group theory
abstract algebra
LaTeX
superior algebra
superior algebra unsolved

Let N_1 and N_2 be normal subgroups of a group G

is the following true?
(N_1N_2)/(N_1 intersection N_2) = N_1/(N_1 intersection N_2) x N_2/(N_1 intersection N_2) where / denotes quotient group, x denotes direct product and intersection as in setwise sense.

are you saying that is true? (I believe you if you say so...)
what about my expression?

Actually ignore what I said, I was thinking about some stupid approach. Just define an isomorphism between them.

$\phi : N_{1}N_{2}/(N_{1}\cap N_{2}) \mapsto N_{1}/(N_{1}\cap N_{2}) \times N_{2}/(N_{1}\cap N_{2})$

As,

$\phi (n_{1}n_{2}(N_{1}\cap N_{2})) = (n_{1}(N_{1}\cap N_{2}), n_{2}(N_{1}\cap N_{2}))$

I am sure this maps is well-defined, not in the mood to show that. And it seems like this is a homormophism. So it seems like it is true.

In fact, we never need the fact that $H_{1}\cap H_{2}$. Rather we can say the following, a stronger statement.

"If $N$ is a normal subgroup of both subgroups $H_{1}$ and $H_{2}$ then,
$H_{1}H_{2}/N \simeq H_{1}/N \times H_{2}/N$"

[quote="sit.think.solve"]thanks for that info.

how do you put math symbols as you did? are the codes same as that in latex? and do i just type it in?

let me try:

$a+b$[/quote]

Yes same as LaTeX just use "dollar symbol" to start and end.


There are 6 differently coloured balls placed in a row.  A new arrangement cannot be made if two balls that were together in the last arrangement, are together again.  How many new arrangements can be made?

[hide]only the numbers like 123456 and 135246 are possible. we get trapped on any other number after this. [/hide]


If $a,b,$ and $c$ are digits for which \[ \begin{tabular}{cccc} & 7 & a & 2 \\ - & 4 & 8 & b\\ \hline & c & 7 & 3 \end{tabular} \] then $a+b+c =$

$\text{(A)} \ 14 \qquad \text{(B)} \ 15 \qquad \text{(C)} \ 16 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 18$

[hide]This problem is easier to do one column at a time. For the ones column, 3 is greater than 2, so we know that 2 must have borrowed 10 from a, leaving 12-b=3. [b]b=9[/b]. a is now 1 digit less than before, so whatever digit we get for a we must add one to. In the tens column, 7+8=15, so we find that a's current digit is 5 and it borrowed from the seven to its left. Adding 1 to 5 leaves [b]a=6[/b]. In the hundreds column, 7 became 6 after a borrowed from it. 6-4=[b]c=2[/b]. a+b+c=9+6+2=[b]17[/b].[/hide]

Oh no I messed up  :(  It supposed to be d) 17

[quote="akimatsu"][hide]
12-b=3, b=9
(10+(a-1))-8=7, a=6
(7-1)-4=2

a+b+c=15

(B)

[/hide][/quote]

you screwed up on your last steps.

About half of these people really don't know how to add. 9+6+2=17 not 15 people. the answer is 17

[hide="check ur additions, guys"]
using common mathematical sense, you're gonna realize that $a+b+c=17$, which is $D$.[/hide]

[quote="dogseatcheese"][hide]The answer should be D) 17 because the only number fit for c is 9 and the only number fit for b is 6 and a is left with 2.[/hide][/quote]

o and by the way, ur answer is correct, but the numbers correspondent to the variables are wrong  :huh:

[hide]
Obviously b=9 because 12 - 9 = 3 and then we get 7A2 - 489 = C73 so a=6 and c=2. a+b+c=17. [b]D[/b]
[/hide]


There are 20 kids in your class and each is randomly assigned a country to report on by random picking from a hat. It goes like this: the teacher puts the names of 20 countries into a hat and goes to each student one-by-one to randomly choose a country. Once a country is chosen, it cannot be chosen again. You really want to pick United States because that means no extra research is required :D

The question is, is it better to go first, last, or somewhere in the middle?

I kind of did like the first three, and you can see that all of them has a chance of $ \frac{1}{20}$ But, I prefer to go third. :D

How will this change if you have more than one country you want to report on? How about if you have a country that you [b]definitely[/b] do not want to report on?

First one uh depends if they have the country I want. :P Jk, lets see, experiment again but this time there is casework.(Use complementary counting) you find that the probability that you don't pick it is 9/10, so 1-9/10=1/10.
Second one same thing as first one.

should be the same.... but... i prefer middle.


I'm having trouble getting past level 11. I put tons of stone towers in the earlier levels (well like 3) because otherwise I can't get past the land creeps but then on level 11 I don't have enough to buy air towers to kill the spaceships. I think the only way is to get several Stick towers in the beginning and sell them off later when you have better towers...

umm you didnt post a link...
SILLY ALBERT TRIX ARE FOR KIDS

Yeah oops... It's called Shock Defense.

http://www.addictinggames.com/shockdefence.html

hmm, pretty good, but not very hard
played it a few times, beat it the second time i actually knew what was going on
i only had 1 life left though :D
i had kinda messed up earlier
my score was 2629
i can describe my strategy if you want

Yeah I just suck at stuff like this... describe your strategy please.

So I put 3 stone towers in the beginning and upgrade them ASAP for level 6 (the fast legions). I also put a few stick towers cuz they attack land + air and I need them for level 1-10 and then for level 11. But I don't have enough money for more anti-air units...

here's how i did it:
first i started with about 5 upgraded stones in the middle of the loopy section.  i then got a few lightning towers in that same place
i sold a few of the stones when the flyings started coming, but kept 2 (this is where i lost some life, i dont think i kept enough stones)
against the flyings i used flying towers and lightning towers upgraded twice (the third was too expensive, dont know how good it is)
then, i put a poison in the left loop
eventually, i added some fires to the mix, though they werent a big part of my evil plan
once i started getting to the 30s i got an ultimate tower
by the end of the last level i had two ultimate towers, one upgraded (those things are awesome)
i also put a few more poisons around the place to slow the people down in the final levels (cuz the ultimates have a huge range)

oh yeah there are 35 levels total

I just beat level 33 as of right now... and I have 13 lives left.

I forgot there were ultimate towers...  :blush:

EDIT: Darnit... I invested 2 of my tributes into +percent so now I can't get ultimates :( . And also, DO NOT get the thunder 3rd upgrade... it increases damage from 360 to 400 :( .

okay, a few more things i've learned
1:  DO NOT KEEP YOUR OLD TOWERS
2:  YOU MUST UPGRADE THE ULTIMATE TOWER OR YOU WILL NOT WIN

I'm on level 30 right now, and I have an upgraded rocket tower thingy.

I think they should be banned.

Let's see how far people get without a Shock Rocket Tower. I got to level 34 :) .

Oh yeah, and for the bonus levels there are 40 enemies each.

Yay. I beat the game.

Score: 3450.

Geez, they should give like lives * 100 because lives left are way more important than money or whatever.

Stone then lightning then rocket
Quite easy to beat! :D
Score: 3155

ehhh
i got a 3317

pretty cool that 3317 is an anagram of 1337

I finished the game with 20 HP, 18000 points and 14000 money

easy strategy

1. levels 1-7
build 1 stone tower ,upgrade twice, dont spend any more money, use 1st tribute on lightning tower

2. levels 8-14 kill stone tower, build lightning tower, upgrade once
use tribute on interest rate, dont spend any more money

3. levels 15-21 upgrade lightning tower 2nd time, dont spend any money, use tribute on interest rate

4. levels 22-28 , build 1 more lightning tower, upgrade twice, use tribute on interest rate

5. levels 29-34 , build adjacent lightning towers so they all shoot at the target at the same time , upgrade twice (not 3 times)
3rd upgrade is a waste of money, no range diff, and damage diff is only 40 points, whilst you spend 720\$ on it

6. spend half your money on twice upgraded lightning towers to finish the last level

important note
put towers on corners
so they shoot in circle and reach many places at once

I fail to comprehend how it is possible to beat level 6 with only 1 stone tower.

put another stone tower

but!! they must shoot at the same target at the same time

I can only get 16039 by 4 rocket towers
Seems upgrading to rocket tower is not a good strategy

How do you get a lot of points? I only got 4300 or so, even though I had 19 lives left.

your points are the same as your gold
influinced by interest

if you only upgrade between levels

and after level 7 get rid of the small towers (ALL OF THEM)

and put only lightning towers

you can win

also until certain level, you can keep on only 2 towers, also dont upgrade the towers more than twice, its a waste of money

how do u beat the level with the 20000 hp ships

Have lots of lightning towers :) .

And the level after that is 30000 hp ships.

Easy.

Score:  31356

I would have higher, but I started selling off my towers at the end too late.  I estimate I could have gotten several thousand more if I had started selling them sooner (and, I built a few too many, but whatever).

I beat it. After you get the advanced towere then the things get easy. after you have a super tower, sell all of youre non advanced towers. Use this monny to buy supertowers and upgrade them.


geometry
trigonometry
trig identities
Law of Sines
Law of Cosines
area of a triangle

$\triangle{ABC}$. $\angle{A}=60^{o}$, $AC=16$. The area of the triangle is $220 \sqrt{3}$.
Find $BC$.

By the Law of Sines,
$[\triangle ABC] =\frac{AC \cdot AB \cdot \sin \angle A}{2} = 220 \sqrt3$
So $AB = \frac{55}{4}$
Using Law of Cosines,
$BC = \frac{\sqrt{3601}}{4}$

Since the answer seems so weird, I am pretty sure that I did something wrong  :?

[quote="Palytoxin"]really?
[hide]i got $BC=49$ :D[/hide][/quote]
Bingo!  :D

[quote="Priapus"]i dunno how, i got ab=55[/quote]
[hide]Yep. AB=55, BC=49.[/hide]

[quote="deej21"][hide="bornformath"]
try recalculating $AB$
[/hide]

[hide="answer"]
BornForMath's method is correct, but calculating gives us $AB = 54$

Then using law of cosines, $BC=2\sqrt{577}$
[/hide][/quote]

Thanks for pointing that out.  :)  But after I redid it, I got $AB = 55$

Weird.  I even redid that two or three times to make sure I did it right.  For some reason, I didn't catch that $\frac{20}{4} = 5$

Meh, I guess that's what I get for doing math late.

sorry, this is a little off topic, but is the area of a triangle $(\frac{1}{2})b)(c)(\sin{A})$?
I used this formula and the law of cosines to obtain 49 for BC also, so I assume it's correct, but it could be luck.

I have never expected that this problem will recieve such a result.
It seems that lots of you misread the problem or made calculation mistakes......
But after all, it is not a difficult problem and all you need is to be careful.


search
projective geometry
geometry unsolved
geometry

This is an old ( > 50 years ) well known problem, but I don't remember the proof. I must search my archive about it and I will post here a solution, if will be not any reply with references.

By the way, I would like to present another interesting and no so well known ( I think ) problem in this configuration.

[b][size=100][color=DarkBlue]We  denote as $A',$ the intersection point of $BE_{2},$ $CF_{1}$ and as $A'',$ the intersection point of $BE_{1},$ $CF_{2}.$

Similarly we define the points $B',$ $B''$ and $C',$ $C''.$ Prove that:

1). -  The segment lines $AA',$ $BB',$ $CC',$ are concurrent at one point, so be it, $P.$

2). -  The segment lines $AA'',$ $BB'',$ $CC'',$ are concurrent at one point, so be it, $Q.$

3). -  The segment lines $A'A'',$ $B'B'',$ $C'C'',$ are concurrent at one point, so be it, $R.$

4). - The points $P,$ $Q,$ $R,$ are collinear.[/color][/size][/b]

Kostas Vittas.

PS. In my drawing, the points $D_{1},$ $E_{1},$ $F_{1},$ are nearer the vertices $B,$ $C,$ $A,$ respectively.

[quote="AkiM"]A circle meets three sides BC, CA, AB of triangle ABC at points D_1,D_2;E_1,E_2 and F_1, F_2 in turn.Then, line segments D_1E_1 and D_2F_2 intersect at point L, line segments E_1F_1 and E_2D_2 intersect at point M, line segments F_1D_1 and F_2E_2 intersect at point N.Prove that three lines AL, BM, CN are concurrent.[/quote]

Really well-know!

I have a solution with Pascal Theorem and Desargues Theorem, but is basically the same of Darij!

See here: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1926634524&t=24378


Euler AMC AIME articles Gauss downwithcloning

I think clonning is a wonderful breakthough for science but it needs to be kept in check for now and as experiments only.  Like the sheep "Dolly".  I do not think it should be avaliable for a long time.

Cloning is absolutely vital to making a new breakthrough in science, but this power must be kept in a TIGHT check system.  I personally am all for cloning human organs... However, I am opposed to cloning human beings because there are tons of problems that come with that...

Just to play devil's advocate, what are those "tons of problems"?

Think about it.  What would happen if someone tryed to clone Einstein.  Its completely possible.  We have his DNA.  What would happen if we did clone him.  Think of the problems it would cause.  Those are part of the "tons" of problems with clonning if there is not tight messures suronding it.

ah...forgot to put what I thought...sorry about that....I think cloning has its good parts and its bad parts. I'm pretty much in the middle. It has the potential to help people, but it could ultimately turn out to be something that has serious consequences....

[quote]Think about it. What would happen if someone tryed to clone Einstein. Its completely possible. We have his DNA. What would happen if we did clone him. Think of the problems it would cause.[/quote]

You haven't actually supplied a single problem. As far as I am concerned, lets clone Einstein. The only problem I see is possible fraud due to having the same retinal scan or fingerprint or dna. But Einstein is dead, so thats not the case. I think the reason that human cloning is banned now is that cloning is sucessful in only a small percentage of cases, and in the process of cloning a human many fetuses would die. However, if the sucess rate was 100%, I am all for cloning a human. You mention Einstein. First, I have to say that the environment this clone will live in is so different from that the real Einstein lived in that they probably will mentally be completely different. If not, whats wrong with having another genius?

[quote]I don't see any problem if we clone Einstein, but I see a lot of problem if we clone Hitler.[/quote]

I dont. What if the young Hitler had been born in, say, the US. Do you think that he would have become the same evil person? I dont. It all boils down to a nature vs. nurture debate, but what you need to note is that it would take 20 years before the clone Hitler reached maturity. Worst case scenario, he is the same evil person as the real Hitler, he could be prevented from doing anything due to the scientific attention on him. From a purely scientific point of view, I think cloning Hitler would be quite interesting.

Same here, but a lot of people would protest if you clone Hitler.

I don't think its so much as who you clone, but what the clone can be used for. I doubt cloning Einstein would be much different than cloning Hitler; Einstein was a scientific genius, and Hitler was a military genius. The only true difference was that Hitler was extremely addicted to many narcotics and extremely racist, but if we cloned him, I doubt he would be as bad as we think him to be :p.

I think the biggest problem with cloning is the whole imposter idea. You can easily create another Koffi or GWB and have both people be genetically identical. Now, in the hands of these crazy terrorists, that would be a bad thing, yes? Cloning is more of a weapon than anything else.

I'm sure there are other things wrong with cloning too...perhaps clones, although genetically human, would not be treated as humans. And then this whole civil rights issue with clones would be a pain to deal with as well. There's also this whole "playing God" issue, but I stand neutral on that.

yes, you're right.
 i dont think cloning humans should be allowed unless it's absolutely essential as a test or something; even then it should be very heavily regulated

[quote="Ragingg"]The entire judicial system could be tricked.... Say I created 13 identical tetrahedr0ns all at once.  Then one of them commits a serious crime, how do the police know which tetrahedr0n to get?  Everyone of them has the same DNA....

Am I right?[/quote]

By the time we are able to clone people, we would also have some really really really advanced science that's totally beyond DNA.

While Euler may not have been doing olympiad problems, I am sure he could have the potential to easily do them.

Genes have to have something to do with smartness.  Everyone is probably born with some degree of math ptential.  Whether or not they reach that potential is decided by the environment.
Cloning would be very beneficial to scientific experiments.  For example,  if scientists wanted to test a new drug theyve created,  they could test it on clones, which allows the scientists to reduce the amount of variables they have to be concerned with.

The genes definitely help, but they aren't the only thing that is important. My father has a degree in mathematics, and I don't remember him ever forcing me to do math or even encouraging it very much (or discouraging it either). It may have helped save me from the fate of many of my friends, who are forced to study engineering or something boring like that rather than math.

On the other hand, I don't think Gauss's family contained any mathematicians at all, and still he was undoubtably one of the greatest mathematicians ever. Apparently his father couldn't even add properly (he had Gauss do his arithmetic for him when Gauss was maybe 4 years old, or so the story goes). His mother apparently didn't believe that he was that exceptional at math until one of his friends or teachers (I forget which) told her that he would be the greatest mathematician in Europe when he was older.

I should add that the Bernoulli family wasn't exactly known for being an especially close-knit one. Jakob and Johann (who were brothers) fought bitterly and eventually were not on speaking terms. Johann kicked his son out of the house when Daniel was co-awarded a prize with Johann. So I don't know how much the members of that family wanted to imitate each other.

Unless Jesus was thrown into a peat bog or glacier, they're not going to get any DNA from something two thousand years old.

[quote=Robomania_534]I personally believe that a clone of one's self is inferior until its value surpasses the original donor.[/quote]

well how exactly do you define value?

[quote=Clemmie]Are we bumping an 18 year old thread?![/quote]

sigma penguin

clonign is scary in the case of consciousness

[quote=pezzeepenguinstalker][quote=SirErnest]What if some scientists found DNA of some religious person in the Bible, like Moses or Jesus?  What would people think about cloning them?  I dont know if this would offend anyone, i do on go to church regularly.[/quote]

If we could clone Jesus back from the dead, that would be amazing. However, there are ethical issues with this.[/quote]

Not really, because jesus would just be a regular guy (probably) seeing as they are technically different people with the same dna.

[quote=callie313][quote=Ragingg]What if I grew all the tetrahedr0ns to be exactly alike, and besides that, even if they were different what if everyone denied it?

Secondly, I am not quite sure, but I don't think identical twins have the exact same genetical code.... They don't come out exactly alike all the time (not sure).

If I grew 1,000 Eulers all at once, would that be fair to other people?  How are the normal people supposed to get into the AIME or MOSP?  What would happen if other countries raised 1,000 Eulers and we competed at the IMO?[/quote]

Again, these "tetrahedr0ns" would not all be born at once.  The original is guaranteed to be older, and that difference would likely be noticeable.  Furthermore, maybe one of the clones got a scar on its knee when it was ten from a bad fall.  Maybe one of them was raised differently and became very distinguishable mentally.  The list goes on and on.  These clones, even if born the same, will
1) not be the same age
2) will develop differently[/quote]

valid points!

btw long time no see!

Cloning would be amazing!!!

However, it should only be done with close security, so that people don't abuse their power.

eg. clone a bunch of children to work on a farm.

@2above somehow this is strangely addicting?!

@above, well, if somebody had enough money to buy clones I doubt they would use it for menial labor or work.... but I do think abuse of power would be a definite outcome of cloning


Harvard
college
search
geometry
geometric transformation
reflection
AMC

[quote="mulan2890"]I'm just wondering what kind of person gets accepted to Harvard. Like what kind of extra curricular activities do you do, how high are your ACT and SAT scores, how high is your GPA, what are your advance classes, and that type of stuff. Just wondering...
 :)[/quote]

Hello mulan2890,

First than all, someone just discussed this a few days ago, you can look for the conversation with the search tools.

On the other hand, you don't have to post three times the exact same post to get someones attention, probably with just one is enough.

Finally, Harvard in general is (as far as I have seen) is pretty worried about extracurricular activities (I am not there), and most of the people that I know who got in, were first perfect or almost perfect scores in SAT, IB, high GPA (the four people that I know got GPA of around 3.9 or so) and were part of the school goverment or some other clubs.

[b]Moderators:[/b] Could you please delete the other two posts!

Best,

Since I got my letter today, I guess I can go ahead and tell you.

SAT I- 750 - Verbal, 800 - Math
SAT II -
800 - Chem
720 - Physics
800 - Math II
770 - Math I
760 - Writing
780 - Literature
(As far as Harvard Applicants go, I think that scores above 700 and even 750 are pretty standard for everything though).
US Hist, Chem, and Calc AB exams - 5s
and currently taking US Gov, Phys B, Eng Lit, Euro Hist, and Stat AP.
GPA of 4.0 unweighted, 2nd in a class of 700.

As you said though, relatively high numbers are standard at places like Harvard.
My real hook was in the extracurricular stuff, like

Music -
Percussion - 4 years All-state band, 2 years National Honor Band, World Percussion Championship and youngest member of the Cadets of Bergen County Drum and Bugle Corps + 11 Years of Piano (Guild Member)
Eagle Scout
Membership in Model UN - (Which is a blast by the way, and not too taxing)
NHS VP, Math Team VP, 7th in National French Contest, State Winner of Reflections Music Composition/photography Award
and Various Academic teams/awards
USNCO Finalist, USABO semifinalist
AIME Qualifier, placed at various math comps.

As you probably can tell, my record in pure [b]mathematics[/b] does not compare with many of the others here (got a 4 on the AIME this year :? ).  However, I think I was able to get because of my jack-of-all-trades persona.

So I would highly recommend beefing up your extracurricular resume as much as possible (try to find things that you enjoy), and listing [i][b]everything[/b][/i] that could possibly set you apart from the average applicant (which for Harvard University, is probably hardly average).  In addition, get an interview and all that jazz.  Third, write a crazy essay that is real far out, to catch their attention (I wrote mine on the importance of sanitation, making it sound like war propaganda).

I would also recommend that you start applying for scholarships as early as possible so that you can accumulate enough money to attend, as I might not be able to this year for financial reasons (though I believe that they offer a good deal of financial aid).

miraculouspostmaster: did you have any sports?  i've heard that you really have to have a sport to get into harvard unless youre extremely lopsided on something like westinghouse 1st place or IMO.  people are telling me to just join a team, even if i won't be any good at it.  is this at all true?

The general word on the street is that athletics is enormously helpful for Ivy League admissions--after all, the Ivy League is a sports conference, and Harvard has more Division I NCAA sports than any other college. Most of the local Harvard alumni who attended the regional information session in my town a year and a half ago were college athletes--the only one who was not has a physical disability, and he was a member of the marching band. One book about college admission advice seriously suggests taking up a sport to strengthen the application for a bright, well-rounded kid without a superimpressive extracurricular activity.

For all that, there are counterexamples of people with no sports background who get into Harvard. Those who do are frequently "lopsided" up to the international level in some kind of nonsport extracurricular activity, for example the International Mathematical Olympiad.

There are lots and lots of different types of people who get into Harvard.  Sure, some atheletes get into Harvard, and there are also plenty of people here who are unathletic.  I certainly didn't participate in organized sports in high school.  There isn't any reason to think that being atheletic will help you get into college (any college) more than anything else you do, if you aren't being recruited.  Certainly, don't join a sports team that you won't enjoy being a member of --  spend your time doing something that interests you.

The fundamental truths of college admissions: there is no formula for getting in, and there are more qualified individuals than there are spots at any one college.  Their corollaries: do the things that you will enjoy, not just the things that will pad your resume, and don't base your life around getting in to Harvard -- if you're good, you'll get in somewhere good.

Someone from Collegeconfidential got into Harvard with a 1300(He was an Asian male too!!). Thus high SAT isn't necessary; Harvard is rather arbitrary looking for people who would fit in best with the campus.

Plus Nathan Pflueger got a Harvard reject and he was pretty damn qualified.

I think it's much easier to find a formula for Caltech admission rather than Harvard admission because Caltech looks purely at merit rather than arbitrary factors.

I think Harvard would characterize that as that it looks for a broader variety of forms of "merit" than does Caltech. I agree with the basic point that it is probably easier to figure out a plan for a high probability of getting into Caltech than a plan for a high probability of getting into Harvard--although it might be HARDER to find a plan for getting out of Caltech with an undergraduate degree (in view of its rather high drop-out rate) than for completing a degree program at Harvard.

No school organized sports, but anyone who participates in drum corps knows that the rigors of a summer on tour are just as physically demanding as any high school sport (I lost about 60 pounds over a combined six months).  The man who interviewed me seemed to be interested in physical activity though.

I've never heard that you need a sport to get into Harvard...I think what the person who told you this means is that unless you're well-lopsided, you need to be very well-rounded and thus participate in a variety of activities (but not necessarily sports).

I know a few people who got in as an undergrad, and none of them was particularly good on any sport (actually, at least one of them was REALLY BAD in any sport).


let $f$ be a maximal solution of the (pendulum) equation:
$y''+ay'+bsin(y)=0$ where $a>0$. Prove that $f$ is defined over $R$.

What is a maximal solution in your opinion?

a maximal solution $f$ is a solution defined on an intervalle $I$ such that there does not exist a greater intervalle $J$ and a solution $g$ defined on $J$ with $f=g$ on $I$.

It is an immediate consequence of the following general theorem: if $X'=F(X)$ is a system of differential equations ($X(t)\in \mathbb R^n$, $F:\mathbb R^n\to\mathbb R^n$ and $F$ is Lipschitz in the entire space, the solution of any initial value problem exists and is unique on the entire real line.

why do you mean by Lipschitz ? locally Lipschitz or globally ?

[quote="Myth"]He wrote "$F$ is Lipschitz in the entire space". I think it means globally, i.e. with the same constant.[/quote]
Yeah, that's what I meant. Sorry for not making it clear the first time...


I'm sorry if this isn't the right forum, for this, but I need an answer for my homework.

What eating habits do people from Venezuela have?

--Eating Times
--Breakfast foods
--Lunch foods
--Dinner foods

Thanks in advance!

Um... use the internet.... search engines are your friend.

I don't think this there is an appropriate forum on AoPS for this, AoPS is not a place to seek answers to homework...

I assumed this was a discussion Venezuela's political status and Hugo Chavez, but, I found a request for eating habits of Venezuelans.    :o


calculus
integration
trigonometry
function
limit
complex analysis

Evaluate ${ \int_{0}^{\infty}\frac{1}{x^{n}+1}dx}$ if $ n$ is an integer with $ n \ge 2$.

(In particular, try to find a nice form)

If anyone can find a proof that does not use complex analysis, I would be interested to hear.

Since you would actually prefer to see a proof that does not use complex analysis, I can give you a one-liner:

Let $ u=x^{n}\rightarrow dx=\frac{du}{n u^{\frac{n-1}{n}}}$.  So we have:

$ \int_{0}^{\infty}\frac{1}{x^{n}+1}\hspace{1 mm}dx = \frac{1}{n}\int_{0}^{\infty}\frac{u^{\frac{1}{n}-1}}{u+1}du=\frac{1}{n}B(\frac{1}{n},\frac{n-1}{n})=\frac{1}{n}\Gamma(\frac{1}{n})\Gamma(\frac{n-1}{n})=\frac{\pi}{n \sin(\frac{\pi}{n})}$

Okay, it would be a one-liner if AoPS did not wrap text so early in its buffer  :lol:   But in the above I just used an alternative definition of the Beta function, the relation between Beta and Gamma, and one of the Gamma identities.

Using complex analysis, one would consider a contour in the shape of a wedge containing the positive real axis from $ 0$ to $ R$, the arc of the circle of angle $ \frac{2 \pi}{n}$, and the radius of the circle from the end of the arc back to $ 0$ oriented counterclockwise when $ n$ is odd, and change the angle to $ \frac{\pi}{n}$ when $ n$ is even.

You only have one residue here at $ z=e^{\frac{\pi i}{n}}$.

$ \text{Res}(\frac{1}{1+z^{n}},e^{\frac{\pi i}{n}}) = \lim_{z\rightarrow e^{\frac{\pi i}{n}}}\frac{z-e^{{\pi i}{n}}}{1+z^{n}}=\frac{1}{n e^{\frac{(n-1) \pi i}{n}}}$

Manipulation of the integrals and limiting estimates will give:

$ \int_{0}^{\infty}\frac{1}{x^{n}+1}\hspace{1 mm}dx = 2 \pi i \frac{1}{n e^{\frac{(n-1) \pi i}{n}}}\frac{1}{1-e^{\frac{2 \pi i}{n}}}= \frac{\pi}{n}\frac{2 i}{e^{\frac{\pi i}{n}}-e^{-\frac{\pi i}{n}}}= \frac{\pi}{n \sin(\frac{\pi}{n})}$

I don't consider that a proof without complex analysis- unless you have a proof of those identities that also uses complex analysis.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=21316]this thread[/url], which contains a full version of the wedge argument.


find x,y,z in this equation that x,y,z are positive integers.
(x+y)(1+xy)=2^z

Because $ x\plus{}y$ is power $ 2^t$ we can write $ x\equal{}2^{t\minus{}1}\plus{}a,y\equal{}2^{t\minus{}1}\minus{}a$, then $ xy\equal{}2^{2t\minus{}2}\minus{}a^2$.
Therefore $ 2^{2t\minus{}2}\plus{}1\minus{}2^{z\minus{}t}\equal{}a^2$. It give $ z\minus{}t\equal{}t,a\equal{}\pm (2^{t\minus{}1}\minus{}1)$ or $ (x,y,z)\equal{}(2^t\minus{}1,1,2t),(1,2^t\minus{}1,2t).$

Posted before, take searches next time. But especially do never use such useless titles again! (see http://www.mathlinks.ro/viewtopic.php?t=135914 )


Let $ ABCD$ be a rhombus with the length of each side is $ 1$.  Denote $ M\in BC$ and $ N\in CD$ such that $ MC\plus{}CN\plus{}NM\equal{}2$ and $ 2\angle {MAN}\equal{}\angle {BAD}$. Find the the values of $ \angle ABC$, $ \angle BAD$.

Construct ray $ Ax$ such that $ AM$ is the bisector of $ \angle BAx$. Let $ G$ be a point on $ Ax$ such that $ AG\equal{}AB\equal{}AD$.
Because $ \angle BAD\equal{}2\angle MAN$ then $ \Delta ABM\equal{}\Delta AGM, \Delta ADN\equal{}\Delta AGN$
$ \Rightarrow MG\equal{}MB, NG\equal{}ND$.
Since $ MC\plus{}NC\plus{}MN\equal{}2\equal{}BC\plus{}BD$ we get $ MN\equal{}BM\plus{}DN\equal{}MG\plus{}NG \Rightarrow M,G,N$ are collinear.
$ 2\angle ABC\equal{}\angle ABM\plus{}\angle ADN\equal{}\angle AGM\plus{}\angle AGN\equal{}180^o$ therefore $ \angle ABC\equal{}90^o\equal{}\angle BAD$.


Δίνεται η συνάρτηση $ f(x)\equal{}x^2\minus{}1$ στο διάστημα [-3,3]

Να κατασκευαστεί μια ακολουθία κλιμακωτών συναρτήσεων που συγκλίνει ομοιόμορφα στην f

(Μια συνάρτηση s ορισμένη στο [α,β] λέγεται κλιμακωτή αν υπάρχει μια διαμέριση {lo, l1, l2,...ln} του [α,β]

τέτοια ώστε η s να είναι σταθερη σε κάθε διάστημα (li-1, li) για κάθε i ανήκει στο {1,2,3,...n}.)

Ummm... υπάρχουν πολλοί τρόποι να το κάνει κανείς. Αν σε δυσκολεύει και βαριέσαι να γράψεις, κοίτα την απόδειξη του πυκνού των κλιμακωτών συναρτήσεων :)

Cheerio,

Durandal 1707

Που βρισκεται αυτή η απόδειξη φίλε Durandal?
Δεν την ξέρω μπορείς να μου πεις λίγα πράγματα γι΄αυτό γιατί έψαξα σε βιβλία και δεν το βρήκα
πως την βρίσκουμε αυτήν την ακολουθία?

[quote="spinos"]Δίνεται η συνάρτηση $ f(x) \equal{} x^2 \minus{} 1$ στο διάστημα [-3,3]

Να κατασκευαστεί μια ακολουθία κλιμακωτών συναρτήσεων που συγκλίνει ομοιόμορφα στην f.[/quote]

Να βάλω μία εύκολη λύση μιας και κανείς δεν έχει γράψει τίποτα και είναι κρίμα να μένουν εύκολες ασκήσεις αναπάντητες...

Επειδή βαριέμαι να γράφω πολλά, θα δουλέψουμε με την $ f(x) \equal{} x^2$ στο $ [0,1]$.

Θεώρησε μία διαμέριση $ a_0,a_1\ldots a_n$ του διαστήματος $ [0,1]$ όπου το $ a_k$ είναι λύση της εξίσωσης $ f(x) \equal{} \frac{k}{n}$ (δηλ. $ a_k \equal{} \sqrt{\frac{k}{n}}$).

Tότε, αν $ f_n(x) \equal{} \frac{k}{n}$ για $ x\in[a_k, a_{k\plus{}1})$, $ k\equal{}0,1\ldots n\minus{}1$, θα έχεις $ \|f_n\minus{}f\|\leq \frac{1}{n}$ και, συνεπώς $ f_n\to f$ ομοιόμορφα.

Cheerio,

Durandal 1707


I have a TI-84 Silver...and when I push the y= button on the top left of my calculator, I get the y1, y2, y3... and when i put, for example, one linear function in y1, and one linear function in y2, only one of the graph's actually graph when I push graph!!  I tried pressing the = button to highlight it but only one "=" will be highlighted at a time...I also tried pushing enter on the little figure on the left of the y1, y2, y3's....but they won't return to the figure of " \ "....it's an arrow with a vertical line behind it right now...I don't know how to change it back to be able to graph two graphs at a time!!

Please help!
Thanks!

nvmm...i just resetted all the RAM...and it fixed it at the cost of my programs...but oh well!

You probably had an application running that changed some of the settings.

I think the symbol of an arrow with a vertical line belongs to the Transformation Graphing application.


A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of
the vertices of the square lie on the hypotenuse and the other two vertices of the
square lie on the legs of the triangle, what is the length of a side of the square?
Express your answer as a common fraction.

Please tell me how to draw that so that I can answer another person's question. Thanks!

Once you know the answer, you can just "compute" the coordinates of all vertices:
[asy]size(150);
draw((0,0)--(15,0)--(0,8)--cycle);
real a=960/409, b=15*a/8;
draw((0,a)--(b,0)--(b+a,b)--(a,a+b)--cycle);[/asy]
Note that I didn't try to divide 960 by 409 on a calculator and write in the resulting long decimal but just asked Asymptote to do every relevant numerical computation.

Just for your fun, here is a function that, given three pairs $ A,B,C$ inscribes a square into the triangle with 2 vertices lying on $ BC$ and draws the result (I used the foot command from the Olympiad package in the code; it is automatically imported on the Forum anyway but you'll need to add [i]import olympiad;[/i] to your code if you want it to work on your home computer).
[code]
void f(pair A,pair B, pair C)
{
pair P=foot(A,B,C);
real a=abs(B-C), h=abs(P-A), t=h/(a+h);
draw(A--B--C--cycle);
pair P=A+t*(B-A), Q=A+t*(C-A), R=foot(Q,B,C), S=foot(P,B,C);
draw(P--Q--R--S--cycle);
}
[/code]
Here is a picture (for some other triangle):
[asy]size(150);
void f(pair A,pair B, pair C)
{
pair P=foot(A,B,C);
real a=abs(B-C), h=abs(P-A), t=h/(a+h);
draw(A--B--C--cycle);
pair P=A+t*(B-A), Q=A+t*(C-A), R=foot(Q,B,C), S=foot(P,B,C);
draw(P--Q--R--S--cycle);
}
f((0,0),(3,-1),(1,5));[/asy]


For how many digits C is the positive three-digit number 1C3 a
multiple of 3?

Because 1+c+3 must be divisible by 3, c=2,5,8. Thus there are 3 digits C can be.


linear algebra
matrix
induction
algebra
polynomial
linear algebra unsolved

How can i prove that \[ A^{n}\equal{}O_{n}
\] if \[ A\in M_{n}(R)
\] is a nilpotent matrix ?

Cayley-Hamilton offers some possibilities for the proof, but let's try this in a more elementary style.

Theorem: a nilpotent matrix (over any field) is similar to an upper triangular matrix with zeros on the main diagonal.

Since $ A$ is nilpotent, there must be some $ v\ne 0$ such that $ Av \equal{} 0.$ (Just take any nonzero $ x$ and look at the sequence $ x,Ax,A^2x,\dots.$ The last nonzero term in that sequence can be taken as $ v.$) Extend $ v$ to a basis, meaning find an invertible $ P$ with $ v$ as the first column. Then we have the following block structured result:

$ B \equal{} P^{ \minus{} 1}AP \equal{} \begin{pmatrix}0 & * \\
0 & A'\end{pmatrix}.$

By matrix multiplication, we get that $ B^k \equal{} \begin{pmatrix}0 & * \\
0 & (A')^k\end{pmatrix}.$

Hence, $ (A')^k \equal{} 0$ for large enough $ k$ and $ A'$ is itself nilpotent. Continue the process. Eventually (essentially this is an induction on the size of the matrix) we get that $ A$ is similar to an upper triangular matrix with zeros on the main diagonal.

The rest is just matrix multiplication.

If $ A \equal{} \begin{bmatrix}0 & * & * & & \\
0 & 0 & * & * & \\
0 & 0 & 0 & * & * \\
& & & \ddots & \\
0 & & & & 0\end{bmatrix},$ then

$ A^2 \equal{} \begin{bmatrix}0 & 0 & * & & \\
0 & 0 & 0 & * \\
0 & 0 & 0 & 0 & * \\
& & & \ddots & \\
0 & & & & 0\end{bmatrix}.$

Continuing, we get a new diagonal of zeros with each power, until we run out of room, so $ A^n \equal{} 0.$

This is one of the possibilities:
$ A^k\equal{}O_{n}\Rightarrow\lambda_{i}^k\equal{}0$, where $ \lambda_{i}$ is an eigenvalue of A;$ i\equal{}\overline{1,n}$. So all the eigenvalues are equal to 0. Then the characteristic polynomial $ p\equal{}(X\minus{}\lambda_{1})\cdot(X\minus{}\lambda_{2})\cdot...\cdot(X\minus{}\lambda_{n})\equal{}X^{n}$.
It is known that $ p(A)\equal{}O_{n}$(Cayley-Hamilton Thorem), so you get $ A^n\equal{}O_{n}$.

thanks for the solution...i didn't tried to solve it with eignvalues, it's a good idea  :D


search logarithms trigonometry geometry calculus AMC AIME

I stumbled upon this website while doing a google search for some mathy stuff... I'm a math major at a pretty big state school here in the....  I just have to say, I cannot believe that those high school guys are solving such complex stuff...  I read up on the students' solutions to:

How many solutions are there to
x + 2y + 3z = 100
edit:  x,y,z in N

And was shocked at the complexity of their answers.

So do you guys study algebra on your own?  I certainly did not learn any abstract algebra in high school - I went to one of the best high schools in my state and the highest level math we offered was multivariable calc for college credit...

Anyways, it's awesome to see high school students solving problems that I am just now encountering in college!

edit for clairty on that example pblm above..

In response to a post below, I'm pretty competent in quantum mechanics, so I guess that does make me feel a little better :)

I doubt many, if any, of us learn much math as part of the high school curriculum. However, it is true that most of us study further topics on our own by getting more advanced books or using this forum as a resource.

For the question, x can be anything, y can be anything, and z is just the result when you plug in an x and a y.

Do you mean natural numbers?

Then that's much harder.

Try going to the basics forum. If you can solve everything there, you are stuck; like me! :(

Welcome to the world of mathlinks, jgb!

I am also a college freshman, but many fourteen year olds on this site know more of number theory and graph theory (let alone geometry) than me. Now I've got used to it!

Something similar happened in college some days ago. I read up the official statement of the Yang Mills theory problem, and next day in college, told my friends that the mass and velocity of a particle are noncommuting operators operating on a Hilbert space! :rotfl: That day, during lunch, when two seniors heard two of my friends discussing this, their mouths dropped open! :D

Wanna stay ahead of these kids? Read up stuff like quaternions and lambda calculus and complex analysis and quantum mechanics and game theory, as I am now trying to do. :(

[quote="bubka"]Wanna stay ahead of these kids? Read up stuff like quaternions and lambda calculus and complex analysis and quantum mechanics and game theory, as I am now trying to do. :([/quote]

Sorry. I'm studying game theory this summer.

I'm curious: What's the solution to the posted problem? Could someone post a link?

Currently I'm trying to study quantum mechanics as much as I can :harhar:

[quote="bubka"]Why? I thought it was for statisticians alone. :maybe:
Anyway, either you are way above fourteen, or you have missed at least one of those I mentioned. :)[/quote]

Neither. :D

[hide="partial solution"]For a natural number x, we need to find the number of solutions to the diophantine equation 2y+3z=100-x.

I could find a pattern, but I'm too tired right now.[/hide]

[quote="7h3.D3m0n.117"]Many people in advanced mathematics in high school like to take classes outside of their own curriculm (not classes at school, but maybe looking at stuff they are not learning yet). For example, I'm only in 8th grade but I am reading up on calculus. My 10th grade friend left high school and started doing linear algebra.

For your question, many of us do study math on our own. For some reason, many people at my school would rather go see movies instead of doing logarithms or trig  :roll: . Those silly people.

Actually, school math curriculms are notoriously slow and wasteful. As proof, chess64 and I skipped about 2-3 years worth of math in less than one school year because we were bored of the way we were taught. Our teachers would focus more on the people who needed help or wouldn't pay attention rather than the people who are looking for a challenge (in my opinion they should treat both groups of people equally).

Because of the sluggish math at school, many of us are obligated to feel that we need to take our own education in our own hands.[/quote]

Yeah, that's pretty much what I did.

Most people would rather see a movie than do math?! :o :D

LOL, what about doing math WHILE watching a movie? (Numb3rs, etc.)

[quote="tjhance"][quote="mathnerd314"]Most people would rather see a movie than do math?! :o :D[/quote]

It's hard to believe buts its true. I don't understand why either. :huh:[/quote]

Yeah. My friend asked me if I could go with him with some girls to the movies to watch something and I said no cuz I had to read my calculus book  :P .

...btw guys, not everybody is interested in doing math for fun - do you guys go to the weight room for fun?

[quote="7h3.D3m0n.117"]Yeah. My friend asked me if I could go with him with some girls to the movies to watch something and I said no cuz I had to read my calculus book  :P .[/quote]

man you fail at life

you shoulda told them to go to http://www.senhorcarteiro.blogspot.com/

unless it was 300, then you should just plain go

[quote="Ignite168"][quote="7h3.D3m0n.117"]Actually, school math curriculms are notoriously slow and wasteful. As proof, chess64 and I skipped about 2-3 years worth of math in less than one school year because we were bored of the way we were taught. Our teachers would focus more on the people who needed help or wouldn't pay attention rather than the people who are looking for a challenge (in my opinion they should treat both groups of people equally).[/quote]

Obviously you didn't skip English and seeing as neither you nor chess64 received anything higher than a 4 on the AIME, I hardly consider that proof.[/quote]
i thought chess64 was a genius? he has 5 stars.  :lol:

[quote="Ignite168"][quote="7h3.D3m0n.117"]Actually, school math curriculms are notoriously slow and wasteful. As proof, chess64 and I skipped about 2-3 years worth of math in less than one school year because we were bored of the way we were taught. Our teachers would focus more on the people who needed help or wouldn't pay attention rather than the people who are looking for a challenge (in my opinion they should treat both groups of people equally).[/quote]

Obviously you didn't skip English and seeing as neither you nor chess64 received anything higher than a 4 on the AIME, I hardly consider that proof.[/quote]

Competition math is different from book math, and I thought that you, of all people would know.

1) My teacher won't let me take AMC, so I wonder why I didn't take the AIME hmm?
2) English is different. The curriculm is VERY flexible. You could be prepping for what you think you will be learning in 12th grade English but be doing something else. Math, however, is very strict.

Yeah competition maths totally sucks (partly because they involve problem solving, which is no maths at all, but mostly because i am no good at competition maths).

But as I now sadly realize, so does analysis. And algebra.


Princeton
college
HMMT
Princeton Math Comp.

Welcome to the Princeton Math Competition forum. Princeton University will be hosting a high school math competition for the first time in the 2006-2007 school year. More information will be posted soon!

You're amazing, Nathan. Congratulations on getting the competition set up and best of luck in the future.

Looking forward to hearing more about it!  I wonder if it's going to run concurrently with the HMMT...being able to go to both would be nice.

We haven't fixed the date yet, but we're going to make sure it doesn't conflict with HMMT or any other big competitions in this region. We're also trying to pick a good date that avoids SAT weekends and other such tomfoolery. We should have the date, schedule, and test format available for your viewing pleasure by early next week.


I thought that it would be a good idea to start a topic like this one, to ask questions about this book.
I will start.

On page 14, Engel says on the last line that $ 2^k(mod)n \equal{} 0(mod)n$, schouldn't that be $ [2^k\cdot a](mod)n \equal{} 0(mod)n$ ?

I couldn't edit, but PSS stands for Problem solving strategies.

Can someone please take a look in the book, I will be very thankful.


abstract algebra
group theory
linear algebra
linear algebra unsolved

Prove that a finite abelian subgroup in $GL_n(Z)$ has at most $ 6^n$ elements.

harazi,

The biggest abelian subgroup of $\mbox{GL}_{n}(\mathbb{Z})$ I found has $2^{n}3^{n/3}$ elements.

Do you have an explicit abelian subgroup whose order is $6^{n}$?

No, actually one can prove that the biggest has $2^{[(n+1)/2]}\cdot 3^{[n/2]}$ elements.


inequalities
real analysis
advanced fields
advanced fields unsolved

Suppose $X$ is an $\mathbb{R}^n$-valued random variable with finite second moments. (In real analysis terms, $X\in L^2.)$ Notationally, treat $X$ as a (random) column vector.

Define $C=\mathrm{Cov}(X)=E(XX^T)-E(X)E(X^T).$

$C$ is obviously symmetric. It's a fairly easy standard argument to show that it is positive semidefinite, and positive definite if the components of $X$ are [i]linearly[/i] independent, which is a far, far, far weaker condition than probabilistic independence.

In what follows, assume that $C$ is invertible ([i]i.e.[/i] positive definite.)

Define $Q(X)=(X-E(X))^TC^{-1}(X-E(X)).$

Prove that for all $\alpha>0$, $P(Q(X)\ge \alpha^2)\le \frac{n}{\alpha^2}.$

(Obviously, this is of interest only if $\alpha>\sqrt{n}.$)


MATHCOUNTS AMC USA(J)MO USAMO AIME videos

You and Bob are at the same level in math competitions. During the summer, you study math competitions on your own and Bob goes to a math camp like Awesome Math... Is Bob going to beat you next year?

How many people would have the motivation and discipline to study on his own as hard as someone who is surrounded by others interested in math and gets lectures and problem sets planned? Is there a limit to what you can do on your own?[hide="Why I Ask"]

I'm wondering because in my middle school there was an enormous ability difference between the people who went to Awesome or other math camps and the next highest level students ...  I might be overgeneralizing but from what I remember last year in middle school was that there were several "tiers" of ability which became apparent during mathcounts practices.

-Highest level students (2 or 3 of them)
All went to math camps and all (I think) qualified for USAMO in middle school. Scores in mathcounts practices were roughly 100 points or more higher than everyone else's out of only 400 or 500 points.

-Next to highest level (around 5 of them)
All had extensive tutoring for several years. Some of them qualified for AIME, close to perfect scores on the SAT math section

-The rest (my group)
Became interested in math competitions in middle school, no significant training in elementary school.
[/hide]

I'm just going to do stuff myself this year, but I might go next year to a camp.
The reason I'm choosing to stay home is because I can do a lot of practice stuff, and have days off and breaks and stuff, while at camp, you have a schedule to follow.

I'd probably fit in the last group.

On one hand, teaching yourself allows you to analyze and examine your own weaknesses and strengths. On the other, being taught by others can allow you to see things in a different view and you can potentially learn a lot.

There are many advantages as well as disadvantages to both. I believe you need to analyze your own personality as to which you should chose. If you are lazy and procrastinating, then teaching yourself isn't a good idea. If you prefer open discussions about problems to benefit your learning, then attending a math camp or something along those lines might help.

When improving in mathematics, there are three things I work on: raising ceiling performance, raising floor performance, and raising dexterity/reliability. These are general ideas relating to each:

[color=darkred][b]Raising ceiling performance:[/b][/color] For this, going to a summer camp can be very effective. You will gain new perspectives, learn new methods/approaches, and perhaps new material altogether. However, there is plenty to be learned on the AoPS forum and such. Self study and playing around help plenty. Also, trying to make your own problems can allow you to feel the mindset of a problem writer, which is good to be familiar with. While trying to write problems, you yourself will gain a better understanding of the concepts as you try to make them fit together nicely.

[color=darkred][b]Raising floor performance:[/b][/color] For this, you need to improve clarity of thought and health. If you keep your body in good condition, you will be more alert and conscious, and less likely to make a stupid mistake or to blank out. You will also have more energy to last through long tests. State of mind comes into play here, too. Exercise, meditation, playing music, art, etc, or genuine relaxation (not video games) have helped me here.

[color=darkred][b]Raising reliability/dexterity:[/b][/color] While practice and problem writing can't be trumped, there are many other ways to improve your mental dexterity that have little to do with math competitions. These include playing chess, art, writing, debate/mock trial/etc, and more. To put things into perspective, this year, I barely learned any contest math material, and barely practiced for contests, but thanks to very involved mock trial work, have seen huge increases in almost all my contest scores.

So now, (if you believe my theory), it's up to you to decide where you want to put your attention  :)

Also influencing your decision should be opportunity cost -- summers are very valuable.


a,b,c are positive real numbers.prove that:
bc/a^2 +2bc +ca/b^2+2ac +ab/c^2+2ab<=1

from cauchy $ (\sum a^2\plus{}2bc) \sum \frac{a^2}{a^2\plus{}2bc}\geq (a\plus{}b\plus{}c)^2$ so  $ \sum \frac{a^2}{a^2\plus{}2bc}\geq 1$


calculus
integration
logarithms
search
real analysis
real analysis unsolved

show :D

$ \int_0^1\;\int_0^1\;\;\dfrac{\textbf dx\;\textbf dy}{\left(\minus{}\,\ln\,(x\,y)\right)^{\frac{3}{2}}}\;\;\equal{}\;\;\boxed{\sqrt\pi}$

use the substitution $ u\equal{}xy,v\equal{}{\frac {x}{y}}$we have that Jacobian $ J\equal{}\minus{}\frac{1}{2v}$:
$ I\equal{}\minus{}\int _{0}^{1}\! {du} \int _{u}^{\frac{1}{u}}\! {dv}\frac{1}{2v}\frac{1}{(\minus{}\ln{u})^\frac{3}{2}}\equal{}
\minus{}\int _{0}^{1}\! \frac{du}{(\minus{}\ln{u})^\frac{3}{2}}\int _{u}^{\frac{1}{u}}\! \frac{dv}{2v}\equal{}
\minus{}\int _{0}^{1}\! \frac{du}{(\minus{}\ln{u})^\frac{3}{2}}\ln{u}\equal{}\int _{0}^{1}\! \frac{du}{(\minus{}\ln{u})^\frac{1}{2}}$
Then we use substitution $ t\equal{}\minus{}\ln{u}$ and:
$ I\equal{}\int _{0}^{1}\! \frac{du}{(\minus{}\ln{u})^\frac{1}{2}}\equal{}\int _{0}^{\infty}\! \frac{\exp{t}}{t^\frac{1}{2}}dt\equal{}\Gamma \left( \frac{1}{2} \right)\equal{}\sqrt{\pi}$


Let $S = \{1, 2, \ldots , 10\}$. Find the number of unordered pairs A, B, where A and B are disjoint non-empty subsets of S. (http://www.kalva.demon.co.uk/aime/aime02b.html)

[hide]

For each of the $10$ members of S, there are $3$ categories it can fall under: Set A, Set B, or Neither. So there are $3^{10}$ ways to assign the distinctions. Order does not matter in this case, so every set except the one where every number is part of Neither is double counted.

$\frac{3^{10}-1}{2}=29524$. We threw out the empty set (meaning A and B are empty, Neither has all 10 values) which we didn't want, and accounted for the double counting. But this includes the cases where either Set A or Set B is empty, while the other letter Set is not empty. To find the number of such cases, think of each value of S being assigned one of $2$ distinctions: part of Set A or part of Neither (we do not have to do this again for B since we are not concerned with order).

So there are $2^{10}-1$ such cases (we subtracted the case where all $10$ values were part of Neither since that case was thrown out in the first step).

Final answer equals: $29524-1023=28501$

[/hide]


Let $ a,b,c$ be positive real numbers. Prove that

$ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}\geq\frac{2}{3}(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\minus{}\sqrt{(a^2\plus{}b^2\plus{}c^2)(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})}$

[quote="Ligouras"]Let $ a,b,c$ be positive real numbers. Prove that

$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq\frac {2}{3}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \minus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})}$[/quote]

easy  proven

$ \sqrt {(a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})}\geq3$

only  prove
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \plus{} 3\geq\frac {2}{3}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$

or
$ (3a^2c^2 \plus{} (bc^2 \minus{} 2b^2c \plus{} 3c^3)a \plus{} 2b^2c^2 \plus{} 3b^4 \minus{} 4b^3c \minus{} bc^3 \plus{} 3c^4)(a \minus{} b)(a \minus{} c) \plus{} ((b \plus{} c)(3b^2 \plus{} 2bc \plus{} 3c^2)a \minus{} bc(3b^2 \plus{} 2bc \plus{} 3c^2))(b \minus{} c)^2\geq 0$
from the inequality  assume $ a \equal{} \max{(a,b,c)}$

then
only prove  while  $ a \equal{} \max{(a,b,c)}$

$ 3a^2c^2 \plus{} (bc^2 \minus{} 2b^2c \plus{} 3c^3)a \plus{} 2b^2c^2 \plus{} 3b^4 \minus{} 4b^3c \minus{} bc^3 \plus{} 3c^4\geq 0$
and
$ (b \plus{} c)(3b^2 \plus{} 2bc \plus{} 3c^2)a \minus{} bc(3b^2 \plus{} 2bc \plus{} 3c^2)\geq 0$

[quote="Ligouras"]Let $ a,b,c$ be positive real numbers. Prove that

$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq\frac {2}{3}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \minus{} \sqrt {(a^2 \plus{} b^2 \plus{} c^2)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})}$[/quote]
need to prove
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \plus{} 3\geq\frac {2}{3}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$
$ \Longleftarrow 3\plus{}3(\frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}) \geq 2(\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{a\plus{}b}{c})$
$ \Longleftarrow 3\plus{}(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2 \geq 2(\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{a\plus{}b}{c})$
$ \Longleftrightarrow \sum(\frac{a}{b}\minus{}1)^2 \geq 0$


va mai propun o problema interesanta inainte sa plec :)
fie triunghiul $ABC$ si $M,N,P$ sunt centrele cercurilor exinscrise, opuse lui $A,B,C$ respectiv. perpendicularele din $N$ pe $AB$ si din $P$ pe $AC$ se intersecteaza in $A_{1}$.
aratati ca $AA_{1}\cap BB_{1}\cap CC_{1}\neq \emptyset$.

O să rezolv următoarea problemă:

[color=blue]Fie $ABC$ un triunghi, $O$ centrul cercului circumscris, $U,V,W$ simetricele lui $O$ faţă de $BC,CA,AB$ şi $D,E,F$ picioarele înălţimilor din $A,B,C$ pe $BC,CA,AB$.

Demonstraţi că $DU, EV, FW$ sunt concurente.
[/color]

[hide="Rezolvare cu numere complexe"]
Considerăm cercul circumscris $\triangle ABC$ drept cercul-unitate din planul complex. Vom nota cu litere mici afixele corespunzătoare punctelor.
Avem $|a|=|b|=|c|=1$, $d = \frac{a(a+b+c)-bc}{2a}$, $e = \frac{b(a+b+c)-ca}{2b}$, $f = \frac{c(a+b+c)-ab}{2c}$, $u = b+c$, $v = c+a$, $w = a+b$ ($d,e,f$ necesită destul de multe calcule). De asemenea, $\overline d = \frac{-a^{2}+\sum bc}{2abc}, \, \overline u = \frac{b+c}{bc}$ etc.

Ecuaţia dreptei $DU$ este \[| \begin{matrix}d & \overline d & 1 \\ u & \overline u & 1 \\ z & \overline z & 1 \end{matrix}| = 0\]  \[\, \Longleftrightarrow \, \, ( 2bc-a^{2}-\sum bc ) z+bc (-a^{2}+\sum bc ) \overline z = 2 ( bc-a^{2}) ( b+c ) .\] Celelalte două ecuaţii ies prin permutări ciclice.
Faptul că cele trei drepte sunt concurente e echivalent cu faptul că cele trei ecuaţii formează un sistem ce are soluţie nenulă.
Altfel zis, noi trebuie să demonstrăm că

$| \begin{matrix}2bc-a^{2}-( ab+bc+ca ) & bc (-a^{2}+ab+bc+ca ) & 2 ( b+c ) ( bc-a^{2}) \\ 2ca-b^{2}-( ab+bc+ca) & ca (-b^{2}+ab+bc+ca ) & 2 ( c+a ) ( ca-b^{2}) \\ 2ab-c^{2}-( ab+bc+ca) & ab (-c^{2}+ab+bc+ca ) & 2 ( a+b ) ( ab-c^{2}) \end{matrix}| = 0$

(scădem prima linie din celelalte două)

$\, \Longleftrightarrow \, \, 2 | \begin{matrix}2bc-a^{2}-( ab+bc+ca ) & bc (-a^{2}+ab+bc+ca ) & ( b+c ) ( bc-a^{2}) \\ 2c(a-b)+(a-b)(a+b) & abc(a-b)+c (ab+bc+ca)(a-b) & (a-b) ( c^{2}+2 \sum bc-ab ) \\ 2b(a-c)+(a-c)(a+c) & abc(a-c)+b (ab+bc+ca)(a-c) & (a-c) ( b^{2}+2 \sum bc-ac ) \end{matrix}| = 0 $

$\, \Longleftrightarrow \, \, 2(a-b)(a-c) | \begin{matrix}2bc-a^{2}-( ab+bc+ca ) & bc (-a^{2}+ab+bc+ca ) & ( b+c ) ( bc-a^{2}) \\ a+b+2c & abc+c (ab+bc+ca) & c^{2}+2 \sum bc-ab \\ a+2b+c & abc+b (ab+bc+ca) & b^{2}+2 \sum bc-ac \end{matrix}| = 0 $

(scădem a doua linie din a treia)

$\, \Longleftrightarrow \, \, 2(a-b)(a-c) | \begin{matrix}2bc-a^{2}-( ab+bc+ca ) & bc (-a^{2}+ab+bc+ca ) & ( b+c ) ( bc-a^{2}) \\ a+b+2c & abc+c (ab+bc+ca) & c^{2}+2 \sum bc-ab \\ b-c & (b-c) (ab+bc+ca) & (b-c)(b+c)+a(b-c) \end{matrix}| = 0 $

$\, \Longleftrightarrow \, \, 2(a-b)(a-c)(b-c) | \begin{matrix}2bc-a^{2}-( ab+bc+ca ) & bc (-a^{2}+ab+bc+ca ) & ( b+c ) ( bc-a^{2}) \\ a+b+2c & abc+c (ab+bc+ca) & c^{2}+2 \sum bc-ab \\ 1 & ab+bc+ca & a+b+c \end{matrix}| = 0 $

(înmulţim ultima linie cu $-bc$ şi o adunăm la prima, apoi o înmulţim cu $-c$ şi o adunăm la a doua)

$\, \Longleftrightarrow \, \, 2(a-b)(a-c)(b-c) | \begin{matrix}bc-a^{2}-( ab+bc+ca ) &-a^{2}bc &-abc-( b+c ) a^{2}) \\ a+b+c & abc & ab+bc+ca \\ 1 & ab+bc+ca & a+b+c \end{matrix}| = 0 $

(înmulţim a doua linie cu $a$ şi o adunăm la prima)

$\, \Longleftrightarrow \, \, 2(a-b)(a-c)(b-c) | \begin{matrix}0 & 0 & 0 \\ a+b+c & abc & ab+bc+ca \\ 1 & ab+bc+ca & a+b+c \end{matrix}| = 0, $

relaţie care este evident adevărată.

Acum mai rămâne de verificat faptul că punctul de intersecţie este finit. Minorul din stânga sus iese $-(a-b) (a+b)^{2}c (a+c)(b+c) \neq 0$ (pe ăsta nu l-am mai calculat de mână, dar am încredere în soft-ul ăsta).

$\blacksquare$
[/hide]

$ABC$ este triunghiul ortic al lui $MNP$, iar $A_{1},B_{1},C_{1}$ sunt simetricele centrului cercului circumscris triunghiului $MNP$ faţă de $NP,PM,MN$. Aceste afirmatii reies dintr-un [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=long&t=50503]exerciţiu pe care l-am postat mai demult[/url].

Acum trebuie doar să aplicăm ce-am demonstrat mai sus triunghiului $MNP$ şi iese problema.


Suppose $ p$ is a prime gretaer than 3. Find all pairs $ (a,b)$ of integers satisfying the equation $ a^2\plus{}3ab\plus{}2p(a\plus{}b)\plus{}p^2\equal{}0$

We have $ p \equal{} \minus{} (a \plus{} b)\pm\sqrt {b(b \minus{} a)}$. So $ b(b \minus{} a)$ must be a perfect square.

If $ b \equal{} 0,\ a \equal{} \minus{} p$.

Suppose $ b\neq0$. Then* any prime that divides $ b$ must divide $ a$; hence $ b\mid a$, say $ bc \equal{} a$. $ \therefore\ b(b \minus{} a) \equal{} b^2(1 \minus{} c)\ \Rightarrow\ 1 \minus{} c$ is a perfect square $ \Rightarrow\ 1 \minus{} c \equal{} d^2\ \Rightarrow\ a \equal{} b(1 \minus{} d^2)$.

So we have $ p \equal{} \minus{} [b(1 \minus{} d^2) \plus{} b]\pm bd \equal{} b(d^2\pm d \minus{} 2)\ \Rightarrow\ b\mid p\ \Rightarrow\ b \equal{} \pm1$ or $ b \equal{} \pm p$. In any case, $ d^2\pm d \minus{} 2 \equal{} d(d\pm1) \minus{} 2$ is an even number, contradicting the fact that $ p$ is an odd prime.

*See my next post below.

Why is $ b|a$ ? For instance, $ 4(4\minus{}3)$ is a perfect square, but $ 4\not|3$.

[quote="bertram"]Why is $ b|a$ ? For instance, $ 4(4 \minus{} 3)$ is a perfect square, but $ 4\not|3$.[/quote]
It seems I have carelessly missed some things there. :blush: What I said would be true if both $ b$ and $ b \minus{} a$ are greater than 1 (in absolute value). I’ve checked the cases $ b \equal{} \pm1$ and found that they are impossible. However, $ a \minus{} b \equal{} \pm1$ is possible: e.g. $ ( \minus{} 3, \minus{} 4)$ for $ p \equal{} 5$.

[quote="Jane_Bennet"]...So it appears that $ ( \minus{} p,0)$ is the only solution.[/quote]

$ p\equal{}7$, $ a\equal{}\minus{}5$ and $ b\equal{}4$ works. So you're wrong.

The equation may be written $ (a\plus{}b\plus{}p)^2\equal{}b(b\minus{}a)$

So $ b(b\minus{}a)$ is a perfect square and we have $ b\equal{}ku^2$ and $ a\equal{}ku^2\minus{}kv^2$ fore some $ k,u,v$.

Then $ a\plus{}b\plus{}p\equal{}\epsilon_0kuv$, which may be written $ 2ku^2\minus{}kv^2\plus{}p\equal{}\epsilon_0kuv$ (where $ \epsilon_0$ is $ \minus{}1$ or $ \plus{}1$)

So $ p\equal{}kv^2\minus{}2ku^2\plus{}\epsilon_0kuv$ $ \equal{}k(v\minus{}\epsilon_0u)(v\plus{}2\epsilon_0u)$

Since $ p$ is prime, we have three cases :
$ 1)$
$ k\equal{}\epsilon_1$
$ v\minus{}\epsilon_0u\equal{}\epsilon_2$
$ p\equal{}\epsilon_0\epsilon_1\epsilon_2(3u\plus{}\epsilon_0\epsilon_2)$
This is equivalent to :
$ k\equal{}\epsilon_1$
$ v\minus{}\epsilon_0u\equal{}\epsilon_0\epsilon_1$
$ p\equal{}3u\plus{}\epsilon_1$
And so :
$ p\equal{}3u\plus{}\epsilon_1$
$ a\equal{}\minus{}2u\minus{}\epsilon_1$
$ b\equal{}\epsilon_1u^2$

$ 2)$
$ k\equal{}\epsilon_1$
$ v\plus{}2\epsilon_0u\equal{}\epsilon_2$
$ p\equal{}\minus{}\epsilon_0\epsilon_1\epsilon_2(3u\minus{}\epsilon_0\epsilon_2)$
This is equivalent to :
$ k\equal{}\epsilon_1$
$ v\equal{}2\epsilon_1\epsilon_2u\plus{}\epsilon_2$
$ p\equal{}3u\plus{}\epsilon_1$

And so :
$ p\equal{}3u\plus{}\epsilon_1$
$ a\equal{}\minus{}\epsilon_1(3u^2\plus{}1)\minus{}4u$
$ b\equal{}\epsilon_1u^2$

$ 3)$
$ v\minus{}\epsilon_0u\equal{}\epsilon_1$
$ v\plus{}2\epsilon_0u\equal{}\epsilon_2$
$ p\equal{}\epsilon_1\epsilon_2k$

Which gives :
$ p\equal{}k$
$ a\equal{}\minus{}k$
$ b\equal{}0$

*******************************
As a conclusion

For any prime $ p$ we have the solution $ (\minus{}p,0)$

For any prime $ p\equal{}1\pmod{3}$, we also have the two solutions :

$ (\minus{}\frac{2p\plus{}1}{3},\frac{(p\minus{}1)^2}{9})$

$ (\minus{}\frac{p(p\plus{}2)}{3},\frac{(p\minus{}1)^2}{9})$

For any prime $ p\equal{}2\pmod{3}$, we also have the two solutions :

$ (\minus{}\frac{2p\minus{}1}{3},\minus{}\frac{(p\plus{}1)^2}{9})$

$ (\frac{p(p\minus{}2)}{3},\minus{}\frac{(p\plus{}1)^2}{9})$


[quote="Arvind_sn"]In countdown round (which i'm pretty sure that im going to make), what is the format? they choose top ten (right?) and then what? how many coundtowns do they do? is it elimination or is it something else? :huh:[/quote]
It depends on your state.
I know that in Georgia they have all of the top ten people come down, and the put the questions up. The whoever buzzes in first and gets the question right, gets a point. It's best out of 80.

In some states, like mine, it's official.

They choose the top 10, then they put #9 against #10, and the winner of that faces #8, and the winner of that faces #7, and so on.
They ask questions, and the first  person to get 2/3 right goes on.

whatever you have i have 1/10 of not good...
see you at state

EDIT: since it isn't official if someone "had a coughing contest" or started say umm to much they wouldn't be dqed or something would they?

If you get first on the written test, you are guarenteed nationals right? Because CountDown is really not my thing...  :(

If the countdown is official, it works exactly the way the [b]old[/b] countdown at nationals worked.  Top ten come up, pair 10 vs. 9, winner vs. 8, and so on.  It's best of 3 with a sudden-death tiebreaker until the last three matches, which are first to three points.  Official countdown rounds everywhere except nationals are required to follow this format.  Also, if you're top 3 on written, you're guaranteed a spot at nats.

If the countdown is unofficial, anything goes.  I think Florida, or at least some Florida chapters, have a system in which every student participates, and teams get a score.  California, or some chapters here, have a giant tournament for all the students.  Minnesota used to do a single-elim tournament for the top 16, but changed it a few years back to top 12--except the four students who get byes are picked randomly to avoid divulging the results before the awards ceremony later.  The format of an unofficial countdown is limited only by the availability of questions and the imagination of the organizers.

[quote="diophantient"][quote="Sly Si"]I think Florida, or at least some Florida chapters, have a system in which every student participates, and teams get a score.[/quote]
No, in all the chapters in Florida it's official. (unless I'm mistaken)

The round you are thinking of, is ciphering, which has questions similar to countdown. It's very very cool.[/quote]

It's official in everything, including state.  The problem is the questions are not that great, but oh well.  It does not affect team standings or anything of that nature.  It's just an extra, fun round where the winning team gets a trophy.

[quote="gtf"]In Georgia, the countdown round is only for a trophy. The state team is chosen based on the points from the Sprint and Target round combined.

See you tomorrow!
Fulton[/quote]
do you know if you can be dqed; and if so can that mean if you did make it to nats you can't go?

[quote="Arvind_sn"]ok, so now i have a 6/17 chance of making nationals...not good. :lol:[/quote]

Where did you get that number from?

at the actual nationals competition, the format is top twelve people.
top four get a bye, and the other 8 are paired such that the sum of their ranks is 17.  Next round is like an eight-person single elimination

I rock at the countdown round, but I never do good enough on the written round to qualify. :)

[quote="gtf"]I guess that it would depend on what you are DQed for. If it was for cheating, for instance, I would assume that you would be DQed from the whole competition and thus any nationals spot would be forfeited. If was for something stupid during the Countdown  round, I would hope that you would still have your spot.

What types of DQs are possible?

This is all my opinion only. Any official decision would need to come from the organizers.

Fulton[/quote]

I am not certain about this, but I remember hearing a while back that if a student has to take the state competition on an alternate day (usually for religious reasons) and can't attend on the day the countdown is held, they forfeit any countdown round matches that they would have participated in.  Thus if countdown is official, they drop one spot automatically, but no more.  This is why official countdowns are required to follow that format, so a student unable to attend could still make nationals by finishing in the top 3 in the written round.

at r state comp we have a tourney bracket, but its unofficial, so we really just dont care

Is the round official, for trophies, or just for fun?

[quote="ckck"][quote="magixter"][quote]The round you are thinking of, is ciphering, which has questions similar to countdown. It's very very cool.
[/quote]
How does the ciphering round work?[/quote]
One person from each school comes up at a time. A question comes up and everyone tries to answer it under a minute. You have this kind of machine that you can type in your answer. After a minute has passed, they see who got it in first. The first person that got it right gets 5 points. The next person that gets it right gets 4 points and so on. Theres four rounds for each of your team mates to do. And each round has 4 questions.[/quote]
When do they do the ciphering round?

In response to magxiter's sig: NEVER!!!!

[quote="erdogankerem123"][/quote][quote="ckck"][quote="magixter"][quote]The round you are thinking of, is ciphering, which has questions similar to countdown. It's very very cool.
[/quote]
How does the ciphering round work?[/quote]
One person from each school comes up at a time. A question comes up and everyone tries to answer it under a minute. You have this kind of machine that you can type in your answer. After a minute has passed, they see who got it in first. The first person that got it right gets 5 points. The next person that gets it right gets 4 points and so on. Theres four rounds for each of your team mates to do. And each round has 4 questions.[/quote][quote="erdogankerem123"]
When do they do the ciphering round?

In response to magxiter's sig: NEVER!!!![/quote]

The ciphering round is only done in some states, like Florida.

[quote="erdogankerem123"][/quote][quote="ckck"][quote="magixter"][quote]The round you are thinking of, is ciphering, which has questions similar to countdown. It's very very cool.
[/quote]
How does the ciphering round work?[/quote]
One person from each school comes up at a time. A question comes up and everyone tries to answer it under a minute. You have this kind of machine that you can type in your answer. After a minute has passed, they see who got it in first. The first person that got it right gets 5 points. The next person that gets it right gets 4 points and so on. Theres four rounds for each of your team mates to do. And each round has 4 questions.[/quote][quote="erdogankerem123"]
When do they do the ciphering round?

In response to magxiter's sig: NEVER!!!![/quote]
Yea, I think Florida is the only one that does it because they let parents video tape the whole thing. But, its right before countdown.

[quote="ckck"][/quote][quote="erdogankerem123"][/quote][quote="ckck"][quote="magixter"][quote]The round you are thinking of, is ciphering, which has questions similar to countdown. It's very very cool.
[/quote]
How does the ciphering round work?[/quote]
One person from each school comes up at a time. A question comes up and everyone tries to answer it under a minute. You have this kind of machine that you can type in your answer. After a minute has passed, they see who got it in first. The first person that got it right gets 5 points. The next person that gets it right gets 4 points and so on. Theres four rounds for each of your team mates to do. And each round has 4 questions.[/quote][quote="erdogankerem123"]
When do they do the ciphering round?

In response to magxiter's sig: NEVER!!!![/quote][quote="ckck"]
Yea, I think Florida is the only one that does it because they let parents video tape the whole thing. But, its right before countdown.[/quote]
Louisiana and this one other state have similar rounds, but the problems are different.


[size=150]The horizontal axis is marked in degrees. Find a+b+c where one period of the sinusoid
y=a+bsinc(x-45^0 )begins at (45^0 ,11) and ends at (135^0 ,11) with horizontal tangent lines at (67.5^0 ,16) and (112.5^0 , 6)

a) 21.5
b)21
c)20.5
D) 20
e)none of these [/size]

Thanks!


Euler
number theory
prime numbers
number theory unsolved

We define $\delta(n)$ is the sum of divisors of $n\in N$.Prove that $n$ is producting of twin prime numbers if and only if $\varphi(n) \delta(n)=(n-3)(n+1)$.($\varphi$ is the Euler function).
  [hide]when $p,q$ are prime and $|p-q|<2$ then we say $p,q$ are twin prime numbers.[/hide]

I think you meant p,q are twin primes iff |p-q|=2, right? :)

This seems hard at first, but when you really tackle it, it's not that bad:

Let P(n) denote the number of distinct primes that divide n. We try to prove that that equation is true iff P(n)=2 and that the two primes which divide n are twin primes.

The "if" part is trivial (just write n=pq, and calculate). So I'll only prove the "only if" part:

First when P(n)=1, it's easy to see that (n-3)(n-1) > Phi(n)*Epsilon?(n).
Now if P(n)>=3, we write n=abk, where a,b are primes and k a positive integer. Then write k=Product (i=1 to s) p_i^q_i. Then let A = Epsilon(n)*Phi(n)= Epsilon(ab)*Epsilon(k) * Phi(ab)*Phi(k)= (a^2-1)(b^2-1)*Product (i=1 to s) (p_i^q_i+1 - 1) < (a^2-1)(b^2-1)*k*Product (i=1 to s). Now you can show that A < n^2 - 2n - 3 for k > 1. And we are done.  :D


Can anyone help me with these two problems:

Consider the function defined on 1..10 by the following: S(10) = 1; S(n) = n + S(n + 1) for 0 < n < 10 What is the value of S(7)?

Consider the language on the alphabet {a,b} defined by the grammar:
<S> ->ab<S> | a
a. What is the shortest string in this language?
b. List all strings of length 7.

Er... are they really that simple or is there something I've missed here?

1:
S(10) = 1
S(9) = 9 + S(10) = 10
S(8) = 8 + 10 = 18
S(7) = 7 + 18 = 25

2:
Your language quite obviously looks like this:
{a, aba, ababa, abababa, ...}

(a) The shortest string is "a".
(b) The only 7-letter string is "abababa".


prove that for every triangle $ABC$,we have:
$\frac{3\sqrt{3}}{2cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}+8sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\geq 5$

it's equivalent with : $p^{2}\leq  \frac{108R^{4}}{(5R-2r)^{2}}$, and if you really want to end it use $p^{2}\leq  4R^{2}+4Rr+3r^{2}\leq  \frac{108R^{4}}{(5R-2r)^{2}}$.

I don't understand,Can you post full solution?


group theory
abstract algebra
superior algebra
superior algebra unsolved

It's sort of nice:

Let K be a field (commutative) and $t\in K\setminus \{0,1\}$. We write K as $A \cup B$, where $A \cap B= \emptyset$ . Show that there are distinct $a, b, c$ in the same set (all in A or all in B) s.t. $a=tb+(1-t)c$.

[amfulger] :Some moderator edits. I could not have deleted the posts you can read downwards. They are much too funny to be deleted  :D

Are there other conditions ? because it's trivial for any a=b=c  !!

The problem may ask that the numbers are distinct, but then for t=0 or 1 we need a=b or a=c. So what is the complete problem? :D

Sorry! I didn't think about that. Take t=/=0 or 1, and a, b, c distinct.

Are you sure that's possible to write K as A \cup B,where A \cap B = \emptyset ?I don't think so.((K,+)can't be written as a reunion of 2 distinct subsets).

You're thinking about writing a group as the reunion of 2 proper SUBGROUPS, while I'm talking about writing it as a reunion of 2 SUBSETS. Nobody said anything about the 2 subsets being rings or groups or anything like that.

For example: K={0} \cup K* and {0} \cap K*= \emptyset .

Don't you think that we can partition any set in 2 subsets?

I think some posts need to be erased :).

Very interesting problem. Visualizing this just for $ \mathbb{C}$ would give you the solution. I'm not going into details.

Let $ a_{1},a_{2},a_{3}\in A$ be different points. If the problem's claim was't right then $ b_{1}\equal{}ta_{1}\plus{}(1\minus{}t)a_{2}\in B, b_{2}\equal{}ta_{3}\plus{}(1\minus{}t)a_{2}\in B, b_{3}\equal{}ta_{1}\plus{}(1\minus{}t)a_{3}\in B$. applying this process again:

$ a_{5}\equal{}\dfrac{b_{3}\minus{}(1\minus{}t)b_{2}}{t}, a_{4}\equal{}\dfrac{b_{1}\minus{}(1\minus{}t)b_{2}}{t}$ where $ a_{4}, a_{5}$ are the solutions of $ tx\plus{}(1\minus{}t)b_{2}\equal{}b_{1},\: tx\plus{}(1\minus{}t)b_{2}\equal{}b_{3}$ and therefore they must lie in $ A$ otherwise the problem would be true again. But now it's easy to check that

$ ta_{5}\plus{}(1\minus{}t)a_{4}\equal{}a_{1}$ where $ a_{1},a_{4},a_{5}\in A$ also they are different ( easy to check ).

some little details must be filled out here. How would we guarantee that $ b_{1},b_{2},b_{3}$ are different? how would we know if either of $ A$ or $ B$ has three elements ( and in fact the problem for fields with cardinality less than 6 is not true....actually I am not sure, I haven't checked it. ) and once you know one of these sets has more than three elements choosing those intial elements is very cruial so that $ b_{1},b_{2},b_{3}$ be different. by our simple selection ( $ a_{1},a_{2},a_{3}$ are different ) it is obvious that  $ b_{1}\neq b_{3}, b_{1}\neq b_{2}$ but for having $ b_{2}\neq b_{3}$ we should have $ t(2a_{3}\minus{}a_{2}\minus{}a_{1})\neq a_{2}\minus{}a_{3}$ and we can have this if those intial elements have this property that $ 2a_{3}\minus{}a_{2}\minus{}a_{1}\equal{}0$ which would be again this problem for $ t\equal{}1/2$ ( when char($ K$)$ \neq 2$ ). but this time when $ t\equal{}1/2$ you can check that in fact $ b_{1},b_{2},b_{3}$ are different and you can find such $ a_{1},a_{2},a_{3}\in A$ that $ 2a_{3}\minus{}a_{2}\minus{}a_{1}\equal{}0$ and then starting with these elements for each $ t$,  $ b_{i}$s would be different.


The dial on the lock of a school locker has 40 numbers. If no combination uses the same number all three times, how many three-number combinations are possible on the lock?

Since it is only the 3rd number that has to be different, the number of combinations is $ 40\times40\times39\equal{}\fbox{62400}$.

No.  Your way does not allow combinations of 1-2-x, where x is any number.  It only allows 39 of them, when all 40 should be allowed.

Instead, we use complementary counting to obtain $ 40^3\minus{}40\equal{}\boxed{63960}$.


[color=indigo]Given $\triangle ABC$, point $D$ lies on $\overline{AC}$ such that $\overline{AB}=\overline{CD}$. If $\measuredangle{\text{ }}A = 2x,{\text{ }}\measuredangle{\text{ }}C = 3x,{\text{ }}\measuredangle{\text{ }}ABD = x$, find $\measuredangle\ x$[/color]

You won't receive from me more than a trigonometric solution:  :)
From the sin law we have:
1) $\frac{BC}{\sin{3x}}=\frac{CD}{\sin{6x}}$
2) $\frac{BC}{\sin{2x}}=\frac{AB}{\sin{3x}}$.
By dividing them we obtain: $\sin{2x}\cdot \sin{6x}=\sin^{2}{3x}$

$\iff 2\sin{2x}\cos{3x}=\sin{3x}\iff \sin{5x}-\sin{3x}=\sin{x}$

$\iff 2\sin{x}\cos{4x}=\sin{x}\iff \cos{4x}=\frac{1}{2}$

$\iff x = \frac{\pi}{12}$.

:rotfl:

Is not bad proving different ways, but at least, each problem that I've posted, has an elementary solution.

Greetings  :lol:

And the answer is correct.

[color=darkblue][b][u]Proof I (synthetical).[/u][/b] Observe that $m(\widehat{BDC})=3x$, i.e. $BD=BC$. Denote the middlepoints $E$, $F$ of the segments $[CD]$, $[AB]$ respectively, i.e. $FA=FB=ED=EC$.  Thus, $EB\perp EA$, i.e. $EF=\frac{1}{2}\cdot AB$. Therefore, $EF=\frac{1}{2}\cdot CD$, i.e. $FD\perp FC$. From $FE=FA$ obtain $m(\widehat{FEA})=2x$ and from $EF=EC$ obtain $m(\widehat{DCF})=x$. In conclusion, $\widehat{DCF}\equiv\widehat{DBF}$, i.e. the quadrilateral $FBCD$ is cyclically. Thus, $\widehat{DBC}\equiv\widehat{DFC}$, i.e. $BD\perp BC$. From $BD=BC$ obtain $3x=\frac{\pi}{4}$, i.e. $x=\frac{\pi}{12}$.

[b]Remark.[/b] $FA=FB=FE=ED=EC=EB$. The triangle $ABC$ is "[u][b]sliced[/b][/u]" in three triangles :

one $F$- isosceles $\triangle AFE$, one $E$- right isosceles $\triangle BEC$ and one equilateral $\triangle BEF$.

[b][u]Proof II (trigonometrical).[/u][/b] $\sin 2x=\sin \widehat{BAE}=\frac{EB}{AB}=\frac{EB}{DC}=\frac{1}{2}\cdot\frac{EB}{ED}=\frac{1}{2}\cdot\tan\widehat{BDE}=\frac{1}{2}\cdot\tan 3x$ $\implies$ $2\sin 2x\cos 3x=\sin 3x$ $\implies$

$\sin 5x-\sin x=\sin 3x$ $\implies$ $\sin 5x-\sin 3x=\sin x$ $\implies$ $2\sin x\cos 4x=\sin x$ $\implies$ $\cos 4x=\frac{1}{2}$ $\implies$ $x=\frac{\pi}{12}$.[/color]


Let $ A$ be an $ n\times n$ real symmetric matrix.  Show that

\[ \mathbb{R}^{n}\equal{}\ker\left(A\right) \plus{}\textrm{Im}\left(A\right).\]

Hint: you can make use of the fact that $ A\equal{}U^TDU$ for orthogonal $ U$ and diagonal $ D.$ Can you solve the original problem for $ A$ diagonal?

$ <x,Ay>\equal{}<Ax,y>\equal{}0$ if $ x\in{ker(A)}$.


function
inequalities unsolved
inequalities

Let $ x \geq 1$. Find the minimum value :
$ A \equal{}$ $ \sqrt {x \minus{} 1}$ + $ \sqrt {2x^{2} \minus{} 5x \plus{} 7}$

hello, our function has no maximum since
$ A'(x)\equal{}\frac{1}{2\sqrt{x\minus{}1}}\plus{}\frac{4x\minus{}5}{2\sqrt{2x^2\minus{}5x\plus{}7}}$ is allways positive for all real $ x$ with $ x>1$.
Sonnhard.

[quote="Nam Luu"]Let $ x \geq 1$. Find the maximum value :
$ A \equal{}$ $ \sqrt {x \minus{} 1}$ + $ \sqrt {2x^{2} \minus{} 5x \plus{} 7}$[/quote]

maximum value???


The mean of two positive single-digit integers is obtained by placing a decimal point between the two numbers. What is the sum of the two numbers?

$ \frac{x\plus{}y}{2} \equal{} \frac{10x \plus{} y}{10}$

$ 20x \plus{} 2y \equal{} 10x \plus{} 10 y$

$ 5x \equal{} 4y$

So $ x \equal{} 4$ and $ y \equal{} 5$, therefore $ 4 \plus{} 5 \equal{} \boxed{9}$.


On my printscreen I have this image. Does anyone has a solution? Thanks a lot !

Solution:
Let $ PY$ be second tangent line to the given circle,$ Y$ is the point of tangency.
Then $ \widehat{YAB} \equal{} \widehat{PYB} \equal{} \frac {\widehat{BOY}}{2} \equal{} \widehat{POB}$,hence $ \boxed{AY\|PO}$ and quadrilateral $ CYDB$ is [b]harmonic[/b].
Consider pencil $ A(C,Y,D,B)$ and its intersection with $ PO$,we obtain new harmonic four:$ (EF,O\infty)$(since $ AY\|PO$),hence $ OE \equal{} OF$.
$ \blacksquare$

Let M be the orthogonal projections of  O to CD, N is the point that MN ∥ AD,MN meet AB at N,CN meet DA at K.
we can aquire N,B,M,C  are concircle  O,M,B,P are concircle   CN ∥ OP  CN/OE=AN/AO=NK/OF  but we have CN=NK
Thus we are done


calculus
derivative
integration
real analysis
calculus computations

I have Kaplan's Prep book but I"m wondering how I can find a good book for AP Calc BC that can guarantee me five. I'll have to study on my own so.. Thank you!

Speaking from general terms, I'm going to assume you know the material well and just want to know it *really well*.

In that case, [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=8]you need a book of tough problems.[/url]

That really should be plenty, but if you want more problems maybe the second book on [url=http://www.skylit.com/beprepared-calc.html]this[/url] page is the way to go.

Well at my high school curriculm for math it says that students take two calculus courses for their junior and senior year. AP Calc AB and BC or whatever...whats actually the difference between AB and BC or whatever? Is one half about applications of the calculus and one about the basic concepts?

sorry if this is off topic. i have a book that i would highly recommend but i don't think it falls under the "BC" category

Calc AB covers differentiation and an introduction to integration; Calc BC also covers techniques/applications of integration and sequences/series.

Back when I was learning calculus, I found Thomas/Finney and Larson/Hostetler to be both readable and precise texts.  That was 20 years ago, though, so the current editions may be less good.  I would avoid most of the currently-popular textbooks (such as Stewart) for self-study for the AP.

Calculus BC covers more material than Calculus AB and also can involve problems that are harder versions of Calculus AB material.  The exact differences should be covered (if vaguely and generally) on the College Board website.

I took AP Calculus last year; I practically had to teach myself the BC material, but my teacher gave me previous multiple choice and free response questions.  The best way to prepare for the AP Calculus Exam (AB or BC) (or any AP exam, really) is to do past problems.

I agree with Zakary about doing past problems. Look into Barron's AP. Barron's always gives challenging problems.

Do the prep books use ACTUAL past problems?  Or do they make up ones that are supposed to be similar to them?  My advice is to get one's hands directly on problems from the exams...

I remember that Barron's AP Physics used variants of past problems. I don't know about other books. But I can confidently state that if you master Barron's then you will get a 5 on the exam no sweat. However Barron's is much tougher than the exam but nonetheless prepares you for it in a way that learning Rudin does not prepare you to take the AP Calc exam. So it's still good.


What does the word $arbitrary$ mean as a mathematical term?

is it supposed to be spelled "arbitrary"?  Or not, since it is spelled as "arbitraty" in both the topic title and your post.  If it's arbitrary, shobber is right, it means random.  If it actually is "arbitraty" then I have no idea, and it's probably a typo.

[quote="random_mathematician"]e.g. take $arbitrary$ triangle $ABC$

i.e. $arbitrary$ means $any$[/quote]

So your name would be arbitrary_mathematician?


PROVE OR DISPROVE

For all positive integer $k$ there exist prime $p$ such that $2,...,k$ are not perfect square modulo $p$

[quote="vnmathboy"]PROVE OR DISPROVE

For all positive integer $k$ there exist prime $p$ such that $1,2,...,k$ are not perfect square modulo $p$[/quote]

what about $1$ man ? :D . If $a^2\leq k\leq (a+1)^2$ then $a^2$ is square modulo $p$ for may be all $p$ which $(p,a)=1$ :D


What number is the most biggest    $ (1\minus{}\frac{1}{365})(1\minus{}\frac{2}{365})\cdot...\cdot(1\minus{}\frac{25}{365})$      and   $ \frac{1}{2}$

[quote="maksmaks24"]What number is the most biggest    $ (1 \minus{} \frac {1}{365})(1 \minus{} \frac {2}{365})\cdot...\cdot(1 \minus{} \frac {25}{365})$      and   $ \frac {1}{2}$[/quote]
Use $ \ln(1\minus{}x)\leq \minus{}x$ for all $ x\in[0,1).$

You know to what this calculation hints! It is the well known that $ 25$ people have a better chance than odds on for two of them to share a same birthday (day of the year only, not the year too).


Are any of you guys eagle scouts?

I'm probably going to be one my end of freshman year (or beginning of sophmore), and have a question

I have asked this to many people, but here it goes

How much will eagle help you get into college?

Heres the summary of answers I have...

Well...it really depends. If you're going to Westpoint or Norwich, it matters a lot, while if it is Harvard or Yale, it doesn't matter to much.

What are your ideas?

-jorian

I'm an eagle scout

if you're getting eagle as early as sophomore year then you haven't really had the full experience (just saying!)

no clue how it matters at all in college

National figures on how many Eagle Scouts there are each year would help show how rare a distinction that is. My educated guess is that it probably hardly helps at all for any Ivy League college--so go for Eagle it it's personally important to you, but not so much for a college admission boost.

ill probably feel bad in 40+ years if i don't get eagle...i'll say, "why didn't i get it...i was soooo stupid"

and the reason the numbers for eagles are so low is because most people aren't in it :P

i met someone at norwich university who got eagle at 13 :D

-jorian

[quote="tokenadult"]It hasn't been established yet that the number of Eagle scouts is low. Maybe it's high: what kind of number of new Eagle scouts each year does Scouting see in the United States?[/quote]

A random Google search turned up this:

http://usscouts.org/usscouts/eagle/numbers.html

Anything that 30,000 boys a year do is not an Ivy-meaningful distinction. It may still be a good idea to finish the Eagle requirements, but that wouldn't impress the admission officers you are trying to impress.

no, i was just asking how much certain schools care

the eagle rate for kids who try in my troop is like 85% :)

-jorian

[quote="jhredsox"]
i met someone at norwich university who got eagle at 13 :D
[/quote]

that's not anything noteworthy. The rate at which people get eagle varies wildly between troops; in my troop, no matter how good you are you won't get eagle before 16; even then, most don't get it until 17 or 18 (I got it at 18).

There's a good point to that, too; a 13 year old eagle scout is nothing compared to a 16 or 17 or 18 year old eagle scout. He just does NOT have the experience to justify getting the eagle distinction. Eagle is so much more than just doing the requirements  in the book -- it's going through the entire scouting experience and actually being a part of it long enough to participate and enjoy all the stuff it has to offer. A guy who rushes through it and gets it at age 13 just does not have that at all.

Unfortunately, there's no real way of determining who should and shouldn't be Eagle -- Of course, the Eagle board is fairly standardized, but they'll push anywhere with a decent plan through. The real barrier is the Scoutmaster, and there are plenty of troops even in my city that are eagle factories and will push out 13-14 year old eagle scouts ALL the time. It's unfortunate there isn't some sort of way to discourage pushing people through just for the badge.

[quote="jhredsox"]ill prob get it at 15

-jorian[/quote]

ok!!!!

[quote="jhredsox"]ill prob get it at 15

-jorian[/quote]

Quite impressive!  I got it at 17.

And as for college admissions, I don't think it matters that much in itself, but the fact that you remained in scouts that long would probably show dedication towards an activity, as one of my interviewers noted.  Of course, it didn't hurt that he was an Eagle himself.


Here is a problem my friend and I came up with.  We had fun solving this, using two different methods.  Care to try?

Assume we have a block of mass $ m$ attached to a spring with spring constant $ k$.  This system is attached to a point such that the system forms a pendulum.  That is, we have two oscillations: the block on the spring, and the motion of the pendulum.  Find an expression for the kinetic energy of the system.

Feel free to introduce new variables for length of the pendulum, angle formed with the vertical,...

I am depressed that no one can get this... :(

Well here are the equations of motion:

\[ \ddot r=\dfrac{kl_0}{m}+g\cos\varphi -\dfrac{k}{m}r\]
\[ \ddot \varphi=-\dfrac{g}{r}\sin\varphi\]

Here $ r$ is the distance of the mass from the point(of course this changes), and $ \varphi$ is the angle with the vertical, $ l_0$ is the length of the spring, pendulum when the spring is loose.

Now at this moment I dont see how you guys managed to get the kinetic energy of the system (The problem is finding $ \dot r^2+r^2\dot\varphi^2$), since this system of coupled DE-s doesnt look solvable like this, in fact, we cant even solve the simple pendulum equation for the time function.
Even if we try solving it for some other function $ E(\varphi,r,t)$ I dont think we can. Maybe there is some simple method, which I cant see at the moment.

If we take the simplifing assumptions that, the angle is small, that is $ \sin\varphi \approx \varphi$, and $ \cos\varphi \approx 1$, then we get the following:

\[ \ddot r=\dfrac{kl_0}{m}+g -\dfrac{k}{m}r\]
\[ \ddot \varphi=-\dfrac{g}{r}\varphi\]

Now this looks much simpler. in fact we can solve the first equation for $ r(t)$. Its just a simple nonhomogenous second order so:
\[ r(t)=A\cdot \cos\left(t\sqrt\dfrac{k}{m}\right)+B\cdot \sin\left(t\sqrt\dfrac{k}{m}\right)+l_0+\dfrac{m^2g}{k}\]
plugging this into the other equation:
\[ \ddot \varphi +\dfrac{g}{A\cdot \cos\left(t\sqrt\dfrac{k}{m}\right)+B\cdot \sin\left(t\sqrt\dfrac{k}{m}\right)+l_0+\dfrac{m^2g}{k}}\varphi=0\]

Theoretically this is solvable for $ \varphi(t)$, and then we can get the kinetic energy, but just now I dont see if it can solved analiticaly...

I would be very curious how you guys managed, to solve, please post your solution JRav  :D


function
absolute value
number theory
relatively prime
combinatorics unsolved
combinatorics

A is a set of all integer sequence and \emptyset :A--->Z is function satisfying these condition:
i)for any s,t  \in A, we have  \emptyset (s+t)= \emptyset (s)+ \emptyset (t)  {s+t is a sequence which ith member of it is ith member of s+ith member of t}
ii)for any n   \emptyset (0,0,....,0,1,0,0,0,...)=0 {1 is placed on nth place}
a)prove  \emptyset (1,2,4,8,16,....)=0
B)PROVE  \emptyset =0

Let {Ok} be the zero sequence (all terms are equal to zero). Since for any {ak} the sum {ak}+{Ok}={ak} holds,
\emptyset({ak}+{Ok})=\emptyset({ak}) and \emptyset({ak}+{Ok})=\emptyset({ak})+\emptyset ({Ok}) by property i, then
\emptyset({ak})=\emptyset({ak})+\emptyset ({Ok}) implies that  \emptyset({Ok})=0.

Lemma 1: For any sequence {sk} where every term is zero except at k, in which sk=m, then \emptyset({sk})=0.
Proof: Suppose m\neq0. \emptyset({sk})=\emptyset(0,...,0,m,0,...)=m*\emptyset(0,...,0,1,0,...)=m*0=0 by properties i and ii, where m is at the nth place.
\emptyset(0,...,0,m,0,...)=m*\emptyset(0,...,0,1,0,...) by property i, you might want to show this mentally, it's easy.
If m=0, {sk}={Ok} so that \emptyset({sk})=\emptyset({Ok})=0.
End of lemma.

a) For the sequence {ak}={2k-1}, define the sequence {sk}=(0,...,0,2k-1,0,...) for all k\geq1 so that {ak}=\sumj=1\infty {sj}
Then \emptyset({ak})=\emptyset(\sumj=1\infty {sj})=\sumj=1\infty \emptyset({sj})=0 by our lemma.

b) As in part a, define the sequence {sk}=(0,...,0,ak,0,...} for all k\geq1 so that {ak}=\sumj=1\infty {sj}
Then \emptyset({ak})=\emptyset(\sumj=1\infty {sj})=\sumj=1\infty \emptyset({sj})=0 by our lemma.

Edit PS: Sorry guys, I am still finding a solution.

[quote="3X.lich"]
a) For the sequence {ak}={2k-1}, let the subsequence {skn}=(0,...,0,2n-1,0,...} for all n\geq1. Then
\emptyset({ak})=\sumj=1\infty (2j-1)*(\emptyset({skj}))=0 by our lemma.

b) As in part a, define the subsequence {skn}=(0,...,0,an,0,...} for all n\geq1. Then
\emptyset({ak})=\sumj=1\infty (aj)*(\emptyset({skj}))=0 by our lemma.[/quote]
It is serious mistake!
We can't state that \emptyset({ak})=\sumj=1\infty (aj)*(\emptyset({skj})).
3X.lich! Are you student?

For a).
\emptyset(1,2,4,8,16,...)= \emptyset(1,0,0,...)+\emptyset(0,2,4,8,...)= 0+2*\emptyset(0,1,2,4,8,...)= 2(\emptyset(0,1,0,0,...)+\emptyset(0,0,2,4,8,...))= 4*\emptyset(0,0,1,2,4,8,...)= and so on. Thus number \emptyset(1,2,4,8,...) is divisible by any power of 2 ==> \emptyset(1,2,4,8,...)=0 (!!!).

:( My bad, looks like my post needs tons of corrections  :(
To Myth: yeah I'm a student, so I can't make mistakes?

\emptyset(1,2,4,8,16,...)= \emptyset(1,0,0,...)+\emptyset(0,2,4,8,...)= \emptyset(1,0,0,...)+2*\emptyset(0,1,2,4,8,...)= \emptyset(1,0,0,...)+2*\emptyset(0,1,0,0,...)+2*\emptyset(0,0,2,4,8,...)= \emptyset(1,0,0,...)+2*\emptyset(0,1,0,0,...)+4*\emptyset(0,0,1,2,4,8,...)=
=\sumj=1n (2j-1)\emptyset(0,...,0,1,0,...)+(2n-1)\emptyset(0,...,0,1,2,4,8,...),
where 1 is at jth place for the first \emptyset and where 1,2,4,8,... starts at nth place for the second \emptyset.

PS: I think I understand where the mistake is. I have assumed in the first place that \emptyset=0 and working with infinities are pretty dangerous. Anybody have any suggestions, like correcting my post, or an alternate approach?  :D

Myth's idea is the key:

If (a_n) is an unbounded sequence we have :
 \emptyset ((a(1),a(2),..,a(n),..)) = a(k) * \emptyset((0,0,(k-1 times),1,a(k+1),...) so a(k) divides \emptyset(a) for each k. Therefore it is divisible by arbitrarily large numbers whose absolute value will exceed its absolute value if it's not zero. So \emptyset(a)=0.

If (b_n) is a bounded sequence then c_n = b_n + 2^n and d_n=2^n are unbounded and thus \emptyset(c_n-d_n)=\emptyset(c_n)-\emptyset(d_n)=\emptyset(b_n)=0.

What's wrong about it? I'm only using the fact that if a number is divisible by arbitrarily large numbers then it must be 0 (a divides b ===> |a|<=|b|). Please explain.

[quote="iandrei"]
\emptyset ((a(1),a(2),..,a(n),..)) = a(k) * \emptyset((0,0,(k-1 times),1,a(k+1),...)[/quote]
This is wrong identity.

[quote="Gillian_Seed"]Here is an answer (thanks to VK ;))
Let x be an element of Z^N. As 2 and 3 are relatively prime integers, there exists
two sequences y and z s.t. $x(n)=2^ny(n)+3^nz(n).$
$\phi(2^ny(n))=\phi(3^nz(n))=0$ with the same technique as above, so $\phi(x)=0.$
Nice problem.[/quote]
Indeed, very nice problem! Who is VK?