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Hello,

at the risk of yet again creating a controversial topic, and probably the worst ever, I would like to bring up a question and I would be most interested in everyone's opinion.

This is NOT the same as the culture clash topic.

My question is, how strong is the argument in general of freedom of speech. I mean : how concrete was it before this cartoon issue arose?

Is it alright for me to make a racist joke on television because that is freedom of speech?  I guess not, racism is an exception.

Is it alright for me to go on television and offer a reward for those who capture or kill a man like Saddam Hussein, like Bruce Willis did?  Is it alright for me to go on television and offer a reward for those who kill the cartoonist who created the original Danish cartoon, like I saw a middle eastern religious leader do a few days ago?

Can I put a list on the internet of the current locations of people who have taken part in a witness relocation program?

And finally I cannot forget this issue :
Armenian genocide/Jewish holocaust

It is NOT my intention to fight out here once and for all who deserves to be punished and who is right, I just wanna understand the logic.

Recently Orhan Pamuk was almost sentenced to jail for claiming the Armenian genocide took place.  Also, there was an issue last month when Turkish authorities felt offended that Belgian school books contained the genocide, they wanted it removed again.  In general, the armenian genocide is not really a big thing here, and it is still treated,
when it pops up in media, as a case to be examined.

Denying the Jewish Holocaust is illegal in the following countries :  Austria, Belgium, Czech Republic, France, Germany, Israel, Lithuania, Poland, Slovakia and Switzerland.

The reason for posting this now is that David Irving, historian and holocaust denier, is now sentenced to prison in Austria.
I heard of Turkish people in the Netherlands doing a counterdemonstration recently when Armenians were trying to earn recognition of the genocide.  So those Turkish people, what is the difference between them and Irving then??

On what ground is the Jewish Holocaust so much more true, that not only it is definitely true, but also punishable, against freedom of speech, by law?  Is there more proof?  Is it because the crime is seen as worse (for the motives or the numbers)?
The answer I expected to hear is the first, abundance of proof, or am I wrong?

It just struck me, Abu Jahjah,of Lebanese descent and now Belgian, leader of Arab European League, brought this up in a debate about the cartoons, why can't we deny or diminish the holocaust?

And how exactly should Holocaust denial be punished and how far does this extend?  If I tell my children every day it didn't happen,and when they're eighteen they'll go yell it a town square, should they be arrested?
If I influence people with low intelligence, not much education, or even mentally disabled people, so they start saying the holocaust isn't true, should they be arrested?
Or should I be arrested?  And who then decides if that person didn't 'make' that opinion on his own?
Recently, when I saw the president of Iran saying he wished to organise the convention to examine the existence of the holocaust ,and I said to my sister I don't see anything wrong with that, should she now call the police on me?

Once more, I am not looking for trouble, I would just like to see someone or more people explain how exactly freedom of speech is defined then?  So if you understand I will be happy to listen.  If you don't, please tell me why you think this is a weird logic.

You always create such interesting topics! I like how you actually type them out instead of leaving one sentence non-sequiters!

Anyway, [url=http://en.wikipedia.org/wiki/Freedom_of_speech]this[/url] might be interesting.

a couple of thoughts
First:  you can say racist things on tv.  the only issue is that there are economic problems because the public will be outraged and may boycott the channel you were on so the people in charge of the channel will not let you say racist things.  there are lots of examples of racism being allowed in america under free speech (American Nazis, KKK) in fact the aclu once defended the american nazis right to have a speech in a town made up of mostly holocaust survivors.

Second: freedom of speech is limited by the concept that you cant shout fire in a crowded theater.  That is you couldnt talk about witness relocations because that puts someones life in danger.

Third:   If you dont allow people to deny the holocaust or talk about the armenian genocide then you dont have true freedom of speech.  Part of the reason most of europe does not allow denial of the holocaust is because after the war one of the allied generals required soldiers to go to the camps to see the horrors this meant that at the time lots of people learned about it.  In addition Jews have enough political influence that a country which denied the holocaust would have trouble negotiating with the US (this is only that countries don't deny it rather than not allow others to deny it).  The reason the countries passed laws against denying it may be because most of europe also has other undemocratic laws of similar nature.  For example in poland it is illegal to have a communist or nazi party.  The reason the armenian genocide is not well known is that Turkey attempts to cover it up and there are not enough people outside of turkey who know about it.

Thanks for your reactions , both.
Keep them coming

non sequiter/sequitur (latin? : he doesn't follow?) what exactly do you mean?

The article on wikipedia was interesting but did not cover holocaust denial unfortunately.

About shouting fire in a theater and witness relocating, that is what I expected, but how well defined is putting someone in danger.  And that brings me back to offering rewards on tv for capturing/killing people.
Also knowing how much people would actually die (christians and muslims have died in many countries now) isn't that a reason to ban the cartoons then?  And who is to blame?  Those who made them?  Those who travelled to the middle east (and are under suspicion of falsifying the material, adding much unrelated images).

So jehr314pi , you (partially) agree with the argument of muslims in europe and outside that israel holds a very influential position in the west and that this is a case of favouring one side?

[quote="fredbel6"]Thanks for your reactions , both.
Keep them coming

non sequiter/sequitur (latin? : he doesn't follow?) what exactly do you mean?

[/quote]

Wikipedia usually always has the answer:
http://en.wikipedia.org/wiki/Non_sequitur_%28absurdism%29

Holocaust denial: http://en.wikipedia.org/wiki/Holocaust_denial

Like many interesting issues, this issue is not simple.  I'm not aware of any important issue that is really simple in detail.  However, if you read this too long posting, you will see that acting in love will bring clarity to this complex mess.

The first constraint is the law under which the statements are made.  The United States is fairly unique in granting it's people right to speak (It's located in the First Amendment to the US  Constitution and has been incorporated into the Fourteenth Amendment to the US Constitution [which is how it applies to state governments, in the United States.]  This says that, in the United
States, the government can't limit what it's citizens, and most others say, if they are regulating the content of the speech [writing is counted as speech].

This means that the government can't prohibit discussion, based on it's content, for the most part.  However, offensive things (porno, swearing, playing stolen songs or movies) are restricted.  Further, a person may well have to answer to others for what they say [slander, libel, breach of contract, intential inflection of severe emotional distress and the like].

One can't broadcast their own TV station without a license, or stop traffic, interfere with public enjoyment of parks, just to say something.  (Businesses may allow public statements in their place of work or ban all, but can't just ban pro labor statement.)

The national laws about denial of holocaust are typically on the books of nations that feel a particular guilt over the historical event.  For example the Swiss asked the Germans to mark the passports of their Jewish citizens so the Swiss would know which bank accounts were not likely to be claimed.  They also purchased from the German government millions of wedding rings, melting them down and providing the German governement a good source  of money during the war.  These types of actions give rise to the legal restrictions in countries to denials of events the governments feel a particular responsibility.

Beyond the technical legal requirements, it seems that everyone should have a moral right to say what they want.  However, as Paul said to Christians, in a different context, just because the law says you can do something, donesn't mean it is right to do this.  You must not exercize your freedom if it will cause harm to another (body or faith).  You are instructed to follow the law of love.

I am ashamed of the activities of Christians in Nigeria (killing about 80 muslims, burning corpses, buring mosques, forcing their Moslem brothers to leave the area, saying, "This is Igbo land. No more Muslims can live here."  This was the loving Chrisitan answer to similar crimes against Christians living in majority Muslim areas in northern Nigeria.  This is just wrong.  I agree with the expression attributed to Rev. Joseph Ezeugo (Nigeria), "What has become of us?  This cannot be Nigeria today.  We have been living side by side with our Muslim brohers for so long.  Why should a cartoon in Denmark bring us to civil war?"  I am sure that many Muslim beleivers are equally ashamed of the actions of evil people claiming to act in defense of Islam.

Sorry for going on so long, but I need to put my comment that one has a moral obligation to exercize his "right" of free speech in a way that is constructive and follows the law of love, one of the two laws choosen by Jesus as the greatest of the approximately 638 Jewish laws at the time, "Love your neighbor as you love yourself" (the other law dealt with man's obligations toward God).  He followed with the story of the good Samaritian to demonstrate that all of mankind is our neighbor.  I just don't know how anyone can see this type of evil in the world and not feel both anger that it occurs and shame that we are all part of mankind that has choosen to do such things.

This may be more than you wanted, but this sums up my current feelings about the freedom of speach.

I believe in absolute freedom of speech, except for calls for illegal acts.  Bruce Willis or anyone offering money to commit murder would be banned in my view.

But, I believe in passing laws to allow all people to speak freely.  Currently in the Western world, the mass media are completely monopolized.  For example, in america, virtually all the media have been bought off by larger companies, and now almost everything is owned by either Time Warner, Viacom, Disney, and FoxNews.  As such, the only views or news that are broadcasted are those that go along with the views of the CEOs of these 4 huge conglomerates.

So, I would make a law requiring these media giants to be broken up into perhaps one thousand smaller groups.  Then, I would pass a law requiring these media sources to give airtime to every political/ideological/philosophical views: the time allocated to each view can perhaps depend on the percentage of the population that supports each view.  For example, if there are 5% of the population who supports the Ralph Nader/green party view, then 5% of the media airtime would have to be given to the green party types.  If 5% support the Libertarian Party, then 5% of media go to them.  And if 1% of the population supports some "racist" views, like the organization of Stormfront.org, then the media would be legally required to give them 1% of their airtime.

There can also be a government funded media as well, maybe something like PBS but with more tax funding, and same rules of airtime would have to implemented: air time for everyone depending on the percentage support of the population.

Based on this, anti-muslim cartoonists would get their airtime based on what percentage of people are anti-muslim.  Same thing with anti-semites, anti-Whites, anti-Blacks, anti-Christians, and so for.  but only so long as no one is directly encouraging illegal acts.  For example, Jerry Falwell would [b]not[/b] be permitted to go on television and ask Christians to shoot muslims, feminists, gays, atheists, and Asians ( :D ).  But, if he says that Christians, through the democratic process, should vote for politicians who will vote for laws requiring the shooting of muslims, feminists, gays, atheists, and Asians ( :( ), then that would be okey since he is not encouraging any illegal act.

These are my personal political views, others may disagree.

Cheers!

The problem with freedom of speech is that it is based on the ideal that everyone -- as they should be -- is entirely relativistic and has respect for someone's opinion as a valid opinion. Seeing things from a dogmatic and narrow-minded point of veiw causes all of the problems today. While staying entirely relativistic is near impossible at times, it must be done in order to keep any frustration or slight angers toward another's narrow-mindedness down.

For example, being a mathematically based community, we all most likely accept all to nearly all laws and facts the scientific community has put in place. But, ideally, we should place as much respect for the perfectly valid opinion "that the moon is made of cheese" as we do for any other opinion. Most of us will probably dismiss the opinion as insane, but in reality, [i]it still has the possibility to be true.[/i] We, as humans, don't actually no any perfect truth becuase we are not infallible. It still remains a possibility that the next apple that falls from a tree falls up. Though the possibility is nearly zero seeing how much collected data we have on how it will fall down. All in all, the problem today is that people take their opinions beyond what the are -- one man's way of defining the world. Which is as perfectly valid as any other man's definitions of the world and universe. Once your opinion is taken to the point that it affects your relationship with others, you fail.

The main big reason I support more or less absolute free speech  is because there may be an idea or truth that is very useful to us, or just truthful, but may be illegal to state because it is politically incorrect, or controversial, or iconoclastic.  I consider people like Socrates, Galileo, and so forth who had truth to say but could not because of anti-free speech laws.

What if David Irving had some historical truth to say, but it is simply politically incorrect to say so?  Why is it that all history is open to study and refutation, but the questioning or studing the history of World War II is illegal in most places in the world?  Could it be that these "revisionists" are today's Galileo's and Socratese?  We will never know now since they are banned and arrested. First Historian Ernest Zundel for questioning the history of WWII, now David Irving.  And 8,000 people are in Germany's jail right now for questioning WWII.   Why?

All of these are interesting remarks.

About the last two : yes the moon could be made of cheese, or the moon could be some projection of earthlings onto the sky.  In order to believe that you must accept that astronauts, engineers, ... have all been lying.

So why can't it be done with the holocaust.  There is a lot of proof, but I myself have not seen the holocaust as I am a bit too young.

Many people say : come on that's true.  Okay then, but does that immediately imply that I have to be jailed??

If I would have to make a million euro bet now, I'd say it was true.

But it wouldn't be the first time a government does everything to cover things up (like the Armenian genocide like I said).

I still have the same opinion : holocaust deniers are treated as complete exceptions in the system.

Asian-American's posting is interesting.  He suggests that he is in favor of freedom of speach, however, his posting supports anything but that.

He supports equality of time, based on voting.  Suppose, I create a party that through divine intervention receives 10% of the vote.  I advocate some strange political agenda, focusing on having actual equality of soilders killed in Iraq to the racial profile of America.  This means that soilders will have their tasks re-assigned based on results to try to assign the death burdens from the war against terrorism to be equal to racial makeup.  Supposed further that I don't want to advocate time on TV, as my position is considered stupid by all that don't immediately embrace the position.  The national election has presented my platform as much as I desire.  Under Asian-American's plan, I will be forced to participate in what I don't want.

Further, some government agency must decide how much time "should" be spent.  I think Asian-American's position has confused equality of opportunity with equality of result.  This is similar to the illogic of affirmative action.  If it were really beleived, we would hear an outcry that too many women are now in universities and far too many minorities are on certain sports teams.  Don't we really want equality of opportuinty.  Statistics show that clustering occurs in truly random situations, so attempting to have equality of results can only occur when some oppresive force, such as occurs in schools forces the results.  Would anyone support the idea that the starting quarterback, or school president should alternate based on the gender or racial makeup of a school.  (Sorry Fred, this is the year for the other race/gender, we won't allow you to earn the position by skill or popularity.)

Sorry, your posting hit one of my areas of concern.  This idea of equality of results has captured many of good intentions and caused them to advocate discrimintory results, to the harm of many.

[quote="wizard2378"]  (Sorry Fred, this is the year for the other race/gender, we won't allow you to earn the position by skill or popularity.)

[/quote]

I don't quite follow here.
What exactly do you mean?

Could I sum your post up as a plea against doing everything equal regarding gender/race?
And what do you mean in the schools?

[quote="wizard2378"]Asian-American's posting is interesting.  He suggests that he is in favor of freedom of speach, however, his posting supports anything but that.[/quote]

Semantics.  Define "freedom of speech."

[quote]Supposed further that I don't want to advocate time on TV, as my position is considered stupid by all that don't immediately embrace the position.  The national election has presented my platform as much as I desire.  Under Asian-American's plan, I will be forced to participate in what I don't want.[/quote]

You would not be forced to say anything on the 10% of media time your party gets.  But other members of your party would be given that option.  If they all refuse, then fine, your party will be eliminated in the air time percentage breakdown.

[quote]Further, some government agency must decide how much time "should" be spent.[/quote]

Whatever: government statisticians, private statisticians, or whoever will break down the media time. What's your point?

[quote]I think Asian-American's position has confused equality of opportunity with equality of result.[/quote]

No: I know what the difference is.

[quote]This is similar to the illogic of affirmative action.[/quote]

My view does not involve any "abosolute" logic.  It is simply a subjective political view that air time should not be monopolized by a few CEOs.

[quote]If it were really beleived, we would hear an outcry that too many women are now in universities and far too many minorities are on certain sports teams.[/quote]

Your arguments have no logical superiority over mine, because both our beliefs are subjective.  You are supporting a pure capitalist view that all media can be bought off by a few men and then used as they see fit.  My view is that the media should not be monopolized and all ideas should be brought to the attention of the public.

[quote]Don't we really want equality of opportuinty.[/quote]

"[b]We[/b]"?    Why are you speaking for all of humanity?  Should you not be saying "I want equality of opportunity"?

[quote]Statistics show that clustering occurs in truly random situations, so attempting to have equality of results can only occur when some oppresive force, such as occurs in schools forces the results.[/quote]

All laws require [b]oppressive force[/b], that is why we have cops with guns: to force people to follow laws.  So what's your point?

[quote]Would anyone support the idea that the starting quarterback, or school president should alternate based on the gender or racial makeup of a school.[/quote]

Invalid extrapolation.  Why are you mixing anti-media  cartel laws with affirmative action?  They are two different topics, two different policies.  Why do you think there must be some type of structured pattern to societal infrastructure, instead of people looking at things from an issue by issue basis?  If I support aportion, does that mean I must also oppose guns in order to follow the Liberal structural line?  What if I told you that I support abortion, but also support radical gun rights?  Is there some absolute law of nature that dictates I could not do such a thing?

[quote]This idea of equality of results has captured many of good intentions and caused them to advocate discrimintory results, to the harm of many.[/quote]

What?

[quote="Asian-American"]I believe in absolute freedom of speech, except for calls for illegal acts.[/quote]

Not to be a spoil-sport, but this is meaningless.  If the Congress were to pass a law against voting for the Green Party, would you then be willing to accept a ban on speech advocating votes for the Green Party?   I realize that this wasn't your point, but the standard you've chosen (illegality) is clearly not the standard you want -- perhaps try again?

Press alt f4, I believe you're bringing two different discussions together: the possibility of knowing the truth or being able to call some ideas right and others wrong is a very different question from the question of freedom of speech.  (You and Asian American might read a little J.S. Mill's "On Liberty," if you haven't.  He has a good discussion of the value of freedom of speech in achieving the truth, even given the falliability of people and so on.)
Even the acceptance of falliability and limits on knowledge doesn't necessarily support your argument, however:  the total equality of all ideas is only legitimate if people are *entirely falliable.*  If people are right even a tiny bit better than chance, we can still determine the truth to arbitrarily good probabilities.  For instance, enough calculations of the moon, if they all result in the conclusion of non-cheesiness, will be good enough for me to correctly say "The moon is almost certainly non-cheesy" and be right, even if any of the individual calculations, observations and whatnot could have been flawed.

A-A, Holocaust denial is a crime in a very small number of countries, not "most of the world."  Also, I assure you that David Irving is not the next Socrates, although whether putting him in jail for 3 years is a reasonable punishment for his being terribly, terribly wrong is a valid question.  Holocaust denial is not particularly unique in its status, however: Britain just passed a law banning the "glorification of terrorism," whatever that means; many (most?  all?) European countries have laws that would never stand under the First Amendment.

wizard: What A-A is suggesting is a program designed for equality of opportunity.  (I don't particularly like A-A's suggestion at all, actually, as it's much to far shifted to the advantage of majority opinion (thus, [i]away[/i] from "equality of result").  I would prefer a system in which every single group gets equal time -- not what A-A was suggesting, but what you were complaining about.)  I don't know any political groups that *mind* getting free air time, but I fail to see how a suggestion that would offer your group 10% of airtime would be impinging on your rights -- you could, after all, just not take advantage of the offer.  You have defined opportunity versus results in a way that doesn't make any sense at all in this context.
I'm also not sure why you think the decision about splitting up the time is difficult.  He told you exactly how to do it: we hold a big national referendum, say once a year.  People can write in whatever views they think should be given airtime, and the available airtime is divided up according to the vote count.  Where's the problem in this?
Basically, I don't think any of your arguments make sense.  In particular, I have no idea whatsoever what affirmative action has to do with this discussion or why you want to make it part of this discussion.

I believe in freedom of speech to the fullest extent safety will allow. That is, I believe in the American ideal of free speech- everyone gets to say what they want, even if they're wrong, and no one gets hurt. With that said, yes, of course, there should be limits to free speech, however, was it not Ben Franklin that said (and I paraphrase) "those that give up liberty for safety deserve neither"? By "liberty," in this context, I mean free speech. People will say things that we don't like (that the Holocaust didn't happen, that the moon is made of cheese, or that all Muslims are terrorists), but we must allow them to say these things, because by inhibiting THEIR freedom of speech, we in turn inhibit our own. Denying one person freedom to say what they believe takes the principle out of the 1st Amendment and makes it completely meaningless. At the moment, we have three main "ideals" I guess, that curb freedom of speech. Intentionally malicious libel/slander that hurts the reputation of a public figure is illegal: however, to sue for damages, the "victim" must prove that the libel/slander was malicious, intentional, and reputation-harming. The second "rein" in on freedom of speech is that you can't shout fire in a crowded theatre. That's the idea: you can't say something that would bring others to harm (notice that's not "you can't say something that would cause others to perform illegal acts"). Most of the time, this too is decided on a case by case basis, and it would have to be proven that what was said caused, or had a high probability of causing, harm to others. With that being said, if any of you are planning on broadcasting something that MIGHT be harmful to others, I would advise against it. The third curb is about obscenity. I like the way this law is done. Basically, you can't publish anything that has no scientific/artistic/etc value AND is considered "obscene" by [i]your community[/i]. This is clever and a very important point, because there are some things that are considered incredibly obscene in some parts of the States but in others are "no big deal."
So that's a break-down of how free speech works in the U.S. For the most part, I agree with these laws. I do not agree with   some of the curbs made by the Patriot Act, however. But yes, I would fully support allowing hate speech (with no calls to violence) of any sort, even though I definitely DON'T agree with it, because I believe in the American ideal.

[quote="Asian-American"]So, I would make a law requiring these media giants to be broken up into perhaps one thousand smaller groups.  Then, I would pass a law requiring these media sources to give airtime to every political/ideological/philosophical views: the time allocated to each view can perhaps depend on the percentage of the population that supports each view.  For example, if there are 5% of the population who supports the Ralph Nader/green party view, then 5% of the media airtime would have to be given to the green party types.  If 5% support the Libertarian Party, then 5% of media go to them.  And if 1% of the population supports some "racist" views, like the organization of Stormfront.org, then the media would be legally required to give them 1% of their airtime. [/quote]
I'm not sure yet how I feel about the principles behind this idea, but I believe that implementing it fairly would be extremely difficult if not impossible.

First of all, how do you decide who to cover?  For example, say you dedicate 30% of your coverage to the Republican Party on the immigration issue.  But how do you break up this coverage?  Within the Republicans there are a wide range of views.  If you want to do it based on the positions of the leaders, then which leaders?

Secondly, you said that this law would require all major media sources to comply with the law.  How do you define a "major media source"?  Say, for example, the newspaper of a midsize city.  Requireing such a newspaper to measure out the exact percentages, in my opinion, would be unusually burdensome and time consuming.  And what would you do about media sources that are inherently biased?  For example, the Christian Science Monitor.  Forcing them to give coverage supportive to, say, atheist groups would be contradictory.

In addition, what would you do about parties and ideas that basically have "guaranteed" newstime?  Say there are 20 anti-abortion rallies over the year, but only 2 pro-abortion rallies.  (I know that's unlikely, but that's not the point.)  Naturally, the anti-abortion activists will receive more media coverage, even if the proportion of people supportion abortion is larger.

Basically, what I feel about your proposed law is that it takes away the media's power of agenda setting - one that I believe is crucial for a free press.

Note:  I'm typing this late at night.  I had quite a few other ideas that I could have put here and could definitely improve my explanations and writing, but I'm tired and about to sleep.

[quote="coolmathematician"]TO fredbel6:
what you say about the Turkish governments' act on Pamuk issue is not right...
In Turkey there are free courts and there is a constitution ...if you swear Turks you are asked why you did that...
also I think lying and blaming the people is not a FREEDOM...
you need to search some history without prejudices...it is not a genocide and we will never accept a LIE...[/quote]

I assume you are from Turkey.  It doesn't matter, I will respond.  Did you walk around in all of Turkey in 1915 checking up on everyone?  No?  Then you cannot confirm or deny an Armenian genocide for sure.  It's that simple.
What you are saying is just in my eyes incomprehensible : you can say everything in Turkey unless it's a lie or insult?  Who decides when someone is insulting or lying?  The Turkish government?
Is the world supposed to believe that it isn't true, BECAUSE it is illegal to write it's true?  If it's really not true, why can't it be discussed?


function
geometry
geometric transformation
linear algebra
matrix
algebra
domain

For implicit function theorem,

can someone show me the translation explicitly such that the function is centered at the origin??
Also, why can we assume that the jacobian matrix to be the identity matrix??

thx for the help

Please show us your implicit function theorem.

ops, sorry,

Let [b][i]g[/i][/b]: Rn x Rm -> Rm be C1, and let [b]x0[/b] in Rn, and [b]y0[/b] in Rm such that [i][b]g[/b][/i]([b]x0[/b], [b]y0[/b])=0. Assume that the Jacobian determinant doesn't not equal to zero at ([b]x0[/b], [b]y0[/b]). Then there exist numbers [b]a[/b]>0 and [b]b[/b]>0, and a unique function [i][b]F[/b][/i]: (a [b]a[/b]-neighborhood of [b]x0[/b]) -> (a [b]b[/b]-neighborhood of [b]y0[/b]), with [i][b]F[/b][/i]([b]x0[/b])=[b]y0[/b], and [i][b]g[/b][/i]([b]x[/b], [i][b]F[/b][/i]([b]x[/b]))=0 for all x inside the domain of [i][b]F[/b][/i].

[quote="math beginner"]For implicit function theorem, can someone show me the translation explicitly such that the function is centered at the origin??[/quote]

What do you mean?

[quote="math beginner"]why can we assume that the jacobian matrix to be the identity matrix??[/quote]

Jacobian determinant doesn't not equal to zero, I do not understand your question.

1) I mean, why we can assume [b]x0[/b]=0, [b]y0[/b]=0, and[i][b] g[/b][/i](0,0)=0 instead of arbitrary  [b]x0[/b] and[b] y0  [/b]?? I know it's by translation of coordinate, but I'm not sure how to write it out explicitly.

2) In the general implicit function theorem, the Jacobian matrix is an arbitrary invertible matrix. But in the restricted case ([b]x0[/b]=0 and [b]y0[/b]=0), why can we assume the Jacobian matrix to be the identity?
Is it simply because any invertible matrix has non-zero determinant? Which is the same as the identity matrix??

Hope it is clear enough....??

The translation? Let $ h(x,y)\equal{}g(x\plus{}x_0,y\plus{}y(0))\minus{}g(x_0,y_0)$. Run this version of the argument, finding $ F_h$ such that $ h(x,F_h(x))\equal{}0$. Now we want $ F_g$ so that $ g(u,F_g(u))\equal{}g(x_0,y_0)$; this is given by $ F_g(u)\equal{}y_0\plus{}F_h(u\minus{}x_0)$.

Adjusting the derivative? Let $ g(0,0)\equal{}0$ and suppose $ A\equal{}\frac{\partial g}{\partial y}(0,0)$ is invertible as stated. Define $ h(x,y)\equal{}g(x,A^{\minus{}1}y)$: $ \frac{\partial h}{\partial y}(0,0)\equal{}A^{\minus{}1}\frac{\partial g}{\partial y}(0,A^{\minus{}1}0)\equal{}A^{\minus{}1}A\equal{}0$. Once we find $ F_h$ with $ h(x,F_h(x))\equal{}g(x,A^{\minus{}1}F_h(x))\equal{}0$, define $ F_g(x)\equal{}A^{\minus{}1}F_h(x)$.


Prove that, for any positive real numbers $a,b$ and $c$ such that $abc=1$, \[23+3(a^{2}+b^{2}+c^{2})\ge 4(1+a)(1+b)(1+c)\]

Yes, I solved it by mixing variables!
WLOG, assume that $a=min\{ a,b,c \}$
Let $f(a,b,c)=15+3(a^{2}+b^{2}+c^{2})-4(a+b+c)-4(ab+bc+ca)$
We have $f(a,b,c)-f(a,\sqrt{bc},\sqrt{bc})=(\sqrt{b}-\sqrt{c})^{2}(3b+3c+6\sqrt{bc}-4a-4)\geq 0$
So it suffices to prove that $f(a,b,b)\geq 0$ with $ab^{2}=1$ and $0<a \geq b$
But it’s very easy!!
$15+3a^{2}+6b^{2}-4a-8b-4b^{2}-8ab\geq 0$
$\Leftrightarrow$ $15+\frac{3}{b^{4}}+2b^{2}-\frac{4}{b^{2}}-8b-\frac{8}{b}\geq 0$
$\Leftrightarrow$ $(b-1)^{2}[2b^{2}(b-1)^{2}+3(b+1)^{2}]\geq 0$

Schur's is also good. Will it be a simpler solution?

[quote="phamduyhiep"]Yes, I solved it by mixing variables!
[/quote]

That's exactly my solution too. The same approach can also be used to prove Vasc's problem.

[quote="hunter_cvp_ic"][quote="Vasc"]The below inequality is sharper \[18+2(a^{2}+b^{2}+c^{2})\ge 3(1+a)(1+b)(1+c)\] [/quote]
No one solve this problem???
I want soln[/quote]

hunter_cvp_ic, just follow phamduyhiep's and think about arqady's remark.

[quote="Vasc"]The below inequality is sharper \[18+2(a^{2}+b^{2}+c^{2})\ge 3(1+a)(1+b)(1+c)\][/quote]
___________
Problem is equivalent to:
$ a,b,c>0,abc=1\Rightarrow 12+2(a+b+c)^2\ge 3(a+b+c)+7(ab+bc+ca) $
We note $ a+b+c=s,s\ge 3 $
From Schur's inequality we obtain:
$ (a+b+c)^3+9abc\ge 4(a+b+c)(ab+bc+ca)\Leftrightarrow ab+bc+ca\le\frac{s^3+9}{4s} $
It is sufficient that:
$ 2s^2+12\ge 3s+\frac{7(s^3+9)}{4s}\Leftrightarrow (s-3)(s^2-9s+21)\ge 0 $
whici is true!
_________
Marin Sandu


A multidigit number contains the digit 0.If you strike it the number becomes 9 times smaller.At which position is this 0 located . Find all such numbers.

does it contain only one zero or if it contains many zeros should we strike everything?

i'll assume that there is only one zero
this zero cant be in the last digit obviously. since the last digit of x and 9x is same it is 5
let the next to last digit be k
$ \implies 9k\plus{}4$ (carry over 4) should end in 4 which is not possible so k should be 4 and thus the next to last digit is zero.
the smallest number is 405


For which positive integers n do there exists positive integers x and y so that

1/x+1/y=3/n?

First rearrange using simon's favorite factoring method:

1/x+1/y=3/n
3xy=nx+ny
9xy-3nx-3ny+n^2=n^2
(3x-n)(3y-n)=n^2.

So there exist the desired x and y iff we can factor n^2 into a*b with a :equiv: b :equiv: -n mod 3.

If n = 3t, then we can factor n^2 as (3)(3t^2), yielding the solution x = t+1, y = t (t+1).

If n = 3t+2, then we can factor n^2 as (1)((3t+2)^2) yielding the solution x = t+1, y = (t+1)(3t+2).

If n  :equiv: 1 mod 3, then n^2 must have a divisor congruent to 2 mod 3 in order for there to be a solution, as was explained above. If p is a prime divisor of n with p :equiv: 2 mod 3, then we find the solution x=(n+p)/3, y=(n)(n+p)/(3p).These are integers because 3|n+p and p|n. If no such prime divisor exists, then all prime divisors of n^2 are congruent to 1, and so every divisor of n^2 is congruent to 1, and hence no solution exists.

So a solution x,y exists for every n unless n is a product of primes all congruent to 1 mod 3.


analytic geometry
geometry
rectangle
combinatorics unsolved
combinatorics

I hope it is understoodable...
I forgott to mention: the other $N-1$ points are inside the square, $(i,j)$, $1 \leq i,j \leq N-1$.
For any 2 points $A_k$ $(i_k,j_k)$, $A_l$ $(i_l,j_l)$ we must have: $i_k \neq i_l$, $j_k \neq j_l$

Someone show me in a second how trivially is this...  :blush:
The maximized area is exactly $\frac{N(N+1)}{2}$.

[quote="heartwork"]Consider the square of some given length $N$ and vertices $(0,0)$, $(0,N)$, $(N,N)$, $(N,0)$.
Consider $N$ points with integer coordiantes as it follows: one in $(0,0)$ and some other $N-1$ of the form $(i,j)$ such that any 2 points does not have any common cordinate - $(N-1)!$ cases.
For each case take a cover of the huge square with $N$ rectangles starting from one corner in such a point and the edges parallel to the axis - to the right and to the top of the huge square. The rectangles are not allowed to overlap. For each case take a cover that maximizes the sum of the rectangles area.
Prove that no matter what case we will choose the maximized area is at least $\frac{N(N+1)}{2}$.[/quote]

What if we consider the same problem ... instead of $N$ points, less points - $M$ with the same rule: any 2 points does not have a common cordinate. ($1 < M \leq N$)
Cover the huge square with $M$ rectangles starting from one corner in such a point and the edges parallel to the axis - to the right and to the top of the huge square. The rectangles are not allowed to overlap. For each case take a cover that maximizes the sum of the rectangles area.
Prove that no matter what case we will choose the maximized area is at least $\frac{N(N+1)}{2}$.


An integer sequence $ (a_1, a_2, . . . )$ is said to be white, if for all n $ > 2008$, an is equal to the total number of those indices $ i, 1\leq i \leq n\minus{}1$ for which $ a_i \plus{}i \geq n$. An integer $ L$ is an important element of the sequence $ (a_1, a_2, . . . )$, if $ a_j \equal{} L$ for infinitely many different indices$ j$. What is the maximal possible number of important elements of
a white sequence?

Answer: 2
But i can't prove, with value's 3, no exist

November problem of : http://www.fen.bilkent.edu.tr/~cvmath/Problem/problem.htm

Solution is

http://www.fen.bilkent.edu.tr/~cvmath/Problem/0811a.pdf


calculus
integration
logarithms
calculus computations

[quote="sos440"]Let $ n$ be a positive integer. Evaluate $ \int_{0}^{\infty} \frac {\ln x}{(1 \plus{} x)^{n \plus{} 2}} \, dx$

[hide="Answer"]$ \minus{} \frac {H_n}{n \plus{} 1}$, where $ H_n \equal{} \sum_{k \equal{} 1}^{n} \frac {1}{k}$[/hide][/quote]

$ J\equal{}\int_{0}^{\infty} \frac {\ln x}{(1 \plus{} x)^{n \plus{} 2}} \, dx\equal{}\minus{}\frac{1}{n\plus{}1}\frac{\ln x}{(x\plus{}1)^{n\plus{}1}}\bigg|_0^\infty\plus{}\frac{1}{n\plus{}1}\int_0^\infty \frac{1}{x(1\plus{}x)^{n\plus{}1}}\, dx\equal{}\frac{1}{n\plus{}1}\int_0^\infty \frac{1}{x(1\plus{}x)^{n\plus{}1}}\, dx$
Let $ I_n\equal{}\int_0^\infty \frac{1}{x(1\plus{}x)^{n\plus{}1}}\, dx$ then if $ x\equal{}\frac{1}{t}\implies \frac{1}{x^2}dx\equal{}\minus{}dt$ so we got $ I_n\equal{}\int_0^\infty\frac{t^n}{(1\plus{}t)^{n\plus{}1}}dt\equal{}\minus{}\frac{1}{n}\frac{t^n}{(1\plus{}t)^n}\bigg|_0^\infty\plus{}\int_0^\infty\frac{t^{n\minus{}1}}{(1\plus{}t)^{n}}dt\equal{}\minus{}\frac{1}{n}\plus{}I_{n\minus{}1}$ So $ I_n\minus{}I_1\equal{}\minus{}(\frac12\plus{}\cdots\frac{1}{n})$
So $ J\equal{}\frac{1}{n\plus{}1}I_n\equal{}\minus{}\frac{1\plus{}\frac12\plus{}\cdots\frac{1}{n}}{n\plus{}1}$


Prove: If BD and CE are altitudes of triangle ABC, then DE and BC are antiparallel with respect to AB and CA.

Prove: If circles with diameters the medians of two vertices are drawn in an acute triangle, the vertex of the triangle is collinear with the intersection points of the two circles.

Prove: If squares ABDE and BCKM are in the exterior of triangle ABC, then DM is twice the length of the median of the triangle from B.

Given: AC is a diameter of circle O. AD, DE, and CE are tangents to the circle. Prove: M is the midpoint of BT and that BT is perpendicular to AC.

This looks like homework, so I'm going to be brief.

[quote="amylopectin"]Prove: If BD and CE are altitudes of triangle ABC, then DE and BC are antiparallel with respect to AB and CA.[/quote]

Depending on your definition of antiparallels, you need either a circle or equal angles. If it's a circle, then find one - it's easy. If it's equal angles, then find the circle first. ;)

[quote="amylopectin"]Prove: If circles with diameters the medians of two vertices are drawn in an acute triangle, the vertex of the triangle is collinear with the intersection points of the two circles.[/quote]

Exercise to radical axes? Prove that two powers are equal.

[quote="amylopectin"]Prove: If squares ABDE and BCKM are in the exterior of triangle ABC, then DM is twice the length of the median of the triangle from B.[/quote]

How do you obtain - geometrically - twice the length of the median of the triangle from B? For instance, by reflecting B in the midpoint of CA. This gives you a parallelogram... Do you see some congruence now?

[quote="amylopectin"]Given: AC is a diameter of circle O. AD, DE, and CE are tangents to the circle. Prove: M is the midpoint of BT and that BT is perpendicular to AC.[/quote]

Huh??

By the way, please don't post many problems in one topic unless they are closely related to each other (all being from one homework sheet doesn't count as such), and please post problems of this level into Pre-Olympiad.

  Darij

Thank you for the help, darij grinberg. Next time I will post in the right section.


[b]   f,g R      R    Find all of the fuction for real x,y such that  f(x+yg(y) )=xf(y)-yf(x)+g(x)[/b]

$y=0 \Longrightarrow f(x)=xf(0)+g(x)\Longrightarrow g(x)=f(x)-xf(0)$
$g(x)=f(x)-xf(0) \Longrightarrow f(x)+y(f(y)-yf(0)))=xf(y)-yf(x)+f(x)-xf(0)$
$y=1 \Longrightarrow f(x+f(1)-f(0)) = x(f(1)-f(0))$ , so $f(x),g(x)$ have  the form $f(x)=ax-a^2 , g(x)=f(x)-xf(0)=(a+a^2)x-a^2$
Now I replace in the initial équation and obtain $a=0,a=-1$.
Finally the solutions are $(1)f(x)=g(x)=0 ,\forall x\in R  ;; (2) f(x)=-x-1 ,\  g(x)=-1 ,\forall x\in R $

 :cool:


calculus
derivative
function
logarithms
real analysis
real analysis unsolved

What is the derivative $ dx$ of:

$ f(x)\equal{} (a^x\plus{}b^x\plus{}c^x)^{\frac{1}{x}}$?

Just calculate $ \frac {d}{dx}\ln f(x)$.


induction
real analysis
real analysis unsolved

\[ \left(1\plus{}x\plus{}\frac{x^2}{2!}\plus{}\cdots\plus{}\frac{x^n}{n!}\right)\left(1\minus{}x\plus{}\frac{x^2}{2!}\plus{}\cdots\plus{}(\minus{}1)^n\frac{x^n}{n!}\right)\equal{}1\plus{}(\minus{}1)^n\left(\frac{x^n}{n!}\right)^2\]

how to prove the equation ?
thanks

There is no way: it is false. Just take $ n\equal{}3$ and open the parentheses. In general, when you see something like that and it is correct, induction is your best friend.


In a jar there is a minimum of six US coins of every denominations ( pennies, nickels, dimes, quarters). In how many ways can you take 6 coins from the jar?

p.s. I am looking for original ways to solve this problem.

This is the same as giving $ 4$ distinguishable items to $ 6$ indistinguishable items if duplicates are allowed, which has $ \dbinom{4\plus{}6\minus{}1}6\equal{}\boxed{84}$ (i.e. number of solutions to $ p\plus{}n\plus{}d\plus{}q\equal{}6$).

How would you explain this to someone who never heard of Combinations?
Is it possible to solve this problem at the middle school level? (without listing all possibilities)
Is there a geometric way?

 o o o o o o
 o o o o o o
 o o o o o o
 o o o o o o

o o | o |o o |o

 9
 C  = 84
 3

math154 - right on !


What is the difference of the angle formed by the second hand and the minute hand and the angle formed by the minute hand and the hour hand at $2: 26: 49$?

[hide] Firstly, let's measure the angle between the seconds hand and the 12:00 point.  Since each second moves the second hand 6 degrees, we have that 49 seconds moves it 294 degrees.   Therefore, the minor angle is 360-294=66 degrees.  Now, look at the hour hand.  Since the hour hand is past to 2:00, we can say it is more than 60 degrees past the 12:00 (since each hour moves the hour hand 30 degrees).  Now, note that the hour hand moves for every minute and second porportionally.  Since there 60 seconds in a minute, you could say that 26 minutes has 1560 seconds; hence 26 minutes and 49 seconds has 1609 seconds.  Now, there are 3600 seconds in an hour, so 26 minutes and 49 seconds is 1609/3600 of an hour.  Since every hour is 30 degrees, we have that 26 minutes and 49 seconds makes an angle of (1609/3600)*30=1609/120=13+49/120 degrees with the 2:00 point.  Therefore 2:26:49 is 60+13+49/120=73+49/120 degrees away from the 12:00 point.  Now, since the seconds hand and hour hand are on opposite sides of the hour hand, we can say that the angle between them is 73+49/120+66=$139\frac{49}{120}$, which is our answer.  [/hide]

Because all the hands move proportionally, the answer your'e going to come up 73 +49/120+66 which gives you  139 49/120 as your answer.

I'm not sure what you people were solving for, but in the picture, I want to know what $\angle{a}-\angle{b}$ is

Oh!  Sorry  :blush:.  We were calculating the angle formed between the hours and second hand.  Here is the solution to your question:
[hide] Since each minute contains 6 degrees, we have that 26 minutes contains 156 degrees ("contains x degrees" means that it is x degrees away from the 12:00 point).  We also know that 49 seconds is 49/60 of a minute, which would make it go an extra (49/60)*6=4.9 degrees.  Therefore, 26 minute and 49 seconds contains 156+4.9=160.9 degrees.  Now, in the previous solution, we noted that the hour hand contains $73\frac{49}{120}$.  Therefore, we know that $b=160.9-73\frac{49}{120}$.  Now again, we can note from the previous problem that the second hand contains a minor angle of 66 degrees with the 12:00 point.  We can say that The minute hand contains 360-160.9=199.1 degrees with the 12:00 point (because we can look at the reflex angle instead of the minor angle).  Thus, $a=199.1-66=133.1$.  Therefore, $a-b=133.1-(160.9-73\frac{49}{120})=133.1-160.9+73\frac{49}{120}=73\frac{49}{120}-27.8=73\frac{49}{120}-\frac{278}{10}=\frac{8809-278(12)}{120}=\frac{5473}{120}$, which is our answer. [/hide]

formula 4 finding the angle measure between second and minute hand is:   60m-59s in absolute value divided by 10

60(26)-59(49) over 10

1560-2891=-1331 which in absolute value would b positive 1331

1331/10=133.1

formula for the angle measure of the minute and hour hand is:
the positive value of 60h-11m divided by 2

60(2)-11(26) over 2
120-286=-166 which again in absolute value is positive 166
166/2=83

the answer probably should be 50.1


For any positive integer n, Prove $ 117^n \plus{} 600^n \minus{} ( \minus{} 8)^n \minus{} ( \minus{} 275)^n$ is divisible by 1000.

[hide]$ 1000\equal{}2^35^3$

since $ \gcd(2^3,5^3)\equal{}1$

we have only to prove

$ 117^n\plus{}600^n\minus{}(\minus{}8)^n\minus{}(\minus{}275)^n\equiv0\mod(2^3)$

$ 117^n\plus{}600^n\minus{}(\minus{}8)^n\minus{}(\minus{}275)^n\equiv0\mod(5^3)$

$ 117^n\plus{}600^n\minus{}(\minus{}8)^n\minus{}(\minus{}275)^n\equiv5^n\minus{}(\minus{}3)^n\mod(2^3)$

for $ n\equiv1\mod2$ we have $ 5^n\plus{}3^n\equiv\minus{}3^n\plus{}3^n\equiv0\mod(2^3)$

for $ n\equiv0\mod2$ we have $ 5^n\plus{}3^n\equiv3^n\minus{}3^n\equiv0\mod(2^3)$

now we will do the same work for $ 5^3$

we have $ 117^n\plus{}600^n\minus{}(\minus{}8)^n\minus{}(\minus{}275)^n\equiv(\minus{}8)^n\plus{}(\minus{}25)^n\minus{}(\minus{}8)^n\minus{}(\minus{}25)^n\equiv0\mod(5^3)$[/hide]


algebra
system of equations
superior algebra
superior algebra unsolved

Hello,
consider the finite field $ GF(q^2)$ and let $ \bar{x}$ denote the unique involutive automorphism.

Consider this system of equations :

$ a \bar{c}\equal{}\bar{a} c$
$ b \bar{d}\equal{}\bar{b} d$
$ a d \minus{}b c \equal{}1$
$ a \bar{d}\minus{} \bar{b} c\equal{}1$

Prove that $ a,b,c,d$ are all fixed by the involutive automorphism.
It's supposed to be pretty obvious but I'm afraid I'm making things too hard. :huh:

First show $ b$ is fixed by multiplying the fourth equation by $ b$ to eliminate $ \bar{d}$.

Now $ a\bar{d}\equal{}ad$ and $ b\bar{d}\equal{}bd$. Since $ a,b$ not both 0, $ d$ is fixed.

We can now obtain the equations
\[ da\minus{}cb\equal{}1,d\bar{a}\minus{}\bar{c}b\equal{}1\]
(The second follows from applying the automorphism to the first.) By similar reasoning, $ a,c$ are fixed.


Solve for $ x$ in terms of $ k$: $ 2^x\plus{}2^x\plus{}2^x\plus{}2^x\plus{}2^x\plus{}2^x\equal{}3\cdot2^k$

let's combine the terms on the left side of the equation

$ 6\cdot2^x\equal{}3\cdot2^k$

we'll now divide by 3

$ 2\cdot2^x\equal{}2^k$

now, let's focus on the left side of the equation: $ 2\cdot2^x$

can we simplify that any more?
we can

we know that $ 2$ is simply $ 2^1$
we also know that if we want to multiply numbers with the same base but different exponents, we simply add the exponents.

so the left side becomes $ 2^{x\plus{}1}$

now we rewrite the equation

$ 2^{x\plus{}1}\equal{}2^k$

since the base is the same, that means that for the equation to work, the exponent must be the same as well.

so, $ k\equal{}x\plus{}1$ and we are done

We can change this equation into
\[ 6(2^x) \equal{} 3\cdot2^k\implies 2(2^x) \equal{} 2^k \equal{} 2^{x \plus{} 1}
\]
In this case, for this equation to be true, their exponents must be equal.  Therefore:
\[ x \plus{} 1 \equal{} k\implies x \equal{} \boxed{k \minus{} 1}
\]
BEATEN, but it was asking for $ x$, not $ k$...


[quote="Mathias_DK"][quote="pifi"]$ a,b,c > 0$

$ (a^5 \minus{} a^2 \plus{} 4)(b^5 \minus{} b^2 \plus{} 4)(c^5 \minus{} c^2 \plus{} 4)\geq3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$[/quote]
$ a^5 \minus{} a^2 \plus{} 4 \ge a^3 \plus{} 3 \iff (a^3 \minus{} 1)(a^2 \minus{} 1) \ge 0$ is true.
So $ (a^5 \minus{} a^2 \plus{} 4)(b^5 \minus{} b^2 \plus{} 4)(c^5 \minus{} c^2 \plus{} 4) \ge (a^3 \plus{} 3)(b^3 \plus{} 3)(c^3 \plus{} 3)$ $ > 9(a^3 \plus{} b^3 \plus{} c^3) \ge 3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$. There is never equality.[/quote]
This can be improved as

$ (a^5 \minus{} a^2 \plus{} 4)(b^5 \minus{} b^2 \plus{} 4)(c^5 \minus{} c^2 \plus{} 4)\geq 6\,(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca),$

but $ (a^5 \minus{} a^2 \plus{} 4)(b^5 \minus{} b^2 \plus{} 4)(c^5 \minus{} c^2 \plus{} 4)\geq 7\,(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$ is not true.


Ich hab gehört es soll ganz lustig gewesen sein.
Könnt ihr, die die da waren, ein paar Sätze sagen wie es für Euch war? Würd mich mal so persöhnlich interessiern. Okay ich  muss gestehen ich hab unteranderem Angst davor Friedrich dabei helfen zu müssen einen Bericht darüber in unser Jahrbuch zu schreiben, weil es gibt bei uns so Lehrer die wollen des und ich weiß wie unlieb Friedrich Aufsätze schreibt und ich mach sowas halt noch ganz gerne wenn des Thema gut ist-aber nur mit entsprechend aussreichenden Informationen.
Und hat jemand ein Bild von allen Preisträgern einschließlich Darij?
Wenn ja könnt ihr mir des mal schicken oder hier on stellen und villeicht mit Namen versehen? weil auf dem Bild das ich gesehen hab konnte ich nur Peter, Annika, Friedrich und Anne identifizieren. (Darij hätte ich glaub auch noch gewusst wenn er dagewesen wäre)

Idril

Hier findest du Bilder der Bundessieger.

http://www.math.uni-goettingen.de/mo/demo05/900/sieger

Im dritten Bild:
rechts hinten z.B. ist Darij daneben Christian Sattler und ganz rechts müsste Peter Scholze sein. Den Rest kenne ich vom Namen nicht.
Achja vorne das zweite Mädchen ist ein Mädchen aus der sechsten Klasse. Die herrausragend war schon alleine deswegen weil sie in Klasse acht gestartet ist.

Die Aufgaben aus 11-13 fand eigentlich leicht. Sonst war es ganz cool. Man konnte Tischtennis spielen und halt die Natur besichtigen weil die Jugendherberge am Rand von Saarbrücken lag.
Falls du noch was konkretes wissen willst, dann frag einfach.


Six trees are equally spaced along one side of a straight road.  The
distance from the first tree to the fourth is $ 60$ feet.  What is the
distance in feet between the first and last trees?

From the first to fourth tree, there are 3 spaces of 20 feet (60/3=20).  From the first to last tree, there are a total of 5 spaces so the total distance is $ 20\times 5\equal{}100$.

Please explain how you got those numbers...

60 for 4 trees so 60/4=15 per tree. 6 trees so 6x15=90.

Oh BETWEEN TREES! Thanks JRav (even if you are Mr. L)!


Numbers $ 1$ through $ 10$ are written on a board. At any time, numbers $ a$, $ b$, $ c$
may be replaced by
\[ \frac{2(b \plus{} c) \minus{} a}{3} ,
\frac{2(c \plus{} a) \minus{} b}{3} ,
\frac{2(a \plus{} b) \minus{} c}{3} \]
Can a number greater than $ 20$ eventually appear on the board?

[hide="Hint 1"]
Notice that
$ \frac {2(b \plus{} c) \minus{} a}{3} \plus{}\frac {2(c \plus{} a) \minus{} b}{3} \plus{}\frac {2(a \plus{} b) \minus{} c}{3}\equal{}a\plus{}b\plus{}c$
thus the sum of all the numbers is invariant an equal to $ 45$.
[/hide]
[hide="Hint 2"]
WLOG $ a \le b \le c$
Similarly with the differences taking two by two:

$ \frac {2(b \plus{} c) \minus{} a}{3} \minus{}\frac {2(c \plus{} a) \minus{} b}{3}\equal{}b\minus{}a$
$ \frac {2(c \plus{} a) \minus{} b}{3} \minus{}\frac {2(a \plus{} b) \minus{} c}{3}\equal{}c\minus{}b$
$ \frac {2(b \plus{} c) \minus{} a}{3} \minus{}\frac {2(a \plus{} b) \minus{} c}{3}\equal{}c\minus{}a$
[/hide]

[hide="huge hint"]Take the sum of the squares.[/hide]

[quote="mszew"][hide="Hint 1"]
Notice that
$ \frac {2(b \plus{} c) \minus{} a}{3} \plus{} \frac {2(c \plus{} a) \minus{} b}{3} \plus{} \frac {2(a \plus{} b) \minus{} c}{3} \equal{} a \plus{} b \plus{} c$
thus the sum of all the numbers is invariant an equal to $ 45$.
[/hide]
[hide="Hint 2"]
WLOG $ a \le b \le c$
Similarly with the differences taking two by two:

$ \frac {2(b \plus{} c) \minus{} a}{3} \minus{} \frac {2(c \plus{} a) \minus{} b}{3} \equal{} b \minus{} a$
$ \frac {2(c \plus{} a) \minus{} b}{3} \minus{} \frac {2(a \plus{} b) \minus{} c}{3} \equal{} c \minus{} b$
$ \frac {2(b \plus{} c) \minus{} a}{3} \minus{} \frac {2(a \plus{} b) \minus{} c}{3} \equal{} c \minus{} a$
[/hide][/quote]
That's what I tried; I failed. How did you continue along with this proof?

On that note, we know that 20 is an upper bound. How would we attempt to find the smallest upper bound of the terms of the sequence?


Let P be a polynomial with integer coefficients, let a be its leading coefficient. It is known that P has degree n, and also n roots in the interval (0,1), and also, not all its roots are equal. prove that |a|>=2^n+1

A cute little problem :).

We have $|P(0)P(1)|\ge 1\Rightarrow a^2\prod x_i\prod (1-x_i)\ge 1$. On the other hand, $x_i(1-x_i)\le \frac 14$, with equality iff $x_i=\frac 12$. We combine these to get $a^2\ge 4^n$, with equality only if $x_i=\frac 12,\ \forall i\in \overline{1,n}$, which is impossible. This shows that $|a|>2^n\Rightarrow |a|\ge 2^n+1$.


In a far-off land three fish can be traded for two loaves of bread
and a loaf of bread can be traded for four bags of rice. How many
bags of rice is one fish worth?  Express your answer as a common fraction.

since 2 loaves = 3 fish and 1 loaf = 4 rice, 3 fish = 8 rice, making one fish $ \boxed{8}{3}$ bags of rice.

[quote="Ebob"]since 2 loaves = 3 fish and 1 loaf = 4 rice, 3 fish = 8 rice, making one fish $ \boxed{\frac{8}{3}}$ bags of rice.[/quote]

Fixed... <-- this is text of my own, but apparently not


trigonometry vector geometry parallelogram

Given that reals $a,b,c,d$ satisfy $a^2 + b^2 = c^2 + d^2 = 1$ and $ac + bd = \frac{1}{2}$, determine the value(s) that $ad-bc$ can take.

[quote="shobber"]You can try trig substitution.

I got: $[-1,1]$.[/quote]

are you sure ? By trigo substitution

say $a=\sin\alpha$ , $b=\cos\alpha$ ,$c=\sin\beta$ , $d=\cos\beta$ then we have

$\cos (\alpha-\beta)=\frac{1}{2}$ . And we wish to find $\sin (\alpha-\beta)$

But we know that $\sin^2(\alpha-\beta)+\cos^2(\alpha-\beta)=1$ . So $\sin(\alpha-\beta)=\pm\frac{\sqrt 3}{2}$ .

Is it correct ?

basically what the subject says.
how would you do the trig substitution?

never mind. I agree with the trig substitution person.

Geometric interpretation: Two vectors of unit magnitude with inner product $1/2$ - how does this relate to the (signed) area between them?

From the first relation, we have $|a\cdot b|=|a||b|\cos\theta\implies \cos\theta=1/2$

But we also have $|a\times b|=|a||b|\sin\theta$

But the cross product is the area of the parallelogram spanned by two vectors, so the result follows easily.

Yes.  I had in mind 3 solutions -  trig, vector, and this one

$(a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2$

This is another problem but I didn't know how to make it a separate post.
Find the maximum value of x-y if 2(x^2 + y^2) = x + y


If $2$ shirts and $3$ pairs of socks cost $36$, and $3$ shirts and $5$ pairs of socks cost $55$, find the cost of $1$ shirt and the cost of $1$ pair of socks.

Those kind of problems are becoming popular on the forum
[hide]$2x+3y=36\rightarrow y=-\frac{2}{3}x+12$
$3x+5y=55\rightarrow y=-\frac{3}{5}x+11$ $\Rightarrow -\frac{2}{3}x+12=-\frac{3}{5}x+11 \rightarrow x=15$, therefore the socks would be $2(15)+3y=36 \rightarrow y=2$
Shirts=15
Socks=2 ;) [/hide]

[quote="lotrgreengrapes7926"]We should start a marathon about money. :D

A hexagonal coin fits exactly inside a circular coin of radius 1. The circular coin fits exactly inside another hexagonal coin. What is the positive difference in area between the 2 hexagonal coins? Express in simplest radical form. :D[/quote]
[hide]
I got the side lengths of the hexagons (assuming they're regular >.>) as $\frac{2\sqrt{3}}{3}$ and $1$... so the areas are $2\sqrt{3}$ and $\frac{3\sqrt{3}}{2}$ respectively so the difference is $\boxed{\frac{\sqrt{3}}{2}}$?[/hide]

[quote="Knuth's_check"]I am a bit too lazy to think the whole thing through. Anyway, I think that there is 0.545 difference between the circular coin and the hexgonal one inscribed in it. :|[/quote]
No calculators. And I got 0.544 for that answer. $\pi-\frac{3\sqrt3}{2}\approx0.5435$.
You're correct chinesechess. If you want to do a marathon, you can post the next problem.

Oh. ok.
Hmm...money...
A coin has radius 10 mm. Another has radius 6 mm. They are both 1 mm thick. If the smaller coin is stacked concentrically (?) onto the larger coin, what is the surface area of the resulting figure?

Yeah. But if you used the method where you found the area of the top of the large coin that isn't covered by the smaller coin, then it would matter.

Find the area of the coin and subtract that twice from the combined surface areas of the coins.

[quote="fireemblem13"][hide]$168\pi$  divide the entire surface area into the two lateral areas and the two faces of the larger coin and add the areas.

it wouldn't matter whether or not the smaller coin was stacked concentrically, right? [/hide][/quote]
[hide] Don't the faces of the larger coin add up to $200\pi$ already? Then I got $32\pi$ for the lateral area. :? $\boxed{232\pi}$ :huh: [/hide]

I flip a coin 4 times. What is the probability that the coin will show up heads exactly 2 times?

[hide]There's $2^4=16$ ways for the order of the coins, and there's 6 ways to have 2 heads. Answer $\frac{6}{16}=\frac{3}{8}$.[/hide]


Solve the inquation: $\left(\frac{1}{3}\right)^{\log_{\frac{3}{2}}\left[\log_{\frac{1}{3}}\left(\frac{x^{2}}{2}+2^{\log_{2}x-1}\right)+3\right]}\ge 1$


\[ \tan 3x \minus{} \tan 2x \equal{} 2\]

hello, your equation
$ \tan(3x)\minus{}\tan(2x)\minus{}2\equal{}0$
can be written in the following form
$ \frac{\csc(\frac{\pi}{4}\minus{}x)\csc(\frac{\pi}{4}\plus{}x)\sec(x)(\cos(5x)\plus{}\sqrt{2}\sin(\frac{\pi}{4}\minus{}x))}{2(\minus{}1\plus{}2\cos(2x))}\equal{}0$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your equation
$ \tan(3x) \minus{} \tan(2x) \minus{} 2 \equal{} 0$
can be written in the following form
$ \frac {\csc(\frac {\pi}{4} \minus{} x)\csc(\frac {\pi}{4} \plus{} x)\sec(x)(\cos(5x) \plus{} \sqrt {2}\sin(\frac {\pi}{4} \minus{} x))}{2( \minus{} 1 \plus{} 2\cos(2x))} \equal{} 0$
Sonnhard.[/quote]

Is it a solution????

then I propose following equivalent:

$ \tan(3x) \minus{} \tan(2x) \equal{} 2\Longleftrightarrow \frac {\sin 3x}{\cos 3x}\minus{}\frac {\sin 2x}{\cos 2x}\minus{}2\equal{}0$

$ \Longleftrightarrow \sin 3x\cos 2x\minus{}\sin 2x\cos 3x \minus{}2\cos 3x\cos 2x\equal{}0$ and $ \  \cos 3x\cos 2x\ne 0$

$ \Longleftrightarrow \sin (3x\minus{}2x) \minus{}\cos 5x\minus{}\cos x\equal{}0$ and $ \  \cos 3x\cos 2x\ne 0$

$ \Longleftrightarrow \sin x \minus{}\cos x\minus{}\cos 5x\equal{}0$ and $ \  \cos 3x\cos 2x\ne 0$

$ \Longleftrightarrow \sqrt {2}\sin \left (x\minus{}\frac {\pi}{4} \right)\minus{}\cos(5x)\equal{}0$ and $ \  x\ne (2n\plus{}1)\frac {\pi}{4},\ (2n\plus{}1)\frac {\pi}{6},\ n\in \mathbb I$


linear algebra
calculus
calculus computations

I am taking ODE and Linear Algebra course and I couldn't solve a few questions. I am posting them here.

The question was $ (3x\plus{}y)dy \equal{} (x\plus{}3y)dx , y(0) \equal{} 1$

As it is a homogeneous differential equation I substituted $ y\equal{}vx$ and I can't solve it further than the following expression :-

$ \frac{x}{dx} \equal{} \frac {(1\plus{}2v)}{(3\plus{}v)dv}$]

Is there any mistake till here?

Please simplify it for me.

Divide out by dx and move terms over:
$ (x\plus{}3y)\minus{}(3x\plus{}y)\frac{dy}{dx}\equal{}0$
Now this looks like an exact equation but it isn't (why?) so see if you can make it an exact equation with integrating factors and then solve it like any other exact equation.


inequalities
search
inequalities theorems

Just a question, i am new to inequalities n wish to know wat this notation means
   ∑
cyclic

Its a summation sign but wif the word cyclic written below it.

Any help provided will be much appreciated.

I love search option.  :)

[url]http://www.artofproblemsolving.com/Wiki/index.php/Cyclic_sum[/url]


Say a cycle on the vertex set $ [n]\equal{}\{1,2,\ldots,n\}$ is non-nesting if there are no $ a<b<c<d$ such that $ \{a,d\}$ and $ \{b,c\}$ are edges. Start at $ 1$, choose a direction and traverse the cycle returning to $ 1$. Let $ u$ be the number of increasing steps and $ d$ the number of decreasing steps. If $ n \geq 2$ prove or disprove that $ |u\minus{}d|$ is $ 0$ when $ n$ is even and $ 1$ when $ n$ is odd.

First, we notice that the parity of $u-d$ is the same as $n$, and thus it's enough to prove that $|u-d|\leq 1$.

Let prove the statement by induction on the cycle length ($n$). For lengths $2,3$ the statement trivially holds. Let's assume it holds for any length $<n$ and prove it for length $n$.

Let $C$ be a non-nesting cycle on $[n]$. Consider 3 cases depending on whether edges $\{n-1,n\}$ and $\{n-2,n\}$ belong to $C$:

1) both edges $\{n-1,n\}$ and $\{n-2,n\}$ are in $C$, then one of them is increasing, while the other is decreasing. Let $C'$ be obtained from $C$ by contracting these edges into a single vertex $n-2$. Then by induction we have $|u(C)-d(C)| = |u(C')-d(C')|\leq 1$.

2) $\{n-1,n\}$ is not in $C$. By adding to $C$ two copies of $\{n-1,n\}$, we can split it into two cycles, say $C_1$ and $C_2$, each having one copy of $\{n-1,n\}$, with direction of these copies imposed by the direction of $C$. In either case, one of these edges is increasing, the other is decreasing. Clearly, both $C_1$ and $C_2$ are non-nesting.

2.1) If at least one of $C_1$ and $C_2$, say $C_1$, has even length, then by induction $u(C_1)-d(C_1)=0$ and $|u(C_2)-d(C_2)|\leq 1$. Then $|u(C)-d(C)| = |u(C_1)-d(C_1) + u(C_2)-d(C_2)| \leq 1$.

2.2) If both $C_1$ and $C_2$ have odd length, then contracting edges $\{n-1,n\}$ in them into vertex $n-1$, we get non-nesting cycles $C'_1$ and $C'_2$ of even length, for which by induction $u(C'_1)-d(C'_1)=0$ and $u(C'_2)-d(C'_2)=0$. It follows that $u(C)-d(C) = u(C'_1)-d(C'_1) + u(C'_2)-d(C'_2)=0$.

3) $\{n-1,n\}$ is in $C$, but $\{n-2,n\}$ is not. In this case, we simply exchange labels $n-1$ and $n-2$, keeping $C$ non-nesting and preserving values of $u(C)$ and $d(C)$, and obtaining case 2 already considered above.

QED


[b]Problem.[/b] [i]Find the number of real solutions to the equation $x^{7}+|x|+x-12=0$.[/i]

i) $x>0$, then $x^{7}+2x-12$; one sign change, so precisely one positive root by descartes' rule of signs
ii)$x<0$, then $x^{7}=12$; no negative solution

1 real solution

[hide]if $x$ is negative, we have $x=\sqrt[7]{12}$ but $x<0$, so no solution in this case. otherwise note that the function $f(x)=x^{7}+2x-12$ is strictly increasing, so we can have at most one real solution. so the answer is one. [/hide]


geometry
AMC
AIME
calculus
LaTeX
AMC 8
percent
\/closed

The Art of Problem Solving Fall Schedule is available [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].

Each class meets once a week from 7:30 PM - 9 PM ET (4:30 PM - 6 PM PT).

Here are the classes we will offer this fall:

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*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=35]Introduction to Algebra[/url].  Thursdays, Oct 16 - May 7.
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Classes marked with a (*) have an accompanying textbook.  If you plan to enroll in one of these courses, you should do so soon, so that you receive the text before the class starts.

There are no classes from Nov. 24 through Nov. 28, and no classes from Dec. 20 through Jan. 4.  Many more details about each class can be found by clicking the appropriate class [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].

All of our courses offer full transcripts for each class, so if you're going to miss a class or two, you'll still have access to everything that happened in class.

We will host FREE Math Jams to discuss the classes.  The Math Jam schedule can be found [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]here[/url].

If you have any questions about a specific class, please contact us at classes@artofproblemsolving.com.

If we have already enrolled in a class thats going to start soon, how do we enter the course?

This is explained in the Course Information document that you can access on your class homepage.  You can access the class homepage by clicking My Classes, and then clicking your course.

mr. Ruscyck this is a little off topic but how long do you consider a good amount of time to finish aops V2. I know your saying it depends on the students, but say for the averagely good AoPser?

That's impossible to answer. It depends on how much you do each day, how intense you are, whether you do every problem, etc. Also, the "average AoPSer," is basically impossible to define. There are people with completely different abilities on this site, it makes no sense to find the average.

I have to agree with xpmath -- very hard to answer this question!  I know outstanding students who have taken a year or two to work through it, because they've started very young or they're doing other things.  I know others who have digested it in a few months (and others who finish it in a few months without doing many problems, which means they probably haven't learned much -- have to do the problems to really learn the math!)

yea i do all the problems- i think itll take like 5 months. Would you recommend the AIME series?
When is the sign up date?

You just have to sign up before the start date, which is (checks rrusczyk's post) erm... december 2.

Mr. Rusczyk, how do you pronounce your last name?

You can enroll up to 10 days after the first class.  For the AIME class, the last day that you can enroll is December 12.

From the [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=30]course description page[/url]:
[quote]Who should take this class

This class is appropriate for students who are pretty confident they will pass the AMC 10 or AMC 12, and are preparing for the AIME. If a student already consistently scores above 10 on the AIME, this class is probably not necessary, and if a student is unlikely to answer more than 1 or 2 questions correctly, then that student should start with some of our Introduction series of classes.[/quote]
Richard Rusczyk's last name is pronounced "russ-sick" where the s's kind of flow one into another.

[quote="isabella2296"]By any chance, Mr. Rusczyk, are you Polish?  :maybe:[/quote]

Dad's family is (and probably a little Russian, too).

I'm using my daughter's log in to ask a question for her. First some background:   She took Intro to Algebra last year and overall enjoyed the experience.  She loves the book, but towards the last half of class she felt a little behind.  She turned in all her challenge sets on time, with all problems attempted excluding the last set where she left off a couple of the most challenging.  She spent a fair amount of time learning to use LaTex, and how to compose answers.  (She was 9 during this class).  Our plan was to this year work through some material on our own to avoid the worry of feeling 'behind'.

We spent the rest of spring after the Algebra class, doing various problem solving materials and occasionally some challenging problems from her algebra book.  She enthusiastically began Intro to Geometry at the end of August.  We have been alternating chapters with it and AOPS V1.  She now says she misses class, the challenge sets, etc. So we are considering whether to let her do Intro to Geometry or the MathCounts/AMC8 basics class.  She did the AMC8 last year (scored 13 which is pretty good considering she could only get to about 12 problems in 40 minutes during our practice run, she's good but slow!).  Which class would be of more benefit?  She is 10, so it is not like there is any hurry to finish Geometry now.  She does love, love geometry though, and is highly motivated to do it now. Any thoughts on which class would be more appropriate?  Would it be insane to do both?  She has completed 5 chapters of Intro to Geometry already, so she will be a bit ahead for a short while.

Thanks!
Marti

this is impressive for a 10 year old- although there are some on this forum who are even more beast

since no one is replying, i'll just say this:
No it would not be insane to take both
Yes, it would be better if she took them one after another.
Yes, it would idea to buy the intro books

The biggest mistake i made in my whole life is not ordering all the intro books at the same time. If i had done that, i would be doing a lot better than i am now. And, i did it in the exact order your daughter is doing right now and i too did it when i was 9. But, i did it on my own and when i couldn't do the percent chapter, i felt like burning my book :P
EDIT: i'm still 11 so i can remember i experience vividly

I'm not sure, but given the information you provided, I think it might be best to take the MathCounts/AMC8 Basics class, but study Intro to Geometry independently. Since Geometry is probably a more challenging class than the MathCounts/AMC8 class (visualizing things at a young age is generally difficult, and the material is more difficult anyway), she would be less likely to feel left behind, while still getting the challenge sets, etc. and learning geometry. After that, you should probably just let her work on problem solving as a whole...if she ever gets bored you can start Number Theory or Counting and Probability.

She has *plenty* of time, so there's no rush to put her into/through MATHCOUNTS/AMC study.  I would probably have recommended Introduction to Number Theory next.  My main concern is that the geometry class requires a lot more proof writing and development, which few students are really ready for at 10 (though there are some who are).  If she's producing good results on the proofs you've had her do in the first 5 chapters, then she should be OK in the geometry class.   Both the number theory and the geometry courses will offer more in-depth study than the MATHCOUNTS/AMC class.  Use the latter as a survey course to prepare for actual competition.  The subject classes should form the main study for her.

Your plan for algebra is a good one -- make sure she spends time with the Challenge Problems on her second pass through.

Thanks!
Actually when I told her you suggested Number Theory next, she jumped at it.  Much better when it is not a mom suggestion.  She is content to wait a bit on geometry.  She is doing a good job with the proof writing, but it is hard work for her.  I'm all for learning the value of hard work, but I also want to preserve her love of math.  Right now she loves, loves, loves it, so as long as she's cool with waiting I think that is the best thing to do.  We'll play with some geometry in her math club and that should satisfy her.

We already have the text as I purchased all the intro texts when she was halfway through algebra.  When she hugged the book to her and said she 'loves' it, I assumed the other texts would be a safe buy. LOL

Thanks again for the work you do here.  When our daughter was tested at age 8, the tester told us to have her take the SAT and look toward getting her to the university very soon.  You have given us a much better option!!

Indeed, I agree that this is a much better option than university -- it would be better for her to wait for university, and attend when she can go to a university with her intellectual & social peers.  Many students who attend university very early get neither, as most top-tier schools won't admit students 3-4 years early.

its just a horrible experience to go into a university early (if you're taking a full blown year). I took a math course (trig) at a univ and it was one of the one of the worst experiences i ever had. and, if you're talking about doing freshman univ in when she's like 10 or something just don't.

No worries there - she's staying busy with AOPS and all her other fun activities.  She may end up taking some individual classes at the local state U in her teens for dual enrollment, but only if she wants to do so.


Let $ x,y,z,w$ be real numbers.

If
     $ x \ge y$
     $ z \ge w$

does iz follow $ xz \ge yw$? if so prove it else give a counter example.

[hide="solution"] $ x\equal{}1$,$ y\equal{}\minus{}2$,$ z\equal{}1$,$ w\equal{}\minus{}4$[/hide]

Ok, I came up with this,

If $ y \in R$ and $ \plus{}$  then

$ x/y \ge 1$

and if $ z \in R$ and $ \plus{}$  then

$ 1 \ge w/z$

so this becomes,

$ x/y \ge 1 \ge w/z$ $ \implies$ $ xz \ge wy$

Now try to pick $ x, y, z, w$ such that $ x \ge y$ and $ z \ge w$ ALSO satisfying that $ y, z \in R$ and positive.

I couldn't, sadly :/

the solution is basically to find a counter-example
[hide]
how to set up a counter example: (this is one method, not the only method)
make $ xz$ negative and $ yw$ positive
lets start with making $ yw$ positive
this means $ y$ and $ w$ have the same sign.
if they are both positive, then $ x$ and $ z$ will also be positive (but this makes $ xz$ positive)
if they are both negative, then the signs of $ x$ and $ z$ can still be either positive or negative

now look at $ xz$
if this product is negative, then $ x$ and $ z$ have different signs.
in this example, we will make $ x$ positive and $ z$ negative

so any ordered quadruple $ (x,y,z,w)$
with $ x$ positive and $ y,z,w$ negative will not satisfy the condition given $ (xz \ge yw)$

[/hide]
if you are interested, the condition that could be added so that $ xz \ge yw$ for any numbers would be
$ y,w \ge 0$
but this is not a biconditional


trigonometry
calculus
derivative
parameterization
parametric equation

Two things:
this looks like homework
parametric would go under intermediate

Otherwise, I'm still learning about parametric equations, so I can't really help you.  :D

[quote="mb17889"]Chris and linda are standing 78 ft. apart.  At the same time they each throw a softball toward each other.  Linda throws her ball with an initial veolcity of 45 ft./sec. with an angle of inclination of 44 degrees.  Chris throws her ball with an initial velocity of 41 ft./sec. with an angle of inclination of 39 degrees.

a)  Find two sets of parametric equations that represent a model of thie problem situation.  (I know how to do this... I'm just showing the way the problem wants it to be done)

b)  Find after what time the balls will be closest.[/quote]
don't quite understand the first part. so you want a set parametric equation for each ball to represent, what?

Here's what I did for the parametric equations:
x1=31.9t
y1=-16t 2 + 25.8t

x2=78-32.4t
y2=-16t 2 +31.2t

I did 78-32.4t to keep chris at the origen, then have linda throwing the other direction.

ok, this is the way i'd do this.
for the first part all you need to do is to say the distance for the ball in the x-direction is equal to:
$x=v_0\cos{t}$
and the distnace in the y-direction is equal to the
$y=-\frac{1}{2}gt^2+v_0t$
[b][u]assuming your parametric equations are correct[/u][/b]
the distance between the two balls at any given time such as $t$ would be:
$\sqrt{(63.3t-78)^2+(5.4t)^2}$
now you have to use some calculus, to find the minimum value for $t$. all you have to do is to find the derivative of the aboves expression, then set it equal to zero.
the derivative would be:
$\frac{2(63.3t-78)(63.3)+2(5.4t)(5.4)}{2\sqrt{(63.3t-78)^2+(5.4t)^2}}$
now set is equal to zero:
$\frac{2(63.3t-78)(63.3)+2(5.4t)(5.4)}{2\sqrt{(63.3t-78)^2+(5.4t)^2}}=0$
solve for $t$
you get $t=1.2233$ seconds.
this is your answer.


[quote="Treething"]In a grocery store, the ratio of apples to bananas is 2:1. While playing with the fruits, Joe realizes that there is a square number of apples and a cubic number of bananas. If there are under 1000 bananas, what is the largest possible number of apples in the store?

[color=white]immature innuendo galore[/color][/quote]

Edited
[hide=":)"]
The only answer I'm getting is 1024 and that's more than 1000, so could someone please give a hint?

EDIT: nevermind, like splashD said, he said less than 1000 banana's so 1024 is right after all!  [/hide]

he said under 1000 bananas not apples so your answer is right

[hide="Hint"] The smallest answer is $8$ with $4$ bananas. [/hide]

[hide="More helpful hint"] Write $\frac{a^{2}}{b^{3}}=2$, solve for $b<10$. [/hide]

[hide]Number of bananas is a cube under 1000. We want the largest cube under 1000 such that muliplying it by 2 gives a square. 10^3 and 9^3 don't work. $2\times8^{3}=2\times2^{9}=2^{10}$ so 1024 is our answer.[/hide]

[hide="No trial and error way :P "]Let $A$=# of apples, $B$=# of bananas.
Also, let $A=a^{2}$, $B=b^{3}$. Since $A: B=2: 1$,

$\frac{a^{2}}{b^{3}}=2$
$a^{2}=2b^{3}$

Since $a^{2}$ contains two equal factors, so must $2b^{3}$. $b$ MUST contain a factor of $2$ (if it didn't, then $2b^{3}$ will not be square). So let $b=2c$:
$a^{2}=2(2c)^{3}$
$a^{2}=16c^{3}$
$a=4c\sqrt{c}$
This implies that $c$ must be a square number. So let $c=d^{2}$:
$a=4d^{3}$
So we have these equalities:
$b=2d^{2}$
$B=8d^{6}$
$a=4d^{3}$
$A=16d^{6}$
For ANY integer value of $d$, $B$ will be a cube, $A$ will be a square, and $\frac{A}{B}=2$.

Now all we have to do is make sure $B<1000$:
$B<1000$
$b^{3}<1000$
$b<10$
$2d^{2}<10$
$d^{2}<5$
Since d is an integer, d=2.
$B=8d^{6}=512$
$A=16d^{6}=1024$

Therefore, $B=512$, $A=1024$.[/hide]

[hide]you know that the answer must be a power of 2 because:

If the cube had more than one factor, we could prime factorize it as x^3*y^3. If you multiply this by 2, it cannot be a perfect square since the powers would only change at best the power of one of the factors since 2 is prime.

If the cube has any other prime factors besides two, we can only factor it as x^3 since it is prime. When you multiply this by two, you get 2*x^3 which is not perfect square.

Thus, it should be a power of two.

Since we know that the number of bananas' cube root must be under 10, we try the largest cube root that is a power of 2 which is 8. Since 8^3=512=2^9,then 2*8^3=1024=2^10 which is a perfect square.

Therefore, the number of apples is 1024[/hide]

[quote="Treething"]Here's an extension. What if there's under 3000 bananas?[/quote]

Isn't the answer still the same?

Nope, it's a little bit different this time.

The smallest different one I found was [hide]$(11664, 5832)$.[/hide] :maybe:

disregard what i said it is the same i'm not thinking straight

Well, if there were more than 1000 bananas, THEN THE PROBLEM IS A PARADOX. :rotfl:

[hide="guess and check gives n immediately"]where n=1024[/hide]


Prove that
\[\sum_{i=1}^{n}\sin^{2}(2k-1)\theta =-\frac{n}{2}-\frac{\sin 4n\theta}{4\sin 2\theta}\]

[hide="Hints"]$\sin^{2}\alpha=\frac{1-\cos 2\alpha}{2}, \cos m\beta=\Re(\mathrm{cis}^{m}\beta)$[/hide]

I get $\frac{n}{2}-\frac{\sin 4n\theta}{4 \sin 2\theta}$...??? Could the problem be wrong?

Their result is wrong, yours is correct. Just check theirs by putting $n=1$ - it doesn't hold.


[quote="Astronum"]Prove that if $b > 0$ then $b|a \Longleftrightarrow a \mod b = 0$

 :yoda:[/quote]

[hide]If $b|a$, then $a=kb$ for some positive integer $k$.

$a=kb\equiv 0\pmod b$[/hide]

This is an iff proof. Therefore, you must prove it the other way as well. :P
[hide]If $a\equiv0(mod b)$, then $a=bq$ for some integer q. If $a=bq$, then $b\mid a$.[/hide]

[quote="Astronum"]Prove that if $b > 0$ then $b|a \Longleftrightarrow a \mod b = 0$

 :yoda:[/quote]

Hmm, I might be wrong, but I think that's just a specific case of the [u]definition[/u] of modulo. \[a\equiv b\mod{n}\ \Longleftrightarrow\ n|(a-b)\]


trigonometry
real analysis
real analysis unsolved

could someone help me with this problem.

find the inverse laplace transform of :

$ \frac{s}{(s\plus{}1)^{2}(s^{2}\plus{}1)}$

after PFD, we have $ \frac{\minus{}1}{2}\frac{1}{(s\plus{}1)^{2}}\plus{}\frac{1}{2}\frac{1}{s^{2}\plus{}1}$.

(i think)?

The second term becomes $ \frac{1}{2}\sin (1\cdot t)$ but the answer is $ \frac{1}{2}(\cos t\minus{}te^{\minus{}t})$

Thanks,
jli