analytic geometry
combinatorics open
combinatorics

One of my friends asked me a question, actually i know the answer but just dont know how to tell it :D need some help:

Suppose we have a set of persons. If  we can find the similarity (as number) between each possible pairs of this set,then how can we show this numbers in graph? Is there any space in mathematics that we can use it for ploting the  similarity between each pair. Based on my imagination, the  possible $ X$ and $ Y$   axes are the  set of persons.

What do you mean by "similarity"?  Do you just mean some arbitrary metric on people?

[quote="t0rajir0u"] Do you just mean some arbitrary metric on people?[/quote]

somethin' like that!!!

If the metric is the distance in some coordinate system (say, if we measure along various dimensions - age, height, political orientation), then plotting in that coordinate system is straightforward.  Alternately, a [url=http://mathworld.wolfram.com/WeightedGraph.html]weighted graph[/url] serves the same purpose but seems to have less applications (to me).

Note that measuring people using many dimensions we run into the [url=http://en.wikipedia.org/wiki/Curse_of_dimensionality]curse of dimensionality[/url].

Well, are the similarities are always computable ?

We can have lot of other constructs.

Suppose a few like coffee, few like tea and a few like fruit juice. We create three sets $ S_c,S_t,S_f$ and a person can belong to any set. We have sieve methods to solve problems of this kind.

Suppose we are looking for people who have crush on some one else. We define a ordered pair ($ a,b)$ if a has a crush on $ b$. For every $ b$ we can make a list of all the $ a$'s (they are similar in sense that they have crush on the same person)

Generalizing it, we can define ordered pairs of k-tuples of people. I feel that we have all these structures and more .. simple graphs, matchings, hypergraphs ...

Either we have a well known structure to solve a concrete problem or we need to come up with them as the problem inspires.


geometry Support inequalities ratio videos

In the topic about whether or not the usa empire will fall, i was wondering, what is considered the greatest non us power nowadays?
Now I am not implying whether or not usa is the most powerful country, so if you think it is, give me your number two, if you don't give me number one

Please elaborate after doing the poll
because the poll can never give all countries
i just gave a few that have a high chance of including your choice

Some of them have great historic influence, others have high population and have been chosen for that

By the word "power", do you mean which country has the greatest potential to be a superpower?  Or which country is strongest militarily?  Or economically?

It is common knowledge that in the next 50 years China will become the world's new super power, along with India if it ever manages to have its industrial revolution.

[quote="G-UNIT"]It is common knowledge that in the next 50 years China will become the world's new super power.[/quote]I really really do not belive this :)

I mean how many members does the Bush family have? Not enough to make presidency for 50 years I tell ya :D

Well i meant in general superpower, mix of economically, military..

I do mean : today, nut in the near future

The question is slightly ill-phrased since the US is the only nation satisfying the conventional definition of "super power" at the moment.

But as for what non-US nation is most powerful [i]today[/i], I think it is Japan.  They have not been using their power in aggressive ways; indeed, they barely impose any demands on other nations at all.  They don't even want immigrants; [url=http://www.economist.com/world/asia/displaystory.cfm?story_id=5323427&no_na_tran=1]they're happier building robots to take their place[/url].  However, they clearly have the capacity to do more than any other non-US nation right now if they set their mind to it.  For example, while Japan possesses no nuclear weapons, they could assemble one in four weeks if they needed to.

In a few decades this might change to China or India.  But there are a lot of things that could go wrong with either nation (especially given the legacy of communism in China).

Why did you put Pakistan on that list?  Because it's a nuclear power?  Besides for that fact, I think there are several nations more powerful than Pakistan not on that list...

Well Pakistan is a nuclear power and it has got a huge population
Some people, i hope they voted seriously, actually chose pakistan
There is still the other option but i admit that is not a very good alternative :blush:
nothing however stops you from elaborating in a reply

I always wondered how japan managed to win against china, russia ,korea,.. especially since china has got a lot more people.

My uncle keeps telling me not to underestimate the chinese, also economically
when i ask him, much of the stuff they sell, like technology (like obscure brands of mp3 players) aren't that good (my sister had experiences with that), he tells me (dunno if it is true) japan went through the same thing : toyota at first wasn't much quality, and now it is top notch I hear

Are japanese essentially different than chinese? or are chinese catching up?

Here we have the cliché of japanese that they are hard working, intelligent, serious people, who set their minds to it when they get started, and that they greet the flague of their company each morning

Maybe i should have included  israel and turkey , iran, thailand , vietnam and philippines as well (judging from population and nuclear power), but hey...  I couldn't copy the entire list of nations on earth

[quote="fredbel6"]I always wondered how japan managed to win against china, russia ,korea,.. especially since china has got a lot more people.[/quote]

When they decided to modernize in the late 19th century, they did it more efficiently than any other nation on Earth.

This extremely impressive accomplishment is still worth studying today.

[quote="fredbel6"]My uncle keeps telling me not to underestimate the chinese, also economically
when i ask him, much of the stuff they sell, like technology (like obscure brands of mp3 players) aren't that good (my sister had experiences with that), he tells me (dunno if it is true) japan went through the same thing : toyota at first wasn't much quality, and now it is top notch I hear[/quote]

The Toyota anecdote is accurate.

[quote="fredbel6"]Are japanese essentially different than chinese? or are chinese catching up?[/quote]

The genetic distance between Japanese and Chinese is roughly the same as that between Anglo-Saxons and Germans.  Certainly not large enough to expect the Chinese to lack the potential to catch up, given their 10x population advantage.

The environmental/cultural difference between the nations was much greater until ~1980.  Deng Xiaoping makes my personal list of top 5 individuals in human history because he did so much to change this.  I hope China continues following in his footsteps.

[quote="fredbel6"]Maybe i should have included  israel and turkey , iran, thailand , vietnam and philippines as well (judging from population and nuclear power), but hey...  I couldn't copy the entire list of nations on earth[/quote]

Your list was fine.  I would have just listed China, Japan, Russia, and "Other".  (India would make the list if we were speaking of the year 2056.)  Maybe Saudi Arabia could be thrown in for thought provoking purposes.

Anyone who's read Michael Crichton's [i]Rising Sun[/i] will know that no nation stands a chance against Japan - that is, if the book's background information is entirely true. You never can tell with Crichton's books...

Michael Crichton does not have a reputation as a responsible provider of meaningful social analysis.

dog of justice, you certainly speak words that i recognise
especially if you consider that before their drastic revolution , they didn't have trains or anything
(belhium had trains forty years earlier than japan) (i'm a rurouni kenshin fan, sometimes a bit over thte top, but it still provides a lot of information about japan and the meiji revolution)

About Deng Xiaoping, I recently read an interview , i believe with robert kaplan, and there it was also claimed that deng xiaoping is underrated.  He actually opposed mao didn't he?
[img]http://en.wikipedia.org/wiki/Image:Destroy_liu-deng.jpg[/img]

Well anyway i am not surprised to see china on top in the poll, but to see only one person vote for russia i had not expected

I'll say that China and india really will take the world by storm within the next 50 years.[China's already begun]
  And India is very much well into it's Industrial revolution! What G-Unit just said is not quite the way things are going here.

The biggest non US super power......in my opinion...... is the European Union.

I would say that Japan [i]used to be[/i] one of the most powerful countries in the world, but we aren't as "powerful" as we used to be. Other Asian countries such as China and Korea are catching up with Japan; in fact, some of the recent economical/technological developments show that these countries have already started to have more power than Japan in some fields.

[quote="fredbel6"]Here we have the cliché of japanese that they are hard working, intelligent, serious people, who set their minds to it when they get started, and that they greet the flague of their company each morning [/quote] It is true that the Japanese worked hard, but that might not be the only reason. Take the economical growth in the recent 50 years (especially 1950s-70s) for example. It was helped by the benefits from the special procurement (for Korean War, Vietnam War, etc.) as well as the cheap price of oil at that time. It was not just about the "diligence" of the Japanese :|

I don't know what is going to happen in the future, but I'm sure that the power of eastern Asian countries will grow enormously anyway. China will become very powerful, and other Asian countries as well.  Now the Japanese need to work hard again :roll:

[quote="frt"]I would say that Japan [i]used to be[/i] one of the most powerful countries in the world, but we aren't as "powerful" as we used to be. Other Asian countries such as China and Korea are catching up with Japan; in fact, some of the recent economical/technological developments show that these countries have already started to have more power than Japan in some fields.[/quote]

Yes, in relative terms Japan was more powerful 20 years ago than it is now.  Nevertheless, after spending most of my winter break this year in Beijing and Shanghai, and having been in Japan before, I don't think China has caught up to it quite yet (even taking its much greater size into account), though my answer may change in 10 years.  And I think every other non-US country except for possibly Russia (because of its raw destructive power) is clearly behind Japan and/or China at the moment.

Clearly I am being massively outvoted, though.  I'd be interested in one of the China voters providing a justification of why they think China is already ahead of Japan.

Hmm i have a couple of notes about that

first of all what i meant when i started this poll, was biggest power on military, economical, etc level
but my main question was, after soviet union collapse, did another nation take its place
you see, when you look at democracy and freedom of speech, human rights, i am not so sure one could call the soviet union a super power (on the other hand, usa does not go along with everything about human rights, they do executions, the first bush election is still controversiel....)

what is eastern turkistan, is there a relation to turkmenistan? guess not?

How central is the government of china?  How autonomous are those provinces?  Do they have governors like the united states of america have?  And is it true that major politicans have roots in Beijing (Deng Xiaoping was not)

Then one final question : as I noted before, there is an issue about one of China's most famous projects, the Three Gorges Dam, that Taiwan might have claimed it has the capability of destroying that dam (with rockets?)  Are tensions really that high??

Do taiwanese consider all chinese their enemies or just the communist party?  Guess not seeing many still seek reunification

[quote="tokenadult"][quote="Valentin Vornicu"][quote="G-UNIT"]It is common knowledge that in the next 50 years China will become the world's new super power.[/quote]I really really do not belive this :) [/quote]

I do not believe that either. China doesn't even have free elections, a free press, or other attributes of truly powerful countries. I'd say it's rather doubtful that anything called "China" will even have effective territorial sovereignty over all the regions now combined under that name by fifty years from now. In particular, I expect to live to see a free Tibet and independent Eastern Turkistan, and Taiwan generally acknowledged to be a de jure independent country.[/quote]

It is true that China has severe limitations on civil liberties, but there is no reason why that would disqualify it as a superpower. Keep in mind that democracy in the modern sense did not come about until 300 years ago, and it was not a feature of many well-run states before then, including the Roman empire. Here in the U.S., we automatically assume that all countries should imitate our government, but this is just not always true. The best possible government depends on the current situation, and current governments will adapt to what the situation is.

But my point is that it really makes a country stronger to have good internal civil liberties. That makes the government of the country more aware of what is really going on, and it makes the citizenry more eager to fight to preserve the country. That's why I think smaller population countries with more civil liberty are generally stronger in world-politics power (as they often are economically also) than larger countries with less civil liberty.

China's future dominance is far from a foregone conclusion at this point.  It may look easy for them to just "continue on their current course", which is in fact, broadly, all they need to do.  Other East Asian nations have shown that authoritarianism is not actually incompatible with several decades of high economic growth.  But every China enthusiast should listen to tokenadult when he says there are A LOT of things that could go wrong along the way.

I see China as the big wild card in world history right now. It can be anything from a friendly competitor that drives the US and Japan to correct many of their inefficiencies, to the opponent in a new Cold War, to a Russia-style economic basket case.

But this is not something we have no control over.  Quite the contrary -- maintaining US economic and moral leadership will go a very long way to ensuring China turns out okay.

[url=http://www.unpo.org/member.php?arg=21]Eastern Turkestan[/url]

[url=http://www.uygur.org/]Eastern Turkistan Information Center[/url]

[quote="fredbel6"]interestingly, someone from china just called the eu the greatest superpower
i didn't list that because i meant nations
we aren't a single nation and we surely aren't going to become one very soon, we just tried to enforce a constitution about it
even before it was fully written there was debate as to whether it should contain christian references as europe is a base for christianity( while belgium is in the stage of removing christian crosses from court houses...) , end in many countries referenda about it failed

also, right now the eu  is expanding eastwards, this is going to be a setback i am afraid
people like valentin have said that themselves.  Here in belgium we feel the effect already of cheaper polish, etc.. employees, like truck drivers

as i said, we consider, economically, china as the super power in eu right now

on a military level, we don't have a european army...[/quote]

what about Eurocorps? :P

To answer the question: China.  It has a huge population and is moving in the right direction.  It is by no means a sure thing, though.  The EU is already about as powerful as the US economically, and I believe they still are moving towards more unity, even if it is going slowly.

I would say that currently, world powers are:

US>EU>China>Japan>Russia>India.

I don't actually know, of course.

capybaralet, eurocorps what?

Well economically, as i already mentionned sometimes, what about textile and car industry moving to china?  Factories are closed down and thousands of people fired as almost simultaneously several factories in china are announced.

[quote="shobber"][quote="LordoftheMorons"]If China didn't have such terrible leadership (In other words, their leaders are more concerned about maintaining their own power than the country's interests,) than they could easily become a super-power in a matter of years.[/quote]
You have no evidence supporting this kind of opinion, I believe.

[quote]US>EU>China>Japan>Russia>India. [/quote]
As is known to all, Japan>China.[/quote]

Please...About LordoftheMorons statement...this is completely false...do you know what Hu did in the years he got into power...

China like most other countries is a complex case...Its leader...esp. new leader must make a passive peaceful transition of power...ppl were actually afraid that jiang ze ming was not going to give up his seat as the commander of the army when Hu came out..."In other words, their leaders are more concerned about maintaining their own power than the country's interests"...this statement is meaningless...can you name one successful leader in the world that didn't view the preservation of their own power before others...since if they don't have the power they can't do no good so called "their ppl"...

China is improving in condition of civl rights and so call "government cares about its ppl"...before Hu...trapped miners, lacking bezene, hostages, and all this other stuff that relate to the ppl would never make it to TV...the national assembly would never be live...

I understand this is a forum and we r free to express opinions...but please don't give out rash and unsupportabled info...

Now about the real issue...about superpowers....China is definitely second...if and only if compared in all areas one by one with each of the other countries...(for example there can't be like a 5 on 1 war or an universal embargo)

China's economical reform have brought results...one of the few things haven't happened to the old communism countries of Soviet past...The economical improvements of china is increasing in double digits in GDA 3 and probably soon to be 4 years straight...something no country in the world had ever done...With the entry into the WTO and the "full membership deal"...have increased fabric trades in one quarter of what use to be 52 years...China is a country with high potentials of overtaking america...

Military wise...china is not as powerful as the Russia, EU, or the USA technologies, but i believe China can get the technologies from favorable deals from the EU...EU is not second nor it is third...I would probably place it 4th in world power after Russia...explain that one later...EU on the outside is a strong union but there all more concerns with the economical reforms going on inside EU....the first constitution failed to pass..some of the wealther countries are concerned of cheap "foreign labor" and the poorer countries all afraid of foreign monoplies...therefore there is no true union...

On the otherhand China provides the favorable economical breeding grounds for foreign business since after all 1.3 billion ppl is a good consumer amount...China is also buying Airbus, magnet rails, military equipments, etc from booming European companies...in the billions...China have money...a huge saving....of trillions that can be thrown in paper tactics...to get favorable deals...

China also have strong support in Africa for China owns most of the oil and natural gas reserves in Sudan (China is protecting Sudan from UN sanction with that no vote)...and other african countries...China have also invested in much money in building a dry dock in Pakistan to issure oil can be transported to pakistan and transported by land pipelines to China...this provent blockades in the Strait of Malacca and the chain of island including Taiwan in the pacific used by USA to stunt China's growth...China also is using its UN seat to help Iran which is going to be an interesting case as USA embargoes Iran and China moves in to secure much needed oil supply currently under american companies...

China had also successfully found a pest in the american backyard as 2005 was a political landslide in South America...as Venezula, Bolivia,Cuba, Chile, Brazil, Argentine,etc turned socialist and blocked Bush's plan to create a American trade organization...and pushed him to form only a mini version in Cafto...Central American free trade organization.

I stated earlier that Russia was strong than the EU...from the natural gas issue in the end of 2005 Russia had regained some national pride and a stricter control over ex-Soviet countries that r western leaning policies...With about half of the EU's energy source in Russia...Russia can easily crush the EU economically and a bitter winter that prevent direct attack...Russia is pretty strong...

As for Japan...it would be 5th for its strong economical achievements...yet it is still weak since its island landscape deprive it of a large population that would offset issues....but also that the island is resource deprived and without a successful import...it is doomed...as if a early blockade action against the tiny yet strong Japanese Navy...if successful...Japan is probably doomed...currently if a war does start you r not going to have 4 weeks to make that atomic bomb...you need it right away...for most likely wars r different now....

All the above are just my two cent....now to prep for AMC ....yawn...sleep...zzzzz

One thing that really scares Chinese social scientists about the future of China is the degree of regional economic inequality, and urban-rural economic inequality, which has increased greatly during the last two decades. The ratio of income inequality region to region now exceeds that found in Yugoslavia before Yugoslavia disintegrated.

http://www.mtholyoke.edu/courses/sgabriel/economics/china-essays/6.html

http://ideas.repec.org/a/aea/aecrev/v89y1999i2p306-310.html

http://www.gwu.edu/~econ270/Taejoon.html

http://ideas.repec.org/p/dgr/kubcen/200153.html

http://repositories.cdlib.org/are_ucb/977/

http://www.wider.unu.edu/publications/rps/rps2004/rp2004-051.pdf

http://www.zonaeuropa.com/20050621_2.htm

[quote="shobber"]
As is known to all, Japan>China.[/quote]

*sniff* i smell sarcasm

I personally think if US did fall (not that I am implying it would) then alot of countries would suddenly break out into civil war:
I suspect that many labrorers would revolt because of mass layoffs (due to lack of demand from US for China) and would cause extreme unrest so I don ot think China is stable enough to be the next Super power.

On the other hand: I expect North Korea will be the first to seize power;
If u have not seen the 60 minutes artile on the insgiht on North Korea: please do and see the video also and [u]I think [/u]it is pretty obvious that from what 60 minutes got that North Korea is ready to jump at every opportuity.
[url=http://www.cbsnews.com/stories/2006/01/13/60minutes/main1209741.shtml]www.cbsnews.com/stories/2006/01/13/60minutes/main1209741.shtml[/url]

Yet After this preliminary attack to seize power; I firmly believe Russia will reestablish the Soviet Union and came on top due to recent advances in the 20th century in spy sattelite technology.
Though I expect some form of alliance with Saudi Arabia .

Thus, to sum it up I believe Russia is still the next superpower but is weak without allied countries

I think you missed an important possibility here: when the U.S. "falls" (as it inevitably will... someday), it's not necessarily going to be a sudden, devastating catastrophe. The U.S. isn't just going to go "poof" one day and disappear. More likely, it would be a gradual decline that takes several stages. Undoubtedly, it will put the world somewhat off-balance, but I doubt there will be any "mass layoffs" the way you describe it. I still think China is probably the primary candidate for the next "superpower," but one can not discount countries like Japan and Russia. In response to tokenadult's post: China doesn't have the ethnic tensions or whatever that Yugoslavia had. While the inequality is still an issue, I wouldn't consider it a major one. We must remember that a nation doesn't have to meet all our ideals of living to be a strong one, even if meeting those ideals would be helpful. And I think China needs that economy boom to try to fix its problems in the long run.

[quote]In response to tokenadult's post: China doesn't have the ethnic tensions or whatever that Yugoslavia had.[/quote]

I think that is a very interesting open question. To be sure, most Tibetans and Uygurs are distinctly cool about the remaining part of China, and while they are small in number, their traditional territory makes up a LARGE part of the territory of China.

The post-Qing Dynasty rulers of China, whether the KMT or the CPC, have tried hard to promote a sense of Han nationalism that embraces all the different Sinitic-language-speaking peoples of China, but I regularly hear comments from Chinese people about Chinese people from different regions that make me wonder how readily China might disintegrate. It's a historical fact that China has had many periods of regional separatism (some dynastic periods are named after the numerous regional kingdoms that arose in those eras), so it wouldn't be a new phenomenon if, for instance, the Cantonese-speaking regions of current China decided they would be better off under their own national administration than under an administration run from Beijing. A lot depends, as you note, on how well the common people of China make economic progress (and, I might add, personal liberty progress) during the next several years. It would be better for most people in China to make a Hungary-style transition to multiparty democracy than a Romania-style transition, but it would be worse than either of those choices, and more weakening to China's position in the world, not to make the transition at all.

AFTER EDIT: My edit was just to fix a spelling error.

Tibet shouldn't be much a problem for China since during the transitional period of the past 50 years, most of tibet should have at least cooled down from the separation deal....

China have employed methods to slow or even prevent future separation radicals from happening like the Go Western program which on the outside is a wonderful program to sent just out of college people into the West, or where many so called other ethnic groups were "forced" into China  throughout history. The program offer many benefits for ppl who goes and millions have gone since this prrgram started during the late 1990s...which is one way to melt the ethnic  tension down through future cross marriages, cultural blending, etc....plus there is always the army...Saddly China had killed many of these resistance ethnic group down to a managable size during the 50s and 60s....

This is my guess since from history when Nixon visited China he told De Xiao Ping to open up more chances for the China to immigrate to other countries (trying to get radicals to absorb western idea and change stuff at home)...then De said how about i give you 100 million Chinese...The point was culture was on China's side...with 5000 years of it...and a sudden 100 million ppl whos kids will likely inter-marry and will soon cause the american culture to diffuse into the Chinese ones...

China have been conquered before - Mongols, parts by Japan....but honestly the Mongols failed to  culturally conquer China in their 200 years, but was instead deeply influenced by the Chinese which is a good cause to the creation of the Inner Mongolia Province of China...And the Japanese failed...1) due to american forces in the pacific...2)even without one...China could have outlasted the Japanese...3) Japan took a time where China was weak...There was a fake puppet emperor created by the Dowager who resistanted Western influence...the betrayal of general Yuan and his own "empire"...Mr. Sun with his western influenced "radicals"...local warlords who took the chance to gain power....bandit everywhere....8 western forces just owned China in the Opium War (twice)...Portugal pirates on the open sea...etc...that is not today...so China should be the next Superpower


Given that:
x1 + 4 x2 + 9 x3 + 16 x4 + 25 x5 + 36 x6 + 49 x7 = 1;
4 x1 + 9 x2 + 16 x3 + 25 x4 + 36 x5 + 49 x6 + 64 x7= 12;
9 x1 + 16 x2 + 25 x3 + 36 x4 + 49 x5 + 64 x6 + 81 x7= 123.
Find 16 x1 + 25 x2 + 36 x3 + 49 x4 + 64 x5 + 81 x6 + 100 x7.

I spent like an hour on this working and working out equations and got it and now i don't even remmember what all steps i used to do this :mad:  :mad: How do you know what to start with in this equation and how to proceed?

See #8 here: http://www.upal.ca/adeel/aime/1989.pdf

[hide="different approach"]notice that we're trying to get the last line from some combination of the other 3.
also notice that all the coefficients are squares.  thus, we have
$a(n-1)^2 + b(n)^2 + c(n+1)^2 = (n+2)^2$.
$an^2-2an+a + bn^2 + cn^2+2cn + c = n^2 + 4n + 4$
match coefficients
$n^2(a+b+c) = n^2 \Rightarrow a+b+c = 1$
$n(2c-2a) = 4n \Rightarrow c-a = 2$
$a + c = 4$
From these 3 equations, we get $c = 3, a = 1, b = -3$
so $3(123) - 3(12) + 1 = \boxed{334}$[/hide]

This problem was also discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57745]here[/url].

My solution is fairly similar to alan's.


Find all funtions F from N to N such that
[tex]F(F(n))=n^2[/tex]

Dear Peter,
Sorry but I don't think that function is good.I assume that you said:
[tex] F(a_{n}^{2^m})=b_{n}^{2^m} \forall m \in N [/tex]
So  take $ m=1 $,we have:
$ f(a_{n}^2)=b_{n}^2 $
Take $a_{n}^2=b_{m}$ ,we have $F(b_m)=a_{m}^2$
Hence $b_{n}=a_{m}$ :)

[quote="Master"]Take $a_{n}^2=b_{m}$[/quote]

We cannot do that since we are told that $a$ and $b$ are both sequences that do not contain any perfect squares.

The proof given is quite nice, but is missing a small indicator that F(1)=1 to provide completeness for the way F is defined.

Example: $F(F(4))=F(F(2^2))$
If 2 is in $a$, then $F(2)$ is some element in $b$ such that $F(2)=b_n$ and $F(2^2)=b_n^2$ and $F(b_n^2)=2^{(2^2)}=16$

yeah, 1 is in neither of the sequences since 1 is a square but is not some power of a number which is not a square, so i didn't define it, that's true. we must have $F(1)=1$ since $F(1)=F(F(F(1)))=F(1)^2$. i assume 0 is not a natural number, anyway that case isn't hard to handle either.

Peter

[quote="Peter Scholze"]there are many such functions and they are the following:
divide the set of natural numbers which are not a perfect squares into two two sequences $A={a_1,a_2,...,a_n,...}$ and $B={b_1,b_2,...,b_n,...}$ such that the intersection of A and B is empty and the union of A and B is the whole set of natural numbers which are not a perfect square. then, set $F(a_n^{2^m})=b_n^{2^m}$ and $F(b_n^{2^m})=a_n^{2^{m+1}}$. it is easy to see that all these function satisfy the equation and that every function satisfying the equation is of that form.

Peter[/quote]
sorry i don't read it carefully


solve:
$ a\plus{}b\plus{}c\minus{}ab\equal{}0$
$ a,b,c\in N$

equivalent transformation yields: $ c\equal{}(a\minus{}1)(b\minus{}1)\minus{}1$ ie. the solution space for the tripel $ (a,b,c)$

over $ \mathbb{N}\setminus\{0\}$ is given by $ \{(u,v,(u\minus{}1)(v\minus{}1)\minus{}1)\,|\,1<u,v\in\mathbb{N},4<u\plus{}v\}$ and
over $ \mathbb{N}$ is given by $ \{(u,v,(u\minus{}1)(v\minus{}1)\minus{}1)\,|\,1<u,v\in\mathbb{N}\}$

P.S. I assume $ 0\in\mathbb{N}$

SOLUTION: $ (a; b; c) \equal{} (a; b; ab \minus{} a \minus{} b); \ \ a,b\in \mathbb{N}^*, \ \ a,b \ge 2$.

[quote="jarno"]Solve $ a \plus{} b \plus{} c \minus{} ab\boxed c \equal{} 0$, where $ a,b,c\in N$[/quote]
I think that above is correctly, i.e. $ a\plus{}b\plus{}c\equal{}abc\ !$

[quote="jarno"][quote="Virgil Nicula"][quote="jarno"]Solve $ a \plus{} b \plus{} c \minus{} ab\boxed c \equal{} 0$, where $ a,b,c\in N$[/quote]
I think that above is correctly, i.e. $ a \plus{} b \plus{} c \equal{} abc\ !$[/quote]

Yeap. I overlook one letter.[/quote]
Sorry but $ (abc)!$ or $ ab(c!)$ but it really quite easy for two case ? What is your problem?

It should be:
$ a\plus{}b\plus{}c\minus{}abc\equal{}0$


How many different 5-digit zip codes are possible if the first digit must be even and the fifth digit cannot be zero?

Shouldn't the first digit technically be only 2, 4, 6, or 8, because zero is nothing; it is neither even nor odd. Because of that, my answer was 36,000.

Um you do realize he will never check that cause he posted that 9 years ago right?
 :oops_sign:


rotation
Asymptote
geometry
3D geometry
tetrahedron
congruent triangles

[quote="aanjohri"]Here are some good questions my math teacher gave us as optional homework a few months ago:

1. What is the largest number of regions into which you can divide a circle with one line? Two lines? Three lines? Four lines? Describe a pattern.

2. Find at least three ways to arrange six congruent toothpicks to obtain four congruent triangles.

3. Where on earch could a man leave his house, then walk 3 miles north, 3 miles west, 3 miles south, and find himslef back home, at the same point where he started (this is apparently a famous problem).

4. Define a Pythagorean triple.

5. Draw a regular pentagon and its five diagonals. How many triangles are formed (include overlappping triangles)?[/quote]

1.  1 Line - 2 regions, 2 Lines - 4 regions, 3 Lines - 7 regions, 4 Lines - 11 regions;
Formula:
$ \frac {n(n \plus{} 1)}{2} \plus{} 1$,   n - number of lines.

1.[u]
region--|1 |2 |4 |7 |11|16[/u]
lines ---|0 |1 |2 |3 |4  |5

the regions are all one more than the sum of consecutive numbers:
$ \frac{n^2\plus{}n}{2}$

2.one way is to make a 1x1x1 triangle with three tootpicks and use the remaining three to conect each of the tree corners to the middle of the opposit side

another way is to start with a 1x1x1 triangle and put a toothpick half way up, rotate it so the next corner is up and put another toothpick halfway, and rotate it to the last corner and put another toothpick half way up

the third way is to make a star by overlapping two traingles when one is rotated 60 degrees, the tips are the congruent triangles

3. one spot is the south pole

also if there was a circle on the north pole that had a circumference of 3mi then he could live 3mi south of any point on the circle

4.a pythagorean triple is any set of inteegers (a,b,c) that satisfy $ a^2\plus{}b^2\equal{}c^2$

number 3:
or, in general if there is any circle parallel to the equator and north of the equator with circumference 3/n km where n is a natural number, then he could be anywhere on THAT circle

number 4:
to be more general, a set of 3 numbers that can be the sides of a right triangle
(yes this implies a^2+b^2=c^2)

number 5: ah a very famous problem
see attachment

ugh i should learn asymptote
ignore the inaccuracy
5+5+10+5+5+5=35

#2: anotehr possibility (mightve been said)

tetrahedron

here are some planar ones that im sure will be easier to understand

the angles ARE NOT ACCURATE the triangles will be congruent ief the lengths and angles are accurate

for questions number 2, all the triangles must be congruent to each other

Which makes tetrahedron correct. (Funny how most people's minds are stuck in 2D with these types of problems...)

And for number three, here are all the possible places:

a) South Pole

b) Any point 3 miles south of the line of latitude with a circumference of 3 mi.

c) Any point 3 miles south of the North Pole (This one has a few technicalities. First, when he travels west, he won't go anywhere, and Second, he would have to walk south on the same longitude as he walked north. But the question said "could"... :wink: )

In case you don't want to follow the link...

A Pythagorean Triple is an ordered pair (a,b,c) such that $ a^2\plus{}b^2\equal{}c^2$.

Also, if (a,b,c) is a Pythagorean Triple, then so is (na,nb,nc).
It's therefore always useful to memorize some of the smaller Triples like (3,4,5) and (5,12,13), because you can multiply each number by n to get a larger Pythagorean Triple. (A lot of contest problems have this type of shortcut, so be on the lookout!)

And don't forget your 45-45-90 and 30-60-90 Triples: $ (1,1,\sqrt{2})$ and $ (1,\sqrt{3},2)$, respectively.

the Pythagorean Triple resembles the units of the base of a triangle (squared) plus the units of the side of the triangle (squared) is equal to the hypotinuse *can't spell  :oops_sign:* (squared)

[quote="aanjohri"]4. Define a Pythagorean triple.[/quote]
Could you be more specific? Do you mean, what is a pythagorean triple (sounds too easy), or what is the general formula for pythagorean triples (sounds too hard, involves diophantine equations and the such)?

I think it's the first one, because he only said to define it. In other words, read the 3 posts before yours.

[quote="fishythefish"]he[/quote]
I thought aanjohri was a she. Anyway, you're not really saying anything, you're just repeating the question.

Yeah, I am a she.

By defining a Pythagorean triple, I mean explaining what it is. It is supposed to be a simple 1 sentence answer. (hope that helps!)

4. A set of numbers $ (a,b,c)$ such that the numbers are the corresponding lengths of sides in a right triangle.  And for this to be true, $ a^2 \plus{} b^2$ must equal $ c^2$.


In a parallelogram, draw a line segment connecting each vertex to the midpoint of the two non-adjacent sides, to construct a total of 8 segments. What fraction of the area of the parallelogram is the region bounded by these segments?

Affine it into a square.

Double count some ares, then subtract to get what you want.

Sorry if this is vague.  :P

rofler post a solution i dont understand.
also could u please define "affine"


A reaction is described by this:

Equal volumes of 0.1-molar sulfuric acid and 0.1-molar potassium hydroxide are mixed.

What would the net reaction equation be?

It would be:

$H^{+}+OH^{-}\rightarrow H_{2}O$

correct???

Both sulfuric acid and potassium hydroxide are strong, so they ionize (almost) completely...  and an acid-base neutralization reaction would occur...

That is correct.

You still have $HSO_{4}^{-}$ in solution though, since only sulfuric acid is a strong electrolyte, but that needn't be in the net equation as you posted.


geometry
parallelogram
geometry proposed

Let $ABCD$ be a convex quadrilateral, let $E$ and $F$ be the midpoints of the sides $AD$ and $BC$, respectively. The segment $CE$ meets $DF$ in $O$. Show that if the lines $AO$ and $BO$ divide the side $CD$ in 3 equal parts, then $ABCD$ is a parallelogram.

[quote="carlosbr"]Let $ABCD$ be a convex quadrilateral, let $E$ and $F$ be the midpoints of the sides $AD$ and $BO$, respectively. The segment $CE$ meets $DF$ in $O$. Show that if the lines $AO$ and $BO$ divide the side $CD$ in 3 equal parts, then $ABCD$ is a parallelogram.[/quote]

Hi, are you sure F is the midpoint of BO as you wrote?..i was just thinking F might be the midpoint of BC instead, i'm not sure i haven't tried the problem yet, but i was jus reading it :blush:

Let $T$ the intersection beetween $BO$ and $DF$. Now we know that $2[TDB]=[TBC]=2[TFB]$ and so $[TDB]=[TFB]$. But this implies $DO=OF$. Similary $OE=OC$ and so $DEFC$ is a parallelogramm because the diagonals bisects each other. But also $ABCD$ is a parallelogramm because $DC//EF//AB$.

[quote="Simo_the_Wolf"]Let $ T$ the intersection beetween $ BO$ and $ DF$. Now we know that $ 2[TDB] \equal{} [TBC]$ [/quote]
How do we know this? And yes, I think T is O
Anyways, we can also use Menelaus to show OD=OF and OE=OC, again showing DEFC is a parallelogram and thus so is ABCD.

I still don't quite how you can prove synthetically that $DO = OF$.

DO = OF due to OT is the midline of the triangle DFS, because T is midpoint of DS and OT $ \parallel $ FS
When T is BO $ \cap $ DC and S is AO $ \cap $ DC
Same strategy we can prove that EO = OC, so DEFC is parallelogram...


geometry
geometric transformation
reflection
homothety
ratio
trigonometry

Q) Let ABC be a triangle with AB = BC, and angle BAC = 30.  Let A' be the reflection of A in the line BC; B' the reflection of B in the line AC; C' the reflection of C in the line AB. Show that A', B', C' form the vertices of an equilateral triangle.

I drew the diagram.. and ended up having the three points collinear!!

Help...

[hide=Index][/hide]

Moreover the problem is from RMO 1998. So, cant expect the problem to be incorrect by any means!!

It's easy to prove that:
$ \Delta ABC\equiv \Delta ABC'\equiv \Delta AB'C\equiv \Delta A'BC$
$ \angle BAB'\equal{}\angle BAC\plus{}\angle B'AC\equal{}30^0\plus{}30^0\equal{}60^0$
moreover $ AB\equal{}AB'$
Hence $ \Delta ABB'$ equilateral
We shall prove that:
$ \Delta AB'C'\equiv \Delta BA'B'$ (*)
$ \angle C'AB'\equal{}30^0\plus{}30^0\plus{}30^0\equal{}90^0\equal{}$
$ 75^0\plus{}75^0\minus{}60^0\equal{}\angle A'BC\plus{}\angle ABC\minus{}\angle ABB'\equal{}\angle A'BB'\implies$
$ \angle C'AB'\equal{}\angle A'BB'$ (1)
$ A'B\equal{}BB'\equal{}AC'\equal{}AB'$ (2)
(1) and (2) proves (*)
Hence $ B'C'\equal{}A'B'$
Similarly it can be proven that:
$ B'C'\equal{}A'C'$
Thus $ \Delta A'B'C'$ equilateral.
QED

So the error is that in the sum its given the base angle to be 30 degrees, but rather the vertical angle should be equal to 30 degrees.
Thanx...

[quote="keshav_57"]So the error is that in the sum its given the base angle to be 30 degrees, but rather the vertical angle should be equal to 30 degrees. [/quote]
OH. I didn't notice that actually. anyways it's wrong with the base angle. the three reflections are not only collinear, but also two of the coincide :P .

Proof is good man.
But you can really improve the presentation.
I hope you go LEMMA by lemma and prove each of them for Geometry and write better statements.
That would make it really elegant though  it may not really be.
Sorry if I hurt you or something since you easily solved 5 INMO problems last year , me on the other hand, only 2 1/2.

[quote="Aravind Srinivas L"]Sorry if I hurt you or something[/quote]
Not at all. I need your advice. I'm trying to improve my proof-writing-presentation, actually. Thanks.
You are a real genius. clearing INMO in 8th  :omighty:  . Forget how many problems ppl does. You got through INMO, that matters. doing 5 problems doesn't.

One question: How do I find out solved problems, in a way a contestant would submit his paper in a contest?
In books(I don't have many books, maybe pre-college mathematics) there are problems solved, but that's very outline-style...
In this site there are loads of solved problems, but no one bothers to write things in that manner, only the outline, which I can get myself in many cases.

Referring to [url=http://www.mathlinks.ro/viewtopic.php?t=230664]this[/url] topic of yours, I'm sorry, I can't help you, I've never heard of this book.

Anyways, How did you get to know about me :huh:  :?:

IS aravind srinivas Ganesh Ajjaganande because i think the only 8ther who cleared the INMO last year was ganesh ajjaganande from karnataka. :rotfl:  :rotfl:  :rotfl:

Akashnil, i think i saw some of your solutions for last year's INMO paper.  :oops: For the most part you omitted "obvious steps" in the sense u went to the most difficult case in a problem without answering for the trivial cases. For example your combinatorics solution says it all.  :| Moreover for the homothety problem i think you should have showed that the corresponding sides are parallel and the ratios are equal , without directly writing that the 2 triangles are homothetic. :huh:  And yes u are right , the number of sums one does doesn't matter , the presentation matters. Hope u make it to the IMOTC this year. :trampoline:  :first:  :D

Akashnil! You didn't read my reply properly. I said I cud solve 2 problems. I didnot clear INMO as rightly said by befuddlers.

Yeah. See I don't Pre college mathematics offers much help in proof writing. But I attend Olympiad classes where my sir stresses on proof writing.

Actually I am not a genius and all man. You are 100 times better than me. I dont think I will qualify RMO this year. You did 6/6 problems in INMO in 9th. I won't even write INMO in 9th.
Can you tell me how you prepare for the olympiads on a daily basis as u must be in 10th now and the school pressurizing you for the boards?
Please. It will help me to take advice from a genius like you.

Well for proofwriting, you can go to AOPS "How to wite a Proof" section also.
As befuddlers said, your combi and homothety solution would have let u down in INMO.
But I am sure you will go to the camp.

Aravind i did not mean to take away the credit given to you by akashnil :rotfl:

u are a genius in ur own right
u cleared the NTSE with such ease
heard u are state topper in NTSE in tamilnadu

yes,he was TN topper in 8th NTSE

Welcome back arvind(after a long time)

I was unaware of fact that akashnil is in class 10  :)
genii are on AoPS(i couldn't solve,in fact draw the diagrama for ques 5 in INMO,just solved 2-3 ques  :oops: )

How do you ppl (befuddlers & Aravind) know about me outside AOPS, let alone seeing my solutions of last year's INMO :o

@befuddlers:
You are right, I actually enjoyed finding the Idea behind the solutions during the contest and neglected the specifics. :(
One more question: Do you not live in India or you generally stay up at 2:08 AM? :alien:

@Arvind:
Dunno how to manage the school pressure, it will always be there, and I was 45th in class IX and now i'm 27th in my class, so for you who might be regular toppers, I can't really help you much. But to tell you one thing, the more you grow up, it'll bother you further. So try to utilize your time now in 9th. I actually spent a month for olympiad problems before last RMO, and got totally black out of school studies(It was end of part I exams), finished the previous' years' RMO papers. Then After RMO I was 85% sure that I wouldn't make it, so concentrated on school studies. However I don't recommend this style, it's not a good idea to stop the school studies totally for some period and forget about previous lessons. Also remember that a peer group will help you greatly. It's a bliss you can attend Olympiad classes. I don't even know of one that I can avail now. If you want to keep touch in regular basis, get some problems and fix some resolutions, "I'll solve x problems in this week". You can find problems in the site's [url=http://www.mathlinks.ro/resources.php]Resources[/url] section. All the problems might not be appropriate, use some countries' regional problems like Russian. They have many levels for example [url=http://www.mathlinks.ro/resources.php?c=143&cid=61&year=2004]this[/url]. I know the link contains a huge surprise, I wish I had seen this first problem on the page before 20th Jan (joking only... I don't want to have unfair advantages even if offered). Not only this one. I've heard a few other problems of the last INMO were not original too. Don't expect things like this, but still these might help you through practice. However I don't find enough time to do these problems from the "Resources" section.

Anyways, I failed to find the "How to write a Proof" section. Could you give the link?

Akashnil i was just browsing through the olympiad section unsolved problems in combinatorics. Here is a link to write proofs in a methodical manner. :rotfl:
[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php[/url]
Hope this helps you. :rotfl:

http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php is the link which I wanted to give.
I don't know if befuddlers gave the same.
No harm befuddlers but how do you know I cleared NTSE?
@ Akashnil
Why don't you go for IIT JEE coaching in West Bengal?
Well, see I try my best to be the best in all three academics, iit coaching and olympiad but practically speaking , I prepare for olympiads in my house only before the exams. That's too bad!
Anyways, are you writing AMTI (NMTC) Maths Olympiad 2nd level?

Yes Aravind it is the same website i wanted to post. Anyways going by what you said you should devote more time to math olympiad i guess because it sort of helps you in IIT maths as well. But this is just a personal opinion , you can decide what u wanna concentrate on. By the way i just wanted to know where you are going for olympiad coaching and IIt coaching in TN.  :P

@ befuddlers
 I go to Mr. Sadagopan Rajesh for olympiad maths and Mr. KK Anand for IIT.

hey are u in tenth grade right now.....???
and in which place????

Thanks a lot to Arvind and Befuddlers for the link.

[quote="Aravind Srinivas L"]Why don't you go for IIT JEE coaching in West Bengal?[/quote]
Isn't it too early for that? i'm not sure, i'm only in tenth... never thought of that...

[quote="Aravind Srinivas L"]I prepare for olympiads in my house only before the exams. That's too bad![/quote]
Same case with me, I told you, last year I too prepared for olympiad since only a month before olympiads...

[quote="Aravind Srinivas L"]Anyways, are you writing AMTI (NMTC) Maths Olympiad 2nd level?[/quote]
Never heard of it... what's that?
I know about some AIMT a bit

[quote="anurags92"]hey are u in tenth grade right now.....???[/quote]
I believe so...

[quote="anurags92"]and in which place????[/quote]
place=?
Region: West Bengal
Institution: Ramkrishna Mission Boys' Home Rahara, high school.

Yeah cool. So there's another Indian tenth grader other than me here. Cool.

Well sorry for this late post.
I just saw the topic when I realised the problem could also be done by using cosine formula and getting the relations for A'B', B'C', C'A' and proving them that they are equal.


Let $ A\in M_{n}(\mathbb{C})$ such that $ A^{4}\equal{}A^{2}.$
Prove that if $ \mathrm{rank}\left(A^{2}\right)\equal{}\mathrm{rank}\left(A\right)$ so we have $ A\equal{}A^{3}.$

thanks!

[hide="hint"]
Note that $ A^4 \equal{} A^2$ implies $ A^2(A^2 \minus{} I) \equal{} 0$, here $ I$ denotes the identity matrix. Now, write $ A^2 \minus{} I \equal{} [{v}_{1}|{v}_{2}|{v}_{3}|\cdots|{v}_{n}]$, then we can claim that $ A^2{v}_{i} \equal{} 0$ for all $ i \in${$ 1,2,3,\cdots,n$}. Here, $ 0$ denotes the zero vector. Because $ rank(A^2) \equal{} rank(A)$, then $ dim(Ker(A)) \equal{} dim(Ker(A^2)$ by rank-nullity theorem. Because we know that $ Ker(A)$ is a subset of $ Ker(A^2)$, then by the property above (in the problem given), we can claim that $ ker(A) \equal{} ker(A^2)$, which means $ A{v}_{i} \equal{} 0$ for all $ i \in$ {$ 1,2,3,\cdots,n$}, here $ 0$ denotes the zero vector, then we can conclude that $ A(A^2 \minus{} I) \equal{} 0$ and the conclusion follows.
[/hide]
I am sorry if my answer is wrong.

Hi Qruni, your proof is correct.
We can say also: according to the theorem of decomposition of kernels  (TDK) $ ker(A^2)\bigoplus{k}er(A^2\minus{}I)\equal{}\mathbb{C}^n$ or $ ker(A)\bigoplus{k}er(A^2\minus{}I)\equal{}\mathbb{C}^n$. Again according the TDK,  $ A(A^2\minus{}I)\equal{}0$.


calculus
integration
function
vector
trigonometry
calculus computations

Hey

I wanted to check over if the way I solved a problem was correct

The question asked

Evaluate $ \int_{C}(x+y)^{2}dx-(x-y)^{2}dy$, where $ C$ is defined by $ y=\abs{2x}$ from $ (-1,2)$ to $ (1,2)$

What I did was to break up $ C$ into two parts, $ C_{1}$ & $ C_{2}$ with $ C_{1}$ being $ y=-2x$ and $ C_{2}$ being $ y=2x$. I redefined the two equations parametrically:
$ C_{1}: x=t-1, y=-2t-2$ and $ C_{2}: x=t, y=2t$ both having t values $ 0 \leq t \leq 1$

Then I plugged all that information in and my final answer was $ \frac{7}{3}$

Is what I did correct and if so, did I end up with the correct answer?

The other question I had confused me. It said

Evaluate $ \int_{C}y^{2}dx+2xy dy$, where $ C$ is the path along $ y=\sqrt{1-x^{2}}$ from $ (-1,0)$ to $ (1,0)$
using
a) the fundamental theorem of line integrals
b) the property of path independence

From what I know, using the fundamental theorem of line integrals means I find the potential function, f, where the conservative vector field is the gradient of f. Then I plug in the end values into f and evaluate.

However, I do not know what it means to use the property of path independence. If someone could explain the methods or point me in the direction so I can get an idea of what formulas to use, that'd be great

Thanks

To use path independence, find a nicer curve between the two endpoints, and calculate the integral around it.

By nicer curve, do you mean parametrically such as

$ x =-\cos (t)$
$ y = \sin (t)$

and evaluate from $ o \leq t \leq \pi$
?

Once I get the equation in parametric form, do I set that as $ r(t)$ and the initial function as $ F$ and evaluate $ \int_{C}F \cdot dr$?

No, that's the same curve. I mean a completely different curve, such as the straight line.

I don't get how my solution will be any different for part B?

The endpoints are the same and it is conservative so I know the answers are supposed to be the same, but am I just supposed to re-write part a?

Or is there another formula to use?


\[ pf(x)\plus{}qf(y)\plus{}rf(z)\plus{}(p\plus{}q\plus{}r)f( \displaystyle \frac{px\plus{}qy\plus{}rz}{p\plus{}q\plus{}r}) \geq (p\plus{}q)f( \displaystyle \frac{px\plus{}qy}{p\plus{}q})\plus{}(q\plus{}r)f( \displaystyle \frac{qy\plus{}rz}{q\plus{}r})\plus{}(r\plus{}p)f( \displaystyle \frac{rz\plus{}px}{r\plus{}p})\] where f is convex on the interval I, $ p,q,r>0$ ,si $ x,y,z$ are in the interval I.

This inequality is well known as Popoviciu's...a very very nice one :D .If f is strictly convex the inequality becomes equality if $ x\equal{}y\equal{}z$
We supose that $ x \geq y \geq z$.We have $ x \geq (x\plus{}y\plus{}z)/3 \geq z$ we have to compare $ y$ with $ (x\plus{}y\plus{}z)/3$.We have 2 cases. [b]CASE 1[/b] We obtain the inequality by aplying Karamata to $ (x,y,\frac{x\plus{}y\plus{}z}{3},\frac{x\plus{}y\plus{}z}{3}, \frac{x\plus{}y\plus{}z}{3},z)$ and $ (\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},\frac{x\plus{}z}{2},\frac{x\plus{}z}{2},\frac{y\plus{}z}{2},\frac{y\plus{}z}{2}).$
We can aply Karamata's because we have the same sum of the components like:
\[ x \geq \frac{x\plus{}y}{2} \Longleftrightarrow x\geq y\]
\[ x\plus{}y\geq\frac{x\plus{}y}{2}\plus{}\frac{x\plus{}y}{2} \Longleftrightarrow x\plus{}y \geq x\plus{}y\]
\[ x\plus{}y\plus{}\frac{x\plus{}y\plus{}z}{3} \geq \frac{x\plus{}y}{2} \plus{} \frac{x\plus{}y}{2} \plus{} \frac{x\plus{}z}{2} \Longleftrightarrow y\geq \frac{x\plus{}z}{2}\]
\[ x\plus{}y\plus{}\frac{x\plus{}y\plus{}z}{3} \plus{}\frac{x\plus{}y\plus{}z}{3} \geq  \frac{x\plus{}y}{2} \plus{} \frac{x\plus{}y}{2}\plus{} \frac{x\plus{}z}{2} \plus{}\frac{x\plus{}z}{2} \Longleftrightarrow y \geq {x\plus{}z}{2}\]
\[ x\plus{}y\plus{}\frac{x\plus{}y\plus{}z}{3}\plus{}\frac{x\plus{}y\plus{}z}{3}\plus{}\frac{x\plus{}y\plus{}z}{3} \geq \frac{x\plus{}y}{2} \plus{} \frac{x\plus{}y}{2}\plus{}\frac{x\plus{}z}{2}\plus{}\frac{x\plus{}z}{2}\plus{}\frac{x\plus{}z}{2} \Longleftrightarrow y\geq z\]
[b]CASE 2[/b] $ y<\equal{}(x\plus{}y\plus{}z)/3 \Longleftrightarrow y<\equal{}(x\plus{}z)/2.$We obtain Karamata's for $ (x,\frac{x\plus{}y\plus{}z}{3},\frac{x\plus{}y\plus{}z}{3},\frac{x\plus{}y\plus{}z}{3},y,z)$ and $ (\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},\frac{x\plus{}z}{2},\frac{x\plus{}z}{2},\frac{y\plus{}z}{2},\frac{y\plus{}z}{2})$ and the inequality can be aplied because the conditions are verified...


geometry
circumcircle
power of a point
radical axis
cyclic quadrilateral
geometry unsolved

Points $P$ and $Q$ on side $AB$ of a convex quadrilateral $ABCD$ are given such that $AP = BQ.$ The circumcircles of triangles $APD$ and $BQD$ meet again at $K$ and those of $APC$ and $BQC$ meet again at $L$. Show that the points $D,C,K,L$ lie on a circle.

Let:
$(APD)=o_{1}$
$(BPD)=o_{2}$
$(APC)=o_{3}$
$(BQC)=o_{4}$
Let $M$ be the point of the intersection of line $DK$ with line $AB$. Because $M$ lies on a radical axis of $o_{1}$ and $o_{2}$ it's easy to conclude from the given relation that $M$ is a midpoint of $AB$ (since $|MP|\cdot|MA|=|MQ|\cdot|MB|)$. Analogically, $M$ lies on a radical axis of $o_{3}$ and $o_{4}$, and it's obvious that it lies also on the radical axis of $o_{1}$ and $o_{3}$ and on the radical axis of $o_{2}$ and $o_{4}$. So it's clear that $M$ is a radical center of these $4$ circles and in particular we have:
$|MK|\cdot|MD|=|ML|\cdot|MC|$
which implies that $D, K, C, L$ is a cyclic quadrilateral (from the converse of the Power of the Point theorem).

We denote as $(O_{1},$ $(O_{2},$ $(O_{3},$ $(O_{4}),$ the circumcircles of the triangles $\bigtriangleup APD,$ $\bigtriangleup BQD,$ $\bigtriangleup APC,$ $\bigtriangleup BQC,$ respectively.

So, we have $K\equiv (O_{1})\cap (O_{2}),$ $L\equiv (O_{3})\cap (O_{4})$ and let $E,$ $F$ be, the intersection points of $(O_{2}),$ $(O_{3})$ $($ In my drawing the point $P,$ is between $A,$ $Q$ and $F,$ between $E$ and side line $AB$ of $ABCD$ $).$

Let $M$ be, the intersection point of $AB,$ from the segment line $DK,$ as the radical axis of $(O_{1}),$ $(O_{2}).$

So we have that $(MP)\cdot (MA) = (MK)\cdot (MD) = (MQ)\cdot (MB)$ $,(1)$

From $(1)$ we conclude that the point $M,$ lies on the segment line $EF,$ as the radical axis of the circles $(O_{2}),$ $(O_{3}).$

Let $M'$ be, the intersection point of $AB,$ from the segment line $CL,$ as the radical axis of $(O_{3}),$ $(O_{4}).$

So we have that $(M'P)\cdot (M'A) = (M'L)\cdot (M'C) = (M'Q)\cdot (M'B)$ $,(2)$

From $(2)$ we conclude that the point $M',$ lies also, on the segment line $EF.$

From $(1),$ $(2)$ $\Longrightarrow$ $M'\equiv M$ and that the segment lines $DK,$ $EF,$ $CL,$ are concurrent at one point, here the point $M,$ lies on the side line $AB.$

So, we have that $(MK)\cdot (MD) = (ML)\cdot (MC)$ $,(3)$

From $(3),$ we conclude that the points $C,$ $D,$ $K,$ $L,$ are concyclic and the proof is completed.

Kostas Vittas.

[quote="Umut Varolgunes"]more generally $AP=BQ$ isn't necessary.[/quote]
Yes!
Doing inversion to $A$ the radius isn't necessary for this problem hence the problem becomes like this [url=http://goo.gl/mc9cY][img]http://b1110.hizliresim.com/11/10/15/6349.jpg[/img][/url] and we know that $D',E',F',C'$ is concyclic because  $DC$ becomes a circle by inversion. Then we will show that $K'\in{B'L'}$ we know that $K'E'.K'D'=K'F'.K'C'$ because $E',D',C',F'$ are concyclic and $K'$ has equal power to circles and $BL$ is the radical axis of the circles so $K'\in{B'L'}$ line hence $K',B',L'$ is collinear so $A,B,K,L$ are concyclic.
[color=#FF0000][u][b]EDIT: [/b][/u][/color]
In the orginal problem $P$ is $E$, $Q$ is $F$, $A$ is $D$, $B$ is $C$


calculus
function
derivative
real analysis
real analysis unsolved

$ g'(x)=(xf'(x)-f(x))/x^{2}$. The function $ xf'(x)-f(x)$ is equal to $ 0$ at $ x=0$, and its derivative is $ xf''(x)$.

A more geometric argument: $ f$ is strictly convex, so it lies strictly below its secant lines. If we had $ a>b>0$ with $ \frac{f(a)}{a}\le \frac{f(b)}{b}$, then $ (b,f(b))$ would be above the line segment between $ (0,f(0))$ and $ (a,f(a))$, contradicting convexity.

Very Nice solutions!I will present to you and my solution;
[quote]Let $ f:$  $ [0,+\infty)\rightarrow R$  differentiable on the $ [0,+\infty)$.   $ f(0)=0$ and also $ f''(x)>0$ $ \forall x \in (0,+\infty)$.
Proove that $ g(x)=\frac{f(x)}{x}$ is  increasing on the $ (0,+\infty)$.[/quote]

As $ f''(x)>0$ $ \forall x \in(0+\infty)$ and $ f'$ is continuous on the $ [0+\infty)$ (because $ f$ is twice differentiable) , therefore
$ f'$ is increasing on the $ [0+\infty)$.We have ;

$ g'(x)=\frac{f'(x)x-f(x)}{x^{2}}$  $ \forall x \in(0+\infty)$

Applying the Mean Value theorem on $ f$ on the $ [0,x]$ we take that exists $ \xi \in(0,x)$ such as
$ f'(\xi)=\frac{f(x)-f(0)}{x-0}$$ \Longleftrightarrow$ $ f(x)=xf'(\xi)$

so $ g'(x)=\frac{f'(x)x-xf'(\xi)}{x^{2}}=\frac{f'(x)-f'(\xi)}{x}$. $ x>0$
Now it's enough to proove that $ g'(x)>0$ $ \forall x \in(0,+\infty)$ or  $ f'(x)-f'(\xi)>0$ $ \Longleftrightarrow$ $ f'(x)>f'(\xi)$ $ \Longleftrightarrow$  $ x>\xi$ (because $ f'$ is increasing) which is true :P

[quote="giannis"] therefore
$ f'$ is increasing on the $ [0+\infty)$
[/quote]

I don't understand how you realized that, function as it is convex but for the you can't easily conclude that is increasing.
Can you explain please?

the problem says that  $ f''(x)>0$ $ \forall x \in (0,+\infty)$ $ \Longrightarrow$ $ f'$ is increasing.

It's the same as $ f'(x)>0$ $ \Longrightarrow$ $ f$ is increasing. :)


(1)
How many ways are there to put $ 2n\minus{}2$ bishops on a $ n\times n$ chess field so no two can capture each other (that is, no two are on the same diagonal)?

(2)
Let $ n\ge 2$ and suppose you have $ n$ positive integers written on a blackboard. In one move, you may wipe out two integers and replace them both by their sum. For which $ n$ can this result in $ n$ equal positive integers?

1) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225452

2) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225457


How can the ketone $ RCOR'$ be converted into the ketone $ RR'CHCOCH_3$?

Maybe:

1. Ylide $ \ce{(Ph)3P\plus{}C^{\minus{}}HCH3}$

2. Hydroboration $ \ce{BH3}$ followed by $ \ce{H2O2/OH\minus{}}$

3. Oxidation $ \ce{CrO3}$ and $ \ce{H3O\plus{}}$

Yes, very good, that's one possible way. Can you propose other ways?


function
integration
real analysis
real analysis unsolved

Let $f$ be a real-valued function of bounded variation over $[0,1]$. Show that \[\sup_{0<h<1}\frac{1}{h}\int_{0}^{1-h}|f(x+h)-f(x)| dx < \infty\]

[quote="mlok"]Well, a function of bounded variation must be bounded. There is probably a mistake in the displayed formula.[/quote]
I've modified the post.

If $f$ is increasing, then \[\int_{0}^{1-h}|f(x+h)-f(x)|dx=\int_{1-h}^{1}f(x)dx-\int_{0}^{h}f(x)dx\le 2h\sup|f|\] A general function of bounded variation can be written as $f=f_{1}-f_{2}$, where $f_{1}$ and $f_{2}$ are increasing bounded functions. (Indeed, $f_{1}(x)=V_{0}^{x}f$ works.)


[b]Descripción[/b]

Se tienen dos pilas de fichas,  [i]A[/i] puede sacar cualquier cantidad de fichas de la primera o segunda pila, o puede sacar igual cantidad de fichas de ambas pilas, luego [i]B[/i] hace lo mismo. Así continuarán hasta q se queden sin fichas. Gana el único jugador que saca la última ficha.

[b]Ejemplo de jugada permitida[/b]

Se tienen dos pilas de fichas: 3 fichas en la primera y 5 en la segunda, [i]A[/i] puede sacar 1, 2 o 3 de la primera pila, 1, 2, 3, 4 o 5 de la segunda, o si quiere sacar de ambas pilas puede sacar 1, 2 o 3 de cada una.

[b]Por conseguir[/b]

Suponiendo que las posiciones son pares $(x_{i}, y_{i})$
Hallar las posiciones perdedoras.

Las primeras posiciones perdedoras para quien tenga el turno son (1,2) (3,5) (4,7) (6,10) ... ello nos da una idea de que las posiciones perdedoras forman una especie de sucesion que probablemente creo se demostraria por induccion  :oops:


$x=1,-1$

There are probably more but I was too lazy  :sleeping:

[quote="Palytoxin"]easy, we can apply the so-called AM-GM inequality (if you doubt what is AM-GM, please ask here). so assuming $x$ is positive and $x\neq 0$, we apply AM-GM and get
\[\frac1{x}+\frac1{x^3}+\frac1{x^5}+\frac1{x^7}+\frac1{x^9}+\frac1{x^{11}}+\frac1{x^{13}} \geq \frac7{x^7}\]
so the original inequality and this inequality can happen simultaneously iff we have
\[\frac1{x}+\frac1{x^3}+\frac1{x^5}+\frac1{x^7}+\frac1{x^9}+\frac1{x^{11}}+\frac1{x^{13}}=\frac7{x^7}\]
but AM-GM also says that equality is achieved when all terms are equal, and so we find that $x=1$ is one solution. now suppose $x<0$. then we multiple both sides of original inequality by -1 and reverse inequality:
\[\frac1{-x}+\frac1{-x^3}+\frac1{-x^5}+\frac1{-x^7}+\frac1{-x^9}+\frac1{-x^{11}}+\frac1{-x^{13}} \geq \frac7{-x^7}\]
now all the terms are [b]positive[/b] and so this is equivalent to the AM-GM, then all numbers $x<0$ satisfies this inequality.

solution is $x=1$ and $x<0$[/quote]

nope, solution is $x=1$ [b]OR[/b] $x<0$   :lol:

:blush: i can never quite understand logic  :blush:

I "know" what AM-GM is, I just don't see how you applied it.
I'm not experienced with it, like all I would know is

$\frac{1+x^2+x^4+x^6+x^8+x^{10}+x^{12}}{7}\ge{x^{42}}^{\frac{1}{7}}$

$\frac{1+x^2+x^4+x^6+x^8+x^{10}+x^{12}}{7}\ge x^6$

Hm.. Oh ok, I think I see where you got it..

Bleh, made so many small latex errors

[quote="Iversonfan2005"]Palytoxin, you are so good at proofs :omighty:[/quote]
ty but the thing that will really make me happy is if someday i can see a good proof from you Iversonfan2005, start doing proofs!!!  :wow:

if you got more questions about this problem, just ask me WindSlicer  :)

[quote="Jan"]I would like to learn AM-GM, does anybody know a good site where I can learn it? I imagine it will be quite a lot to type, so any links please?[/quote]

Jan, you can use the search that is on this site type AM-GM and there will be [b]a lot[/b] of search results just for you :)

Search is at top screen, under the big 7 icons, next to FAQ and Members, if you are using MathLinks layout...

[quote="WindSlicer"]Ok, I understand your proof of the equality part, but you lost me on the $x<0$. You stated because of the AM-GM, but that's kind of vague, so could you state more specifically why?

Thanks![/quote]
hmm let see how to start off...

ok let me go over step-by-step the second part

ok since we finished checking for $x>0$ now we need to check if there are solutions for $x<0$

now multiply -1 to both sides of this $x<0$ and you get $-x>0$

ok now back to the big original inequality:

\[\frac1{x}+\frac1{x^3}+\frac1{x^5}+\frac1{x^7}+\frac1{x^9}+\frac1{x^{11}}+\frac1{x^{13}} \leq \frac7{x^7}\]

we are going to do the same thing and multiply -1 to both sides of original inequality and get

\[\frac1{-x}+\frac1{-x^3}+\frac1{-x^5}+\frac1{-x^7}+\frac1{-x^9}+\frac1{-x^{11}}+\frac1{-x^{13}} \geq \frac7{-x^7}\]
name this inequality (*)

i hope you are following me here

ok, now we can rewrite (*) in this form

\[\frac1{(-x)}+\frac1{(-x)^3}+\frac1{(-x)^5}+\frac1{(-x)^7}+\frac1{(-x)^9}+\frac1{(-x)^{11}}+\frac1{(-x)^{13}} \geq \frac7{(-x)^7}\]

because $-x^\text{odd power}=(-x)^\text{odd power}$. call this new inequality (**)

but, now we have all the terms in (**) [b]positive[/b]

now here is where we bring in AM-GM. ok you know that for a seven variables AM-GM $a,b,c,d,e,f,g$ all [b]positive numbers[/b], then

\[a+b+c+d+e+f+g\geq 7\sqrt[7]{abcdefg}\]

this inequality is satisfied by all positive numbers that you can select for $a,b,c,d,e,f,g$

so if you let
\[a=\frac1{(-x)},b=\frac1{(-x)^3},c=\frac1{(-x)^5},d=\frac1{(-x)^7},\]
\[e=\frac1{(-x)^9},f=\frac1{(-x)^{11}},g=\frac1{(-x)^{13}}\]
 then
\[7\sqrt[7]{abcdefg}=7\sqrt[7]{\frac1{(-x)^{1+3+5+7+9+11+13}}}\]
\[=7\sqrt[7]{\frac1{(-x)^{49}}}=\frac7{(-x)^7}\]
so inequality (**) is equivalent to AM-GM!!!

time to make a comparison:
(**)
\[\frac1{(-x)}+\frac1{(-x)^3}+\frac1{(-x)^5}+\frac1{(-x)^7}+\frac1{(-x)^9}+\frac1{(-x)^{11}}+\frac1{(-x)^{13}} \geq \frac7{(-x)^7}\]
AM-GM 7 variables
\[a+b+c+d+e+f+g\geq 7\sqrt[7]{abcdefg}\]

they have the same form!
since all the conditions for AM-GM are satisfied and since there are no restrictions on $x$ other than $x<0$ then inequality is satisfied for all $x<0$. hope this helps (what a long post! im going to be yetti's apprentice for long posts :D )  :thumbup:  :pilot:

Ah, great post!

I completely understand now. Thanks for your time.

[quote="chess64"][quote="Palytoxin"]but, now we have all the terms in (**) [b]positive[/b][/quote]

:huuh: they look negative to me  :?[/quote]

thats because you are skimming thru my post, you have to pay attention to details

let me quote myself:
[quote="Palytoxin"]now multiply -1 to both sides of this $x<0$ and you get $-x>0$[/quote]

ahh... I didn't think carefully about $x$ negative  :wallbash:

haha i like kunny's proof.. very original


Hi,
This question was on a comprehensive exam last year.
Show that the meromorphic function [e^(1/z) + (1/(1- e^(1/z))] on C\{o} does not omit any value at z=0.

$w=e^{1/z}$ attains all nonzero values in any neighborhood of $z=0$. It remains to analyze the map $g(w)=w+1/(1-w)$. Being rational, $g$ is onto $\mathbb C\to \mathbb C$. So we only need to check that $g(0)=1$ is covered. And it is: $g(2)=1$.

The only way to prove the first statement (without working from elementary principles--which could take a while) is to invoke Big Picard, correct?  Is there a way to prove this without big Picard?

Big Picard may not be necessary when you have an explicit function. Any nonzero complex number can be written as $re^{i t}$ with $r>0$ and $t\in \mathbb R$. We have $e^{1/z_{k}}=re^{i t}$, where $z_{k}=\frac{1}{\log r+i(t+2\pi k)}$, where $k\in\mathbb Z$ is arbitrary. As $k\to\infty$, $z_{k}\to 0$.


trigonometry
invariant
complex numbers
number theory proposed
number theory

prove that if $n\in N$ isn't a power of a prime number then there is permutation $(i_1,i_2,...,i_n)$ of 1,...,n s.t :$\sum_{k=1}^{n}k.\cos (\frac{2\pi i_k}{n})=0$

dear sam-n
i remember trying to solve this problem  quite sum time back and dint make any headway
mind giving me a small hint (as a private messges probably ) and let the others give it a try
thanx

I will prove more: let $\zeta= e^{\frac{2 \pi i}{n}}$, then we want to show that $\sum_{k=1}^n \zeta^{i_k}=0$ (by just looking at the real part, this solves the problem).

Let $n=pq$ with coprime $pq>1$ (exist since $n$ is not power of a prime).
Lets set $\xi_{s,t} = \sum_{k=1}^q (pk+t) \zeta^{pk+s} = \sum_{k=1}^q pk \zeta^{pk+s}$, where the last equality comes from $\sum_{k=1}^q t \zeta^{pk+s} = t \cdot \zeta^{s-p} \cdot \sum_{k=0}^{q-1} \zeta^{pk} = t \cdot \zeta^{s-p} \cdot \frac{\zeta^{pq}-1}{\zeta^{p}-1}=0$ (here we need that $p>1$).
As we have seen, $\xi_s : = \xi_{s,t}$ is invariant under changing of $t$. Now $\zeta^{lq} \xi_s = \xi_{s+lq}$. Thus $\sum_{l=1}^p \xi_{s+lq} = \xi_s \cdot \sum_{l=1}^p \zeta^{lq} = 0$ like above (now $q>1$ is needed).
Now we know enough about these sums to conclude:
$0= \sum_{l=1}^p \xi_{lq} = \sum_{l=1}^p \sum_{k=1}^q (kp+(l-p)) \zeta^{lq+kp}$.
Note that every value from $1$ to $n$ is archieved exactly once by $kp+(l-p)= (k-1)p+l$. Additionally, the value of $lq+kp$ gets every value $\mod n$ exactly once by the chinese remainder theorem. Thus we get that
\[ \sum_{l=1}^p \sum_{k=1}^q (kp+(l-p)) \zeta^{lq+kp \mod n}=0 \]
is a sum of the desired type.


function
limit
geometry
rectangle
analytic geometry
real analysis
real analysis unsolved

Prove or disprove that each function $f:\mathbb{R}\to\mathbb{R}$ has at most countable number of discontinuos points of first kind ($x_0$ is a point of discontinuos of first kind iff there exist $\lim_{x\to x_0-0}$ and $\lim_{x\to x_0+}$).

Define new function $ g: \mathbb{R} \rightarrow \mathbb{R} $ such that
$ g(x_{0}) $ equals to $ f(x_{0}) $ if $ x_{0} $ is not a discontinuous points of first kind,
and to the average of $ lim_{x \rightarrow x_{0} - 0} f(x) $ and
$ lim_{x \rightarrow x_{0} + 0} f(x) $, if it is. It is easy to see that all
discontinuous points of first kind of $ f $ are also of $ g $. Now for every
$ x_{0} $ which is discontinuous points of first kind of $ g $, we can take
a small rectangle on the 2-dim plane which contains a point with coordinates
$ (x_{0},g(x_{0})) $ and it's intersection with the graph of $ g(x) $ is the only
point $ (x_{0},g(x_{0})) $. We can of course also take the vertices of the rectangle
such that their coordinates are rational numbers. There are at most countable
number of possibilities for such a rectangles, so $ g(x) $ has at most countable
number of discontinuous points of first kind, so $ f(x) $.

This problem apears also in Walter Rudin 's principles of mathematical analisys as an exersize, (with a very strong hind above)

[quote="grobber"]It's true.

For each $n$, let $A_n$ be the set of points of discontinuity of the first kind in which the oscillation of $f$ is $\ge\frac 1n$. Let $x\in A_n$. Then there is an $\varepsilon>0$ s.t. both conditions $x<y_1<y_2<x+\varepsilon$ and $x-\varepsilon<y_1<y_2<x$ imply $|f(y_1)-f(y_2)|<\frac 1n$. This means that in the interval $(x-\varepsilon,x+\varepsilon)$ there are no other points from $A_n$, which shows that $A_n$ is a discrete, and thus countable set. Since the set of discontinuity points of the first kind is $\bigcup_n A_n$, we're done.[/quote]

I'm sorry, but if we follow to the definition given by Myth, then this proof is not correct, since the set of discontinuity points of the first kind [b]is not[/b] equal to $\bigcup_n A_n$. What we should do with the points of discontinuity with $\lim_{x \to x_0-0}=\lim_{x \to x_0+0}$?

[quote="Remike"]
I'm sorry, but if we follow to the definition given by Myth, then this proof is not correct, since the set of discontinuity points of the first kind [b]is not[/b] equal to $\bigcup_n A_n$. What we should do with the points of discontinuity with $\lim_{x \to x_0-0}=\lim_{x \to x_0+0}$?[/quote]

For those points the oscillation is still greater than $0$, namely $|f(x_0)-\lim_{x\to x_0}f(x)|$.

[quote="grobber"]For those points the oscillation is still greater than $0$, namely $|f(x_0)-\lim_{x\to x_0}f(x)|$.[/quote]

Yes, you are right. Exuse me.

But I have one more problem concerning this one.

We will say that a point $x_0$ is a point of discontinuity [i]of semifirst kind[/i] iff $x_0$ is a point of discontinuity and there exist at least one of the limits $\lim_{x \to x_0-0}$ and $\lim_{x \to x_0+0}$. Is it true that each function $f \colon \mathbb{R} \to \mathbb{R}$ has at most countable number of discontinuity points of semifirst kind?

Only a tiny refinement of the argment above is needed to see that this question also has positive answer:

Let's look at the points of discontinuity for which the left limit exists. Again, we divide them up into sets $A_n$ consisting of those points with this property where the function has oscillation $\ge\frac 1n$. For each such point $x\in A_n$ there is an $\varepsilon_x>0$ s.t. $(x-\varepsilon_x,x)$ contains no points from $A_n$, and if to each $x\in A_n$ we associate a rational $q_x\in(x-\varepsilon_x,x)$, it's clear that this is an injection from $A_n$ to $\mathbb Q$, meaning that $A_n$ is countable.

Can one find a function $f : \mathbb R \to \mathbb R$ such that all $x \in \mathbb Q$ are points of discontinuity of the first kind?

Let $(r_n)_n$ be a enumeration of rationals form $[0,1]$

$f(x)=\sum_{r_n<x}2^{-n}$.
It is increasing, and discontinuous in every rational point from $[0,1]$. You can easily construst a function on whole real line.

How do you prove that it is discontinuous in every rational point? And how do you make it work on the real line?

I don't have a proof. Probably it is very long. This example was given by Kent in another topic, and maybe he has a proof. To construct it on the entire real line, remark that $f(1)=1,f(0)=0$. And set $f(x)=f(\{x\})+[x]$ for $x\in \mathbb{R}$.


geometry
projective geometry
geometry unsolved

Let $ p(O)$ be a circle and A a point outside it. Denote by $ B,C$ the points where the tangents from A wrt $ p(O)$ meet the circle, D be the point on $ p(O)$, for which $ O,A,D$ are collinear, X the foot of the perpendicular from $ B$ to $ CD$, $ Y$ the midpoint of the line segment $ BX$ annd by $ Z$ the second intersection of $ DY$ with $ p(O)$. Prove that $ \angle AZC \equal{} 90^\circ$.

Let $ M$ be the midpoint of $ BC$. Consider the inversion through pole $ O$, power $ k \equal{} R^2$, where $ R$ is the radii of the circle. We have $ \mathcal {I}(O,k):$ $ A\mapsto M$, $ D\mapsto D$, $ Z\mapsto Z$. Hence $ \mathcal {I}(O,k): (ADZ)\mapsto (MDZ)$. Note that $ BD$ $ \mapsto (OBD)$ through $ \mathcal {I}(O,k)$. We claim that $ BD$ is the tangent at $ D$ of $ (ADZ)$.

Indeed, since $ D\equiv OA\cap (O)$ $ \Longrightarrow \angle {DCO} \equal{} \angle {CDO} \equal{} \angle {DBO}$. Hence $ CD$ is the tangent at $ D$ of $ (OBD)$, $ (1)$. Since $ ZDCB$ is concyclic $ \Longrightarrow$ $ \angle YZB$ $ \equal{}$ $ \angle DCB$ $ \equiv$ $ \angle XCB$. But $ M,Y$ respectively are the midpoints of $ BC,BX$ $ \Longrightarrow MY\parallel{}CX$ $ \Longrightarrow \angle YMB \equal{} \angle XCB$. Therefore, $ \angle YZB \equal{} \angle YMB$. Hence $ ZYBM$ is concylic $ \Longrightarrow \angle ZMB \equal{} \angle XYD$. But $ \angle DMB \equal{} \angle MYX \equal{} 90^{\circ}$ $ \Longrightarrow \angle DMZ \equal{} \angle DYM \equal{} \angle XDY$ $ \left( CX\parallel{}MY \right)$, which implies $ DX$ is the tangent at $ D$ of $ (DZM)$, $ (2)$. From $ (1),(2)$, we obtain $ \overline {CDX}$ is the common tangent at $ D$ of $ (DZM)$ and $ (OBD)$, which implies $ (DZM)$ and $ (OBD)$ tangent to each other at $ D$.

Therefore $ BD$ tangents $ (ADZ)$ $ \Longrightarrow \angle BDZ \equal{} \angle DAZ\equiv \angle MAZ$. But $ \angle BDZ \equal{} \angle BCZ$. Hence $ \angle MAZ \equal{} \angle BCZ\equiv \angle MCZ$, which implies that $ AZMC$ is concyclic. But $ \angle AMC \equal{} 90^{\circ}$ $ \Longrightarrow$ $ \angle AZC \equal{} 90^{\circ}$ $ \square$

$ \bullet$ [b]REMARK-[/b] In the figure below, I have taken $ T\equiv X$.

I would like solution without inversion,especially the one using harmonic division

$ \bullet$ [b]REMARK-[/b] The problem can be solved by harmonic division also. I think that the full proof of this problem should be solved with $ 2$ cases of the positions of $ D$. The $ 1^{st}$ [b]case[/b]: $ D\in$ the smaller arc of $ BC$ has been solved in my previous post. Now here is the solution for the $ 2^{nd}$ [b]case[/b]: $ D\in$ the bigger arc of $ BC$. This solution is offered by [b]Cosmin Pohoata[/b]

Let us call $ H \equal{} CO\cap (O)$. Thus $ DC\bot DH$ , so $ DH \parallel{}BX$ . Because $ Y$ is the midpoint of $ BX$ , we deduce that the division $ (BYX\infty)$ is harmonic, so also is the pencil $ D(BYXH)$ and by intersecting it with $ (O)$, it follows that the quadrilateral $ HBZC$ is harmonic. Then, the pencil $ C(HBZC)$ is harmonic, so by intersecting it with the line $ HZ$ , it follows that the division $ (A'ZTH)$ is harmonic, where $ A' \equal{} HZ\cap CC$ and $ T \equal{} H Z\cap BC$ . So, the line $ CH$ is the polar of $ A'$ with respect to $ (O)$, but $ CH \equal{} BC$ is the polar of $ A$ as well, so $ A\equiv A'$ , hence the points $ H, Z, A$ are collinear, therefore $ Z A\bot ZC$ $ \square$

Could someone write here explanation (or PM me) why is pencil $ D(BYXH)$ harmonic(these points aren't on a line or a circle), and especially why is quadrilateral $ HBZC$ harmonic.

I only prove this problem in case $ D\in [OA]$
Let $ M$ be the midpoint of $ BC$.
$ MY//XC$ then $ \angle YMB \equal{} \angle XCB \equal{} \angle YZB$
So $ BYZM$ is cyclic.
$ \Rightarrow \angle BZM \equal{} \angle BYM \equal{} 90^o$
We get $ \angle ZBM \equal{} \angle ZMA$
But
$ \angle ZBM \equal{} \angle ZCA$ then $ \angle ZMA \equal{} \angle ZCA$ therefore $ AZMC$ is cyclic or $ \angle AZC \equal{} 90^o$ (QED)

This problem is interesting because we note the two possible positions for $ D$ gives the same result. I believe that there is an underlying reason for this I believe is that the intersection of line AO and the circle works either way. The other parts of the diagram have no betweenness conditions. It seems that algebraically we should be able to prove that they are equivalent.


[b]Hi
is it a rule in IMO that a contestant must be with the nationallity of the team he will join  :?:
please answer .

[/b]


On the page where it talks about submitting solutions as a pdf. file using Latex, it also says that this other file must be sent in with the solutions file.  Does that mean we make a folder and put the solution pdf and the must attatch pdf inside it? It was slightly confusing and I didnt find anything about it on the FAQs page.  Thanks

I [i]think[/i] with the new uploading thing, you only have to attach the .pdf, and not worry about the LaTeX source code.

This should be moved to the USAMTS Forum.

Moved this to the USAMTS section.  Should've looked before posting......

[quote]Students using LaTeX to typeset their solutions are strongly encouraged to use the template files provided by the USAMTS. Click here for the template solutions file. This is the file you should use to input your solutions. Instructions for how to use it are included in the file. [b]You must also download the USAMTS include file. This file must be in the same folder as your solutions file.[/b][/quote]

This is from http://www.usamts.org/Rules/U_RulesSubmitting.php

As using their template to submit my solutions but it says I need download this include file.  Do I just like paste that at the bottom of my latex code and then build the file?  Do I build both the files and put the pdfs in folders?  What does it mean it has to be in the same folder as my solution file?

Download both files, make sure they're in the same folder, and read the solutions file. Once you read the instructions in the file it isn't confusing.

[quote]Students using LaTeX to typeset their solutions are strongly encouraged to use the template files provided by the USAMTS. Click here for the template solutions file. This is the file you should use to input your solutions. [b]Instructions for how to use it are included in the file.[/b] You must also download the USAMTS include file. This file must be in the same folder as your solutions file.[/quote]

If you are using LaTeX, then you must [b]download[/b] (from our website to your computer) both the solutions template file and the include file.  Both files need to be in the same folder on your computer.

When you [b]upload[/b] your completed solutions (from your computer to our website), you only need to upload a single PDF file.  (In fact, that's all you will be allowed to upload.)

And, I'll say this again because some people might still be confused, you are [b]not[/b] required to use LaTeX for the USAMTS.  You can use any program you wish to type up your solutions, so long as you can create and upload a PDF file.  Or you can handwrite your solutions, scan them into a PDF file, and upload it.  Or you can mail your solutions.  See the complete details on the USAMTS web site.

[quote="Ketvoman"]So that means that the include file is not in the pdf that we uploaded? Since we only upload a single pdf and that would be our solutions file?  Sorry Im slow and a first timer and just want to clarify. Thanks :D[/quote]

It's ok, its the first time this new uploading thing has been used  :D .

From DPatrick's post, is seems that you are right. Anyways, just download everything you see on the site!

[quote="mathemonster"]can USAMTS solutions be handwritten then mailed in????[/quote]
Yes. Read the website.

It's LaTeX, by the way. Just to be picky  :wink:

The include file is an extra $ \text{\LaTeX}$ file that allows the solutions file to work properly.  This just sits in the same folder as your solutions file.  When you are ready to submit your solutions, compile the solutions file to get a .pdf.  This year, you just upload the .pdf.

And to be even pickier, its $ \text{\LaTeX}$.


integration
limit
calculus
real analysis
real analysis solved

Let $f: [1, \infty) \to R$ and \[ f(x) = x \int_{1}^{x} \frac{e^{t}}{t} dt - e^{x}.\]

a) Prove that f is increasing.

b) prove that \[\lim_{x \to \infty} f(x) = + \infty\].

c) prove that there exists and it is only one number $a>1$ so that \[ \int_{1}^{a} \frac{e^{t}}{t} dt = \frac{e^{a}}{a}.\]

b) has already been discussed in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40293[/url].

As to a), just diffferentiate and use the Fundamental Theorem of Calculus; you'll get $f'(x)=\int_1^x \frac{e^t}t\,dt+x\frac {e^x}x-e^x=\int_1^x \frac{e^t}t\,dt>0$ if $x>1$. As to c), note that $f$ is continuous, $f(1)=-e<0$, $f\to+\infty$ as $x\to+\infty$, so, by the Intermediate Value Theorem, there must exist $a$ such that $f(a)=0$. Since $f$ is strictly increasing, there may be only one such $a$.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be a non-abelian simple group and P a Sylow p-subgroup of G. Is normalizer N_G(P) of P in G a maximal subgroup?

Why should it be?
Let $ P$ be a Sylow 2-subgroup of $ A_6$.
Then $ P$ is also a Sylow 2-subgroup of $ A_7$, as $ 2\nmid|A_7: A_6|\equal{}7$.
But then $ 7$ is a fixed point of $ P$ in $ \{1,2,3,4,5,6,7\}$, and for any $ g\in N_{A_7}(P)$
we have $ g(7)$ also fixed point of $ gPg^{\minus{}1}\equal{}P$.
So if $ g(7)\not\equal{} 7$, then $ P$ would also have a fixed point in $ \{1,2,3,4,5,6\}$,
which gives $ P$ conjugate to a subgroup of $ A_5$. (the stabilizer of $ 6$ in $ A_6$)
This contradicts $ 2\mid |A_6: A_5|\equal{}6$, and therefore $ |P|\nmid |A_5|$.

So we must have $ 7$ a fixed point of $ N_{A_7}(P)$, and therefore $ N_{A_7}(P)\le A_6$.
But $ A_6$ is simple, so $ P$ is not normal in $ A_6$, which gives $ N_{A_7}(P)\equal{}N_{A_6}(P)\lneq A_6\lneq A_7$.
So $ N_{A_7}(P)$ is not maximal in $ A_7$.


$ 3|x| \plus{} 2|x\minus{}1| \leq 4|2\minus{}4x|$. is the way to solve this is by categorizing them in cases? like $ |x|$= $ x$ when $ x\geq 0$ and so forth? or is there an easier solution? thanks! im not asking for the answer btw, im just checking for a simpler solution...

maybe squaring first of all may help you in many inequalitites with abolute values.

Note that squaring only works if both sides are positive, as $ f(x)\equal{}x^2$ is only increasing on $ (0,\infty)$.


summer program
PROMYS
pigeonhole principle
quadratics

Prove that for $p\in\mathbb{P}$ with $p>2$, there exists $x,y\in\mathbb{N}$ such that \[x^{2}+y^{2}\equiv-1\bmod p.\]

[hide="an attempt"]

Case 1: if $-1$ is a square in $\mathbb Z_{p}$ -> obvious.

Case 2: if $-1$ is not a square, we first will proove

[b]Lemma:[/b] $\exists x, y\in\mathbb Z_{p}^\ast$ such that $x^{2}+y^{2}$ is not a square.

[b]Proof:[/b]
Assume the opposite : $\forall x, y\in\mathbb Z_{p}$, $x^{2}+y^{2}$ is a square. Choose $x\in\mathbb Z_{p}^\ast$, then there exist $x_{1}$, $x_{2}$, $...$ $\in\mathbb Z_{p}$ such that:

$2x^{2}=x^{2}+x^{2}=x_{1}^{2}$,
$3x^{2}=x^{2}+x_{1}^{2}=x_{2}^{2}$,
$...$

Hence all multiples of $x^{2}$ are squares. Since multiplication by $x^{2}$ is an automorphism of $\mathbb Z_{p}$, this implies that all elements of $\mathbb Z_{p}$ are squares, which is false. (/lemma)

Now fix $x$ and $y\in\mathbb Z_{p}^\ast$ such that $x^{2}+y^{2}$ is not a square. Consider the set $E=\{(zx)^{2}+(zy)^{2}|z\in\mathbb Z_{p}^\ast\}\subset\mathbb Z_{p}^\ast$.

I claim that $-1\in E$. Indeed, $\text{card}(E)=\frac{p-1}2$. Furthermore, the set of squares $S=\{x^{2}|x\in\mathbb Z_{p}^\ast\}$ has same cardinality. Hence $\mathbb Z_{p}^\ast$ is the disjoint union of $E$ and $S$.

Since, by assumption, $-1\not\in S$, we conclude that $-1\in E$, i.e. $\exists z|(zx)^{2}+(zy)^{2}=-1$. QED
[/hide]
---
zpz

This was in a PROMYS problem set, but I'm too lazy to remember the proof.  I remember it uses pigeonhole on the fact that there are exactly $\frac{p-1}{2}$ quadratic residues.  The $p \equiv 1\mod{4}$ case is easy, because then -1 is a quadratic residue, and the $p \equiv 3\mod{4}$ case is easy if we assume the opposite, because we then get fewer than $\frac{p-1}{2}$ quadratic residues.

[hide]If $p \equiv 1 \bmod 4$ just select $y=0$,
if $p \equiv 3 \bmod 4$  Let $z$ be the smallest number which is not a square   (obviously $z \ge 2$  so $z-1$ can be written as $x^{2}$
But it's easy to see that $-z$ is quadratic residue  (-1 is non residue and (-1)(-1)=+1) .
we have $x^{2}+(-z) =-1$ (and -z is a square)
QED [/hide]

Problem:

Given a prime $p$ and an integer $k$, consider the equation $x^{2}+y^{2}\equiv k \mod p$.
a) show that it has always a solution (thus all residues $\mod p$ are sum of two squares).
b) Find a necessary and sufficient condition such that there is a solution with $x,y \not\equiv 0 \mod p$.
c) Find the number of solutions in terms of $k$ and $p$ (also allowing $x, y \equiv 0 \mod p$; but that doesn't make a big difference).


What does "QED" mean?  Some people write this after the end of proofs...

It comes from Latin: "quod erat demonstrandum", which literally means "that which was to be demonstrated". So you can write QED at the end of a proof to show that the result required for the proof to be complete has been obtained. In other words, you can write QED so that you sound cool :P

QED stands for Quod Erat Demonstratum, or "that which has been demonstrated" (or something like that).  It's, um, Latin.

Gah.  TripleM beat me to it by a few seconds.  Well, if you want to be fancy, you can also do [tex]\blacksquare[/tex].

Or you can do W5, which stands for Which Was What We Wanted.

In some class handouts, I used to put a smiley-face at the end of proofs.  It seemed like the right emotion. :)

One of my friends told me that he put a :) at the end of correct proofs,  a :( at the end of incorrect proofs, and  a :? at the end of proofs he wasn't sure about.

Our math class has only ten students in it, and one week we decided we would all use our own creative end of proof symbols -- one person used a stick figure with a voice bubble, and inside the voice bubble was a black square.  I modified my black square into a television that was showing "Q.E.D."

Finally! I've always wanted to know what that meant. Thanks.

Ahh.. I love  how latin lives on in today's world.  It makes me think that there is a point to taking it in high school... jk... i like latin a lot even though it is "dead"


function
inequalities unsolved
inequalities

let $ a;b;c>0$ and $ a\plus{}b\plus{}c\equal{}3$; prove that:
$ (3a\plus{}\sqrt{9a^2\plus{}16})(3b\plus{}\sqrt{9b^2\plus{}16})(3c\plus{}\sqrt{9c^2\plus{}16})\leq 512\equal{}8^3$

I think Karamata works for funtion $ F(x)\equal{}ln(3x\plus{}\sqrt{9x^2\plus{}16})$ because $ (a,b,c)\succ(1,1,1)$ so $ F(a)\plus{}F(b)\plus{}F(c)\le 3F(1)\equal{}3ln(8)\equal{}ln(512)$


Hi

I would like to make some tables in LaTeX. But I  do not really know how to do so. Therefore I am requesting for you help to do so. And I would like tables with more than two horizontal boxes.

All the best

Visit this site's LaTeX guide at http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#tables which tells you all about tables.

If you'd like a visual table editor for Windows then have a look at [url=http://www.g32.org/latable/]LaTable[/url]. It makes designing LaTeX tables very easy indeed.

Ok thank you very much, your help is mostly appreciated.


let $ z_i\in \mathbb{C}\ ,\ 1\le i\le n$ and $ \sum_{i \equal{} 1}^nz_i \equal{} 0$ prove that for all$ z\in \mathbb{C}$:
$ n\le \sum_{i \equal{} 1}^n|z \minus{} z_i|$

there is probably sth missing in the problem statement, set $ z_i\equal{}0$, then $ n\leq \sum_{i\equal{}1}^n|z\minus{}z_i|$ iff $ |z|\geq 1$, ie. it is not fulfilled $ \forall z\in\mathbb{C}$


The ineq is equivalent to

$3(ab+bc+ca)(\frac{1}{2a^2+3bc}+\frac{1}{2b^2+3ac}+\frac{1}{2c^2+3ab})\geq 4$

which is a well-known ineq posted by Vasc I think

[quote="SuperMath"]Let $\large a;b;c > 0$. Prove that:
$\large \sum \frac{a(3b+3c-2a)}{2a^2+3bc} \geq\ 1$[/quote]
At last, it's first your true and correct inequality.


Given any three triangles $ \Delta_1,\Delta_2,\Delta_3$. Let $ H_1,H_2,H_3$ be the orthocenters;$ G_1,G_2,G_3$ the centroids;$ O_1,O_2,O_3$ the circumcenters. Prove that the centroids of the triangles $ H_1H_2H_3,G_1G_2G_3$ and $ O_1O_2O_3$ are collinear and their circumcenters are likewise collinear.

It is easy to prove the collinearity of the centroids of the triangles $ \bigtriangleup H_{1}H_{2}H_{3},$ $ \bigtriangleup G_{1}G_{2}G_{3},$ $ \bigtriangleup O_{1}O_{2}O_{3},$ as an application on a well known theorem, already has been posted before in this forum.

Please see at the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinearity of orthocenters[/url].

As for the collinearity of the circumcenters of the above triangles you mentioned, I think that it is not true, as I can see by the drawing and I will check more carefully before a definite answer.

Kostas Vittas.


In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other?

We can use complementary counting. Let's find the number of ways they can be arranged if Wilma and Paul were sitting together. If we view them as one person, there are $ 6!$ ways to arrange those 6 people, times $ 2!$ ways to arrange Wilma and Paul. This gives $ 720 \cdot 2 \equal{} 1440$ ways with Wilma and Paul sitting together.

With no restrictions, there are $ 7!$ ways to arrange all the people. Thus our answer is $ 5040 \minus{} 1440 \equal{} \boxed{3600}$.

[hide=My Solution]Without restrictions there are $7!=5040$, and the amount of cases where Wilma and Paul sit next to each other are: $$W, P,3,4,5,6,7$$
$$1,W,P,4,5,6,7$$
$$1,2,W,P,5,6,7$$
$$1,2,3,W,P,6,7$$
$$1,2,3,4,W,P,7$$
$$1,2,3,4,5,W,P$$
And since we know each seat (1,2,3,4,5,6,7) has $5!=120$ different options so $120(5)$, and since W, P is can also be represented as P, W, (which are distinct cases), so we have $5040-120\times 12 = 3600$.


:?: an alphabet consists of 3 letters.some sequence of letters
of length 2 0r more are prohibited,and all of the prohibited seq. have
different lengths.a word is a seq.of letters of any length.a correct
word does not contain any prohibited seq.prove-there ara correct words
of any length.(kvant)

Let $A(n)$ be the number of correct words of length $n$. We want to show that $A(n)>0$, but a much stronger statement is true: $A(n+1)>2A(n)$. Since $A(1)=3, A(2)=8$ this implies $A(n)>2^{n}$.

Now we will prove $A(n+1)>2A(n)$. To do this, consider words of the form $x_{1}x_{2}\ldots x_{n}l$, where $x_{1}\ldots x_{n}$ is a correct word of length $n$ and $l$ is an arbitrary letter. It is easy to see that any correct word of length $n+1$ must be of this form, since its first $n$ letters must themselves form a correct word. Besides, there are $3A(n)$ words of this type ($A(n)$ ways of choosing $x_{1}\ldots x_{n}$ and $3$ ways of choosing $l$). So we will be done if we can prove that, from these $3A(n)$ words, less than $A(n)$ are incorrect.

To do this, we can assume inductively that $A(n)>2A(n-1), A(n-1)>2A(n-2),$ etc. Now, why would a word of the form $x_{1}\ldots x_{n}l$ be incorrect? In order to be incorrect, it must contain a prohibited sequence, but since the first $n$ letters form a correct word, no prohibited sequences can be found among them; so the prohibited sequence must be of the form $x_{i+1}\ldots x_{n}l$ for some $i$.

Thus, an incorrect word of this type would be of the form $x_{1}\ldots x_{i}B$, where B is a bad sub-word $n-i+1$ letters long. How many words like this can exist? For a given $i$, $n-i+1$ is fixed, so there's only one choice for $B$ (because there's only one prohibited sequence of each length); and there are $A(i)$ choices for the first $i$ letters, so at most $A(i)$ words of this type can exist. Since $i$ can take the values $1, 2, \ldots, n-1$, the number of incorrect words is at most $A(n-1)+A(n-2)+\ldots+A(1)+1$ (the $+1$ at the end is for the case when B is $n+1$ letters long).

Now, using that $A(j+1)>2A(j)$, the above is less than $(0.5+0.5^{2}+...)A(n)<A(n)$. Thus there are less than $A(n)$ incorrect words among the ones we considered, and more than $2A(n)$ of them are correct.

Some more things I've found: it turns out that the recursion $A(1)=3, A(n+1)=3A(n)-A(n-1)-\ldots-A(1)-1$ gives a minimum of $A(n)=(n+2)2^{n-1}$, and this can be achieved explicitly by prohibiting the sequences $ac, abc, abbc, \ldots$.


I currently play piano and violin.

Actually my pizzing is pwn because during tuning everyday (school orchestra:which takes 2 min. per string?????) I get bored and pizz a piano song on the violin.
Been going on for 2 years.

Anyone else do that?

I am currently playing: (ordered from most important  to least)
Viola (5th year, pretty good)
Piano (9th year, still not that good ((It's not my main instrument)))
Violin (1 year)

Oh does anyone know about any viola competitions or things that will look good on a resume in some way. That aren't very intense?

[quote="pascal12"]our teacher yells at us if we do anything other than tuning while the rest of the orchestra is tuning.[/quote]

Oh my gosh. We have two teachers, one volunteered teacher and one paid teacher. The volunteered teacher:
-Makes innocent children cry (my friend)
-hits the baton against a stand many times to get the orchestras attention (almost breaks it)
-gives very LOOOOOOOOOOOOOOOOOOOOOOOOOOOONNNNNNNNNNGGGGGGGGGGGGG stories right before we practice the piece.
-Has to sit when he conducts
- Is mean and evil!

[quote="pascal12"]our teacher yells at us if we do anything other than tuning while the rest of the orchestra is tuning.[/quote]

My teacher accused us of "talking" not "tuning" that's because I tune in at most 10 seconds, don't need 5 years of tuning a string :mad:  :mad:

Blah my friends don't know how to tune.

there is a song for string orchestra that is just pitz

it is very fun :lol:

I find pitz complicated if it's fast and boring if it's through the whole piece

There's a song for string orchestra where the whole orchestra uses only one bow.  Pretty epic.  (It gets passed around.)


function
calculus
derivative
calculus computations

This is a multiple choice question, though I'd appreciate if someone can explain how you get the answer.

___

If f(5)=9, f'(5)=2, g(9) = 3, g'(9)=4 then we know that

a.) (f * g)'(5)=30

b.) (f+g)'(9)=6

c.) (f 0 g)'(9)=6

d.) (g 0 f)'(5) = 8

NOTE: the "0" in "(f 0 g) and (g 0 f) represents the composite sign, I just can't figure out what to use. So, for example, c can be read as f(g'(9))=6

Thank you.

Basic differentiation rules:

$\frac{d}{dx}[f \cdot g] = f'g+fg'$

$\frac{d}{dx}[f+g] = f'+g'$

$\frac{d}{dx}[f \circ g] = (f' \circ g) \cdot g'$.

Funny, I wasn't taught that in my course.

Anywho, I tried that, and I can't seem to get any answer. Which one is the correct answer? Also, can you explain how the differentiation of (f 0 g) would work out? I don't understand what you got for that.

That is the chain rule, and it is explained in every calculus textbook.


Find $a;b \in Z$ such that $a^{b}-b^{a}=1$

[hide="if a,b are positive integers"] http://mathworld.wolfram.com/CatalansConjecture.html[/hide]

a=2 and b=1 that is the solution
I think maybe they are another solution


calculate 10 modulus 5 and give me the answer

[quote="Jan"]10 MOD 5 = x
that means that 10=5q+x with q a natural number.
=> q=0 and x=0[/quote]
Actually q=2, but that's probably what you meant anyways.

sufyan, do you know what modular congruence is? It's a little like taking the remainder after division, instead fractions or decimals. You would say $a\equiv b\bmod{c}$ ($a,b,c\in\mathbb{Z}$) if $a=b+cx$ for some integer $x$.

For more information, go here:
http://mathworld.wolfram.com/Modulus.html
http://mathworld.wolfram.com/Congruence.html