LaTeX
probability
arithmetic sequence
geometric sequence

Welcome to ERS AoPS Version!

This game is played only with the "slapping" part of the game.

Here's how it works. No sign ups needed. Slap in whenever you want. I will be using my own shuffled deck. Quite often, I will post the next card I draw. Whenever you see a
-Red 10
-Sandwich
-Ladder
-Double
-Arithmetic Sequence
-Geometric Sequence
post and say slap!
(I made up the last 2)

If you are the first to slap, and you do it correctly, then you collect as many points as cards were in that pile. Then I put down another card to start another round. Match continues until 100 points are given out. The winner of each match will get a "super point" for the master game. Scoreboard will be available. Hope this doesn't phail.

Starting now!

Seven of Spades
Two of Diamonds
Four of Diamonds
Six of Spades

What's a ladder?

Can others also post cards?

I post the cards.

Red 10- just a red 10
Ladder- 4,5,6 or 9,8,7, or J,Q,K or anything else like that
Sandwich- 5-6-5, 3-8-3, J-4-J, etc.
Arithmetic, Geometric- self explanatory, aces count as 1's
Doubles- two of same card in a row

Ace of Diamonds

[hide]lotsofmath- 4[/hide]

Ace of Diamonds
Seven of Clubs
Five of Hearts
King of Hearts
King of Clubs

Jack of Spades

[hide]crazypianist1116- 5
lotsofmath- 4[/hide]

[hide=" :) "]9 $ \spadesuit$[/hide]

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
fortenforge- 4
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

[hide=" :) "]9 $ \spadesuit$[/hide]
[hide=" :lol: "]2 $ \spadesuit$[/hide]

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
fortenforge- 4
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

Smileys don't count as cards, do they? :)

Slapper-time is approaching.....   *rubs hands together in anticipation*

Fine, I won't hide this card. :)

[hide=" :) "]9 $ \spadesuit$[/hide]
[hide=" :lol: "]2 $ \spadesuit$[/hide]
9 $ \heartsuit$

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
fortenforge- 4
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

2 $ \diamondsuit$

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
fortenforge- 4
lotsofmath- 3
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

2 $ \diamondsuit$
K $ \heartsuit$
A $ \heartsuit$

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
fortenforge- 4
lotsofmath- 3
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

2 $ \diamondsuit$
K $ \heartsuit$
A $ \heartsuit$
K $ \spadesuit$

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
fortenforge- 4
lotsofmath- 3
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

9 $ \spadesuit$
6 $ \diamondsuit$

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
chenhsi- 4
fortenforge- 4
lotsofmath- 3
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

9 $ \spadesuit$
6 $ \diamondsuit$
3 $ \clubsuit$

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
musicnmath- 6
chenhsi- 4
fortenforge- 4
lotsofmath- 3
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]

[hide="Scoreboard"]
[hide="Overall Championships"]
ROUND ONE: 1. [b]r15s11z55y89w21[/b] 2. mathfreak123 3. lotsofmath
[/hide]
[hide="Round 2"]
ksun48- 8
BOGTRO- 8
chenhsi- 7
musicnmath- 6
fortenforge- 4
lotsofmath- 3
[/hide]
[hide="Round 1"]
r15s11z55y89w21- 26
mathfreak123- 16
lotsofmath- 15
PhireKaLk6781- 8
yaofan- 8
mz94- 6
abcak- 5
budi713- 5
1=2- 5
crazypianist1116- 5
ytrewq- 4
eggylv999- 3
[/hide][/hide]


怎么中国的国家队这么强？
我来了加拿大4 年了，一年不入一年啊？
中国同胞都怎么学习阿，给传授传授经验。

哈哈,经验就是不要命的作题拉,虽然我没经验,可是我们学校有一个牛人今年拿了一块金牌回来,我那个羡慕呀!听说他把能看到的参考书都做了......


FTW geometry perimeter probability LaTeX MATHCOUNTS AMC

I agree with alexhhmun...

Sometimes if it's a really easy question, I can't help but *remember* what the answer is.
I don't try to memorize it, but seriously, I can't help it sometimes...

It just comes to me.

What I suggest for memming is: maybe FTW should put the same [i]type[/i] of problem, but different values. Like, "How many diagonals does a 145234-gon have" instead of, say, an octagon. It isn't to hard to put different values of things on most problems.
Another example might be, "How many four-digit integers greater than [x thousand] are there for which the thousands digit equals the sum of the other three digits?"
This method doesn't currently seem to be in use in FTW, and will prevent, for the most part, accidental memming.

No. If you are sitting in front of your computer, and you see

a question, which you answer correctly, then it gets stored

in your mind. When you see the question again, then You know the

answer. It isn't your fault. If you know the answer, you aren't just

going to sit there, just because you memmed (Unless you are very

stupid). My answer: No.

A lot of people will disagree with this, but if can say that you have

NEVER memmed an answer, you have never played FTW in your

life. In a sense, everyone is a memmer. This doesn't mean they

TRY to mem, or try to ruin the game, it just happens.

So, for whoever answered yes and doesn't like this post,

Get over it. You've memmed before, everyone has. So, Get over it.

Memmers have been, are, and always will be in FTW, the number

of memmers will be equal to the number of players.

Nothing can stop people from memorizing an answer. If you see

an answer, you're going to remember it. Duh.

See about 50 other complaints if you want to complain. It has

already been said that memming cannot be stopped. If you want

to, talk to the administrators about random numbers

(Another post has been made about this)

But still, the answer is no.

EDIT: Valentin, That would be cool.

[quote="Valentin Vornicu"]Don't count on memming. We will have a way in the future to scramble number results, so memming now will get you really confused in the future :)[/quote]

That happened to my friend. There is this one parcel question which has like the last sentence differnet that throws off the answer. He memmed it as "1.06" when it asks in terms of something else. :P

There was this problem that didn't have a diagram, and it asked for the perimeter, and the number 48 just came to me, which was the right answer.

Sometimes, I get a problem in a game, and then I get the exact same problem two games later. It's weird.

They whould just get more problems so the probability of a problem reappearing is very low.

I need to post an addendum to the future scrambling of problems.

This is still a very difficult thing to do as one has to make sure the problems that are either generated randomly or are changed statically by someone will have to be verified to work for all possible values.  This is an exceedingly time consuming process, and certainly would not work for many of the problems that we currently have in the database, especially those that involve images.

In addition, if we made the problems random, we'd have to generate a separate image file for EVERY problem displayed.  For example, say we had the problem $ \text{Simplify } x \plus{} y$ where we chose x to be a random number between 1 and 10 and y to be a random number between 1 and 5.  This would generate 50 different LaTeX image files.  Now, make the random range bigger or apply it to thousands of problems, and you can see how unwieldy this would get.  For those who would reference saab as being able to do it, remember that their problem set is quite limited and they do not need to generate separate LaTeX files for each problem - which we need to do in order for typesetting the problems and making them work in Flash.

This is not to say that this is impossible and won't happen, but rather it is a time consuming process with some significant logistical issues, which means you shouldn't expect to see it anytime in the immediate future.

Also, in reference to adding more problems.  We currently have over 2800 problems.  If anyone knows of a book which has (a) more problems of the same nature as the type that are found in FTW and  (b) has problems we can use without violating copyright, please let us know.

What I find nice is changing the last sentence, often to change the term in wihch something has to be expressed. Then people have to actually read teh problem

I don't understand how memorization harms FTW. If you see a problem on a test and remember the answer is $ \pi$, are you seriously going to resolve the whole problem every single time you see the exact same problem on a test in the future?

(yes, I memorize. I'm not as hardcore a memmer as, say, SamE was back in the beginning. But I do mem, and will unless it's forbidden.)

[quote="Valentin Vornicu"]Don't count on memming. We will have a way in the future to scramble number results, so memming now will get you really confused in the future :)[/quote]

Woohoo, then I'll get the chance to lose even MORE badly!  YAY!

I agree with bleuwiki.  That's all.  I think that people should NOT look through the answers of a game to find answers, but to learn.  For anyone who says this sounds corny, that's too bad for you.

[quote="zephyredx"]I had the same problem appear consecutively once, and I got it right the first time. What could I do?

It's true that memming the whole time could get you good ratings, but in the end, what benefit does it give? It's pointless for mathcounts, doesn't make you any faster, and sooner or later you'll screw up because you didn't read a detail. So I think memming as a strategy is not beneficial in the long run.

I do mem sometimes, but mostly on problems with bugs and those with memorable diagrams (you can't really help with those). However, I would much rather try and lose than win by memming. It is the sensation of speed coming from your head that makes ftw fun.[/quote]Agreed, but everyone here will still ignore that fact.

I don't really think memmers cause that big of a deal. youre playing ftw because you either want to get better or just for the fun of it. memmers are nto going to hurt you from improving your cd skills,  and if youre playing for fun, i don't think a person gettign the answer right in 2 seconds will ruin it. they can't mem all of them  :P

Memming concerns only the people who are playing to win (*hypocrites). I personally hate making distinctions between mem and fast solving, also, as several times I have seen the conversation "Did you mem?" "No" "Yes you did" and I don't think the defendant is lying (especially sometimes on the part of alexhhmun and #H34N1 (and me)).

Next, as many people have said on this thread, memming is natural. So while people can try to prevent memming, it'll always prevail.

Finally, meming as a strategy is the worst idea I have ever heard of; FTW is supposed to help your math skills (which it did a lot for me). If you mem every question, what good is FTW doing for you? So don't mem, appreciate.

lol who cares its not like FTW will really determine your true mathematical abilities.  Do it for yourself (judge winning and losing relative to yourself) so try to work on every problem to really see how good you are.  If you just memorize it, you are not actually benefiting and you will not improve.  Yea you get your username on the leaderboard, but who cares?...if you make National Mathcounts youll get your real name on the mathcounts homepage, same with USAMO and many other math competitions.  (and people actually know mathcounts, USAMO,...)
^hehe didnt realize the person above me made the same argument

Doesn't it depend on how you mem?  Because if you see the same problem twice in a row, would you work it again?  Also, I think that the only bad way of memming is looking at the answers after the game.

I think sitting down and actually reviewing questions to simply MEMORIZE them is despicable, which may sound a bit harsh, but it is cheating, after all! Sometimes I remember the answer, but not on purpose, it just sticks to my brain. What am I supposed to do, solve it again while knowing the answer?

But, like Alex said, you should review to learn, not to memorize so you can cheat(and no, it's not corny.)

I(like everyone) also mem, but I think that memming is fine as long as you actually solved the problem, as opposed to looking at the answers after the game. If you solved a problem once, and you get it again, is it really going to help your math skills to resolve the problem? If you actually solved the problem, you will mem it. It just happens. So, no, I don't think memmers ruin FTW, unless that is their strategy for winning, in which case they seem to miss the reason FTW is fun.

Super --

Well said.  :clap2:

(And in case of any mix-ups, no, I'm not being sarcastic.)

you can mem by accident but you should not try to lik sometimes people play by themselves to mem answers

theres nothing wrong with reviewing answers....i do it, to find errors, but NOT to mem...also to learn, which is the point of FTW

[quote="alexhhmun"]I try very hard not to m3m.  And THANKS A TON for thinking I'm a hypocrite, I think I already know that...

Please argue your case, bearing in mind that we all m3m SOMETIMES.

Yes, even #H34N1, no matter how he may deny it.[/quote]
If it's easy I tend to do it or if i cant remember how to solve it.
So basically it's a "last resort" thing

I never TRY to mem, but it always happens..........

[quote="myyellowducky82"]Try not to mem :wink:[/quote]

[hide="My only solution for not memming."]Take a break from FTW.  I've done it before.[/hide]

i've done that b4 too
during school, i don't really get to go on as often, so when i did go on again, i didn't remember as much of the probs


In triangle $ABC$ prove that:
$\sum{asin{\frac A2}} \geq p$
$a,b,c$ are the sides of triangle and $p=\frac{a+b+b}2$

with Cauchy+Schwarz inequality we have:$(\sum asin\frac{A}{2})(\sum \frac{a}{sin\frac{A}{2}})\geq 4P^2$
so it is enough that we prove:$  (\sum \frac{a}{sin\frac{A}{2}})\leq4p$
or:$\sum a.AI\leq 4S$(I is cross of bisectors and S is surface.)
we know:$AI=\frac{bc}{p}\cdot cos\frac{A}{2}$
and we know:$S=\frac{abc}{4R}$
so it is enough that we prove:$\frac{p}{R}\geq \sum cos\frac{A}{2}$
we know:$\frac{\sqrt{3}}{2} \geq \sum cos\frac{A}{2}$
so we want to prove :$\frac{p}{R} \geq \frac{\sqrt{3}}{2}$
or:$a+b+c\geq R\sqrt{3}$
we know:$a=sin A \cdot 2R$
so it is enough that we prove:$\sum sin A \geq \frac{\sqrt{3}}{2}$
and it is true

[quote="amparvardi"]Chejur shode in postaye ghadimi ro zende kardin ? :)[/quote]
be ghadimi boodanesh kari nadashtam baraye in ke kasi be in soal javab nadade bood in soal ro javab da :lol: dam


$ 1.)$ Prove that $ \forall M$ a point ,  $ \triangle{ABC}$ is a triangle and $ O$ - its circumcenter , there is the identity :

\[ ( \sin{2A} \plus{} \sin{2B} \plus{} \sin{2C}) \cdot \vec{MO} \equal{} \sin{2A} \cdot \vec{MA} \plus{} \sin{2B} \cdot \vec{MB} \plus{} \sin{2C} \cdot \vec{MC}\]

This can be proved quite easily with mass points (barycentrical coordinates). The Idea is that $ P \equal{} \alpha A \plus{} \beta b \plus{}\gamma c$ with $ a\plus{}b\plus{}c\equal{}1$ iff $ \alpha: \beta: \gamma\equal{}[BCP]: [CAP]: [ABP]$.


logarithms
calculus
calculus computations

find the sum :D

$ \frac{5}{1\cdot 2\cdot 3}\plus{}\frac{7}{2\cdot 3\cdot 3^2}\plus{}\frac{9}{3\cdot 4\cdot 3^3}\plus{}\frac{11}{4\cdot 5\cdot 3^4}\plus{}\frac{13}{5\cdot 6\cdot 3^5}\plus{}\;\ldots$

$ S \equal{} \sum_{n\equal{}1}^{\plus{} \infty}\frac{2n\plus{}3}{n(n\plus{}1)3^n}$.

Let's find the sum of $ T \equal{} \sum_{n\equal{}1}^{\plus{} \infty}\frac{(2n\plus{}3)x^n}{n(n\plus{}1)} \equal{} 2\sum_{n\equal{}1}^{\plus{} \infty}\frac{x^n}{n\plus{}1} \plus{} 3\sum_{n\equal{}1}^{\plus{} \infty}\frac{x^n}{n(n\plus{}1)}$.

From $ \sum_{n\equal{}1}^{\plus{} \infty}x^n \equal{} \frac{x}{1\minus{}x}$ we get $ \sum_{n\equal{}1}^{\plus{} \infty}\frac{x^{n\plus{}1}}{n\plus{}1} \equal{} \minus{}x\minus{}\ln|x\minus{}1|$ and so $ \sum_{n\equal{}1}^{\plus{} \infty}\frac{x^n}{n\plus{}1} \equal{} \minus{}1 \minus{} \frac{\ln|x\minus{}1|}{x}$.

From $ \sum_{n\equal{}1}^{\plus{} \infty}x^{n\minus{}1} \equal{} \frac{1}{1\minus{}x}$ we have $ \sum_{n\equal{}1}^{\plus{} \infty}\frac{x^n}{n} \equal{} \minus{} \ln|1\minus{}x|$, and so $ \sum_{n\equal{}1}^{\plus{} \infty}\frac{x^{n\plus{}1}}{n(n\plus{}1)} \equal{} (1\minus{}x) \ln|1\minus{}x| \plus{} x$. Hence,

$ \sum_{n\equal{}1}^{\plus{} \infty}\frac{x^n}{n(n\plus{}1)} \equal{} \frac{(1\minus{}x) \ln|1\minus{}x|}{x} \plus{} 1$.

Therefore, $ S \equal{} 2\left(\minus{}1 \minus{}3 \ln\frac{2}{3}\right) \plus{} 3\left(2 \ln\frac{2}{3} \plus{} 1\right) \equal{} 1$.


function
integration
calculus
real analysis
real analysis unsolved

Let $f : \mathbb R \to \left( 0, \infty \right)$ be a Riemann integrable function such that $\int_{0}^{1}f(t) \, dt = \int_{0}^{1}f^{2}(t) \, dt$.
Prove that there are $x_{1}, x_{2}\in \mathbb R$ such that
\[\int_{x_{1}}^{x_{2}}\frac1{f(t)}\, dt = 1 . \]
[i]Atila Abdula[/i]

So far, I've proved that $\int_{0}^{1}f\le1$ and that $\int_{0}^{1}f\le\int_{0}^{1}\frac1f$.

Neither of those seems to be the full answer, since I want $\int_{0}^{1}\frac1f\ge 1.$

By Cauchy, $\int_{0}^{1}f\cdot \int_{0}^{1}\frac1f\ge \left(\int_{0}^{1}\sqrt{f}\cdot\sqrt{\frac1f}\right)^{2}=1$. That gives $\int_{0}^{1}\frac1f\ge 1$ when combined with your first result.

OK, that's it. That first result was also from Cauchy-Schwarz:

${\int_{0}^{1}f\le1\cdot\left(\int_{0}^{1}f^{2}\right)^\frac12}= \left(\int_{0}^{1}f\right)^{\frac12},$

Then square and divide.

Once we have $\int_{0}^{1}\frac1f>1,$ then we use the fact that the integral is a continuous function of its endpoints, the fact that the integral over a single point is zero, and the Intermediate Value Theorem to finish it off.

[quote="perfect_radio"]How do we know that $\int_{0}^{1}\frac1{f(t)}\, dt$ exists?[/quote]
We don't, and we don't have to. Since $f>0,$ it is unambiguously defined as an improper integral - either as convergent or as $+\infty.$ Either way, the final intermediate value argument follows.


Find the general term of the recursive sequence:
$G_{0}= a$
$G_{1}= b$
$G_{n+2}= G_{n+1}+G_{n}$

[hide="A word to the wise"] You can solve this general case with the characteristic polynomial, but before you try, consider a simple induction. ;) [/hide]

[hide="hint"]if we calculate a few terms, then we see the fibonacci sequence emerge!![/hide]

i guess $a_{n+2}=a_{n+1}+a_{n}$ doesnt give it away..


Let ABC be a triangle.Prove that
$ \frac{1}{sin\frac{A}{2}}\plus{}\frac{1}{sin\frac{A}{2}}\plus{}\frac{1}{sin\frac{C}{2}}$≥$ 4\sqrt{\frac{R}{r}}$

At first we have: $ \frac {R}{r} \equal{} \frac {1}{4\prod {sin\frac {A}{2}}}$.

The inequality is equivalent to:

$ \sum {\sqrt {\frac {sin\frac {B}{2}.sin\frac {C}{2}}{sin\frac {A}{2}}}}\ge 2$

$ \Leftrightarrow \sum {\sqrt {\frac {p \minus{} a}{a}}}\ge 2$, $ (*)$

Denote $ (p \minus{} a,p \minus{} b,p \minus{} c) \equal{} (x,y,z)$.

$ (*)\Leftrightarrow \sum {\sqrt {\frac {x}{y \plus{} z}}}\ge 2$
But the above one is obviously true due to the $ AM \minus{} GM$.
Our proof is completed then.


[code]private static void sort(List<Comparable> theList){
	Iterator<Comparable> itr = theList.iterator();
	PriorityQueue<Comparable> pq = new PriorityQueue<Comparable>();

	while (itr.hasNext()){
		pq.add(itr.next());
		itr.remove();
	}

	while(!pq.isEmpty()){
		theList.add(pq.remove());
	}
}[/code]

If the parameter is a LinkedList in a sorted order, which data structure operation dominates the running time of the sort? (refer to the above code)

$ \textbf{(A)}\ \texttt{pq.add}\qquad
\textbf{(B)}\ \texttt{itr.remove}\qquad
\textbf{(C)}\ \texttt{pq.isEmpty} \qquad
\textbf{(D)}\ \texttt{pq.remove}\qquad
\textbf{(E)}\ \texttt{theList.add}$

The answer is D, but I don't get how they got that answer. I thought adding to a LinkedList was $ O(n)$ and removing from a PriorityQueue was $ O(\log n)$?

Adding to a linked list is O(1) as long as you're adding to the back of the list (which in this case you are).

So just for practice, I decided to find the Big-O for each of the choices. Are A, B, C, and E: $ O(1)$ and D: $ O(\log n)$?

Of course it always depends on the implementation of your linked list and priority queue. But generally your answers are correct. Except for (A), which I'd expect to be a log N operation. (This is true for heap or red-black tree implementations.) However, you can achieve O(1) time if you use Fibonacci heap.

[quote="bus"]However, you can achieve O(1) time if you use Fibonacci heap.[/quote]

I think Java's PriorityQueue implementation uses a Fibonacci heap, so it would be O(1).


induction
calculus
calculus computations

Let $ x_0\equal{}0, x_1\equal{}1$, and $ x_{n\plus{}1}\equal{}\frac{x_n\plus{}nx_{n\minus{}1}}{(n\plus{}1)}$ Show that the sequence $ x_n$ converges and find its limit.

Why don't you compute the values for the first few n's, say for n = 3, 4,5. This is usually done for two reasons: to see if the sequence is monotone and also to see if the sequence is bounded and then one uses induction to prove those two things and then one cites the Monotone Convergence Theorem for one part of your question while the other part boils down to simple algebra. Sometimes, a sequence is simply constant after some fixed n.

Except that this sequence is not monotone, and nothing you've said applies here.   :roll:

Actually, computing a few terms of this sequence is a useful exercise, but that's because it's not terribly hard to guess an explicit formula, which can then be proved by induction.

So I do exactly what akech asks for. Only I'm using a spreadsheet (very friendly to recursively defined sequences) so I run it out to around $ n \equal{} 30$ or so.

The quick observations:

1. It's not monotone. Rather, it alternates between larger and smaller numbers.

2. These larger and smaller numbers seem to be squeezing together. Convergence is slow, but it looks like there should be a limit somewhere near $ 0.7.$

3. These numbers look .... [i]familiar.[/i] Say, maybe, partial sums of the alternating harmonic series?

[b]Conjecture:[/b] $ x_n \equal{} \sum_{k \equal{} 1}^n\frac {( \minus{} 1)^{k \minus{} 1}}{n}$

Carry on ...

Well, the explicit formulae are great but this is an analysis subforum, not an algebra one :P. So, let's generalize a bit to make it a pure analysis problem. Consider any sequence $ a_n\ge 1$ such that $ \sum_{n\ge 1}a_n^{\minus{}1}\equal{}\plus{}\infty$ and define $ x_{n\plus{}1}\equal{}\frac {x_n\plus{}a_n x_{n\minus{}1}}{a_n\plus{}1}$. Show that $ x_n$ converges.
[hide="Hint"]
If $ x_{n\minus{}1}$ and $ x_n$ are the endpoints of some interval, then $ x_n$ and $ x_{n\plus{}1}$ are endpoints of a slightly shorter subinterval.
[/hide]

Experimentally, it appears that if you use a sequence such that $ \sum a_n^{\minus{}1}<\infty$ in fedja's definition, then the sequence $ x_n$ tends to a stable 2-cycle.

$ x_{n \plus{} 1} \equal{} \frac {x_n \plus{} nx_{n \minus{} 1}}{(n \plus{} 1)}$

$ \Longleftrightarrow (n\plus{}1)!x_{n\plus{}1}\equal{}n!(x_n\plus{}nx_{n\minus{}1})$

$ \therefore x_{n\plus{}1}\minus{}x_n\equal{}\frac{n!}{(n\plus{}1)!}(x_n\minus{}x_{n\minus{}1}).$

I got that $ d_n\equal{} x_{n\plus{}1}\minus{}x_n \equal{} ((\minus{}1)^n/(n\plus{}1))*d_0$ Now what? How do I finish?

Just sum them up:

$ x_{n\plus{}1}\minus{}x_n\equal{}\frac{(\minus{}1)^n}{n\plus{}1}$

$ \sum^ k _{n\equal{}0}\left(x_{n\plus{}1}\minus{}x_n\right)\equal{}\sum^ k _{n\equal{}0}\frac{(\minus{}1)^n}{n\plus{}1}$

$ x_{k\plus{}1}\minus{}x_0\equal{}\sum^ k _{n\equal{}0}\frac{(\minus{}1)^n}{n\plus{}1}$

$ x_{k}\equal{} \minus{} \sum^k _{n\equal{}1}\frac{(\minus{}1)^n}{n}$


For an abitrary natural n \geq 2 we define x_1 = n, Y_1 = 1, x_(i+1) = li(1/2*(x_i + y_i)), y_i = li(n/x_(i+1)) where li(x) is the greatest integer \leq x, i = 1,2,...,n-1. Prove that min(x_1,....,X_n) = li(n^0.5).

Let me rewrite the statement as I understood it.

For an abitrary natural $n \geq 2$ we define $x_{1}= n$, $y_{1}= 1$, $x_{i+1}= \left[\frac{x_{i}+y_{i}}{2}\right]$, $y_{i+1}= \left[\frac n{x_{i+1}}\right]$, $i = 1,2,...,n-1$.
Prove that $\min(x_{1},....,x_{n}) = [\sqrt n]$.

Note that if $x_{k}>[\sqrt n]$ then $y_{k}\leq [\sqrt n]$. so $x_{k+1}<x_{k}$. It means that we are only to prove $x_{k}\geq [\sqrt n]$. Indeed, if $x_{k}=[\sqrt n]+l$ then $y_{k}=\left[\frac{n}{[\sqrt n]+l}\right] \geq [\sqrt n]-l$. Thus $2x_{k+1}=x_{k}+y_{k}\geq 2[\sqrt n]$.


geometry
3D geometry
function
calculus
integration
algebra
factorization

Is there a simple example of a field $K$ with the following property: for any $N$ one can find $x$ in $K$ which is non-zero and  cannot be expressed as a sum of $N$ squares of elements of $K$?  What about written as $x_1^{k_1}+...+x_N^{k_N}$ where $k_1,...,k_N$ are in a fixed finite set and $x_i$ in $K$?

hm.. i think it depends on your meaning of "simple"..
i think (thought i didn't write anything nor thought long about it) $\mathbb{Q}(x)$ should work: i don't think $x$ can be expressed as a sum of powers :)

I hope I'm not wrong, but as far as I remember, any rational can be written as a sum of 3 cubes, which are rational functions in that number (I even think Alekk posted the huge identity), therefore I believe the example does not work.

Already in $\mathbb Q$ you cannot express $-1$ as a sum of squares or any even powers. :)

In ma_go's example, $x$ is not a rational number. But it can be expressed as a sum of cubes of elements of $\mathbb Q(x)$: Play with rational linear combinations of $1^3$, $x^3$, $(x+1)^3$ and $(x+2)^3$ and observe that a rational linear combination of these elements can be transformed into an integral linear combination of other cubes and that can be transformed into a sum of cubes. Since this argument does not use any specific property of $x$ (except being a field element) it shows a way to express any rational as a sum of cubes:
$x = (-1)^3 + 36 \left(\left(\frac{x+2}{6}\right)^3 + 2 \left(-\frac{x+1}{6}\right)^3 + \left(\frac{x}{6}\right)^3\right)$
So $x$ can be expressed as a sum of 145 cubes. :-)

LiMa

Does someone know any reference about this problem? I mean, someone must have asked it before me...


function
calculus
integration
derivative
real analysis
real analysis unsolved

let $ I\subset\mathbb{R}$ given interval,
$ f,g\in C(I)$ and for every  $ h\in C^1(I)$ such that $ h(p)\equal{}h(q)\equal{}0$, we have $ \int_I (f(t)h'(t)\plus{}g(t)h(t)) dt\equal{}0$
show that $ g\equal{}f'$.

[quote="primenumber"]let $ I\subset\mathbb{R}$ given interval,
$ f,g\in C(I)$ and for every  $ h\in C^1(I)$ such that $ h(p) \equal{} h(q) \equal{} 0$, we have $ \int_I (f(t)h'(t) \plus{} g(t)h(t)) dt \equal{} 0$
show that $ g \equal{} f'$.[/quote]

What are $ p$, $ q$ ?

If $ I\equal{} (p;q) \subset \mathbb{R}$ and $ \int_I (f(t)h'(t) \plus{} g(t)h(t)) dt \equal{} 0 \, , \, \forall h \in C^\infty _c (I)$  then $ f'$ exists in the distribution sense.

yes, $ p,q$ - endpoints of $ I$

[quote="brahman"][...] then $ f'$ exists in the distribution sense.[/quote] I know nothing about "derivatives in the distribution sense" :( it is an introduction to real analysis

Hint: let $ g(t)\equal{}f'(t)\plus{}r(t)$ for some function $ r$ and show that $ r\equiv0$.

$ f'$ is in the thesis, so one should prove that it exists and is equal $ g$

Let $ \beta(x) \equal{} \begin{cases} 1\minus{}|x| & |x| < 1 \\ 0 & \text{otherwise} \end{cases}$ so that $ \beta$ is a compactly supported continuous function whose integral over the real line equals to 1.

Fix $ a \in (p, q)$ and let $ h_{\varepsilon}(x) \equal{} 1 \minus{} \frac{1}{\varepsilon} \int_{p}^{x} \beta\left( \frac{t\minus{}a}{\varepsilon} \right) \, dt$. Then $ \int_{p}^{q} (f(t)h_{\varepsilon}'(t)\plus{}g(t)h_{\varepsilon}(t)) \, dt \equal{} 0$.

Taking $ \varepsilon \to 0$, we have $ \minus{} f(a) \plus{} \int_{p}^{a} g(t) \, dt \equal{} 0$. (This is what you have to prove!) Since this is true for any $ p < a < q$, we see that $ f$ is differentiable and $ f' \equal{} g$.

but $ h_{\varepsilon}(p)\equal{}1\minus{}0\equal{}1\neq 0$ ...

[quote="primenumber"]but $ h_{\varepsilon}(p) \equal{} 1 \minus{} 0 \equal{} 1\neq 0$ ...[/quote]

You're right, but this is not a critical flaw. We can get around this by slightly modifying the definition of $ h_{\varepsilon}$. For example, we can use $ h_{\varepsilon}(x) \equal{} \frac{1}{\varepsilon} \int_{p}^{x} \left[ 2\beta\left( \frac{t\minus{}p}{\varepsilon} \right) \minus{} \beta\left( \frac{t\minus{}a}{\varepsilon} \right) \right] \, dt$ instead.

 Then $ h_{\varepsilon}(p) \equal{} h_{\varepsilon}(q) \equal{} 0$ for small $ \varepsilon$, so $ \int_{p}^{q} (f(t)h_{\varepsilon}'(t)\plus{}g(t)h_{\varepsilon}(t)) \, dt \equal{} 0$. Taking $ \varepsilon \to 0$ as usual, we have $ \int_{p}^{a} g(t) \, dt \equal{} f(a) \minus{} f(p)$.


Όπως πρότεινε ο Αλέξανδρος , σε αυτό το topic  θα συζητάμε για βιβλία , ηλεκτρονικά ή μη , ανταλλαγές βιβλίων και ότι έχει σχέση με τον εμπλουτισμό της ολυμπιακής βιβλιοθήκης.Γράψτε ,  αν κάποιο νέο βιβλίο έπεσε στην αντίληψή σας .
 Ένα βιβλίο είναι σαν ένα γυμναστήριο : σε κάνει αθλητή , όχι πρωταθλητή. Το βιβλίο από μόνο του σου ανοίγει δρόμο για να τρέξεις. Όμως εσύ είσαι ο δρομέας που θα κάνεις το  ρεκόρ και θα φέρεις το μετάλλιο.
Λοιπόν , ας  δώσουμε στους μαθητές και όλους τους φίλους των μαθηματικών όλες τις πληροφορίες που αφορούν τα βιβλία και τρόπους για να τα αποκτήσουν.

 [color=red]Τη λίστα με τα βιβλία - ξένα και ελληνικά - για το Γυμνάσιο και το Λύκειο θα τη συνεχίσουμε στο άλλο topic .Εκεί γράψτε [b]μόνο[/b] βιβλία για τη λίστα.[/color].
 Ευχαριστώ -
 [u]Μπάμπης[/u]

γεια σε ολους.βλεπω αυτη την υπεροχη κινηση να γινεται εδω και χαιρομαι πολυ.,κ Μπαμπη μου προτεινετε βιβλια που ουτε καν ηξερα οτι υπηρχαν... μπραβο ας βοηθησουν ολοι λοιπον .ειναι αστειο γιατι για θεωρια αριθμων στο λυκειο διαβαζα απο βιβλια πανεπιστημιου τα μονα που ηξερα, ενω καλα καλα δεν τα καταφερνα με τα σχολικα μαθηματικα. :mad:  :(

Κώστα γεια !

  Θα προσπαθήσω να δώσω στους μαθητές μας ,αλλά και σε άλλους πιθανόν αναγνώστες που μπαίνουν σιγά σιγά στο χώρο ,όσο περισσότερες πληροφορίες μπορώ σχετικά με κατάλληλα βιβλία για ολυμπιάδες. Είμαι σίγουρος ότι θα βοηθήσουν και άλλοι φίλοι και κυρίως παιδια που πέρασαν από αυτό το στάδιο ή βρίσκονται ακριβώς τώρα στην πιο καλή φάση για να ασχοληθούν με μαθηματικά. Θέλω να φτιάξουμε σιγά σιγά καλλύτερη ολυμπιακή ομάδα. Υπάρχει ταλλέντο και παιδιά που μπορούν να κάνουν τρομερές διακρίσεις. Θέλουν όμως καθοδήγηση και ενημέρωση. Νομίζει κανείς ότι η Κίνα που πήρε 6 χρυσά μετάλια αρκέστηκε στο να επιλέξει τους καλλίτερους μαθητές;Πίσω από αυτά τα μετάλια υπάρχει οργανωμένη δουλεια πολλών χρόνων. Έχουν γραφεί ειδικά βιβλία , γίνονται μαθήματα από ειδικούς trainer , καθηγηταράδες με παγκόσμια ακτινοβολία κλπ.
Πιστεύω ότι μετά από λίγα χρόνια και η Ελλάδα θα πάει καλλλίτερα γιατί οι μαθητές θα προεοιμάζονται καλλίτερα. Φυσικά δεν φτάνουν μόνο τα βιβλία, αλλά πρέπει να αρχίσουμε από κάπου.
 Για μένα δυστυχώς τα χρόνια πέρασαν και δε μπορώ να οργανώσω πανελλαδικά μαθήματα για προικισμένα παιδιά στα μαθηματικά. Αυτό μπορούν όμως να το κάνουν νεότεροι. Εξ΄άλλου άργησα και εγώ να ανακαλύψω τη μαγεία των ολυμπιάδων.Παροτρύνω όμως τους μαθητές που τρέφουν αγάπη για τα μαθηματικά να ασχοληθούν από νωρίς σοβαρά με το αντικείμενο. Να μάθουν καλά αγγλικά , την -εύκολη -αγγλική μαθηματική ορολογία και να ασχολούνται διαρκώς με τα μαθηματικά θέματα. Αν ο καλός μαθητής ασχολείται μέχρι την τρίτη λυκείου διαβάζοντας μέρα νύχτα για να γράψει 20 στο δίκαιο , στα θρησκευτικά και την κοινωνιολογία , καλό είναι να μην καταπιαστεί με τα μαθηματικά. Για τα παραπάνω μαθήματα αρκεί να παρακολουθήσει κανείς το σχολείο του και έχει στην τσέπη το 19 , με 20 λεπτά διάβασμα το μήνα. Τα μαθηματικά είναι μια ειδική επιστήμη. Σε θέλουν αποκλειστικά δικό τους. Αν μοιράζεσαι σε 15 μαθήματα , απλά για να λέει η μαμά ότι το παιδί της είναι άριστος μαθητής, τότε η ελπίδα να μάθεις από νωρίς μαθηματικά πάει χαμένη.
  Τέλος πάντων , δε θέλω να υποβαθμίσω κανένα μάθημα αλλά αυτό που λέω είναιτο εξής.  Διάλεξε: ή πολύ καλά μαθηματικά(μαζί με τη φυσική και τη γλώσα βέβαια, γιατί χωρίς αυτά δε μπαίνεις πανεπιστήμιο ,αν δεν πάρεις μετάλιο) και στα άλλα το ελάχιστο ή όλα και από λίγο. Αλλά με όλα και από λίγο - ή έστω στα μαθηματικά λίγο περισσότερο- διάκριση δεν έρχεται. Αυτό πρέπει ννα το καταλάβουν όλοι , κυρίως οι γονείς.Γιατί  τα μαθηματικά θα σε ανταμείψουν και μάλιστα καλά. Κυρίως όμως θα σε κάνουν να σκέφτεσαι σωστά.Και αυτό είναι το μεγαλύτερο αγαθό!
  Μπάμπης

το παθος για να μαθεις μαθηματικα δυσκολα ειναι σπουδαιο,αλλα και γενναιο.δεν ξερω για αλλους αλλα στην πολη μου αρκετα παιδια που περασαν αρχιμηδη  και με ταλεντο δεν τους αφησαν οι γονεις τους  να πανε λογω αποστασης.για να μην πω μαθηματικους που δεν αφηναν τα παιδια τους να πανε σε μαθ.διαγ. οι αλλους συναδελφους σας που χαρακτιριζαν τους διαγωνισμους ...με χοντρους χαρακτιρισμους... ετσι ειναι η ζωη ομως . :maybe:

Κώστα ,
 δυστυχώς οι επισημάνσεις σου είναι σωστές.Ο κόσμος δεν αγαπάει τη μαθηματική σκέψη γιατί τη φοβάται. Δυστυχώς , εμείς οι μαθηματικοί , δεν πείθουμε με τη στάση μας για το τι είναι μαθηματικά.  Τα μαθηματικά νομίζουμε ότι είναι έξω από τη ζωή μας , αλλά κάθε στιγμή κάνουμε μαθηματικά. Πώς να το δείξουμε αυτό ,  πρώτα στους μαθητές μας και μετά στην υπόλοιπη κοινωνία ; Για τους γονείς τα Μ. είναι μια δοκιμασία που πρέπει τα παιδιά τους να την πληρώσουν με ακριβά ιδιαίτερα , για να μπουν στην 3/θμια εκπαίδευση. Για μας τους καθηγητές, δημόσιους ή μη είναι μόνο ένας τρόπος απόκτησης των μέσων επιβίωσης. Κακοπληρωμένοι δάσκαλοι στα σχολεία θα καταφέρουν με μαθηματική ακρίβεια να κάνουν τους μαθητές να μισήσουν τα μαθηματικά.
    Προτείνω και σε σένα και σους αναγνώστες το βιβλιο << Οι μαθηματικοί >> του Bell.Είναι η ζωή των μεγαλύτερων μαθηματκών ,από την αρχαιότητα έως τώρα. Μαθηματικά δεν μπορούν να κάνουν όλοι , μπορούν όμως όλοι να μάθουν περισσότερα από όσα ξέρουν και κυρίως να απολαύσουν τη στοιχειώδη μαθηματική δημιουργία που χαρίζουν στην ανθρωπότητα τα λίγα και ξεχωριστά μυαλά που γεννάει η φύση.
   Καλό καλοκαίρι σε όλους ! Φεύγω την Παρασκευή . Ελπίζω το λάπτοπ να δουλεύει και στην εξοχή , εκτός και αν η γυναίκα μου έχει πάρει τις ...αποφάσεις της και του αφαιρέσει κανένα καλώδιο!!!
 Μπάμπης


linear algebra
matrix
vector
analytic geometry
calculus
algebra
polynomial

In the finite dimensional case, rank is just the number of nonzero rows in the row reduced echelon form of a matrix.  :)

I'd say to stick with it, you will (eventually) learn something.

Here's an exerpt from my syllabus for a course I taught recently. In our numbering system, it's a sophomore level course, required for math majors, math minors, and computer science majors, and occasionally taken by other science, engineering, or economics majors. I was using David Lay's textbook, but there are dozen other books that would work as well.

---

Systems of linear equations; vector and matrix equations; solution sets; linear independence; matrix operations and inverses.

(Exam I)

Partitioned matrices and matrix factorizations; determinants; subspaces of $\mathbb{R}^n$; null spaces, column spaces; more on linear independence; bases, coordinate systems and dimension.

(Exam II)

Rank; change of basis; eigenvalues and eigenvectors; characteristic equations; complex eigenvalues; applications of eigenvalues and eigenvectors.

(Exam III)

Numerical considerations, iteration; inner product spaces and orthogonality.

(Final exam)

MATH ___ is an “intermediate level” mathematics course. It is intermediate in both sequence and nature between the computational and algorithmic lower division courses, such as the calculus sequence and introductory programming courses, and abstract and theoretical upper division mathematics or computer science courses. It has its own algorithmic core. We will learn the single most important computational technique of the course - Gaussian elimination - on the very first day of class. Linear and matrix algebra is also pervasively applicable to a wide variety of fields.

The part of this course that students may find the most difficult - the part that is most different from previous courses, the part that will require the largest share of your work and attention - is mathematical vocabulary. You will need to know the precise definitions and uses of a large number of words, including, but not limited to, the following:

[i]Linear combination
Span
Linearly independent (or linearly dependent)
Linear transformation
Basis
Dimension
Vector space
Subspace
Rank (of a matrix)
Null space[/i]

All of these words have careful mathematical definitions. A mathematical definition is not at all like a normal dictionary definition, in which words are like other words and meanings come in various shades and emphases. A mathematical definition is full of carefully placed little words or phrases, such as “if .. then”, “and”, “or”, “for every”, and “there exists” which are indispensable to the meaning and cannot be rearranged or replaced. You must memorize these mathematical definitions, but more than that, you must understand these mathematical definitions.

---

There are plenty of the users of this board who might ask where such subjects as these are: vector spaces over arbitrary fields; infinite dimensional vector spaces; Jordan canonical form; rational canonical form; the Cayley-Hamilton theorem; minimal polynomials; convex sets; and many other topics. Well, this was a first course, and for a fairly broad audience. But even in such a course, one should explore the vocabulary. One of the reasons for picking Lay's book instead of some of the others is so that by the 2nd or 3rd week, I should already be talking about linear combinations of columns (or rows) of a matrix, and already defining what it means for those columns to be independent.

Alright, but a lot of the stuff still seems a little easy. I took Diff Eq, and we spent A LOT of time covering matrix theory. So I learned about linear dependence, eigenvalues/vectors, etc. I guess I'll stick with it though.

hi i am student(1 kurs or freshman as u say I think)  of National Uzbekistan University of mathematics faculty and i want learn more then the teachers are educating me. How can I find some books in Enflish about Linear algebra and something like that :?:

[quote="vinny919"]Ok so I'm a high school senior taking Linear Algebra (Junior level course) at the local university. Now, what should one normally be covering by the 2nd or 3rd week in such a course? Right now, ALL we do is convert some systems of equations into matrix form and solve them using row manipulations - not very interesting. Also, it doesnt' look as if the course will ever be interesting since the professor keeps making the worst errors (like says there are 3 variables in an equation when there are actually only 2 variables but 3 rows in the matrix since it's overdetermined), and then taking quite a while to find out what he did wrong. My guess is that he's not dumb, its just that its not his field (I guess this happens a lot, since my science mentor, a grad student, is now teaching a class he took once but is not in his field of expertise).

Anyway, I'm considering dropping it since it's so easy. What should one be covering in a class like this, and if we continue at this rate, will we learn all the cool stuff like "inner product" and "rank" which I've heard thrown around a lot?[/quote]vinny,

My experience with college courses, I taken a lot of them, is that they are forums to help you learn things that you can learn on your own but it would just be more difficult to do so. Also your professors are guides to help you learn, but because they are experts in the fields of their dissertations doesn't mean that they know everything about everything even in their own disciplines. In my opinion, if you are not learning from the professor who is teaching your course it is better to drop out and take it again later from someone with whom you have better rapport.

[quote="vinny919"]Ok so I'm a high school senior taking Linear Algebra (Junior level course) at the local university. Now, what should one normally be covering by the 2nd or 3rd week in such a course? Right now, ALL we do is convert some systems of equations into matrix form and solve them using row manipulations - not very interesting. Also, it doesnt' look as if the course will ever be interesting since the professor keeps making the worst errors (like says there are 3 variables in an equation when there are actually only 2 variables but 3 rows in the matrix since it's overdetermined), and then taking quite a while to find out what he did wrong. My guess is that he's not dumb, its just that its not his field (I guess this happens a lot, since my science mentor, a grad student, is now teaching a class he took once but is not in his field of expertise).

Anyway, I'm considering dropping it since it's so easy. What should one be covering in a class like this, and if we continue at this rate, will we learn all the cool stuff like "inner product" and "rank" which I've heard thrown around a lot?[/quote]

Linear algebra courses are usually  geared towards two types of audiences: math majors and non-math majors. The non-math major linear algebra course  is really watered down. Maybe you should switch to the linear algebra course for math majors. I'd be surprised if there wasn't such a course. If there isn't such a course just stick with this course and get credit for it. But you never know, maybe towards the end of the course there might be things that would challenge and interest you.


algebra
polynomial
superior algebra
superior algebra unsolved

I want to show that $ \sqrt [3]{6}$ is not contained in $ \mathbb{Q}(\sqrt [3]{2})$.

OTC assume it is. Then I can get,

$ a \equal{} b\sqrt [3]{2} \plus{} c(\sqrt [3]{2})^2$

where a,b,c are rational. That has got to be a contraction, but I don't know how to prove it. It seems really obvious I know.

EDIT: better yet, can you prove that a finite sum of nonzero irrational numbers must be irrational?

It's pretty hard to prove that the sum of $ \sqrt{2}$ and $ \minus{}\sqrt{2}$ is irrational  :P

For the first question, show $ t^3\minus{}2$ is irreducible so any polynomial with $ \sqrt[3]{2}$ as a root must be divisible by $ t^3\minus{}2$.

Here is the complete proof.

If $ \sqrt [3]{6}$ is contained in $ \mathbb{Q}(\sqrt [3]{2})$, then

$ \sqrt [3]{6} \equal{} a \plus{} b\sqrt [3]{2} \plus{} c(\sqrt [3]{2})^2$

$ 6 \equal{} a^3 \plus{} 3 a^2 b 2^{1/3} \plus{} 3 a b^2 2^{2/3} \plus{} 3 a^2 c 2^{2/3} \plus{} b^3 2 \plus{} 12 a b c \plus{} 3 b^2 c 2^{4/3} \plus{} 3 a c^2 2^{4/3} \plus{} 3 b c^2 2^{5/3} \plus{} 4 c^3$

That reduces to

$ 6 \equal{} d \plus{} e\sqrt [3]{2} \plus{} f(\sqrt [3]{2})^2$

where d,e,f are rational and

$ e \equal{} 3 a^2 b \plus{} 6 b^2 c \plus{} 6 a c^2$
$ f \equal{} 3 a b^2 \plus{} 3 a^2 c \plus{} 6 b c^2$

So I guess I do not have a complete proof as I need to show that that e and f are not both 0. That looks rather difficult. Any ideas?

If you're talking about solving the equations, consider $ b\cdot \frac{e}{3}\minus{}a\cdot \frac{f}{3}$ as this gets a $ c$ factored out.

[quote="-oo-"]have you read scorpius119's answer? ..[/quote]

Yes. In fact, my last post was using scorpius199's answer although I did not explain it well. I was trying to get a contradiction of what scorpius199 wrote by showing that $ \sqrt[3]{2}$ was a zero of a non-constant polynomial with degree less than 3. To do that I need to show that e and f are not both zero. I am just trying to figure out what the restrictions on a,b,c are that make that true. Does that make sense?


geometry
3D geometry
number theory unsolved
number theory

Find positive integers $ a$ and $ b$ such that

$ (a^{\frac{1}{3}} \plus{}b^{\frac{1}{3}} \minus{}1)^{2}\equal{}49 \plus{}20(6)^{\frac{1}{3}}$.

[quote="AndrewTom"]Find positive integers $ a$ and $ b$ such that

$ (a^{\frac {1}{3}} \plus{} b^{\frac {1}{3}} \minus{} 1)^{2} \equal{} 49 \plus{} 20(6)^{\frac {1}{3}}$.[/quote]

$ 49\plus{}20\sqrt[3]6\equal{}(\minus{}1\plus{}2\sqrt[3]6\plus{}2\sqrt[3]{36})^2$ and so :

either $ \sqrt[3]a \plus{} \sqrt[3]b\minus{} 1 \equal{} 1\minus{}2\sqrt[3]6\minus{}2\sqrt[3]{36}$ and so $ \sqrt[3]a \plus{} \sqrt[3]b\plus{}\sqrt[3]{48}\plus{}\sqrt[3]{288}\equal{}2$

either $ \sqrt[3]a \plus{} \sqrt[3]b\minus{} 1 \equal{} \minus{}1\plus{}2\sqrt[3]6\plus{}2\sqrt[3]{36}$ and so $ \sqrt[3]a \plus{} \sqrt[3]b \equal{} \sqrt[3]{48}\plus{}\sqrt[3]{288}$

The question asks for "Find positive integers" and not "Find all positive integers", hence one answer $ \boxed{(a,b)\equal{}(48,288)}$

Once again, thanks pco. Is it possible to find all positive integers $ a$ and $ b$ here?


Please list what you predict for each series and in the poll say who you think will win the world series.

Cardnails vs. Astros
4-2
White Sox vs. Angels
4-3
Cardnails vs. White Sox
4-2
Cardnails win!!!!! :D

I won't try to predict; I'll just root for the Angels, since they're the only West Coast team left.

If you go by runs scored and runs allowed, the better team won in every series, except for Houston-Atlanta in which the teams were essentially equally good.

By that measure, the three best teams in the AL were Cleveland, Oakland, and LAA.

I might as well go with "better team wins": this would be Cards over Angels in the WS.

I have to agree with jmerry and root for the Angels.

However, I don't think they have it in them.

Predictions:
Angels over White Sox 4-3
Cardinals over Astros 4-2
Cardinals over Angels 4-1

Cardinals Win.


154 marks physics - 45 chemistry - 36 maths - 73

Hey, I think I got good marks
Total-263
M:89
P: 95
C:79

FTSE Rank: 14

my frends got only ftse percentile.........

the marks increased of 1 student from 149 to 174 and percentile=96.3.......

Do you think that the IIT-JEE Rank Projections given by FIITJEE are correct?
.. :o

TO A GREAT EXTENT.........
probably.........
 :P


I am trying to solve a math related problem i have and as i am no mathematician by any means, any help would be very much appreciated!

With the following numbers:

9  -   11  -   3   -  3.5  -   50

i was wondering if anyone could come to either or both of the following answers:

44.89796     or     36.73469

If you struggle like me then there is a chance that the numbers could be although unlikely.

8  -   10 -   2  -   2.5  -   50

Thanks

What operations are allowed?

Also, couldn't you write a program to do this?

First, what is the problem or question being tackled? What exactly are you trying to do with the numbers?

Any operations are allowed. What do you mean by program to work this out? How would i go about doing that? The reason for this is i need the formula for an excel sheet i am trying to make. Currently i have lots of sheets working it out by a trial an error basis, but the problem with that is its very cumbersome and requires tons of info to work out what a formula would be able to in one cell.


Since recently (about a week ago) I am having problems with hidden content of posted messages. When clicking on a link to the hidden text, the browser (MS Explorer) just scrolls to the top, instead of revealing the hidden content. This happens on Win2000, WinXP, various computers, even though I make sure that I am logged in. Right now, it is happening on a PC running Win2000 with MS Explorer 5.0, but I cannot reproduce it on 2 other PCs next to me, one running MS Explorer 6.0 on Win2000, the other MS Explorer 6.0 on WinXP. I have seen it on my home computer (WinXP) as well. It does not happen always, but once its starts, it happens every time. Logging out and in does not help. Exiting all instances of the browser, restarting and logging in again does not help either. Any idea what is going on ?

We don't support IE 5! :) Try using a different browser, such as Firefox.


I am assuming that the avatar ranks are the names of the seven mysterious math problems not yet solved (the Poincare Conjecture is believed to finally be solved by Grigori Perelman).  Could someone tell what they are all mean?

http://www.claymath.org/Millennium_Prize_Problems/

that should explain it

I'm just wondering.  Is this what the seven wonders of the world means?

usually it refers to this: http://ce.eng.usf.edu/pharos/wonders/list.html

no 7 wonders of the world is a bunch of trig inequalities

Speaking of 7 wonders, you often hear about such-and-such a place being "one of the seven wonders of the natural world".  Fact is, there is no such list.

By the way, "7 wonders of the world" refers to seven places from the ancient world, whereas those unsolved problems are quite modern in comparison.


Functional Analysis
real analysis
real analysis unsolved

Give an example of orthogonal set $ (g_n) _{n \equal{} 1} ^{\infty}$ in $ L_2 (A)$ for which $ \sum _{n \equal{} 1} ^{\infty} g_n \equal{} g \in L_2(A)$.

($ A$ can be an interval of your choice)

(additional problem: generalize to any Hilbert space)

[Edited: see below]

Isn't it true that in a Banach space the series $ \sum g_n$ converges if and only if the series $ \sum \parallel{}g_n\parallel{}$ converges.
But if you indeed want it to be orthonormal, not just orthogonal, then each $ \parallel{}g_n\parallel{} \equal{} 1$ and the series always diverges and it is not possible to take such a basis in an infinite Hilbert space.

You're right. Sorry for posting nonsense. I've changed the thesis. Hope it now makes more sense.


group theory
abstract algebra
algorithm
logarithms

El-Gamal Cryptosystem can be implemented in any subgroup <a> of a finite multiplicative group G.
as follows:
let  B in <a>, and define (a,B) to be the public key.  log B is the private key.
planitext space is <a> and e(x) = (a^k,x*B^k) where k is random.

now assume i have an ORACLE that gets an encryption text (y1,y2) and {x1,x2}
and tells whether (y1,y2) is a valid el gamal encryption of x1, x2, both or none.

i should use the oracle to write an algorithm that decides for every B,C,D in <a>
whether log B * log C = log D, where the logarithm is the discrete logarithm in base a.

i have managed to do the follows:
y1 = B
y2 = D
then apply the oracle on {1,1}
if it says "NO" then the relation of the logarithms doesn't hold.
if it says "yes" then i know  D = B^t for some t
by doing the same with C and D, I can get that D = C^s for some s.
but now i'm stuck.

Am i going in the right direction?  what should I do?

Thanks

If I remember correctly both Applied Cryptography (Schneier) and Introduction to Cryptography (Buchmann) have nice sections on ElGamal. If you have time you might want to try to obtain these books and study what they have to say about ElGamal implementation. It might be helpful.


function
limit
calculus
derivative
calculus computations

For $n=1,2,...$ and $x\in\mathbb{R}$, put
\[f_{n}(x)=\frac{x}{1+nx^{2}}. \]
Show that $\{f_{n}\}$ converges uniformly to a function $f$, and that equation
\[f'(x)=\lim_{n\to\infty}f_{n}'(x) \]
is correct if $x\not =0$, but false if $x=0$.

The serie converge iff it is fitted in Cauchy´s criteria  so:
$|f_{n}(x)-f_{n-1}| =\frac{x}{1+nx^{2}}-\frac{x}{1+(n-1)x^{2}}=\frac{x(1+nx^{2}-x^{2}-1-nx^{2})}{(1+nx^{2})(1+(n-1)x^{2})}=\frac{-x^{3}}{1+nx^{2})(1+(n-1)x^{2})}<\epsilon$
since we can do n arbitrarly large, so the sequence uniformly converges.
Now, the second part, just derivate f_n and you will get:
$f'_{n}(x)=\frac{1-nx^{2}}{(1+nx^{2})^{2}}$
For any x different of zero, since n tends to infinity, the function f is the constant function with value=0 so the derivative is 0 everywhere, what is true also for f_n (x) different of 0. Since  if we put 0 in our equation above, we will get f(x)=1, then, we showed what you asked us.
A question: what about the points such like $x=\sqrt{1/2n}$ or anything like that? f_n will give very bad behaved derivatives

[quote="Brambila"]The serie converge iff it is fitted in [color=blue]Cauchy´s criteria  [/color]so:
$|f_{n}(x)-f_{n-1}| =\frac{x}{1+nx^{2}}-\frac{x}{1+(n-1)x^{2}}=\frac{x(1+nx^{2}-x^{2}-1-nx^{2})}{(1+nx^{2})(1+(n-1)x^{2})}=\frac{-x^{3}}{1+nx^{2})(1+(n-1)x^{2})}<\epsilon$
since we can do n arbitrarly large, so the sequence uniformly converges.
[/quote]
You are wrong, because this is Cauchy criterion:
The sequence of functions $\{f_{n}\}$ defined on $E$, converges uniformly on $E$ iff for every $\epsilon >0$ there exists an integer $N$ such that $m\geq N,n\geq N,x\in E$ implies \[|f_{n}(x)-f_{m}(x)|\leq\epsilon.\]

Ouch, sorry for this dumb mistake, just substitute n-1 by n+a for any ainteger higher than 0, n>N and choose a nice value for epsilon and the conclusion will follow easily. What about the second part? I think it´s okay  isn´t?

But Cauchy criterion not works in this problem, i think so. Can you concrete?

Ok! This is my solution.
We have $\lim_{n\to\infty}\frac{x}{1+nx^{2}}=0\forall x\in\mathbb{R}$ and  $\sup_{x\in\mathbb{R}}{|f_{n}(x)|}\leq \frac{1}{2\sqrt{n}}\to 0$ when $n\to\infty$. Also $f_{n}\to 0$ uniformly on $\mathbb{R}$.


Ai có cuốn này có thể photo cho mình một bản được không? Hoặc nếu mua hộ mình thì càng tốt. Cám ơn nhiều.( Tất nhiên sẽ có hậu tạ  :)  :P )

Anh có, biết địa chỉ anh rồi ấy nhỉ? :D Cứ chuyển tiền đến đó, cho dùng luôn bản chính! :P

Lâu quá em quên mất địa chỉ anh rồi, anh cho em xin lại được không? :blush:  Mà anh có tài khoản ATM không, em chuyển tiền luôn cho tiện :D

Cuốn đó giá 43000, trừ cho chú 3000 vì anh dùng rồi :D Như vậy gửi tiền sách cho anh + tiền chuyển sách từ Hạ Long đi Đà Nẵng! Không biết là bao nhiêu ấy? Chú tính xem!  :maybe:

Giảm giá chút đi anh, dùng rồi trừ 13 nghìn luôn cho nó tiện, cộng phí bưu điện coi như em mua sách đúng giá :D  :lol:

Chậc, thương chú nhà nghèo nhưng hiếu học, anh đồng ý :D PM cho anh để lại địa chỉ chính xác, anh sẽ cho chú số tài khoản, nhớ phải có số điện thoại nhà riêng nữa nhé!  :wink:

Anh Tuan co thi mua em cuon luon nha .Anh de lai dia chi em con gui tien nua chu  :D

Có mỗi một cuốn , mà đang trao đổi với Bách mất rồi, hạ giá rồi nhưng xem gia vẫn khó :P

Hì, anh để lại em cuốn đó nhé! Đừng trao đổi với người khác đấy :lol:  :D

Anh Tuân sao mãi vẫn chưa thấy liên hệ lại thế nhỉ?

Chú chuyển tiền theo địa chỉ sau: Nguyễn Trung Tuân, THPT chuyên Hạ Long, phố Chu Văn An, thành phố Hạ Long, tỉnh Quảng Ninh.

Nhớ rằng, tiền thiếu là không thấy sách đâu đấy! :P

Thế thì chuyển thường thôi, ko phát nhanh, có thể 2 tháng sau mới đến :D .

cuon nay noi ve cai gi vay nhi?
to hop roi nhung ma danh cho dai hoc hay hoc sinh cap 3 ma danh nhau ghe nhi
sach viet nam ma kho kiem the sao?

Sách nói về cái mà tên nó nói :D Cuốn này nằm trong bộ sách Cao học của Viện Toán học,  ai ở gần HN thì mua dễ , nhưng các bạn đây toàn là xa HN thôi, nên cần người giúp.  :wink:


probability
number theory
relatively prime

A jar has 10 red candies and 10 blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

[hide="Solution"]
There are three possible cases.

Case 1: Both get 2 red candies

$\frac{\binom{10}{2}}{\binom{20}{2}}\cdot \frac{\binom{8}{2}}{\binom{18}{2}}=\frac{14}{323}$ (after lots of cancellations...)

Case 2: Both get 2 blue candies

$\frac{14}{323}$

Case 3: Both get 1 red and 1 blue

$\frac{10\cdot 10}{\binom{20}{2}}\cdot \frac{9\cdot 9}{\binom{18}{2}} =\frac{90}{323}$

$\therefore \frac{14}{323}+\frac{14}{323}+\frac{90}{323}=\frac{118}{323}$

Hence $m+n=\boxed{441}$
[/hide]

[hide="Answer"]If Terry picks two reds - $\frac{1}{2}\cdot \frac{9}{19}=\frac{9}{38}$ chance - then Mary has a $\frac{4}{9}\cdot \frac{7}{17}=\frac{28}{153}$ chance of picking the same combination.

If Terry picks two blues - again, $\frac{9}{38}$ chance - then Mary again has a $\frac{28}{153}$ chance of picking the same combination.

If Terry picks one red and one blue - $2\cdot \frac{1}{2}\cdot \frac{10}{19}=\frac{10}{19}$ chance - then Mary has a $2\cdot \frac{1}{2}\cdot \frac{9}{17}=\frac{9}{17}$ chance of picking the same combination.

Therefore, the total is $2\cdot \frac{9}{38}\cdot \frac{28}{153}+\frac{10}{19}\cdot \frac{9}{17}=\frac{118}{323}$ chance, and the answer is $118+323=\boxed{441}$.[/hide]

[hide]My solution:

Combined probability that Terry/Marry will each pick $2$ pieces, each of the same color, for both red and blue:

$2(\frac{10}{20})(\frac{9}{19})(\frac{8}{18})(\frac{7}{17}) = \frac{90}{323}$

Combined probability that Terry/Mary will pick 2 of seperate colors $(R/B or B/R)$.

$4(\frac{10}{20})(\frac{10}{19})(\frac{9}{18})(\frac{9}{17}) = \frac{28}{323}$

Total probability of both scenarios:

$\frac{90}{323} + \frac{28}{323} = \frac{118}{323} = \frac{m}{n}$

It is evident that $323$ is not a product of any composite numbers and $118$ is $2(59)$. They are coprime.

Thus $m + n = 118 + 323 = \boxed{441}$[/hide]

Quick question: For the 3rd case, you have 2 red's and 2 blue's, so why can't you do $\frac{\binom{10}2*\binom{10}2}{\binom{20}4}$?

[hide]There are three cases:
Case 1:Both got red.
1/2*9/19*8/18*7/17=14/323.
This is also true for the second case, in which they get both blues.
Case 3: They both get one blue and one red.
The first does not matter, but the second must be 10 of the remaining 19 colors, so it is 10/19*9/17=90.
We have 90+14+14=118/323, so our answer is 118+323, or 441. [/hide]


[url=http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf]The entirely mixing variables method[url]?[/url]


Let $ 1 = x_1 < x_2 < x_3 < ... $ be an infinite strictly increasing sequences of positive integers, with $ x_{n+1} \le 2n $ for each n :ge: 1.

Prove that for each positive integer $k$, there are integers $i$ and $j$ such that $x_i - x_j = k$.

Darn! I saw this while at school, and realized how stupid the idea I began with was :). We can prove even more (I think you did): we can find $i>j,\ i,j\in[1,k+1]$ s.t. $x_i-x_j=k$.

[quote="Myth"]Suppose contrary. Let $k\in\mathbb{N}$. Unite 1, 2, ..., $2k$ into $k$ pairs: $(1,k+1)$, $(2,k+2)$, ..., $(k,2k)$. Each of these pairs contains at most 1 number from sequence $\{x_n\}$. So these pairs totally contains at mosts $k$ terms of $\{x_n\}$ - contradiction, since $x_{k+1}\le 2k$.

I have a sense that something is wrong, since solution is too simple.[/quote]

Well I did say "pretty easy," didn't I? :)

[quote="grobber"]I don't really have the time right now (I have to go to school :)), but I think the idea is that if we assume the contrary, we can find $t$ s.t. $x_{n+1}\in [2n-t,2n],\ \forall n\ge 1$.[/quote]
It was my first idea too. :)

I think the main thing about this problem that might make it hard or time consuming is overcomplicating it, which is pretty easy to do.


calculus
geometry
integration
calculus computations

I am having some trouble on my calculus homework, and was wondering if you couyld help. Find the Area of the surface generated by revolving the curve y-x^3 between (1,1) and (2,8) about the x-axis. If you could please show or try to explain you work.

Do you mean to work out the area between the curve and the x-axis?
$areas=\int_1^{2}x^3 dx =2^4/4-1^4/4=3.75$
Am I right?

No, I'm pretty sure they wanted the surface area

I think what it is asking for is if you revolve the curve around the axis the figure becomes three-dimensional, and they want the surface area of the 3-d object, not the two dimensional drawing on paper

:blush: My english is poor.I didn't get your means.
 :idea:
First,we can divide the interval(1,2) into n small intervals,like $(1,x_1),(x_1,x_2),(x_2,x_2,x_3)..........(x_{n-1},2)$
 each interval's length is 1/n,we mark it as "h".
second,the hight of these points are $1,x_1^3,x_2^3,x_3^3........2^3$
the surface is near $2*pi*h*(1+x_1^3+x_2^3+x_3^3........+2^3)$
The "h" is  more small ,the surface is more exactly.
So :
$S=2*pi*\int_1^{2}x^3 dx=2*pi*3.75$

Am I right?


combinatorics unsolved
combinatorics
Additive combinatorics

Find the number of sets $A$ containing $9$ positive integers with the following property: for any positive integer $n\le 500$, there exists a subset $B\subset A$ such that $\sum_{b\in B}{b}=n$.

[i]Bogdan Enescu & Dan Ismailescu[/i]

Let $a_i=2^{i-1}$.

Each $a_i$ is in the set $B$ if and only if the $i^{th}$ digit (counting from the decimal point) of $b$ in base $2$ is a one.

This isn't very hard at all - in fact, compared to the problems you usually put up, ehsan, it's trivial.

Guess what: you have to find [b]all[/b] such sets $A$ :). You have just shown that the set of powers of $2$ with exponent  between $0$ and $8$ works.

Oops...  :blush:

I think I'm going to go home before I embarass myself any more tonight - I knew that looked too easy...

Which of the integers have to be positive?

I presume when you say 'every integer $n\le500$', you mean positive integers - but do the nine integers in the set have to be positive?

As I wrote ,$n\in Z$  and about your second question

[quote]do the nine integers in the set have to be positive?[/quote]

no, the nine integers in the set $\in Z$  ;)

The original statement is the following (I should know, because I proposed this problem):
Find the number of sets $A$ containing $9$ positive integers with the following property: for any positive integer $n\le 500$, there exists a subset $B\subset A$ such that $\sum_{b\in B}{b}=n$.
[color=#FF0000][mod: original post has now been edited to this][/color]

[quote="enescu"]The original statement is the following (I should know, because I proposed this problem):
Find the number of sets $A$ containing $9$ positive integers with the following property: for any positive integer $n\le 500$, there exists a subset $B\subset A$ such that $\sum_{b\in B}{b}=n$.[/quote]

well,try to solve it ,plz  :lol:

OK, here's the solution:
Let $A$ be such a set. Then, if $x,y,z$ are distinct elements of $A$, we cannot have $x=y+z$. Indeed, if the previous holds and $B$ is an arbitrary subset of $A$, then the sets $B\cup \{x\}$ and $B\cup \{y,z\}$ have the same sum of their elements, hence the number of distinct sums obtained from the nonempty subsets of $A$ is at most $511-2^6=447<500$, a contradiction. In a similar way it results that no element of $A$ can be written as a sum of three or four other (distinct) elements of $A$.
Now, clearly $1\in A$ and $2\in A$. Since $3=1+2$, we have $3\notin A$ and hence $4\in A$. The numbers $1,2,4$ generate (by subset sums) all positive integers up to $7$, hence $8\in A$. The set $\{1,2,4,8\}$ generates $1,2,\ldots , 15$, so $16\in A$. The set $\{1,2,4,8,16\}$ generates $1,2,\ldots , 31$, but $31$ can only be written as $31=1+2+4+8+16$ (that is, as a sum of [b]5[/b] elements of $A$, thus it is possible that $31\in A$. It follows that the sixth element of $A$ is $32-a$ for some nonnegative integer $a$. The set $\{1,2,4,8,16,32-a\}$ generates $1,2,3,\ldots , 63-a$, therefore the next element of $A$ is $64-a-b$ and this set generates $1,2,\ldots , 127-2a-b$. Continuing in this way, we conclude that
\[A=\{1,2,4,8,16,32-a,64-a-b,128-2a-b-c,256-4a-2b-c-d\}
\]with $a,b,c,d\ge 0$ and this generates
\[1,2,3,\ldots ,511-8a-4b-2c-d.\]
Because $511-8a-4b-2c-d\ge 500$ we obtain $8a+4b+2c+d\le 11$. Thus, the problem reduces to find the number of systems of nonnegative integers $(a,b,c,d)$ such that $8a+4b+2c+d\le 11$.
It is not difficult to count them, since we have only two choices for $a$. If $a=0$, we might have $b=2$ (6 solutions) or $b=1$ (20 solutions) or $b=0$ (42 solutions). If $a=1$ we have $b=0$ and $2c+d\le3$ yields 6 more solutions.
Finally, there are $74$ such sets.
Co-author of this problem was prof. Dan Ismailescu.


analytic geometry
geometry unsolved
geometry

Given that $P$ is a point lying outside circle $O$, $PS$ and $PT$ are tangents to circle $O$. Through $P$ draw a line meeting circle $O$ at points $A$ and $B$, let $PAB$ meet $ST$ at $C$. Show that
$\frac{1}{PC} = \frac{1}{2}\left(\frac{1}{PA} + \frac{1}{PB}\right)$

meh, just crunch your way through with coordinates, it's not THAT messy.... :D

Here is my confusion note: i converted the problem to this form 2$PA$.$PB$=$PC$.$AP$+$PC$.$PB$ after we have
2$PA$.$PB$=$PC$.$AP$+$PA$.$PB$+$AC$.$PB$ now we have $PA$.$PB$=$PC$.$AP$+$AC$.$PB$=$AP^2$+$AC$.$PA$+$AC^2$+$AC$.$PA$+$AC$.$CB$. Also we open $LHS$ we write $PB$ as $PA$+$AC$+$CB$ then by canceling some terms we have $AP$.$CB$=$AC$.$PB$. And the last $ONE$ is impossible i think...

Davron

The division $(P,A,C,B)$ is harmonically, i.e. $\frac{PA}{PB}=\frac{CA}{CB}\ !$ (the line $ST$ is the polar of the point $P$ w.r.t. the given circle). See the characterization $A9$ of a harmonical division from the [u]lemma[/u] mentioned in the my reply at http://www.mathlinks.ro/Forum/viewtopic.php?t=46146

Stewart Theorem also works for this problem.


We [as in one of the coaches] should do this...you pretty much ask the admins for a private forum.

Who really goes to the Illinios forum other than ARML people? I think that most people Illiniois that know about AOPS know about ARML.

Obviously, we could use this forum, but the main idea is that a private forum would spur people to do something; it offers a direct reason to join aops/post at the forum.

I'll mention the private forum idea to Coach Greenspan.

Never trust a coach who offers a prize :P

coach fritz

Since no progress has been made and I don't see anything that will happen soon...I wrote the letter...send to one of the Admins or post it here:
[url]http://www.artofproblemsolving.com/Forum/index.php?f=242[/url]

[i]Dear AoPS Admin,

As a Chicago ARML Coach, I request a private forum for the Chicago ARML team. We have a fair number of active members on the AoPS forum including: Altheman, nateharman1234, cincodemayo5590, besttate, djshowdown2, jackb1115, winluo2, and gotztahbeazn. The addition of this forum would be helpful for us to distribute extra problems, to encourage discussion among our students, and to get other students to join your great problem solving community! Please make myself, ilg, the moderator of this forum. Also make Altheman a moderator because he is our most active member on AoPS.

Thanks,
I. Greenspan[/i]

About that last sentence...just think, I will be active; I am on aops daily.

Thanks Alex (altheman) for taking the lead. I think that a Chicago ARML section would be a good idea, because getting Chicago ARML kids involved in AoPS would definitely be a plus. Also, it really lets ARML members grow together outside of the practices in addition to the problem sets. Just my two cents.

P.S. (I'm Ric Best (D Team) in case any of you were wondering.)

[quote="probability1.01"]Dude I actually emailed one of your coaches but he never replied. I was thinking vaguely about doing ARML but looks like it's not gonna happen :(. Oh well scheduling is questionable anyways.[/quote]

well i will be like:

hey we can get an imo person on our team

we also lost tony liu

he had some conflict during the tryout [prom?]

so yea

we got like 5th last year

we need to step our game up

fritz was like we always do meh on power...so +2 moppers is never a bad thing

i mean like if mop is going to qualify you for usamo, then mop should qualify you for chicago arml team

[man...i will get the #8 this year if it is not counting...i see many problems that are kind of like short olympiad problems that could be put into an arml format...like if they have a geo problem...or algebra/trig...i am all over that...i have to work through the rest of math olympiad challenges by the time that arml comes around....but at least after ap tests, i will have 3 less classes to work in...that way i can step my math up...i also should do a book of old problems...the arml things can be a bit contrived and they repeat problems...] [woo did not mean to type that much]


function
integration
inequalities
real analysis
real analysis unsolved

Let $ S$ be the set of all real $ C^1$ functions $ f$ on $ [0,1]$ such that $ f(0)=0$ and $ \int_0^1 f^'(x)^2 dx \leq 1$. Define $ J(f)=\int_0^1 f(x)dx$. Show that the function $ J$ is bounded on $ S$, and compute its supremum. Is there a function $ f_0 \in S$ at which $ J$ attains its maximum value ? If so, what is $ f_0$ ?

integrate by parts and use the schwarz inequality to get
\begin{eqnarray*} J(f) & = & \int^1_0 f(x)dx \\
& = & (x - 1)f(x)|^1_0 - \int^1_0 (x - 1)f'(x)dx \\
& = & \int^1_0 (1 - x)f'(x)dx \\
& \leq & \left(\int^1_0 (1 - x)^2 dx\right)^{1/2}\left(\int^1_0 [f'(x)]^2dx \right)^{1/2} \\
& \leq & \frac {1}{\sqrt {3}}\cdot 1 \end{eqnarray*}
for equality we want $ \alpha(1 - x) = f'(x)$, $ f(0) = 0$, and $ \int^1_0 [f'(x)]^2dx = 1$.  can this happen simultaneously?


I may be missing something, but the last identity seems not to hold.

hello, from the second equation we get
$y=-\frac{6x^2+3-12x}{6x-13}$ plugging this in the first equation and factorizing we get
$(x-2)(3x-2)(24x^2-118x+149)=0$
from here you will get all solutions.
Sonnhard.

[hide="Solution"]
Let $a=x+y+\frac{1}{x+y}$ and $b=x-y$. Then we have
\[\begin{aligned}
a+b=\frac{13}{3}\\
3a^2+b^2=\frac{103}{3}
\end{aligned}\]
The second equation is obtained from
\[\[\begin{aligned}
3a^2+b^2&=3\left(x^2+2xy+y^2+2+\frac{1}{(x+y)^2}\right)+x^2-2xy+y^2\\
&=4xy+4(x^2+y^2)+\frac{3}{(x+y)^2}+6=\frac{85}{3}+6=\frac{103}{3}
\end{aligned}\]
Solving the first equation for $3b$, we have $3b=13-3a$ and substituting it to the second equation yields
\[\begin{aligned}
3a^2+b^2=\frac{103}{3}&\rightarrow 9a^2+3b^2=103\\
&\rightarrow (13-3b)^2+3b^2=103\\
&\rightarrow 12b^2-78b+66=0\\
&\rightarrow 6(2b-11)(b-1)=0
\end{aligned}\]

Thus, we need to consider two cases.
[b](1)[/b] If $b=x-y=1$, then $a=\frac{13}{3}-1=\frac{10}{3}\rightarrow x+y+\frac{1}{x+y}=\frac{10}{3}$, which gives $x+y=3$ and $x+y=\frac{1}{3}$. Thus we have the solutions $(x,y)=(2,1),\left(\frac{2}{3},-\frac{1}{3}\right)$.
[b](2)[/b] If $b=x-y=\frac{11}{2}$, then $a=\frac{13}{3}-\frac{11}{2}=-\frac{7}{6}\rightarrow x+y+\frac{1}{x+y}=-\frac{7}{6}$. Letting $c=x+y$, we get the equivalent quadratic equation $6c^2+7c+6=0$, which has no real roots.

Therefore, $(x,y)=\boxed{(2,1),\left(\frac{2}{3},-\frac{1}{3}\right)}$.
[/hide]


[color=darkblue]Let $ABC$ be a triangle with the centroid $G\ .$ Prove that

$BC+GA=CA+GB=AB+GC\Longleftrightarrow AB=BC=CA$ ([b][u]Marius Mainea[/u][/b]).

See http://www.arhimede.ro/first.html [/color]

Let $AA', BB'$ be the medians. Then

$AA'^{2}={b^{2}+c^{2}\over 2}-{a^{2}\over 4}\qquad(*)$
$BB'^{2}={a^{2}+c^{2}\over 2}-{b^{2}\over 4}$

giving

$(AA'-BB')(AA'+BB')={3\over 4}(b-a)(b+a)\qquad(1)$

From the given condition

\begin{eqnarray*}GA-GB=CA-CB=b-a\iff{2\over 3}(AA'-BB')=b-a\iff AA'-BB'={3\over 2}(b-a)\qquad(2)\end{eqnarray*}

Plugging $(2)$ into $(1)$:

${3\over 2}(b-a)(AA'+BB')={3\over 4}(b-a)(b+a)$

Assume $a\neq b$. Then the last equation yields

$AA'+BB'={1\over 2}(b+a)$

Together with

$AA'-BB'={3\over 2}(b-a)$

we get

$AA'=b-{a\over 2}$

Plugging that into $(*)$ we get

$b^{2}-ab+{a^{2}\over 4}={b^{2}+c^{2}\over 2}-{a^{2}\over 4}\iff c^{2}=(b-a)^{2}\iff c=|b-a|$

but that's impossible by the triangle inequality. Therefore $a=b$. Similar argument for $b=c,c=a$. QED


number theory
relatively prime
Pythagorean Theorem
geometry

Let $\overline{CH}$ be an altitude of $\triangle ABC$. Let $R$ and $S$ be the points where the circles inscribed in the triangles $ACH$ and $BCH$ are tangent to $\overline{CH}$. If $AB = 1995$, $AC = 1994$, and $BC = 1993$, then $RS$ can be expressed as $m/n$, where $m$ and $n$ are relatively prime integers. Find $m + n$

I know this is an old thread, but I would really like an answer to this one. Thanks.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=1993+aime&t=57036[/url]

see my solution near the bottom

[hide="Solution"]
The diagram is easily obtained. Now, by the Pythagorean Theorem, $ (1993 \minus{} h \plus{} 2s)^2 \plus{} h^2 \plus{} 1993^2$, and $ (1994 \minus{} h \plus{} 2r)^2 \plus{} h^2 \plus{} 1994^2$; subtracting the first from the second gives $ (1994 \minus{} h \plus{} 2r)^2 \minus{} (1993 \minus{} h \plus{} 2s)^2 \equal{} (1 \plus{} 2(r \minus{} s))(1994 \plus{} 1993 \minus{} 2(h \minus{} r \minus{} s))$
$ \equal{} 1994^2 \minus{} 1993^2 \equal{} 1994 \plus{} 1993$. Since the base of this is $ 1995$, $ 1995 \equal{} 1993 \minus{} h \plus{} s \plus{} s \plus{} r \plus{} 1994 \minus{} h \plus{} r \equal{} 1993 \plus{} 1994 \minus{} 2h \plus{} 2s \plus{} 2r$
$ \implies1995 \minus{} 1993 \minus{} 1994 \equal{} \minus{} 2(h \minus{} r \minus{} s)$, so plugging this into the previous equation gives $ (1 \plus{} 2(r \minus{} s))(1994 \plus{} 1993 \minus{} 1993 \minus{} 1994 \plus{} 1995) \equal{} 1995(1 \plus{} 2(r \minus{} s))$
$ \equal{} 1995 \plus{} 2\cdot1995(r \minus{} s) \equal{} 1993 \plus{} 1994\implies r \minus{} s \equal{} \frac {1994 \minus{} 2}{2\cdot1995} \equal{} \frac {1992}{2\cdot1995}$
$ \equal{} \frac {332}{665}\implies m \plus{} n \equal{} 332 \plus{} 665 \equal{} \boxed{997}$.
[/hide]

[hide]Note that in $\Delta AHC$, $RH=\frac{CH+AH-1994}{2}$. Similarly, in $\Delta BCH$, we find that $HS=\frac{CH+BH-1993}{2}$. The value of $RS=|RH-HS|=|\frac{AH-BH}{2}-\frac{1}{2}|$. Since $BH=1995-AH$, we also derive that $RS=|AH-998|$. We will first find the value of $\cos A$. The Law of Cosines on $\angle CAB$ in $\Delta ABC$ yields $1994^2+1995^2-2\cdot 1994\cdot 1995\cos A=1993^2$, or $\cos A=\frac{1994^2+1995^2-1993^2}{2\cdot 1994\cdot 1995}$. To simplify this computation, we will denote $1994=x$. The fraction is then $\frac{x^2+(x+1)^2-(x-1)^2}{2x(x+1)}=\frac{x+4}{2(x+1)}$, so $\cos A=\frac{999}{1995}=\frac{333}{665}$. In $\Delta AHC$, $\cos A=\frac{AH}{1994}$, so $AH=\frac{333\cdot 1994}{665}$. Our desired $RS=|\frac{333\cdot 1994}{665}-998|=|\frac{333\cdot 1994-665\cdot 998}{665}|$. To simplify this, let $x=997$ and $y=333$. Our fraction is then $\frac{2xy-(2y-1)(x+1)}{665}=\frac{x+1-2y}{665}=\frac{332}{665}$. The greatest common divisor of $332$ and $665$ can easily checked to be $1$, so our answer is $332+665=997$.[/hide]

It is obvious that the answer is of the form ((n-1)/2)/n. Using adihaya jayasharmaramankumarguptareddybavarajugopal's lemma, we get that n = 672+1586-1586=665.

Let the incircle of $(ACH)$ is $r_1$ and incircle of $(BCH)$ is $r_2$.

Let $r_1$ meets $AC,AB$ at $P,T$ respectively and $r_2$ meets $BC,AB$ at $Q,M$ respectively. $AC=AP+PC=AT+CR=AH-r_1+CH-r_1=AH+CH-2r_1$; same reason, we have $BC=BH+CH-2r_2$. So $AC-BC=AH-BH+2(r_2-r_1)=1$

Some horrible computation tells that $BH=\frac{662673}{665}, AH=\frac{664002}{665}$ so $r_1-r_2=\frac{\frac{1329}{665}-1}{2}=\frac{332}{665}$ to $\boxed{997}$


Let $ l$ be a line having no common points with triangle $ ABC$. Let $ L,M,N$ be the projections of $ A,B,C$ onto $ l$. Let $ LX,MY,NZ$ be the perpendiculars from $ L,M,N$ to $ BC,CA,AB$ respectively. Prove that $ LX,MY,NZ$ are concurrent.

use the carnot's theorem that says that LX, MY, NZ are concorrent iff $ BL^2 \minus{} LC^2 \plus{} CM^2 \minus{} MA^2 \plus{} AN^2 \minus{} NB^2 \equal{} 0$ but

$ BL^2 \minus{} LC^2 \plus{} CM^2 \minus{} MA^2 \plus{} AN^2 \minus{} NB^2 \equal{} MB ^2 \plus{} LM^2 \minus{} LN^2 \minus{} NC^2 \plus{} NM^2 \plus{} NC^2 \minus{} LM^2 \minus{} LA^2 \plus{} LA^2 \plus{} LN^2 \minus{} NM^2 \minus{} MB^2 \equal{} 0$

It could be more interesting to prove this:
Let a triangle ABC and a triangle A'B'C', then the prpendicular from A to B'C', from B to A'C', from C to A'B' are concurrent iff the prpendicular from A' to BC, from B' to AC, from C' to AB are concurrent.

Which Carnot are you referring to?

I can see here but unrelated: http://www.mathlinks.ro/viewtopic.php?search_id=1830777513&t=45859

see [url=http://www.cut-the-knot.org/pythagoras/Carnot.shtml]here[/url]  :)


The product of three consecutive multiples of 5 is 26,250. What is the largest of these three factors?

[hide]
Let the numbers be $ (n-5),(n),(n+5)$ respectively. Then

\begin{align*}
(n-5)(n)(n+5)&=26250\\
n^3-25n-26250&=0\\
(n-30)(n^2+30x+875)&=0
\end{align*}

So $ n=30$. This implies that the largest integer is $ \boxed{35}$
[/hide]

[quote="hunter34"][hide]
Let the numbers be $ (n - 5),(n),(n + 5)$ respectively. Then
\begin{align*} (n - 5)(n)(n + 5) & = 26250 \\
n^3 - 25n - 26250 & = 0 \\
(n - 30)(n^2 + 30x + 875) & = 0 \end{align*}
So $ n = 30$. This implies that the largest integer is $ \boxed{35}$
[/hide][/quote]

How do you see the factoring fast enough to do this? Just wondering cuz 'twas a countdown problem.

It was a national countdown. :wink:

And if you're really good, you could memorized the cubes up to $ 30$.

Then, find $ \sqrt[3]{26250}$, which is approx. $ 30$.  Then you could realize that $ 25 \cdot 30 \cdot 35$ was equal to $ 26250$ and you could quickly answer $ 35$. :P

Calculations will be far simpler with a more natural representation:
$ 5(n \minus{} 1)\cdot 5n \cdot 5(n \plus{} 1) \equal{} 26250$
$ 125n(n \minus{} 1)(n \plus{} 1) \equal{} 26250$
$ n(n^2\minus{}1)\equal{}210$
$ n^3 \minus{} n \equal{} 210$

Now, assuming you've studied your cubes, $ n \equal{} 6$ would leap out at you.

$ 5(6 \plus{} 1) \equal{} \boxed{35}$

Using this instead of the previous solution should cut your time in half.

You could factorize it to
5^4 *2*3*7
which you could simplify to (5*5)(5*6)(5*7)
the largest one is (5*7) which is 35
Answer-35


Find all real $x$ satisfying:

$ax=[x]$ with $0<a<1$ a real paramether.

$x=\frac ka$, were k is an nonnegative integer $k<\frac{a}{1-a}$ (k=0,1,...,m, $m=-[\frac{a}{a-1}])$.


algorithm
analytic geometry
projective geometry
advanced fields
advanced fields unsolved

Where can I found an algorithm to transform an elliptic curve to its Weierstrass form?

This is exactly the question I wanted to ask. I have a cubic curve $ X^3\plus{}Y^3\equal{}1$ and I wanted to get the Weierstrass form which is $ y^2\equal{}x^3\minus{}432$ but no where I found the algorithm (there is some very vague procedure described in the book Rational Points on Elliptic Curves,  Joseph H. Silverman and John Tate, pag. 22,23 and 24)
What I now is that one putes $ X\equal{}X/Z$ and $ Y\equal{}Y/Z$ and gets $ X^3\plus{}Y^3\equal{}Z^3$ in the projective space. Then one uses 'projective transformations' and some change of coordinates to get the Weierstrass form. I tried several times without succeeding.

[quote="puuhikki"]Is there a bug in Maple or an error in the Silverman's and Tate's book?
with(algcurves);f := x^3+y^3-1; v := Weierstrassform(f, x, y, x0, y0);
does not output $ y^2 \equal{} x^3 \minus{} 432$ or anything similar.

Äh. Weierstrass form is not unique.[/quote]
It is not wrong...I found also $ y^2\equal{}x^3\minus{}432$ using this method:
We start in the projective space with $ U^3\plus{}V^3\equal{}W^3$. If we define $ u\equal{}U/W$ and $ v\equal{}V/W$ then we get rid of $ W$ and find $ u^3\plus{}v^3\equal{}1$ and  one point (direction)  $ O$ at infinity corresponding to $ W\equal{}0$ namely $ [\minus{}1,1,0]$.

Now the tangent line to $ O$ has equation $ u\plus{}v\equal{}0$. So according to the book we set $ Z: \equal{}u\plus{}v$. Since we work with an inflection point we know that there are not intersections of the tangent line with the curve. So for $ V$ and $ W$ we can choose any independant 'linear' expressions.
For example $ U\equal{}1$ and $ V\equal{}u\minus{}v$. Remember: we are taking 'linear' transformations in the projective space.
Now we have $ x\equal{}\frac{U}{W}\equal{}\frac{1}{u\plus{}v}, y\equal{}\frac{V}{W}\equal{}\frac{u\minus{}v}{u\plus{}v}$.
These last  two expressions give expressions for $ u$ and $ v$ in terms of $ x$ and $ y$.
After some calculations you get $ 6y^2\plus{}2\equal{}8y^3$ so $ 6y^2\equal{}(2x)^3\minus{}2$.
Since we want the coefficient of $ y$ to be $ 1$ we must multiply the whole equation with some number $ b$ such that $ 6b$ is a square and such that $ b$ is a third power. The number $ b\equal{}2^3*3^3$ satisfies the two conditions and we get
$ 216*6*y^3\equal{}216*x^3\minus{}216*2$
so
$ (36y)^2\equal{}(2*3*2x)^2\minus{}432$
this gives
$ y_1^2\equal{}x_1^2\minus{}432$

By the way I would like to 'prove' that at $ O$  we indeed have an inflection point. Also I know that $ u\plus{}v\equal{}0$ goes through $ O$ but is it also the tangent line at that point?


a rod of mass m and length l is kept on a horizontal plane. It rotates with an angular velocity w and the velocity of its centre of mass is u such that u=wl/2.
A is the farther end and B is the nearer end. Find the DISTANCE travelled by B when the rod completes a full rotation.

I don't know which end is the farther , but both follow a cycloid path, whose period length (hump) is 4L, L being the rod length. (what for the mass m?)


If $x_i > 0$ and $x_iy_i - z_i^2 >0$ for $i\leq n$, then

\[\frac{n^3}{(\sum_{i=1}^{n} {x_i})(\sum_{i=1}^{n} {y_i})-(\sum_{i=1}^{n} {z_i})^2} \leq \sum_{i=1}^{n} {\frac{1}{x_iy_i-z_i^2}}\]

Prove this inequality for n=2, and then also generally.


Could anyone be kind enough to send me the 2007 Mathcounts State Solution set? I have the 2007 State Questions, but I have no way to check my answers. I've tried to find it from the old places but they've all been shot down by Mathcounts.

Thanks in advance!


algebra
polynomial
superior algebra
superior algebra solved

Let k be an algebraically closed field. Then, why does every homogeneous polynomial in k[x,y] factor as a product of linear homogeneous factors?

Set $y=1$ (thus dehomogenize) factor into linear factors in $x$, homogenize the factors again, voila.


Consider a circle C(O,R) and three chords AB,CD,EF (without be intersect each other) such that AB=CD=EF=R.If K,L,M are the middles of the chords BC,DE,AF respectively prove that the triangle KLM is equilateral. :)


Putnam
absolute value
Irrational numbers

Let $a, b, c$ be integers, not all zero and each of absolute value less than one million. Prove that \[\left\vert a+b\sqrt{2}+c\sqrt{3}\right\vert > \frac{1}{10^{21}}.\]

[quote="Peter"]This is from Putnam 1980 though: http://www.kalva.demon.co.uk/putnam/psoln/psol804.html[/quote]

You're right.


function
inequalities
algebra proposed
algebra

Determine all functions $f: \mathbb N\to \mathbb N$ such that for every positive integer $n$ we have: \[ 2n+2001\leq f(f(n))+f(n)\leq 2n+2002. \]

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76767 (and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84789 ).

  darij

nice probelm for sure ...
no solutions for this problem untill now ... so here is mine , feel free to correct me if i am wrong.
let
$Q(n)= 2n+2001 \le f^2(n)+f(n) \le 2n+2002$
awe clearly have $f(n) \le 2n+2002$ so
$2001+2n \le f^2(n)+f(n) \le 3f(n)+2002$ so
$ \frac{2n}{3} - \frac{1}{3} \le f(n) \le 2n+2002$
now define the sequance of reals $a_i,b_i,c_i,d_i$
such that
$a_i n + b_i \le f(n) \le c_i n +d_i$
now find a recursive relation for $a_i,b_i,c_i,d_i$
such that
$c_{i+1}= \frac{2}{a_i +1}$.
$d_{i+1}=\frac{2002-b_i}{a_i +1}$.
$a_{i+1}=\frac{2}{c_i+1}$.
$b_{i+1}=\frac{2001-d_i}{c_i+1}$
clearly $a_{i+1},b_{i+1},c_{i+1},d_{i+1}$ also work in the statement we have.
now we have that
$a_i=\frac{2}{c_{i+1}}-1=\frac{2}{c_{i-1}+1}$
first we clearly have $c_i \ge 1$  now we prove $c_{i-1} \ge c_{i+1}$ assume for contradiction that
$c_{i-1} < c_{i+1}$
now we have
$\frac{2c_{i-1}+2}{c_{i+1}}=c_{i-1}+3$ so
$\frac{2}{c_{i+1}}>c_{i-1}+1>2$ so $1> c_{i+1}$ contradiction , so $c_i$ is decreasing and bounded below so it converges , clearly it converges $1$ working with $a_i$ is simillar it also converges $1$
now we use this , clearly $d_i >0$
now we find this for $d_i$
.
$2002 - d_{i+1}(a_i+1)=\frac{2001-d_{i-1}}{c_{i-1}+1}$
let $i$ be large enough to assume $a_i=c_{i-1}=1$ (i mean its really close not actually equal)
so $\frac{2003+d_{i-1}}{4}=d_{i+1}$ so clearly
$d_i$ converges to $\frac{2003}{3}$ so we have
$f(n)\le n+\frac{2003}{3}$ also bd $f(n)$ is natural we have
$f(n)\le n +\frac{2001}{3}$
so $2001+2n \le f(f(n))+f(n) \le 2n+2001$ so
$f(n)=n+667$
for every positive integer which clearly works ...
.
@below
the comments that are not nessesary can be given in massages as this is a contest tread. also i have looked the links Mr(or Mis)Darij has sent , but i didn't realised the second post is a solution. anyways , another solution wouldn't hurt right :D ?

[quote=Mr.C]
no solutions for this problem untill now[/quote]
There is a solution in Darij's link

the following proof is by the user "enescu".
i just posted it here so that the his/her solution be on the official tread. (and nobody makes the same misstake as me not seeing this beautiful solution)
.
We define the sequence of positive integers $\left( a_{n}\right) _{n\geq0}$
as follows: $a_{0}$ is an arbitrary positive integer and $a_{n+1}=f\left( a_{n}\right) ,$ for all $n.$ Let
\[ c_{n}=a_{n}-a_{n-1}-667,  \]
for all $n\geq1.$ Then
\[ c_{n+1}+2c_{n}=a_{n+1}+a_{n}-2a_{n-1}-2001,  \]
therefore for all $n\geq1,$ $c_{n+1}+2c_{n}$ is an integer, satisfying
\[ 0\leq c_{n+1}+2c_{n}\leq1.  \]

Suppose that $c_{1}>0;$ then $c_{1}\geq1$ and $c_{2}\leq-2c_{1}+1\leq-1.$
Moreover, $c_{3}\geq-2c_{2}\geq2.$ We claim that $c_{2k+1}\geq2^{k}$ and we prove it by induction$.$ If $c_{2k+1}\geq2^{k}$ holds, then $c_{2k+2} \leq-2c_{k+1}+1\leq-2^{k+1}+1$ and $c_{2k+3}\geq-2c_{k+2}\geq2^{k+2} -2\geq2^{k+1},$ and the claim is proved.

Observe that
\[ a_{2k+2}-a_{2k}-1334=c_{2k+2}+c_{2k+1}\leq-2^{k}+1,  \]
therefore, for $k\geq11,$ we have
\[ a_{2k+2}<a_{2k}.  \]
This is a contradiction, since all $a_{k}$'s must be positive integers.

If $c_{1}<0,$ then $c_{2}\geq-2c_{1}>0$ and we repeat our argument until we obtain
\[ a_{2k+3}<a_{2k+1},  \]
for all $k\geq11,$ another contradiction.

We conclude that $c_{1}=0,$ that is, $a_{1}=a_{0}+667.$ It follows that
\[ f\left( n\right) =n+667,  \]
for all $n.$

[b][u]Claim 1: $f(n)>n ~~\forall n \in \mathbb{N}.$[/u][/b]
[i]Proof.[/i] Assume the contrary. Let $a\in \mathbb{N}: f(a)\leq a$ and let $f(b)$ be it's least value. Now letting $c=f(b) \implies$ \begin{align*} f(c)+c \leq 2k +2002 \text{ and } b+c\geq 2a+2001 \\ \implies f(c)+2a+2001-b \leq 2b+2002 \\ \implies f(c)\leq 3b-2a+1<b<a<c, \end{align*} a contradiction. $\blacksquare$
----------------

[b][u]Claim 2: $f(n)=n+667~~\forall n \in \mathbb{N},$ which fits.[/u][/b]
[i]Proof.[/i] Let $f(n)-n=g(n)>0,$ and let $g(m)$ be it's least value. Now setting $e=f(m)\implies$ \begin{align*} 2g(m)+g(e)\geq 2001 \\ \text{ and } 2g(e)+g(f(e))\leq 2002 \\ \implies g(m)\geq 667.\end{align*} But $2g(n)+g(f(n))\leq 2002 \implies g(n)=667,$ thus we're done. $\blacksquare$


ratio
LaTeX
algebra
polynomial
Rational Root Theorem

Find the sum of all the solutions of $x$:

$\sqrt{1+\sqrt{1+x}}=x$

[hide]$\sqrt{1+\sqrt{1+x}}=x$

$x^4-2x^2-x=0$

$x=\frac{\sqrt5+1}{2}, ...$ No other solutions work in the original equation.

Sum of $x= \frac{\sqrt5+1}{2}$

Can someone explain the method to why some solutions (after making exponents in whole ratios) work in radical equations while others don't [as, why do you have to plug the solutions back into the equation to make sure they work]? Or maybe a way of telling which solutions will work...[/hide]

$\sqrt{1 + x} = {x^2} + 1$
$1 + x = {x^4} + {2x^2} + 1$
${{x^4} + {x^2} - x} = 0$
${x}({x^3} + x - 1) = 0$

Which, as far as I can tell, is unfactorable (ignoring imaginary solutions)
Thus, ${x = 0}$, and $\boxed{s = 0}$

[quote="surge"][hide]Can someone explain the method to why some solutions (after making exponents in whole ratios) work in radical equations while others don't [as, why do you have to plug the solutions back into the equation to make sure they work]? Or maybe a way of telling which solutions will work...[/hide][/quote]

Some solutions take the negative of the square root to make the equation true.

For example, in your last equation, $x=-1$ too, but if you plug it back, it doesn't work since $\sqrt9=3$ and not $-3$.

If you substitute $x=-1$ back into the original equation, you get $\sqrt1$=-1, which is taking the negative square root, but you can't do that, so that's why it doesn't work.

BTW, how do you do the not-equal sign in Latex?

[quote="robinhe"][quote="riddler"][quote="robinhe"]
Some solutions take the negative of the square root to make the equation true.
For example, in your last equation, $x=-1$ too, but if you plug it back, it doesn't work since $\sqrt9=3$ and not $-3$.
If you substitute $x=-1$ back into the original equation, you get $\sqrt1$=-1, which is taking the negative square root, but you can't do that, so that's why it doesn't work.
[/quote]
I dont see how that cannot work :?  :(  $(-1)^2=1$
So, $\sqrt{1}$ could be $-1$ or $1$ ????
Why is the answer then what surge said and not 1 lesser than that?[/quote]

No, I meant $\sqrt{1}\not=-1$... :?[/quote]
Why not?

For convenience and brevity, we define $\sqrt{x}$, where $x$ is a positive real number, to be the positive square root.  That is, we have two logical possibilities for the square root, but we choose the positive one to define $\sqrt{x}$, and write $-\sqrt{x}$ for the negative one.  You can find a similar explaination at [url]http://mathworld.wolfram.com/SquareRoot.html[/url].

[quote="kokosai"][quote="LordoftheMorons"]$\sqrt{1 + x} = {x^2} + 1$[/quote]I think you should have subtracted $1$ from the righthand side, not add it to it.

robinhe: not= is the Latex for not equal to: $\not=$.[/quote]

>_< That's what I get for running back and forth to the computer during commercial breaks.

[hide]Start squaring both sides:
$\sqrt{1+\sqrt{1+x}}=x\Longrightarrow$ $1+\sqrt{1+x}=x^2$
$\sqrt{1+x}=x^2-1\Longrightarrow$ $x+1=x^4-2x^2+1\Longrightarrow$ $x^4-3x=0$
Sum of the roots=$\frac{-b}{a}$
$x^4+0x^3=0x^2-3x+0=0\Longrightarrow$ $\frac{-b}{a}=\frac{0}{1}=\boxed{0}$
This includes complex roots.[/hide]

[hide="solution"]
$\sqrt{1+\sqrt{1+x}}=x$
$x^2-1=\sqrt{1+x}$
$x^4-2x^2+1=1+x$
$x(x^3-2x-1)=0$
Using rational root theorem we find that -1 is a root.
$x(x+1)(x^2-x-1)=0$
Notice that $\frac{1-\sqrt{5}}{2}$ cannot be a root as well as -1.
So the sum is $\boxed{\frac{1+\sqrt{5}}{2}}$ since 0 doesn't affect the sum.
[/hide]

Ahhh. I forgot extraneous roots. Silly me :P

[quote="kokosai"][quote="ragnarok23"][hide]$x+1=x^4-2x^2+1\Longrightarrow$ $x^4-3x=0$.[/hide][/quote]I don't understand this step.[/quote]
He just subtracted x+1 from both sides.

i don't think you understand it because it's wrong :rotfl:

if $x+1=x^4-2x^2+1$ then $x^4-2x^2-x=0$


I don't know if this has already been suggested (Actually, I sort of do, I did a search and it didn't turn up anything), but should there be a Purple Comet sub forum of Contests & Programs » Other US Contests & Programs? Or in the International competition section? Other online competitions have their forums, why not Purple Comet?

I thought about suggesting this, but at the time I thought it wasn't a big enough competition to have its own subforum. If you consider the number of schools that participate each year, however, it seems like a good idea to have a subforum for it to discuss problems and such.

Yeah, it would be a good place to discuss results with other competitors.

...which, if you think about it, is the main point of a competition forum.

Yes, that's why it's strange that we don't have one for Purple Comet... then again, if you had them for every contest, there would be a lot of subforums...


[color=blue][size=150]XVII CONO SUR Mathematical Olympiad.[/size][/color]
[color=blue]TEAM SELECTION TEST.[/color]

[color=blue]Problem 1[/color]
Find all the pairs of positive numbers such that the last
digit of their sum is 3, their diference is a primer number and
their product is a perfect square.

[color=blue]Problem 2[/color]
Let $AA_1$ and $BB_1$ be the altitudes of an acute
triangle non isosceles $ABC$, $A_0$ y $B_0$ are the midpoints of
$BC$ and $CA$, respectively. The line $A_1B_1$ cuts the line
$A_0B_0$ in $C'$. Prove that $CC'$ is perpendicular to the line
that joins the orthocenter and the circuncenter of the triangle
$ABC$.

[color=blue]Problem 3[/color]
 The set $M= \{1;2;3;\ldots ; 29;30\}$ is  divided in $k$
subsets such that if $a+b=n^2, (a,b \in M, a\neq b, n$ is an
integer number), then $a$ and $b$ belong different subsets.
Determine the minimun value of $k$.

[color=blue]Problem 4[/color]
All the squares of a board of $(n+1)\times(n-1)$ squares
are painted with three colors such that, for any two different
columns and any two different rows, the 4 squares in their
intersecctions they don't have all the same color. Find the
greatest possible value of $n$.

[i]Lima, April 1st - 2006.[/i]

[Translated by M4RIO]
$\text{\LaTeX}{}ed$ by Carlos Bravo 2006 - MathLinks AoPS

[color=blue]WELL HERE
THE pdf VERSION and the latex resource!!

ALSO, YOU CAN FIND ITS SPANISH VERSION  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=82074]HERE[/url]
[b]The Spanish Communities[/b]

Carlos Bravo ;)
Lima - PERU[/color]


Do you know where I could find a list of application deadlines for colleges?

I'm trying to get a sense of which of your high school senior year accomplishments will appear on your college application and which can not.

It varies a bit.  The majority have the the deadline right after January 1.  If you have a College Board account, the deadlines for the colleges you choose are listed in the calender on the account page.
However a large parts of the applications and its requirements must be completed a while before the deadline.

I don't know of any comprehensive list; it's usually better to just go through the schools you're interested in individually and compile it yourself.

In general, as mentioned above, the deadline in January 1 for regular action. December 15 is also a popular regular action deadline. If you're applying early, it's usually November 1. If you're applying to any of the UC's, the deadline is November 30.

There will be very little time to accomplish much in your senior year, before applying to colleges. If something impressive happens after you apply, you can send an email update. If you are wait listed, you can certainly send updates - in fact, it is an excellent idea!

Who would you email an update to, lfm? How and why would that person you email be able to contact with the people in charge of admissions?

It sounds like a good idea. I wonder if a senior at my school who got a 13 on the AIME did that this year...

[quote="aufha"]Who would you email an update to, lfm? How and why would that person you email be able to contact with the people in charge of admissions?

[/quote]

You send it to admissions! If you have an admissions counselor assigned to you, then you would go through them.


[quote="GameBot"]The Wonderlich family has three children. At least one of them is female. If having a male or female child is equally likely, what is the probability that at least one of the three children is a male?[/quote]

The probability of having no boys is $ \frac{1}{8}$. The complement of that is $ \frac{7}{8}$, our probability.

one kid is already a girl.  probability of no boys is 1/4.  so complement is [b]3/4[/b]
EDIT: FTW has a different answer though 6/7

If we ignore the "At least one of them is female" statement, we get the following possibilities, where B is a boy and G is a girl:

GGG
GGB
GBG
GBB
BGG
BGB
BBG
BBB.

However, the possibility BBB violates "At least one of them is female", so we are left with 7 possibilities:

GGG
GGB
GBG
GBB
BGG
BGB
BBG.

Of these, 6 have at least one boy. So, the answer is 6/7.


Harazi and Myth play the following game: Harazi writes at least three consecutive numbers on the black board. The first and the last number are zero and all other numbers are positive. Denote the sum of all numbers written on the board by $S$.
Now Myth clears on of the postive numbers $z$ from the board, the sum of the numbers left to $z$ are denoted by $L$ and the numbers on the right hand of $z$ are denoted by $R$. Myth wins when both, $L$ and $R$, are not greater than $\frac{1}{2} \cdot S$. Otherwise Harazi wins.

a.) Show that Myth can always win.
b.) Rules are changed that Myth wins exactly iff $L$ and $R$ are both smaller than $\frac{1}{2}.$ Can Myth always win in this case ?

[quote="Myth"]I don't like if someone write my name with small first letter. :/[/quote]

It seems the first letter of your name is a capital letter...  :P