[url]http://www.dailywaste.com/cool-stuff/one-handed-3x3x3-rubiks-cube-solve-in-20-seconds.html[/url]

Something to work toward, eh?

Of course David can do that.  And if not, he'll spend a few hours tomorrow, and learn how.  He'll definitely be able to show you by the time he gets to Waterloo.  Too bad I won't be there to see it.

[quote="killmesoft"]man.. girls must love him![/quote]

Korean girls only love guys who play Starcraft!  In particular, to quote a famous Korean player, "Girls only sleep with guys who play terran!"

:P

[quote="Elyot"]

Korean girls only love guys who play Starcraft!  In particular, to quote a famous Korean player, "Girls only sleep with guys who play terran!"

:P[/quote]

I played as zerg back in the day.

I guess that explains a lot.

I still play Starcraft all the time... so far I'm doing pretty well in the current Pro Gamer's Tour season... but I still get raped by the elite Korean SC gosu masters.

FYI, I play as Random.  But my favourite matchup is definitely ZvZ.

I don't have the patience to get good at Rubik's cube.  SC is far more interesting.

I got good enough so that I was beating everyone I knew from school, but never ended up being able to stand a chance against the koreans on bnet.

[quote="killmesoft"]pls stop playing starcraft already!!!! starcraft is 10 years old. it makes me sick now.[/quote]

SC is 10 years old, but its an awesome game.  No other RTS game has ever accomplished what SC has.  There are at least 5 major SC leagues worldwide (PGT, WGT, WCG, PLU, and many Korean leagues), and international tournaments are held almost every week, many with sponsors and prize money.  SC even has its own TV channel in Korea.  It was inducted into the hall of game last year, along with 5 other game franchises, including Zelda and Final Fantasy.  There are still tens of thousands of people who play on BNet, and hundreds of thousands of games played each day.  Don't mess with that.

I still play SC frequently and watch the replays of the top games in international tournaments.  I can beat many people on BNET (actually, a lot of people on BNET really suck), but there are some really good players on PGT, and the games are way more intense.  I've played well over 2500 games in my lifetime.  I :love: SC!!

And I mean... hey... what else is there to do when you've got 3 free weeks in summer, and nobody's around because everyone at Waterloo has gone home?

After a while, I learned how to get a rubix cube to the last three corners. I still haven't figured out the series of moves to finish it off  :blush:

Comming up to UW for two days and two nights this week. Dunno if I'll have time too drop anyone a line. I can only rescue so many people from bordom....  :lol:

[quote="jpark"][quote="elyot"]Korean girls only love guys who play Starcraft! In particular, to quote a famous Korean player, "Girls only sleep with guys who play terran!" [/quote]

...wrong..  :P[/quote]

Really?  Jenn, I never knew... who is the lucky Zerg or Protoss player?

[quote="Elyot"]And I mean... hey... what else is there to do when you've got 3 free weeks in summer, and nobody's around because everyone at Waterloo has gone home?[/quote]

I don't want to get into details too much because of the nature of this forum, but it seems like a lot of people have made the jump from online gamer to online poker player pretty seamlessly, and no matter how good the money is in SC tourneys or whatever, it's immeasurably better playing poker.

and yes, online poker is definitely beatable - it's like printing money.

[quote="Elyot"]Definitely true.  I play some online poker myself... though I don't really play for cash.  The truth is though, playing poker is rather boring and requires a lot of patience, whereas gaming is really fast and exciting.  It's fun to take people's money from them, but idunno... I just don't find it that interesting.[/quote]

Fair enough.  I found that once I was making 100+/hr and playing like 500-800 hands/hr, it started to become a bit of a grind.  I don't think it's boring, however - there is the 'fun' aspect of winning (and to some extent losing), but even though the decisions start to be standard, there are a lot of interesting situations from a game theoretic/mathematical point of view.  I mean, every hand is different, and playing well is easy but playing optimally is nearly impossible, and it's a challenge to try and do so :)

[quote="blahblahblah"]Well, betting to protect your hand is only good if you don't pay people off when their hand hits.  One of the biggest pots I've ever won was with QhTh on an AsAhJh flop, where I shoved the river and got calls from, like, A9 and QQ, which were so obviously not good given how the hand played out.
[/quote]

How rotten of you to not tell me how the hand DID play out...  :mad:

Wouldn't somebody with A9 on that kinda hand make a big bet on the flop to force you to fold?  I understand them slowplaying it if it was a rainbow, but isn't a little protection betting warranted here?  QQ is OK, but I think in a full 10-person ring game, odds are that somebody has an ace, even when there are 2 showing on the flop, yes?  Maybe if this was just a 6-person table, QQ is good enough... but still... calling all-in with that is pretty rediculous.

So I guess you hit a straight?  a flush?  A royal flush??? ;)

[quote="blahblahblah"]The guy with the ace bet half pot on the flop, I called, QQ called, he made the exact same bet size on the turn, I called, QQ guy called, river was a low heart, guy with an ace checks, I bet 3/4 pot, QQ guy raises all in, ace guy calls, I call.

I don't know what do to here with A9, since you're against a better ace so often.  This is why I don't play weak aces (e.g, anything but AK/AQ/sometimes AJ) unless people fold around to me on the button/one off the button.[/quote]

But you're not against a better ace very often when there are 2 aces on the board after the flop (only 18/47 chance that the 4th ace was even dealt, and given that, only 15/46 chance his kicker is better than yours).  So A9 is best after the flop about 88% of the time in this case (ignoring pocket jacks).  The risk is pretty low.  12% of the time he loses, 3% of the time, he ties, and if he calls somebody's semi-bluff, he wins the vast majority of the time (even if the flush is hit, he still has 7 outs: jacks, nines, and the 4th ace).

I think raising half the pot is too low... probably 80% of the pot is enough to make any flush draw fold (maybe not in your case, since you had a straight-flush draw).  Of course, that also depends on whether the guy with the QQ calls to sweeten the pot for you.

hey blahhh! tell me which site you use for real-money poker.

wow, this is becoming a poker thread  :lol:

To killmesoft, make money at online poker is not as easy as blah said. I guess he is just very talented. You better get some serious practice before play for money. A good place to get those practice would be join those sit&go games or free tournament with cash prizes. People there are more serious and you can get some practice (probably not enough). Also, be prepared to lose a hundred bucks before you start winning :-)

I play at partypoker.com, I enter tournaments, but really play in real money games

hey weileaf! i made an account from party poker and got 1000 play balance. can this be transferred to real money balance?

no :( else, everybody would be a millionaire :-) if you lost those 1,000, they give you another 1,000 play money

PartyPoker's play money games suck.  People just go all-in like idiots, call with crap, bluff their asses off, and essentially don't play very well at all.  Since when you lose your 1000, you can just get 1000 more, there is no incentive to play well.

If you try mansionpoker.net, their play money system is much better... when you run out of play money, you're done for 24 hours, so people actually value their play money chips.

[quote="Elyot"]I think raising half the pot is too low... probably 80% of the pot is enough to make any flush draw fold (maybe not in your case, since you had a straight-flush draw).  Of course, that also depends on whether the guy with the QQ calls to sweeten the pot for you.[/quote]

Even if he bets pot I call, since my EV is def. high enough with three in the hand, and since at low limits most people are willing to pay you off if the flush hits (implied odds).

[quote="weileaf"]Also, be prepared to lose a hundred bucks before you start winning[/quote]

The usual number people are throwing out these days is 30 buyins, which means that you *should* need about 750 to start playing NL25 (the lowest limit at Party, although they have beginner NL10 tables).  This definitely isn't the case, since most sites have lucrative signup bonuses (half of that would be fine for NL25), but sick downswings are a part of gambling, and it's so easy to lose 4-5 buyins in an hour.  It took me 4 months before I started killing NL100, and I still don't think I'm killing NL200 yet.

[quote="blahblahblah"]
Even if he bets pot I call, since my EV is def. high enough with three in the hand, and since at low limits most people are willing to pay you off if the flush hits (implied odds).
[/quote]

Well, as usual, you'd have to take their playing styles into effect... if it was likely that the third guy would call, and they'd pay you off if your straight/flush hit, then it might be a good call.

In my boredom tonight, I came 2nd in a 300-person multi-table NL tournament... :D.  Fun fun fun.

1) ROI of higher than 10% in $\$$100+ SNGs is possible, and is well documented.  I think alot of people (including blahblahblah, Elyot) are being too narrow-minded in what is and what isn't "good play."*  With good game selection, good seat selection, datamining, correct stereotyping of players, [b]good (read: not weak tight) early-sng play[/b], and more knowledge, a > 10% roi is possible because even a turbo SNG is more than just pushbotting.

2) People that play alot of tables are not necessarily the biggest winners.  General development among a good, adequately rolled poker player, in SNG or cashgames, goes like this:

 - splash around in tiny games, win/lose
 - play ultra weak/tight and be a solid minor winner at small stakes
 - play ultra laggy/fast and discover that more pots are winnable; use things like c'bets to protect monster hands ; being ultra-agressive (too agressive) in 'bubble' situations
 - absorb both the concepts of quality preflop hand selection and ability to set up bluffing on later streets to be better / use controlled agression during 'bubble' situations, including folding to the shorty, for higher +$\$$EV in SNGs
 - get better, while improving number of tables played to multiply $\$$/hr rate
 - move to higher stakes, and understand game selection and how this is a major focus of high winrates at any game
 - discard playing many tables at a time in favor of selecting very favorable games at higher limits
 - move up to highest stakes, and realize the ecology of the poker world and where all the money comes from, including only playing HU/shorthanded poker with huge fish (when playing at the highest stakes)

of course, these are all waypoints.  a good player is very adaptable, is capable of any move/play, and understands all concepts (and advanced concepts) of poker.  good poker players also understand untechnical concepts of poker, like "where is the money coming from at this table", etc.  Example: When I was playing 30/60 limit, i'd often raise preflop with rags against players that are capable of folding for one more bet.  If you look at this call (example, with 56s) in a bubble (that is, by itself), alot of players might say that is incorrect and that I am losing over 30 dollars.  But in 30/60 limit I win.  Here, it is arguably correct - also i am getting "shania" value where people think my raising range is higher than it actually is, getting me more action on my good hands, etc.  It could be correct or not correct - who knows.  Of course, better players will have a better idea of whether questionable plays are good or not.  This again relates to my main first point (*) - it is arguable what is and what isn't good play.  [b]Good play isn't just raising your good hands, raising medium hands in position, playing TAG, and pushbotting appropriately.  There is alot more to the poker world and poker play than playing taggy.  I am sure if you play 50/100NL or higher, you (or I for that matter) will lose badly on average.  That is the reason that there is still alot to learn in poker, stuff that alot of people (including me) dont know.[/b]

---------------------

[quote="weileaf"]Also, be prepared to lose a hundred bucks before you start winning [/quote]

Wrong.  Unless you are strong with money management, and consistently never play above your head, be prepared to lose most of, if not your whole bankroll at one point.  This can mean anything from 100 dollars to 1 million dollars.  (Yes, every week you see some new fish / shark that thinks he is the best   that wants to play 1k/2k NL, and most of the time he loses 1M+)   Personally, I lost ~50k from my roll playing poker real fast, and it really sucks to move from 555 sngs to 6.5 sngs, or 100/200 limit to 50c/1 limit in one day, but whatever - young people take shots, and everyone has their different opinion of what is a good idea / good bet / acceptable risk tolerance.  [b]Poker's not for everyone[/b], and NL cash has some huge variance.  If you don't like making 5 figure calls when you are 53% to win, stay out.  Of course, you always hear stories of people (such as Daniel Negranu) that consistently played over their head, won, and now they have millions of dollars and don't play over their head.

However, if you are just having fun, then of course big tournaments with low buy-ins are fun.  [b]I recommend low buyin poker to anyone.[/b]

Most SnGs don't let you select your seat... generally the ones I've played at wait for a table of players to sign up, then randomly assign seats.

mansionpoker.com offers tournaments every day with a 1 penny buy-in and a 500 dollar prize pool.  Typically there are only a few hundred players playing each game.  It is just a way to suck you in to making your first deposit, but its a rediculously good and cheap way to play in a real money tournament.

And Alex, it is a perfectly viable strategy to play like an idiot for a few hands to get people to think you're a dumbass.  But you've gotta be able to stick around for long enough to make it pay off, and you'll never be able to use those kinda tactics on the same player twice.

There aren't many people with an ROI of higher than 10% in the 109+s, and even less in the 225s, 530s, etc.  Obviously datamining is important, as it is pretty much anywhere, but SNGs seem to be a game of fairly small edges, whereas cash games are significantly less so.  Admittedly I don't really know what I'm talking about here, as I don't really play SNGs unless I am on mega tilt.

I don't know why you're lecturing me about being a tag, I'm like 30/22 in PT?

I also don't necessarily agree with your comment about the biggest winners not playing the most tables.  Sure in the short run the highstakes players on prima/ftp can win ridiculously big (e.g, Benyamine winning over 1m from 8/14-8/24 just recently), but the Party 10/20 regulars like bld and lolo probably average between 100k-200k a month, which is pretty cushy.

As for NL cash having huge variance, what, is NL cash the new PLO?  Shorthanded limit and PLO both have significantly higher variance than NL cash.

I don't have particularly high aspirations for poker, but I know that 6-tabling NL200 at Party is like printing money, and if I never move beyond NL400 or w/e, who cares?

what limits/where do you play?

Elyot:  playing differently from TAGgy is not "playing like an idiot."  It is difficult to explain, but essentially you are putting pressure on other people and trying to force them into mistakes.  Also read my (*) point: giga is not playing taggy in the sngs, and he is clearly not an idiot.  He plays for maximum edge and squeezes edges where most people cannot find them.

blahblah:  SNG is a game of small edges.  But what about Gigabet when he was pwnting the 1k sngs?  And have you seen that 15k sng HH?  People can think that they have SNGs all figured out, but then be blown away reading that HH because of new concepts that Gigabet was putting to the table YEARS ago. (*)

Also, maybe PLO > NL cash in terms of variance, but I meant that NL cash was variance high anyways.  You are right about lolo pwnting 10/20 for alot, but that is a special case.  Most very very strong players cannot acheive spectacular winrates at 10/20 when playing many tables to get the kind of $\$$/month lolo gets.  However, as much as lolo makes, there are even bigger fish that are making more money, and I am not talking about benya one-timing the huge $\$ \$$.  I'm talking about consistent winners at extremely large stakes, and also people that shop for tricky, more unusual games (read: not NLHE) at extremely high stakes where they have potentially HUGE edges.

I mainly play at party, but also some splashing around at stars.  I was going to start playing again, but I just went on this HUGE roadtrip and wasted a HUGE (UNBELIEVABLE) amount of money partying, .. also, university is coming up in a week.  I intend to start playing soon, but I want to wait and see how university is treating me.  I've played virtually every limit.  $\$$27 turbo sngs used to be my bread and butter, but I am thinking about different stuff like NL cash.  I've come off some sick bad beats at 5/10 NL though, so I don't know how I feel about that.  I am fairly confident I am +$\$$EV in any turbo 9 handed SNG (including 555) on stars with proper game selection - there are some pretty bad donks on those sometimes.

eh, I still disagree about NL cash having high variance, at least until you get to 5/10.  Who do you consider to be 'bigger fish making more money than lolo', who was at like +800k through the first 3 months of the year?

Gigabet isn't killing the SNGs anymore, is he?  Anyways there's probably more money in NL400 or NL600 than SNGs, which is why a lot of SNG players I know are moving in the NL cash direction, although it requires a different skill set.

Benyamine is reputedly the biggest winner in the big game, and aba on 2p2 says he's the best omaha player he's ever played with, and aba is stupidly good, so it's probably not a one time thing.

Do you play any games besides holdem?  PLO at Party, or wherever, is such a soft game it's ridiculous.  How big was your roll?


analytic geometry
algebra solved
algebra

we choose a segment of each line in R^3 arbitrarily. prove that there exist a subset of point  of union of those segments such that form a traingle?

This is a strange problem...what is your magazine level? Maybe is there another solution more in an olympiad-style....

Lemma :
Any set of points in the plane with cardinality $c$ (the cardinality of the reals) has an accumulation point $P$ such that in any neighbourhood of $P$ there are an uncountable number of points from the set.

Proof :
Let's cover the plane with open disks centered at rational points (points with both coordinates rationals) and each with radius $1$. The set has cardinality $c$ and each of its elements belong to the union of the disks which is a countable union. Since a countable union of countable sets is countable, it follows that at least one of the disks, say $D_1$, contains an uncountable number of elements from the set.
Now use the same reasoning with the disks centered at rational points which belong to $D_1$ and each with radius $ \frac {1} {2}$. And so on...you will construct a sequence of disks $D_1, D_2,....$ with respective centers $Q_1, Q_2,...$ and radii $ \frac {1} {2}, \frac {1} {4},....$.
Now the sequence of centers is clearly a Cauchy sequence in the plane, so that it converges to a limit point $P$ which has the desired property.

Now, we will prove the desired result but in the plane! Using any plane of the space, it will remains true in higher dimensions.

Let's give an orthogonal system of coordinates.
Let $ \theta$ be a real number in $ [0; \pi[$. Now consider the set of all the lines of the plane which made an angle $ \theta$ with the $x$-axis.
The set $S$ of these lines can be written as a countable union of sets $S_i$ for $i = 1,2,3,....$ where $S_i$ denotes the set of the lines whose chosen segments have lengths at least $ \frac {1} {i}$.
As above, since a countable union of countable sets is countable and since $S$ has cardinality $c$, it follows that at least one of the $S_i$ has cardinality $c$. That is :  there is a length $l( \theta ) > 0$ such that the set of the lines whose segments-lengths is at least $l( \theta )$ has cardinality $c$.
Now let's consider the midpoints of each of these segments. It leads to a set of points in the plane with cardinality $c$. From the lemma above, this set has an accumulation point, say $P( \theta )$ such that any neighbourood of $P( \theta )$ contains an uncountable ($c$) number of these midpoints.

The construction above can be done for each $ \theta \in [0; \pi[ $, which has cardinality $c$.
Using the same arguments as above about countable unions, we deduce that we may find a real number $L > 0$ such that $l( \theta ) \geq L$ for a set $ \Omega $ of values of $ \theta $ with cardinality $c$.
It leads to a set with cardinality $c$ of points $P( \theta )$ for $ \theta \in \Omega $.
Using the lemma again, we may consider an accumulation points of these $P( \theta) $, say $Q$. Then :
Any neighbourhood of $Q$ contains the midpoints of segments from a set with cardinality $c$, each of them with length at least $L$, and in each of "$c$ different directions".

Now select three of these directions, say $ \theta_1, \theta_2, \theta_3$. Every neighbourhood of $Q$ contains the midpoints of $c$ segments , each of them with length at least $L$, in each of these three directions.
Taking the neighbourhood $N$ sufficiently small (according to $L$ and the chosen directions), it will ensure that if $ i \neq j$ then a segment with direction $ \theta_i$  meets a segment with direction $ \theta_j$ whenever both have their respective midpoints in $N$.
In particular, we may choose three segments, one for each of the $ \theta_i$, which pairwise meet to form a triangle, and we are done.

Pierre.

I think it's a generelization of problem from Erdos.
The problem of Erdos was in R^2

Have you some reference?
In another hand, if the problem is to find a triangle, and that it is useless to work in the plane or in higher dimensions, we can't call it a generalization...

Pierre.


If $ a, b, c$ are positive numbers with $ a \plus{} b \plus{} c \equal{} 1$, prove the inequality \[ \frac{a^3}{a^2\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}a^2} \geq \frac{1}{2}\]

[quote="ozgurkircak"]If $ a, b, c$ are positive numbers with $ a \plus{} b \plus{} c \equal{} 1$, prove the inequality
\[ \frac {a^3}{a^2 \plus{} b^2} \plus{} \frac {b^3}{b^2 \plus{} c^2} \plus{} \frac {c^3}{c^2 \plus{} a^2} \geq \frac {1}{2}
\]
[/quote]
Nice inequality :)
 we have:$ \frac{a^3}{a^2\plus{}b^2} \equal{}a\minus{}\frac{ab^2}{a^2\plus{}b^2} \ge a\minus{}\frac{b}{2}$
Similar,we have Q.E.D
Vasc had a stronger inequality: $ \frac{a^5}{a^4\plus{}b^4}\plus{}\frac{b^5}{b^4\plus{}c^4}\plus{}\frac{c^5}{c^4\plus{}a^4} \ge \frac{1}{2}$ :)


What kind of score do you need to qualify? At school I got this packet but the questions look extremely trivial, which slightly concerns me...(since you would need almost all of the questions, little room for mess ups)

Dude even i saw the 07 paper and felt over-confident and so did badly on 09 test(atleast not upto my expectations :( ).

my teacher told the cut off for last year was around 20

it couldn't have been, there was a kid at my school who got a 16.75 and made the next rd. i think the cutoff was like 14 or something.

Are you kidding??? These problems are like, what is the acceleration (you have a graph of velocity v time with labels and stuff, and you approximate, and choose between the correct answer among a few ridiculous ones)

No, it isn't nearly that trivial.  The entire exam can't be characterized by the first few questions.

Yes but the last few questions were not much harder, so I think it's reasonable to assume that the majority of the test is.

Did you get the last question on the 08 prelim test? Maybe I'm a bit dumb, but it seems less than trivial, considering the time control.


I think this is well-known problem but in any case I don't understand the solution in my book so

[b]Problem[/b] Find all pairs of natural numbers $a$ and $b$ s.t. $a|b^2+1$ and $b|a^2+1$.

I am not so inteligent so let solution be more understandable and complete :D

That's equivalent to find any couple $(a,b)$ of positive integers such that $\frac{a^2 + b^2 + 1}{ab}$ is an integer itself, and this is done by Fermat's method of infinite descent! I suppose it has been discussed a lot of times in the past...

have you seen [b]I am not so intelignet[/b] part... whole solution please or link :D


Let a, b, c be positive real numbers such thta $ abc\equal{}1$. Prove that
$ \frac{ab}{ab\plus{}a^5\plus{}b^5}\plus{}\frac{bc}{bc\plus{}b^5\plus{}c^5}\plus{}\frac{ca}{ca\plus{}c^5\plus{}a^5} \le 1$

it is a well known inequality,
here is the solution(not mine)
[i][b]Lemma :[/b][/i] $ a^5 \plus{} b^5 \ge a^2\cdot b^2(a \plus{} b)$
this is easily proved by power mean inequality,

so the expression $ \frac {ab}{ab \plus{} a^{5} \plus{} b^{5}} \plus{} \frac {bc}{bc \plus{} b^{5} \plus{} c^{5}} \plus{} \frac {ca}{ca \plus{} c^{5} \plus{} a^{5}} \le \sum_{cyclic} \frac {ab}{a^2\cdot b^2(a \plus{} b)\plus{}ab}$ = $ \sum_{cyclic}\frac {1}{ab(a \plus{} b) \plus{} 1}$
                                                                       = $ \sum_{cyclic}\frac {1}{ab(a \plus{} b \plus{} c)}$
                                                                       = $ \sum_{cyclic}\frac {c}{a \plus{} b \plus{} c} \equal{} 1$
hence proved
i think it is used to prove an IMO inequality

[quote="geniusbliss"]
i think it is used to prove an IMO inequality[/quote]

This inequality is [b]IMO Shortlist 1996[/b]

Ah! now i remember if i am correct u posted a similar inequality also an IMO but it was weaker.

Wow. This is a really easy problem with rearrangement inequality. But your solution is much more beatiful than mine.  :P


inequalities induction trigonometry vector

Prove
\[\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^{2}\leq \left(\sum_{k=1}^{n}a_{k}^{2}\right) \left(\sum_{k=1}^{n}b_{k}^{2}\right) \]
using induction.

The base case is trivial. Assume that the inequality is true for $n$ pairs of variables. Then we have to prove it for $n+1$ pairs, i.e.
\[\left(\sum_{k=1}^{n+1}a_{k}b_{k}\right)^{2}\leqslant \left(\sum_{k=1}^{n+1}a_{k}^{2}\right) \left(\sum_{k=1}^{n+1}b_{k}^{2}\right)\]
which is equivalent to
\[a_{n+1}^{2}b_{n+1}^{2}+2a_{n+1}b_{n+1}\left(\sum_{k=1}^{n}a_{k}b_{k}\right)+\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^{2}\leqslant \]
\[\leqslant a_{n+1}^{2}b_{n+1}^{2}+a_{n+1}^{2}\left(\sum_{k=1}^{n}b_{k}^{2}\right)+b_{n+1}^{2}\left(\sum_{k=1}^{n}a_{k}^{2}\right)+\left(\sum_{k=1}^{n}a_{k}^{2}\right) \left(\sum_{k=1}^{n}b_{k}^{2}\right).\]
From the assumption, it sufficies to prove
\[a_{n+1}^{2}\left(\sum_{k=1}^{n}b_{k}^{2}\right)+b_{n+1}^{2}\left(\sum_{k=1}^{n}a_{k}^{2}\right)\geqslant 2a_{n+1}b_{n+1}\left(\sum_{k=1}^{n}a_{k}b_{k}\right).\]
This inequality is equaivalent to
\[\sum_{k=1}^{n}\left(a_{n+1}b_{k}-b_{n+1}a_{k}\right)^{2}\geqslant 0\]
which is obviously true.

i hope i understood you correctly... it just means 'greater than or equal to'

Yes, $\geqslant \ =\ \geq$ and $\leqslant \ =\ \leq$.

[quote="neelnal"]So, this is an induction problem?[/quote]
What do you exactly mean? Induction is one of many possible ways to prove Cauchy-Schwarz, but the shortest is writing the inequality in the form $\left|\vec{a}\right|\cdot \left|\vec{b}\right|\geqslant \left(\vec{a}, \vec{b}\right)$. Then we just notice that $\left(\vec{a},\vec{b}\right)= \left|\vec{a}\right|\cdot \left|\vec{b}\right|\cdot \cos \varphi$, where $\varphi$ is the angle between vectors $\vec{a}$ and $\vec{b}$, and $\cos\varphi\leqslant 1$.


I only know how to do this one because of reading [b]Discrete and Combinatorial Mathematics[/b], which I highly reccommend to anyone involved in contest math.  The book gives an example problem just like this one (unfortunately I am without a diagram).

8 clay targets have been arranged in vertical columns (3 in the first column, 2 in the second, and 3 in the third).  In how many ways can they be shot (one at a time) if no target can be shot until the target below it has been shot?

The diagram is similar to this:

-----------
x...x....x
x...x....x
x........x

All the AIME problems (along with a thread designated to each of them) can be found on the Resources page.

[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1990]1990 AIME[/url]

But I agree, a very cool problem if you don't know the fundamental idea behind it already (i.e. you learn it as a result of doing the problem).

guess what...I did this problem a year ago when I'm still dumb.  I brute force with DYNAMIC PROGRAMMING!!! HAHAHA!!! I still finish it in 20 min. Tho...

Yep.  It's probably hard for people to whom the thought of "hey... they're just strings" never occurred.

[quote="pkerichang"][quote="ShoeFactory"][hide]Call the first column A, the 2nd column B, and the 3rd column C.
We are just looking for the # of permutations of AAABBCCC.
$\frac{8!}{3!2!3!} = \frac{(8)(7)(6)(5)(4)}{(2)(6)} = \boxed{560}$

This is a #8?!?![/hide][/quote]

What's that comment...I don't know how to do it...[/quote]
You didn't one year ago. I bet if you looked at it again then you'd be like "THIS IS NOOB"


Support
geometry
MIT
college
geometric transformation
reflection
search

a) What team do you support for the next football world cup???.

b) Who do you think that is going to win (no who do you want, that was  question "a)", I'm asking who do you think)...

a) Argentina!!!!!!!!!!!!!!!!!.

b)Favorites to win: Brazil, Germany, Italy, The Netherlands, Chezc Republic, England, Argentina. Emmmmmm.....yes....... I would choose Argentina (watch out with Messi!!!).

A) Germany

B) Brazil,England,GErmany

Sanjith

[quote="Kent Merryfield"]Some of us are baseball fans.

There's something new this year, and it's about to start: the World Baseball Classic. The U.S.A. should probably be a slight favorite in that, but I would say that both the Dominican Republic and Venezuela should have real chances to win the event.[/quote]

Yes, probably USA... Is Cuba taking part or not???

Cuba is participating. The games of this event have already begun. Last night, Korea defeated Taiwan.

Only english people think england are going to win. Like Spain, they've good players on paper. But as some clever guy said "Paper teams win paper cups!". I'd like to see an outsider win, like Greece at Euro 2004, as my home team somehow failed to qualify.
b) probably Brazil, maybe Italy

[quote="leg"]Only english people think england are going to win.[/quote]

Jajajaajajj!!!!!!

[b]Ecuador is my beloved team...I believe that anything can and will be possible...yet, Brazil, Germany, and Argentina should be the heavy favorites...can't ignore Mexico..they are due to surprise the world...and yes, Ecuador can be very upsetting in the first round...remember they finished 3rd in the South American elimination round.[/b]

[quote="José"][quote="wisdomseeker"][b]Ecuador is my beloved team...remember they finished 3rd in the South American elimination round.[/b][/quote]

Yes, but when they play as the local team, they play in a pitch over 2600m high, where all the other teams never play, only 1 time in 4 years!!! Buy in Germany, they won't play there...[/quote]

A true fact that Quito is a very high place...however, they won games outside of Quito too...it is also true that Ecuador is new to the World Cup scene...and as the years go by they will acquire the truly necessary experience...I am rooting for Ecuador...and may the best team win...and that all of us fans get to enjoy a glorious world cup in Germany.

[quote="wisdomseeker"]all of us fans get to enjoy a glorious world cup in Germany.[/quote]

Yes, without doubt.

a) Germany
b) Brazil, The Netherlands, Argentina, Germany could be any one of these four depending on injuries, i think

(a) I would like to see the Netherlands win

(b) Germany, Brazil, or the Netherlands probably, depending on luck and injuries.

I for one am huge fan of Messi so I hope Argentina will get 42  :D

[quote="José"]Some of you are looking for excuses to forgive Zidane. He made a very bad thing. It doesn't matter what did Materazzi told him, he's a professional player, and he couldn't do that. So, please, don't try to forgive him. He did a nasty thing, and, for me, he has to pay a huge amount of money, in order to give an example of fair play.[/quote]

Typically lol-worthy post?

He headbutted someone and got sent off.  Frankly, I think it's about the funniest thing ever, but can you tell me exactly what is unfair about it?  Materazzi is classless trash, he got what was coming to him, Zidane got punished as he should have within the rules (Materazzi should have been sent off too!), this is, according to the rules of soccer, 'fair play'.  If someone did what Materazzi did on the street, the same thing would have happened to them.  Zidane may have a temper, but Materazzi is a worthless human being.

Besides, soccer is about the least fair and least sportsmanlike of all major professional sports, so I think they have a bit more to worry about than an isolated headbutt (e.g match-fixing, rampant, deplorable behavior by fans, racism and other generally disgusting behavior by players, etc).

[quote="blahblahblah"](Materazzi should have been sent off too!)[/quote]
I think that would increase the number of red cards with more than 70% in soccer. This because in (almost) every match something stupid is said.
I guess most derbys would in half time only have 5 players on the pitch (ofcourse, they would break the game long before, but in theory).

Wikipedia's article on Marco Materazzi:
[url]http://en.wikipedia.org/wiki/Marco_Materazzi[/url]

FIFA Hears Marco's side of the story:
[url]http://www.theglobeandmail.com/servlet/story/LAC.20060715.FIFA15/TPStory/TPSports/OtherSports/[/url][/url]

[quote="blahblahblah"](Materazzi should have been sent off too!)[/quote]

I disagree. They were both insulting each others, but if you show a red card for every discussion, the match won't even inish the 1º half. Then, when Zidane made what he made, he was sent-off correctly

Any geek off the street who knows anything about soccer can tell you it should have been Materazzi, not Zidane, who deserved to be sent off.  After all, Materazzi is the one with the foul mouth, Zidane was simply reacting to this unprevoked, appalling display of uncourteous behaviour and, cosequently, defending the pride of his mother.  Such a headbut in this situation is a very understandable response, and, I think, typical of most human beings.

Whath out for Romania...Even it was not present in the last two editions i think i will have a strong chance in 2010.... :P

I've heard you'll have a strong team next time. Maybe top 10  :P

There's a game at addictinggames.com where you're the guy that headbutted the other guy in that one match and you have walk around headbutting people.

[quote="shobber"]See this short video cast: http://www.6rooms.com/watch_fs.php?v=22877[/quote]

Jajajaja! Nice video! Here you can see the huge imagination of people

I'm bumping this up (sorry, but it seems Round Table is getting filled with spam and that is of course the least acceptable)

I saw a documentary yesterday about England supporters and their behavior at the World Cup.  They were caught on tape by an undercover BBC crew in several cities like Munchen,Stuttgart,...

Some things they did :

- singing songs mocking German casualties of war in WWII
- making the Hitler salute in public (ironic isn't it)
- requesting to be on a picture with two naive Brazilian girls and then persistently requesting their nudity
- grabbing the left breast of a Brazilian female supporter passing by with a flag
- harassing a Tunisian man because they mistook his flag for a Turkish flag
- singing a song saying that they would rather be a "Paki" than a German
- waving an English flag in a public place in broad daylight while wearing nothing except perhaps shoes
- stealing the pair of glasses from a Turkish man(who was alone) and then scaring him off

If it were up to me, I wouldn't organize events like that.  It's an "organized disaster". :|

[quote="fredbel6"]
- grabbing the [b]left[/b] breast of a Brazilian female supporter passing by with a flag[/quote]
It seems to me that you memorised some details. :roll:

[quote]
- singing a song saying that they would rather be a "Paki" than a German
[/quote]

what's Paki?

[quote]what's P#ki?[/quote]
It's a derogatory term (mostly used in Britain) and is derived from "Pakistani".

[quote="deimos"][quote="fredbel6"]
- grabbing the [b]left[/b] breast of a Brazilian female supporter passing by with a flag[/quote]
It seems to me that you memorised some details. :roll:
[/quote]
Ahem  :| she was hanging out of a car on the lefthand side waving a Brazilian flag, so it's not hard to remember which one they grabbed. :oops:

I find little or no joy in such foul behavior, I thought the British were the kind of gentlemen insisting on letting a lady ascending in a staircase first so they could catch her if she fell :huh: .

I'm afraid that they used "Paki" even though it is a derogatory term, but seeing they beat up Turks and Tunisians too and make Hitlersalutes... I think that was the point. :!:

[quote="fredbel6"]I saw a documentary yesterday about England supporters and their behavior at the World Cup.  They were caught on tape by an undercover BBC crew in several cities like Munchen,Stuttgart,...[/quote]
I think I saw the same documentary. (How is that possible?  :P ) They didn't only scare the Turkish man of, but made his nose bleed and then began taking pictures of him. Also, there were really young kids who were thaught some songs about German bombers. Sad.

Anyway, when I went to Germany last year, I was told that making the Hitler salute in public is a crime over there.

Let's not forget the German barkeepers kept serving beer...

[quote="fredbel6"]- waving an English flag in a public place in broad daylight while wearing nothing except perhaps shoes[/quote]
Yes. I saw that too  :D


number theory
least common multiple
greatest common divisor
AMC
USA(J)MO
USAMO
symmetry

$a,b,c$ are three arbitrary positive integers.

Prove that:

$\frac{(a,b,c)^{2}}{(a,b)(b,c)(c,a)}= \frac{[a,b,c]^{2}}{[a,b][b,c][c,a]}$

[hide="Notations"] $(a,b,.......,n)$ denote the greatest common divisor of $a,b,.......,n$

$[a,b,.......,n]$ denote the lowest common multiple of $a,b,.......,n$[/hide]

[quote="kunny"]Is this problem from USAMO?[/quote]

Not quite sure. I found it in a book at my school library. I posted it because I solved it and think it's cool.

\[\\a=\prod p_{i}^{\alpha_{i}},b=\prod p_{i}^{\beta_{i}}\\\implies \gcd (a,b)=\prod p_{i}^{\min\{\alpha_{i},\beta_{i}\}}\wedge \text{lcm}(a,b)=\prod p_{i}^{\max\{\alpha_{i},\beta_{i}\}}\]

[quote="t0rajir0u"]The technique boxedexe gives is expounded upon [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]here[/url].

[hide="Solution"] Consider an arbitrary prime $ p$ and let $ x, y, z$ be the greatest exponents such that $ p^x | a, p^y | b, p^z | c$.  WLOG $ x \le y \le z$.  Then the given identity is equivalent to

$ 2x \minus{} x \minus{} y \minus{} x \equal{} 2z \minus{} z \minus{} y \minus{} z$ [/hide][/quote]

I have a question regarding this method. Does it not assume that the a is the product of the minimum exponent for each prime and c is the product of the maximum exponent for each prime? What if that's not the case? Does the argument still hold?

"WLOG" is shorthand for "without loss of generality," which means "the problem statement has a symmetry with the property that every possible case can be relabeled to become a case in which this assumption is true."  The idea that really makes this work out nicely is that one can perform this relabeling [b]for each prime[/b], since to prove the equality it's enough to prove the equality "locally," that is, for each prime separately.

[b]Edit:[/b]  In fact, by generalizing this technique it's possible to characterize every possible symmetric identity involving gcd's and lcm's.


$ {2}^n > {n}^2$ is true for?

1) no integers $ n\ge 0$
2) all integers $ n\ge 0$
3) all integers $ n>4$
4) all integers $ n\ge 4$

1) False, $ n\equal{}0$ works.
2) False, $ n\equal{}2$ produces $ 4>4$ which is false.
3) True. Testing for $ n\equal{}5$ works and from there, $ 2^n$ will increase at a faster rate than $ n^2$.
4) False. When $ n\equal{}4$, we obtain $ 16 > 16$ which is false.


Prove that there exists a complex number w s.t. w is not a root for all polinomials of the form X^n+r,for every n and r real.

Hmmm...just choose $z  = e^{i}$.

Pierre.

Or take $w=\frac{2+i}{2-i}$ (this is from a Romanian textbook). Since $|w|=1$, if for some $n$ we have $w^n+r=0,$ then $r=\pm 1.$
If $r=-1$, we obtain $(2+i)^n=(2-i)^n$, or
\[(2-i+2i)^n=(2-i)^n
\]
\[(2-i)^n+\binom{n}{1}(2-i)^{n-1}2i+\ldots+(2i)^n=(2-i)^n.
\]This can be written as
\[(2i)^n=(2-i)(mi+n)
\]for some integers $m,n.$ Taking  absolute values and squaring, one gets
\[4^n=5(m^2+n^2),
\]a contradiction. If $r=1$, then we have
\[a_n=(2-i)^n+(2+i)^n=0.
\]It is easy to see that $a_{n+2}=4a_{n+1}-5a_n,$ with $a_0=2,a_1=4$. Checking the remainders of the $a_i$'s upon division by $5$, we find that the equality $a_n=0$ is impossible.

[quote="pbornsztein"]Hmmm...just choose $z  = e^{i}$.

Pierre.[/quote]
Are u sure in the case $n=\pi$ and $r=1$ ?

[quote="Kondr"]
Are u sure in the case $n=\pi$?[/quote]

I'm quite sure that it is not a polynomial in that case  :D

Pierre.

Oh, you are right.  Thus there is a mistake in the task because it should be polynomial and it must be proven for each REAL n.


First post! Actually, I guess I should apologize for making the first post in this forum something so esoteric, but unfortunately I'm not currently working on anything that would be "typical" for this forum...

Anyway, I was hoping that someone could explain to me a bit about characteristic factors. Suppose we are given a probability space $ \Omega$ and a measure-preserving transformation $ T$ and we want to study the effect of $ T$ on $ \Omega$. But $ \Omega$ is large and complicated. So we might find some factor map $ f : \Omega \to \Omega'$, where $ \Omega'$ is a simpler space. Then we can study the behavior of $ S \equal{} f \circ T$ on $ \Omega'$, the idea being that this new space is simpler, but still captures the important aspects of the behavior under the operator.

I can understand this as a concept, but I really don't see how to put it into practice. I had a problem about this on the midterm of my analysis class last fall where we had an intro to ergodic theory, and I completely didn't get it. Can someone please explain this (with an example, preferably) or give me a reference that is accessible? (Think 1st year grad student.)

Sorry everyone, I have just discovered that my definition of a factor isn't even exactly right. Suppose that $ (X, \mathcal{F}, \mu, T)$ is a probability space with a measure-preserving invertible operator. We want to study the effects of the operator on the space. So a factor of the space is a smaller $ \sigma$-algebra $ \mathcal{F}' \subset \mathcal{F}$ under which $ T$ is still measurable. Then we study the behavior of $ T$ on the simpler space $ (X, \mathcal{F}', \mu)$ (noting that $ X$ and the measure are the same, it's just the $ \sigma$-algebra that's changed).

Additionally, I think that I'm about to read a decent example of factors in action. If you are interested, read my blog over the next 2 weeks or so.

I would not be offended if a moderator deleted this thread, as it's kinda pointless now, unless someone else wants to discuss factors of probability spaces.


Let a,b,c>0.Prove that:
\[ \sum {a^5  \plus{} \sum\limits_{cyc} {2b^3 c^2  \ge \sum\limits_{cyc} {ab^4 } }  \plus{} \sum\limits_{cyc} {2a^4 b} }
\] :)

jiji

[hide]It is equivalent to $ \sum{a(a \minus{} b)^2(a^2 \plus{} b^2)}\ge 0$. We could make a little stronger one by expanding $ \sum{a(a \minus{} b)^4}$.
[/hide]

:)


Let M be a metric space and X $ \subseteq$ M. Show that X $ \cap$ K is closed in K for every compact subset K $ \subseteq$ M, then X is closed in M.

Let $ x\in\bar{X}$,

                       there is a sequence $ \{x_n\}$ such that $ \lim_{n\to\infty}{x_n}\equal{}x$

                        let $ K\equal{}\{x_n\vert\; n\in N\}\cup \{x\}$,

                        $ K$ is compact so $ X\cap K$ is closed in $ K$

                        $ \Rightarrow$ $ x\in {X\cap K}$ $ \Rightarrow$ $ x\in X$

                        $ \Rightarrow$ $ X\equal{}\bar{X}$

 :wink:

Knowledge is a shield, truthfulness is might, ignorance is abasement, understanding is glory, generosity is success, good behavior causes friendship, he who has knowledge about his time, ambiguous things do not attack him.


For given $n\in\mathbb N,n>1$ is it true that every set which contains those $x$ s.t $x|n^x-1$ has at most finitely many prime divisors (prime divisors of its elements)?

No, for n=4 it follows from one of 2000 IMO problems that the result is false.

At least ,for $n=2$ we know that the set consists of only $x=1$.

Speaking about IMO 2000, has anyone found a solution for the nice problem: is the density of those n such that $n|2^n+1$ equal to 0?

Yes,you are right,harazi.Thanks in advance for remind me,and please forgive me for I bothered you with stupid questions.I don't know,it's really bad day to me.
To tell the truth,my initial question is that wherether the cardinality of the set of natural exponent of some primes  $x=p^k$s.t$x|n^x-1$ is finite for given $n\in\mathbb N,n>1$.
I've made series of mistake  so I bet this question is trivial again :( .

No, It was really nice question, I had rapid reaction since I was trying (for how many times till now...) to solve the problem with the density. I imagine the answer to this question is also no, but I have no contrexample. And I do not think it is trivial.

In fact I do, take $n=4$ and x powers of 3.

harazi,please hear me,my question comes from the fact that $2^{n+2}|3^{2^n}-1$,and I meant wherether te set of primes $p$ s.t $p^k=x|n^x-1$ is finite?

We have $p\mid n^{p^k}-1$ and $p\mid n^{p^k-1}-1$, so $p\mid n-1$.


i agree. the color scheme of this page in general is a bit nervewracking for me. the bright blues and reds and greens and the black and yellow...a lot of the word/pictures are difficult to read, like this blue on red problem i can see right now and this sign that says "Community".

We'll be messing around with the colors in a few months.  I increased the point size on posts you're reading (since that's easy)...

On the same note, can you change the font?

Yeah, if its not too much trouble for you, to a more typical Times New Roman or Arial.

I agree. When the font is enlarged, it kind of looks...well, stupid. (and I don't mean that in an offensive way)

[quote="fairyofwind"]The black and yellow color theming is really confusing and is difficult to read... maybe we can change to a less exotic color theme as such of collegeconfidential.com with a LIGHT background with black text... the forum is very hard to browse.[/quote]

This issue has come up before, namely [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=347]here[/url].  I believe Mr. Rusczyk is busy conducting AoPS classes, writing out scripts for the classes, writing out homework for the classes, answering questions for the classes, etc.  But maybe in a few months...


AMC
AIME
trigonometry
geometry
cyclic quadrilateral

A circle has center on the side $ AB$ of the cyclic quadrilateral $ ABCD$.  The other three sides are tangent to the circle.  Prove that $ AD \plus{} BC \equal{} AB$.

yes. this is trivial but it does take me a while to actually prove it. (as the world always says, I suck at olympiad math.) :(

By the way, earlier IMO problems are usually much, much easier than today's olympiad problems. But they are probably a good way to start your study.

Yeah that's exactly what I do. I start with the easiest IMO problems to prepare for AIME. It's probably a good start to mastering Olympiad math I guess?

Well, you probably should study for the AIME by doing AIME problems or reading AoPS 1 and 2.
If you want to start studying for olympiad problems you should look for good resources. Some are books (AoPS isn't so much help here, although it gives you a good foundation; ACoPS is good, Engel is rather difficult but it's chock-full of problems, etc.) but you might also want to talk to great math people? That is one of the reasons that summer camps exist - not just so you learn stuff, do stuff, etc. but also to form relationships with the good people. I don't know who you know, so I can't advise you there.

[quote="gauss1181"]lol that's why i talk to some great math people like:

Aryth
pythag011
not_trig
The QuattoMaster 6000
...etc.

 :D[/quote]

Why am I in the list? I fail at math.

But anyways, to prepare for aime it probably would be helpful to read aops 1 and 2... and probably acops as well.

Engel is good, though it is mostly problems...

Actually, some countries (Canada) have easy olympiads...

Another thing: Problems are useful.

:huh:  :huh:  :huh:

oh well you're there anyway cuz you're awesome to me =)


vector
Functional Analysis
IMC
college contests

Let $ \mathcal{H}$ be an infinite-dimensional Hilbert space, let $ d>0$, and suppose that $ S$ is a set of points (not necessarily countable) in $ \mathcal{H}$ such that the distance between any two distinct points in $ S$ is equal to $ d$. Show that there is a point $ y\in\mathcal{H}$ such that
\[ \left\{\frac{\sqrt{2}}{d}(x\minus{}y): \ x\in S\right\}\]
is an orthonormal system of vectors in $ \mathcal{H}$.

If $ S\equal{}\{x_1,\,x_2,\,\dots\}$ is countable, then define points $ y_n$ as centroids(=circumcentres) of simplexeswith vertices $ x_1,\,x_2,\,\dots,x_n$. It is easy to check that the sequence $ \{y_n\}$ is fundamental and so have a limit, which is desired point.

In general case, consider the generalized sequence (or how is it called?) of all finite subsets of $ S$ and take limit along this g.s.

We may apply a simpler idea. Let $ S_0 \equal{} \{x_1,...,x_n,...\}$ is a countable subset of $ S$. It was shown in the previous post how to find a needed $ y$ for $ S_0$. Now it is easy to exhibit, that any other $ x \in S$ creates a vector $ \frac {\sqrt 2}{d}(x \minus{} y)$, which has the norm $ 1$ too. Really, let $ e_n \equal{} x_n \minus{} y$ for every $ (n \in \mathbb N)$ and $ e \equal{} x \minus{} y$. Then from orthogonality of $ \{e_n\}$ follows $ (e,e_1) \plus{} ... \plus{} (e,e_n) \plus{} ... \le \frac{d}{\sqrt 2}\parallel{}e\parallel{}$, so for any $ \varepsilon > 0$ there exists $ e_n$, for which $ |(e,e_n)| < \varepsilon$. Now, $ \parallel{}e \minus{} e_n\parallel{}^2 \equal{} \parallel{}e\parallel{}^2 \minus{} 2(e,e_n) \plus{} \parallel{}e_n\parallel{}^2$ or $ d^2 \equal{} \parallel{}e\parallel{}^2 \minus{} 2(e,e_n) \plus{} d^2/2$. Hence $ \parallel{}e\parallel{}^2 \in (d^2/2 \minus{} 2\varepsilon,\ d^2/2 \plus{} 2\varepsilon)$ and the statement is proved.

Indeed, I just did not believe that such point for countable subset is appropriate for whole $ S$.

The minor remark: I think, it is not $ \sum (e,e_i)$ which is finite, but rather $ \sum (e,e_i)^2$ (by Bessel's inequality).

Yes, of course. Thank you for the correction.


Given a permutation $\sigma = (a_1,a_2,a_3,...a_n)$ of $(1,2,3,...n)$ , an ordered pair $(a_j,a_k)$ is called an inversion of $\sigma$ if $a \leq j < k \leq n$ and $a_j > a_k$. Let $m(\sigma)$ denote the no. of inversions of the permutation $\sigma$. Find the average of $m(\sigma)$ as $\sigma$ varies over all permutations.

i think i have a solution  :D
please check if it has a boo boo :(
 number of permutations = n!
we choose 2 elements out of n and choose 2 places out of n for them thus creating an inversion and permute the rest as we want
no. of ways to choose 2 elements out of n = n(n-1)/2
no. of ways to choose 2 places out of n = n(n-1)/2
no. of ways to permute the rest = (n-2)!
therefore avg = n(n-1)/4
is it correct?!!! :?:

I too got the same answer, however I adopted an approach using recursions.

Consider $ (a_1,a_2,a_3,...,a_n)$, and $ (a_n,a_{n \minus{} 1},...,a_1)$. Consider the placement of $ k$. Then consider any other number, $ j$. It must occur on the left of $ k$ exactly once in both of the permutations above. Hence there is one inversion between $ j$ and $ k$ there. Hence there are $ n \minus{} 1$ inversions for each $ k$. There are $ n$ $ k$'s, so a total of $ (n)(n \minus{} 1)/2$ since we count each pair $ k$, $ j$ twice.  Also we have to divide by another factor of $ 2$ because we were dealing with two permutations above, not one. This gives $ \boxed{\frac {(n)(n \minus{} 1)}{4}}$.

Sorry for thread necromancy. Call an "antinversion"(couldn't find a better name :P) an ordered pair $ (a_i,a_k) $ such that $ 1 \leq j < k \leq n $ and $ a_k>a_j $.Now call $ I $ the total number of inversions over all permutations and $ A $ the total number of antiversions over all permutations.It is clear that $ I=A $ because by reversing a permutation every inversion becomes an antiversion and vice versa.Also, $ I+A=(n!)\binom{n}{2} $ because this is the total number of pairs over all permutations,and each pair is either an inversion or an antiversion.Thus $ 2I=(n!)n(n-1)/2 \Leftrightarrow I=(n!)n(n-1)/4 $ and the average is $ n(n-1)/4 $.

[quote="AlexanderMusatov"]Wanted to post a different approach.[/quote]  But your solution is the same as the one immediately above it in every important respect :(


function
algebra
polynomial
calculus
calculus computations

Approximate √7 using three iterations of Newton's method starting at X0 = 3

A typical homework problem. Show what you've done so far and we'll give you a hint about what comes next :)

Begin by explaining how Newton's Method works.

First step:

Find an equation of the form $ f(x)\equal{}0$ that has $ x\equal{}\sqrt{7}$ as a root.

It's best that $ f(x)$ be a "nice" function, easy to evaluate and with easy to evaluate derivatives. (Polynomials, especially polynomials of reasonably low degree with integer coefficients, qualify as "nice".)

OK, let me teach you the Newton method then.

Assume that we need to solve $ f(x) \equal{} 0$. Take an initial guess $ x_0$ that you believe is not too far from the root. Now use the linearization $ f(x)\approx f(x_0) \plus{} f'(x_0)(x \minus{} x_0)$ and instead of the difficult (nonlinear) equation $ f(x) \equal{} 0$ solve the easy (linear) one $ f(x_0) \plus{} f'(x_0)(x \minus{} x_0) \equal{} 0$. The result will be some $ x_1$ (if you want, you can easily write the formula for it: $ x_1 \equal{} x_0 \minus{} \frac {f(x_0)}{f'(x_0)}$ but I have never tried to memorize it). Now repeat the process using $ x_1$ instead of $ x_0$ and so on.

Your function is $ f(x) \equal{} x^2 \minus{} 7$, your $ x_0 \equal{} 3$, you are asked to find $ x_3$.

Should I say anything else?

Edit: WOW, three replies at once :lol: Mine took the longest to type obviously...

yes please, continue to the end. .......

I think all of us have said everything we were planning to say. Look at fedja's explanation; everything you need to know is there. Now we're waiting for you to do the calculations.

thank you guys, i got it. It was my stupid.


complex analysis
function
invariant
LaTeX
complex analysis unsolved

If $ f$ is a holomorphic function that maps $ U$ to $ V$, where $ U$ is the unit disk ,$ f(0)\equal{}0$ and $ V$ is an open connected region that is invariant to the map : $ z|\minus{}\minus{}>iz$, prove that $ f(iz)\equal{}if(z)$. THanks in advance.

The latex command "\mapsto" gives "$ \mapsto$".

Are you sure of the hypothesis?

Consider the map $ z\mapsto z^2$ from the disk to itself; it's holomorphic and it fixes 0, but $ f(iz)\equal{}(iz)^2\equal{}\minus{}z^2\neq iz^2\equal{}if(z)$.


Any tips on how to write for partial credit on USAMO? I've never been in a position to write half a proof.

Write down whatever ideas you might have about the proof: some algebra that you've come up with to express a quantity -- even if you don't completely know what to do with it.  Any ideas about how to work with a geometric figure.

Write it all down even if you don't solve the problem (after you write your full solutions up of course).  In fact, it's a great exercise in general to write down all thoughts and facts in general.  It's often looking at the list of facts that leads you to a solution.  It's easier to play around with puzzle pieces when you have [i]as many of them as possible[/i] accounted for.

Not all parts of every question require a proof.  Not all proofs are rigorous.

Round 1, Problem 5.  One of my youngest students (age 11) looked at the drawing and said "I can't prove it, but I know how you guys write problems.  240 is a multiple of 12, and I see a 5-12-13 Pythagorean triplet.  The answer is 260."  Good intuition?  Dumb luck?  The student was awarded 1 point, with a stern note from the grader that mathematics requires proof.

Round 4, Problem 2.  One of my students provided the example, but no proof.  Again, partial credit was awarded.

On the USAMTS, sometimes you can get away without a proof, but the USAMO grading is far stricter, so I think need to have made significant progress in order to get partial credit on the USAMO.


Does someone have any problems using the probabilistic method???
I'll be thankful to who send me some problems!! :D

I don't exactly know what you mean by "probablistic method" (there is one "probablistic method" to prove existence theorems, and there is another "probablistic method" of proving combinatorial identities, and there may be more probablistic methods), but [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=966]here you can probably find something interesting[/url].

  Darij

In download section there is a file with combinatorial problems from math olympiads that can be solved using probabilistics


limit
real analysis
real analysis unsolved

The sequence $ x_{n}$ is given by follow: $ \left(1+\frac{1}{nx_{n}}\right)^{n+x_{n}}=e$. Compute $ \lim_{n\to \infty}n(x_{n}-1)$

I just reasoned this out in my head, so I hope this is not too sloppy and/or incorrect:

Rewriting the first relation I have:

$ (n+x_{n}) \log(1+\frac{1}{nx_{n}})=1$

Use the first term of the Taylor expansion I have:

$ \frac{n+x_{n}}{nx_{n}}\sim 1$

I can conclude two things:

(i) if $ \lim_{n\rightarrow \infty}x_{n}=L$, then the asymptotic estimate says $ L=1$.

(ii) I can rewrite the estimate as:  $ nx_{n}-n \sim x_{n}\rightarrow L=1$.

If this is correct, I wouldn't mind seeing a more rigorous version of these arguments.  If not, then I post the following in advance   :oops:

I think your proof is not correct, although(not necessarily by Taylor) $ \frac{n+x_{n}}{nx_{n}}\sim 1 \Longleftrightarrow \lim_{n\rightarrow \infty}x_{n}=1$, so these are equivalent, so the only thing you can conclude is this. You can easily find a counter-example taking $ x_{n}\equiv 1$ you find $ nx_{n}-n\equiv 0$, although $ \frac{n+x_{n}}{nx_{n}}\sim 1$.
Anyway you can show $ nx_{n}<x_{n}+n$. We will have  $ 0\leq \lim_{n\rightarrow \infty}n(x_{n}-1)\leq 1$ This does not mean i have an answer, yet i am eager to see it.

Lorincz wrote:

[quote] You can easily find a counter-example taking $ x_{n}\equiv 1$ ... [/quote]

Actually that is not a counterexample since we must have $ \left( 1+\frac{1}{nx_{n}}\right)^{n+x_{n}}=e$ for all $ n \in \mathbb{N}$.

Nonetheless I am not defending the second claim in my original post at this point.   :lol:

In fact I think I could only rightly conclude that the limit we are looking for must be $ O(1)$ which is not too helpful.

Numerically, it looks like $ \lim_{n\rightarrow \infty}n(x_{n}-1) =\frac{1}{2}$.  It's a very nice result.

You didn't get my point. I know we can't have $ x_{n}\equiv 1$, but you only used this $ \frac{n+x_{n}}{nx_{n}}\sim 1$  to prove that the result is $ 1$.
Anyway, I've found the ("surprising" :P ) result:$ \frac{1}{2}$ ,by Taylor expansion with Lagrange remainder for example:
$ (n+x_{n})\left(\frac{1}{nx_{n}}-\frac{1}{2}\frac{1}{(nx_{n})^{2}}+\frac{1}{3}\frac{1}{(nx_{n})^{3}}\frac{1}{x^{*}^{3}}\right)=1$ with $ x^{*}\in\left(1,1+\frac{1}{nx_{n}}\right)$
$ x_{n}-\frac{1}{2}\cdot\frac{n+x_{n}}{nx_{n}}+\frac{1}{3}\cdot\frac{n+x_{n}}{(nx_{n})^{2}}\frac{1}{x^{*}^{3}}=nx_{n}-n$
$ \lim_{n\rightarrow \infty}n(x_{n}-1)=1-\frac{1}{2}+0=\frac{1}{2}$

Well done Lorincz!

Really Taylor should have been the first thing to try.  There are many more horses out there than zebras.   :)


A line is drawn through the points $ (\minus{}2, 6)$ and $ (\minus{}4, \minus{}2)$. What is the y-coordinate of the point where this line intersects the y-axis?

finding the slope:

-2, 6
-4, -2 -
------
2,8

8/2= 4
thus the slope is 4, we may write this as

$ y\equal{}4x\plus{}b$

now let us substitute -2,6 yielding
$ 6 \equal{} \minus{}8 \plus{} b \equal{}\equal{}>  b\equal{}\boxed{14}$


Many math competitions, especially high school ones, require knowledge of how to use sum.

$\sum^{20}_{i=1} 2+i$

Above is an example of what the sum looks like. This is also commonly expressed using:

$\Sigma^{20}_{i=1} 2+i$

The both sign represents the symbol [b]SIGMA[/b] which means adding all the terms.

Let's look at the first example:

$\sum^{20}_{i=1} 2+i$

What is the value then i=1?

Putting in 1 for i, then 1+2 = 3. So, the first term is 3.

Similarly, 2 is 4. 3 is 5, 4 is 6 and so on.

Then how can we find it the sum?

Well, there is a nice formula if the term starts with one:

$\frac {n(n+1)}{2}$

Where n represents the number of terms. For this one, however, the first term is 3. So, we take 2 away from the each term and get:

$\sum^{20}_{i=1} 2+i\\
= 3+4+5+...22\\
= 2(20) + 1+2+3...20\\
= 40 + \frac {20(21)}{2}\\
= 40 + 210\\
= 250$

So, the answer for the orginial question is 250. I know that there is more deeper sums and it is very complicated. So, I'll leave here as getting started.

Few challenges:

If the first term is 1, second term is 2, third term is 3, and so on, find the sum of first 20 terms:

In another word:

$f(1) = 1\\
f(2) = 2\\
f(3) = 3$

Hint: Use sum to find the function and write it. It would be a lot easier. Of course, you can just imply the formula of n(n+1)/2 in here though. :)

Wouldn't the start term be 2 because you put i=0 on the sigma notation?

Yes.

Fixed. See if you can do my challenging problem though. Define the series using the sigma.

[quote="Gyan"]A small point: To remove confusion, please use "(" and ")" around  the expression inside sigma:

sigma (2 + i)  is not the same as    sigma (2) + i

 Will remove this message (or you can edit/remove it) once you edit your  original  post. Thx.[/quote]

But notice that in Silverfalcon's examples, sigma (2) + i does not make sense.

Point  here was to write   clearly,....  sigma(2 +i)   could be written as
40 +  210, while sigma(20) + i would be simply (40 + i )   (I am assuming the range  for sigma is given as above etc..)

In any case, as I said before, best is  to use a formn which causes no confusion.. Hope this helps.

I would like to make an easier introduction and hopefully supplement Silverfalcon's post!!!

So what does the mathematical formula $\sum_{i=1}^5f(i)$ mean?
It is shortcut way to write a long summation, this means that
\[
\sum_{i=1}^5f(i)=f(1)+f(2)+f(3)+f(4)+f(5)
\]
Components of $\sum_{i=1}^5f(i)$: each of the components of this summation has a name:
$i$ is called the [b]index[/b] of the summation;
$\Sigma$ is called the [b]summation sign[/b] (it comes from the Greek alphabet letter Sigma, uppercase sigma is $\Sigma$ and lowercase sigma is $\sigma$);
$f(i)$ is called the [b]term[/b] of the summation;
the entire summation is also called a series.

But how to interpret this summation? The summation starts from $i=1$ and it will subsequently substitute the next consecutive integer until $i=5$.
As a start this should be a description of what this summation notation means.

That is very nice! :)

Thank you so much!

Like 3X.lich said, the sum is just another way to write out the huge addition.

You might wonder, "why do I need this?" but exception of few small numbers, the finding the sum of many terms is only easy and accurate if you use the sum.

Of course, you can sit down with your cookie and punch the number for hours and get the correct answer, though. :P


Show that, if f is a function from A to B and T is a subset of B, then the image of the inverse image of T under f is contained in T.

I know that I need to let an arbitrary x in the image of the inverse image T, then try to prove that x is in T. But I don't know how would I go from there.
by definition of inverse image, I know that the inverse image of T is a subset of A. Anyone can help me please.

[quote="namelessguy"]I know that I need to let an arbitrary x in the image of the inverse image T, then try to prove that x is in T.[/quote]
Read the problem carefully: you are to prove that the [b]image[/b] of $ x$ is in $ T$, not $ x$ itself.


We define a sequence $ \{x_n\}$ as

$ x_1\equal{}x\in \mathbb{N}$, $ x_{n}\equal{}\left\lfloor \frac{21}{20}x_{n\minus{}1}\right\rfloor$ $ (n\ge 2)$.

If $ 20<x <100$, show that there exists $ i$ such that $ 10|x_i$ .

[quote="Bovy11"]We define a sequence $ \{x_n\}$ as

$ x_1 = x\in \mathbb{N}$, $ x_{n} = \left\lfloor \frac {21}{20}x_{n - 1}\right\rfloor$ $ (n\ge 2)$.

If $ 20 < x < 100$, show that there exists $ i$ such that $ 10|x_i$ .[/quote]

[hide]By the division algorithm, $ \exists q,r\in\mathbb{Z}$ s.t. $ x = 20q + r, 0\le r < 20$. Also $ 20 < x < 100\Rightarrow 1\le q\le 4$.
So $ x_2 = \left\lfloor\frac {(20q + r)21}{20}\right\rfloor = 21q + \left\lfloor\frac {21r}{20}\right\rfloor = 21q + r + \left\lfloor\frac {r}{20}\right\rfloor = 21q + r$.
If $ q = 1$, the sequence eventually reaches $ 40$.
If $ q = 2$ and $ r\equiv 0\mod 2$, the sequence eventually reaches $ 60$. If $ r\equiv 1 \mod 2$, the sequence eventually reaches $ 61$:
If $ q = 3$ and $ r\equiv 1\mod 3$ (which covers the above case when $ r\equiv 1 \mod 2$), the sequence eventually reaches $ 82\Rightarrow$ reaches $ 90$. If $ r\equiv 2\mod 3$, the sequence eventually reaches $ 80$. If $ r\equiv 0\mod 3$, the sequence eventually reaches $ 81$:
If $ q = 4$ and $ r\equiv 1\mod 4$ (which covers the case above when $ r\equiv 0\mod 3$), the sequence reaches $ 101\Rightarrow$ reaches $ 121\Rightarrow$ reaches $ 145\Rightarrow$ reaches $ 166\Rightarrow$ reaches $ 182\Rightarrow$ reaches $ 200$. If $ r\equiv 2\mod 4$, the sequence reaches $ 102\Rightarrow$ reaches $ 122\Rightarrow$ reaches $ 140$. If $ r\equiv 3\mod 4$, the sequence reaches $ 103\Rightarrow$ reaches $ 123\Rightarrow$ reaches $ 141\Rightarrow$ reaches $ 162\Rightarrow$ reaches $ 170$. If $ r\equiv 0\mod 4$, the sequence reaches $ 100$. [/hide]


quadratics
inequalities
algebra
number theory unsolved
number theory

solving the equation with nonzero integers $ x,y$
$ (x\plus{}y^2)(x^2\plus{}y)\equal{}(x\minus{}y)^3$

[quote="delegat"]After simple transformations equation becomes:

$ y[2y^{2} \plus{} y(x^{2} \minus{} 3x) \plus{} 3x^{2} \plus{} x] \equal{} 0$

Now consider equation $ 2y^{2} \plus{} y(x^{2} \minus{} 3x) \plus{} 3x^{2} \plus{} x \equal{} 0$ as quadratic with respect to $ y$ and discriminant must be perfect square and then use some

inequalities and well known fact that there is no perfect square in interval $ (x^{2}, (x \plus{} 1)^{2})$[/quote]
you means that this equation not have root of integers.I don't think so
$ x\equal{}y\equal{}\minus{}1$ is one root of the equation

[quote]Lijepo rješenje.[/quote]

Hvala Al evo ovaj opet ne kuzi.

[quote]you means that this equation not have root of integers.I don't think so
$ x\equal{}y\equal{}\minus{}1$ is one root of the equation[/quote]

No I didn't say that there are not integer solutions
After you find discriminant of quadratic equation, just use some inequalities to find upper and lower bonds of $ x$ and then check for which of them discriminant is perfect square. I hope it is OK now.


Let $x,y,z$ be positive reals with sum 1. Prove that \[ 10xyz\leq x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2. \]

I don't agree with you ,phamduyhiep.
I think $x^2y^2+y^2z^2+z^2x^2\leq \frac{1}{27}$
and my solution is:
$x^2y^2+y^2z^2+z^2x^2\geq (x+y+z)xyz=xyz$
$x^2+y^2+z^2\geq \frac13\geq 9xyz$

:blush:  :blush:  :oops:

I got it as a special case after long calculations. I never realized it was that easy ... :blush:

Anyway, I must've made a miscalculation, because the original inequality would look like this: Denote by $(a,b,c)$ the symmetric sum \[ \sum_{\mathrm{sym}}x^ay^bz^c. \] Then for all $x,y,z\in\mathbb{R}^+$ we have \[ (2,1,1)+2(1,1,1)+(1,1,0)\leq 4(2,1,0)+(2,2,0). \]  (notice that for small $x,y,z$ we don't have $\sum y(1-2x)\leq0.$)

I won't be suprised if this one turns out to be a disaster too...  :dry:


How do you teach a computer to understand and interpret commands? To teach it to understand commands, you'd have to first establish a way of communicating... and that seems only to be possible with previously established commands.

Could someone explain? Is the first step (i.e. the way to establish a simple way of communicating with a computer so that you can begin to teach it commands) a physical step? something that relies are laws of chemistry or physics?

Also, what is a computer? How do you make it actually do anything?

http://en.wikipedia.org/wiki/Computer#I.2FO_devices

Wikipedia is a powerful tool.

But on the most basic level, computers function with "ON" and "OFF" switches, or bits, which work by physics - electrons moving through the switch when it's in the "ON" state. From this, you can make computers do cool things, like add, subtract, etc. There's an interesting book I read, but I forget what it's called, it has to do with circuits and how computers really work and stuff.

BTW, I hope that info is right... I'm not super knowledgeable about the fundamentals of hardware, but that's my understanding.

If you ever get time, try these books --

White, Ron. How Computers Work. Ziff-Davis Press, 1993.

White, Ron. How Software Works. Ziff-Davis Press, 1993.

They're pretty interesting..............

The last word however -- Even though I can say that I sort of understand how a computer works , I'm still in quite a bit of awe as to how to some pieces of plastic and some other metals can together end up doing all the complex stuff that we're doing today with our computers....


How do the current rules concerning emoticons within the text work?  Is it tied to the text length?  If so, how?
 :huh:

The number of emoticons is determined by the length of your message.  However, you can't post more than 10 emoticons in a single post, no matter how long the message is.

I'm not sure about how long you need a message to be for each number of emoticons...

Also,   :?: and  :!: count as 2 because they have  :D and  :lol: embedded in them. :(  :(  :(
I think that is another bug.  :(

Thank you.  It's always better when you can answer a question by referring to another post.

I think there's something wrong with the smile-counting algorithm:
[code]:lol:)[/code]
is counted as 3 smilies.  Maybe the "overlaps" between the smilies are being counted?


Can anyone suggest to me a fairly mathy topic in modern physics? Preferably one  that is interesting enough to keep a high school physics class' attention.

There is not much in high school modern physics and therefore I dont find anything interesting in there.But still I like photoelectric effect and x-rays but there isnt much math in them, I suppose. I have done few problems combining electromagnetism and modern , those I really enjoyed.

Topics don't have to anything [i]taught[/i] in high school, I just have to present the topic to my high school [i]class[/i]. And by modern I mean stuff that has ongoing research right now.

I'm not sure if it's [i]that[/i] modern, but topics on Superconductivity, quantum tunnelling are definitely interesting.  Also, you can do the maxwell-boltzman distribution, and then the Bose-Einstein condensate theories.

It's not so clear to me what you intend to do with the subject.

Maybe something about the Lagrange/Hamilton formalism, i.e. calculus of variations? Stuff that actually has ongoing research is probably way too incomprehensible for a high school class.

How about Einstein-Podolsky-Rosen Paradox [url=http://en.wikipedia.org/wiki/EPR_paradox](EPR Paradox)[/url].

An extremely interesting topic most high-school students [b]may[/b] understand.


algebra
system of equations
algebra unsolved

$ \begin {cases} x\plus{}y\plus{}xy\equal{}10\\y\plus{}z\plus{}yz\equal{}11\\z\plus{}x\plus{}zx\equal{}14 \end {cases}$

Thank you very much. I think we can done it by other way.

With tense this such systems of equations we is generality it goes up and similar demob way is ditto
$ \begin {cases} x \plus{} y \plus{} xy \equal{} a \\
y \plus{} z \plus{} yz \equal{} b \\
z \plus{} x \plus{} zx \equal{} c \end {cases}$

Now, we continue to make this system of equations  :)
$ \begin{cases} \sqrt{3\plus{}2x^2y\minus{}x^4y^2}\plus{}x^2(1\minus{}2x^2)\equal{}y^4\\1\plus{}\sqrt{1\plus{}(x\minus{}y)^2}\equal{}\minus{}x^2(x^4\plus{}1\minus{}2x^2\minus{}2xy^2)\end{cases}$

[quote="GioMott"][quote="zaya_yc"]$ (z \plus{} 1)(x \plus{} 1) \equal{} 14$[/quote]
No, it's =15, so 1848 it's not the correct number  :wink:[/quote]
$ why it 15 not 14 ??????? : roll:$

Just elution lapsue, still way resolve tense is true already  :)
To do an exercise of mine to go     :D


function
integration
calculus
real analysis
LaTeX
number theory
prime numbers

I'm also REALLY interested in it :)

Check this out : http://mathworld.wolfram.com/RiemannZetaFunction.html !

I could be wrong, but I think you're talking about a different function. I'm talking about the Riemann Zeta Function, which for s>1 is $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$.

And I've looked at the MathWorld website, but they don't show the steps to analytically continue the function as far as I remember.

[quote="blahblahblah"]There is also an 'intermediate' analytic continuation to $\text{Re}(s)>0$; it is an exercise in chapter $6$ of Rudin's Principles of Mathematical Analysis.[/quote]
Just use [url=http://mathworld.wolfram.com/SummationbyParts.html]partial summation[/url] for $\text{Re}(s)>1$:
\[ \zeta(s)=\sum_{n=1}^{N}{1*n^{-s}=N*N^{-s}+s*\int_{t=1}^{N}{t^{-s-1}(\sum_{n\leq t}{1})dt} }\]
Then for $N\rightarrow\infty$ and $\text{Re}(s)>1$
\[ \zeta(s)=s\int_{t=1}^{\infty}{t^{-s-1}\left[t\right]dt} \]
Write $t=\left[t\right]+\left\{t\right\}$, where $\left\{t\right\}\in\left[0,1\right)$.
It follows
\[ \zeta(s)=\frac{s}{s-1}-s\int_{t=1}^{\infty}{t^{-s-1}\left\{t\right\}dt} \]
The integral on the right side is convergent in $\text{Re}(s)>0$ and holomorph.
Therefore $\zeta(s)$ can continued analytically on $\text{Re}(s)>0,s\neq 1$.

Hyp

Indeed, and welcome to the forums.  That can also be looked upon as Lebesgue-Stieltjes integration.  By the way, you can use $\cdot$ (\cdot) or $\times$ (\times) to produce a nicer multiplication symbol in $\LaTeX$.

Do you want the globally convergent (entire complex plane except@z=1) continuation of $\zeta(z)$? It may be obtained by first doing the usual continuation of $\sum_{k=1}^{\infty}\frac{1}{k^{z}}$ which converges for Re[z]>1 to $\frac{1}{1-2^{1-z}}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^{z}}$ which converges for Re[z]>0. This can be further continued to $z\in \mathbb{C}\setminus\{1\}$ applying Euler's series transformation to the prior series to obtain

$\zeta (z) = \frac{1}{1-2^{1-z}}\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{cc}n\\k\end{array}\right)(k+1)^{-z}$

which converges $\forall z\in\mathbb{C}\setminus \left\{1\right\}$.

The other global analytic continuation of the zeta function proved by Helmut Hasse (c. 1930) is

$\zeta (z) = \frac{1}{1-z}\sum_{n=0}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{cc}n\\k\end{array}\right)(k+1)^{1-z}$

which converges $\forall z\in\mathbb{C}\setminus \left\{1\right\}$. This representation is better that the previous one in that it can be generalized to the Hurwitz zeta fcn as

$\zeta (z,w) = \frac{1}{1-z}\sum_{n=0}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{cc}n\\k\end{array}\right)(w+k)^{1-z}$

[quote="Jameson"]Can someone explain the steps it takes to extend this function to the entire complex plane or point me to a site that does?

Thanks[/quote]

Here's what I think are the quintessential references:

Ingham, "The Distribution of Prime Numbers"

Titchmarsh, E. C. "The Theory of the Riemann Zeta Function"

Edwards, H. M. "Riemann's Zeta Function"

These references go through the derivation of the following  analytic continuation of zeta to the entire complex plane (except s=1) as a contour integral function originally defined by Riemann:

$\zeta(s)=\Gamma(1-s)\frac{1}{2\pi i}\mathop\int_{\multimap}\frac{z^{s-1}}{e^{-z}-1}dz$

Here's a suggestion for anyone interested:

Split-out the real and imaginary parts of this expression and verify the Cauchy-Riemann Equations of the resulting integral functions (use Leibnitz rule).  Kinda' messy though.


To have a specific column width, you can use the [b] tabularx [/b] environnement in the package of the same name.  You use it the same way that [b]tabular[/b] but you have new possibilities, which is the point after all...  :P

\begin{tabularx}{\textwidth}{X r}
...
\end{tabularx}

The X is the rubber column which will vary to fill the given width.  You need to have one.  Also, in both[b] tabularx[/b] and [b]tabular[/b], you can use "p" to define the width of a specific column.

\begin{tabularx}{\textwidth}{p{7.5cm} X r}
...
\end{tabularx}

If you want to have more control, you can look up this post

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33660]www.artofproblemsolving.com/Forum/viewtopic.php?t=33660[/url]

ledurt

Your error is in the definition of the columns.  You have have 5 columns and you defined only 2.  Also, to have a specific width of column, don't forget to use "p" and you need to have a "X" column.

[b]\rotatebox{90}{[/b]
\begin{tabularx}[b]{\textheight}{p{1.5cm} p{3cm} p{3cm} p{3cm} X}[/b]\hline
\noindent\textbf{Criteria}&\textbf{0 Points}&\textbf{1 Point}&\textbf{2 Points} & \textbf{3 Points}\\\hline
Calendar&The website does not have any calendar&There is a calendar, but it has very little information&The website contains a calendar with a good amount of information for upcoming competitions&There is a calendar with many dates for competitions and tentative team meetings\\\hline
\end{tabularx}[b]}[/b]\medskip

I didn't see the whole table but from what I have, I figured it would be better in landscape so I added the [b]\rotatebox[/b] command and changed the width of the table.

Ledurt


Thinking about taking the Putnam for the first time this year. After checking out some old exams I have a few questions ...

- If the problem doesn't specifically ask for a proof (e.g., "Find a number ... ") is it sufficient to just give the right answer? Or do you have to rigorously prove your solution is correct? Or just give a reasonable justification?

- If you are using a known theorem/fact in a proof -- e.g., something like, if V1 and V2 are finite dimensional subspaces of a vector space V, then dim (V1 + V2) = dim V1 + dim V2 - dim (their intersection) -- I assume it's OK to just state it as fact. You wouldn't have to prove something like that, would you?

- How hard do they really grade? Suppose in proving a proposition you make all the right insights and observations but the proof is a bit handwavey in parts. Does that win significant partial credit or a score closer to a 0?

Thanks.

[quote="waite"]- If the problem doesn't specifically ask for a proof (e.g., "Find a number ... ") is it sufficient to just give the right answer? Or do you have to rigorously prove your solution is correct?[/quote]  No and yes.

[quote="waite"]- If you are using a known theorem/fact in a proof -- e.g., something like, if V1 and V2 are finite dimensional subspaces of a vector space V, then dim (V1 + V2) = dim V1 + dim V2 - dim (their intersection) -- I assume it's OK to just state it as fact. You wouldn't have to prove something like that, would you?[/quote]  Yes and no.  (Edit: Curse the English language for allowing constructions like your second question and thereby making my answer possibly ambiguous -- by "no" I mean that you are correct that you would not have to prove such a thing.)

[quote="waite"]- How hard do they really grade? Suppose in proving a proposition you make all the right insights and observations but the proof is a bit handwavey in parts. Does that win significant partial credit or a score closer to a 0?[/quote]  Many people have been known to over-predict their Putnam scores, sometimes by substantial amounts.  It's rather difficult to answer a hypothetical question like this, unfortunately.  Suffice it to say that you'll probably be better off answering one question thoroughly than answering three sketchily.  You might look through the older threads on the forum to see what other people have said about similar questions.

The grading is extremely (exceedingly) harsh -- far more than on olympiads, if you're familiar with that grading scale. Even a minor error such as forgetting to count some trivial subcase seems to yield a score of around 1 out of 10 (whereas on an olympiad it would probably merit 6 out of 7).


Outline a reasonable synthesis of 2-oxobutanal.

hmm, if this is butanal with a carbonyl in the 2-position (I think that's what oxo means), could one start with but-1-ene, apply catalytic osmium tetroxide and NMO, and then an oxidizing agent such as Collins' reagent ?

That is correct. However, I made a mistake: I wanted to write $ 3\minus{}oxobutanal$.

1-methylcyclopropene acted on by ozone and dimethyl sulfide?

apparently 1-methylcyclopropene can in turn be synthesized from methallylchloride and phenyllithium


geometry
circumcircle
inradius
inequalities
inequalities proposed

[quote="skywalker"]Given triangle $ABC$ with side $a,b,c$ circumradius $R$, inradius $r$ show that $a^{4}+b^{4}+c^{4}\le 32R^{4}-80r^{4}$ and $32,-80$ are the best constant.[/quote]
Easy ! We need to prove that :

$(sin A)^{4}+(sin B)^{4}+(sin C)^{4}+1280[(sin \frac{A}{2})(sin \frac{B}{2})(sin \frac{C}{2})]^{4}\leq\ 2$

Not easy  :mad:  !!!!!!!!!!!!!!

Ok , after replace $R=\frac{abc}{4S}$ and $r=\frac{S}{p}$ Put $a=x+y$ $b=y+z$ $c=z+x$ Expand all , We are done by Muirhead's Theorem ! ( Very stupid solution  :blush: , but easy with computer ! )

Ok an ideal  :D ,I think with your ideal I can do every inequalties for triangle  :D, now we need to prove an inequality in triangle $f(\triangle ABC)\ge 0$ we easily turn it to three side $f(a,b,c)\ge 0$ now by Ravi subtitude we need to prove $f(y+z,z+x,x+y)\ge 0\forall x,y,z>0$ ahhhh now it is really very easy ...by computer (as you said)  :D !!!
Ok can you post deltaily your solution ? or Do anybody have another solution ?  :)

[quote="skywalker"]Ok can you post deltaily your solution ?   :)[/quote]
Do you have computer ?   :lol:
Another idea : After changing the inequality in the form of $(x,y,z)$ , Assume that $z=min(x,y,z)$ so that $\exists a,b \geq\ 0$
and $x=z+a$ and $y=z+b$
We need to prove that ( Oh , where is my computer ?   :lol:   )
$20480(a^{2}-ab+b^{2})z^{10}+............$
( try , It's really fun ! all of coefficient are nonnegative , $z$ is nonnegative , we are done  ! )

[quote="tanpham90"] Very stupid solution  :blush: [/quote]
Oh I agree with you  :D !,
Actually I don't have solution for this, I only have a solution for the general problem, but this especial case is nice hope anybody can post a nice solution!


linear algebra
matrix
linear algebra unsolved

Let $ A \in M_{n} \mathbb{(R)}$ such that :
   $ (A+A^{t}-2008I_{n})^{2009}=I_{n}$
$ I_{n}$ is identity matrix
$ A^{t}$denotes the transposed matrix A
Prove that :
 1) $ \det(A) \geq \frac {2009^{n}}{2^{n}}$
 2)$ \det(e^{A})=\sqrt {e^{2009n}}$
 :)

[hide="Hint"]Show that the symmetric matrix $ B \equal{} A \plus{} A'$ is a multiple of the unit matrix: $ B \equal{} 2009I_n$. For this purpose assume w.l.o.g. (?) that $ B$ is diagonal. Using $ A' \equal{} 2009I_n \minus{} A$ prove that the set of eigen values of $ A$ is preserved by the mapping $ \lambda\mapsto 2009 \minus{} \lambda$ and that $ A$ and $ A'$ commute: $ AA'\equal{}A'A$. Finding appropriate pairs of eigen values (each different from $ \frac {2009}{2}$) should lead you directly to the first claim on the determinants in terms of the eigen values of $ A$. The second claim drops out after some algebraic manipulation starting with $ \det(e^{A\plus{}A'})$.[/hide]


number theory
prime numbers
number theory proposed

Let pi(x) denote the number of prime numbers up to x.

Prove that pi(x) is always less than pi(x + x^1/2) for all x.

Here's a slightly different conjecture:

Prove that for any integer k, pi(x) is always less than pi(x + x^1/k) for sufficiently large x.

The first conjecture is a stronger version of [url=http://en.wikipedia.org/wiki/Legendre%27s_conjecture]Legendre's conjecture[/url], which is an open question.

Isn't the first conjecture is false:
Take x=7, then pi(x)=4 (since 2,3,5 and 7 are prime numbers) however 7 + 7^1/2=9.65 (aprrox), and pi(9.65)=4 as well.

Legrende's conjecture is similiar to saying that
pi(x) is always less than pi(x + 2x^(1/2) +1) for all x.

Adrica's conjecture: If Pn-1 and Pn are the (n-1)th and nth prime numbers, then
(Pn)^1/2 - (Pn-1)^1/2 is less than 1 for any integer n.

If we assume that Adrica's conjecture is true, can we use it to prove Legrendre's conjecture?

I think we can use Adrica's conjecture to prove the equivalent conjecture, that pi(x) is always less than pi(x + 2x^(1/2) +1) for any integer x.


logarithms
geometric sequence
arithmetic sequence
algebra proposed
algebra

[quote="kunny"]Suppose that  $ a,\ b,\ c$ be positive real numbers which are different from 1 and form geometric sequence in this order.

(1) Find $ (a,\ b,\ c)$ such that $ \log_a b,\ \log_b c,\ \log_c a$ form geometric sequence in this order.

(2) Find $ (a,\ b,\ c)$ such that $ \log_a b,\ \log_b c,\ \log_c a$ form arithmetic sequence in this order.[/quote]

Given $ ac\equal{}b^2$

(1) we get $ (\log b)^{3} \equal{} (\log c)^{3} \Rightarrow a\equal{}b\equal{}c\neq {1}.$

[quote="kunny"]Suppose that  $ a,\ b,\ c$ be positive real numbers which are different from 1 and form geometric sequence in this order.

(1) Find $ (a,\ b,\ c)$ such that $ \log_a b,\ \log_b c,\ \log_c a$ form geometric sequence in this order.

(2) Find $ (a,\ b,\ c)$ such that $ \log_a b,\ \log_b c,\ \log_c a$ form arithmetic sequence in this order.[/quote]

Solution (1):

$ \log_a b,\ \log_b c,\ \log_c a$ are in G.P. $ \implies(\log_b c)^2\equal{} \log_a b\cdot \log_c a \equal{}\log_c b\equal{}\frac {1}{\log_b c}$

$ \implies(\log_b c)^3\equal{} 1$

$ \implies\log_b c\equal{} 1$

$ \implies c\equal{} b$

$ a\in\mathbb R^{\plus{}}\minus{}\{1\}, \ c\equal{} b\ne 1$

earldbest's answer is almost correct.  The correct answer is $ (a,\ b,\ c)\equal{}(k,\ k,\ k)\ (k\neq 1\ and\  k>0).$
How about question (2)?

[quote="kunny"]Suppose that  $ a,\ b,\ c$ be positive real numbers which are different from 1 and form geometric sequence in this order.

(1) Find $ (a,\ b,\ c)$ such that $ \log_a b,\ \log_b c,\ \log_c a$ form geometric sequence in this order.

(2) Find $ (a,\ b,\ c)$ such that $ \log_a b,\ \log_b c,\ \log_c a$ form arithmetic sequence in this order.[/quote]

[color=blue]
Solution (2):

$ \log_a b,\ \log_b c,\ \log_c a$ are in arithmetic sequence

Let  $ \ \log_a b \equal{} x \minus{} y,\ \log_b c \equal{} x,\ \log_c a \equal{} x \plus{} y$ where $ \ x\ne\pm y$, and $ \ x\ne 0$

$ \implies b \equal{} a^{x \minus{} y},\ c \equal{} b^{x},\ a \equal{} c^{x \plus{} y}$

$ \implies a \equal{} c^{x \plus{} y} \equal{} (b^{x})^{x \plus{} y} \equal{} b^{x(x \plus{} y)} \equal{} (a^{x \minus{} y})^{x(x \plus{} y)} \equal{} a^{x(x^2 \minus{} y^2)}$

$ \implies {x(x^2 \minus{} y^2)} \equal{} 1\implies y^2 \equal{} x^2 \minus{} \frac {1}{x} \equal{} \frac {x^3 \minus{} 1}{x}$ ----(i)

$ \implies \frac {x^3 \minus{} 1}{x}\ge 0\implies \frac {(x \minus{} 1)(x^2 \plus{} x \plus{} 1)}{x}\ge 0 \implies x\in ( \minus{} \infty,0)\cup [1,\infty)$

$ \implies (a,\ b,\ c) \equal{} (a,\ a^{x \minus{} y},\ a^{x(x \minus{} y)})$

Now apply $ \ b^2 \equal{} ac$

$ \implies a^{2(x \minus{} y)} \equal{} a\cdot a^{x(x \minus{} y)} \equal{} a^{1 \plus{} x(x \minus{} y)}$

$ \implies 2(x \minus{} y) \equal{} 1 \plus{} x(x \minus{} y)$

$ \implies y \equal{} x \minus{} \frac {1}{2 \minus{} x}$

$ \implies y^2 \equal{} x^2 \plus{} \frac {1}{(2 \minus{} x)^2} \minus{} \frac {2x}{2 \minus{} x}$----(2)

from (1) and (2) we get

$ x^2 \minus{} \frac {1}{x} \equal{} x^2 \plus{} \frac {1}{(2 \minus{} x)^2} \minus{} \frac {2x}{2 \minus{} x}$

$ \implies 2x^3 \minus{} 3x^2 \minus{} 3x \plus{} 4 \equal{} 0$

$ \implies (x \minus{} 1)(2x^2 \minus{} x \minus{} 4) \equal{} 0$

$ \implies x \equal{} 1,\ \frac {1 \plus{} \sqrt {33}}{4},\ \frac {1 \minus{} \sqrt {33}}{4}$

$ \implies y \equal{} 1,\ \minus{} \frac {3}{2},\ \minus{} \frac {3}{2}$ respectively, but $ y$ not equal to $ x .$

$ \implies (a,\ b,\ c) \equal{} (a,\ a^{x \minus{} y},\ a^{x(x \minus{} y)})$

Where $ (x,\ y) \equiv \left(\frac {1\pm\sqrt {33}}{4}, \minus{} \frac {3}{2}\right)$ and  $ \  a\in\mathbb{R}^{\plus{}}\minus{}\{1\}$

Thx kunny, I overlooked the first line of the problem.Now I edited my solution. please check it. After going through the first line I got a shorter version the solution also but I already answered the problem so i just edit my old post without changing much.

[/color]

[quote="kunny"]earldbest's answer is almost correct.  The correct answer is $ (a,\ b,\ c) \equal{} (k,\ k,\ k)\ (k\neq 1\ and\ k > 0).$
How about question (2)?[/quote]

Kunny, For (1) I think $ \ a$ need not be equal to $ \ b$ and  $ \ c\ .$

Let $ \ b\equal{}c$

$ \implies  \log_a c,\ \log_c c,\ \log_c a$ are in G.P

As $ \log_a c,\ 1,\ \frac{1}{\log_a c}$ are in G.P

[quote="earldbest"][quote="makar"][quote="kunny"]earldbest's answer is almost correct.  The correct answer is $ (a,\ b,\ c) \equal{} (k,\ k,\ k)\ (k\neq 1\ and\ k > 0).$
How about question (2)?[/quote]

Kunny, For (1) I think $ \ a$ need not be equal to $ \ b$ and  $ \ c\ .$

Let $ \ b \equal{} c$

$ \implies \log_a c,\ \log_c c,\ \log_c a$ are in G.P

As $ \log_a c,\ 1,\ \frac {1}{\log_a c}$ are in G.P[/quote]

but if $ a\neq b,c$, then $ a,b,c$ will not be in geometric progression :D[/quote]

Sorry i forget to read first line of the problem   :oops:


For our answers, can we give an approximate answer if we cannot find an exact value for it?

You can, but you likely won't receive full credit.


I'm solving the problems in the AoPS Vol.2 right now, and it's really annoying.

All of the solutions seem to come out of nowhere, and I'm resorting to just reading solutions instead of trying to solve the problems.  Any tips for me or for other people regarding these problems?  Is the only strategy to just gain more experience and start to see stuff better?

Thanks.

Oh yeah, I'm new here.  :)

Edit: Ok, I'm getting a little bit frustrated.  I know that I have the potential to qualify for USAMO, and I know that I'm fairly decent at math in general...but I think I'm stuck in a rut right now.  Please give me hope and tell me that being able to "see" things really improves as you gain experience.  Would it be ok to just read solutions and analyze them (see what clues could lead to the insights, etc) until I'm confident?  Maybe I'm not a trial and error learner...this goes not only for functional equations but for geometry...

If you don't see how you could come to a solution, then you are wasting your time. You need to slow down and take time to figure out the motivation for the solution. You should not expect to be able to blitz through AoPS vol. 2 like you could have with AoPS vol. 1. I skipped some of the harder problems and came back to them later. I also get the impression that you have the idea that: going through the AoPS books will gurantee that you get to the USAMO. This is not really true. It all depends on how you approach the book. You should be trying to develop your ability to solve problems, not learn solutions to specific problems. If you are able to solve many of the problems in the AoPS books [i]on your own[/i], you should have a good shot at making the USAMO. In summary, you should slow down and take the time to solve things on your own. You will learn more by struggling. Note that there is a difference between struggling and simply making no progress. If you are making no progress, just come back to the problem later. If you still have no idea, maybe glance at the solution or something.


[quote="SuperMath"]Let $\large a;b;c > 0$.
$\large \sum \frac{a}{ \sqrt{a^2+bc}} \leq\ \sum \frac{1}{ \sqrt{2a}}$[/quote]
Certainly wrong:
$a=b=c=2$


[quote="razad"]Find the surface area and volume of a regular hexagonal prism, with a side length of 3.2, and a height of 9.[/quote]

[hide]The area of a hexagon with side 3.2 is $6\cdot \frac{3.2^{2}\cdot \sqrt{3}}{4}$.

To get the volume, muliply that by 9 to get the volume.

Muliply the first quataty by 2 and add $6\cdot 3.2\cdot 9$ to get the surface area.[/hide]


A book with 53 pages has its pages numbered from 1 to 53 on the top of each page, but it also has its pages renumbered in reverse, from 53 to 1, on the bottom of each page.  So, the first page has a 1 on top and a 53 on bottom.  For how many pages do both numbers on the page have the same units digit?

We have $ 1 \plus{} (10a \plus{} b) \equal{} 10a \plus{} (b \plus{} 1)$
$ 53 \minus{} (10a \plus{} b) \equal{} 10(5 \minus{} a) \plus{} (3 \minus{} b)$

Two case:

$ b \plus{} 1 \equal{} 3 \minus{} b\implies b \equal{} 2$

or $ b \plus{} 1 \equal{} 10 \plus{} (3 \minus{} b)\implies b \equal{} 6$ (because 3 - b could be negative)

So the possibilities are any numbers that ends in 3 or 7 in the given interval. I actually missed the second case on my first try, and then realized it... I think this solution is better than the official one, though.


quadratics
Galois Theory
algebra unsolved
algebra

[quote="thanhnam2902"][color=darkblue]Hi every One. Let $ f(x) \equal{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e$[size=150],[/size] $ a,b,c,d,e \in R$. Some times we can factor $ f(x) \equal{} (mx^2 \plus{} nx \plus{} p)(qx^2 \plus{} rx \plus{} s)$, $ m,n,p,q,r,s \in R$. But some times we can't factor it. So, When we can factor $ f(x)$?[/color][/quote]

We can factor f(x) if there are real roots of this equation: $ ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}e\equal{}0.$

That's not quite right; it can happen that both quadratic factors have only non-real roots.

As it happens, it is [b]always[/b] possible to factor a quartic equation into two real quadratics over $ \mathbb{R}$ - this is a restatement of the Fundamental Theorem of Algebra.  Whether those quadratics have "nice" coefficients is a question of [url=http://en.wikipedia.org/wiki/Galois_theory]Galois theory[/url].


[quote="Wikipedia users"]Tacoma-Pierce County has been named one of the most livable areas in the country.[8] Tacoma was also recently listed as the 19th most walkable city in the country.[9] In contrast, the city is also ranked as the most stressed-out city in the country in a 2004 survey.[10] However, in 2006, women's magazine Self named Tacoma the "Most Sexually Healthy City" in the United States.[11][/quote]
Huh. Interesting.

Hmm, do you intend to go stress out by going to Tacoma? Mathcamp is really quite a relaxing place in my mind.


let $P(x)=x^3+(m-5)x^2+(9-5m)x+6(m-1)=0$. What are the values of $m$ for which the ecuation has all the roots narural numbers ?

2 is a root, so we have to examin the roots of:

$P(x)=(x-2)(x^2+(m-3)x - 3(m-1))$

and it must be $m\in Z$ and $n^2=(m-3)^2 +12(m-1) = (m+3)^2 - 12$ with $n\in Z$ and so $m+3=\pm 4$ and so $m=1$ or $m=-7$


I'm interested in learning chinese online (and free). Can anyone here help me out? You can help personally, or by giving helpful links. I have found some, but none satisfactory enough.

Thanks.

Disregarding Pejin's post...

I doubt you'll find a substantial and beneficial program that will allow you to full understand chinese for free. The internet might offer a few out-of-class guides on how to learn chinese but I think that's the best you can do.

[quote="ZhangPeijin"]I do not want to help you. Please go away.[/quote]

That is not nice to say.

All I can think of is an online translator set to pin-yin AND simplified Han...

-prods ZhangPeijin- That's nice.  :o ---At least you said please.
You might try podcasts. I hear you can get to some nice programs, if you search iTunes.
I think there's CLO and Learn Chinese.
Their websites, respectively, are:
http://www.chineselearnonline.com
http://chinesepod.com/

Hm... what about : http://www.livemocha.com/
My dad wants me to learn Chinese...

EDIT: Haha wth... that was dumb meh

[quote="Cyrazeno"]My dad want's me to learn Chinese...[/quote]
First let's work on your English... (want's?)
sorry.. i just had to do that..

[quote="ZhangPeijin"]对不起.我觉得你们可以在网上找到一个好的中文字典.
但是我还不想帮你.你为什么不到一个中文学校学中文呢?[/quote]

Seriously, what the [censored]?

If you're not going to be helpful, don't even post.

Ok fine.
Still, look online, there's a lot of chinese learning texts.
Google is your friend.

[size=42]If everyone had the option of going to a chinese school, would they really come on here and start asking for help?[/size]Nevermind.

Anyways, there's always Rosetta Stone. It's a really good program, but pricey. I think you'd be better off with something free/lower cost.

i think this could help at least a bit:

http://www.internetpolyglot.com/

the best one by far is livemocha.
it really [i]is[/i] scintillating! :lol:

and since I can't read Zhang's posts without a translator, I don't think I can take offense  :)

thanks everyone - keep 'em sites rolling!