [hide]4/9 [/hide] for the 1 blue marble case.

MMM: [hide]4/9?  Whatever happened to the 11 in the denominator?[/hide]

Oopsies what on earth am I doing   [hide]28/55 should be a bit better   [/hide]

oh  :oops: oops i forgot that that there were more than 1 blue marble to choose from :(


Solve the equation $ \left[ \frac{x\plus{}1}{2} \right]\equal{}\left[\frac{2x\plus{}1}{3} \right]$, where $ [a]$ is the integer part of the real number $ a$.

[quote="moldovan"]Solve the equation $ [ \frac {x + 1}{2} ] = [\frac {2x + 1}{3} ]$, where $ [a]$ is the integer part of the real number $ a$.[/quote]

Let ${ \ \lfloor\frac {x + 1}{2}}\rfloor = \lfloor{\frac {2x + 1}{3}}\rfloor=k,k\in\mathbb I$

$ \implies k\le\frac {x + 1}{2}<k+1$ and $ \ k\le\frac {2x + 1}{3}<k+1$

$ \implies 2k-1\le x <2k+1$ and $ \ \frac {3k-1}{2}\le x< \frac {3k+2}{2}$

Now first analyze the case in which there is no solution,there are two such cases

$ 2k+1\le \frac {3k-1}{2}$ or $ \ \frac {3k+2}{2}\le 2k-1$

$ \implies$ For $ \  k\in (-\infty,-3]\cup [4,\infty)$ there is no solution.

$ \implies$ there is a solution for $ k = -2,\ -1,\ 0,\ 1,\ 2,\ 3$ only.

Solving for these values of $ \ k$ we get $ \ \boxed{\boxed{x\in [-\frac 72,\ -3)\bigcup[-2,\ -1)\bigcup [-\frac 12,\ \frac 52)\bigcup [3,\ 4)\bigcup[5,\ \frac {11}{2})}}$


Pentru doua cercuri din plan, $\Gamma_{1}$ si $\Gamma_{2}$, spunem ca cercul $\Gamma_{1}$ [i]injumatateste[/i] cercul $\Gamma_{2}$ daca ele se intersecteaza iar coarda obtinuta din cele doua puncte de intersectie este diametru in $\Gamma_{2}$. Doua cercuri identice se [i]injumatatesc[/i] reciproc.
Fie doua cercuri $\omega_{1}$ si $\omega_{2}$ fixate in plan. Determinati locul geometric al centrelor cercurilor $\omega$ din plan, care le [i]injumatatesc[/i] pe amandoua.

O intrebare mica: carei clase se adreseaza problema??locul geometric presupun ca nu se invata in gimnaziu. :?

multe chestii de la olimpiada nu se invata la scoala, in general. locul geometric nu e o notiune complicata... cred ca stii ce este, nu ?
problema este de la un concurs international din 1999 (nu mai stiu care, am facut`o acum multa vreme), iar rezolvarea pe care eu am gasit`o nu foloseste nici o notiune complicata, ci doar cateva idei. cred ca se poate rezolva la nivelul clasei a 7a  ;)

Bine,..o s-o incerc si eu..dar spune-mi si mie ce e ala $\text{loc geometric}$..Am vazut in carti dar nu explica bine.

pai
cercul acela $\omega$ este variabil in tot planu ala. singura conditie este ca el trebuie sa injumatateasca si pe $\omega_{1}$ si pe $\omega_{2}$.
acum facem un "desen" cu toate centrele cercurilor astea...ce vei obtine ?
uite si cateva exemple clasice de loc geometric
locul geometric al punctelor din plan egal departate de doua puncte $A$ si $B$ este mediatoarea acestui segment.
locul geometric al punctelor din plan egal departate de laturile unui unghi $\angle XOY$ este bisectoarea acestui unghi.
locul geometric al punctelor din plan aflate la o distanta fixa de un punct fix este cercul cu centrul in acel punct.
etc...

Dacă am înţeles bine enunţul, atunci iese aşa:

$\bullet$ Daca $\omega_1,\omega_2$ sunt concentrice, atunci l.g. = tot planul (pt. $r_1=r_2$) sau l.g. = $\phi$ (pt. $r_1 \neq r_2$).

$\bullet$ Dacă nu, atunci l.g. e o dreaptă (linia albastră din desenul de mai jos) $\perp$ pe dreapta ce uneşte cele două centre . Are vreo caracterizare mai specială? Că eu n-o găsesc.

[url=http://img209.imageshack.us/my.php?image=51wh.png][img]http://img209.imageshack.us/img209/1822/51wh.th.png[/img][/url]

Pentru ultimul caz am folosit faptul că $O_1 P^2 + r_1^2 = R^2 = r_2^2 + O_2 P^2$.

[edit][i]Ecuaţia axei radicale este $O_1 P^2 - O_2 P^2 = r_1^2 - r_2^2$, iar ecuaţia locului nostru este $O_1 P^2 - O_2 P^2 = r_2^2 - r_1^2$. Mă gândisem şi eu că ar fi ceva cu axa radicală, dar am văzut că nu ieşea un semn, şi am lăsat-o baltă.[/i][/edit]

@stancioiu sorin: Locul geometric se învaţă în a noua (şi câteva extinderi în a unşpea).

da, asa am facut si eu, este corect
dreapta obtinuta este simetrica axei radicale fata de mijlocul lui $[O_{1}O_{2}]$, iese din ecuatie imediat... dar nici eu nu mi`am dat seama, ca nu stiam ecuatia axei radicale...

Deci nu e o problema de nivel de clasa a VII-a,...Am auzit si eu de axe radicale, dar nu stiu mai nimik de ele. :(


[quote="mdk"]For what value of n does (2^2007-2^2006)(2^1997-2^1996)=2^n ?[/quote]

Right from MML #3 Question 3

[hide]Notice that 2^(n+1)-2^(n)=2^(n) and 2^(a)*2^(b)=2^(a+b)[/hide]

[quote="mdk"]Yeah, this is how I approached it. 4002 right?[/quote]

yeah, thats it

mathgeniuse^ln(x), how did you do 3-4, the one about the triangle?

I'm really stupid in math, but does that mean (2^2006*2^1996)
which gives you 2^1996+2006 or 2^4002?

I did the same thing. Those were some really easy problems.

ARe we allowed to discuss Contest 4 yet.

[quote="mdk"]I'm not sure. I'm not taking the tests and believe it or not I  got a copy of contest 3 by finding it on a random table at a school  where I went for a quizbowl tournament. We should be able to talk about it tomorrow. Did you get all of the questions right?[/quote]

YEah...

[quote="tjhance"]i got them all too :D

but i totally bombed #4 :([/quote]

Wait, you got them all, but you bombed it???

i meant i got them all on #3

i didn't quite catch that you had switched topics from #3 to #4


inequalities
geometry
inradius
circumcircle
rearrangement inequality

Let $a,b,c$ be the side length of $\triangle ABC$. Prove that:

$abc\geqslant(a+b-c)(b+c-a)(c+a-b)$.

Nice, here's my solution.

[hide] Since $a+b-c>0, b+c-a>0, c+a-b>0$

By AM-GM we obtain that $\frac{(a+b-c)+(b+c-a)}{2}\geqslant \sqrt\overline (a+b-c)(b+c-a)$
$\Rightarrow$ $b \geqslant \sqrt\overline (a+b-c)(b+c-a)$

Similarly, $a \geqslant \sqrt\overline (c+a-b)(a+b-c)$
               $c \geqslant \sqrt\overline (b+c-a)(c+a-b)$

Now, times $a,b,c$ together and we had obtain

$abc \geqslant (a+b-c)(b+c-a)(c+a-b)$

                                                                $QED$[/hide]

[quote="stupidkid"][hide="solution"]
Let
$a+b-c = 2x$, $b+c-a = 2y$, $c+a-b = 2z$

Therefore:
$a=x+z$, $b=x+y$, $c=y+z$

Substituting everything in we are left with:
$(x+z)(x+y)(y+z) \ge 8xyz \Longleftrightarrow (x^{2}y+y^{2}z+z^{2}x)+(x^{2}z+y^{2}x+z^{2}y) \ge 6xyz$.

Assume WLOG $x \ge y \ge z$. This implies $xy \ge xz \ge yz$. Using the  Rearrangement Inequality we can prove each part of the LHS is greater than or equal to $3xyz$ and we're done.  :)
[/hide][/quote]

[hide="Or..."]Just AM-GM it after obtaining the equivalent inequality $(x+y)(y+z)(z+x) \geq 8xyz$.[/hide]

[quote="Kurt Gödel"]Just AM-GM it after obtaining the equivalent inequality $(x+y)(y+z)(z+x) \geq 8xyz$.[/quote]
Ahh...I knew was an easier way to prove that, but my mind was blank at the time.  :blush:

Let $S$ be the area of $\Delta ABC$, $r$ the inradius, $R$ the circumradius and $p$ the semiperimeter.
$R \ge 2r$ $\Rightarrow \frac{abc}{4S}\ge 2r \Rightarrow abc \ge \frac{8S^{2}}{p}\ge 8(p-a)(p-b)(p-c)$


trigonometry
inequalities
algebra
polynomial
absolute value
calculus
calculus computations

does the limit exist?

summ(sin^2(n^2)/n)
where n-->infinity

The infinite series $ \sum_{n \equal{} 1}^{\infty}\frac {\sin^2(n^2)}{n}$ is divergent. Indeed, if $ \sin(n^2)$ and $ \sin((n \plus{} 1)^2)$ are less than $ \epsilon$ in absolute value, then there exist integers $ k,j$ such that $ |n^2 \minus{} \pi k| < \frac {\pi}{2}\epsilon$ and $ |(n \plus{} 1)^2 \minus{} \pi j| < \frac {\pi}{2}\epsilon$. This implies
\[ |2n \plus{} 1 \minus{} \pi(j \minus{} k)| < \pi \epsilon
\]
No contradiction here yet. But if $ |\sin((n \plus{} 2)^2)| < \epsilon$ too, then we can replace $ n$ with $ n \plus{} 1$ above and obtain
\[ |2n \plus{} 3 \minus{} \pi(j' \minus{} k')| < \pi \epsilon
\]
for some integers $ j',k'$. Combining the displayed inequalities, we get $ |2 \minus{} \pi l| < 2\pi\epsilon$ for an integer $ l$. This is false if $ \epsilon$ is small enough. Therefore, at least one of any three consecutive elements of the sequence $ \sin(n^2)$ is at least $ \epsilon$ in absolute value. This suffices to conclude that the series diverges.

A generalization: if $ p$ is a nonzero polynomial of degree $ r$ with integer coefficients, then there exists $ \epsilon \equal{} \epsilon(r)$ such that at least one of any $ (r \plus{} 1)$ consecutive elements of the sequence $ \sin(p(n))$ is at least $ \epsilon$ in absolute value.


limit
real analysis
real analysis unsolved

Let $ a_{0}$ and $ b_{0}$ be positive real numbers. For $ n \equal{} 0, 1, 2, \dots$ let

\[ a_{n\plus{}1} \equal{} \frac{\left(a_{n}\plus{}b_{n}/e\right)^2}{\left(a_{n}\plus{}b_{n}\right)^2},\]
\[ b_{n\plus{}1} \equal{} \frac{b_{n}\times\left(a_{n}\plus{}b_{n}/e\right)}{\left(a_{n}\plus{}b_{n}\right)^2}.\]

Determine the values of $ \displaystyle\lim_{n\rightarrow\infty}a_{n}$ and $ \displaystyle\lim_{n\rightarrow\infty}b_{n}$.

[hide="Hint"]Express $ a_{n\plus{}1}$ and $ b_{n\plus{}1}$ in terms of $ c_n \equal{} \frac{a_n}{b_n}$. Determine $ c_n$ from this. Plug this back to the formulas for $ a_{n\plus{}1}$ and $ b_{n\plus{}1}$ and evaluate the limit.[/hide]

$ \frac {a_{n \plus{} 1}}{b_{n \plus{} 1}} \equal{} \frac {a_n}{b_n} \plus{} \frac 1e$ , so $ \frac {a_n}{b_n} \rightarrow \plus{} \infty$, so $ \frac {b_n}{a_n} \rightarrow 0$.
$ 1 \geq a_{n \plus{} 1} \geq \frac {a_n^2}{(a_n \plus{} b_n)^2 } \equal{} \frac 1{(1 \plus{} \frac {b_n}{a_n})^2}$
Finally  $ a_{n} \rightarrow 1$ and $ b_n \rightarrow 0$
:cool:


I have a terminology question for polar coordinates.

Given numbers r and :theta:, we define the point with polar coordinates (r, :theta:) to be the point with rectangular coordinates (r cos :theta:, r sin :theta:).  Given a point (in say rectangular coordinates), there are many infinitely many ways to represent it in polar coordinates.  That's because we can add an arbitrary multiple of 2:pi: to :theta: without changing the underlying point.  Even weirder, we could add :pi: to :theta: and negate r.  The origin is even weirder; we could let :theta: be any number as long as r = 0.

To make the situtation a little nicer, we could define the "canonical" polar representation of a point to be the polar representation with r :ge: 0 and :theta: between 0 (inclusive) and 2:pi: (exclusive).  Actually we still need to handle the origin in some unique way (not sure what is best).

Anyway, my question has to do with polar equations.  For example, let's say we have an equation like :theta: = 3:pi:/2.  Is the solution set to such an equation the set of points that have at least one polar representation that satisfies the equation?  Or is the solution set the set of points for which the canonical polar representation satisfies the equation?  In other words, for this particular equation, is the solution set the negative half of the y-axis, the entire y-axis, or maybe the entire y-axis except the origin.

I am writing a question for an SAT II Math practice test, and I wasn't sure what the conventions are for such polar equations.  Thanks.

I would probably be stuck if I saw that problem between assuming that it was the negative y-axis or the entire y-axis, both including the origin.  I think I would probably opt for the negative y-axis including origin, which I think is what you called canonical form.  I have no idea what the "SAT standard" is for that, though.

Does the SAT II cover polar coordinates?

I think the proper way to handle it is to assume that r can be both positive and negative unless otherwise stated. Because students should understand that r can be negative and what it means graphically. When students graph polar functions in class, they are taught to look at all possible values of r for a given theta, regardless of whether it is positive or negative.

That being said though, I think such a question would be unfair to ask on a standardized test because it does have a natural ambiguity to it. The best thing to do would be to specify exactly what values of r you mean to permit, and take that ambiguity out of the question.

Thanks, guys, for your input.  Yes, the SAT II Math 2C test covers polar coordinates.

I should clarify that I'm really just editing a question that someone else at my company wrote.  The original version of his question was "What is the equation of the y-axis in polar coordinates?".  (Intended answer :theta: = :pi:/2.)  Then I mentioned to him my doubts regarding ambiguity.  He changed the question to "What is the equation of the negative half of the y-axis in polar coordinates?".  (Intended answer: :theta: = 3:pi:/2.)  I don't think the revised version answered the objection about ambiguity.  I'll go back to released SAT II tests to see how the ETS handles such issues.

But, independently of the SAT, I was curious how mathematicians treat such polar equations.  So thanks for the feedback.

Ill give my imput on this having taken the math2c test, but this is nothing you dont already know...

What is the equation of the y-axis in polar coordinates?
(Intended answer  = pi/2.)

This is correct.  This is the equation for the positive AND negative y axis.  Yes, there is more than one answer;theta= pi/2 + 2kpi or theta =3pi/2+2kpi , but this is a multiple choice test, so even if there are many correct answers only 1 of them will be A, B, C, D, or E.

To be tricky like ETS loves to do you could put one of those questions like:

Which of the following is the equation for the y-axis in polar coordinates?

I. theta=pi/2
II. theta= 9pi/2
III. theta = 3pi/2

A. I only
B. II only
C. I and II
D. II and III
E. I, II, and III

"What is the equation of the negative half of the y-axis in polar coordinates?". (Intended answer:  = 3pi/2.)

I would say he is wrong, theta=3pi/2 is the equation for the whole y-axis.  (In canonical form of course its different).

You are taught in precalc exactly what you described in the first paragraph of your post.  I would not assume "canonical form" b/c this is not standard, you have to say, "In canonical form..." (but most students would not understand it because this is not a common precalc vocab word I dont think).

SAT tests want you to understand more than one aspect of the topic, so thats what they aim to test.  So the question "What is the equation of the negative half of the y-axis in polar coordinates?". (Intended answer:  = 3pi/2.) would actually be a common mistake a student would make, one who doesnt know that r can be negative.

Thanks for your comments.  I decided not to be tricky.  I rewrote the question and answer choices so that, no matter whether students used the general interpretation or the canonical interpretation, they would get the same answer.  Here's the revised question and answer.

"Which of the following is an equation for the negative half of the y-axis in polar coordinates?"
Intended answer: :theta: = 3:pi:/2 and r > 0.


trigonometry
logarithms
inequalities
abstract algebra
real analysis
real analysis unsolved

Discuss the convergence or divergence of

$ \sum_{n\equal{}1}^{\infty}\frac{1}{n}\left(\frac{2 \plus{} \sin n}{3}\right)^{n}$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]here[/url]. I am not aware of any solutions appearing elsewhere.

is         the  following     right?

$ \sin n \leq 1 \minus{} \frac {3 \ln n}{n}$.  $ \forall n \in \mathbb{N}$

No. There is a constant $ c$ such that $ \sin n > 1\minus{}\frac cn$ infinitely often.

[quote="jmerry"]No. There is a constant $ c$ such that $ \sin n > 1 \minus{} \frac cn$ infinitely often.[/quote]

Thanks!   but  how  to   prove  it?

See posts #6-9 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php&t=90103]here[/url].

$ |n \minus{} m\pi \minus{} \frac {\pi}2| < \frac {c}{n}$,  so
 $ \cos (n \minus{} m\pi \minus{} \frac {\pi}2) > \cos \frac {c}{n}$, so
  $ |\sin n | > \cos \frac {c}{n}$
$ |\sin n| > 1 \minus{} \frac {c^{2}}{2n^{2}}$.   by Bernoulli's inequality  ( http://en.wikipedia.org/wiki/Bernoulli%27s_inequality) , we  have
$ |\sin n|^{n}>1\minus{}\frac{c^{2}}{2n}$,  so  $ |\sin n|  > 1\minus{}\frac{c^{2}}{2n}$
am  I   right?

The proof of the convergence should be here:  http://www.siam.org/journals/problems/downloadfiles/99-005s.pdf

I have not checked it.

solution is correct , but  there  we do not need Hata's result, there is enough to   know that irrationality measure of $ \pi$ is  [b]finite[/b]. this can we do because:
Suppose irationality measure of $ \pi$ is less than some finite number $ c$ then choose smooth  kernel with fourier coeficients $ a_{n} \approx const \varepsilon^{\minus{}1}  (\varepsilon n )^{\minus{}c\minus{}100}$ . where $ n \neq 0$. this will be enough

That is interesting. Thank you

[quote="Extremal"]solution is correct..........[/quote]


algebra
function
domain
inequalities unsolved
inequalities

Minimize the expression
$a(a(b^{2}+c^{2})+2(b+c))$
subject to constraints $a, b, c > 0$ and $abc = 1$.

I think this one is stupid. But I still want to see how you might approach it.

edit: fixed the constraint (so now it is not quite as stupid as before ;) )

umm. it's obviously always positive and can get arbitrarily close to 0 ... (minimum doesn't necessarily exist since the domain isn't a compact space, and if we consider the closure instead, the minimum really is 0, so uh... ?)

I made a mistake in the constraint. I will edit.

But It still can goes arbitrarily close to 0.
Let $a=k^{2},b=c=\frac{1}{k}$
$k \to 0$

Ok I don't know what went wrong in my mind whilst making that but at least my intuition that it was dumb was correct.  :wink:
Sorry.  :oops:


inside the earth there is small spherical oil patch of radius $R$. The distance of the center of the oil patch and the surface of the earth is $d$.
How does this change the gravitational accelaration on the surface of the earth?

This should lower the acceleration, if the density of the oil is not greater than the average density of the Earth (and vice-versa). I believe you can imagine a sort of 'negative-massed' sphere at the place of the patch (while considering the Earth to be homogenous); this negative mass will supply a repelling gravitational field and lower the acceleration. But you would need the density of the oil, so...  :o


algebra
system of equations
superior algebra
superior algebra unsolved

Let p>=2 be a prime. A ring (A,+,*) is called a p-ring if:

a) p=0;
b) xp=x.

Show that  any p-ring is commutative.

[i]Gheorghe Titeica contest/2001/11'th and 12'th grade (team contest)[/i]

I think amfulger mentioned this problem (he was in the 11'th grade then).

[quote="grobber"]b) xp=p.[/quote]

I think you mean x p =[b]x[/b]

[quote="grobber"]I think amfulger mentioned this problem (he was in the 11'th grade then).[/quote]

Yes I did.

You should really try to get some sleep after 1 am  :D .

Thanks, I've edited the post.

I did get some sleep: from 1 AM to about 3 AM :D

[quote="Moubinool"][quote="grobber"]Let p>=2 be a prime. A ring (A,+,*) is called a p-ring if:

a) p=0;
[/quote]

p \geq 2
p=0
Not clear[/quote]

p is prime in N. In A we have p*1=1+1+...+1 (p terms)=0 (i.e. A has characteristic p). Now clear.

Ok, let's start something here:

In A x2=0 =>x=0 because x2=0=>xp=0=>x=0.

Let x be an element s.t. x2=x (idempotent). Then for any y in A we have (xyx-xy)2=(xyx-yx)2=0 (easy to calculate) => xyx-xy=xyx-yx=0 => xy=yx, so any idempotent element is in Z(A).

For x in A we have (xp-1)2 =
x2p-2=xp-1, so xp-1 is idempotent, so xp-1 is idempotent, so xp-1 is in Z(A) for all x, so y*xp-1=xp-1*y for all x and y in A.

This is pretty much what I have up until now. What next? :)

Looks like nobody liked it, huh? :)

I'm not sure about this, but here it goes anyway :):

I've shown that $x^{p-1}\in Z(A),\ \forall x\in A$. In this case, $(x-y)^{p-1}\in Z(A)\ \forall x,y\in A$. At the same time, $(x-y)^{p-1}=x^{p-1}+x^{p-2}y+\cdots +xy^{p-2}+y^{p-1}$. We replace $y$ with $1,2,\cdots p-1$ and solve the system of equations $x^{p-1}+x^{p-2}i+\cdots+xi^{p-2}=a_i\in Z(A),\ i\in \overline{1,p-1}$ with $x^k$ as unknown quantities. We write the corresponding determinant and see it's non-zero (it's Vandermonde-type) and then write the solutions according to Cramer's rule, and find that $x^k\in Z(A)\ \forall k\in \overline{1,p-1}$. In particular, this means that $x\in Z(A)\ \forall x\in A$.

Think it's correct?


linear algebra
matrix
combinatorics unsolved
combinatorics

Let $M$ be a $m\times n$ matrix, with positive integers entries (including 0).
$M[1][1] = 0$ and $M[i][j]$ is the least positive integer(including 0) which does not appear on the line $i$ or on the column $j$.
Find a formula for $M[i][j]$.

this is a well known problem.it is closely related to the game of Nim.
as for the answer, write both $i$ and $j$ in binary notation, and perform a Nim addition,i.e.$1+0=1=0+1,0+0=1+1=0$(i.e. binary addition without carryovers);call this $\oplus$. then the $(i,j)$th entry is $i\oplus j$.

well.. I know the answer.. but I want the proof if possible...

one way to see this is to first restrict to $2^n\times 2^n$ grids, for each $n$. Then try to show that the $2^{n+1}$ grid
 arises from the $2^n$ grid in the following way:if $A$ denotes the $2^n\times 2^n$ grid, then the enlarged grid simply looks like \[\begin{array}
                  {cc}(2^n+A)&A\\A&(2^n+A)\end{array}\].(this may be  done inductively)

then the result follows from this.

I think you meant
$\left(\begin{array}{cc}
A & 2^n+A\\
2^n+A & A
\end{array}\right)$
thank you

this doesn't seem as natural proof to me.
or maybe i am too stupid to think of this but then again that might
be my problem... :huh:


Two players are throwing a six-sided die. If the first player rolls 6 he loses and the game is over. If not, second player throws the die. If he rolls 6 he loses and the game is over, if not the first player throws again. The game is not finished until one of the players rolls 6 when he loses, and the other player wins. What is the probability that the second player wins?

We set up the recursion $ p\equal{}\frac16\plus{}\left(\frac56\right)^2p\implies\frac{11}{36}p\equal{}\frac16\implies p\equal{}\boxed{\frac6{11}}$.

Hehe, that was quick. But what if the problem was generalized to $ n$ players?

:huh: Could you explain your solution, please?

It's different for $ n$ players.  If the first person loses, then what? Do the other $ n \minus{} 1$ players win? Do we keep on going until there is only $ 1$ person left?

And for the original question....
The probability that the first player loses is:
$ \frac {1}{6} \plus{} (\frac{5}{6})^{2} \bullet \frac {1}{6}... \equal{} \frac {1}{6}(1 \plus{} (\frac {5}{6})^2...) \equal{} \frac {1}{6}(\frac {36}{11}) \equal{} \boxed {\frac {6}{11}}$

I can give you another solution.

1st player wins in one of this cases:
$ ab'$, $ abab'$, $ ababab'\cdots$ where $ a$ denotes first player's throw which didn't roll 6, $ b$ denotes second player's throw which didn't roll 6, $ a'$ denotes first player's throw that rolled 6 and $ b'$ denotes second player's throw that rolled 6.
Probabilities for $ a$ and $ b$ to happen are $ p\equal{}\frac{5}{6}$ and for $ a'$ and $ b'$ are $ q\equal{}\frac{1}{6}$.

Probability that first player wins is $ p_A\equal{}qp\plus{}qp^3\plus{}qp^5\plus{}\cdots$

2nd player wins in one of this cases:
$ a', aba', ababa'\cdots$

Probability that second player wins is $ p_B\equal{}q\plus{}qp^2\plus{}qp^4\plus{}\cdots$

We can notice that $ p_A\equal{}p_B\cdot p$ and $ p_A\plus{}p_B\equal{}1$. Solving the system gets us solution.

Notice how this is a little bit long, but this is good basis for generalization.

@Z-coli the game keeps going until only one player is left

In the general case with $ n$ people is a much harder problem, because at each stage of the game, you can have a different person lose.  Not only that, but the order in which they lose combined with the times at which they lose makes this problem very challenging. Complementary counting doesn't do much, either, as the probability that the $ n$th person wins is also a matter of caswork: (the probability that he loses in the first round, the second round...)$ \plus{}$(the probability that he is still in the game and loses after one person has lost)..etc.


induction
algorithm
number theory
Euclidean algorithm

Prove that any two consecutive Fibonacci numbers $ F_k$ and $ F_{k\plus{}1}$ are relatively prime.

1 and 1 are relatively prime.

Let $ (F_n,F_{n \plus{} 1}) \equal{} 1$

Then $ (F_{n\plus{}1},F_{n \plus{} 2}) \equal{} (F_{n\plus{}1},F_n \plus{} F_{n \plus{} 1}) \equal{} (F_{n\plus{}1},F_{n}) \equal{} 1$

Shouldn't that be

$ (F_{n\plus{}1},F_{n\plus{}2}) \equal{} (F_{n\plus{}1},F_n \plus{} F_{n\plus{}1}) \equal{} (F_{n\plus{}1}, F_n) \equal{} 1$?

The induction step doesn't work properly otherwise.

Can someone please explain this problem to me step-by-step?  I'm not too good with induction.  Thanks.

[hide]Here is a sketch of the proof

Base Case: $ (F_1,F_2) \equal{} (1,1) \equal{} 1$.

Inductive Step: Assume that $ (F_n,F_{n \plus{} 1}) \equal{} 1.$ (Aside: Then by like Euclidean Algorithm or whatever, you know that $ (m \plus{} n,n) \equal{} (m,n)$) Using this logic you see that $ (F_{n \plus{} 2},F_{n \plus{} 1}) \equal{}(F_{n \plus{} 1} \plus{} F_n,F_{n \plus{} 1}) \equal{}  (F_n,F_{n \plus{} 1}) \equal{} 1$ and so you win[/hide]

[hide="Not induction"]
Assume that $ F_k$ and $ F_{k\plus{}1}$ have a common factor other than one. Then their sum and difference would have that factor and so all Fibonacci numbers would but clearly that's false.
[/hide]

That is still pretty much using induction...here is a stupid proof but I suppose that it illustrates an interesting idea...

Let $ S \equal{} \{d|d \equal{} (F_n,F_{n \plus{} 1}),d > 1,n\in \{1,2,...\}\}$

That is to say, $ S$ is the set of all common factors of adjacent terms of the Fibonacci sequence that are not one.

Assume for sake of contradiction that $ S$ is non-empty. Let $ n$ be the minimal positive integer such that $ (F_n,F_{n \plus{} 1}) >1$. But $ (F_n,F_{n \plus{} 1}) \equal{} (F_{n \minus{} 1},F_{n})$. So if $ n > 1$, then $ n \minus{} 1$ is a positive integer that is smaller than $ n$ with the property. This contradicts the minimality of $ n$. Hence $ n \equal{} 1$, but $ (F_1,F_2) \equal{} 1\not\in S$, which is a contradiction. Hence $ S$ is empty, and the result follows.

[quote="caffeineboy"][hide="Not induction"]
Assume that $ F_k$ and $ F_{k \plus{} 1}$ have a common factor other than one. Then their sum and difference would have that factor and so all Fibonacci numbers would but clearly that's false.
[/hide][/quote]

[u]Then their sum and difference would have that factor and so all Fibonacci numbers would[/u]

I don't get why this is true.

To clarify what caffeineboy wrote:

[hide="Proof"]
Assume that there exists two consecutive Fibonacci Numbers, $ F_n$ and $ F_{n\plus{}1}$, that have a common divisor $ d$, where $ d>1$. Therefore, there difference $ F_{n\plus{}1}\minus{}F_n\equal{}F_{n\minus{}1}$ will also be divisible by $ d$. However, we know that $ F_1$ is not divisible by $ d$. A contadiction.[/hide]

This is our homework for Mathcounts.  :dry:


Μπορει η παρακατω προταση να ειναι θεωρημα??

Εαν α δεν ειναι σημειον συσσωρευσεως εις το πεδιον ορισμου της..........$ f: A\subseteq R\rightarrow R$ .τοτε η f εχει οριο στο α?

Νομίζω ότι δεν έχει νόημα το όριο . Όχι δεν υπάρχει . Είναι σαν να ρωτάς αν ο  $ \frac{2}{3}$ είναι άρτιος ή περιττός?

[quote="τρικλινο"]Μπορει η παρακατω προταση να ειναι θεωρημα??

Εαν α δεν ειναι σημειον συσσωρευσεως εις το πεδιον ορισμου της..........$f: A\subseteq R\rightarrow R$ .τοτε η f εχει οριο στο α?[/quote]

Πολύ σωστά απάντησε ο κ. Μπόρης και δεν χρειάζονται άλλες εξηγήσεις.

Για διδακτικούς λόγους όμως είναι χρήσιμο να προσθέσω ότι:

Στα σχολικά μαθηματικά ο ορισμός του ορίου στο α απαιτεί η συνάρτηση να ορίζεται τουλάχιστον σε ένα σύνολο της μορφής (β,α)U(α,γ). Σε πιο προχωρημένα μαθηματικά, ο ορισμός αυτός είναι κάπως περιοριστικός. Π.χ. ο σχολικός ορισμός δεν μας επιτρέπει να μιλάμε για όρια όταν το πεδίο ορισμού της f είναι το σύνολο των ρητών, ενώ θα θέλαμε να μελετάμε και τέτοιες περιπτώσεις. Οπότε αυτό που κάνουμε είναι να "χαλαρώσουμε" λίγο τον ορισμό για να περιέχει πιο γενικές περιπτώσεις.
Το αμέσως επόμενο στάδιο είναι να υιοθετήσουμε ορισμό που το α να είναι σημείο συσσώρευσης.
Γιατί δεν κάνουμε τον ορισμό ακόμη πιο χαλαρό; Γιατί απλούστατα χάνουμε αντί να κερδίζουμε. Π.χ. αν το α δεν είναι σημείο συσσώρευσης, δηλαδή αν είναι μεμονωμένο σημείο, τότε υπάρχει δ>ο τέτοιο ώστε το πεδίο ορισμού της f δεν έχει κανένα στοιχείο στο σύνολο (α-δ, α)U(α, α+δ). Δηλαδή η χρήση του εψιλοντικού ορισμού δίνει κενό σύνολο από x για να ελεγχθεί η σχέση
|f(x) - L| < ε, οπότε δεν έχει νόημα να μελετάμε τέτοιες περιπτώσεις.
Το σημείο συσσώρευσης είναι επαρκές για την μελέτη ορίων και εκεί μας οδηγούν φυσιολογικά τα πράγματα.

Ελπίζω να βοήθησα.

Φιλικά,

Μιχάλης Λάμπρου.

Σας ευχαριστω και συγνωμη για την καθυστερηση.

Και βεβαια το λεχθεν οριο δεν ειναι συμβατο με τον ορισμο του οριου που  [b]προυποθετει a[/b] να ειναι σημειον συσσωρευσεως.

Αλλα αν και υπαρχουν αλλοι ορισμοι οριων που δεν προυποθετουν a να ειναι σημειον συσσωρ.( π.χ ενας ορισμος που θελει το a σημειον επικολλησεως ( point of adherence,βλεπε σελιδα 38 στο βιβλιο του SERGE LANG ANALYSIS I) ) και ετσι μπορουν να ικανοποιησουν το παραπανω οριο,[b]δεν ειχα αυτο υποψιν μου[/b]

Εγω ειχα υποψιν μου το εξης:

Εαν a δεν ειναι σημειον συσσωρ. εις το πεδιον ορισμου της $ f: A\subseteq R\rightarrow R$ ,ΤΟΤΕ να αποδειχθη οτι:

.......................................υπαρχει m τετοιο ωστε.............................................¨

...................για ολα τα ε>0 υπαρχει r>0 ετσι ωστε για ολα τα χ που ανοικουν στο Α ΚΑΙ .

..................................0|x-a|<r ,τοτε..........|f(x)-m|<ε................................................................................

...........................................και σε σyμβολα..........................................................................

............................................υπαρχει m τετοιο ωστε...............................................................................

..........$ \forall\epsilon [\epsilon>0\rightarrow\exists r(r>0\wedge\forall x( x\epsilon A\wedge 0<|x\minus{}a|<r\rightarrow |f(x)\minus{}m|<\epsilon))]$


[quote="happyme"]Gavin is driving at a constant rate. He passes a milepost that displays a two-digit number. Exactly one hour later, he passes a second milepost that displays the same two digits in the reverse order. One hour later, he passes a third milepost with the original two digits separated by a zero. What is his rate in miles per hour?

 :D[/quote]

[hide]Let the first number be $10a+b$
The second number then is $10b+a$
The third number is $100a+b$

($a$ and $b$ are digits. In other words, $1 \le a \le 9$ and $1 \le b \le 9$ and both are integers.)

Since he is driving at a constant rate,
$10b+a-10a-b=100a+b-10b-a$
$9b-9a=99a-9b$
$b-a=11a-b$
$12a=2b$
$6a=b$

Then this is only satisfied when $a=1$, $b=6$.

Therefore, $\text{Rate}=9b-9a=\boxed{45}$ mph.

Check:
First post: $16$
 -- 45 miles during one hour --
Second post: $61$
 -- 45 miles during one hour --
Third post: $106$[/hide]

[quote="happyme"][quote="vishalarul"][quote="happyme"]Gavin is driving at a constant rate. He passes a milepost that displays a two-digit number. Exactly one hour later, he passes a second milepost that displays the same two digits in the reverse order. One hour later, he passes a third milepost with the original two digits separated by a zero. What is his rate in miles per hour?

 :D[/quote]

[hide="Nice problem..."]Let the first milepost be $ab$. Value=$10a+b$
So the second one is $ba$. Value = $10b+a$
The third one is: $a0b$. Value: $100a+b$.

In 1 hr, he goes $9b-9a$, which is the same as: $99a-9b$.

Putting the equations,
$9b-9a = 99a-9b$
$b-a=11a-b$
$2b=12a$
$b=6a$

So the mileposts are:
$16$
$61$
$106$

He goes $45$ miles per hour.[/hide][/quote]

I don't see how you get the mileposts from $b=6a$

What I did was:

[hide] Since
$b=6$
$a=1$
are the only integer values for a and b that satisfy that a and b are digits between 1 - 9 inclusive, you can plug that into the first equation to get:

$10(1)+6+x=10(6)+1$
$16+x=61$
$x=45$

My original equation had:

Let $x =$ speed Gavin is driving at (mph)

$10a+b+x=10b+a$

and my second equation was

$10b+a+x=100a+b$

Did I do something wrong?[/hide][/quote]

[hide="I think"]
I think that is fine except that $x$ should not be speed Gavin is driving at in mph, but rather the distance Gavin travels in an hour (which is numerically equal to his mph).

vishalarul should have skipped figuring out 1 and 6 since that is very obvious knowing $a$ and $b$ are the digits.[/hide]


This problem can be split into two problems. The area of the semi-annulus and the area of the rectangles. The area of the rectangle is $ 2(4\cdot2)\equal{}16$. The area of the annulus is $ \frac{4^2\pi}{2}\minus{}\frac{2^2\pi}{2}\equal{}8\pi\minus{}2\pi\equal{}6\pi$ Therefore, the area of the magnet is $ \boxed{16\plus{}6\pi}$


Βάζω ένα hint (από το σημείο που την αφήνω και πέρα τελειώνει πολύ εύκολα)
[hide]Έστω $ p$ ο ελάχιστος πρώτος διαιρέτης του $ n$ . Τότε $ 3^{2n}\equiv 1(\mod p)$ . Έστω ο ελάχιστος φυσικός $ d$ ώστε $ 3^d\equiv 1(\mod p)$

Τότε $ d|2n$ και $ d|p \minus{} 1$ άρα $ d < p$ άρα ο $ d$ δεν διαιρεί τον $ n$ άρα ο $ d|2$ άρα $ d \equal{} 1$ ή $ d \equal{} 2$ από όπου $ p \equal{} 2$ άτοπο κτλ  :)[/hide]

silouan pws vgazeis oti $ d| p\minus{}1$?

Εγω δεν καταλαβα σχεδον τπτ (ισος επηδη απο θεωρια αριθμων το μονο που εχω διαβασει μεχρι τωρα ειναι το βιβλιο της ΕΜΕ μεχρι τη μεση)... Απο που προκειπτουν τα: $ 3^{2n}\equiv1(mod p)$, και το οτι ο d διαιρει τους 2n και p - 1?

ΥΓ: Ασχοληθηκα κανα 3ωρο με την ασκηση και το μονο που καταφερα να αποδειξω (Με το μικρο θεωρημα του Φερμα) ειναι οτι ο n δεν μπορει να ειναι πρωτος... απο κει και περα τιποτα αλλο...

Edit: νομιζω οτι το $ d|p \minus{} 1$ προκειπτει απο: $ 3^d\equiv1\equiv3^{p \minus{} 1}(modp)$ (Απο μικρο θεωρημα Φερμα)

nick afou o p diairei to 3^n +1
isxyei
$ 3^n\equiv \minus{} 1 {modp} < \equal{} > 3^{2n}\equiv( \minus{} 1)^2 \equal{} 1{modp}$
pantws απο το fermat's little theorem nomizw nick oti isxyei
$ 3^{p \minus{} 1}\equiv1(modp)$ (kai oxi $ 1\equiv3^{p \minus{} 1}(modp)$) to
opoio pragmati simainei oti d|p-1 afou d o elaxistos fysikos.. ara m lithike i aporia i dikia m.
nick ousiastika exeis
$ 3^{p \minus{} 1}\equiv1(modp)$
$ 3^{2n}\equiv1(modp)$
$ 3^{d}\equiv1(modp)$
kai epeidi d o elaxistos fysikos gia ton opoio isxyei i sxesi ayti epetai
$ d|2n$ kai $ d|p\minus{}1$

[quote="Athinaios"]nick afou o p diairei to 3^n +1
isxyei
$ 3^n\equiv \minus{} 1 {modp} < \equal{} > 3^{2n}\equiv( \minus{} 1)^2 \equal{} 1{modp}$
pantws απο το fermat's little theorem nomizw nick oti isxyei
$ 3^{p \minus{} 1}\equiv1(modp)$ (kai oxi $ 1\equiv3^{p \minus{} 1}(modp)$) to
opoio pragmati simainei oti d|p-1 afou d o elaxistos fysikos.. ara m lithike i aporia i dikia m.
nick ousiastika exeis
$ 3^{p \minus{} 1}\equiv1(modp)$
$ 3^{2n}\equiv1(modp)$
$ 3^{d}\equiv1(modp)$
kai epeidi d o elaxistos fysikos gia ton opoio isxyei i sxesi ayti epetai
$ d|2n$ kai $ d|p \minus{} 1$[/quote]

apo Ferma exoyme $ 3^{p \minus{} 1}\equiv1(modp)$, ara kai $ 1\equiv3^{p \minus{} 1}(modp)$ (simmetriki idiotita twn isotimiwn)

Gia ta ipolipa tora, einai pou me theoria arithmon den exo poliasxolithei, 2-3 mines exo pou diabazo pou kai pou to biblio tis EME kai akoma me diskolevoun arketa pragmata eno kapoia alla den ta ksero kan... gia paradigma den gnoriza kan oti $ 3^{k}\equiv3^{d}(modp), d<k  \Rightarrow d|k$, (to sigkekrimeno isos tha eprepe na ipirxe stous kanones dieretotitas tou vivliou pou diabazo, alla den to exo dei grammeno pouthena opote kai den to gnoriza mexri tora.)

[quote="Athinaios"]niko signwmi gia tn kathisterisi, imun diakopes k dn eixa proxeiro internet. nomizw oti vrika mia ikanopoiitiki apodeiksi gia tn parapanw protasi pou sizitousame
estw oti isxyei gia tixaious akeraious thetikous a,m
$ a^x \equiv 1{modm}$
kai
$ a^d \equiv 1{modm}$
opou d o elaxistos fysikos wste na isxyei i parapanw sxesi ipswnontas se katallili dynami kai pollaplasiazontas perneis eykola
$ a^{gcd(x,d)}\equiv 1{modm}$
apo orismo gcd perneis
${ gcd(x,d)}\leq d$(1)
kai afou d o elaxistos gia ton opoio isxyei i sxesi mas
$ d\leq gcd(x,d)$(2)
ara d=gcd(x,d) kai epetai ws porisma oti kathe x gia ton opoio i sxesi mas einai pollaplasio tou d.
elpizw na mn xanei kapou.[/quote]

Πιστευω πως ειναι ενα λημμα που μπορει να σε βοηθησει να λυσεις πολυ δυσκολες ασκησεις.
Πολυ καλη η σκεψη σου για αποδειξη με $ gcd$ αλλα νομιζω οτι υπαρχει ενα λαθος :( . Γνωριζουμαι οτι για καθε δυο ακαιρεους $ a,b$ υπαρχουν ακεραιοι $ k,m$ με $ gcd(a,b) = ka + mb$, οι ακεραιοι $ k,m$ ομως δεν ειναι υποχρεωτικα θετικοι, οποτε υψωνοντας την ισοτιμια σε καποιο αρνητικο αριθμο, προκειπτει μια ισοτιμια με μη-ακεραιους αριθμους μεσα, κατι που φυσικα δεν οριζεται, κοιταξε το λιγο αν μπορεις και πες μου γιατι ισος να κανω και εγω λαθος καπου η ισος να μη καταλαβα κατι καλα...

oxi safws exeis dikio ipirxe keno stin apodeiksi.
nomizw pantws pws ayto to limma isxyei mono otan gcd(a,m)=1[sthn periptwsi tis askisi mas to m einai prwtos].
ayto to lew giati sto vivlio tou titu andreescu vrika to exis xrisimotato limma
apo bezout id pernume oti $ gcd(x,y)\equal{}rx\minus{}uy$ opou u,r mi arnitikoi.
kai isxyoun oi sxeseis
$ a^{rx}\equiv 1 {(mod m)}$
$ 1 \equiv a^{uy} {(mod m)}$
pollaplasiazontas tes loipon pernume
$ a^{rx } \equiv a^{uy} {(mod m)}$(1)
twra loipon simfwna tin idiotita twn mod gia a,b,c akeraious c diaforo tou midenos kai m thetiko
$ ac\equiv bc {(modm)}\equal{}>a\equiv b{(mod \frac{m}{gcd(c,m)})}$(2)
gia a=a^{rx},b=1,c=a^{uy} kai gcd(a^{uy},m)=1 exoume
$ (1)\equal{}> a^{gcd(x,y)} \equiv a^{rx\minus{}uy}\equiv 1 {(mod m)}$

Γενικά το ord διαιρεί το φ(ν) οπότε αυτό εξηγεί το d|p-1 και γενικεύει... :wink:

[quote="galois01"]Βρείτε όλους τους περιττούς ακέραιους $ n\geq1$ για τους οποίους ο $ n$ διαιρεί τον $ 3^{n}\plus{}1$[/quote]
διαφορετικά [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=855221#p855221]εδώ[/url]


The square root of $ x$ is greater than 3 and less than 4. How many integer values of $ x$ satisfy this condition?

hello, from the condition we get
$ 3<\sqrt{x}<4$,squaring both sides we have $ 9<x<16$, therefore the numbers $ 10,11,12,13,14,15$ fulfill our inequation.
Sonnhard.


Let $ \ A\in M_{4}({Z}), A^3\equal{}A\plus{}I_4$. Then $ det(A^2\minus{}I_4)\equal{}1$. Can you find an example for A? What about the eigenvalues of A ?

$ A^3\plus{}A\plus{}I\equal{}0$ has no solutions in $ \mathcal{M}_4(\mathbb{Q})$.


inequalities
function
Online Math Open
Putnam

Έστω $ x,y,z$ θετικοί πραγματικοί με άθροισμα $ 3$. Να βρείτε το $ \min$ του $ x^2\plus{}y^2\plus{}z^3$.

By Cauchy Schwarz inequality, $ (1^2\plus{}1^2\plus{}1^2)(x^2\plus{}y^2\plus{}z^2)\geq (1\cdot x\plus{}1\cdot y\plus{}1\cdot z)^2$, yielding the minimum value is $ 3$ when the equality holds when $ x\equal{}y\equal{}z$ and $ x\plus{}y\plus{}z\equal{}3\Longleftrightarrow x\equal{}y\equal{}z\equal{}1.$

$ z$ is cubed though, not squared :wink:

Cheers,

Durandal 1707

Δεν ξέρω εαν αυτό που θα δώσω θα βοηθησει αλλά ίσως κάπου να οδηγεί:
Έστω 3 αριθμοί θετικοί $ a,b,c$ με $ a \plus{} b \plus{} c \equal{} 3$
$ (x^2 \plus{} y^2 \plus{} z^3) (a^3\sqrt {x} \plus{} b^3\sqrt {y} \plus{} c^3) (a^3\sqrt {x} \plus{} b^3\sqrt {y} \plus{} c^3) \ge (a^2x \plus{} b^2y \plus{} c^2z)^3$.
Άρα :
$ (x^2 \plus{} y^2 \plus{} z^3) \ge \frac {(a^2x \plus{} b^2y \plus{} c^2z)^3}{(a^3\sqrt {x} \plus{} b^3\sqrt {y} \plus{} c^3)^2}$.
Έστω $ a^2 \equal{} b^2 \equal{} c^2 \equal{} k^2$ τότε $ (x^2 \plus{} y^2 \plus{} z^3)\ge \frac {(k^2 3)^3}{(a^3\sqrt {x} \plus{} b^3\sqrt {y} \plus{} c^3)^2}$.
Από εδώ και πέρα πρέπει κάπως να εκμεταλευτείς την ισότητα του holder για να μπορέσεις να βρεις το $ k$ και να εκφράσεις τα $ a,b,c$ συναρτήσει των $ x,y,z$.........

We will first of all assume that if $ x\plus{}y\plus{}z\equal{}1$, then the least value of $ x^2\plus{}y^2\plus{}z^2$ is 3. Also, we already

know that $ x^2\plus{}y^2\plus{}z^3$ can assume the value 3. So we wish to see if it can acquire a value less than 3. To

this end, we write:

\[ x^2\plus{}y^2\plus{}z^3\equal{}x^2\plus{}y^2\plus{}z^2\plus{}(z^3\minus{}z^2)\]

which reveals that the only chance for $ x^2\plus{}y^2\plus{}z^3$ to have a value less than 3 is if $ z<1$. So fix $ 0<a<1$
and let's investigate what happens if $ z\equal{}1\minus{}a$, in which case $ x\plus{}y\equal{}2\plus{}a$. In this case:

\[ x^2\plus{}y^2\equal{}(x\plus{}y)^2\minus{}2xy\equal{}(2\plus{}a)^2\minus{}2xy\]

so $ x^2\plus{}y^2$ will be least if $ xy$ is maximum. The maximum for $ xy$ occurs if $ x\equal{}y\equal{}a/2\plus{}1$ and in this case,
we have:

\[ x^2\plus{}y^2\equal{}(2\plus{}a)^2\minus{}2(a/2\plus{}1)^2\equal{}\dfrac{a^2}{2}\plus{}2a\plus{}2\]

So for a fixed value of $ a$, the minimum for $ x^2\plus{}y^2\plus{}z^3$ is:

\[ (1\minus{}a)^3\plus{}\dfrac{a^2}{2}\plus{}2a\plus{}2\equal{}3\minus{}a^3\plus{}\dfrac{7}{2}a^2\minus{}a\]

This expression as a function of $ a$ has two critical points and because of its shape the minimum occurs at

\[ a\equal{}\dfrac{7\minus{}\sqrt{37}}{6}\]

It turns out that this minimum is indeed less than 3 and by approximating it we can easily find
numbers $ x,y,z$ that give $ x^2\plus{}y^2\plus{}z^3<3$. In particular $ x\equal{}1.075$, $ y\equal{}1.075$ and $ z\equal{}0.85$ produce:

\[ x^2\plus{}y^2\plus{}z^3\equal{}2.925375\]

I am sure that there has to be a more elegant solution but I was curious to see if the quantity $ x^2\plus{}y^2\plus{}z^3$
can ever get to be less than 3.

By the way: Γιατί μιλάμε αγγλικά;

[quote="Jondrette"]

By the way: Γιατί μιλάμε αγγλικά;[/quote]

Aπλώς ο kunny είναι από την Ιαπωνία....
Γι'αυτό και Durandal του απάντησε στα αγγλικά..... :P  :P

[quote="Jondrette"]Και πως στο καλό διάβασε το πρόβλημα του Γιώργου; Για την πλάκα του ψάχνει τα Ελληνικά μαθηματικά φόρουμ; Φίλε kunny, μήπως μπορείς να μας λύσεις την απορία; 日本人ですか[/quote]

Τι ψάχνεις να βρεις και συ τώρα ρε Jondrette :P . Επι τι ευκαιρία, θα ήθελες να μοιραστείς μαζί μας το όνομά σου :?:

Πάλι πίσω στη λύση της άσκησης και για να ολοκληρώσω την άρχική απάντησή μου. Χρησιμοποιώντας το γεγονός ότι για το συγκεκριμένο $ a$ που βρήκα, ισχύει  $ 3a^2\minus{}7a\plus{}1\equal{}0$, μπορούμε με αντικαταστάσεις να προσδιορίσουμε την
ελάχιστη τιμή της παράστασης $ x^2\plus{}y^2\plus{}z^3$ ως:

\[ \dfrac{541\minus{}37\sqrt{37}}{108}\approx 2.925349874\]

Ή έχω κάνει κάποιο αριθμητικό λάθος (το πιο πιθανό) ή δεν υπάρχει τελικά μια εξαιρετικά κομψή λύση για το πρόβλημα αυτό. Μπορείς να μας βοηθήσεις λίγο Γιώργο; Που την ξετρύπωσες αυτή την άσκηση;

[quote="Jondrette"]
Ή έχω κάνει κάποιο αριθμητικό λάθος (το πιο πιθανό) ή δεν υπάρχει τελικά μια εξαιρετικά κομψή λύση για το πρόβλημα αυτό. Μπορείς να μας βοηθήσεις λίγο Γιώργο; Που την ξετρύπωσες αυτή την άσκηση;[/quote]

omorfo apotelesma dn uparxei. tin asksisi tin psarepsa apo to vivlio tou hung-secrets in inequalities

oso anafora ti methido lusis akolouthoume ta eksis vimata|

an paroume $ a,b$ thetikous pragmatikous, tote apo am-gm exoume

$ x^2\plus{}a^2\geq 2ax$

$ y^2\plus{}a^2\geq 3ay$

$ z^3\plus{}b^3\plus{}b^3\geq 3b^2z$

ara $ x^2\plus{}y^2\plus{}z^3\plus{}2(a^2\plus{}b^3)\geq 2a(x\plus{}y)\plus{}3b^2z$

opote prepei $ 2a\plus{}b\equal{}x\plus{}y\plus{}z\equal{}3$ kai $ 2a\equal{}3b^2$.... prakseis kai kataligoume sto apotelesma.

P.S i parapano methodos einai gnosti os balancing coefficients kai polles fores polu xrisimi

[quote="Jondrette"]Έκανες τις πράξεις για να βρείς τη λύση, κι αν ναι συμπίπτει με αυτή που βρήκα εγώ; Αρχίζω να πιστεύω ότι οι αριθμητικές μου πράξεις είναι σωστές. Σ' ευχαριστώ για τις πληροφορίες για τη μέθοδο λύσης. Στην προκειμένη περίπτωση φαίνεται να μη χρειάζεται αλλά είναι πολύ χρήσιμο να την ξέρει κανείς.[/quote]

variemai na kano tis prakseis :P . opoios thelei omos mporei na sunexisei apo edw-->

to $ \min$ einai $ 6a\minus{}2a^2\minus{}b^3$ opou $ a\equal{}\frac{19\minus{}\sqrt{37}}{12}$ kai $ b\equal{}\frac{\minus{}1\plus{}\sqrt{37}}{6}$

Jondrette, και εγώ την ίδια λύση βρήκα με διαφορετική μέθοδο.

Χρησιμοποιώντας Lagrangians, θέλουμε να ελαχιστοποιήσουμε το $ x^2 \plus{} y^2 \plus{} z^3 \minus{} \lambda(x\plus{}y\plus{}z\minus{}3)$. Υπάρχει ελάχιστο όταν $ \lambda \equal{} 2x \equal{} 2y \equal{} 3z^2$. (Ακριβώς ότι βρίσκει και ο Γιώργος. Οι πράξεις πράγματι δίνουν ακριβώς την ίδια απάντηση που βρήκε και ο Jondrette διαφορετικά.)

[u]Just an an observation:[/u]

Θα αρκούσε να βρούμε την ελάχιστη τιμή της $ x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} 2(x^{2} \plus{} y^{2} \plus{} z^{2})$

[color=red]EDIT: ΛΑΘΟΣ ΔΕΙΤΕ ΤΟ ΕΠΟΜΕΝΟ ΠΟΣΤ ΜΟΥ[/color]

[quote="jitter"][u]Just an an observation:[/u]

Θα αρκούσε να βρούμε την ελάχιστη τιμή της $ x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} 2(x^{2} \plus{} y^{2} \plus{} z^{2})$[/quote]

Huh? :huh:

Παρόμοια προσέγγιση με του Jondrette απλά λίγο πιο άμεση. Η όλη ιδέα είναι να κάνουμε την παράσταση συνάρτηση μιας μεταβλητης.
Έχουμε $ x^2\plus{}y^2\geq\frac{(x\plus{}y)^2}{2}\equal{}\frac{(3\minus{}z)^2}{2}$ . Επομένως αρκεί να βρούμε το ελάχιστο της

 $ f(z)\equal{}z^3\plus{}\frac{(3\minus{}z)^2}{2}$ , $ 0\leq z\leq 3$ . Παραγωγίζοντας έχουμε $ f'(z)\equal{}3z^2\plus{}3z\minus{}3$ . Έ'υκολα τώρα με τις ρίζες βλέπουμε ότι έχει ελάχιστο στο $ z_0\equal{}\frac{\sqrt{37}\minus{}1}{6}$ , και αν το βάλουμε παραπάνω βρίσκουμε ότι το ελάχιστο είναι $ \frac{541\minus{}37\sqrt{37}}{108}$

Ισότητα έχουμε ότι $ x\equal{}y\equal{}\frac{3\minus{}z_0}{2}$ και $ z\equal{}z_0$

[quote="Demetres"]Χρησιμοποιώντας Lagrangians, θέλουμε να ελαχιστοποιήσουμε το $ x^2 \plus{} y^2 \plus{} z^3 \minus{} \lambda(x \plus{} y \plus{} z \minus{} 3)$. Υπάρχει ελάχιστο όταν $ \lambda \equal{} 2x \equal{} 2y \equal{} 3z^2$.[/quote]

Dem κι εγώ έτσι το πήγα - you can't teach an old new tricks and all that :P Αν βάλεις το extra ansatz $ x\equal{}y$, οι πράξεις δεν είναι τραγικές...

Sil, πολύ κομψός ως συνήθως :)

Cheerio,

Durandal 1707

[quote="jitter"]Ζητώ συγγνώμη για το λάθος αλλά είχε πήξει το κεφάλι μου, ξημερώματα. Η σκέψη μου ξεκίνησε κάπως αλλιώς. Σκέφτηκα ότι [color=red]όταν έχουμε μια συνάρτηση και θέλουμε να την ελαχιστοποιήσουμε, αν καταφέρουμε με κάποιο τρόπο να δείξουμε ότι όταν η συνάρτηση ελαχιστοποιείται όλες η μεταβλητές σε αυτή είναι ίσες τότε αρκεί να βρούμε το ελάχιστο της νέας συνάρτησης που προκύπτει αν πάρουμε το κυκλικό άθροισμα της πρώτης συνάρτησης[/color]. Δεν νομίζω ότι μια τέτοια μέθοδος έχει εφαρμογές αλλά τουλάχιστον είναι σωστή σαν ιδέα. Τώρα κάπως-πως εκείνη την ώρα, λόγω κούρασης και βιαστικότητος «ανακάτεψα» αυτή την ιδέα με την άσκηση και κατέληξα στο να κάνω αυτή τη λανθασμένη διατύπωση.[/quote]

Αυτό που αναφέρεις γενικά σε κόκινο θυμίζει πολύ την εφαρμογή της αρχής του Sturm σε ανισότητες. Μπορείς να βρεις περιγραφικά και αναλυτικά στο Putnam and Beyond και ένα παράδειγμα χρήσης της [url=http://www.mathlinks.ro/viewtopic.php?p=1503839#1503839]εδώ[/url]. Αν και στο παράδειγμα που σου δίνω ο σιλουανός το σκότωσε όπως έκανε και εδώ... :roll:


trigonometry
integration
floor function
algebra
polynomial
function
calculus

Sa se demonstreze ca pentru $x \in (0,\pi)$ avem:

$\frac{sin{x}}{1}+\frac{sin{2x}}{2}+...+\frac{sin{nx}}{n}> 0$

Vezi [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Jackson&t=20812]aici[/url] o alta demonstratie.

[quote="perfect_radio"]Vezi [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Jackson&t=20812]aici[/url] pentru o alta demonstratie.[/quote][b][color=green]
Va multumesc ! [/color] [/b] Adesea am punctul de vedere (nu stiu daca este cel mai bine  ales) ca pentru a demonstra o inegalitate de forma
$A> 0$ (in real),  este "benefic de a reprezenta " pe $A$ " ca si (e.g. ) o   suma de "cantitati " pozitive.

Beneficiul ar fi ca am putea, " in astfel de situatii " , sa studiem asa numitul "rest".
[b] Cu alte cuvinte sa demonstram inegalitatile prin(cu ajutorul )... intermediul  [u]egalitatilor[/u].[/b]

Călcat-up inegalitate $ \sum_{k\equal{}1}^{n}\frac{\sin{kx}}{k}>x\left(1\minus{}\frac{x}{\pi}\right)^3$ deţine pentru $ x \in (0,\pi).$

Vezi, de asemenea, aici : http://www.mathlinks.ro/viewtopic.php?t=54360

[quote=flip2004][quote]
[b]TEOREMA (L.Fejer-1910,  D. Jackson-1912, T.J.Gronwall- 1912).[/b][i] Pentru $x\in (0,\pi)$ au loc inegalitatile
\[(1)\; \; \; \; \; \; \begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin{kx}}{k}> 0 \\ \\ \hline \end{array}\; \; ,\; \; \forall n \in{\mathbb N}. \]
[/i]
[/quote]

[b][color=blue] DEMONSTRATIA I . [/color][/b]
Prin efectuarea unor calcule elementare se constata
\[\frac{d}{d\varphi}\left\{ \frac{\sin{2k\phi}}{(\sin{\phi})^{2k}}\right\}=-2k\frac{\sin{(2k-1)\phi}}{(\sin{\phi})^{2k+1}}\; . \]
 Efectuand calculele precum si substitutia $\phi \leadsto \frac{x}{2}\; ,\;$ prin integrare  avem
\[\frac{\sin{kx}}{k}=2\left(\sin{\frac{x}{2}}\right)^{2k}\int_{\frac{x}{2}}^{\frac{\pi}{2}}\frac{\sin{(2k-1)\phi}}{(\sin \phi)^{2k+1}}\; d \phi \; \; ,\; k\in{\mathbb N}. \]
Insumand pentru $k\in \{1,2,...,n\}$ se obtine
\[\sum_{k=1}^{n}\frac{\sin{kx}}{k}=2\int_{\frac{x}{2}}^{\frac{\pi}{2}}\sum_{k=1}^{n}\left[r(x,\phi)\right]^{k}\frac{\sin{(2k-1)\phi }}{\sin{\phi}}\; d\phi \]
unde \[r(x,\phi): =\left(\frac{ \sin\frac{x}{2}}{\sin{\phi}}\right)^{2}\in [0,1] \]
pentru $0\le \frac{x}{2}\le \phi<\frac{\pi}{2}\; .$ Egalitatea de mai sus implica
\[\sum_{k=1}^{n}\frac{\sin{kx}}{k}=\int_{x}^{\pi}\sum_{k=1}^{n}\left[r(x,\frac{\psi}{2})\right]^{k}\frac{\sin{\left(k-\frac{1}{2}\right)\psi }}{\sin{\frac{\psi}{2}}}\; d\psi\; . \]
Aplicand identitatea lui Abel ("'insumarea prin parti"), adica
\[\sum_{k=1}^{n}A_{k}B_{k}=\sum_{k=1}^{n-1}\left(A_{k}-A_{k+1}\right)\sum_{j=1}^{k}B_{j}+A_{n}\sum_{j=1}^{n}B_{j}\; , \]
si notand
\[\left\{\begin{array}{rcl}{\mathbf r}&: =&r(x,\frac{\psi}{2})\; \; , \; \;{\mathbf r}\in [0,1]\; \; \; \mbox{vezi}\; \; ()\\{\mathcal F}_{n}(\psi)&: =&\sum_{k=1}^{n}\sin{\left(k-\frac{1}{2}\right)\psi}\; \; \; , \; \;{\mathcal F}_{n}(\psi)\ge 0 \; \; \mbox{dac'a}\; x\in [0,\pi]\; \;-\mbox{vezi L. Fej\'er}\end{array}\right.\; , \] din identitatea lui Abel g'asim
\[\sum_{k=1}^{n}\frac{\sin{kx}}{k}=\int_{x}^{\pi}\left\{(1-{\mathbf r})\sum_{k=1}^{n-1}{\mathbf r}^{k-1}{\mathcal F}_{k}(\psi)+{\mathbf r}^{n}{\mathcal F}_{n}(\psi)\right\}\frac{d\psi}{\sin{\frac{\psi}{2}}}\; . \]
Deoarece ${\mathcal F}_{n}(\psi)=\frac{1-\cos(n\psi)}{2\sin{\frac{\psi}{2}}}\; ,$ concludem cu
\[\begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin{kx}}{k}=\frac{1}{2}\int_{x}^{\pi}\left\{(1-{\mathbf r})\sum_{k=1}^{n-1}{\mathbf r}^{k-1}\left(1-\cos(k\psi)\right)+{\mathbf r}^{n}(1-\cos(n\psi))\right\}\frac{d\psi}{\left(\sin{\frac{\psi}{2}}\right)^{2}}\\ \\ \hline \end{array}\; . \]
Aceasta reprezentare completeaza demonstra'tia I.

[b][color=blue]DEMONSTRATIA II. [/color][/b]
Fie $P_{n}(x)$ polinomul lui Legendre de gradul $n$, adic'a
\[\begin{array}{c}P_{n}(x)=\frac{1}{n! 2^{n}}\left[(x^{2}-1)^{n}\right)^{(n)}={}_{2}F_{1}\left(-n,n+1;1;\frac{1-x}{2}\right)=\\ \\ =\frac{1}{2^{n}}\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}{n\choose 2k}(-1)^{k}\frac{(n+1)_{n-2k}(2k)!}{k!(n-k)!}x^{n-2k}\end{array}\]
unde
\[\begin{array}{c}{}_{2}F_{1}(-n;b;c;z): =\sum_{k=0}^{n}(-1)^{k}{n\choose k}\frac{(b)_{k}}{(c)_{k}}z^{k}\\ (d)_{k}: =d(d+1)\cdots (d+k-1)\; ,\; \; k\in{\mathbb N}\; ,\; (d)_{0}: =1 . \end{array}\]
Se cunosc urmatoarele:
-- radacinile lui $P_{n}(x)$ sunt reale,distincte, situate in $(-1,1)$;
-- $|P_{n}(t)|\le 1 \; ,\; \; \forall t\in [-1,1] .$
Demonstratia a II-a (vezi [16] precum si comentariile lui R.Askey )
se bazeaza pe identitatea:
\[\begin{array}{|c|}\hline \\ \sum_{k=1}^{n}\frac{\sin(k\cdot\arccos{x})}{k}= \frac{\sqrt{1-x}}{2}\int_{-1}^{x}\frac{1-P_{n}(y)}{1-y}\frac{dy}{\sqrt{x-y}}\\ \\ \hline \end{array}\; ,\; x\in (-1,1). \; .\]

[b]Observatii.[/b]

1) Inegalitatea (1) a fost conjecturata de catre Leopold (Lipot) Fejer. Ulterior a fost demonstrata de catre D.Jackson-[14] si T.H.Gronwall-[12].
In prezent se cunosc peste 100 de demonstratii. Am ales pentru a Va prezenta pe cele pe care subsemnatul le considera "mai simple".

 Se spune ca pana la moasrte L.Fejer a cautat sa gaseasca noi  demonstratii a lui (1). L.Fejer a predat si la Universitatea din Cluj (sub numele de L.Weiss).
2)Desi simpla la prima vedere, inegalitatea (1) a dat de furca matematicienilor.
3) Inegalitatea (1) intervine in urmatoarele domenii: Serii Fourier (fenomenul lui Gibbs - vezi [13]), Polinoame ortogonale ([1]-[5],[18] ),  Functii Complexe (Demonstratia conjecturii lui Bieberbach,functii  univalente, [4],[6]), Teoria Aproximarii ([5]).

[b]BIBLIOGRAFIE.[/b]
[1]R. Askey , [i] Orthogonal Polynomials and Special Functions[/i],
Regional Conf.Lect.Appl.Math., vol.21,SIAM,Philadelphia,Pa., 1975.
[2]R.Askey [i]Positive quadrature methods and positive polynomial sums[/i], 'in  Approximation Theory V, Academic Press, 1986.
[3]  R. Askey and J. Fitch , [i]Integral reprezentations for Jacobi
polynomial amd some applications [/i],J.Math.Anal.Appl., 26 (1969)
411-437.
[4]R. Askey and G. Gasper , [i]Positive Jacobi polynomial sums,(II)[/i],
 Amer.J.Math., 98 (1976) 709-737.
[4]R. Askey and G. Gasper , [i]Inequalities for polynomials}, 'in "The Bieberbach Conjecture", Proc.of the Symposium on the Occasion
of the Proof,[/i] Mathematical Surveys and Monographs,  21, Amer.Mathematical Society, 1986, 7-32.
[5]H. Bavinck , [i] Jacobi Series and Approximation[/i],
Mathematical Centre Tracts 39, Mathematisch Centrum Amsterdam 1972.
[6] L. de Branges , [i]The Story of the Verification of the Bieberbach Conjecture[/i], 'in " The Bieberbach Conjecture", Proc.of the Symposium on the Occasion of the Proof, Mathematical Surveys and Monographs,  21, Amer.Mathematical Society,  1986, 199-203.
[7] L. Fejer , [i]Sur les fonctions bornees et integrables,[/i]
 C.R.Acad.Sci.Paris 131 (1900) 984-987.
[8] L. Fejer , [i]Sur le develpopment d'une fonction arbitraire suivant les fonctions de Laplace, [/i] C.R.Acad.Sci.Paris , 146 (1908) 224-227.
[9]L. Fejer , [i]Ueber die Laplacesche Reihe[/i], Math. Ann. (1909)76-109.
[10] L. Fejer , [i]Einige Saetze , die sich auf das Vorzeichen einer ganzen rationalen Funktion beziehen u.s.w.,[/i] Monta.fuer Math. und Phys., 35 (1928) 305-344.
[11]L. Fejer , [i] Gesammelte Arbeiten (I)-(II),[/i]  Birkhauser Verlag,  Basel,  1970 .
[12]T.H. Gronwall , [i]Ueber die Gibbsche Erscheinung und die trigonometrischen Summen    $\sin{x}+(1/2)\sin{2x}+...+(1/n)\sin{nx}$ [/i] ,  Math.Ann., 72 (1912) 228-243.
[13] E. Hewitt and R.E. Hewitt , [i]The Gibbs-Wilbraham phenomenon: an epsiode in Fourier analysis[/i], Arch.Hist.Exact Sci., 21 (1979)  129-160.
[14]D. Jackson , [i]Ueber eine trigonometrische Summe,[/i] Rend.Circ.Mat.Palermo 32(1911)  257-262.
[15]A. Lupas , [i]Advanced Problem 6517,[/i]   Amer.Math.Monthly (1986) p. 305 ; (1988) p.264.
[16]A. Lupas , [i]Advanced Problem 6585,[/i]  Amer.Math.Monthly
(1988) p. 880 ; (1990) p.859-860.
[17]L. Lupas , [i]An identity for ultraspherical polynomials,[/i] Revue d${}^{,}$Analyse numerique et de Theorie de l'approximation, tome 24 , no.1-2 (1995) 181-185.
[18]G. Szego , [i] Orthogonal Polynomials ,[/i] Amer.Math.Soc.Colloq. Publications  vol.23, fourth ed.,Amer.Math.Soc., Providence, R.I.,   1975.[/quote]

Very complete bibliographie


If a = 1, l = 2, g = 3, e = 4, b = 5, and r = 6, what is the value of a + l + g + e + b + r + a?


We consider the polynomial $f=X^{3}+3X+3$ and denote by $a$ the only real root of $f$.
Denote by $Q(a)=\{g(a)/g \in Q[X]\}$.
Prove that,if $\alpha \in Q(a)^{*}$ then $\frac{1}{\alpha}\in Q(a)$.

If $\alpha =g(a)\in Q[a]\not =0$, then exist $p(x),q(x)\in Q[x]$, suth that $g(x)p(x)-q(x)f(x)\equiv 1$,
therefore $\frac{1}{\alpha }=p(a)\in Q[a]$.

Because $a$ is algebraic over $\mathbb{Q}$ therefore $\mathbb{Q}[a]$ is a field, and that's all.